ABBOU IB PHYSICS 1

IB Physics

ABBOU IB PHYSICS 2

TABLE OF CONTENTS

1. MEASUREMENTS, UNCERTAINTIES, AND GRAPHS (5H) ...................................................................... 5

1.1. MEASUREMENTS IN PHYSICS ............................................................................................................................................................. 5 1.2. UNCERTAINTIES AND ERRORS ........................................................................................................................................................... 7 1.3. ABOUT GRAPHS .................................................................................................................................................................................... 9 1.4. VECTORS AND SCALARS ..................................................................................................................................................................... 10

2. MECHANICS (22H) ........................................................................................................................................... 12

2.1. MOTION ............................................................................................................................................................................................... 12 2.2. FORCES ................................................................................................................................................................................................ 17 2.3. WORK, ENERGY AND POWER .......................................................................................................................................................... 19 2.4. MOMENTUM AND IMPULSE............................................................................................................................................................... 23

3. THERMAL PHYSICS (11H) ............................................................................................................................ 27

3.1. THERMAL CONCEPTS ......................................................................................................................................................................... 27 3.2. MODELLING A GAS ............................................................................................................................................................................. 29

4. WAVES (15H) .................................................................................................................................................... 33

4.1. OSCILLATIONS .................................................................................................................................................................................... 33 4.2. TRAVELLING WAVES .......................................................................................................................................................................... 35 4.3. WAVE CHARACTERISTICS ................................................................................................................................................................. 37 4.4. WAVE BEHAVIOUR ............................................................................................................................................................................. 40 4.5. STANDING WAVES .............................................................................................................................................................................. 45

5. ELECTRICITY AND MAGNETISM (15H) .................................................................................................... 48

5.1. ELECTRIC FIELDS ............................................................................................................................................................................... 48 5.2. HEATING EFFECTS OF ELECTRIC CURRENTS .................................................................................................................................. 54 5.3. ELECTRIC CELLS ................................................................................................................................................................................. 61 5.4. MAGNETIC EFFECTS OF ELECTRIC CURRENTS ............................................................................................................................... 65

6. CIRCULAR MOTION AND GRAVITATION (5H) ....................................................................................... 68

6.1. CIRCULAR MOTION ............................................................................................................................................................................. 68 6.2. NEWTON’S LAW OF GRAVITATION................................................................................................................................................... 70

7. ATOMIC, NUCLEAR AND PARTICLE PHYSICS (14H) ............................................................................ 72

7.1. DISCRETE ENERGY AND RADIOACTIVITY ........................................................................................................................................ 72 7.2. NUCLEAR REACTIONS ........................................................................................................................................................................ 78 7.3. THE STRUCTURE OF MATTER ........................................................................................................................................................... 80

8. ENERGY PRODUCTION (8H) ........................................................................................................................ 87

8.1. ENERGY SOURCES ............................................................................................................................................................................... 87 8.2. THERMAL ENERGY TRANSFER .......................................................................................................................................................... 91

9. (AHL) WAVE PHENOMENA (17H) .............................................................................................................. 97

9.1. SIMPLE HARMONIC MOTION ............................................................................................................................................................. 97 9.2. SINGLE-SLIT DIFFRACTION ............................................................................................................................................................... 99 9.3. INTERFERENCE ................................................................................................................................................................................ 100 9.4. RESOLUTION .................................................................................................................................................................................... 102 9.5. DOPPLER EFFECT ............................................................................................................................................................................ 104

10. (AHL) FIELDS (11H) ................................................................................................................................ 107

10.1. ABOUT GRAVITATIONAL AND ELECTRIC FIELDS ......................................................................................................................... 107 10.2. FIELDS AT WORK ............................................................................................................................................................................. 114

11. (AHL) ELECTROMAGNETIC INDUCTION (16H) ............................................................................. 115

11.1. ELECTROMAGNETIC INDUCTION ................................................................................................................................................... 115

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11.2. POWER GENERATION AND TRANSMISSION ................................................................................................................................. 117 11.3. CAPACITANCE .................................................................................................................................................................................. 121

12. (AHL) QUANTUM AND NUCLEAR PHYSICS (16H) ......................................................................... 124

12.1. THE INTERACTION OF MATTER WITH RADIATION ..................................................................................................................... 124 12.2. NUCLEAR PHYSICS........................................................................................................................................................................... 132

13. (OPTION A) RELATIVITY (15H/25H) ............................................................................................... 137

13.1. THE BEGINNINGS OF RELATIVITY ................................................................................................................................................. 137 13.2. LORENTZ TRANSFORMATIONS ...................................................................................................................................................... 138 13.3. SPACETIME DIAGRAMS ................................................................................................................................................................... 145 13.4. RELATIVISTIC MECHANICS (HL ONLY) ........................................................................................................................................ 150 13.5. GENERAL RELATIVITY (HL ONLY) ............................................................................................................................................... 151

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WARNINGS:

NOS information will mostly be found (and should therefore be read) in the 2014 edition of the Oxford IB Physics textbook. The process of understanding the concepts of Physics requires to do many practice exercises. They should come from various sources (from the proofs and exercises included in this guide, from the textbook, from past papers…). Some Labs are mentioned in this guide, and others aren’t. Nevertheless, all of them are part of the Physics course and should be understood, known and revised just as the usual lessons.

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1. Measurements, uncertainties, and graphs (5h)

1.1. Measurements in Physics

1. Physical quantities and units Definitions:

A Physical quantity is the property of an object that can be measured with an instrument. A Unit is a particular amount (of time, length…) that is used as a standard for measuring

Remarks: A unit has to be as stable as possible, and as precisely known as possible The unit for one quantity changes over time and space (foot, pouce…)

The definition of a particular unit changes over time (definition of the second)

2. NOS General remarks:

Quantities enable to order and compare physical properties. The establishment of a common international unit system enables to improve international collaboration. The improvement of the precision of a unit in order to narrow its definition necessitates to improve: apparatus and instrumentation

replication and comparability of experiments Evolution of the definition of the second:

from “1/86400 th of a solar day” to “the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the Caesium 133 atom”

3. Fundamental units and derived units

Quantity Mass Distance Time Electric Current

Temperature Amount of substance

Fundamental unit

Kilogram Meter Second Amp Kelvin Mole

kg m s A K mol

Quantity Velocity Force Power Energy … Derived unit Newton Watt Joule

m.s-1 N = kg.m.s-2 W = kg m2 s-3 J = kg m2 s-2 Useful website: http://www.bipm.org/

Remark: The definitions of the SI units should be known! They are in the textbook (p3).

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4. Scientific notation and metric multipliers

Scientific notation: 112 m = 1.12 * 102 m 0.00234 µm = 2.34 * 10-3 µm

Metric multipliers (data booklet p 2):

Unit Kilometer Centimeter Millimeter micrometer nanometer picometer femtometer

Symbol km cm mm µm nm pm fm

In meters

103 10-2 10-3 10-6 10-9 10-12 10-15

5. Significant figures The writing of a result reflects its precision. The significant figures (sf) of a result are the digits that are known with certainty. It is better to use the scientific notation to write any result.

Examples: The radius of the Earth is 6,400 km The precision of this value is of about 100 km The first 2 figures are significant (regarding precision) It should be written 6.4 * 103 km 40 m = 4 * 101 m 1 sf less precise than 4.0 101 m 0.000568 mm = 5.68 * 10-4 mm 3 sf Exercises: Do the following calculations and express the results to the most appropriate number of significant figures. 1.2 * 36.1; 6.4 * 10-4 * 4.85 * 101 / 0.030 2.53 * 104 + 3.2 * 102

6. Orders of magnitude Definition: The order of magnitude of a number is the power of ten closest to it.

Example: Tree: 25 m = 2.5 * 101 m order of magnitude: 101 m Range in Universe:

From To

Distance nucleus and sub nuclear particules

10-15 m

(known) Universe 1026 m

Mass electron 10-30 kg

(known) Universe 1053 kg

Time passage of light across nucleus 10-23 s

age of Universe 1017 s

Video: “powers of ten” (0:30 to 4:10 and 5:50 to 8:20)

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Comparison of order of magnitudes between 2 objects: Height of Eiffel Tower: 324 m = 3.24 * 102 m Length of DVD jacket: 20 cm = 2.0 * 101 cm 1/ put the to quantities in the same unit

324 m 0.20 m 2/ Divide the bigger one by the smaller one 324/0.20 = 1620 = 1.6 * 103 3/ Compare

Closest power of ten: 103 There are 3 orders of magnitudes of difference between the lengths of these objects

Lab 1

1.2. Uncertainties and errors

7. NOS General remarks:

All scientific knowledge is uncertain (apparatus, human judgment, fluctuations, quantum mechanics…) Exact values do not exist: it doesn’t mean anything to talk about “exact” or “perfect” values. One should only talk about “accepted” values.

8. Random and systematic errors How far from the expected (literature) value is the best estimate value (which is derived from a series of measurements)?

Random error Systematic error

Definition All the measurements differ from one another. Some of them are greater than the expected value, others are smaller than the expected value.

The measurements are all greater (or all smaller) than the expected value

Sources Equipment not precise enough. Fluctuations in the surroundings between measurements. Observer.

Wrong calibration of the equipment (offset) Always the same reading error Clock too slow…

Reduce Taking many measurements reduces the random error (the best estimate value will be closer to the expected one)

Difficult because it is not easy to spot when you don’t know the expected value Taking many measurements doesn’t change anything

Related words

Precision Accuracy

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9. Absolute, fractional and percentage uncertainties

Every measurement is uncertain. The sources of uncertainty are: A piece of equipment which is not very precise The fluctuations in the surroundings between measurements The experimenter Definitions: m = mbest estimate ± Δm

Δm: absolute uncertainty it takes into account: Range of measurements/2 Precision of equipment Precision of the experimenter

Fractional uncertainty: Δm/mbest estimate

Percentage uncertainty: Δm/mbest estimate * 100 Remarks: Taking many measurements does NOT reduce the absolute uncertainty, but gives a better idea of how precise the mean value is. The absolute uncertainty of the measure of a time period T can be reduced by taking many measurements: The measure of 10 Time period T

If the absolute uncertainty (due to the clock, the experimenter…) of a measurement is of 1s, then the absolute uncertainty for 10T is 1s. Therefore, the absolute uncertainty for T is 0.1s.

Using a more precise piece of equipment will reduce the absolute uncertainty

“best estimate” should not be too precise compared to the absolute uncertainty experimental absolute uncertainty are usually written with 1 or 2 sf

1.01 cm ± 0.04 cm is OK 2.53 m ± 0.12 m is OK 5.91 kg ± 2 kg is not OK best estimate is too precise

Small Random error Small Random error Large Random error Large Random error

Small Systematic error Large Systematic error Small Systematic error Large Systematic error

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10. Propagation of uncertainty (Lab 2)

Definitions:

The absolute uncertainty of A + B and A - B is: ΔA + ΔB k * A is: k * ΔA (k is constant)

The percentage uncertainty of A * B and A / B is: ΔA/ABest Estimate + ΔB/BBest

Estimate Aα is α * ΔA/ABest Estimate

Analysis of uncertainties will not be expected for trigonometric or logarithmic functions in examinations

1.3. About Graphs

11. Uncertainty of gradients and intercepts (Lab 2) Important notions derived from Lab 2:

Error bars Best fit line (should go through the most error bars, all of them if possible) Finding best fit, Min and Max lines (done by eye) Determination of a gradient and a y-intercept (with their uncertainties)

12. Linearization

Linear relationships

Some relationships between physical quantities in physics are linear. Y = a * X + b Example: F = k * x (cf 2.3 force exerted by a spring)

The constants a (gradient) and b (y-intercept) can be found by plotting the Y vs X graph (cf 11)

Non linear relationships

Some relationships between physical quantities are not linear: Y = a * Xb Example: EK = ½ m * v2 (cf 2.3 kinetic energy) Y = a * exp(b*X) Example: A = A0 * exp(-λ*t) (cf 7.1 radioactive decay graph) Definition: The linearization of a relationship consists in defining new variables Y’ and X’ (which are functions of Y and X) so that the relationship between them is linear.

Goal: The constants a and b can be found by plotting Y’ vs X’ Exercise: 1°/ An experiment enables to measure EK and v of an object. What linearized graph could be drawn in order to find m? 2°/ Linearize the two following relationships and find which graphs enable to derive the constants a and b. Y = a * Xb Y = a * exp(b*X)

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1.4. Vectors and scalars

13. NOS Vectors: Very useful tools because the world is 3D

First explicit mention in a scientific paper: 1846

14. Vectors and scalar quantities Definition: A vector u has: a magnitude (units depend on the physical quantity it represents) a direction and a sense Examples in Physics:

Scalar Vector

Speed Velocity

Mass Force (need a magnitude and a direction)

Temperature Magnetic field

15. Combination and resolution of vectors Two vectors can be added (sum) or substracted (difference) A vector can be multiplied or divided by a scalar

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It can be useful to resolve a vector into perpendicular components:

v = vx + vy:

vx = v * cos(α) vy = v * sin(α) v2 = vx

2 + vy2

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2. Mechanics (22h)

2.1. Motion

16. NOS Fundamental to many areas of physics (astrophysics…)

17. Frame of reference Definition: A frame of reference is made of:

a solid object a set of axis attached to the object an origin (O) attached to the object a clock

18. Distance and displacement Frame of reference: the ground A cat goes from O to A in a straight line

and then from A to B in a straight line It takes T = 2.0 s overall

Definitions: The distance D between O and B is the distance travelled to get from O to B. It is a scalar (positive quantity) In the example, the distance between O and B is 8.0 m The displacement between O and B is a vector d: d has a magnitude: 7.0 m d has a direction and a sense Animation: Addition Vecteurs

19. Speed and velocity Definitions: The speed (a scalar) between O and B is

v = D/T (In the example, v = 8.0/2.0 = 4.0 m.s-1)

The velocity between O and B is a vector: v = d/T unit: m.s-1

Example in 1D: An object moves along a straight line: 15 m to the left (from O to A); 5.0 m to the right (from A to B); it lasts 5.0 s.

Distance: 15 + 5 = 20 m Speed: 20/5.0 = 4.0 m.s-1 Displacement: -15 + 5 = -10 m

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Velocity: -10/5.0 = -2.0 m.s-1 The sign of the displacement (vector) gives the sense of the vector

20. Relative motion Exercise: An object is moving at velocity u in the frame of reference S. S’ is a frame of reference moving at v relative to S. u and v are represented on the diagram to the left. Draw on the diagram to the right u’, the velocity of the object relative to S’.

Example: Velocity of running man (Obj) relative to rain (S'): v' v: man/ground (relative to S) u: rain/ground (relative to S)

Velocity combination: u (OBJ in S) = u’ (OBJ in S’) + v (S’ in S)

21. Instantaneous velocity In a reference frame (S) an object moves along a random path:

It goes from M(t) to M(t + δt):

short displacement between M(t) and M(t + δt): OM(t+δt) - OM(t) = δd short duration between M(t) and M(t + δt): δt Definition: the instantaneous velocity at M(t) is the vector: v = δd/δt Vector v: magnitude: instantaneous speed at M direction: tangent to the trajectory at M sense: the sense the object in moving when in M Notations: ΔX means: Xfinal – Xinitial Δ is for big differences δ is for thiny differences Useful website: http://www.ostralo.net/3_animations/swf/vitesse.swf

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22. Acceleration

Definition: The acceleration of an object moving is a vector: a The acceleration at point M is a = δv/δt Units: m.s-2

Remark: a = 0 if NEITHER magnitude NOR direction of v change

23. Vocabulary A motion with constant velocity v is called

a uniform motion the trajectory is a straight line constant acceleration a is called

a uniformly accelerated motion the trajectory does NOT have to be a straight line

24. Graphs describing motion Preliminary remarks: The graphical analysis done in points 23 and 24 are 1D motions only. Nevertheless, it can be useful for 3D motions considering the fact that all 3D motions can be resolved into three independent perpendicular 1D motions: motion along x: x, vx, ax motion along y: y, vy, ay motion along x: z, vz, az The notations d, v, and a are used for displacement, velocity and acceleration instead of d, v, and a because it is a 1D motion (there is only 1 direction and the sense of the vector in given by the sign of d, v, and a)

Displacement Vs Time graphs

Instantaneous velocity at 1 point: gradient: v = δd/δt

Remark: instantaneous speeds can be derived by taking the gradient of Distance Vs Time graphs

Velocity Vs time graphs Acceleration at 1 point:

gradient: a = δv/δt

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Displacement between 2 points: area under graph between these 2 points

Acceleration Vs Time graphs Δv between 2 points: area under the graph between these 2 points

25. Equations of motion for uniform acceleration

Uniform acceleration: when the acceleration a is a constant (constant magnitude AND constant direction)

Equation of the graph:

a = constant (a = 3.0 m.s-2) Area under the graph between:

t = 0 and t: Δv = v – u = a * (t – 0)

therefore: v = u + a * t u: velocity at t = 0

Velocity Vs Time graph Equation of the graph:

v = u + a * t

Area under the graph between t = 0 and t: Lower rectangle: u * (t – 0)

Upper triangle: ½*(v - u) * (t - 0) Therefore the displacement between t = 0 and t is the total area under the graph:

s = (v + u) * t / 2

s = u * t + ½ a * t2

if we want to get rid of time in these equations:

s = (v + u) * t / 2 leads to t = 2s/(v + u) combined with v = u + a * t it leads to v – u = at = 2as/(v+u)

and therefore to v2 – u2 = 2as

Animation: Topic 2 Calculus grapher (http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html)

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26. Projectile motion

Definition: A body is in the conditions of free fall (near the earth) when its acceleration a is constant with a = g The magnitude of g is g = 9.8 m.s-2

The direction of g is vertical, downwards

Lab 3 Exercise: 1°/ Show that the x(t) and y(t) coordinates of an apple initially at x = 0 m, and y = 4.2 m are as follows: x(t) = 0 y(t) = 4.2– 4.9t2

2°/ Show that the x(t) and y(t) coordinates of tennis ball, thrown upwards (along the y axis) with an initial speed of 10 m.s-1, which location is x = 1.0 m and y = 1.2 m at t = 0, are as follows: x(t) = 1.0 y(t) = 1.2 + 10t – 4.9t2

3°/ Find the x(t) and y(t) coordinates of cannon ball, thrown upwards at an angle of 35° relative to the horizontal line, with an initial speed of 6.5 m.s-1, which location is x = 0 m and y = 8.0 m at t = 0.

Derive the trajectory of the cannon ball (the y = f(x) equation). Draw the trajectory on graph paper.

Help: the x motion and the y motion are independent 1D motions.

27. Fluid resistance and terminal speed

When there is a resistance inside the fluid there is no longer free fall The cannon ball goes:

Less high Less far The ball eventually:

moves in the y direction only reaches a constant velocity (called

terminal velocity)

Lab 4

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2.2. Forces

28. Objects as point particles Remark: Objects are 3D things. In a first approach it is convenient to consider them as point particles.

29. Forces Definition: An external force is an external action on a system by something outside of the system. What does a force do? Puts into motion Stops the motion A force changes velocity Changes the direction

Remark: When an object is not considered as a point particle, a force can also deform or rotate a system.

Characteristics of a force: Magnitude (in Newtons N) Direction A force is represented by a vector Sense Remark: When an object is not considered as a point particle, the point of application of the force is important to take into consideration (contact forces, action-at-a-distance forces) Examples: The weight on Earth of an object of mass m (W = m * g) is an action at a distance. The force exerted by a racket on a ball when it hits the ball is a contact force.

30. Free-body diagrams Exercises: For the three following examples: List the external forces acting on the system. Draw the free-body diagrams (forces labelled or named, properly scaled vectors, acting from the point of application). Guess what the motion of the system is (or could be).

Example 1: A ball on a table (system: the ball) Example 2: An free falling apple (system: the apple) Example 3: A hockey puck in a frictionless horizontal motion on ice (system: the puck)

31. Solid Friction

A book is on an inclined plane. The angle of inclination is small enough that the book is at rest relative to the table. External forces acting on the book:

W: Weight Rn: Normal reaction force, normal to the plane FS: Static friction force (exists as long as FS ≤ µSR)

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Example: Concrete/Rubber (Dry: µS = 1.0 , Wet: µS = 0.3)

The angle is increased until the book starts to slide down External forces: W: Weight R: Normal reaction force, normal to the plane Fd: Dynamic friction force

Fd = µdR Remarks: Often µd < µs

µd and µs depend on the nature of the materials in contact with each other.

32. Newton’s first and second laws of motion Remark: We need to add all the forces (VECTORS!!!) in order to predict motion. Definition: The net force (also called resultant force) Fnet = ∑F is the vectorial sum of all the external forces.

Definition: A body is said to be in translational equilibrium when the net force acting on the body is equal to 0 Exercise:

A picture of mass m is hanged on a wall. The angle between one string and the horizontal is α and the angle between the other string and the horizontal is β.

Find the tension forces provided by both string.

Definition: The linear momentum p of an object of mass m and velocity v is p = m * v Remarks: Newton's second law of motion also writes Fnet = Δp/Δt (cf paragraph 43)

Newton's first and second laws of motion are only valid in an inertial frame of reference.

Translational equilibrium (the first law of Newton)

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction (same velocity) unless acted upon by a net force. OR Every object in a state of uniform motion tends to remain in that state of motion unless acted upon by a net force.

The second law of motion

An object of mass m, is acted upon by a net force Fnet. The second law of Newton states that:

Fnet = m * a

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Comment on units: R is in N m is in kg v is in m.s-1 p is in kg.m.s-1 a is in m.s-2 So 1 N = 1 kg.m.s-2

33. Free Fall Definition: A body is in the conditions of free fall when the only external force acting on it is its weight (near the Earth it means that its acceleration a is constant with a = g). Remarks:

In a free fall situation, │vy │increases forever which is physically impossible (cf option A). The air resistance acts against the constant increase of │vy │. The air resistance eventually equals the weight and therefore │vy │reaches a limit called terminal speed (cf section 27) and the acceleration becomes equal to 0.

34. Newton’s third law of motion

Examples: Starting of a car (A: wheel B: ground) Bumping into a wall (A: me B: wall) Motion of a space ship (A: space ship B: gas released by ship) Earth/me (A: Earth B: me) Earth/moon (A: Earth B: Moon)

2.3. Work, Energy and Power

35. NOS Energy is a quantity that is conserved and can be converted from one form into another:

Car: from chemical energy to kinetic energy Nuclear power plant: from nuclear energy to electrical energy

Energy is a notion which has evolved over time (recognition of a relationship between mass and energy)

36. Kinetic energy The damage done by an object crashing on a wall is all the more important as: the mass of the object is important the velocity of the object just before the crash is important Definition: An object of mass m, and of velocity v has an energy due to its motion called translational kinetic energy EK:

EK = ½ m * v2 Units: v (m.s-1) m (kg) EK (Joules: J) 1 J = 1 kg.m2s-2

Remarks: EK is a scalar EK is always > 0

The third law of motion If two bodies A and B interact and

A exerts a force FA/B on B then B exerts a force F B/A on A and FA/B = - F B/A (same magnitude, along same line, opposite sense)

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37. Work done by a force

Work along a small displacement An external force F is acting on a system. The system is moving which causes small variation of displacement δd. The work done by F along δd is:δW = F.δd = F * δd * cos(α)

Units: F (N) δd (m) δW (J) 1 J = 1 N.m = 1 kg.m2.s-2

Remarks:

δW is positive if α < 90° and δW is negative if α > 90°. Work is a mode of energy transfer. Work along a large displacement A system travels from A to B (1). The path can be broken down into many small displacements (2). An external force F is acting on the system when it goes from A to B. The force is not constant (3).

The work done by F along AB is:

WA to B (F) = Σ F.δd

If F is constant along AB (4), then:

W = Σ F.δd = F.Σδd = F.Δd The work done by a constant force

F along AB = Δd is (5)

WA to B (F) = F * Δd * cos(α) Remark: in a 1D motion, the work done by a force F on an object going from x1 to x2 is the area under the F Vs x graph.

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38. Gravitational potential energy

The damage done by an object crashing after a free fall is all the more important as: the mass of the object is important. the height from which it was let go is important. Definition: An object of mass m, positioned at a height H (from an origin O) has a reserve of energy due to gravitation called gravitational potential energy Epp. Epp is defined through the work done by the gravitational Force between points 1 and 2.

W1/2 (Fgravitationnal force) = Epp,2 – Epp,1.

Epp = m * g * H

Units: g (m.s-2) m (kg) H (m) Epp (J)

Remarks: Epp can either be positive or negative. Δ Epp = m * g * Δ H (Δ Epp is the change of Epp)

39. Force exerted by a spring - Elastic potential energy

An object is on frictionless horizontal

rails (forced to move in only 1D).

A spring is attached at one end to a wall and at the other end to the object.

The equilibrium position is where the object is at rest (the spring doesn't exert any force: x = 0).

The object is pulled at x (the spring

exerts a force F directed towards the equilibrium position). Hooke’s law: F = - k x

k: spring constant (unit N.m-1) x: displacement (m)

At t = 0 s, the object is let go.

Work done by F between x1 and x2 is the area under the F vs x graph. As the magnitude of F is equal to k*x, the area under the graph between x1 and x2 is equal to:

W1/2 = ½ k * (x22 - x1

2)

Definition: The elastic potential energy stored at x is Epe = ½ k x2

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Consequence: W1/2 = Epe,2 – Epe,1.

40. Power Definition: The power P is the rate of energy transfer: P = δE/δt

Unit: Watt (W) 1 W = 1 kg.m2.s-3

The power of a force

P = δW/δt = F. δd/ δt = F.v = F * v * cos (α). P is max when the angle between F and v is equal to 0.

41. Principle of conservation of energy When an isolated system undergoes transformations, its total energy is conserved. When the external forces f1, f2, f3… are exerted on a (non-deformable) solid which undergoes a translational motion and which center of mass moves from point A to point B, the principle of conservation of energy is:

ΔEK = EK, B – EK, A = WA to B (f1) + WA to B (f2) + WA to B (f3) + … Animation: Topic 2 Energy transfer (http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html) Remarks: Without friction, there is transfer between EK and potential energies (EP, EPe, …).

Examples

42. Efficiency

When a system converts a form of energy (input) into another form which can be easily used (output), there is some energy loss to the surroundings.

Total work in: work (or energy) input in the system. Useful work out: useful work (or energy) going out of the system. Losses: energy going out of the system in a useless form.

Forms of energy

Kinetic energy

Gravitational potential energy

Thermal energy

Chemical energy

Electric energy

Em wave energy

Transformations of energy

From To

Gravitational potential Kinetic

Kinetic Electric Windmills (wind machines)

Kinetic Thermal Friction

Chemical Electric Batteries

Em wave energy Electric Solar panels

Electric Thermal kettle

Chemical Gravitational potential

Climbing up stairs

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Example:

Definition The efficiency (η) of the machine is:

η = Useful work out / Total work in η = Useful power out/ Total power in

Remark: 0 ≤ η ≤ 1

2.4. Momentum and impulse

43. Definitions Definition 1: The linear momentum of a system of mass m velocity v is:

p = m * v (units: kg.m.s-1)

Remark: EK = p2/2m Definition 2: An impulse is a change in linear momentum:

Δ p = m * Δ v (if m doesn't change)

Exercise: an atom strikes a wall with a velocity v at an angle θ to the direction normal to the wall. It bounces off the wall with the same speed but in a different direction (angle – θ).

Draw the diagram of the situation. Show that the impulse is: 2 m * v * cos(θ).

44. Second law of Newton The second law of Newton can also be written: Fnet = Δp/Δt (cf paragraph 31) Therefore, Δp = Fnet * Δt

ABBOU IB PHYSICS 24

45. Impulse and force-time graphs

Remark: the impulse Δp is the area under the Fnet vs t graph (if Fnet doesn't change directions ie 1D motion). A ball is dropped from the top of the Eiffel

tower: At first, the net force is first equal to

W = mg. Then it decreases because of air friction. The impulse along this path is the area under the F vs t graph.

46. Conservation of linear momentum If a system is a collection of objects (obj1, obj2, obj3...) If no external force is acting on this system (Fnet = 0) There can be forces between the objects of the system, but no external forces Then, According to the Second Law of Newton:

Δp/Δt = 0 which leads to p is constant total momentum of the system doesn't change p = p1 + p2 + p3: it is a constant.

47. Collisions and explosions Definition: A collision between 2 objects is elastic when EK is conserved. Otherwise, it is called inelastic.

Collisions

There is a collision between 2 objects (they form an isolated system). Before collision After collision Object 1: p1 p'1

Object 2: p2 p'2

Conservation of linear momentum: p’1 + p’2 = p1 + p2

Elastic collision: EK1 + EK2 = E’

K1 + E’K2

Exercise: Object 1 (mass m1) is moving at v1. It collides with Object 2 (mass m2) initially at rest. After the collision, both objects move along the direction of motion of Object 1 before the collision (ie it is a 1D motion problem). The collision between the two objects is elastic. 1°/ Use conservation of momentum to show that:

(v1 – v’1) = m2 v’2/m1. 2°/ Use the fact that the collision is elastic to show that:

(v12 – v’12) = (v1 – v’1) (v1 + v’1) = m2 v’22/m1.

3°/ Show that: v’1 = (m1-m2) v1/(m1+m2). v’2 = 2 m1 v1 / (m1 + m2).

ABBOU IB PHYSICS 25

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Explosions

There is an explosion when a system initially at rest parts into 2 different objects. Before collision After collision Big object p = 0 Object 1: p'1

Object 2 p'2

Conservation of linear momentum: p’1 + p’2 = 0. Remark: EK is not conserved during an explosion

Exercise: Explain how the conservation of momentum does account for the motion of a space ship initially at rest in outer space (and also for the take-off of a space ship) Animations: Collision Balles (www.scienceslycee.fr)

Topic 2 Collision-lab (http://phet.colorado.edu/sims/collision-lab/collision-lab_fr.html) The “elasticity” is the coefficient: (v2’ – v1’)/v1

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3. Thermal Physics (11h)

3.1. Thermal concepts

48. NOS Thermal physics was developed during the 19th century and most of it only dealt with macroscopic quantities (energy, heat, work, temperature…).

Statistical mechanics provides a deep understanding of thermal physics by linking macroscopic quantities to microscopic ones.

49. Molecular theory of solids, liquids and gases

Solid Liquid Gas

Density Tightly packed Close together Well separated

Arrangement of particles

Can be regular (crystals)

Not regular Not regular

Motion of particles Vibrate about a fixed place

Vibrate and move about

Vibrate and move freely (high EK)

Animation: Topic 3 States of matter (http://phet.colorado.edu/fr/simulation/legacy/states-of-matter-basics)

50. Temperature and absolute temperature Definition: The temperature measures how cold or hot a system is.

It is a measure of the average kinetic energy per particle of the random motions of the particles of a system.

Units: Celsius (°C) Kelvin (K) Fahrenheit (°F)…

Remarks: The absolute temperature is the temperature in K. The absolute zero temperature (0 K = -273.15 °C) is the point where particle motion is at its minimum. “K” = “°C” + 273.15

51. Internal energy and thermal energy Definitions: The internal energy (U) of a system is the sum of the total intermolecular potential

energy and the total random kinetic energy of the molecules.

U = (Ep + EK) Ep: potential energy between molecules (electric potential energy…) EK: kinetic energy (translational, and rotational)

The thermal energy (Q) is a mode of energy transfer between a system and its surroundings. It is a “non-mechanical” mode of transfer.

Remarks: Q is a mode of transfer, not a measure of any storage of energy. Work (W) done on (or by) a system and Q can change the U of a system. When 2 systems of different temperatures are in contact, a positive Q takes place from the hottest system to the coldest system until they reach thermal equilibrium (equal temperatures).

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52. Specific heat capacity Definition: The specific heat capacity of a substance is the thermal energy required to raise the

temperature of one kilogram of the substance by one degree Kelvin. Property: The thermal energy needed to raise a substance of mass m from T1 to T2 is:

Q = m * c * (T2 – T1) = m * c * T

m: mass of the substance (kg) T: temperature of the substance (K) c: specific heat capacity of the substance (J.K-1.kg-1)

Remark: C = m*c is called the heat capacity of a certain mass of a certain substance (in J.K-1)

53. Specific latent heat Definitions: The latent heat of fusion (Lf) of a substance is the thermal energy required to

change the substance from a solid at its melting point completely to a liquid at its melting point.

The latent heat of vaporisation (Lv) of a substance is the thermal energy required to change the substance from a liquid at its boiling point completely to a gas at its boiling point.

Property: The thermal energy needed to change a substance of mass m from one phase to

another phase, at the phase change temperature is:

Q = m * L

m: mass of the substance (kg) L: Latent heat (J.kg-1)

Exercise: A 200g ice cube is at T = -7.5°C. It is put in a calorimeter together with 400g of liquid water at 45°C. Determine the final temperature reached by the water inside the calorimeter (it is assumed that no thermal energy transfer occurs between the inside and the outside of the calorimeter).

54. Phase change Property: Pure substances change phase (from liquid to gas, from solid to liquid..) at specific

constant temperatures which are specific to each substance particular.

Pure liquid Water Pure solid Water Pure gas Water Pure liquid Ethanol

c (kJ.K-1.kg-1) 4.18 2.11 2.08 2.44

Pure Water Pure Ethanol

Lf (kJ.kg-1) 334 108

Lv (kJ.kg-1) 2260 855

Pure Water Pure Ethanol Pure Gold

Melting point (°C) 0 -114 1063

Boiling point (°C) 100 79 2970

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Example: pure water is heated at a constant energy rate

Remark: Throughout the whole process, the internal energy of the water increases

Remarks: Evaporation takes place at temperatures below the boiling point.

Evaporation takes place as soon as the molecules of higher speed reach the surface of the liquid with enough EK to overcome intermolecular potential energy. This process results in a slight decrease in the temperature of the liquid because the average kinetic energy of the molecules in the liquid decreases as the faster molecules go into the gaseous phase. The boiling point is reached when all the energy input is used to vaporize and none is used to increase the temperature of the substance. The temperature (boiling point) is therefore constant.

Lab 5

3.2. Modelling a gas

55. Pressure Definition: The pressure P due to a force F exerted on a surface of area A is equal to:

P = F┴/A

F┴ is the component of F perpendicular to the surface (in N). A: surface area (in m2). P: pressure in Pascals (1 Pa = 1N.m-2). Remark: Pressure is a force per unit surface.

From -40°C to 0°C Molecules vibrate more and more about a fixed position.

At 0°C All the energy added is used to separate the molecules which increases their potential energy (at constant T ie constant EK)

From 0°C to 100°C Molecules move with increasing EK (still almost in contact with each other).

Evaporation occurs.

At 100 °C All the energy added is used to increase the molecules’ potential energy so that they become further apart: liquid turns into gas.

From 100°C to 130°C EK of molecules increases.

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56. Mole, molar mass and the Avogadro constant

Definitions: One mole of X contains 6.02 * 1023 X entities. (unit: mol) NA = 6.02 * 1023 mol-1 is called the Avogadro constant.

It represents the number of atoms of 12C inside exactly 12g of 12C. The molar mass of X is the mass of exactly one mole of X (unit: g.mol-1) Property: N molecules of X corresponds to n moles of X: n = N/NA

Exercises: Calculate M CO2. (MC = 12,0 g.mol-1 and MO = 16,0 g.mol-1) Calculate n CO2 of a sample of 1.2 g of CO2.

57. Equation of state for an ideal gas: the gas laws Definition: A gas will be considered as an ideal gas if

There are no interactions between the molecules. The volume of the molecules is negligible compared to the whole volume of the gas. The collisions between the molecules and the container of the gas and between the molecules themselves are elastic.

Graphical illustrations of the ideal gas law

The ideal gas law

n moles (mol) of an ideal gas of volume V (m3), at pressure P (Pa) and at temperature T (K) follow the ideal gas law:

P * V = n * R * T

R: ideal gas constant R = 8.31 J.K-1.mol-1

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Derived gas laws (for a constant amount of gas: n is constant):

At constant V: P/T is constant At constant P: V/T is constant At constant T: P * V is constant

58. Kinetic model of an ideal gas In an ideal gas, N molecules of various masses, move in random directions and with random velocities (v1, v2….). All the molecules do not have the same velocity. The temperature of the gas is a measure of the average kinetic energy per molecule of the random motions of the molecules. Definitions:

The Root Mean Square speed (rms speed) is an average of the speeds of all the molecules inside the ideal gas which have the same mass. There is one rms speed for every type of molecules inside the ideal gas:

N

v

v i

i

2

The average kinetic energy per molecule is: 22

1vmEK

m: mass of one molecule (in kg)

It has the same value for all molecules inside the gas, whatever their mass. It is related to the temperature T of the gas through the following law:

TkE BK

2

3 = T

N

R

A2

3

kB : Boltzman constant (kB = 1.38 10-23 J.K-1) T: temperature of the gas (in K)

Remarks: The molecules inside an ideal gas at temperature T:

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Can have different masses. Have all sorts of different v. Have an rms speed which depend on their mass.

Have the same KE .

Exercise: Air is made mostly of O2 and N2 (MN = 14,0 g.mol-1, and MO = 16,0 g.mol-1)

Calculate the average kinetic energy per molecule of air at 25°C. Calculate the rms speed of O2 molecules and of N2 molecules.

59. Differences between real and ideal gases In any real gas, molecules have sizes, and interact with each other. Therefore any real gas is NOT an ideal gas and does not follow the ideal gas law. However, under low pressure, moderate (or high) temperature and low density, a real gas approximates to an ideal gas. If P increases and V decreases too much, then the gas is no longer ideal. An ideal gas cannot be liquefied!

Lab 6

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4. Waves (15h)

4.1. Oscillations

60. NOS The study of oscillations is key to the understanding of:

Natural phenomena: Pendulum, tides, car suspensions…. Parts of Physics: Electromagnetism, waves, motion…

Oscillations are used to define the time units.

61. Simple harmonic motion (SHM) Definition: An oscillation is a repetitive variation (often a displacement) about a value (often an equilibrium value).

Examples: heartbeat, pendulum, yo-yo … Definition: A system undergoes a simple harmonic motion when:

Its displacement x(t) verifies the following defining equation of a SHM:

a(t) = - ω2 * x(t)

The acceleration a(t) (or the net force) is always pointing towards the system’s equilibrium position.

Consequence: x(t) = Amax * sin(ω*t + φ) Amax: maximum amplitude is a constant ω: angular frequency is a constant φ: phase at the origin is a constant The phase: ω*t + φ (unit: rad)

The frequency: f = ω/2π (unit: Hertz; 1Hz = 1s-1) The time period: T = 1/f (units: s)

Examples: simple pendulum, spring, guitar string…

Definition: An isochronous oscillation is an oscillation for which T does not depend on Amax. Many examples (graphs, oscillators…) on this website:

http://www.sciences.univ-nantes.fr/sites/genevieve_tulloue/Meca/Oscillateurs/Index_Oscillat.html Examples: 4-Pendule élastique vertical (λ = 0, change v0, ω, and x0)

8-Période du pendule pesant (change A the amplitude, and measure T)

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62. Example of a SHM

An object of mass m is on a frictionless horizontal surface. It is attached to a spring (spring constant k) which other end is tied to a wall. According to Hooke’ law: F = - k x According to Newton’s second law: F = ma a = - k/m * x = - ω2 x F is pointing towards the equilibrium position. The object therefore undergoes a SHM.

63. Graphical analysis

Displacement-Time

x(t) = Amax * sin(ω*t + φ). φ creates an offset on the t axis

Acceleration-Displacement

a(t) is proportional to x(t) with a negative slope (-ω2).

x(t), v(t), and a(t)

x(t) and v(t) have a π/2 phase difference. x(t) and a(t) have a π phase difference (phase opposition).

Energy-Time

There are energy changes during one cycle. The kinetic energy (EK) and the elastic potential energy (Epe) vary. The total energy remains constant (EK + Epe). The time period for EK and Epe is equal to T/2.

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4.2. Travelling waves

64. NOS Many different phenomena which exhibit common patterns are modelled as waves (sound, light…) The pattern they have in common is to carry energy without permanently disturbing the medium they travel through.

65. Definition and characteristics Definition: A travelling wave is both:

the propagation of a disturbance (perturbation) in a medium. the propagation of energy.

Example: a stone thrown in a pond creates travelling waves at the water surface. Remark: A travelling wave varies both in space and in time. Animation: Caractéristiques Onde (scienceslycee.fr). Use the « continu » mode

Time period

At one point in space, a property of the medium is modified because of the wave travelling through it. The amount of disturbance which is often called “displacement” of the perturbation, varies in time. The perturbation can exhibit a regular (periodic) pattern (first graph). The Time period (T in s) is the duration it takes the perturbation at one point in space to complete to a full oscillation (duration between the nearest crests or troughs on the graph).

Wavelength If the wave is a periodic one, at any moment in time (time is frozen, like in a photography), the medium through which the wave travels exhibits a regular pattern. At one moment in time, the “displacement” of the disturbance varies in space. The wavelength (spatial period), (λ in m) is the distance between the nearest crests or troughs.

Remarks: Frequency: f = 1/T units: Hz (or s-1)

Speed of propagation: c = λ/T = λ*f units: m.s-1

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The “displacement” of the perturbation A (M, t) both depends on the time and the position. Example of a sine shaped wave: A (x, t) = Amax * sin [2π*f*t – (2π/λ)*x + φ]

66. Transverse and longitudinal waves Definitions:

When a longitudinal wave passes through a particle of a medium, its direction of oscillation is the SAME AS the direction of propagation of the energy transfer.

When a transverse wave passes through a particle of a medium, its direction of oscillation is perpendicular to the direction of propagation of the energy transfer. Animations: Onde Longitudinale (www.scienceslycee.fr) Use the « sinusoïdal » mode Caractéristiques Onde (www.scienceslycee.fr) Use the « continu » mode

Remark: transverse waves cannot be propagated through gases.

67. The nature of electromagnetic (EM) waves Definition: An electromagnetic wave is a disturbance of the electric (E) and magnetic fields (B).

It can travel both through vacuum and various media. It consists of oscillating electric and magnetic fields.

Remarks: An electromagnetic wave is a transverse wave. Visible light are electromagnetic waves (very small wavelength range) Speed of electromagnetic waves in vacuum: c = 3.00 * 108 m.s-1.

68. The nature of sound waves Definition: A sound wave is a disturbance of matter which can only travel through matter (solid, liquid or gas). Animations: Onde Longitudinale (www.scienceslycee.fr) Press « sinusoïdal », and « affichage des micros »

Remarks: A sound wave is a longitudinal wave. When a sound wave passes through a medium, the particles from that medium oscillate back and forth about their fixed position. There is no matter propagation.

The propagation of a sound wave results in a succession of compression (high pressure) and rarefaction (low pressure) inside the medium. Speed of sound:

in air: c = 340 m.s-1 in liquid water: c = 1.5 * 103 m.s-1

The higher the frequency of a sound wave, the “higher the note”.

Lab 7

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4.3. Wave characteristics

69. Wavefronts and rays Definitions: A point source which can produce waves free to travel in all the 3 dimensions of a homogenous space emits spherical waves.

A wavefront is a surface or line in the path of the wave motion on which the disturbances at every point have the same phase. The rays are lines extending outward from the source, representing the direction of propagation of the wave. The rays are perpendicular to wave fronts.

Rays and spherical wavefronts

70. Amplitude and intensity Definition: A wave of amplitude A(M, t) propagates some energy. The intensity I(M, t) of a wave is the energy (or power) transferred per unit surface (of the wavefront).

Properties: I(M, t) is proportional to A(M, t)2

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I is proportional to 1/r2

71. Huygens principle Huygens principle: Every point in space reached by a wavefront behaves like a secondary point

source of spherical waves (wavelet). Animation: http://www.walter-fendt.de/ph14e/huygenspr.htm

Consequences: Large planar sources create planar wavefronts Many wave behaviours can be easily accounted for (refraction, diffraction…)

72. Superposition Two waves are emitted from the two ends of a string. They travel through the string and they cross: Wave 1 creates A1 (M, t) Wave 2 creates A2 (M, t) Principle of superposition: at any point M in space, at any time

A (M, t) = A1 (M, t) + A2 (M, t) Consequence: the total displacement of a particular point of the string: x = x1 + x2

Animation: Croisement Ondes (www.scienceslycee.fr)

73. Polarization

The propagation of an EM wave disturbs the E and B fields in directions perpendicular to each other which are both perpendicular to the direction of propagation of the wave. For example, if an EM wave travels along the Oz direction, the perturbation of E can be along the Ox axis, and the perturbation of B along the Oy axis. Animation: http://www.amanogawa.com/archive/wavesB.html (H in the animation should be understood as B)

If we only focus on E, we see it can be resolved into 2 perpendicular components (Ex and Ey) both perpendicular to the direction of propagation (Oz): E = Ex + Ey

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Definitions:

Light is polarized if E (z, t) has the same direction for all the rays.

Light is unpolarized if all the rays have random polarizations.

A polarizer only allows waves with one specific polarization of E to go through it.

74. Polarization using polarizers Remark: The axis of polarization of a polarized light can be found using a second polarizer

called an analyser.

Property: The light that comes through the analyser is polarized along the axis of the analyser.

The magnitude of E that comes through is: E = E0 * cos θ. The intensity of the light that comes through is (Malus’s Law):

I = I0 * (cos θ)2

I0: intensity of the incoming polarized light Remarks:

if unpolarized light goes through a polarizer, whatever the direction of the polarizer, I = I0/2 if unpolarized light goes through 2 polarizers with perpendicular axis of polarization, no light

goes through

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75. Polarization by reflection When light hits a non-metallic plane surface (water, glass…) separating two media (of refractive indices n1 and n2), it is both reflected and refracted. The reflected light is partially polarized. E can be resolved into 2 components: E1 parallel to the reflecting surface. E2 perpendicular to E1. E1 is better reflected than E2: the reflected light is partially polarized. When reflected light and refracted rays form a 90° angle (cf diagram), E2 is NOT reflected at all (reflected light is therefore totally polarized): this happens for a unique incident angle

called the angle of Brewster (B):

tan (B) = n2/n1 Remarks:

The glare from the reflection off the surface of the sea (or a window) is partially polarized. Polarized sunglasses can be used to get rid of this glare.

4.4. Wave behaviour

76. NOS Newton believed that light was made of particles. Huygens believed that light was a wave. Both theories could explain some of light’s behaviours but contradicted others. As is often the case in science:

theories which have known flaws can be used theories contradicting each others can be both used

77. Reflection

Reflection of pulses A pulse is sent through a string which other end is fixed: the pulse is inverted (3rd law…)

A pulse is sent through a string which other end is free: the pulse is not inverted (the free end is made to move up and down. This movement sends back a wave that goes up and down) Animation: Réflexion Onde (www.scienceslycee.fr)

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Reflection of electromagnetic waves When an electromagnetic ray hits a reflective surface (mirror) the ray bounces back on the surface according to the following law: The reflected ray belongs to the incident plane (the incident plane is defined by the incident ray and the perpendicular to the mirror) i1 = i2

Animation: Réflexion Réfraction (www.scienceslycee.fr) http://www.walter-fendt.de/ph14e/huygenspr.htm

78. Refraction Properties: Electromagnetic waves have different speeds depending on the medium they are progressing through (the speeds can also sometimes depend on the wavelengths of the electromagnetic wave).

Examples: vacuum c = 3.00 * 108 m.s-1 liquid water v = 2.26 * 108 m.s-1 liquid ethanol v = 2.21 * 108 m.s-1 Definitions:

Refraction occurs when a wave goes from a medium into another medium and changes direction of propagation.

The refractive index of a medium (n) is equal to: n = c/v

Examples: vacuum n = 1 liquid water n = 1.33 liquid ethanol n = 1.36 Snell’s law : When a wave goes from a medium of refractive index n1 (where the speed of the wave is v1) into a medium of refractive index n2 (where the speed of the wave is v2), the ray changes direction according to the following laws: The refracted ray belongs to the incident plane (The incident plane is defined by the incident ray and the perpendicular to the surface)

n1 sin (i1) = n2 sin (i2)

or sin (i1)/v1 = sin (i2)/v2

Remark: When n1 > n2, there is a critical angle:

icritical = asin(n2/n1).

if i1 > icritical, no refraction can occur, and all the rays are reflected: this phenomenon is called total internal reflection. The wavelength of a wave changes when the wave goes from one medium into another. If the wavelength in vacuum is λ, then the wavelength in a medium of refractive index n is:

λ’ = λ/n

Lab 8

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Example: The speed of water waves if greater in deep waters than in shallow water (near the shore)

vdeep waters > vshallow waters

(ie λdeep > λshallow) Therefore, according to Snell’s Law: ideep > ishallow

Animation: http://www.walter-fendt.de/ph14e/huygenspr.htm

79. Diffraction through a single-slit and around objects

Diffraction of water waves

A ripple tank creates planar water waves. The wavefronts go through a single slit (of aperture b). The wavefronts are parallel to the aperture.

Exercise: Draw the wavefronts after the slit.

Diffraction of electromagnetic waves Laser light (wavelength λ) is sent through a single slit (of aperture b) The rays are perpendicular to the aperture.

Animation: Interférences Diffraction 1 (www.scienceslycee.fr): select « 1 fente ».

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Remarks: The smaller the aperture, the greater the diffraction. There is a central maximum, and secondary maxima. The intensity is maximum at the center of the central maximum.

When the slit is square: θ = λ/b circular: θ = 1.22 λ/b

If θ is small enough: θ = d/2D An object of the same dimensions as the slit and surrounded by vacuum creates identical

diffraction patterns. The diffraction pattern can be explained with Huygens principle.

80. Double-slit interference

Planar waves travel perpendicular to two small apertures distant of d.

Diffraction occurs at each of the apertures, creating two circular waves.

There are some points in space where the two circular waves are in phase (2 crests or 2 troughs meet): the amplitude of the wave varies between 2Amax and – 2Amax: constructive interference occurs.

There are some points in space where the two circular waves are in phase opposition (a crest meets a trough). The amplitude of the wave is equal to 0: destructive interference occurs. Animation: Interférences Diffraction 1, Interférences, Interfranges 1, Interfranges 2 (www.scienceslycee.fr)

ABBOU IB PHYSICS 44

81. Path difference

Mathematical expressions of the amplitudes created by the two waves after the slits are:

Wave 1: A1(d1, t) = Amax*sin(2π*f *t – (2π/λ)*d1) Wave 2: A2(d2, t) = Amax*sin(2π*f*t – (2π/λ)*d2)

d1 : distance travelled by the wave after the aperture 1. d2 : distance travelled by the wave after the aperture 2

At any point in space, according to the principle of superposition, A = A1(d1, t) + A2(d2, t) At any point in space, there will be: Constructive interference if (2π/λ)*│d1 – d2│= 2π*m │d1 – d2│= mλ m: integer

Destructive interference if (2π/λ)*│d1 – d2│ = π+ 2π*m │d1 – d2│= (m+ ½) λ m: integer Animation: Interférences Diffraction 1, Interférences, Interfranges 1, Interfranges 2 (www.scienceslycee.fr)

Definition: │d1 – d2│is called the path difference between the 2 circular waves

The distance between 2 consecutive dark fringes or 2 consecutive bright fringes is called the interfringe: s = λD/d

Remarks on interference:

Interference can happen with all types of waves (em, mechanical…). In the case of EM waves, 2 sources will interfere if they are coherent (phase difference

between them remains constant). Laser light is monochromatic AND coherent.

General remark on waves: the characteristics of a wave can change: its speed can change (refraction: v = c/n) its amplitude can change (interference…) its direction can change (interference, refraction…) its wavelength can change (refraction λ’ = λ/n)

But whatever happens to a wave its frequency DOESN’T change

ABBOU IB PHYSICS 45

82. Diffraction and interference

A slit of aperture b is made of an infinite number of points, which act as secondary spherical point sources. Therefore, although the incident wave has a single direction of propagation, waves crossing the slit travel in all directions.

Let’s consider all the rays (ray 1 to ray 9) coming out of the slit, going in the same direction (represented by the angle θ). They will meet at infinity (ie a few meters away from the slits…) and interfere.

The path difference between the upper ray (ray 1) and the lower one (ray 9) is equal to b* θ (because the angles are very small).

Therefore, the path difference between the upper ray

(ray 1) and the one coming out from the middle of the slit (ray 5) is equal to b* θ/2. And such is the path difference between rays 2 and 6, 3 and 7… and so on.

If this path difference is equal to λ/2, then all the pairs

of rays destructively interfere, leading to the absence of light in this direction θ.

The first minimum of the diffraction pattern (absence

of light) occurs at an angle θ such that b* θ/2 = λ/2, which leads to the condition θ = λ/b (point 78).

4.5. Standing waves

83. The nature of standing waves When two travelling waves of same frequency and same maximum amplitude, travelling in opposite directions meet, they produce a standing wave. For example, this situation happens when an “incident” sine wave travelling along a string is reflected when it reaches the end of the string. The reflected wave travels in the opposite direction, and has the same frequency and the same maximum amplitude (is no losses occur). Animation: Réflexion Onde (www.scienceslycee.fr) f = 1.50 Hz; v = 2.0m/s; « sinusoïdal » ; « un obstacle »

Properties:

Contrary to travelling waves, the amplitude of the oscillations is not the same for all the points in space:

some points (called nodes) do not oscillate. some points (called antinodes) have a maximum amplitude.

Standing waves do not transfer energy.

All the points in space are either in phase, or in phase opposition (π phase difference).

ABBOU IB PHYSICS 46

Remarks: The frequency of the standing wave is the same as the frequency of the travelling waves which created it. The wavelength (λ) corresponds to the distance between nodes 1 and 3, or nodes 2 and 4…

84. Boundary conditions When a waves is reflected, the resulting wave is the combination of the “incident” wave and the reflected wave (in accordance to the principle of superposition).

Reflection at an end At a free end: when reflection happens at a free end, the reflected wave is not inverted (see reflection of a pulse: point 77). Therefore, the amplitude of the oscillations at that free end is always maximum. A free end always corresponds to an antinode. At a fixed end: when reflection happens at a fixed end, the reflected wave is inverted (see reflection of a pulse: point 77). Therefore, the amplitude of the oscillations at that fixed end is always equal to 0. A fixed end always corresponds to a node. Animation: Ondes Stationnaires 1 (www.scienceslycee.fr)

Remark: a free end can correspond both to: a string free to go up and down an open pipe where air vibration is free (it is called an open end)

a fixed end can correspond both to: a string fixed to a wall. a closed pipe where air vibration is impossible (it is called a closed end)

Boundary conditions When a boundary is closed (pipe)/fixed (string), it corresponds to a node (amplitude is equal to 0). When a boundary is open (pipe)/free (string), it corresponds to an antinode (amplitude is maximum).

85. Harmonics

Property:

Oscillating systems (a pipe: clarinet…or a string: guitar…) produce standing waves. The frequencies they can produce:

Depend on the velocity of the waves v Depend on the length of the oscillating system L

The only frequencies an oscillating system can produce are the ones which obey the laws stated in the table below. These frequencies are called harmonics.

Boundary conditions First harmonic Other harmonics (n: integer)

2 open ends (open pipe) OR 2 closed ends (guitar string)

f1 = v/2L λ1 = 2L

fn = n * f1 (λn = λ1/n)

1 open end and 1 closed end (closed pipe) f1 = v/4L λ1 = 4L

fn = (2n + 1) * f1 λn = λ1/(2n + 1)

Animation: Ondes Stationnaires 2 (www.scienceslycee.fr) Animation: Réflexion Onde (www.scienceslycee.fr)

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« deux obstacles » ; « sinusoïdal » ; v = 12.0 m/s. f = 1.50 Hz, and then f = 3.00 Hz, and then f = 4.50 Hz, and then f = 6.00 Hz and then f = 3.51 Hz

Exercise : Use the boundary conditions and the diagram below to derive:

the formula for λn the formula for fn

LAB: Melde’s experiment

Animation: http://www.sciences.univ-nantes.fr/sites/genevieve_tulloue/Ondes/ondes_stationnaires/melde.php

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5. Electricity and magnetism (15h)

5.1. Electric fields

86. Charge An action-at-a-distance force, different from the gravitational force is at work in certain situations: Pieces of paper are attracted to a plastic ruler (after it has been rubbed against a cloth). Hair stand up like spikes on one’s head after it’s been thoroughly combed. … This behaviour is due to a new property of matter called charge:

There are positive charges and negative charges. Like charges repel, and opposite charges attract. The unit for charge is the COULOMB (C).

Examples: A proton is a particle with a positive charge: e = 1.6 10-19 C

An electron is a particle with a negative charge: - e = -1.6 10-19 C Definition: e is called the elementary charge Any charge Q is a multiple of e : Q = N * e (N integer)

87. Conductors and insulators Some materials have electrons (called free electrons) that don't belong to any atom in particular and can travel across the whole material: they are conductors Example: metals, graphite (pencil)... Some material don't have such electrons: insulators

Example: glass, plastic...

88. Charging of an object

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89. Measuring the charge of an object

A positively charged rod is brought close to an electroscope. Therefore, negative charges

gather in the upper part of the electroscope, leaving the bottom part positively charged. The two bottom parts are both positively charged and therefore repel each other. The greater the charge in the rod, the greater the repulsion.

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90. Coulomb’s law

Two point objects of charges Q and q which are distant of r (distance between the centres of

the spherical objects), in a given medium (air, vacuum, water…) exert forces called electrostatic forces on each other.

These forces are: Of equal magnitude F, and direction. Along the same line (the line joining the centres of the objects). Of opposite senses. If Q and q have opposite signs, it is an attractive force. If Q and q have the same sign, it is a repulsive force.

Magnitude of the electric force

F = k Q q/r2

Units: Q and q in C r in m F in N

Definitions: k = 1/(4πε) ε : permittivity of the medium k = 1/(4πε0) = 9.0 109 SI ε0 : permittivity of free space (vacuum)

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91. Electric field

A point object of charge Q situated at point O creates an electric field E (P) (it is a vector) at any point P (which is at a distance rP away from O)

Magnitude of the electric field at point P

E (P) = k Q/rP2

Property: any charge q, situated at P feels an electrostatic force F = q * E (P) Units: F in N q in C E in N.C-1 Definitions: the electric field strength at a point is the force per unit charge experienced by a

small point positive charge at that point. Created by an object of charge Q its magnitude is:

E (P) = k Q /rP2

Two object are in O1 and O2:

Object 1 creates an electric field at P, E1 (P). Object 2 creates an electric field at P, E2 (P).

The electric field at P is therefore: E (P) = E1 (P) + E2 (P)

Remarks: The electric field inside a conductor at equilibrium is equal to 0. When a conductor at equilibrium is charged, the charges all lie at the surface of the conductor, and none remain inside of the conductor.

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92. Electric field lines A point particle Q creates an electric field is all of space. A map of this electric field can be drawn using field lines. Properties: E is parallel to a field line. A field line is oriented: it has the same sense as E. When field lines are close to each other, the magnitude of E is strong. Field lines never cross and never touch. Field lines cross a conducting surface with an angle of 90°.

93. Potential difference Property: When a charge (q) is located at point M, in an electric field E (M), it feels a force:

F = qE. Definitions: The electric potential difference between M1 and M2 is the work done per unit charge to move the charge from M1 to M2:

ΔV = V2 – V1 = W1 2/q The electric potential V(M) at point M is the work done per unit charge (W/q) to move a test charge (q) from infinity (V = 0) to M.

Units: (Volts) V = J.C-1 Properties: The change in electric potential energy (ΔU) when a charge (q) moves between 2 points of different electric potentials (ΔV = V2 – V1) is:

ΔU = q * ΔV

Defintions: The electric potential energy (U) stored in a charge (q) at a point of electric potential V is:

U = q * V

The magnitude of the change in electric potential energy of an electron (charge – e) moved across a ΔV = 1V is called an electronvolt: |ΔU| = e * 1 = 1,6 * 10-19 J = 1 eV

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94. Electric current in a metal

Remark: Conductors have free electrons (charge: - e) that can all start moving in the same direction under an electric field (or under an electric potential difference ΔV, often called voltage V)

Definitions: The electric current (I) flowing through a conductor (wire) is the amount of charge that goes through its cross section per unit time:

I = ΔQ/Δt Units: (Amps) A = C.s-1

Direct current (dc) means current which only flows in one direction. Remarks:

In an electric circuit, I flows from points of high V to points of low V.

When a voltage is applied across a conductor, the charges (q) all move inside the conductor (cross section Σ of surface area A), in the same direction, at a drift speed v.

The charges which will cross Σ between t and t + Δt are all inside a small cylinder of length v * Δt.

The volume of this small cylinder is therefore A * v *

Δt.

If n is the number of free charges per unit volume of the conductor, then the number of charges which will cross Σ between t and t + Δt is: n * A * v * Δt.

The total charge which will cross Σ between t and t + Δt is: ΔQ = q * n * A * v * Δt.

Therefore, the direct current (dc) inside the conductor

is:

I = ΔQ/Δt = n * A * v * q Exercise: Calculate the drift speed of electrons inside a copper wire of 1.0 mm diameter inside which a current of 2.0 A is flowing.

Density of copper: 8.94 g.cm-3 1 free electron per atom Molar mass of copper: 63.5 g.mol-1

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5.2. Heating effects of electric currents

95. Circuit diagrams

A circuit diagram is a simple representation of a real electric circuit.

Universal symbols are used to represent all the parts of an electric circuit (see page 4 of the data booklet).

Examples:

96. Current and Voltage Electric current flowing inside any component (light bulb, wire…) is represented by an arrow inside an electric circuit.

It can be:

Positive (when the arrow representing the current and the flow of electrons have opposite senses: I1) Or negative (when the arrow representing the current and the flow of electrons have the same sense) It is measured with an ammeter plugged in series inside the circuit (it measures ± I1 depending on how it is plugged) Voltage across a component is a potential difference:

V1 = Va – Vf V1 = Va – Vf is sometimes represented by an arrow going from f to a It can be

Positive (V1 > 0 voltage across a cell) Negative, Or equal to 0 (across a wire: V2 = Vb – Va = 0)

It is measured with a voltmeter plugged in parallel (it measures ± V1 depending on how it is plugged)

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Remark: V1 = Va – Vf is sometimes represented by an arrow going from f to a: And therefore Vf – Va is represented by an arrow in the opposite sense:

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97. Kirchhoff’s circuits laws Kirchhoff’s first law: at any junction in a circuit Σ I = 0 Kirchhoff’s second law: in any closed loop in a circuit Σ V = 0

I: current towards the junction V: voltage (potential difference) across an electric component

Example:

Kirchhoff’s first law: I1 + (-I2) + (-I3) = 0 I1 = I2 + I3

Kirchhoff’s second law: Loop 1: V1 - V2 - V4 = 0 Loop 2: V2 - V3 = 0

98. Heating effect of current Current flows inside an electric circuit because electrons move inside conductors. The energy carried by electrons is supplied by the power supply. When electrons travel through an electric component, they transfer some energy to the component: Light energy and thermal energy to a lamp. Thermal energy to a wire or an electric resistor. … The power (energy per unit time) supplied to an electric component is equal to:

P = V * I

I: current flowing through the electric component (Units: A) V: voltage across the electric component (Units: V) P: Power supplied to the electric component (Units: W)

99. Resistance Definition: The electric resistance of an electric component is:

R = V / I

I: current flowing through the electric component (Units: A) V: voltage across the electric component (Units: V) R: Electric resistance of the electric component (Units: Ω Ohms)

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100. Characteristics

Definition: The characteristic of an electric component is its the graph of V vs I.

Examples: a: Ohmic device (example: metal wire at a constant temperature)

b: Non ohmic device: filament lamp c: Non ohmic device: diode

101. Ohmic device/ Ohm’s law Definition: An ohmic device (called a resistor) has a constant electric resistance R (it does not depend on I or V). Symbol of an ohmic device:

Ohm’s Law The voltage across an ohmic device (constant R) through which a current I flows follows Ohm’s law:

V = R * I

Power dissipation Ohmic devices turn the electric energy provided by the power supply into thermal energy. This simple diagram represent how a power supply provides energy to a light bulb. The resistor (R) accounts for the wires and electric component which will heat up during the process and therefore dissipate part of the energy supplied. Power provided: Ptotal = V * I Useful power for the light bulb: Puseful = V1 * I Power dissipated (lost) through R: Pdissip = V2 * I Ohm’s Law: V2 = R*I Power dissipated: Pdissip = R*I2 = V2

2/R

Remarks:

The energy supplied (dissipated, used…) by the light bulb during Δt is related to power by the following equation: E = P * Δt

The law of conservation of energy implies that Etotal = Euseful + Edissip and therefore that Ptotal = Puseful + Pdissip

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Lab 10

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102. Combination of resistances

Resistors in series

Kirchhoff’s second law: V = V1 + V2 Ohm’s law: V1 = R1 * I V2 = R2 * I Therefore: V = (R1 + R2) * I The circuit a between points A and B is equivalent to circuit b if:

REQ = R1 + R2. Property: R1, R2, R3…. associated in series, are equivalent to a single resistor of resistance:

REQ = Σ Ri

Resistors in parallel Kirchhoff’s first law: I = I1 + I2 Kirchhoff’s second law: V = V1 = V2 Ohm’s law: V1 = R1 * I1

and so I1 = V1/R1 = V/R1

V2 = R2 * I2

and so I2 = V2/R2 = V/R2

Therefore I = V/R1 + V/R2 = V * (1/R1 + 1/R2)

If REQ is such that 1/REQ = 1/R1 + 1/R2

then I = V/REQ and V = REQ * I

The circuit a between points A and B is equivalent to circuit b if:

1/REQ = 1/R1 + 1/R2. Property: R1, R2, R3…. associated in parallel, are equivalent to a single resistor of resistance REQ such that:

1/REQ = Σ 1/Ri

Lab 10 Exercise: Calculate REQ for the following circuits

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103. Potential divider

A potential divider circuit enables to control the voltage across an electric device. In the

following examples, the electric device is a resistor (of resistance R). The voltage applied across A and B (V = VA – VB), is constant (independent from the rest of the circuit).

Circuit a

VR = R * V/(R + R’)

Circuit b

Definition: The electronic device of total resistance RT is called a potentiometer or rheostat. It is divided into 2 resistors R1 and R2 connected in series such that RT = R1 + R2. The values of R1 and R2 are chosen thanks to a slider, but the total resistance RT is fixed. When connected as shown on diagram b, R is connected in parallel to R1 (see the equivalent circuit).

VR = REQ * V/(REQ + R2) with 1/REQ = 1/R1 + 1/R

Therefore,

1

1

1

1

1

*

**

RRRR

RR

VRR

RR

V

T

R

Properties: The voltage across the electronic device (R in these examples) can be controlled.

Circuit a: R’ is a variable resistor. If R’ varies from 0 to R, then VR varies from V/2 to V. Circuit b: R1 can be varied from 0 to R. Therefore VR varies from 0 to V.

Remarks: Circuit b allows a greater range for VR. It will be used in order to control the voltage applied to an electronic device with the largest possible range. Circuit a is used when R’ is an LDR or a thermistor. A variation in R’ due to a change in temperature (thermistor) or a change in light intensity (LDR) will induce a variation in VR.

The measurement of VR enables to determine the temperature of the thermistor, or the light intensity falling on the LDR. It can also be used as feedback information in another part of the circuit.

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104. Resistivity The resistance R of a resistor depends on its nature and its size. If a resistor is made of a rod of length L and cross section of surface area A, then R is equal to:

R = ρ*L/A L: length of the rod (units: m) A: cross section of the rod (units: m2) ρ: resistivity of the material (units: Ω.m) Exercises: textbook 6 to 12 p 241

105. Ammeters and voltmeters

Voltmeter: A voltmeter is plugged in parallel to an electric device.

In order not to interfere with the behaviour of the circuit, it must not divert any part of the energy, and therefore the electric current going into the voltmeter has to be equal to 0. To that effect, a resistor with a very high resistance is added to the voltage measuring device. An ideal voltmeter is considered to have an infinite resistance. A real voltmeter has a high but not infinite resistance.

Ammeter: An ammeter is plugged in series with an electric device.

In order not to interfere with the behaviour of the circuit, it must not divert any part of the energy and therefore have a voltage equal to 0 (so that there is loss in potential energy between the two terminals of the ammeter). To that effect, its resistance must be equal to 0. An ideal ammeter is considered to have a 0 resistance. A real ammeter has a very small but not 0 resistance.

Lab 11

5.3. Electric cells

106. Cells Definitions: A cell is a simple power supply in an electric circuit. It turns chemical energy into electric energy thanks to a chemical reaction which takes place inside the cell. A battery is a DC supply of voltage and current, usually made of many cells. Symbol:

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107. Characteristic of a cell: internal resistance and emf

Definition: The characteristic of a cell is its V vs I graph.

Important features and definition:

The voltage across a cell (“supplied by a cell) depends on the current.

The terminal potential difference of a source is the potential difference it supplies (V).

The terminal potential difference at I = 0 (open circuit) is called the electromotive force (emf) of the cell (ε). It can also be defined as the total energy per unit charge supplied by a battery (or electrical source) around a circuit (ε = W/q).

V differs from ε because of the cells internal resistance (r) due to the wires… inside the cell:

V = ε – r*I Remarks: The current coming out of the positive terminal of a cell is positive (because the electrons come out of the negative terminal of a cell) An ideal cell has an r = 0.

The power wasted in a real cell is equal to: Plost = r*I2 The total power supplied by a real cell is equal to: Ptotal = ε *I

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108. Discharge characteristic of a cell

An ideal cell provides a constant voltage to a given electrical component. When plugged to a

given light bulb for instance, it delivers a constant current I, and therefore provides a constant terminal potential difference (V = ε – r*I) ie a constant energy supply to the light bulb. When there is no energy left in the cell, its voltage suddenly drops to 0.

A real cell has the following discharge characteristic:

The terminal potential difference of a cell quickly decreases when it is first used. It remains almost constant for a long period of time.

It very quickly drops to 0 when there is very little energy left (when it is almost completely discharged) Remark: A cell (or a battery) contains a certain amount of electric charge it can deliver to a circuit. When all this electric charge has been delivered, its voltage drops to 0. Definition: The Capacity of a cell (or battery) is the amount of charge it can deliver. Its most common unit is the Ah.

Exercises : Convert 1Ah into C

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Textbook: 13 to 16 p 243

109. Primary and secondary cells Primary cells can only be until they are discharged (V = 0: no more energy is stored). A secondary cell is a device which can be recharged: when V = 0, energy can be stored back into the cell (it is recharged). During its recharge, the current inside the cell goes in the opposite direction, in order to reverse the chemical reaction that takes place inside the cell when it is used as a power supply.

Example: Cell phone batteries are secondary cells

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5.4. Magnetic effects of electric currents

110. Effect of a bar magnet on the needle of a compass A bar magnet brought near a compass rotates the needle of the compass. When moved around the needle, it changes the orientation of the needle. The bar magnet :

Exerts a force at a distance on the needle. Creates a magnetic field (B) in the 3D space. Has a North (N) and a South (S) pole. N and S from 2 different magnets attract. N and N or S and S poles from 2 different magnets repel.

111. Magnetic field Magnetic fields B can be created by various devices (magnets, electric wires, solenoids…). It is a vector and has all the properties of a vector (direction sense, direction, vector addition…). The unit of a magnetic field is the Tesla (T).

Magnetic field created by bar magnets:

Materials affected by a magnetic field:

A magnetic field B has an effect on

A magnetic field B has no effect

Magnets Some metals (Iron, Cobalt, Nickel..)

Current-carrying conductors Moving charged particles

…

Some metals (Aluminum, Lead, Gold, Silver…) Plastic, wood, glass

Charged particles at rest …

Example: The Earth’s magnetic field in Paris has a magnitude of 4.7 * 10-5 T

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112. Magnetic field lines

Remark: A map of a magnetic field can be drawn using field lines. Properties: B is parallel to a field line. A field line is oriented: it has the same sense as B. When field lines are close to each other, the magnitude of B is strong. Field lines never cross. Examples:

A bar magnet creates a magnetic field (figure a). A Long straight conductor through which an electric current I flows creates a magnetic field. The field lines are circles perpendicular to the wire (figure b), which centres are located on the wire. The orientation of the field lines follow the right hand rule (thumb: sense of I: four other fingers: sense of the field lines). A solenoid (an electric wire wrapped around an iron core) through which an electric current I flow creates a magnetic field (figure c). The orientation of the field lines follow the right hand rule (thumb: sense of the field lines; four other fingers: sense of I)

Other examples:

Animation: www.sciences.univ-nantes.fr/sites/genevieve_tulloue/Elec/Champs/topoB.html

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113. Magnetic force

Force exerted on a charge moving in a magnetic field

A point charge q, moving at velocity v, at a point P in space where there is magnetic field B feels a force F. The characteristics of F are the following:

Direction: perpendicular both to B and to v Sense: use the right hand rule: Thumb: direction and sense of F Index: direction and sense of qv Middle finger: direction and sense of B Magnitude F = qvB sin (θ) θ: angle between B and v

Units: F in N q in C v in m.s-1 B in T

Force exerted on a current-carrying conductor in a magnetic field A current-carrying conductor (current I) of length L (the vector L has the sense and direction given by I, and a magnitude of L), inside a magnetic field B feels a force F. The characteristics of F are the following:

Direction: perpendicular both to B and to L. Sense: use the right hand rule: Thumb: direction and sense of F Index: direction and sense of L Middle finger: direction and sense of B Magnitude F = BIL sin (θ) θ: angle between B and L Units: F in N I in A L in m B in T

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6. Circular motion and gravitation (5h)

6.1. Circular motion

114. NOS Different types of circular motions (planets around the Sun, car in a roundabout, object attached to a string…) are due to different forces which share one common characteristic: they are perpendicular to the trajectory (they act radially) and act inwardly.

115. Characteristics Characteristics of a uniform circular motion (of radius r): The magnitude of the velocity (the speed v) is constant. Because velocity is always tangent to the trajectory, it is perpendicular to the radius of the circle. Acceleration is centripetal (always pointing towards the centre of the circle). Angular displacement: α = L/r (units: rad) Angular velocity: ω = v/r (units: rad.s-1) Time period: T = 2πr/v = 2π/ω (units: s) Frequency: f = 1/T = ω/2π (units: Hz = s-1)

Property: The magnitude of centripetal acceleration of a uniform circular motion is:

a = v2/r = ω2r = 4π2r/T2

116. Centripetal force and centripetal acceleration Definitions: a centripetal force is a force which always points towards the same point in space. a centripetal acceleration is always pointing towards the same point in space. Examples: the Earth exerts a gravitational centripetal force on the moon. the string of a pendulum exerts a centripetal force on the object of the pendulum. when a car makes a turn, the road exerts a centripetal force (friction) on the wheels. a fixed point charge exerts a centripetal force on any other charge. a magnetic force… Properties: a system experiencing a centripetal net force has a centripetal acceleration:

Fnet = ma = m ω2 r

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a system experiencing a net force perpendicular to its displacement can be in circular motion.

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Exercises:

1°/ A car of mass m = 600 kg is turning in a roundabout on a flat horizontal road at constant speed v = 50km/h. The radius of the roundabout is r = 20 m.

Calculate the friction force exerted by the road on the car. 2°/ An object of mass m = 500 g, attached to a thin rod of length L = 1.3 m, rotates in a vertical plane at the pace of 5.0 rotations per second. The speed of the object is constant. Determine the external forces exerted on the object. Determine the tension force exerted on the object both at the top and at the bottom of the trajectory. When is the rod most likely to snap? 3°/ A train (of mass m = 800 kg) on a roller coaster is circling around a vertical circular loop (of radius r = 20 m) at constant speed. What is the minimum speed it has to reach so that it doesn’t fall of the roller coaster when it reaches the top of the loop?

4°/ A car is circling around a frictionless track which has a banking of 20° relative to the horizontal. Draw a diagram of the external forces exerted on the car. Explain how the banking helps the car turn around.

5°/ Why does a motorcycle turn when it bends towards the ground?

6.2. Newton’s law of gravitation

Newton’s laws of motion are the foundation of deterministic classical physics. Along with Newton’s law of gravitation, they enabled to predict the motion of planets, falling

objects… As most laws in Physics, the law of gravitation does not explain the phenomenon of

gravitation, but only helps describing and analyzing its consequences.

117. Newton’s law of gravitation

Two point objects (or spherical objects of uniform density which masses are concentrated at their centres) of masses M and m which are distant of r (distance between the centres of the spherical objects) exert attractive forces called gravitational forces on each other.

These forces are:

of equal magnitude F. of equal direction. along the same line (the line joining the centres of the objects). of opposite senses.

F = G M m/r2

Units: M and m in kg r in m

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G is the universal gravitational constant: G = 6.67 10-11 SI

118. Gravitational field strength A point object of mass M situated at point O creates a gravitational field at any point P in space:

g(P) = G M/rP2 u(P)

u(P) is the unit vector going from P to O

Definition: the gravitational field strength at a point is the force per unit mass experienced by a small point mass at that point. Properties: Any mass m, situated at P feels a gravitational force F = m * g (P) Two object are in O1 and O2. Each object creates a gravitational field at point P in space (g1(P) and g2(P).

The gravitational field at point P is therefore:

g(P) = g1(P) + g2(P)

119. Uniform circular motion of the Moon around the Earth The Moon orbits about the Earth in a uniform circular motion: The radius of the trajectory is: r The time period is: T = 27.3 days = 2.36 106 s Exercises:

1°/ Show that the radius of the trajectory of the Moon is equal to r = 3.8 108 m. 2°/ Show that the planets circling round the Sun in a uniform circular motion all obey the

following law: R3/T2 = G * MSun/4π2 (R: radius of the trajectory T: time period)

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7. Atomic, nuclear and particle physics (14h)

7.1. Discrete energy and radioactivity

120. Energy transfers on a macroscopic scale When a stone is released from the top of a cliff, it will fall downwards, hit the ground and eventually stabilise there. Energy wise: the stone falls so that it decreases its gravitational potential energy. the stone eventually stabilises on the ground after it has transferred the gravitational potential energy lost during the fall. This energy is dissipated through heat, sound, work done on the ground (or on the stone itself which as a result might break)… throughout the fall, the evolution of the rock’s gravitational potential energy is proportional to its height, and therefore continuous. Overview: the less energy the stone has, the more stable it is. the gravitational potential energy of the stone is continuous. Property: some forms of energies (Epp, Echemical...) transform spontaneously into other forms of energies (Ethermal… ). The lower they are, the more stable the system is.

121. Discrete energy and discrete energy levels Property: Nuclei, atoms, or molecules at rest can be in different states of (potential) energy. These energy levels are quantized, which means they are discrete. A particle (nucleus, atom, molecule…) can’t be in a state of energy which value is in between two energy levels. The lower the energy state of a molecule, the more stable it is. Remark: In the microscopic world, energy is discrete.

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122. Transitions between energy levels Energy is a conservative quantity. Therefore, when a particle goes: From a high to a low energy level (E3 E1), it releases ΔE = E3 – E1 to its surroundings. From a low to a high energy level (E2 E3), it needs an energy supply of ΔE = E3 – E2.

Remarks:

Each particle has its own set of discrete energy levels.

The lowest energy level of a particle (E1) is called the ground state.

The other energy levels are called excited states.

The unit most commonly used are the eV, keV, or MeV

1 eV = 1,6 * 10-19 J 1 keV = 1,6 * 10-16 J 1 MeV = 1,6 * 10-13 J

123. Emission and absorption spectra of common gases The energy (Ewave, f) carried by light (or any EM wave) of frequency f is also quantized (cf Option A). It can only be an integral multiple of a small quantity (E) called a photon: E = h * f = h * c / λ f: frequency of the wave (s-1 or Hz) λ: wavelength in vacuum (m) h: Planck’s constant (h = 6,63 * 10-34 J.s) c: speed of light in vacuum (c = 3,00 * 108 m.s-1) Ewave, f = n * E n an integer An atom can absorb or emit EM waves by absorbing or emitting one photon at a time. The photons an atom can absorb or emit have specific frequencies which relate to the energy levels of the atom. These frequencies f have to verify the following equation:

h * f = Ei - Ej (Ei and Ej are two random energy levels of the atom)

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Example of the emission and absorption of photons by an H atom

An H atom can absorb or emit EM waves (photons) of such frequencies f1 = (E2 – E1)/h f2 = (E5 – E2)/h …. An H atom can’t emit or absorb a photon which energy does not correspond to an energy transition between two energy levels of the H atom. Exercise: 1°/ Calculate the energy transition between E5 and E2 in J. 2°/ Calculate the frequency and the wavelength of the emitted light. 3°/ Determine whether this light is IR, visible, UV….

Emission spectrum An atom or a molecule has its own energy levels. It can only emit a given set of EM waves which correspond to its own energy transitions. The emission spectrum of the atom or the molecule is made of all the wavelengths or frequencies it can emit. Remark: Two different atoms or molecules have their own energy levels and therefore their own emission spectra. Property: A particle has an emission spectrum of its own which is different from all the other emission spectra. A particle can be identified through the study of its emission spectrum.

Study of common gases Collection of an emission spectrum: A gas is submitted to an electric discharge. Its atoms or molecules reach excited states. They return to their ground state by emitting EM waves. The emission spectrum of the gas is obtained by the collection of these waves.

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Examples: H2, O2, He, and N2 are gases at normal pressure and temperature. Other elements (Hg, Na…) can also be studied in the gaseous state.

Remarks: when all the visible spectrum of white light is sent through a gas, it absorbs the exact same wavelengths it was able to emit when submitted to an electric discharge. The resultant absorption spectrum of the gas is show above. The emission lines and the absorption lines of the same element match. Exercise: Use the energy levels of the H atom to find the energy transitions which correspond to the wavelengths of the emission and absorption spectra of the H atom.

124. Composition and stability of a nucleus

Composition

An atom is made of: A nucleus with positively charged particles called protons. neutral particles called neutrons. Electrons which orbit the nucleus.

Notation: 14

6C designates a Carbon nucleus containing: 6 protons (which is what makes it a Carbon nucleus). 8 neutrons.

Definitions:

126C and 14

6C are Carbon nuclei with different numbers of neutrons. They are isotopes. A nuclide is an atom which has a specific number of protons and neutrons. The nuclides

146C and 12

6C are two isotope atoms.

Stability The protons inside a nucleus repel each other because of their positive charges. Therefore

there has to be another force holding the nucleons (protons and neutrons) together. It is called the strong force and is one of the four fundamental forces (cf 7.3).

Some nuclei have so many nucleons that the range of the strong force is not enough to keep

them together. They undergo a nuclear reaction (α decay). Some nuclei have too many protons so that the repulsion between them is too important.

They undergo a nuclear reaction (β+ decay).

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Some nuclei have too many neutrons. They undergo a nuclear reaction (β- decay).

Some nuclei are in an excited state. They undergo a nuclear reaction (γ decay).

125. Natural radioactivity The unstable nuclei transform into other nuclei. In the process:

they can emit particles (α particles, β particles)

they can emit energy (γ rays…)

α decay

(when there are too many nucleons) A

ZX A-4Z-2Y + 42He

23892U 234

90Th + 42He

42He (a He nucleus) is called an α particle

β- decay

(when there are too many neutrons)

A

ZX AZ+1Y + 0-1

e + 00ῡ 14

6C 147N + 0

-1e + 00ῡ

0-1

e is an electron also called a β- particle 00ῡ is an electron antineutrino

β+ decay

(when there are too many protons)

A

ZX AZ-1Y + 01e + 00υ

2211Na 22

10Ne + 01e + 00υ

0-1

e is a positron also called a β+ particle 00υ is an electron neutrino

Gamma decay

Some nuclei are unstable because there are in an excited state (like atoms and molecules, nuclei

also have quantized energy levels). They reach a lower energy level by emitting a photon (a γ ray).

A

ZY* AZY + 0

0γ

Remark: one nucleus can undergo a series of decays (, , or γ)

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126. Radioactive decay

Living things are made of Carbon atoms. Most of them are 12

6C (98,9%) or 136C (1,1%) but there is also a very small

proportion of 146C which remains constant throughout the life

of the organism. When it dies, the proportion of 146C starts to

decrease because of the β- decay of the 146C nuclides.

A is the activity of the sample: the number of decay reactions per second. N is the number of 14

6C nuclides in the sample. The graph of A (or N) vs time is a decaying exponential curve: A (t) = A0 * exp (-λ * t) It can also be written in the following way:

A (t) = A0 * 2/12t

t

Unit: The activity is in Bq (Becquerel). One Bq is equal to one disintegration per second.

Remark:

t1/2 is called the half-life of the 146C nucleus: every t1/2, A (or N) is divided by 2.

Exercise: 1°/ Determine the half-life (t1/2) of 14

6C. 2°/ At a given time t, a sample (from an organism which has died a very long time before t) has a magnitude A0. Determine the activity of the sample at: T ’ = t + 2 * t1/2 T ’’ = t – t1/2 3°/ Find and use the “Décroissance radioactive” (www.scienceslycee.fr) or “CRAB” simulations or any other online radioactive decay simulation.

Remark: all the decay curves (, , or γ) have the same shape.

127. Absorption characteristics of decay particles The particles emitted by nuclear reactions interact with matter. When a particle penetrates matter, it can cut bonds between atoms and create ion pairs This phenomenon can be very dangerous for living tissues.

a few characteristics of these particles

- (or -) γ

Charge

+2e

+ e (or – e)

0

Mass

Of a He nucleus

Of an electron

0

Ionizing property

Strong

(104 pairs/mm)

Moderate

(100 pairs/mm)

Weak

(1 pair/mm)

Penetrative power

Low (cm of air)

Moderate (mm of metal)

Strong (many cm of lead)

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128. Background radiation People on Earth are exposed to both natural and artificial radiations called background radiation. The human exposure to ionizing radiations has various sources:

The ground (Black sand in India contains Thorium nuclides which produce particles).

Air (from Radon gas which produces particles). Cosmic radiations (EM rays of various wavelengths). Food or water (14

6C or 4019K…).

Remark: The worldwide average natural dose is 2,4 mSv per year. Definition: the Sievert (Sv) is a unit which measures the biological effect of radiations on human tissues. (1Sv = 1J.kg-1)

7.2. Nuclear reactions

129. The unified atomic mass unit Definition: The unified atomic mass unit (u) is equal to 1/12th of the mass of a neutral 12

6C atom in its ground state. Remarks: 1 u = 1,661 * 10-27 kg

1 u is very close to (but smaller than): the mass of a proton mP = 1,673 * 10-27 kg = 1,007276 u the mass of a neutron mN = 1,675 * 10-27 kg= 1,008665 u This unit corresponds to the order of magnitude of the masses of nuclei and atoms.

130. Mass defect Property: The mass of any nucleus is smaller than the mass of all its nucleons. Definition: The mass defect of a nucleus AZX of mass mX is defined by:

Δm = (A – Z) * mN + Z * mP - mX

Exercise: The mass of the 56

26Fe nucleus is equal to mFe = 55,934937 u. Calculate the mass defect of 56

26Fe in u and in kg. Property: In a natural nuclear reaction, the mass of the products is smaller than the mass of the reactants.

Exercise: 6027Co undergoes a - reaction in which it transforms into Nickel (Ni).

1°/ Write the equation of this - nuclear reaction. 2°/ The antineutrino being a massless particle, calculate the mass lost during the reaction of one 60

27Co nucleus (in u). What do you think happened to this lost mass? Data (masses of nuclei): m (60

27Co) = 59, 93382 u m(60ZNi) = 59,9307864u

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131. The mass-energy equivalence

Property: Mass and energy can transform into each other. The energy ΔE released by the transformation of a mass Δm into energy or the mass Δm created by the transformation of an energy ΔE into mass are linked by:

ΔE = Δm * c2

ΔE: energy (in J) Δm: mass (in kg) c : speed of light in vacuum c = 3,00 * 108 m.s-1 Exercises:

1°/ Calculate the energy released by the - reaction studied in the exercise above. 2°/ Calculate the energy released by the transformation of 1 u of matter into energy: In J and In MeV Remarks: As E = m * c2, any energy divided by c2 is a mass. 1 J/c2 is a mass. (it is equal to 1/(3,00 * 108)2 = 1/(9,00 * 1016) = 1,11 * 10-17 kg) 1 MeV/c2 is a mass. (it corresponds to 1MeV of energy) Definition: The mass of a particle can be expressed in a new unit called the MeV/c2. Exercise: Show that 1 u = 931,5 MeV/c2 The conservation laws of nuclear reaction: The Number of nucleons is conserved in a nuclear reaction. The charge is conserved in a nuclear reaction. The mass is NOT conserved in a nuclear reaction.

132. Nuclear binding energy Definitions and remarks: The nuclear binding energy Ebinding of a nucleus AZX of mass defect mdefect is: Ebinding = mdefect * c2 It corresponds to the energy released by the following (hypothetical) reaction:

(A – Z) neutrons + Z protons AZX It can be interpreted as the energy which binds the nucleons together in AZX. The nuclear binding energy per nucleon E/A of a nucleus A

ZX is: E/A = Ebinding/A. The higher E/A is, the more stable AZX is. Remarks:

The most stable nuclei are the ones at the top of the graph. Nuclei with high A can undergo fission and form two smaller nuclei of higher stability. Two nuclei with small A can undergo fusion to form a heavier and more stable nucleus.

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133. Nuclear fission and nuclear fusion

Fission A fission nuclear reaction occurs when one heavy unstable nucleus is split into lighter nuclei.

Example (artificial transmutation): 10n + 235

92U 9438Sr + 140

54Xe + 2 10n Remark: this is the reaction mostly used in nuclear power plants

Fusion

A fusion nuclear reaction occurs when two light nuclei join into a heavier and more stable nucleus. Example : 2

1H + 31H 42He + 10n Remark: fusion is the main source of the energy released by the Sun.

4 11H 42He + 2 01e + 20

0υ + 200 γ

Exercises:

1°/ Calculate the energy released by the fission reaction 1 g of 23592U:

in J and in MeV Calculate the quantity of 0°C ice it could melt into 0°C liquid water (Lf = 334 kJ.kg-1). 2°/ Calculate the energy released by the production of one 42He in the Sun.

3°/ The power released by the Sun is equal to P = 4 * 1026W. Calculate the number of 42He created every second in the Sun. The mass of the Sun is equal to M = 2 * 1030 kg.

Assuming all this mass consists in 11H

all these 11H will transform into 42He at a constant rate (P = 4 * 1026W)

Calculate the expected time during which the Sun will produce energy. 4°/ Show that the energy released by a fusion or a fission reaction is equal to:

ΣEbinding (products) - ΣEbinding (reactants) 5°/ Calculate the binding energy per nucleon of 140

54Xe (in MeV) Data (nuclear masses): m(4

2He) = 6,6647 * 10-27 kg m(11H) = 1,007276 u

m(23592U) = 235,0439 u m(94

38Sr) = 93,9154 u m(14054Xe) = 139,922 u

Remark : an approximation (less than 1% off) for the mass of AZX is A * u

7.3. The structure of matter

134. The Rutherford-Geiger-Marsden experiment At the beginning of the XXth century, the atom was known to be neutral and made of both positively charged particles and negatively charged particles. In 1909, Rutherford, Geiger and Marsden were investigating the scattering of alpha particles by a thin gold film. Alpha particles (4

2He nuclei and therefore positively charged particles) were sent at very high speed (3% of the speed of light). Most of the particles got through the film, some of them being slightly deflected from their original path because of the charged particles within the atom. But 1/8000th of the alpha particles bounced back. This amazing fact led Rutherford to the conclusion that the positively charged part of the atom had to be very massive so that it could stop and reverse the motion of a high speed and positively charged alpha particle.

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A first convincing model of the atom soon followed which consisted in: A positively charged nucleus which is very massive and has a size of about 10-15 m. Negatively charged electrons with very small masses, very far away from the nucleus. An overall size of about 10-10 m. This was the first time it was shown that matter is mostly empty space.

135. The Standard model Since Rutherford’s experiment, more than 300 new particles have been found. This called for a new theory (and classification) of these particles which is what the standard model is about.

Elementary particles and interactions

The proton and the neutron are not elementary particles. They are made of quarks:

the Up quark (u) of charge + 2/3 e the Down quark (d) of charge – 1/3 e.

These quarks were discovered in 1968 by firing electrons at protons at very high speeds inside a particle accelerator (deep inelastic scattering experiments). The different directions the electrons (charge – e) bounced off (were scattered by) the protons suggested that the protons were made of smaller particles of different charges.

They have familiar characteristics which account for well-known properties: A mass which accounts for the magnitude of the gravitational interaction. An electric charge which accounts for the magnitude of the electromagnetic interaction. They also have other characteristics which account for other (new) properties: A spin number which has to do with their magnetic properties.

A lepton number, « colour charge », baryon number (and other numbers) which account for other properties (ie the way they combine and interact together).

The standard model Many (more or less stable) particles have been hypothesized and discovered which take part in a new theory of matter and interaction called the standard model. In the standard model: The elementary particles of matter are fermions. The exchange particles (which are bosons) are the elementary particles of the fundamental interactions. (see 138). Definition: An elementary particle is a particle which can’t be split: it does not have a substructure.

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136. Basic presentation of Fermions: Quarks and Leptons

There are two categories of elementary particles of matter which are both fermions:

Quarks and Leptons. The quarks can combine into bigger particles (which are therefore not elementary): The combination of 1 quark and 1 antiquark makes a meson (which is a boson). The combination of 3 quarks makes a baryon (which is a fermion). Baryons and mesons are called Hadrons. There are three generations (or families) of fermions which correspond to the order of magnitude of their masses: Mass (3rd generation) > Mass (2nd generation) > Mass (1st generation). Quarks, leptons and hadrons have various characteristics such as mass, charge, spin, lepton number, baryon number, strangeness… For every particle there is an antiparticle which has: The same mass as the particle. Opposite electric charge, opposite Baryon number… as the particle. When a particle and its antiparticle collide, they transform (annihilation) into other particles. Some particles are their own antiparticle (photon, Z boson…)

Quick overview

QUARKS

Particles Antiparticles

Generation Generation

1st 2nd 3rd 1st 2nd 3rd

Name Charge B Name Charge B

Up (u ) Charm ( c ) Top ( t ) +2/3 e 1/3 u c t -2/3 e -1/3

Down ( d ) Strange ( s ) Bottom (b ) -1/3 e 1/3 d s b 1/3 e -1/3

LEPTONS

Particles Antiparticles

Name Charge L Name Charge L

Electron ( e ) Muon ( ) Tau ( ) -e 1 e +e -1

e 0 1

e 0 -1

Notation: x particle x antiparticle

x neutrino for the particle x x antineutrino for the particle x

e (or e-) is called an electron

e (or e+) is called a positron

Remark: e, μ, τ, (electron, muon, tau) are negatively charged particles:

their lepton number is +1.

e , μ , τ , (antielectron known as the positron, anti-muon, anti-tau) are positively

charged particles: their lepton number is -1.

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137. Examples, definitions and properties

Example

A muon is an elementary particle of the second generation. It has a much higher mass than the electron, a very short half-life, and can decay into an electron and other particles. It can be formed when EM radiations hit the upper atmosphere.

Charge : – e Mass : 207 melectron Half-life : t1/2 = 2,2 µs

Exercise: π+ ( du ), π- ( du ), π0 ( uu or dd ) are pions.

What kind of particles are they (leptons, fermions, baryons….)?

Definitions and properties

A neutrino is a neutral and (almost) massless particle. Flavours: Up, Charm, Top, Down, Bottom, and Strange are the six different types of quarks, called the « flavours » of the quarks. Strangeness: The s quark has a strangeness of -1

The s quark has a strangeness of +1

All the other quarks have a strangeness of 0.

Lepton numbers (L): e, , and their neutrinos have a lepton number of +1.

e , , and their neutrinos have a lepton number of -1.

Baryon number (B): Quarks have a Baryon number of 1/3

Antiquarks have a Baryon number of -1/3

Spin: The spin is a positive number. It is a half integer for fermions and an integer for bosons. A quark has a spin of 1/2.

Exercise: uud and duu are two baryons

1°/ Determine the charge of both of them.

2°/ uud is a nucleon. Is it a proton or a neutron? Derive the name of duu .

3°/ Determine the combination of quarks of the other nucleon.

138. Quark confinement

Quarks never exist on their own and make combinations to form Hadrons: the combination of two quarks forms a Meson and the combination of three quarks forms a Baryon.

Indeed, the attractive force between two quarks does not decrease as they are moved apart from each other but remains constant (and can even

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increase). Therefore, if an attempt is made to separate two quarks within a hadron, the energy involved (work done) quickly gets so important that it is transformed into matter (see Option A) in the form of a quark-antiquark pair which lies in between the two quarks.

139. Gauge bosons and the fundamental forces Four fundamental forces can explain the behaviour (reaction, motion…) of all matter:

The gravitational force The electromagnetic force The strong force The weak force

Quick overview of their effect on matter

Force Range Relative strength effect

Gravitational infinity 1 Motion and shape of planets, stars, galaxies

Electromagnetic infinity 1035 Binding electrons and nuclei in atoms, creation of magnetic fields

Strong 10-15 m 1037 Binding nucleons inside nuclei, fusion processes in stars

Weak 10-18 m 1024 Transmutation of elements ( reactions), breaking up of stars,

transformation of a quark into another quark

Exchange particles

In the standard model’s description of forces, when an interaction exists between two particles, it is thanks to an exchange particle (gauge bosons) which carries the information of the interaction between the two particles. Gauge bosons:

Force Exchange particle Acts on

Gravitational graviton All particles

Electromagnetic photon Electrically charged particles

Strong gluons Quarks and gluons

Weak W+, W-, and Z0 Quarks and leptons

Remarks:

A gauge boson also has a mass, a charge, and various numbers (« colour charge »….). The smaller the range of the force, the more massive the exchange particle: Photons have no mass, and W boson are very heavy.

140. Feynman diagrams and conservation laws

Conservation laws

All the fundamental forces which account for interactions as well as transformations of matter follow conservations laws. In all the interactions, decays, or reactions involving particles:

Mass/energy, Charge, Baryon number, Lepton number are conserved. Strangeness is conserved when there is a strong interaction involved not conserved when a strange (s) particle decays through the weak force.

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Feynman diagrams

All the interactions and transformation of particle can be represented by a simple diagram called a Feynman diagram. It represents: Time (along the vertical direction) (examination papers might use the vertical axis) Space (from left to right) (examination papers might use the horizontal axis) Fermions (before and after the reaction, interaction…) Exchange particles (the point where three lines meet is called a vertex) Properties of conservation: Charge, Lepton number, and Baryon number are conserved at a vertex. Strangeness is conserved at a vertex only when there is strong interaction involved (but it is not conserved when the weak interaction is involved) Examples:

Example of a - reaction

146C 14

7N + 0-1

e + e

It can be seen as the transformation of a neutron into a proton, emitting an electron an electron neutrino

n p + e- + e

Conservation laws:

n p + e- + e

Quarks u d d u u d Charge 2e/3 – e/3 – e/3 = 0 2e/3 + 2e/3 – e/3 -e = 0 Baryon number 1/3 1/3 1/3 = 1 1/3 1/3 1/3 = 1 Lepton number = 0 +1 -1 = 0

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Exercises: 1°/ What do these simple Feynman diagrams represent? The following equations represent interactions or transformations of particle

β+ decay: p n + e + e

p and e- collision: p + e- n + e

2°/ Draw the Feynman diagrams which correspond to theses equations. 3°/ Show that they follow the laws of conservation.

141. The Higgs boson Particles which have masses interact through the gravitational force which exchange particle is the graviton (not yet discovered). The Higgs boson doesn’t account for the gravitational interaction but for the property of having a mass. Particles have a mass because they react to a field (Higgs field) that permeates all space. The stronger the interaction between a particle and Higgs field, the stronger the mass of the particle. Mass is therefore not a primary property of a particle of matter, but a secondary one caused by the interaction between the particle and Higgs field.

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8. Energy production (8h)

8.1. Energy sources

142. Primary and secondary energy sources Definitions: A primary source of energy has not been submitted to any transformation process (Sun, Crude oil, Coal, Wind…) A secondary source of energy is the product of the transformation of a primary energy source: Crude oil is turned into petrol. Wind is turned into electricity…

143. Renewable and non-renewable energy sources Renewable sources of energy: A renewable source of energy comes from resources which are naturally regenerated over a short period of time relative to the human timescale. Most of these sources derive either directly from the Sun (Thermal energy, EM wave energy…) or indirectly from the Sun (Wind, Hydropower, Biomass…) Some come from other natural phenomena (Geothermal energy, Tidal energy…) Non-renewable sources of energy: A non-renewable source of energy exists in limited supply and won’t be replaced if it is used up. Fossil fuels (Coal, Oil, Natural gas) and nuclear fuels (Uranium …) are non-renewable sources of energy.

144. Specific energy and energy density of fuel sources Definition:

the specific energy of an energy source is the amount of energy stored per unit mass (J.kg-1). Its energy density is the amount of energy stored per unit volume (J.m-3).

Fuel Specific energy (MJ.kg-1)

Energy density (MJ.L-1)

Uranium 8 * 107 1.5 * 109

anthracite coal 34

diesel fuel/residential heating oil 48 35.8

gasoline 44.4 32.4

ethanol 26.4 20.9

liquid hydrogen 142 5.6

wood 6 - 17 13

natural gas 55.5 0.0364

Water falling through 100m 10-3 10-3

Exercises: 1°/ A power station has a power output of P and an efficiency of ε. It burns a mass M of coal every second. What is the best estimate of the specific energy of the coal? 2°/ A power station has a power output of 500 MW, an efficiency of 27%. It uses natural gas as a fuel that has a specific energy of 56 MJ.kg-1. Determine the rate of consumption of

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natural gas in the power station.

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145. General feature of a Power plant

A power plant turns a primary

source of energy such as Oil, Coal or Uranium into electricity.

The energy released by the primary source is used to boil water so that the mechanical energy (motion) of the steam produced is transformed into AC current through an alternator.

An alternator is an electronic device which transforms mechanical energy into electricity. It consists of a magnet which rotation inside a circular coil of electric wire produces AC current (cf 11.2). The energy efficiency of an energy transformation or transfer can be illustrated by a Sankey diagram: The thickness of an arrow is proportional to the amount of energy it represents.

146. Fossil fuel power station

A fossil fuel power station uses either Coal, Natural gas or Oil as the primary source of chemical energy. Their combustion releases thermal energy (as well as EM energy) which is used to produce the steam which mechanical energy is turned into electric energy.

147. Nuclear power station

All nuclear power stations use fission reactions to produce energy and most of them use the

fission of 23592U. A possible reaction equation is:

10n + 235

92U 23692U 94

38Sr + 14054Xe + 2 10n

The transformation of one 235

92U requires one 10n and produces two 1

0n which can trigger the fission of two other 235

92U nuclei causing therefore a chain reaction. But only the slower 10n can

turn 23592U into the unstable 236

92U which goes through the fission process. Therefore, a nuclear reactor consists of: Fuel rods made of enriched Uranium (between 3% and 4% of 235

92U). A moderator (a water pool) which slows down the emitted 10n. Control rods which can increase or decrease the rate at which the fission reactions occur.

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The energy released by the rods

heats up the water surrounding them. This hot water is then used to heat a secondary circuit of water, which is outside of the nuclear reactor, through a heat exchanger. The steam produced in the secondary circuit is then used to produce electric energy. Remark: The chain reaction has to be

controlled to avoid catastrophic consequences (Fukushima : thermal meltdown of the reactor …)

self-sustained (if the amount of fuel is smaller than the critical mass, the reaction will die out)

Safety issues:

The most dangerous radioactive wastes consist of the rods after they’ve been used and don’t contain enough 235

92U anymore. They have to be taken care of in a waste processing plant and then stored for a very long period of time (most of the half-lives of the remaining nuclei have an order of magnitude of 105

years and more) All the materials which have been used in the power plant (concrete, water in the primary

circuit, gowns…) have a small radioactive activity and have to be stored (for a much shorter period of time until they and not significantly radioactive anymore).

The extraction of the nuclear fuels in Uranium mines causes great damages to the workers as well as to the environment.

There can be a risk of using the nuclear fuel to make nuclear weapons (which requires to enrich the fuel even more).

148. Wind generator

A wind generator transforms the translational mechanical energy of the wind into rotational mechanical energy of its rotor blades. This rotational mechanical energy is then turned into electric energy through the same similar process involving an alternator.

Assuming the wind transfers all its kinetic energy to the blades (which is very unlikely as it would mean that the air would go still right after it went through the turbine) and there are no energy losses inside the turbine, the power produced by a wind generator is given by the following equation:

Pmax = ½ Aρv3

Pmax : power (in W) A = π * r2 : area swept by the blades of radius r (in m2) ρ : air density (kg.m-3) v: speed of the air (m.s-1)

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149. Pumped storage

Once produced,

electricity has to be consumed and can’t be stored on a large scale.

It is therefore not a very flexible source of energy.

Pumped storage can make it more flexible by turning electricity into gravitational potential energy during periods of low electricity demands, and turn it back into electricity during periods of high demands.

Electric energy is used to pump water from a low reservoir into a high reservoir and is thus transformed into gravitational potential energy. When electric energy is needed, water comes down through the same circuit and the same reversible pump-turbine/motor-generator device.

150. Solar power cell On average over a period of 24h, the solar intensity reaching Earth is equal to 350 W.m-2. A

solar cell is a device made of a semi-conductor material which turns EM radiations into electric energy.

Heating panels also use the energy directly provided by the Sun and turns it into heat which can be used in houses as a heating system, or transformed into other forms of energy (electricity…).

Exercises:

1°/ What surface of solar panels do we need to provide enough hot water to 65 houses? Average power needed per house to heat water: 2.0 kW. Average solar intensity: 450 W.m-2. Efficiency of a solar panel: 21%. 2°/ Are there any disadvantages in using solar power to provide hot water?

8.2. Thermal energy transfer

151. Thermal energy transfers Three different types of thermal energy transfers are studied in this section.

Conduction

A solid bar is held horizontally above a table. A candle is lit up underneath its left end. Although it is not directly heated up, the right end of the bar experiences a gradual increase in its temperature. Microscopic interpretation: Particles (atoms, ions, molecules, electrons…) experience random motion inside matter at any temperature which speed increases as the temperature increases. When the left end of the bar is heated up, the atoms (or molecules) inside the bar tend to vibrate faster about their equilibrium positions. When a “fast” atom hits its slower neighbour, it transfers part of its energy to

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the slow atom which therefore starts to vibrate faster. This process called induction carries on along the bar until it reaches its right end. Remarks:

If the material contains free electrons (electric conductor), they can move freely along the whole bar and will therefore contribute to a quicker heat transfer. Electric conductors are better thermal conductor than electric insulators.

Conduction requires a contact/collision between particles. Therefore, conduction is almost negligible in liquids and gases (relative to convection).

Convection A window separates a 20°C room of a house and the -5°C outside air of a cold winter night. If the window is opened, the temperature of the room will go down very quickly, at a much faster rate than conduction could account for. Indeed, a flow of hot air will go out of the room towards the low temperature air which will make a flow of cold air enter the room. This current of many atoms and molecules is called convection. Remark: Convection requires a translational motion of many particles in the same direction and is therefore a bulk property. In a conduction process inside a solid, the particles vibrate above a fixed position: the energy transfer happens on a microscopic scale between two neighbour atoms. Examples: Hot air is less dense than cold air. Therefore, the air heated up by a radiator goes straight up towards the ceiling of a room, because it is surrounded by colder room temperature air. Sea breeze, magma convection, wind…

Thermal radiation

Two identical cans, one of them painted matt black, are filled with hot water and put on a table in a 20°C room. The cooling curves of these two cans shows that the plain can cools slower that the black can. Interpretation:

The thermal energy lost by the water and the can through conduction or convection is the same for both cans as they are identical (except for their colour) at the beginning (same size, shape, material, weight, same amount of hot water).

The curves are steeper at the beginning because the bigger the difference between the hot system (a can) and its cold surroundings (the room temperature air), the quicker the thermal energy transfers. The difference between the curves is due to thermal radiation. A body of temperature T radiates EM wave energy which amount depends on its temperature (and also on other factors such as colour…).

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Property: A body which is a good emitter of certain EM radiations is a good absorber of the same EM radiations (and vice versa). Remark:

A body can be a good emitter for certain wavelengths and a bad emitter for other wavelengths.

152. Black-body radiation

Definition: A black body is a body which absorbs all the EM radiations that fall on it.

Being a perfect absorber, a black body is also a very good emitter. The EM radiation spectrum emitted by a black body only depends on its temperature T.

Wien’s displacement law links the wavelength of the peak of the curve λmax to the temperature T of the black body: λmax * T = 2.90 * 10-3 m.K. Remark:

Just like ideal gases do not exist but provide a good approximation of real gases at low pressures, black bodies do not exist either.

The curves for real bodies can be of various shapes. Some bodies called grey bodies almost behave like black bodies, others behave rather differently.

153. Emissivity and Albedo

A body at temperature T emits EM waves (the black-body emission curves have been shown in the previous paragraph). The Stefan-Boltzmann law gives the power it radiates through all the wavelengths is emits.

P = ε σ A T4

σ = 5,67 10-8 W.m-2.K-4 Stefan-Boltzmann’s constant P: power emitted (in W)

A: surface area of the body (in m2) T: temperature of the body (in K) ε: emissivity of the body (0 < ε < 1)

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Remark: ε = 1 for a black body as it is a “perfect” emitter. ε is close to 1 for a grey body Definition: the albedo of a surface is the fraction of the EM incident power it scatters: a = PSCATTERED/PINCIDENT

Remark: The albedo depends on the wavelength of the incident EM radiations (although most of the applications in the syllabus do not take that characteristic into account). Earth’s albedo varies daily and is dependent on season (cloud formations) and latitude. The global annual mean albedo will be taken to be 0.3 (30%) for Earth.

154. The solar constant

The power emitted by the Sun is estimated at Pout = 3.9 * 1026W. The intensity of the light (power per unit surface) which reaches the Earth (situated at a distance DSun-Earth

= 1.5 * 1011 m from the Sun) perpendicular to its surface is therefore equal to:

S = Pout/(4πDSun-Earth

2) = 1.4 kW.m-2. Vocabulary: S is called the solar constant. Exercise: Assuming the Sun acts as a black body, prove that the power emitted by the Sun is indeed equal to Pout = 3.9 * 1026W. (RSun = 6.9 * 108 m TSurface of the Sun = 5800 K)

155. Calculations Example 1: Let’s calculate the power emitted and absorbed by a black body of surface area A and at temperature Tbody surrounded by air (emissivity ε) of temperature Tair.

The black body emits EM radiations and the power emitted is Pout = σA Tbody 4 The air surrounding the black body also emits EM radiations. Pout’ = ε σA’ Tair 4

A’ represents the surface area of the air. Therefore, the relevant surface area needed to calculate the power emitted by air and incident on the black body is equal to A. The power coming from the air and absorbed by the black body Pin is equal to Pout’ because by definition a black body absorbs all the EM radiations that fall on it: Pin = ε σA Tair 4. If Pin > Pout the temperature of the black body will increase. If Pin < Pout the temperature of the black body will decrease.

If Pin = Pout the black body is at thermal equilibrium. Therefore, at thermal equilibrium: Tbody/Tair = ε1/4.

Material Emissivity Albedo

Black body 1 0

Ice 0.98 0.60

Coal 0.95

Water 0.65

Dark wet soil 0.05

Ocean 0.06

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Example 2: Let’s calculate the temperature Tbody of a black body located at a distance D from the Sun.

The power radiated by the Sun is: PSun = εSun σ 4π Rsun2 TSun

4 The power received at D over a surface area of A is: Preceived = A/(4π D2)* PSun

Preceived = εSun σ A (Rsun/D) 2 TSun

4 The power absorbed by the black body of surface A perpendicular to the incident rays is

therefore equal to: Pabsorbed = εSun σ A(Rsun/D) 2 TSun

4. The power emitted by this black body is: Pemitted = σ A Tbody

4 If the black body is at thermal equilibrium, its temperature is therefore such that:

Tbody 4 = ε Sun * (Rsun/ D) 2 TSun

4

Remark: if the black body is spherical, the surface area of the object is equal to 4π Robject

2. This value should be used to calculate the power emitted by the object but a different value should be used to calculate the power received by the object. Indeed, half the surface doesn’t receive any EM radiations and most of the rest of the surface does not receive them at right angle. The cross section (A = π Robject

2) of the object is the relevant value. Exercise: Calculate the temperature of the Earth assuming both the Earth and the Sun are black bodies. (DSun-Earth

= 1.5 * 1011 m RSun = 6.9 * 108 m TSun = 5800 K) Exercise: Calculate the temperature of the Earth assuming it has an average albedo of 0.30 and it therefore not a black body, and an emissivity equal to 1. Mean intensity of the sunlight incident on Earth: I = 350 W.m-2.

156. The greenhouse effect

The temperature calculated is the last exercise is clearly not the average temperature experienced at the surface of the Earth which is about 15°C. This is due to the fact that a fraction of the energy emitted by the Earth is absorbed by some gases in the atmosphere and remitted back to Earth. Explanation: The gases of the atmosphere are almost transparent to the mostly visible light sent by the Sun. These visible radiations along with all the other radiations (IR, UV…) are absorbed by the Earth. The radiations emitted towards space by the Earth, a body at a temperature of about 300 K are in the IR range (around 8 µm according to Wien’s law). Some of the gases in the atmosphere (CH4, H2O, CO2, N2O) absorb these radiations, and then reemit them in all directions (section 7.1). Therefore, a large proportion of this energy is sent back to Earth. These gases are called greenhouse gases. Remarks:

The greenhouse effect is a natural phenomenon which is very fortunate since life would probably never have developed on Earth if its temperature had been equal to 256 K. All the greenhouse gases have a natural and a man-made origin: An increase in the average temperature tends to increase the amount of H2O in the atmosphere. The combustion of fossil fuels increases the presence of CO2 in the atmosphere. Global warming is most likely a consequence of the increase in the proportion of greenhouse gases in the atmosphere, due to human activities.

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The distribution of the energy levels of a greenhouse gas accounts for its absorption of some IR radiations. Every radiation absorbed by a gas corresponds to a transition between two energy levels.

Climate Model LAB

157. Energy balance in the Earth surface-atmosphere system

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9. (AHL) Wave phenomena (17h)

9.1. Simple harmonic motion

158. NOS

All periodic oscillations can be described in terms of harmonic oscillators (Fourier analysis): any periodic signal can be broken down into a sum of sine or cosine signals of specific frequencies.

Therefore, harmonic oscillators are found in many areas of physics: sound waves, light, electric circuits…

159. The defining equation of SHM

a (t) = - ω2 x (t)

160. Energy changes Example 1 (cf point 62): Horizontal oscillations

a = - k/m x a = - ω2 x with ω2 = k/m

Displacement: x(t) = x0 * cos(ω*t + φ)

x(t) = x0 cos(ω*t) if φ = 0 Time period: T = 2π/ω = 2π (m/k)1/2 Velocity: v(t) = δx/δt = - x0 * ω sin(ω*t) v2 = x0

2 ω2 sin2 (ω*t) = ω2 x02 [1- cos2 (ω*t)]

v2 = ω2 [x02 – x(t)2]

v = ±ω [x02 – x(t)2]1/2

Kinetic energy :

EK(t) = ½ m v2 = ½ m ω2 [x02 – x(t)2]

Elastic potential energy:

Epe(t) = ½ k x(t)2 Conservation of energy: ET(t) = EK(t) + Epe(t) = ½ m v(t)2 + ½ k x(t)2

= ½ m ω2 [x02 – x(t)2] + ½ k x(t)2

= ½ k [x02 – x(t)2] + ½ k x(t)2

ET(t) = ½ m ω2 x02

ET remains constant (as long as there is no friction)

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Example 2: Oscillations of a pendulum:

Object of mass m String (no mass) of length L

The proper resolution of this problem involves maths outside of the IB syllabus:

It can be proved that: d2θ/dt2 = - g/L * θ d2 θ/dt2 : second derivative of θ (“acceleration a” of θ)

Therefore, if we call θ a displacement x, we get a = - g/L * x a = - ω2 x with ω2 = g/L

Time period: T = 2π/ω = 2π (L/g)1/2

In this case, without friction, it is ET = Epp + EK which is conserved (same graphs as in example 1)

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9.2. Single-slit diffraction

161. NOS Every point of the aperture behaves like a secondary point source of spherical wavelet. Light therefore goes in all directions The amplitudes of the wavelets add up to lead to a non-intuitive light intensity pattern.

162. The nature of single-slit diffraction

Properties: square aperture: θ = λ/b (point 79) circular aperture: θ = 1.22 λ/b First minimum: Zmin = d/2

As Zmin/D = tan (θ) = θ (θ small) Therefore Zmin = D* λ/b

Secondary maxima:

Maximum 1st secondary max 2nd secondary max 3rd secondary max

Light intensity 100 % 5 % 2 % 1 %

Remarks: Zmin increases when b decreases

Zmin is proportional to λ White light contains rays of various λ: diffraction

of white light creates iridescence. All the waves diffract (mechanical and non

mechanical waves) Animation: Interférences Diffraction 1 (www.scienceslycee.fr): select « 1 fente ».

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9.3. Interference

163. Young’s double-slit experiment Two slits of aperture b, separated by a distance d, a distance D away from the interference pattern. The direction of propagation of the waves (of wavelength λ) is perpendicular to the slits.

164. Modulation of two-slit interference pattern by one-slit diffraction effect

Interference pattern: aperture b Interference pattern: very thin aperture b The interference pattern is modulated by the one slit diffraction pattern (aperture b for each slit).

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Animation: Interférences Diffraction 1 (www.scienceslycee.fr): select « 1 fente » and then « 2 fentes »

165. Multiple slit and diffraction grating interference patterns The number of slits is now increased but: The aperture of each slit remains the same (b) The distance between the slit remains the same (d)

The bright fringes: Are much brighter (the light intensity is much higher). Are much thiner (the width of a bright fringe is much

smaller). Keep falling at the same place:

n * λ = d * sin(θ) n: integer If there are a large number of slits (> 100), there is no light except at certain θ:

n * λ = d * sin(θ)

166. Thin film interference

When a light beam (wavelength in vacuum λ0) hits a thin film (thickness d made of a transparent material of refractive index n) perpendicular to the film, part of the incident beam is reflected which produces the first reflected ray, and part of the incident beam follows another path which produces the second reflected ray (see diagram). Consequence: when the first and the second rays combine (principle of superposition), a localized interference pattern is produced which can be seen by an observer looking at the film (from above ie from a location in space where the incident ray is coming from)

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Explanation:

1- The second reflected ray travels an extra distance relative to the first reflected ray. extra distance: 2d induced phase difference: 2d * (2π / λ) λ: wavelength in medium of refracted index n

2d * (2π n/ λ0) λ0: wavelength in vacuum

2- When a ray is reflected on a surface made of a material of greater refractive index, a π phase difference occurs (see diagram reflection 1 and reflection 2) between the reflected ray and the incident ray.

Overview: Therefore, the phase difference between the two reflected rays is φ = 2d * (2π n/ λ0) - π. They will constructively interfere if:

2d * (2π n/ λ0)- π = m * 2 π m = 2n * d/λ0 - 1/2 (m + 1/2) * λ0 = 2n * d m an integer They will destructively interfere if: m * λ0 = 2 n * d m an integer Example: Iridescence (interference) in peacock feathers or soap bubbles.

9.4. Resolution

167. The size of a diffracting aperture

An optics system creates a single image point of an object point. But, when light goes through the optics system (lens, telescope, eye…), it diffracts because

the system has a size (a certain aperture) The point-image is therefore not a point, but has a certain size.

If two object points are very close to each other, their images made by the system may

overlap so that it is impossible to see them separately: the two images are not resolved.

168. The resolution of simple monochromatic two-source systems Two light beams coming from two objects produce two diffraction patterns:

Rayleigh’s criterion of resolution: The maximum of the 1st pattern should not fall closer than the first minimum of the 2nd pattern.

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If the aperture is circular (lens of a telescope, microscope, eye…):

Each central maximum has an angular width of 1,22 λ/b Rayleigh’s criterion means that

θ ≥ 1,22 λ/b

Examples: Two objects are resolved if: D/2L = d/2l = tan(θ/2) ≈ θ/2 ≥ 1,22 λ/2b Telescope: D ≥ 1,22 λL/b

Microscope: d ≥ 1,22 λl/b (and often l ≈ b in a microscope so d ≥ 1,22 λ)

169. The resolvance of diffraction gratings When two rays of different wavelengths (λ1 and λ2, and λ1 > λ2) come from the same source (and therefore from the same direction), it is possible to resolve them (separate the 2 rays) using a diffraction grating. Indeed, the places where the peaks fall:

Depend on the wavelength (sin(θ) = n * λ/d) Are very thin (more slits leads to thiner/brighter peaks and therefore better resolution) Resolvance criterion: two rays (λ1 and λ2, and λ1 > λ2) coming from the same source will be resolved if:

λ/Δλ ≤ N*m Δλ = λ1 - λ2 λ1 ≈ λ2 = λ N: number of slits m: order or diffraction

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9.5. Doppler effect

170. NOS Observation: when a police car is moving (siren on):

towards you : you hear a high frequency note away from you: you hear a low frequency note

The Doppler effect has applications is very different fields (medicine, astrophysics, radars…)

171. Sound waves

When a source S emits a sound wave, it leaves S with a velocity v relative to the medium (the air) which is only determined by the characteristics of the medium (density, T…) and NOT by the speed of the emitter. The source S has a velocity uS. The observer O has a velocity uO.

Notations:

Changes to the wavelength

S is moving so the distance between two consecutive crests (the wavelength) changes

λ’ = λ + uS * T = v/f + uS/f = (v + uS)/f

Animation: Effet Doppler (www.scienceslycee.fr)

Velocity (relative to the medium)

Of S: us

Of sound: v

Of O: uO

Sign

> 0 If S moves away

from O

> 0 If the sound wave goes

from S to O

> 0 If O moves towards S

Relative to S

Relative to O

Wavelength λ λ’

Frequency f f ’

Speed of sound v + uS v + uO

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Changes to the frequency

If O is moving, so the speed of the sound wave relative to O changes: v’ = v + uO Consequence: As v’ = λ’ * f ‘

f ‘ = v’/ λ’ = (v + uO)/[(v + uS)/f] = f * (v + uO)/(v + uS) Therefore f ‘ = f * (v + uO)/(v + uS)

Moving S Fixed O

Fixed S Moving O

Frequency

f ‘ = f * v/(v + uS)

f ‘ = f * (v + uO)/v

Applications: Measure of the speed of blood cells with ultra-sound waves

172. Light waves Doppler effect also happens with light waves. Although there is no medium needed for the propagation of light waves, the change in the perceived frequency is determined by the following equation:

Δf/f = Δλ/λ = v/c

Δf = f – f ‘ Δλ = λ’ – λ

c: velocity of light in vacuum

v: velocity of S relative to O (v > 0 if S is going away from O)

Applications: Radars Δf/f = 2 v/c v : velocity of car relative to fixed S and O

Redshift: A star is moving away from the Earth (v > 0) f – f ‘ > 0 which means that f ‘ < f λ’ – λ > 0 which means that λ’ > λ (λ increases)

The light waves received on Earth have slightly greater wavelengths than expected (Redshift).

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10. (AHL) Fields (11h)

10.1. About gravitational and electric fields In section 10.1, the notions will be:

- Either introduced or illustrated by one specific example. - Given definitions and general properties which therefore apply to all the possible cases.

173. Gravitational force and field

Gravitational force Two point masses m1 and m2 separated by r exert on each other an attractive force FG, called the

gravitational force, of magnitude:

FG = G * m1 * m2/r2

Gravitational field and potential Example:

A point object of mass M creates a gravitational field g (a vector) at any point P (which is at a distance rP away from the point object of mass M). Its magnitude is:

g(P) = G*M/rP2

g(P) is always pointing towards the central mass M.

General properties:

At any point P, g can be created by one or many masses (M1 creates g1, M2 creates g2…)

Therefore, as g is a vector, at any point P: g(P) = g1(P) + g2(P) + g3(P) + …

Any mass m, situated at P feels a gravitational force:

F = m * g (P) Units: F in N m in kg g in N.kg-1 Definitions:

The gravitational field strength at a point is the force per unit mass experienced by a small point mass at that point.

The gravitational potential difference (ΔVg) between two points in space is the work done

(by all the gravitational forces) per unit mass to move a mass from one point to the other: ΔVg = W/m

The gravitational potential Vg(P) at point P is the work done (by all the gravitational forces) per unit mass (W/m) to move a test mass (m) from infinity (where Vg = 0) to P.

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174. Gravitational field lines and equipotential surfaces

Gravitational field lines Example:

A point object of mass M creates a gravitational field is all of space.

A map of this gravitational field can

be drawn using field lines. General properties: g is parallel to a field line. A field line is oriented: it has the same sense as g. When field lines are close to each other, the magnitude of g is strong. Field lines never cross nor touch. Exercise: Draw the gravitational field lines close to the surface of a massive celestial body.

Gravitational equipotential surfaces Example: When a mass m moves along one of the dotted lines, its displacement δd is tangent to the dotted line. As g is always pointing towards M, so is the gravitational force F exerted on m. Therefore F and δd are always perpendicular to each other.

Along such a path:

No work is done on m (W = 0). The gravitational potential doesn’t change:

(ΔVg = W/m = 0). Definition: An equipotential surface is a region of constant gravitational potential Vg. General properties:

A field line is always perpendicular to an equipotential surface. A mass can move on an equipotential surface without work being done on the mass. Equipotential surfaces can’t cross nor touch.

The sense of g always goes from a high Vg to a low Vg. The study of two equipotential surfaces leads to the following trends:

When they get closer to each other, g in that region is quite high. When they get further apart from each other, g in that region is quite low.

Exercise: Determine the equipotential surfaces close to the surface of a massive celestial body. Determine the equipotential surfaces around a spherical mass.

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175. Overview Definition:

The gravitational potential energy of a mass m at a point of potential Vg: Epp = m * Vg (see 173). General properties:

Along a line perpendicular to the equipotential surfaces (ie along a field line), the magnitude (and the sense) of g is determined by:

g = - ΔVg/Δr

Units: m, M: in kg r: in m F: in N

Vg: in N.kg-1.m g: in N.kg-1 Epp: in N.m = J G = 6,67 10-11 N.kg-2.m2

Remarks: The “-“ sign in g = - ΔVg/Δr gives the sense of vector g. For example, Vg created by a point mass increases as the distance r away from the mass increases. Therefore, ΔVg/Δr is positive. The “-“ sign accounts for the fact that g is directed towards the point mass and not away from it. F, g, Vg and Epp are not independent (see diagram) Illustration of the link between F and g:

Exercise: BEFORE reading anything about electric fields: List all the definitions and formula on gravitation. Compare the equations defining the gravitational force and the electrostatic force. Derive a similar list of definitions and formula about electric field, potential….

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176. Electric force and field

Electric force

Two point charges q1 and q2 separated by r exert on each other a force Fe of magnitude:

Fe = k * q1 * q2/r2

If q1 and q2 have the same signs: Fe is repulsive If q1 and q2 have opposite signs: Fe is attractive

Electric field and potential (Go back to 5.1) Charges create electric fields in the 3D space surrounding them.

Examples:

177. Electric field lines and equipotential surfaces

Electric field lines (Go back to 5.1) Exercise: Draw the electric field lines created by 2 identical positive charges separated by d. Draw the electric field lines created by 2 identical negative charges separated by d. Draw the electric field lines created by 2 identical but opposite charges separated by d. Draw the electric field lines created by 2 parallel identical plates bearing opposite charges separated by d.

Electric equipotential surfaces (Go back to 5.1) Example:

When a charge q moves along one of the dotted lines its displacement δd is tangent to the dotted line.

As E is always in the radial direction (in this case pointing away from Q because Q > 0), so is the electric force F exerted on q.

Therefore F and δd are always perpendicular to each other.

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Along such a path:

No work is done on q (W = 0). The electric potential doesn’t change (ΔVe = W/q = 0).

Definition: An equipotential surface is a region of constant electric potential Ve. General Properties:

A field line is always perpendicular to an equipotential surface. A charge can move on an equipotential surface without work being done on the charge. Equipotential surfaces can’t cross nor touch.

The sense of E always goes from a high Ve to a low Ve. The study of two equipotential surfaces leads to the following trends:

When they get closer to each other, E in that region is quite high When they get further apart from each other, E in that region is quite low.

Exercise: Determine the equipotential surfaces for the 4 examples in the Electric field and potential paragraph. Determine the motion of a charged particle (charge q) entering a region between 2 parallel plates bearing opposite charges (potential difference between the plates: V) separated by a distance d. The initial velocity v of the particle is constant and parallel to the plates.

178. Overview Definition:

The electric potential energy of a charge q at a point of potential Ve: Ep = q * Ve General properties:

Along a line perpendicular to the equipotential surfaces, the magnitude (and the sense) of E is determined by:

E = - ΔVe/Δr

Units: q, Q: in C r: in m F: in N

Ve: in N.C-1.m = V E: in N.C-1 Ep: in N.m = J k = 9,0 109 N.C-2.m2

Remark: F, E, Ve and Ep are not independent (see diagram) Example:

The electric potential created by one charge Q at a distance r away from Q: Ve = k*Q/r

Ve = 0 at r = infinity If Q > 0 Ve increases as r decreases If Q < 0 Ve decreases as r decreases

The electric potential energy of a charge q at a distance r from a charge Q is: Ep = k*q*Q/r

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179. Examples

Example 1: Gravitational potential created by one mass M at a distance r from M:

Vg = - G*M/r Vg = 0 at r = infinity Vg decreases as r decreases

Gravitational field created by one mass M at a distance r from M: g = - ΔVg/Δr = - G*M/r2

Gravitational potential energy of a mass m at a distance r from a mass M is: Ep = -G*m*M/r

Example 2: Electric field and potential created by a charge sphere or radius R. Properties (read pages 412 - 413 from textbook):

The charges on a charged sphere reside on the outside of the sphere. The electric field inside a charged sphere is equal to 0.

The electric potential inside a charged sphere is constant. Exercises: Draw the electric field lines created by one charged sphere. May 2012 HL Paper 2 TZ2: A5; B1 part 1 All the exercises from textbook page 426.

180. NOS Reflect on the similarities and differences between electric and gravitational phenomena.

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Reflect on how the mathematical tools allow to make links between different parts of physics.

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10.2. Fields at work

181. Escape speed An object of mass m at a point P (which is at a distance r from the center of a planet of mass M) wants to escape from the gravitational attraction of the planet.

Therefore: It has to reach infinity (where F = 0). It must be given enough kinetic energy at P in order to reach infinity.

Energies: At P EP (r) = m* Vg = - G*m*M/r

At infinity: EP (∞) = m* Vg = 0

Conservation of energy: EP (r) + EK (r) = EP (∞) + EK (∞) EK (r) = - EP (r) + EK (∞)

½ m*v2 = G*m*M/r + EK (∞)

Therefore, the minimum speed that must be given to the point mass at r is such that: ½ m vESC

2 = G*m*M/r vESC

2 = 2G*M/r vESC = (2G*M/r)1/2

Definition: the minimum speed an object located at r from the center of a planet of mass M needs to escape from the planet is called the escape speed:

vESC = (2G*M/r)1/2

182. Orbital motion, orbital speed and orbital energy An object of mass m is undergoing orbital uniform circular motion around a planet of mass M (cf 6.2). Its orbital speed at a distance r away from the centre of the planet is:

vORBIT = (a * r)1/2 = (G * M/r)1/2

Kinetic energy at r EK = ½ mv2

Gravitational potential energy at r EP = m * Vg

Total orbital energy EP + EK

½ G *m* M/r > 0

-G * m*M/r < 0

-½ G * m*M/r < 0

Exercises: Find vORBIT of a geostationary satellite. Find vORBIT, vESC, EK, EP… for a charge particle undergoing orbital uniform circular motion around charged particle Q. Describe what happens to the height and speed of a satellite experiencing a friction force as it goes around the Earth. Exercise: A charged particle (charge q) enters a region of uniform magnetic field B. The velocity of the particle is constant (it travels along the x-axis), and the direction of the magnetic field is out of the page. Assuming the particle will undergo a uniform circular motion, find the radius of the path and draw that path.

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11. (AHL) Electromagnetic induction

(16h)

11.1. Electromagnetic induction

183. Electromotive force (emf) A loop of Copper is connected to a voltmeter (or an oscilloscope) and a bar magnet is moved close to it, with a certain velocity. Observation: A voltage across the loop is measured while the magnet is moving. Property: The relative motion between a conductor and a magnetic field creates an induced voltage ε (electromotive force: emf) across the conductor.

Induced emf across a moving straight wire.

A horizontal straight conductor of length L moves upwards (velocity v) in a region of uniform magnetic field B (out of the page).

If the wire is not connected, its free electrons are at rest because they can’t move along the wire. But they feel a magnetic force Fm = e*v*B.

According to the first law of motion, they have to experience another force which cancels Fm out so that they can be at rest. This force is an electric force Fe.

The existence of this force implies that an electric field E, and

therefore a voltage across the wire ε have been induced by the motion of the wire.

Fm and Fe cancel out so: Fe = e*v*B Fe is pointing to the left Fe is an electric force so: Fe = e*E = e* ε/L

The induced emf across the wire has a magnitude of ε = B * v * L

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184. Faraday’s law of induction

The same experiment is carried out in order to determine whether ε depends on a series of parameters: Magnitude of B. Orientation of the magnet relative to the loop. n: normal to the loop θ: angle between n and B. Area inside the loop (A). Number of loops (N). Rate of change of B. Observations:

ε changes sign if The poles of magnet are inversed Motion changes sense ε increases if The area (A) of the loop increases The angle (θ) between n and B is closer to 0 The magnet is closer to the loop (the magnitude of B is higher) There are two loops instead of one. When neither B, A nor α change, ε = 0.

The quicker the changes, the higher the magnitude of ε.

Consequence: Faraday’s Law ε = - ΔΦ/Δt Definitions: Φ = B * A * cos(θ) the magnetic flux Units: B in T A in m2

When there are N loops, ε = - N * ΔΦ/Δt

Or ε = - ΔΦlink /Δt Φlink = N * Φ The magnetic flux linkage Exercise: Prove that the unit of ε is V. Remark: an induced emf is produced if: A is changed (modification f the size of the loop) B is changed θ is changed (rotation of the magnet or the loop)

185. Lenz’s law A coil in connected to a resistor in series. A bar magnet can move closer or away from the coil. V and I are measured. Observations: When the magnet is moved, a current I and therefore a voltage across R are induced.

When B is increased (magnet brought closer), I < 0 and V < 0.

When B is decreased (magnet moved away), I > 0 and V > 0.

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Interpretation: The coil reacts to the increase in the magnitude of B by creating an induced magnetic field B’ which sense is opposite to the sense of B. When the magnet approaches, it transfers energy to the circuit. Indeed: B’ created inside the coil repels the magnet, and slows it down. The kinetic energy lost if transformed into electrical energy and then heat (lost through R). Lenz’s law: The current induced by a change in the magnetic flux through a loop is such that it

opposes that change.

Exercises: May 2012 HL Paper 2 TZ1: A6 May 2012 HL Paper 2 TZ2: A6 May 2011 HL Paper 2 TZ1: B4 part 2 May 2011 HL Paper 2 TZ2: A6 May 2010 HL Paper 2 TZ2: A4

11.2. Power generation and transmission

186. Alternating current (ac) generators

emf of a rotating coil in a constant B According to Faraday’s law:

ε = - N Δ/ Δt = - N* Δ(B * A *cos (θ))/ Δt B and A are constant

if θ = * t (constant angular speed )

then ε (t) = N* B * A * * sin (*t) Property: The induced emf ε(t) created by a coil rotating

at angular speed in a uniform B is sine shaped. Remark: The rotation of a magnet (Rotor) about a fixed coil (Stator) has the exact same effect. Definitions:

An AC (Alternating Current) generator consists of a magnet rotating () next to a coil. An induced emf ε (t) and therefore an induced current i(t) are created when the generator is connected to an electric circuit. Property: An AC generator is connected in series to a resistor R:

ε (t) = N* B * A * * sin (*t) = εmax * sin (*t)

i (t) = N* B * A * * sin (*t)/R = Imax * sin (*t)

f = /2 the frequency of ε (t) and i(t)

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Remarks: An AC generator can generate many signals (sine, rectangular, triangular…).

When increases:

The frequency f = /2 increases. The amplitudes of ε (t) and i(t) increase. Property: The power delivered by a cell is P = V * I. Therefore, the average power delivered (Paverage) by an AC generator can be calculated by averaging (over time) the quantity ε (t) * i(t). The result of this calculation is:

Paverage = Imax * εmax/2 Remark: A cell delivering a constant voltage (V = εmax/√2) will create a current going through the same resistor R (I = Imax/√2). The power it delivers is therefore equal to P = Imax * εmax/2. Definitions: The root mean square (rms) value of an alternating current (or voltage) is the value of the direct (constant) current (or voltage) that dissipates power in a resistor at the same rate. The rms value is also known as the rating.

Irms = Imax/√2 Εrms = εmax/√2

Therefore, Paverage = Irms * εrms.

187. Transformers A transformer is made of two separate coils inside which there is an Iron core. One coil is the primary circuit: N1 loops, ε1 and i1. The other is the secondary circuit: (N2, ε2, i2)

Thanks to the iron core, the flux is channeled from the primary circuit to the secondary circuit so that

it remains the same: 1 = 2

which leads to Δ1/Δt = Δ2/Δt

As ε = N * Δ/ Δt εmax1/ N1 = εmax2/ N2 εmax1/ εmax2 = N 1 / N2

In an ideal transformer, there is no energy loss from the primary to the secondary circuit. Therefore: P1 = P2

so εmax1 * Imax1 = εmax2 * Imax2 and Imax2/Imax1 = N 1 / N2

Remark: Most of the transformers are not ideal (flux leakage, heating, Eddy currents…)

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Use of step up transformers

Electric power (Pproduced) is transferred from a power plant to a house through transmission lines. The wires of the line have a resistance R. If the voltage applied across the lines is V: Pproduced = V * I

I = Pproduced/V current in the lines. The power lost by the heating of the wires is equal to: Plost = R * I2 = R * (Pproduced/V)2 In order to reduce Plost as much as possible, V has to be very high. That’s the reason why electric power is at very high voltages (400 000 V). Step up transformers are needed to change the voltage produced by the power plant into such very high voltages.

188. Diodes The ideal characteristic of a diode shows that: As long as the voltage across the diode is < 0 V, no current goes through (I = 0 A).

When the current is > 0 A, the voltage across the diode is = 0 V

An ideal diode is an electric component which: Blocks all < 0 currents (in which case the voltage is < 0 V) and brings them to I = 0 A. Lets all > 0 currents go through (in which case the voltage is = 0 V)

189. Diode bridges: half-wave and full-wave rectification

Half-wave rectification circuit An alternative voltage is produced between

A and B.

When Vin < 0: Vdiode < 0 and I = 0 A. When Vin > 0: Vdiode = 0 V and I (> 0) goes through. The voltage across R being equal to Vout = R * I, When Vin < 0: Vout = R * I = R * 0 Vout = 0 When Vin > 0 : Vin = Vout + Vdiode = Vout Vout = Vin

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A half-wave rectification circuit: turns the negative voltages to 0 keeps the positive voltages.

Full-wave rectification circuit

An alternative voltage is produced between A and B.

When Vin > 0: When Vin < 0: Positive current follows the path ACDFEB Positive current follows BEDFCA

Iout > 0 Iout > 0 VCD = VEF = 0 so Vout = Vin > 0 VFC = VDE = 0 so Vout = - Vin > 0

A full-wave rectification circuit: turns the negative voltages into positive ones (absolute value). keeps the positive voltages. Remark: If you had a capacitor to the circuit, Vout is almost constant. The circuit acts like a

“peak detector”.

LAB: diode bridge rectification circuit

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11.3. Capacitance

190. Capacitance A capacitor is an electric device made of: Two parallel plates (conductors of area A)

Separated by d. An insulator (permittivity ε) between the plates.

Charge of a capacitor

When a voltage E is applied between the two terminals of a capacitor, a current (i > 0) flows into the circuit and the electrons travel in the opposite sense. Therefore plate 1 gradually becomes negatively charged (q-) and plate 2 becomes positively charged (q+), creating a charge separation. The current eventually stops when the voltage Vcapacitor across the capacitor reaches E. Properties:

q- = - q+ because q- + q+ = 0 q+ is proportional to the voltage V across the

capacitor. q+ = C * V C: the capacitance of the capacitor units: Farad (F): 1 F = 1 C.V-1

C = ε*A/d Definition: A dielectric is an electrical insulator which is polarized when placed inside an electric field. Various dielectrics can be used in order to increase the capacitance of a given capacitor (example: paraffin has an ε = 2,3 ε0). ε / ε0 is called the dielectric constant of the medium of permittivity.

191. Capacitors in series and parallel

Two capacitors in parallel

In parallel, i = i1 + i2 : As i1 = dq1/dt and i2 = dq2/dt (cf 5.1)

i = dq1/dt + dq2/dt = C1 * dV1/dt + C2 * dV2/dt

In parallel, V = V1 = V2: i = (C1 + C2) * dV/dt

By definition, i = dq/dt dq/dt = d (C * V)/dt if we define C = C1 + C2 q = C * V

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C = C1 + C2 + C3 …is the equivalent capacitance of many capacitors (C1, C2, C3 … ) in parallel.

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Two capacitors in series

In series, i going through both capacitors is the same:

i = dq1/dt = dq2/dt = C1 * dV1/dt = C2 * dV2/dt

In series, V = V1 + V2: dV/dt = dV1/dt + dV2/dt = i * (1/C1 + 1/C2) = i/C

if we define 1/C = 1/C1 + 1/C2 i = d(C * V)/dt By definition, i = dq/dt so q = C * V

The equivalent capacitance C of many capacitors (C1, C2, C3 … ) in series is defined by:

1/C = 1/C1 + 1/C2 + 1/C3 …

192. Charge of a capacitor At t = 0, Vin the voltage between A and B, goes from 0 V to V0. Vin = VR + VC

VR = R * i (t) = R * dq/dt (t) and VC = q (t)/C As Vin = VR + VC,

V0 = q (t)/C + R * dq/dt (t)

Therefore dq/dt (t) + 1/RC * q (t) = V0/RC Properties : τ = R * C is the time constant of the

circuit. If VC (0) = 0, then

VC (t) = V0 * [1 – exp (-t/τ)] VR (t) = V0 * exp (-t/τ)

Therefore,

q (t) = C * V0 * [1 – exp (-t/τ)] q (t) = q0 * [1 – exp (-t/τ)]

i (t) = V0/R * exp (-t/τ) = i0 * exp (-t/τ)

Remark : when the charge of the capacitor is over: The voltage across the capacitor is VC = V0

The energy stored in the capacitor E = ½ C * V02

Property: The energy stored in a capacitor (voltage V and capacitance C) is E = ½ C * V2.

LAB: RC circuit, time constant…

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12. (AHL) Quantum and nuclear physics (16h)

12.1. The interaction of matter with radiation

193. The photoelectric effect

Experiment 1 Set-up:

A Cathode (C) and an Anode (A) are put in a vacuum and plugged to a circuit which can apply and control a voltage V between A and C (V > 0 means that VA > VC).

UV light (only one frequency) is sent on C. Both V and the intensity of the light I can be

changed and controlled. Observations:

1- For certain values of V, a current (i) flows through the circuit because some electrons go from C to A through vacuum.

2- There is a stopping potential (Vstop) below which no current is measured, whatever the intensity of the light.

3- When V increases, i increases. 4- The current i reaches saturation for large

magnitudes of V. 5- The maximum current imax is proportional to the

intensity of the light (I).

Experiment 2

Set-up: The same set-up is used with monochromatic lights of

different frequencies. Observations:

6- Vstop depends on the frequency of the light. 7- The higher the frequency, the higher Vstop. 8- Below a certain f0, no electrons are expelled (no

current measured), whatever the intensity of the light.

Remark 1: Energy is transferred from light to electrons in C which puts them into motion. Remark 2: Two observations are in contradiction with the wave theory of light:

1- Whatever the intensity of light, if f < f0, no electrons are expelled from C. 2- Whatever the intensity of light, and whatever the frequency, there is (almost) no time

gap between the instant light hits C and the occurrence of the current in the circuit. Indeed, any light wave, given enough time, whatever its frequency or intensity, could manage to build up enough energy in C in order to expel some of its electrons.

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Einstein’s postulate 1/ Light (of frequency f) carries energy in small amounts (quanta): E = h * f (photon). 2/ One photon transfers its energy to one electron (one to one interaction). Definitions: Φ = h * f0 Φ is the work function of the C. f0 is the threshold frequency of C. Interpretation of the photoelectric effect (with V = 0) if h * f < Φ No electron is emitted One photon does not provide enough energy to free one electron

If h * f = Φ The energy provided by a photon is just enough to expel the electron from C, without any EK.

if h * f > Φ One photon expels one electron which

leaves with a maximum kinetic energy EK max = h * f – Φ. Remarks: The slope of the │e * Vstop│vs f graph is h. Millikan’s experiment enables to determine h. Φ depends on the material (ΦFe ≠ ΦCu …) Exercise 1: How do Einstein’s hypothesis account for all the observations made in the two experiments? Show how h can be derived from these experiments. Exercise 2: When UV light (λ = 253,7 nm) is sent on a Potassium (K) Cathode, Emax = 3,14 eV. When visible yellow light (λ = 589 nm) is sent on the same electrode, Emax = 0,36 eV. Derive Planck’s constant from these measurements. Derive ΦK Find the threshold frequency of Potassium. Exercise 3: Light of various frequencies is sent on a photoelectric Cathode. The stopping potential is measured for every frequency.

f (1014 Hz) 5.09 5.20 5.49 6.10 6.88 7.41

Vstop (V) 0.20 0.25 0.37 0.62 0.94 1.16

Calculate the maximum EK with which the electrons are expelled when f = 5.49 * 1014 Hz

Draw the Vstop vs f graph. Derive the value of the work function of the Cathode. What is the response of the Cathode to an EM radiation of 680 nm?

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194. Wave-particle duality

Light

Light can do diffraction which is a wave property unaccounted for by the particle theory. The amount of energy carried by light is quantized in integral values of h*f. This behaviour reveals the particle nature of light which is unaccounted for by the wave theory. Wave-particle duality: light cannot be described just as a wave or just as a particle, but has to be described as both.

Matter Experiment: An (very weak) electron beam is sent towards two very close and very narrow slits. The white dots show the impact of the electrons on the screen.

Observations: (a), (b): very few impacts, randomly distributed. (d), (e): many impacts showing an interference pattern Conclusion: The electrons do not behave like particles which would have mostly landed on the screen along the initial direction of the beam. The electrons behave like waves when they reach the two slits. Wave-particle duality also applies to small particles of matter such as electrons (and protons, neutrons…). The wavelength of a particle of mass m, and velocity v (momentum p = mv) is given by the De Broglie equation:

λ = h/p Remarks: In 1924, De Broglie had hypothesized that matter should exhibit the same wave-particle behaviour as light.

The first evidence of electron diffraction was produced in 1927 when an interference pattern was observed after an electron beam was sent on a Nickel crystal.

Exercise: Calculate the wavelength of : A grain of rice (m = 20 mg) thrown at v = 10 m.s-1 An electron orbiting around its nucleus at v = c/100 Calculate the wavelength of an electron with 100 eV of kinetic energy? How does it compare to the atomic scale?

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Animation: video in french “dualité onde corpuscule” (www.scienceslycee.fr)

195. Quantization of angular momentum in the Bohr model for Hydrogen

Bohr’s model for Hydrogen

Rutherford’s model of the atom in which electrons orbit a small nucleus was a great step forward in the understanding of the structure of matter. It took into account the fact that most of matter is in fact empty and consists of pure vacuum. Nevertheless, it still had some flaws:

Electrons were considered like point particles. An electron being an accelerated charged particle, it should emit EM waves and therefore lose energy at such a quick rate that it should collapse on the nucleus very quickly (10 ns). In 1913, Niels Bohr put forward the following assumptions:

1- There are some radii where the electron is in a stationary state (doesn’t emit EM waves).

2- These radii are quantized. 3- The electrons can move from one stationary state to another only by emitting or

absorbing a quantum of EM radiation (photon). Definition and property:

The angular momentum of an electron (mass m) rotating (velocity v) about a nucleus along a circular path (radius r) is L = mvr.

The quantization of the stationary states (assumption 2) was derived from the original assumption that the angular momentum of the electrons was quantized:

L = n * h/2π n: integer h: planck’s constant

Energy level of a Hydrogen atom

Energy of an electron in a circular orbit (r):

The electron (charge –e) orbiting around an H nucleus (charge +e) has an electric potential energy (Epe = 0 when the electron is very far away from the nucleus):

Epe = - ke2/r. The electron also has a kinetic energy:

EK = ½ m v2 = ½ ke2/r.

The total energy of an electron orbiting an H nucleus is therefore E = -½ ke2/r. (1) Quantization of the energy levels:

The angular momentum is quantized mvr = n * h/2π (2) The electron is going in a uniform circular motion so mv2/r = ke2/r2 (3) Therefore: from (3) v = (ke2/mr)1/2 from (2) r = nh/(2πmv) = nh/[2πm(ke2/mr)1/2] (4)

From (4) r2 = n2h2/(4π2mke2/r) and r = n2h2/(4π2mke2) (5)

The quantized radii are rn = n2 * h2/(4π2mke2) From (1) and (5) E = -1/2 ke2/[ n2h2/(4π2mke2)] = -2π2 mk2e4/(n2h2) The quantized energy levels are : En = - 2π2 mk2e4/(h2)* 1/n2 = E1/n2

En = -13.6/n2 in eV

Exercise: 1°/ Show that E1 is indeed equal to -13.6 eV. 2°/ Calculate the first 4 energy levels of the H atom.

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Compare them to the ones used in 7.1.

196. The wave function

Bohr’s model gives good results for the Hydrogen atom (atomic spectrum) but doesn’t succeed in describing atoms or ions containing more than one electron. Moreover, it is not supported by a general theory which could properly account for the wave-particle duality of both light and small particles of matter.

In 1926 Schrödinger introduced the notion of the wave function as a way of dealing with

these small system behaving so strangely. In Newtonian mechanics the state of a particle of mass m is described by its position OM(t), its velocity v(t) (or its momentum p(t)). But such a localized description of a particle (with a definite position and velocity at any time) is not possible for small particles which exhibit wave properties for which the notion of single and unique coordinates is irrelevant (a wave is spread out in 3D space).

The wave function ψ(r) is a complex function which describes the state of a particle at r.

Remarks: ψ has no physical meaning. P(r) = │Ψ2(r│ * ΔV represents the probability of finding the particle at t in a small volume ΔV situated at r. Ψ2(r) represents the probability per unit volume or probability density. Solving Schrödinger’s equation enables to determine the wave functions describing a system.

197. Measurement of the position of a particle Note: Unfortunately, the word particle is used for small systems which can exhibit both wave and particle behaviour.

A particle behaving like a wave does not have a specific position, and the only thing we have access to is its probability P(r) of being at a certain location in a small volume surrounding r. Its position can only be known when it stops behaving like a wave, and exhibits particle behaviour. Therefore, when the position of a particle is measured, an interaction occurs between the measuring tool and the particle which makes it stop behaving like a wave and acquire a precise position.

Interpretation of the electron diffraction experiment Like a wave, one electron goes through both slits. The two parts of the wavefront then interfere on the other side of the slits. When the wavefront hits the screen, it stops behaving like a wave and becomes a particle, located at a precise but random point on the screen. It is only when a large number of electrons have hit the screen that the interaction pattern appears. Indeed, many electrons will have landed at points of high probability P(r) (bright fringe), and very few at points of low probability (dark fringe).

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Remark: when a quantity (x, p…) is measured in a particle, the measurement itself changes the state of the particle: it forces the particle to materialize into one state, at one location.

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198. The uncertainty principle for energy and time and position and momentum

2N identical particles are prepared so that they have the same state (they have the same

wave function Ψ(r)). The position x of N particles is measured, and the momentum p of the other N particles is

measured. Whatever the precision of the apparatus, there will be a range of results (characterized by Δx

for the position in a 1D problem and Δp for the momentum) which reflects the probabilistic nature of these wave behaving particles.

Heisenberg’s uncertainty principle

Heisenberg’s uncertainty principle for position and momentum: Δx Δp ≥ h/4π Heisenberg’s uncertainty principle for energy and time: ΔE Δt ≥ h/4π ΔE being the range of the energy distribution of the particle Δt being the lifetime of the particle

Illustrations

(the following calculations only intend to give rough orders of magnitudes)

The size of an atom roughly represents the uncertainty on the position of an electron I that

atom: Δx = 10-10 m

Therefore the uncertainty in the momentum of the electron Δp will be such that: Δp ≥ h/4π Δx = 5 * 10-25 kg.m.s-1 The momentum of an electron is very small. Indeed:

p = mv << mc = 9.1 * 10-31 * 3 * 108 = 3 * 10-22 kg.m.s-1 One can therefore assume that Δp is very large relative to p. Therefore the

maximum value for p can be assumed to be approximately equal to Δp (p = Δp). The energy of the electron being equal to p2/2m, we finally get E = p2/2m = (5 * 10-25)2/(2*9.1 * 10-31) = 10-19 J = eV Conclusion 1: The order of magnitude of electron energy levels of an atom is 1 eV (cf H energy levels)

Remark: If the electron was confined in the nucleus, Δx would be much smaller (Δx = 10-15 m). With the same reasoning, the order of magnitude of the electron energy levels would be of a few GeV which contradicts all the measurements of atom energy levels. Conclusion 2: The electron of an atom cannot exist within a nucleus.

Remark: The order of magnitude of the lifetime of an electron in an excited state is 10-10 s. Therefore, according to the uncertainty principle, the uncertainty in the value of the energy of the excited state is: ΔE ≥ h/4π Δt = 5 * 10-25 J = 3 * 10-6 eV. Conclusion 3: The fundamental uncertainty with which the energy level of an excited state can be known is of the order of magnitude of 10-6 eV, which is quite small relative to the values of the energies of these states (order of magnitude of 1 eV).

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Exercise: how can the uncertainty principle account for an electron diffraction experiment?

199. Pair production and pair annihilation

Pair production A photon can turn into matter in the form a pair of a particle and its antiparticle. In the process, conservation laws apply:

1- The conservation of mass/energy: h*f = 2 * m0 * c2 + EK (particle) + EK (antiparticle) m0: the rest mass of the particle (and the antiparticle)

2- The conservation of the momentum: p = h/λ Consequences: The most common pair produced is the electron/positron.

Therefore, h*f has to be greater than 2 * m0 * c2 = 1.02 MeV. 1.02 MeV is the threshold energy for pair production

A pair production can’t occur in empty space where conservation of momentum would not be possible (if the electron and the positron are created at rest, the total momentum would go from h/λ to 0). This phenomenon occurs near atomic nuclei which, being very massive, can “absorb” the change in momentum without changing its energy much. For example, pair production is observed when gamma rays enter a solid. Exercise: What part of the EM spectrum can be involved in pair production? What are these rays called?

Pair annihilation When a particle and its antiparticle collide, they can turn into a pair of photon of the same frequency. In the process, conservation laws apply:

1- The conservation of mass/energy: 2 * m0 * c2 + EK (particle) + EK (antiparticle) = 2 * h*f

2- The conservation of the momentum. Consequences: When an electron and a positron meet with very little speeds, 1- becomes

2 h * f = 2 * m0 * c2 so h * f = 0.51 MeV The smallest energy of a photon produced by pair annihilation is 0.51 MeV.

If the kinetic energies of the particle and its antiparticle are small compared to m0 * c2, the total momentum before the collision is very small (almost equal to 0). Therefore, the two photons created will travel in opposite direction so that the total momentum is conserved. If the initial momentum is not negligible, the angle between the paths of the two photons will not be equal to 180°. Remark: Pair annihilation often occurs inside a solid which provides the electrons.

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200. Tunneling, potential barrier and factors affecting tunneling probability

A wave function is continuous. Therefore, there is a probability for a particle to be anywhere in space, although this probability gets very very small far away from the most probable location (see graph in 195). Illustration: An electron is confined within a certain volume by an electric potential energy barrier. If it only behaved like a particle, it could not cross this barrier (just like a ball can’t go over a 10 m high wall if it doesn’t have enough initial kinetic energy). Nevertheless, in certain situations, it manages to cross this barrier Explanation: The electron being a wave, it has a probability of being everywhere, even on the other side of the barrier (even though the wave function and therefore the probability density are modified by the potential barrier) In terms of energy, this means that for a very brief period of time, the energy of the electron has increased so that the barrier can be overcome. Although this sudden increase in energy seems to breach the law of conservation of energy, it can be explained thanks to the uncertainty principle. Provided the lifetime (Δt) of the high energy state the electron reaches when crossing the barrier is very short, there is an uncertainty in the value of this energy state (which mean it can reach a wide range ΔE of values). The electron can very briefly reach an energy greater than the barrier and cross it. Remarks: This phenomenon is known as tunneling. The bigger the mass of a particle the more difficult the tunneling. The bigger the gap between the energy of the particle and the energy of the barrier, the more difficult the tunneling.

12.2. Nuclear physics

201. Rutherford scattering and nuclear radius

Estimation of the Gold nucleus radius High energy α particles (He nuclei) are sent towards a thin gold foil. Most of them go through undeflected, but some of them are scattered in many directions, and very few even bounce back along the initial direction of the particles. This shows that matter is mostly vacuum, and that most of the mass of an atom is concentrated in a very small, very massive and positively charged particle.

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The scattering angle depend on how close the α particle gets to the gold nucleus. An estimate of the radius of the gold nucleus was obtained with the following calculation. A particle comes from a very far away distance from a gold nucleus where:

Its electric potential energy Epe = 0 Its kinetic energy EK = ½ mv0

2 Because of the electric repulsive force, the particle which will bounce back stops at rC away from the center of the gold nucleus: Its electric potential energy is Epe = 2kZe2/rc Its kinetic energy is EK = 0 rc is then determined through the conservation of energy: rc = 4kZe2/(mv0

2) Exercise: Assuming that the initial velocity of the α particle is v0 = 2.0 * 107 m.s-1, determine the value of rC Limitation of the calculation:

It overestimates the value of the gold nucleus radius (accepted value R = 7.3 * 10-15 m) Indeed, the initial velocity of the α particles used by Rutherford is not high enough. When higher energy α particles are used, they come so close to the nucleus that the scattering measurements do not agree with Rutherford’s predictions. Indeed: The strong force between the α particle and the gold nucleus has to be taken into account.

The “collision” changes their paths.

Radii and nuclear density A nucleus is made of nucleons which can be assumed to have the same volume. Therefore, the radius R of any nucleus (Containing A nucleons) can be obtained through the following law:

R = R0 * A1/3

R0 = 1.2 * 10-15 m the fermi radius Exercise: 1°/ Show why R is proportional to A1/3. 2°/ Calculate the radii of 197

79Au 6026Fe

3°/ Show that all the nuclei have the same density (ρ = 2.3 * 1017 kg.m-3) Remarks: The density of matter (liquid water : ρ = 1.0 * 103 kg.m-3; Gold: ρ = 1.9 * 104 kg.m-3) is very small compared to the density of a nucleus (13 to 14 orders of magnitudes of difference).

Neutron stars are only made of neutrons and have the density of nuclei.

202. Electron diffraction and nuclear radius Electron diffraction through thin films is a more accurate way of determining the nuclear radii. Indeed: The wavelength of high energy electrons and the size of a nucleus have the same order of magnitude. Contrary to α particles, electrons do not feel the strong force. Light incident on a small circular aperture diffracts. Electrons behaving like waves (λ = h/p) incident on a small spherical nucleus of diameter D also diffract. The first minimum of the diffraction pattern occurs at an angle θ such that:

sin (θ) = λ/D

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Remarks: For small angles, sin (θ) = θ For most of these experiments, θ is not small enough for this approximation to work. Exercise: 1°/ What is the order of magnitude of the diameter of a nucleus? When a particle goes at a speed which is not negligible compared to the speed of light in vacuum,

its energy can be calculated with the following formula (cf option A) : E2 = (pc)2 + (m0c2)2 p: momentum m0 : rest mass

2°/ Calculate the energy (in MeV) of an electron which De Broglie wavelength has the same order of magnitude as the diameter of a nucleus.

203. Nuclear energy levels A nucleus can emit gamma rays which have the following characteristics:

The number of frequencies produced is discrete. The energies of the photons only depend on the nature of the nucleus. One nuclide emits a discrete set of wavelength which is:

Always the same Different from the set of wavelengths emitted by another nuclide. Interpretation:

When a nucleus is in a high (excited) energy state, in reaches a lower energy state by emitting a photon.

These results can be accounted for by the existence of nuclear energy levels. The order of magnitude of nuclear energy levels is 1 MeV This order of magnitude is much bigger than the one for an atomic energy level (1eV)

204. The neutrino The electron neutrino was first hypothesized before it was even detected. Indeed, the β+ particles emitted in a β+ decay have a continuous energy spectrum. According to the law of conservation of energy (and because nuclei have quantized energy levels) this can only happen if another particle is emitted alongside the β+ particle. An electron neutrino (νe) is a particle which is emitted in a β+ decay:

It only interacts through the weak force. It can be detected through various methods.

The choice of a technique mostly depends on the energy of the neutrino. Method 1: A low energy (0.3 MeV) electron neutrino is absorbed by a Gallium nucleus:

7131Ga + νe 71

32Ge + e- The Germanium nucleus (Ge) produced is not stable and decays back to 71

31Ga through a β+ decay. The rate of this decay enables to:

Prove the presence of 7132Ge and thus the absorption of a νe by a 71

31Ga nucleus.

Determine the amount of νe.

Method 2: A very high speed charged particle (electron or muon) is created in water by a νe.

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The speed of the charged particle is greater than the speed of light in water. It creates a shockwave (a “shock EM wave”) in water in the form of a photon (mostly UV photons). The detection of these photons enables to prove the presence and the number of νe.

205. The law of radioactive decay and the decay constant The graph of A (or N) vs time is a decaying exponential curve:

A (t) = A0 * exp (-λ * t) = A0 * 2/12t

t

A(t) = λ * N(t) is the activity of the sample (Bq) N(t) is the number of radioactive nuclei in the sample. It is derived from the characteristics of any nuclear decay. One nucleus will decay at a time which: Is random which means there is no way of predicting when ONE particular nucleus will decay. Does not depend on its environment (T, P, time, number of remaining radioactive nuclei…) which means that the proportion per unit time of nuclei that will decay is a constant (called λ) which does not depend on T, P….

Proportion per unit time = 1/ Δt * [ΔN(t) / N(t)] = λ N(t): number of radioactive nuclei at t ΔN(t) = N(t)-N(t+Δt): number of radioactive nuclei which will decay between t and t + Δt. Therefore, [N(t)-N(t+Δt)] / N(t) = λ * Δt

[N(t) – N(t) – Δt * N’(t)] / N(t) = λ * Δt N’(t)]/N(t) = - λ Ln(N(t)) = Cste - λ * t N (t) = N0 * exp (-λ * t) Remark : λ ( = 1/ Δt * [ΔN(t) / N(t)]) is the probability that ONE particle will decay in a second. Definition: λ is the decay constant. t1/2 is the half-life. Exercise: Prove that λ = ln(2)/t1/2

Determination of the decay constant (λ) of a sample If t1/2 is not too big (smaller than a few hours), the activity of the sample is measured over a long period of time. Then, the A Vs t graph is drawn and λ can be derived. But IF t1/2 is very important (for 14C, t1/2 = 5.7 * 103 years…), the variations of A(t) over a short period of time (even if the activity is measured for several days) are be too small. The A Vs t graph is (almost) completely flat and therefore λ can’t be derived from it. In such a situation:

The mass m of the sample is measured, The number of nuclei is derived (n = m/M, and N = n * NA). This of course, requires that the

sample is pure (ie only made of one type of radioactive nuclei). The activity A is measured. λ is derived through A = λ * N.

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Equipment: the GM tube

A GM tube consists of a chamber filled with an inert gas (Helium 3, or Boron trifluoride…)

where two electrodes create a very high electric field. When ionizing radiation hit the inert gas, some of the molecules are ionized. The charged

particles thus created (ions and electrons) are detected thanks to the strong electric field. The amount of the ionizing radiations which enter the tube (x) can therefore be measured.

The amount of radiation emitted by the sample (X) is proportional to x. Indeed, the sample emits radiations in all directions. Only a small fraction of these radiations enter the GM tube through its window. If the sample is situated at d from the window of the GM tube, and if the window of the tube has a surface area of s, the following calculation enables to determine X:

x = X * s/(4πd2) If the measurement is carried out during a period of time Δt, the activity of the sample is therefore:

A = X/Δt

Datation Datation consists in measuring the activity of a sample (or its concentration in a certain nucleus X) and link this information to the age of the sample through the decay curve of the radioactive nucleus responsible for the activity of the sample (or for the presence of X in the sample).

Short half-lives: The β- decay of 14

6C is used to determine the time of death of organic material. The half-life of 146C

being equal to 5.7 * 103 years, it can only determine dates ranging from a few hundred of years to 5 * 104 years (approximately 10 t1/2).

Long half-lives: The determination of much longer dates (age of certain rocks, age of the Earth…) uses much longer half-life radioactive nuclei such as 238

92U (t1/2 4.4 * 109 years) which eventually decays into 206

82 Pb after a series of successive nuclear decays (succession of α and β due to unstable

products) called a decay chain.

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13. (Option A) Relativity (15h/25h)

13.1. The beginnings of relativity

206. Frames of reference Definitions: A frame of reference is made of:

A set of axis and an origin (O, x, y, z) A clock (t)

An inertial frame of reference is in uniform motion (constant velocity) relative to another inertial frame of reference (where Newton’s first law is valid) Remark: in this chapter we’ll only consider Uniform motions Occurring in inertial reference frames In 1 direction (x axis)

207. Galilean relativity and Newton’s postulate concerning space and time

S is an inertial frame of reference (O, x, y, z) S' is a frame of reference (O’, x’, y’, z’) moving with constant velocity v (along the x axis) relative to S:

v is constant therefore S’ is also inertial An object is in motion. Its velocity is:

u relative to S u’ relative to S’

Galilean relativity

Galilean transformations: x’ = x – v.t

(y’ = y) and (z’ = z) t’ = t

Therefore: u’ = u – v (cf 2.1) Newton’s postulate: Time is the same in S and S’: time is absolute.

208. Maxwell and the constancy of the speed of light In Maxwell’s equations that describe the propagation of EM waves, the speed of light c is equal to 1/(ε0μ0)1/2, regardless of speed of the source that emitted it (or regardless of the frame of reference). This means that the speed of light is a constant, the same constant in all inertial frames of reference. This statement is in contradiction with galilean relativity. Indeed, if light is emitted with a speed c by a source at rest relative to S, its speed has to be equal to c’ = c – v relative to S’.

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209. The Michelson-Morley experiment

Before Einstein developed special relativity, it was believed that the speed of light in vacuum

was only equal to c = 1/(ε0μ0)1/2 in a specific medium called Aether. This medium was thought to be very very light (almost massless) and present everywhere in the Universe. Planets, stars, galaxies were supposed to be moving relative to this medium.

Between 1897 and 1905, several interferometer experiments were carried out in order to

determine the speed of the Earth relative to Aether. The result was quite surprising since the speed of the Earth relative to Aether was always found to be equal to 0. Indeed, considering the spinning of the Earth about its axis, and the Earth’s revolution about the Sun, such a result was very strange.

Einstein solved the problem by stating that Aether was an unnecessary hypothesis (meaning

it did not exist) which was quite bold at the time since it was firmly believed that any wave needed a medium to travel through. And then, he created the theory of special relativity… Video: Michelson-Morley (“scienceslycee.fr”)

13.2. Lorentz transformations

210. The two postulates of special relativity

1- The laws of Physics (mechanics, electromagnetism…) are the same in all inertial frames of reference.

2- The velocity of light is a constant (c) in all inertial frames of reference regardless of the speed of the source which emitted the light.

Consequence: Durations (time) and lengths (space) are not absolute anymore. If we consider two events happening to a system, the time interval (duration) between two events and the distance travelled by the system between the two events depend on the observer (the inertial frame of reference). Definition: an event is something happening at a particular time and a particular point in space (or a point in spacetime).

211. Einsteinian relativity

S and S’ are two inertial frames of reference. S’ is moving at v relative to S (which means that S is moving at –v relative to S’) along the x-axis.

Clock synchronization When O’ coincides with O, the clocks in S and in S’ are synchronized (are both set to 0). Therefore, when O’ coincides with O, t = t’ = 0.

Lorentz transformation

The motion of an object is studied in S and S’: (x, y, z t) are the coordinates of the object in S.

(x’, y’, z’, t’) are the coordinates of the object in S’.

x’ = γ (x – v.t)

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(y’ = y) and (z’ = z) t’ = γ (t – v.x/c2)

2

2

1

1

c

v

> 1 is called the Lorentz factor

Which leads to the velocity addition formula u’ = (u – v)/(1 – u.v/c2)

Two events are studied both in S and S’:

Δx is the distance, measured in S, between the locations of the two events. Δt is the duration, measured in S, between the two events. Δx’ is the distance, measured in S’, between the locations of the two events. Δt’ is the duration, measured in S’, between the two events.

Δx’ = γ (Δx – v.Δt)

(Δy’ = Δy) and (Δz’ = Δz) Δt’ = γ (Δt – v. Δx/c2)

Remark: Durations (Δt and Δt’) and distances (Δx and Δx’) depend on the frame of reference!

212. Time dilation Two events are studies in S and S’ (S’ moving at v along the x-axis relative to S). The clocks are synchronized and set to 0 (t = t’ = 0s), when O meets O’.

In S, let the two events happen at the same location: Δx = 0 According to Lorentz transformations: In S’, they don’t happen at the same location: Δx’ = γ (Δx – v.Δt) = – γ v.Δt

The duration between the two events is not the same in S and in S’: Δt’ = γ (Δt – v. Δx/c2) = γ Δt

Definition: The time interval between two events happening at the same location (same point) in space is called the proper time interval (Δt0). Property: The time interval (Δt) between two events, measured in a frame of reference moving at v relative to the frame of reference where the two events happen at the same location verifies:

Δt = γ Δt0

Remarks: Δt > Δt0 hence « time dilation » The observer in S’ sees the two events as if they were happening in slow motion (the duration between the two events is greater in S’ than it is in S) Animation: “dilatation du temps” (scienceslycee.fr).

Symmetry of time dilation Peter (P) and Bobby (B) move towards each other in a straight line at constant speed. They both have a clock:

P moves at v relative to B. B moves at – v relative to P.

Point of view of B:

Me and my clock (clockB) are at rest. P and his clock (clockP) are moving at v relative to me. Let’s consider the two following events: Event 1 tick of clockB

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Event 2 tock of clockB These two events happen at the same location in the frame of reference of B. So Δt(according to P) > Δt(according to B) P says: “My clock beats faster than B’s clock”.

“The journey will last longer for me than for B”

Point of view of P: Me and my clock (clockP) are at rest. B and his clock (clockB) are moving at -v relative to me. Let’s consider the two following events: Event 1 tick of clockP

Event 2 tock of clockP These two events happen at the same location in the frame of reference of P. So Δt(according to B) > Δt(according to P) B says: “My clock beats faster than P’s clock”.

“The journey will last longer for me than for P” Conclusion: They are both right because time flows in a different way according to B and to P.

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213. Length contraction

In S, an inertial frame of

reference, an object is moving along the x-direction at velocity v.

S’ is the frame of reference

attached to the object, where the object is at rest. It is inertial because v is constant. O’ the origin of the x’-axis is at the left hand side of the object. The object is studied between the two following events: Event 1: Light is sent from the left end side of the object, along the x-direction (towards the right). The clocks are synchronised then, which means that at that event, O and O’ coincide and t = t’ = 0. Event 2: light reaches the right end side of the object. According to Lorentz transformations: Δx’ = γ (Δx – v.Δt) L’ = γ (L + v.Δt – v.Δt) L’ = γ L

where L is the length of the object measured in S along the x-axis. L’ is the length of the object measured in S’ along the x-axis.

Definition: The length of an object measured in a frame of reference where the object is at rest is called the proper length (L0). Property: The length (L) of an object measured in a frame of reference where the object is moving at v, verifies:

L = L0/γ Remark: L < L0 hence « length contraction »

214. Simultaneity Two events are studies in S and S’ (S’ is moving at v along the x-axis relative to S):

In S let the two events: not happen at the same location Δx ≠ 0 happen at the same time (they are simultaneous) Δt = 0

According to Lorentz transformations: Δx’ = γ (Δx – v.Δt) = γ Δx Δt’ = γ (Δt – v. Δx/c2) = -γ. v. Δx/c2 Δt’ ≠ 0

Conclusion: the two events are simultaneous in S. not simultaneous in S’.

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Example

A man (O) is standing in the middle of the carriage of a train (inertial frame of reference S),

holding a remote control that sends EM waves to both doors located at both ends of the carriage. When an EM wave reaches a door, the door opens. O is at rest relative to S.

The train is moving towards the right at a constant velocity v relative to the ground (the ground is an inertial frame of reference S’). Another man (O’), at rest relative to S, watches the train go by.

S and S’ are both inertial frames of reference because v is constant. Remark: S’ (the ground) is moving at –v relative to S (the train) along the x-axis

Event 1: O presses the remote control. Event 2: Left door opens. Event 3: Right door opens. Remark: because of length contraction occurring in S’, the carriage is smaller in S’ than it is in S. Study in S: Light (emitted by O’s remote control) travels at c in S. Light travels the same length in both directions (half of the length of the carriage). Both doors open at the same time: events 2 and 3 are simultaneous. Study in S’: Light (emitted by O’s remote control) travels at c in S’. Light travels a shorter distance to get to the left hand side door than to get to the right hand side door because the train is moving to the right in S’. The left hand side door opens before the right hand side door: events 2 and 3 are not simultaneous. Exercises: 1°/ Determine Δt’, the duration between events 2 and 3 in S’

(L = 40 m proper length of the carriage, v = 0.90 c)

2°/ Show that if in S, an inertial frame of reference, two events happen: at the same time (simultaneous)

at the same location These two events will be simultaneous in any other inertial frame of reference S’.

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215. The muon decay experiment

When cosmic rays (very high speed protons) enter the atmosphere, they interact with the nuclei of the atoms to form mostly pions which are not stable and form photons, neutrinos or muons. The muons created have very high speeds (0,99 c) and can progress through the atmosphere without interacting much with its atoms. The proper half-life of a muon (its half-life in a frame of reference where it is at rest) is equal to:

t0 = 2.2 µs Experiment: muons of speed 0,99c relative to the Earth, are detected for 1 hour at two locations: At the top of Mount Washington (1907 m above sea level). At sea level. Results: number of muons at the top of Mount Washington: N = 568 number of muons at the sea level: N = 412 Let’s consider the two following events: Event 1: a muon leaves the top of Mount Washington. Event 2: the muon reaches see level. Exercise: 1°/ What is the distance travelled by the muon between the two events:

in the frame of reference of the Earth (S)? In the frame of reference of the muon (S’)?

2°/ What is the duration between the two events: in the frame of reference of the Earth (S)? In the frame of reference of the muon (S’)?

3°/ What is the half-life of a muon: in S? in S’? 4°/ Show that the results are consistent with special relativity and not with Newtonian mechanics. Remark: this experiment was carried out in 1963 and was one of the great confirmations of special relativity.

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216. Forces on a charge or current

Preliminary remark: The two examples studied in this section are using qualitative interpretations based on the results of special relativity presented so far in this chapter. Complex calculations are necessary to calculate properly the forces involved.

Current carrying wire

A charge (frame of reference S’) moves at velocity v relative to a current carrying wire (frame of reference S). The electrons move with a speed u relative to S, and the nuclei (copper nuclei) are at rest relative to S. The charge feels a force exerted by the wire.

Interpretation 1 (in S): a magnetic field B is created around the current carrying wire.

the charge moving in a magnetic field feels a magnetic force: F = qv x B (x represents the vector product)

Remark: The charge distribution in the wire is neutral. The positive charge density (nuclei) and the negative charge density (electrons) are equal. Interpretation 2 (in S’): The wire (and therefore the nuclei) move at –v in S’. Due to length contraction, the positive charge density in greater in S’ than it is in S. The electrons move at u – v in S’. Due to length contraction, the negative charge density is greater in S’ than it is in S, but slightly smaller than the positive charge distribution. Therefore, from the point of view of the charge (ie in S’), the wire is NOT neutral anymore. It is slightly positive. The charge, at rest in S’, feels an electric force due to the positively charged wire. Remark: The magnetic force in S is an electric force in S’. A magnetic field is just a relativistic effect of an electric field. Exercise: Show that both interpretations lead to the same result whether the charge of the particle is positive or negative.

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Two charged particles moving with parallel velocities

Preliminary remark: A charged particle in motion creates a magnetic field. Two identical positively charged particles move in the same direction, and at the same speed relative to S. Two observers (O, at rest in S, and O’ moving at the same speed and in the same direction as the charges: frame of reference S’).

From the point of view of O’: The two charges are at rest. They repel each other because of the electric field they create.

From the point of view of O:

Electric repulsion exerted by the charges increases because of length contraction. The motion of the charges induce a magnetic field. Each charge is now moving in

the magnetic field created by the other one. This induces an attractive force. O and O’ come to the same conclusion (they measure the same force acting on both charges) but from different perspective. Conclusion: The effect of a magnetic field on a moving charge is nothing else than an electric field, in a different frame of reference

217. Invariant quantities Although time and space depend on the frame of reference, some quantities remain constant in all inertial frames of reference:

The speed of light in vacuum: c The spacetime interval: Δs2 = Δx2 - c2 Δt2 = Δx’2 - c2 Δt’2 The proper time Δt0

The proper length L0 The rest mass m0 (cf 13.4 for the definition) The electric charge of a particle (cf 13.4)

Exercise: Show that Δx2 - c2 Δt2 = Δx’2 - c2 Δt’2

13.3. Spacetime diagrams

218. Spacetime diagrams S and S’ are to inertial frames of reference. S’ is moving at v relative to S in the x-direction. Remark 1: x’ = γ (x – v.t) so x’ = γ (x – (v/c).ct)

t’ = γ (t – v.x/c2) so ct’ = γ (ct – (v/c).x)

Lorentz transformations can be rewritten as follows:

x’ = γ (x – β.ct) ct’ = γ (ct – β.x) where β = v/c

Definition and use: A spacetime diagram (also known as Minkowski diagram): has two axis: one is “c.t”, and the other is “x” (which are both distances).

is a tool which enables to visualize the effects of special relativity. enables to plot events.

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is used to study the duration and the distance between events, in various inertial frames of reference.

Remark 2: An event E is placed on the spacetime diagram representing S. If E is on the ct-axis, it means that E happens at x = 0. If E is on the x-axis, it means that E happens at t = 0. The same event E also happens in S’. If E happens at x’ = 0, it should be along a line of equation x – β.ct = 0. This straight line (which has a gradient of 1/β in the S-spacetime diagram) represents the ct’-axis of the spacetime diagram representing S’.

If E happens at t’ = 0, it should be along a line of equation ct - β.x = 0. This straight line (which has a gradient of β in the S-spacetime diagram) represents the x’-axis of the spacetime diagram representing S’. The angle θ between the ct-axis and the ct’-axis is the same as the angle between the x-axis and the x’-axis. Indeed, the coordinates of any point on the ct’-axis lead to tan (θ) = x/ct = β (and the coordinates of any

point on the x’-axis also lead to tan (θ) = ct/x = β). Property: θ = tan-1 (v/c) Remark 3: A random event E is represented in a spacetime diagram. Its coordinates in the spacetime diagram represent the location where E occurred and the time at which it occurred in both S and S’. Remark 4: the axis for S and the axis for S’ have the same origin because of clock synchronisation. Remark 5: Sometimes, the axis on a spacetime diagram are changed into (t; x/c). Sometimes, c is set to 1, a unitless quantity which means that all velocities will also be unitless, and smaller than 1. Warning:

The scale on the (x; ct) axis is DIFFERENT than the scale on the (x’; ct’) axis.

1m on the (x’; ct’) axis is2

2

1

1

β

β

longer than on the (x; ct)

axis.

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219. Worldlines and lines of simultaneity

Definition: A worldline is the “path” of a particle represented in a Minkowski diagram. It represents the series of events that happen to the particle. Example: A photon travels at the speed of light c in all inertial frames of reference. Therefore, whatever the frame of reference, when a duration t (or t’) has passed, it has travelled a length x (or x’) equal to ct (or ct’). The worldline of a photon going in the positive x-sense is therefore the first bisector of any diagram (in S, θ = tan-1(1) = 45°). The worldline of a photon going in the negative x-sense is therefore the symmetrical line of the first bisector relative to the ct-axis (in S, θ = tan-1(-1) = - 45°). Definition: a line of simultaneity is a line on a spacetime diagram representing events happening simultaneously in a given frame of reference.

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220. Applications

Simultaneity

S is the frame of reference of the carriage. S’ is the frame of reference of the ground. S’ is moving at –v along the x-axis. In this example, θ = tan-1 (-v/c) < 0.

Time dilation

Two events E1 and E2 happen at the same location in S (at the same x) and the duration between E1 and E2 is the proper time Δt0.

S’ is moving at v along the x-axis relative to S. The duration between E1 and E2 in S’ is equal to

Δt’ = γ Δt0.

Δt’ > Δt0

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Length contraction

In S, an inertial frame of reference, an object is moving along the x-direction at velocity v. S’ is the frame of reference attached to the object, where the object is at rest. S’ is moving at v relative to S. Event 1: Light is sent from the left end side of the object, in the x-sense. Event 2: Light reaches the right end side of the object. The length L of the object measured in S is smaller than the length L0 (proper length) of the object measured in S’ where it is at rest: L = L0/γ.

Exercise 1: Peter (P) and Bobby (B) are two twins. P stays on Earth as B goes on a space trip. He gets into in a rocket going away from Earth at v relative to the Earth, turns back at some point and come back towards the Earth at v relative to the Earth. When they meet again on Earth it seems that according to the symmetry of time dilation (211) they can both say: “I’m older than you”. This thought experiment is known as the twin paradox. 1°/ Explain why this is indeed a paradox (and why one of them has to be wrong). 2°/ Which of the two twins is right (if any)? 3°/ use a spacetime diagram to resolve the twin paradox Exercise 2: Worked example p 528

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13.4. Relativistic mechanics (HL only) Preliminary remark: a constant acceleration increases the speed at constant rate, which therefore goes towards infinity: this is unacceptable in Special relativity: the laws of motion have to be modified

In this part, we only consider 1D motions: the sense of the vectors is given by the signs.

221. Total energy and rest energy A body of mass m0 (at rest) is moving at velocity u in an inertial frame of reference S

The total energy of the body is: E = γ m0 c2 = m c2

The rest energy is: E0 = m0 c2 The kinetic energy of the body is: EK = (γ – 1) m0 c2

Remarks: m0 is the rest mass of the particle (mass of the particle in a frame of reference where it is at rest): m0 is an invariant. m can be seen as the “relativistic mass” of the body at velocity u. If u = 0 then EK = 0 (because γ = 1). Property: The law of conservation of energy remains the same: the change in kinetic energy is

equal to the work done by the net force (ΔEK = WF). Consequence: the potential difference V necessary to accelerate a particle of charge q to a given speed or energy can be calculated through the following equation: ΔEK = q.V Exercise: 1°/ Show that when u << c then EK = ½ m0 u2. 2°/ Calculate the wavelength of the two photons emitted in the decay reaction of a pion moving at velocity u = c/2.

222. Relativistic momentum

The momentum of the body is: p = γ m0 u [γ = 1/(1-u2/c2)1/2]

Properties: E2 = (pc)2 + E0

2 Momentum conservation remains valid. When the net force acting on a system is equal to 0, the momentum remains constant. Exercises: 1°/ Show that E2 = p2c2 + m0

2 c4 2°/ Collision exercise. (H2 November 2012) Consequences: When u tends towards c, γ tends towards infinity.

If m0 ≠ 0 p (= γ m0 u) tends towards infinity which is impossible. conclusion: therefore, no object can reach c.

If m0 = 0 (photon) p seems undefined (m0 * γ = “∞” * “0”)

p can be calculated through E2 = p2c2 p = E/c = hf/c = h/λ (the De Broglie equation also applies to photons)

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223. New units for mass (MeV.c-2) and momentum (MeV.c-1)

E = γ m0 c2 = m c2 so in terms of units: [m] = [E]/[c]2

New unit for mass: MeV/c2 or GeV/c2 Example: a system with a rest energy of 5,89 MeV has a rest mass m of 5,89 MeV.c-2 pc = (E2 - m0

2 c4)1/2 so in terms of units: [m] = [E]/[c] New unit for mass: MeV/c or GeV/c

Example: a system with a p.c of 5,89 MeV has a momentum p of 5,89 MeV.c-1 Exercise: Calculate m0, m and p of a particle of energy 2.5 * 10-10 J and speed v = 0.60c

224. Particle acceleration

Property: A constant force exerted on a particle produces a decreasing acceleration. Indeed: p = γ m0 u and dp/dt = m0 γ3 du/dt = m0 γ3 a

Fnet = dp/dt = m0 γ3 a so a = Fnet/(mo γ3)

As u increases, so does γ. Therefore, if Fnet is constant, a decreases.

Exercise: Show that dp/dt = m0 γ3 du/dt

13.5. General relativity (HL only) Preliminary remark: general relativity can be seen as the “new law of gravitation”. Indeed, it is Einstein’s attempt to make Newton’s gravitation theory compatible with special relativity’s new description of space and time (spacetime).

225. The equivalence principle Definitions: The inertial mass mi is the property in a body that resists the change in motion. It appears in the second law of Newton (F = mi a). For a given force F, the bigger the mass mi, the smaller the acceleration F, which means the smaller the change. The gravitational mass mgrav is the property in a body which exerts an attractive force on another body which also has a gravitational mass. It appears in the universal law of gravitation (F = G * mgrav,1 * mgrav,2/d2).

Newton’s assumption Newton stated that as far as he could measure (using pendulums made of different materials):

mi = mgrav = m Consequence: When a body is in a gravitational field g, and feels no other force than the gravitational force: According to the law of gravitation Fnet = m g

According to the second law of motion Fnet = m a Therefore m a = m g

and a = g

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Einstein’s principle of equivalence

Thought experiment 1:

An observer is in a closed box without windows. He floats and feels weightless. The observer can’t tell whether: Both the box and himself are in outer space far away from any mass (g = 0) drifting at constant velocity (relative to the very far away Earth for example). He is near the Earth (which is creating g), and the box is moving towards the Earth with an acceleration exactly equal to g relative to the Earth. This situation is what would happen if we were trapped in a free falling elevator. We would be at rest (floating) relative to the elevator, and feeling no force from its walls (…until the crash of course).

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Thought experiment 2:

An observer is in a closed box without windows. He stands on the floor feeling his own weight

The observer can’t tell whether: The box is at rest on the surface of the Earth (Earth is creating g) and he is feeling a

reaction force from the floor of the box (R = - m g). The box is in outer space far away from any mass (g = 0) and moving upwards (relative to the very far away Earth) with an acceleration exactly equal to – g.

Principle of equivalence:

The effect of an accelerating frame of reference are the same as the effects of a gravitational field.

OR Gravitational effects can’t be distinguished from inertial effects.

Interpretation of the thought experiments

Thought experiment 2 The acceleration of the frame of reference in a no gravitation zone

has the same effect on the observer (he feels a reaction force) as The gravitational field in a frame of reference at rest relative to that field.

Thought experiment 1 The observer feels no force because the effect of the gravitational field on the observer is cancelled out by the acceleration of the frame of reference (relative to the field).

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226. The bending of light

A box has a small opening. It is in outer space, very far away from any massive object (no gravitational field). Light is entering the box through a small hole. The path of the light beam (of a photon) is represented both in R, an inertial frame of reference, and R’, the frame of reference of the box. Situation 1: The box is at rest relative to R. Situation 2: The box is moving with a constant velocity relative to R. Situation 3: The box is moving with a constant acceleration relative to R. Observations: In R, light goes in a straight line. In R’, light goes is a straight line in situations 1 and 2. In R’, light bends in situation 3, in a sense opposite to the sense of the acceleration of the box in R. Conclusion: According to the equivalence principle, light will bend when in a gravitational field g in the same “sense” as g.

Experimental evidence : Eddington’s measurement during an eclipse of the Sun in 1919

During an eclipse of the Sun, photographs were taken of the Sun. Stars that were behind the Sun could be seen on the photographs because the sun bent

the light rays coming from the stars.

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227. Notion of spacetime

The bending of light near massive bodies requires a new theory of gravitation and motion: that is

what general relativity is about. Characterisation of spacetime:

Spacetime is a 4D world (x, y, z, c*t) Spacetime determines the motion of mass and energy (objects, EM waves…) Mass and energy determine (bend) the structure of spacetime. Remarks: Space and time are NOT two independent concepts anymore.

Gravitation is not a considered to be a force anymore. Mass (as well as energy) does not affect the motion of things because of the gravitational field it creates, but because it “bends” spacetime and therefore modifies both space and time!

In the absence of forces (gravitation is not a force…), particles (with a mass) and energy (photons, EM waves) follow the shortest path along spacetime called geodesic which is not a “straight line” anymore if a mass has bent (warped) spacetime.

Illustrations

Let’s imagine a 2D spacetime (easier to picture): In outer space, far away from any massive object, spacetime is a 2D flat surface (diagram a). In the absence of any force, the shortest path (geodesic) for an object is a straight line. Near the Sun, spacetime is warped (diagram b). In the absence of any force (again, gravitation is not a force anymore…), the shortest path (geodesic) for an object is not a straight line. The faster the object is travelling, the less its path will differ from a straight line. Examples: The geodesic for a planet is an ellipse.

An object of high speed can be deflected (non-periodic comet)

The geodesic for light is not a straight line (light is slightly deflected)

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228. Gravitational redshift and the Pound-Rebka-Snider experiment

Pound – Rebka - Snider experiment (1960)

A photon of frequency ftop is sent towards the ground from the top of a tower at a height H above the ground. Its frequency fbottom is measured when it reaches ground level. Observation: ftop < fbottom (fbottom - ftop)/ftop = Δf/f = gH/c2 Qualitative interpretation: g has an effect on light in terms of trajectory (it bends light). It also has an effect on its energy which means that a “gravitational potential energy” can be associated to the photon. “Epp“= “m”gH But of course “m” does not mean anything for the photon “Epp“= gH * p/c The speed of the photon being c, “m” is replaced by p/c. “Epp“= gH * h/(λ*c) because for a photon p = h/λ As the photon goes towards the ground its total energy is conserved. Therefore, Energy at the top: h ftop + Epp = hfbottom + 0 Energy at ground level

h ftop + gH *h ftop /c2 = h fbottom So h ftop (1 + gH/c2) = h fbottom

So (fbottom - ftop)/ftop = Δf/f = gH/c2 Conclusion: (fbottom - ftop)/ftop = Δf/f = gH/c2 ftop < fbottom: there is a gravitational blueshift near massive object.

Ttop > Tbottom: time slows down near massive objects. Remark: photons created by massive objects producing a high g (dense stars..) experience a gravitational redshift. Indeed, instead of going towards the massive object, they go away from it. Therefore, the light emitted them is redshifted when it reaches us (a region of smaller g). Property: the EM waves emitted in a region of high gravitational field is redshifted (frequency reduced) when it is observed in a region of smaller gravitational field.

229. Schwartzschild black hole

Escape condition from a massive body in Newtonian physics An object of mass m is at r relative to the center of a massive body (M). The object has:

a kinetic energy: EK = ½ mv2 a gravitational potential energy: Epp = - M*m*G/r (with EPP = 0 at infinity)

The mechanical energy (Em) of the object stays constant if no other force than the gravitational force is exerted on the object:

Em = EK + Epp As the object goes away from the massive body, its Epp increases and its EK decreases.

It will stop before infinity (r = ∞) if EK becomes equal to 0 before infinity. (Em < 0) It will reach infinity (r = ∞) if EK becomes equal to 0 (or more) at infinity. (Em > 0)

Conclusion: the bodies with Em > 0 can escape from the massive body’s gravitational field. the bodies with Em < 0 are trapped by the massive body’s gravitational field.

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Analogy for a photon near a Black hole

The kinetic energy of the photon is:

EK = ½ mc2 (“m” should be replaced by p/c = h/(λ*c). The “gravitational potential energy” of the photon

Epp = - M*m*G/r (“m” should be replaced by p/c = h/(λ*c). The photon near a massive body M will be trapped by the massive body’s gravitational field if: Em < 0 ½ mc2 - M*m*G/r < 0 r < 2GM/c2 Definition: RS = 2GM/c2 is called the Schwarzschild radius. A massive body which gravitational field traps everything including light is called a black hole (a region of spacetime of extreme curvature due to the presence of a mass) Property: Light will be trapped by a black hole (mass M) if it comes closer to it than RS.

230. Time dilation near a black hole

Qualitative considerations

Let’s consider a large light beam coming from the left. All the points perpendicular to the

direction of propagation belong to the same wavefront. Therefore, the direction of the wavefronts (represented in dotted lines on the diagram) changes because of the gravitational effects of the black hole.

While light is bent because of the

gravitational effects of the black hole, the part of the beam further away from the black hole travels a longer distance (dfar) than the part of the beam closer (dclose) to the black hole.

The two parts of the beam travel at the same speed: c. Therefore, Δtclose < Δtfar, the duration of the travel of the light is longer for the outside part of

the beam. Interpretation: Time slows down near a black hole.

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Gravitational time dilation effect near a black hole

An observer A (with his clock) is at a distance r relative to the center of a black hole. Another observer B (with his clock) is very far from the center (at infinity). The two observers are motionless relative to the black hole. The duration between two events happening at A’s location is measured: In the frame of reference of A: Δtclose In the frame of reference of B: Δtfar

Property: Δtclose < Δtfar

Δtfar = Δtclose /(1 – RS /r)1/2 Conclusion: Two observers at two different points in a gravitational field measure different time intervals between the same 2 events. In this example, clock B ticks faster than clock A (Δtclose < Δtfar ). Consequence: Two events occurring at P (r away from the center of a black hole) are separated by a small time interval Δtclose for an observer located at P. The same two events will be separated by a much larger time interval Δtfar for an observer located at infinity from the center of the black hole. The observer situated at infinity experiences the gravitational time dilation effect of the black hole: it sees the two events as if they were happening in slow motion. Definition: an event horizon is a boundary in spacetime. An event taking place on one side of the event horizon can’t affect an observer located on the other side. Illustration: Light emitted inside RS inside a black hole can’t get out of the black hole and won’t affect an observer outside the black hole. An object approaching RS from the outside appears to go in a slow motion (time dilation) to an outside observer. The closer it gets to RS, the greater the effect, until time almost seems to freeze as the object gets very close to RS. According to the outside observer, the object will never quite go through RS (according to the object of course, time goes on as usual, and he does go through RS in a finite amount of time). The region of radius RS around a black hole is a good example of an event horizon.

231. The Hafele–Keating experiment In 1971, the following experiment was carried out. Three clocks were synchronised: The first one remained on Earth. The second one flew westward around the Earth on a plane. The third one flew eastward around the Earth on another plane. Then the clocks were compared: they disagreed with each other. Exercise: Discuss the two reasons which can explain the disagreement between the clocks.

232. Applications of general relativity to the universe as a whole Many topics regarding the evolution and the characteristics of the Universe rely on general relativity: Its lifetime (and the hypothesis of the big bang). Its expansion rate which leads to a calculation of its total mass (predicted mass greater than the mass observed which leads to the hypothesis of dark matter). The existence of black holes …