Example 37.7

a) What is the probe’s velocity relative to earth?

b) What is the scoutship’s velocity relative to spaceship?

Setup: Two frames: SEarth, S’ Spaceship, u = + 0.900c

a) In S’-frame, the probe moves at ' 0.700xv probe c

2

' 0.700 0.9000.982

1 ' 1 (0.700)(0.900)x

xx

v u c cv c

uv c

Example 37.7

a) The scoutship’s speed is given in S-frame (with respect to Earth),

0.950xv scout c

2

0.950 0.900' 0.345

1 1 (0.950)(0.900)x

xx

v u c cv c

uv c

The scoutship’s speed with respect S’-frame is given by the inverse transform,

The Twin Paradox

E

Star

Eartha

Astrid

20 light years

u=0.95c

First consider Time of Travel according to Astrid:

1st leg: distance from earth to star appears to be contracted according to Astrid.

2 2 2Astrid

20dist to star 20 1 20 1 0.95 6.25

lyly u c ly ly

This is proper according to Eartha.

The Twin Paradox

Time of travel = 6.25ly / 0.95c = 6.6 yrs

2nd leg: return trip distance also appeared to be contracted according to Astrid.

Astriddist to star 6.25ly And time of travel = 6.6 yrs

Earthadist to travel 20 2 40ly ly

So, if Astrid started her journey at 20 years of age, when she comes back to earth, she will be 20+2*6.6 ~ 33 years old.

Now, we consider Time of Travel according to Eartha:

And time of travel = 40ly / 0.95c = 42 years

So, if Eartha said good bye to Astrid at 20 years of age, when they meet back, Eartha will be 20+42 ~ 62 years old.

The Twin Paradox

So, according to this description, Astrid will be almost half her twin sister’s age after she comes back from her trip to the star !

Can Astrid claims that all inertial reference frames are equivalent and make exactly the same arguments with the spaceship being stationary and the earth moving so as to conclude that Eartha will be younger at the end of her journey?

Both of these physical descriptions cannot be true but SR seems to suggest that they should ( apparent paradox).

Here is the apparent paradox:

The Twin ParadoxTo resolve this apparent paradox, one need to realize that the following two situations are not symmetric !

E

StarEartha

Astrid

20 light yearsu=0.95c

E

StarEartha

Astrid

20 light years

u=0.95c decelerate

accelerate

Astrid has to switchinert. ref. frame here.

Earth did not acc nor decc.

The Twin Paradox

AdditionalNotes:

• Astrid was not in a single inertial reference frame. She switched inertial reference frame at midpoint.

• Eartha was in a single inertial reference frame throughout the entire journey.

The seemingly symmetric descriptions between Astrid and Eartha are not symmetric at all and their actual physical experience needs not be the same.

And, they were not!

Conclusion:

SR remains correct but one needs to be careful in making claims which are consistent with SR’s assumptions.

Space-Time Diagram

x

ctThe time axis is measured in units of ct.

Space-Time Diagram

x

ctThe time axis is measured in units of ct.

An object moving with speed v in the x direction will trace out a line with slope c/v

ct c t cslope

x v t v

An stationary object (v=0) will trace out a vertical line (slope = ).

v

x

ct

Space-Time Diagram

x

ctThe time axis is measured in units of ct.

An object moving with speed v in the x direction will trace out a line with slope c/v

ct c t cslope

x v t v

An stationary object (v=0) will trace out a vertical line (slope = ).

A light beam will trace out a line with slope = 1 or -1.

45

v

x

ct1slope

Space-Time Diagram

x

ct

The path an object traced out in Space-Time is called a World Line.

Since nothing can go faster than the speed of light

All world lines starting from O must stay within the shaded “light cone”.

O

Space-Time Diagram & Simultaneity

x

ct Since event O is exactly half way between events A and B,

Paul will consider event C to be simultaneous with O.

CC’ O

Paul sends out light beam

Light reflects at C

Light back to Paul

A

B

Similarly, C’ and all points on the x-axis can be considered to simultaneous with O !

w

reflector

The x-axis (red) is also referred to the line of simultaneity with respect to the observer along the world line at O.w

Space-Time Diagram & Simultaneity

x

ct One can also construct the line of simultaneity for another observer S’ who is moving relative to S along the world line .

Similarly, S’ sends out a light signal to a reflector at C and C’, one can infer the midway point O’ to be simultaneous with C and C’.

'w

'ct

'w

'xCO’

A

B

C’reflector

The line joining O’ and C will be a segment of the line of simultaneity (red) for observer S’ at O’.

Symmetry of Time Dilation & Simultaneity

u

S

S’

- Both observers in S and S’ have their own measurement devices and they can also measure his/her partners devices and compare with his/her own.

- Both S and S’ will respectively measure time dilations from the moving clock from his/her partner.

Time Dilation (Space-Time Diagram)

http://www.phys.vt.edu/~takeuchi/relativity/notes/section12.html

Length Contraction

x

t

x’

t’

1

2

2*L0

L

O measure L bet 1 and 2*

O’ measure L0 bet 1 and 2

Ruler is stationary in S’

Length measurement involves simultaneity and L/L0 are measured from two different sets of events.

Both O and O’ must measure the two ends of the ruler at the same time according to their own clocks.

Length Measurement using Laser

t

x’

t’

1

2

2*L0

L

laser signal

P

R

In O’ frame, laser reflects from the mirror at R and since P is half way between the leaving event (1) and arriving-back event (3) of the laser, O’ interprets P being simultaneous with R.

3

ut1

ut2

In O frame, the distance for laser light to travel to mirror is

1L u t and the distance for laser light to return is

2L u t

LO’

O

The Twin Paradox

t

x’ (outgoing)

6.6yrs

Note: Astrid changes his inertial reference frame at turning pt

turning pt

21 yrs

x’ (incoming)

2. Eartha’s ages significantly according to Astrid as she switches her reference frame(here is the deceleration-acceleration effect)

2.1yrs

1. Time dilation is symmetric before Astrid switches his ref frame:

-Eartha will see Astrid’s clock dilates from 6.6yrs to 21 yrs-Astrid will equivalently see Eartha’ clock dilates from 2.1yrs to 6.6yrs(with the same factor)

1c

Eartha

Astrid

Relativistic Momentum & EnergyAs we have seen, time intervals, length intervals, and velocity change according to the Lorentz Transformation depending on the observer’s frame of reference.

( , ) ( ', ') Lorentz Transformationx t x t

Other dynamical quantities (such as momentum, energy, etc.) must also be appropriately expressed so that the laws of physics satisfy the following conditions:

• Laws of physics (e.g., conservation of momentum, conservation of energy, Newton’s laws) apply equally to all inertial observers.

• The modified relativistic dynamical quantities should reduce to the classical ones for u << c.

Relativistic Momentum & EnergyRelativistic Momentum:

2 21

mm

v c

vP v

Relativistic Energy:

22

2 21

mcE mc

v c

Momentum of a particle moving with velocity as measured in the lab frame (S-frame).

v

Total energy of a particle moving with velocity as measured in lab frame (S-frame).

v

Relativistic Momentum & EnergyAll laws of physics remain valid in all inertial reference frames means.

Conservation Laws for relativistic must remain the same!& EP

Experimentally, it has been shown repeatedly that it is ( ) rather than their classical counterparts that are conserved in high energy collisions!

2andm mc v

For v << c, 2 2

11

1 v c

So,

So that for non-relativistic speeds, relativistic reduces to classical .P

P

m m P v v

Relativistic Force

d

dt

PF

Newton’s 2nd Law: same form as in the classical case but with relativistic momentum .

Note: If (no external force), momentum as expected will be conserved in both relativistic & classical regimes !

0F

P

Let use the relativistic force to consider the work-energy theorem:

f f

i i

dPW F dx dx

dt

Work done by F to accelerate a relativistic particle from vi to vf

(for simplicity, F is in x-direction only)

P

Relativistic Work & Energyf

i

dP dPdx

dt dt v dt

f

i

f

i

W v dP

Now, ( ) ( )d Pv vdP Pdv vdP d Pv Pdv

Substituting this into the above equation for W:

( )f f f

i i i

W v dP d Pv P dv 1st term 2nd term

Relativistic Work & Energy

1 term = ( ) ( )f

f f i iist Pv mv v mv v

For simplicity, choose 0 andi fv v v

Then, the 1st term becomes: 2

2

2 2( )

1

f

i

mvd Pv mv

v c

Now, let consider the 2nd term,

2 21

f f

i i

mvP dv dv

v c

We can integrate this by a simple change of variable.

Relativistic Work & EnergyLet

1/ 22 2

1/ 22 2 2

2

2 2

1

11 2 /

2

/

1

s v c

ds v c v c dv

v cds dv

v c

So, 22 2

2 2

/

1

f f

i i

v cmc dv mc ds

v c

2

2 2 2 2 2 2

2 2 2 2

1 1 0

1

f imc s s

mc v c mc c

mc v c mc

2 21

f f

i i

mvP dv dv

v c

Relativistic Work & Energy

Putting everything together,

22 2 2

2

2

211mc v

mvc mcW

v c

2 22

2 2

2 22 2

2 2

1

111

v cmc v c

v

mv

v c cmc

2

2 2

2 2

22

2 211

1

mcmv

v c ccv c

vm

2mv

2 2mc mv 22

2 2 2

2

21 1

mcmc

v cm

vc

c

Relativistic Work & EnergyApplying the work-energy theorem, this amount of work done to accelerate the particle from 0 to v should equal to the change in KE.

(Laws of physics should be unchanged in relativity !)

Since 0, 0

,i i

f f

v KE

v v KE KE

Relativistic Kinetic Energy of a particle moving with as measured in the lab frame. v

22 2 2

2 21

mcKE mc mc mc

v c

W =

21

2mv

Relativistic KE Classical KE

Slow moving particle regime :v c

Using binomial theorem,1/ 2 22 2 2 2

2 2 2 2

1 11 1 1

2 2

v v v vO

c c c c

Substituting this into the equation for Relativistic KE,

1/2 1/22 22 2 2

2 21 1 1

v vKE mc mc mc

c c

This is the classical result for v << c.

22

2

2

11 1

2

1

2

vmc

c

mv

Total Relativistic EnergyLet look at the equation for the relativistic kinetic energy of a moving particle again.

22

2 21

mcKE mc

v c

It separates into two terms: • 1st depends on speed of the particle• 2nd is a constant term independent of motion

KE can be interpreted as the difference between a total energy term depending on motion and a constant rest energy term independent of motion.

2

2 2

2 2

1

mcE

v cKE mc mc

total energy rest energy

Total Relativistic Energy

E is the total relativistic energy for a relativistic particle.

22

2 21

mcE mc

v c

2

2 2 2

2 21

1

mcKE E mc mc mc

v c

and, its kinetic energy KE is given by,

- But, there is a residual Total Relativistic Energy even for a particle at rest. The quantity mc2 is called the Rest Energy.

Total Relativistic Energy

Note 1: for v = 0, and

2 0E mc

Rest Energy (mc2) • Independent of velocity• Proportional to the mass of the particle• Mass is a form of energy

- KE = 0 for a particle at rest is expected…

2 21 1 1 0KE mc mc 1

Total Relativistic Energy

Note 2: Since E = mc2 is the total relativistic energy of the system,

E is conserved in all processes !

- The statement on the Conservation of Total Relativistic Energy is more general

- It combines with the two classical independent conservation laws:

conservation of energy conservation of mass

Total Relativistic Energy

Note 3:

The mass “m” which we have been using is a constant in our analysis. It is called the rest mass (“proper” mass) and is the mass of an object measured by an observer stationary with the object.

The quantity mrel = m is called the “relativistic mass” and is not a constant for a moving object and is measured by an observer not at rest with the object.

Conservation Laws

Relativisity Energy

Relativistiy Momentum

i f

i f

E E

P P

22

2 2

2 2

1

1

mcE mc

v c

mm

v c

vP v

The following set of equations form the generalized conservation laws in Special Relativity.

AND, these conservation laws apply to all processes equally in all inertial reference frames !

Example 37.11: A Relativistic Collision

Initially, both protons move in opposite directions, net linear momentum is zero. The three particles after collision are at rest again with net linear momentum equals to zero. Thus, relativistic momentum is by design conserved.

Question: Find initial velocity of proton.

We now need to consider the conservation of total relativistic energy.

Example 37.11

Before After

2 2 22 2proton proton pionM c M c m c

22 2 2 2 2 2 2

2 21 1

cc v c v c

c v

21 1 1.072 0.360v c c

Note: Mass by itself is not conserved in this process!

28

27

2 1

2.40 101 1 1.072

2 2 1.67 10

proton pion

pion

proton

M m

m kg

M kg