Transformations

Before I get lost in the details, let me share the answer.You have had the freedom to choose axes (reference frames) throughout this course.

If the laws of physics are valid, you must be able to transform from your coordinate system to someone else’s coordinate system, and get the same results.

Relativity deals with transformations between reference systems which are moving with respect to each other.Reference frames which are moving, but not accelerated, with respect to each other, are those in which Newton’s laws may be applied.

Mathematically (after all, that’s what this class is about) how do you transform from one coordinate system to another?Let’s take the simplest example: a point in reference frame x’y’z’ moving with a fixed velocity relative to reference frame xyz, such that the reference frames coincide at t=0, and the relative velocity is along the x-axis.

The coordinates x’y’z’t’ give the location of a point as measured from the fixed frame in terms of its location in the moving reference frame.

(An observer in the fixed frame would say that the point is located at x’y’z’t’ relative to the moving frame.)

This is called the Galilean Transformation.

Newton’s laws are invariant under a Galilean transformation.It is “the common sense relationship which agrees with our everyday experience.” (hyperphysics)

It turns out to be not how the universe works! It is not the law!

Maxwell’s equations are not invariant under the Galilean transformation.

This was recognized by physicists of the 1800’s as a serious problem to be overcome.

Lorentz discovered the transformation under which Maxwell’s equations are invariant.

The Lorentz transformation (simplified version).

This is the Law.

The Lorentz transformation is not intuitive. Why should the laws of nature be invariant under it?

That is not a question of science.

Nature follows Lorentz transformations. Live with it.

I am about to “derive” some equations of relativity using thought experiments. It is appropriate for you to be suspicious of thought experiments.

The same equations can be derived using the Lorentz transformation. No dubious thought experiments. Just pure mathematics describing nature.

Hyperphysics has a table showing the ideas and experiments that DEMAND relativity.

Relativity

What does it mean tobe in motion?

If we both measure the same object with the same tools, should we get the same result?

Should the laws of physics be the same for everybody?

You’re in a spacecraft and a comet zips by. Are you moving or is the comet moving?

Special Relativity

What does it mean toknow something?

What does it mean to measure something?

We have this idea that “physical reality,” whatever that is, ought to be independent of who/when/where/how a measurement is made.

Electromagnetic theory was perfected by Maxwell and others in the late 1800's.

*Imagine the ether attached to this universal reference frame. If you are moving relative to it, you experience an “ether drift.”

Water waves propagate through water, sound waves propagate through air.

It is not critical to electromagnetic theory, but it was believed that electromagnetic waves propagated through the “ether,” relative to some universal reference frame.*

The ether, being ethereal, proved very difficult to detect!

rivercurrent

T1

T2

You swim 200 meters downstream in a river, turn around, and swim 200 yards upstream. It takes a time T1.You swim 200 meters perpendicular to the river bank, turn around, and swim 200 yards back. It takes a time T2.T1 and T2 are different. Newtonian “relativity theory” shows you how to calculate T1 and T2 (add the vector current velocity to your vector velocity).If you make the “same” measurement on light moving through the ether, you ought to get the “same” result, T1 and T2 are different.

Newtonian Relativity Theory

Michelson and Morley built an interferometer capable of making such a measurement.

mirror

mirror

“partial”mirror

detector

lightsource

hypotheticalether drift

A

B

Half the light follows path A.

Half the light follows path B.

The dashed line portions of the paths are oriented differently relative to the ether drift.

If the times to travel paths A and B are the same, the two light beams arrive in phase and interfere constructively.

If the times are different, the beams interfere destructively.

Measurement of changes in interference fringe shifts allow you to deduce the time difference.

hypotheticalether drift

A

B

But wait! you object. It is impossible to make paths A and B exactly the same length.

An observed fringe shift might be due to the path length difference, or it might be due to the different orientations of the path relative to the ether drift.

So you take a measurement, rotate the apparatus 90 degrees in the horizontal plane, and take another measurement.

The difference between the two measurements allows you to very precisely measure the time difference due only to the ether drift.

Michelson and Morley did the experiment in July, 1887. They found nothing.

No ether drift. (Less than 5 km/s; current upper limit is 15 m/s.)

No ether.*

*No universal frame of reference?

They tried it again later, in case during the July measurement the earth was coincidentally at rest with respect to the ether. They got the same results.

I’ve seen sources that say this result wasn’t terribly bothersome, because the “ether” was a conceptual convenience, and was not required to make E&M theory work. I’ve seen other sources that say this was “devastating” at the time. It certainly created a problem.

In fact, if you believe Michelson and Morley and Maxwell, you are forced to conclude that the speed of E&M radiation is the same in all non-accelerated reference frames, regardless of the motion of the radiation source. A bit difficult—no, make that extremely difficult—to accept!

How do you reconcile the lack of a universal reference frame with the idea that everybody's measurement of the same thing ought to produce the same result?

Einstein, 1905, Special Theory of Relativity

We believe the laws of physics work and are the same for everybody.

Let’s make these two things our postulates and see where we are led. The validity of the postulates will be demonstrated if the predictions arising from them are verified by experiment.

Experiment demands that the speed of light be constant.

Special theory of relativity treats problems involving inertial (non-accelerated) frames of reference.

Postulates of special theory of relativity:

the laws of physics are the same in all inertial reference frames

the speed of light in free space has the same value for all inertial observers*

The first “makes sense.” The second is required by experiments but contradicts our intuition and common sense.

*Independent of the motion of the source or relative speeds of observers!

Time Dilation – a consequence of the two assumptions

The time interval between two events which occur at the same place in an observer’s frame of reference is called the proper time of the interval between the events. We use t0 to denote proper time.

Suppose you are timing an event by clicking a stopwatch on at the start and off at the end. In order for the stopwatch to measure the proper time, the “start” and “stop” events must occur at the same place in your frame of reference.

You’ve been chosen to be a timer at a track meet, so you go stand by the finish line. You start your stopwatch when you see the puff of smoke from the starter’s gun at the starting line, and stop it when the first runner crosses the finish line. Did you measure the proper time for the sprint?

Let’s begin with a definition, and then construct a clock.

Now let’s make a clock.

mirror

laser with built-in light detector

tick tock

L0

“It's not that I'm so smart , it's just that I stay with problems longer.”—A. Einstein

How long does a “tick-tock” take?

mirror

laser

L0

time = distance / velocity

t0 = 2L0 / c

“I measure proper time because the light pulse starts and stops at the same place.”

Now put this clock in a transparent spacecraft and observe as it speeds past.

L0

entire clock moves with speed v

tick tock

“I don’t measureproper time because

“tick” and “tock” occurin different places.”

How long does a “tick-tock” take? Let the total time be t.

L0

v

vt/2 vt/2

( L 0

2 + (vt

/2)2

)1/2 ( L

0 2 + (vt/2) 2 ) 1/2

distance = velocity · time

220

tD=2 L + v

2

According to the second postulate of special relativity, light travels at a speed c, so D = ct.

We also know the proper time from our “stationary clock” experiment: t0 = 2L0 / c

220

tD=2 L + v ...(1)

2

D=ct ...(2)

00

ctL = ...(3)

2

Solving (1) and (2) for t and replacing L0 using (3) gives:

0

2 2

tt =

1- v c

Note that (1-v2/c2)1/2 < 1 so t > t0. It takes longer for an event to happen when it takes place (is timed) in a reference frame moving relative to the observer than when in takes place at rest in the observer's reference frame. Time is dilated.* This applies to all clocks.

*How to remember what “dilated” means. Pupils in your eye can dilate or contract. Dilate must be the opposite of contract, so “dilate” must mean take on a larger value.

If I time an event which starts and stops in in my frame of reference, I measure t0.

If I use my clock to time the same event as it takes place in a reference frame moving relative to me, I measure t>t0.

A moving clock ticks slower.

In the latter case, I claim my clock, which measured t, is correct, so that an identical moving clock, which would measure t0 in the moving reference frame, is slow.

An event must be specified by stating both its space and time coordinates.

Example: the Apollo 11 spacecraft that went to the moon traveled a maximum speed of 10840 m/s. An event observed by an astronaut in the spacecraft takes an hour. How long does an earth observer say the event took?

Problem Solving Step 0. Think first!

Always ask: what is the reference frame of the event? Is the observer in this reference frame or moving relative to it?

The event took place in the spacecraft. The proper time t0 is the time measured in the spacecraft. Thus, t0 = 3600 s.

The “observer” in this problem is the person on earth, not the astronaut! The earth observer measures t.

Problem Solving Step 1. Draw (if appropriate) a fully-labeled* diagram. (*Include values of known quantities.)

If it helps you to draw a sketch of the earth, a spacecraft, and a couple of stick figures, do so!

Include values of known quantities. c = 3x108, v = 10840, t0 = 3600.

If you use SI units throughout, your answer will be in SI units, and I only need to see units with your final answer.

If you mix systems of units, show the units at each step.*

You can always show units at each step if it helps you.Sooner or later, if you mix units, you will suffer pain.

Problem Solving Step 2. OSE.

So far, we only have one relativity OSE:

0

2 2

tt =

1- v c

“Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. THAT'S relativity.” – A. Einstein.

Problem Solving Steps 3 and 4. Solve algebraically first, then substitute values.

The algebra is already done in this case.

8 2

3600t =

1- (10840 3×10 )

t = 3600.00000235 s.

Not a big difference, but it is measurable. The actual experiment has been done with jets flying around the earth, and the predicted time dilation has been observed.1

As expected, the earth observer measures a bigger number for the time. The moving clock on the spacecraft measured a smaller number. The moving clock ticks slower.

1J. C. Hafele and R. C. Keating, Science 177, 186 (1972).

What if v>c? It can't happen. We’ll see later that it is not possible to accelerate an object up to the speed of light.

What about time running backwards? Sorry, time always runs forwards.

What about seeing an event before it happens? Can't, because c is finite.However, because of time dilation, events which appear to be simultaneous in one reference frame may not appear to be simultaneous in another reference frame.

“The only reason for time is so that everything doesn't happen at once.” – A. Einstein

On the constancy of c…Recent research suggests that c may not be constant, slowing down by 0.00000000000008% per year over the last 12000000000 years.

If this is true, there will be subtle corrections to special relativity, which may or may not have profound consequences.

"If we knew what it was we were doing, it would not be called research, would it?“—A. Einstein

Newtonian mechanics will still work just fine as long as velocities are not too big. Relativity will work just fine as long as we don’t extrapolate too far into the past or future.Lots of physicists will have nice jobs for a long time to come.

Let’s consider another problem that time dilation helps us solve.

Has anyone here ever felt a muon?

A muon is an elementary particle with a mass 207 times that of an electron, and a charge of either +e or –e. Muons are created in abundance at altitudes of 6 km* or more when cosmic rays collide with nuclei in the atmosphere.

Fortunately, muons interact only very weakly with matter, which is why it is OK that many of them are passing through your body right now.

*This is in the upper reaches of the troposphere, the part of the atmosphere in which we live.

Does anybody even know what a muon is?

Muons travel with speeds of about 0.998 c (fast!) and have an average lifetime of 2.2 s (2.2x10-6 s).

How far can an average muon travel during its lifetime?

d = v td = 0.998 · 3·108 · 2.2·10-6 = 0.66 km.

How can muons get through the 6 or more kilometers of atmosphere between their birthplace and us if they only live long enough to travel 0.66 km?

OK, some will go more than 0.66 km, and some less, but mostly not by very much. So the question stands.

Time dilation!

I say the muon’s clock ticks slow. I say that while the muon thinks* its clock ticks 2.2 s, I observe that it actually ticks -6

2

2.2×10t =

1- (0.998)

t 34.8 s

During this time the muon travels a distance

d = 0.998 · 3·108 · 34.8·10-6 = 10.4 km,

so the average muon will reach me before decaying.

*Of course, a muon doesn’t “think” anything, but we use words like that to help us form a mental image of the process. If you prefer, imagine a nano-human riding on the muon and reporting what he/she sees.

Double-check: what is the event, who is the observer, and who measures the proper time.

The event is the muon “living.”

The event does not take place at a single location in my reference frame, so I measure the dilated time, and the calculation was correct.

One important aspect of relativity is that there is only one reality. If I see the muon arrive at the surface of the earth, the muon must agree that it actually did arrive at the surface of the earth.

“Relativity teaches us the connection between the different descriptions of one and the same reality.”—A. Einstein

Our average muon “says” there is no doubt whatsoever that its lifetime is 2.2 s, and during that time it travels 0.66 km. I say the muon reaches the surface of the earth. The muon says it doesn’t??

“I thought you said time dilation would help us solve the muon problem.” We seem to have created a new problem.

Either we have encountered two different realities, or else there is…

Length Contraction

If two observers in relative motion measure different times for an identical event, what makes us think they should measure the same lengths for an identical object?The formula for length contraction is not terribly difficult to derive. I’ll lend you a book if you are curious. Here is the formula.

2 20OSE: L L 1 v / c

“The faster you go, the shorter you are.”—A. Einstein

The Proper Length, L0, of an object is its length as measured in its own rest frame.

An observer measuring the length of an object moving relative to him will measure a length L less than the length L0 he would measure if he were not moving relative to the object.

The length contraction occurs only along the direction of relative motion. A spacecraft moving past an observer at nearly the speed of light will seem to be very short in length and normal diameter.

Let me demonstrate length contraction using a meter stick…

A muon created at an altitude of 10.4 km would say that during its lifetime it saw an atmosphere of length

2 2 20L L 1 v / c (10.4) 1 0.998 0.66 km

I say the muon gets to earth because its lifetime is longer. The muon says it gets to earth because the atmosphere is shorter. Different descriptions of the same reality.Be careful when you talk about the lifetime of a particle moving with v close to c. You need to specify the reference frame in which the lifetime is measured!

The Twin Paradox

A and B are 20 year old twins. A travels on a spaceship at v = 0.8c to a star 20 light years* away and returns.

B, left behind on earth, says the trip takes 2·20/0.8 = 50 years.B is 70 years old when A returns.

*A light year, y, is the distance light travels in one year. Thus, y = (1 year)·(c). If D is a distance expressed in light years, then the number of years it takes to travel that distance at a speed of v is found from time = (distance) / velocity. Thus:time in years = (distance in light years) / (velocity expressed as a fraction of

c).

B also observes that A’s clock (which is identical to B’s) ticks slowly, and records less time. If the event in question is the ticking of A’s clock, then the 50 years calculated above is the dilated time t (why?).

The proper time, which in this case is amount of time recorded by a clock in the spacecraft, is is found by solving our time OSE for t0:

2 20t =t 1- v c

20t =50 1- 0.8

0t =30

According to B (who was left back on earth), A’s clock only ticked 30 years, so that A is 20 + 30 = 50 years old on return to earth.

At the end of the trip, B, left behind, is 70 years old. A, who made the trip, is 50 years old. Can this be possible?Yes! Absolutely! and it was verified experimentally in the jets-around-the-world experiment mentioned earlier.

Now here’s the paradox. A moving clock ticks slower. This applies to all observers. A, on the spacecraft, sees B move away and then come back.* A says B’s clock ticks slower. A does the calculation presented on the last slide and concludes that at the end of the trip, B is 50 and A is 70.

*Remember, there is no absolute reference frame for specifying motion. Motion is relative! An observer is free to say “I am at rest; you are the one moving!”

That’s the famous twin paradox. It would appear that each twin rightfully claims the other aged less. Have we discovered an example of the existence of two different, mutually exclusive realities?

When you encounter a paradox like this you can be sure that someone has pulled a fast one on you.

In this case, an unwarranted calculation was made.

Special relativity applies only to observers in inertial (non-accelerated) reference frames. A had to accelerate (very rapidly) to leave earth and get up to speed, and again when turning around to head home, and a third time when landing on earth.*

A is not allowed to use the equations of special relativity! B is, and B’s calculation is correct: A comes back 20 years younger.

What’s poor A to do? Doesn’t a moving clock tick slower? Yes, so evidently during A’s period of extreme acceleration, B’s clock (as observed by A) would tick incredibly fast. Isn’t A allowed to use the laws of physics? Yes, but it would have to be general relativity.

If you examine the problem carefully, it’s only the turning around part that causes A trouble.

We won’t have completely eliminated the paradox unless we can find a description for A’s reality that agrees with B’s reality.

A, in the spacecraft, needs to reconsider the distance traveled. During the “out” portion of the trip, A will say that the actual distance traveled was

2 2 20L =L 1- v c = 20 1- 0.8 =12 light years,

and that the back portion was also 12 light years. 24 light years at a speed of 0.8 c takes 30 years so A ages 30 years during the trip, and comes back at age 50.

B tells A “you are younger because your clock ticked slower.”A says “I am younger because the trip covered less distance than you thought.”

Same reality, two different descriptions.

There are a number of famous paradoxes based on relativistic calculations. Typically, someone makes an invalid calculation (usually on purpose, to see if they can trick you).

In another problem, where a very fast runner tries to put a 10 meter pole in a 5 meter barn, a paradox arises because…

Electricity and Magnetism

“What led me more or less directly to the special theory of relativity was the conviction that the electromagnetic force acting on a body in motion in a magnetic field was nothing else but an electric field.”—A. Einstein.

In other words, Einstein believed that what you and I might call a magnetic force is really just an electric force in another inertial reference frame.

The material we’ve been studying is fascinating and thought-provoking, but it is not how Einstein’s theory of relativity came into being.

Consider a conducting wire and a positive test charge.

+ + +

+

+

+ +

+ + ++

-

-

-

---

- - -

-

-

+

What force does the test charge “feel” due to the charges in the wire?

+ + +

+

+

+ +

+ + ++

-

-

-

---

- - -

-

-

+

No net charge inside the conductor.

No electric field outside the conductor.

No force!

Repulsion, because there is a – closest to the test +?

+ + +

+

+

+ +

+ + ++

+

What does the test charge see when an electric field is applied and current flows?

E

-

-

-

---

-` - -

-

-

-

-

-

-

-

-

- -

--

The test charge “observes” that the space between the moving electrons is contracted. There are more electrons in the part of the conductor nearest the test charge!

The test charge “observes” that the moving electrons are closer together than the stationary protons, and therefore "feels" a Coulomb attraction.

A human observer is unable to see the electrons, and attributes the attraction to a “magnetic force” generated by the moving charges.

Same reality, two different descriptions!

And both descriptions are mildly troublesome, as we will see shortly…

It is a fact that electric charge is both conserved and relativistically invariant.

Our thought experiment with the conductor and test charge suggests that a conductor which is electrically neutral in one reference frame might not be electrically neutral in another. How can we reconcile this with charge invariance?My modern physics textbook author claims there is no problem, because you have to consider the entire circuit. Current in one part of the circuit will be balanced by opposite current in another part.

Although the explanation is correct, I don’t find it satisfying.* Maybe the pole-in-barn paradox will help us understand.*It seems logical that if moving electrons are closer together in one part of the circuit, they ought to be closer in other parts of the circuit too, so that the conductor is no longer neutral and charge is not conserved.

The Pole-Barn Paradox

A speedy runner carrying a 10-meter pole approaches a barn that is 5 meters long (short barn!), with open doors at each end. A farmer stands nearby, where he can see both front and back door at the same time.a) How fast does the runner have to go for the farmer to observe that the pole fits entirely in the barn?

b) What will the runner observe?

The answer to a) involves a simple length contraction* calculation.

For the pole to fit in the barn, the farmer must measure a contracted length L = 5 m for the pole of proper length L0 = 10 m.

2 20L =L 1- v /c

2 25=10 1- v /c

The result is v = 0.866 c. If the runner is going that fast, or faster, the farmer observes the pole to fit inside the barn.

Length contraction is often called the Lorentz contraction, named after the scientist who discovered the mathematical transformations which lead to the equation for length contraction.

The answer to b) starts with another length contraction calculation.

The runner is moving …

The runner says the speeding barn has a length equal to

2 20L =L 1- v /c

2L =5 1- 0.866

L =2.5 m.

The pole can’t possibly fit inside the barn.

no, the runner isn’t moving. The runner sees the barn moving towards him at a speed of v = 0.866 c.

How do we explain this paradox? Which observation is the physical reality?

The answer: both observations are correct!

A detailed calculation (I can lend you the book it is in, if you are interested) shows that the runner observes the rear end of the barn arriving* at the front end of the pole long before the front end of the barn arrives at the rear end of the pole. The pole doesn’t fit!

*Remember, the runner sees the barn moving past him.

Events which are simultaneous in the farmer’s frame of reference (front pole arriving at back barn and back pole arriving at front barn) are not simultaneous in the runner’s frame of reference.

Simultaneity is not a “universal physical reality.”

Now I’m no longer worried about the test-charge-plus-conductor example. At a certain instant in time I may observe an excess of moving negative charge in the portion of the circuit nearest me, but does not mean I can claim there is a net excess of moving negative charge in the entire circuit at that instant in time.

Now where were we before this interruption started…

“Because simultaneity is a relative concept and not an absolute one, physical theories that require simultaneity in events at different locations cannot be valid.”—Beiser, Modern Physics.

An observer who doesn’t know about relativity, or even one who knows about relativity but invokes charge invariance, will claim that the conductor has a neutral charge density and invents a “magnetic” force to explain the attraction.

But the “magnetic” force is present only when current is flowing. It is not valid to talk about a separate “magnetic” force. You must talk about the “electromagnetic” force.

What you call “magnetic” force is just a manifestation of the Lorentz contraction and Coulomb’s law, and is not a separate force of nature.

The mathematical transformations which lead to our relativistic equations for length and time were derived by Lorentz to make Maxwell’s equations invariant in inertial reference frames.*

*Part of Einstein’s genius was realizing that Lorentz was on to something big!

Because Maxwell’s equations are invariant in inertial reference frames, special relativity does not demand that we correct them.

On the other hand, when it comes to Newton’s Laws…

Relativistic Momentum

We believe very strongly that momentum is conserved. Let’s see what effect a relativistic calculation has on momentum.Imagine this collision: person A on the ground throws a ball straight at the side of a railroad flatcar moving past him. Person B on the flatcar throws an identical ball straight out at the same speed. The two balls meet head-on and bounce back.

A and B measure different travel times for ball thrown from the flatcar (time dilation). They will calculate different initial and final momenta. They will not agree that momentum is conserved.

A more complete derivation is in my klunky relativity supplementary lecture.

2

2

mm(v) = ,

v1-c

where m is the mass of the ball at rest, and m(v) is the mass it needs to have when it is moving.

Non-conservation of momentum is an alarming idea. What can we do to fix this situation?

It turns out that if the ball thrown by the flatcar observer has a greater mass than the ball thrown by the stationary observer, momentum is conserved:

In the old days* we then said “OK, m is the mass of the object at rest and m(v) is its mass when it is moving. Let’s call m0 the rest mass and m the mass when it is moving (‘relativistic mass’).” This notation is consistent with our equations for time dilation and length contraction, so we have

0

2

2

mm= .

v1-c

*I.E., back in the dinosaur ages, when I studied relativity in college. Also, in the previous edition of my modern physics text.

This was Einstein’s original approach, but later he said it is “not good.”

The “new” approach is to say “Look, mass is mass. We believe it is something fundamental. If we believe in conservation of momentum, we had better change our definition of momentum.”

If we define momentum as 2

2

1= ,

v1-c

where p= mv

Then “mass is mass,” momentum is conserved in our thought experiment (and in real life), and relativistic momentum reduces to classical (Newtonian) momentum in the limit v0.*

*More satisfying than saying “mass changes with velocity.”

Let’s make this new notation official.

OSE: p= mv2

2

1OSE: = ,

v1-c

A consequence of this new definition of momentum:

dp F = =ma

dt

dp d F = = mv

dt dt

In the old days, rest mass was relativistically invariant. Now mass is relativistically invariant. Same reality, just different use of words.

More consequences of this new definition of momentum:

3/22

2F v a= 1-

cm

For finite F, a0 as vc.*

No finite force can accelerate an object having nonzero mass up to the speed of light!

no limit to classical velocity

relativistic momentum increases without bound as vc

When can I use “rest” mass, and when do I have to use

2

2

mm(v) = ?

v1-c

Object v v/c m(v)/m

jogger 10 km/h .000000009 ≈ 1

spaceshuttle

104 m/s 0.000033 1.0000001

electron 106 m/s 0.0033 1.001

electron 108 m/s 0.333 1.061

Mass and Energy

From your first-semester physics course:

d mvKE F ds ds.

dt

Use the definition of and integrate by parts to get

2 2 2KE= mc - mc = -1 mc

2 2mc =mc +KE.

Assuming potential energy is zero (we can always choose coordinates to do this), we interpret mc2 as total energy.

2E=mc +KE.

When an object is at rest KE = 0, and any energy that remains is interpreted as the object’s rest energy E0. 2

0E =mc .

When an object is moving, its total energy is

2

2

2

2

mcE= mc = .

v1-c

This is the closest you’ll come to seeing E=mc2 in this class. In the “old days,” E=mc2 would have been written E=mc2.

This is really just a variation of the OSE on the previous slide.

These equations have a number of interesting implications.Mass and energy are two different aspects of the same “thing.”

Conservation of energy is actually conservation of mass-energy.

The c2 in E0=mc2 means a little mass is “worth” a lot of energy.

Total energy is conserved but not relativistically invariant.Rest (or proper) mass is relativistically invariant.

Mass is not conserved! (But it is for the purposes of chemistry.)

Your lunch: an example of relativity at work in “everyday life.”

Example: when 1 kg (how much is that?) of dynamite explodes, it releases 5.4x106 joules of energy. How much mass disappears?

This is actually a conservation of mass-energy problem. If this material were presented in Physics 23, I would make you start with your conservation of mass-energy OSE and derive the appropriate equations from there.For now, it is sufficient to realize that the problem is just asking “what is the mass equivalent of 5.4x106 joules of energy?”

20E =mc 0

2

Em=

c

6

-1128

5.4×10m= =6×10 kg.

3×10 Conservation of mass is a very good approximation!

If we are to claim relativistic mechanics as a replacement theory for Newtonian mechanics, then relativistic mechanics had better reduce to Newtonian mechanics in the limit of small relative velocities.

2

2 2 2

2

2

mcKE= mc - mc = - mc .

v1-c

Our text shows that for v<<c,

21KE mv .

2

Use Newtonian KE every time you can get away with it! Use relativistic KE only when you must!

If v = 1x107 m/s (fast!) then mv2/2 is off by only 0.08%. Probably OK to use mv2/2. If v = 0.5 c, then mv2/2 is off by 19%. Better use relativity.

When can I get away with using KE = mv2/2, and when do I have to use KE = mc2 - mc2?

2

2

mvp= .

v1-c

2

2

2

mcE=

v1-c

Total energy and magnitude of momentum are given by

22 2 2 2E - p c = mc .

With a bit of algebra, you can show

The quantities on the LHS and RHS of the above equation are relativistically invariant (same for all inertial observers).Rearranging:

22 2 2 2E = mc +p c .

Energy and Momentum

2

2

mvp= .

v1-c

2

2

2

mcE=

v1-c

Is it possible for a particle to have no mass? If m = 0, what are E and p?

For a particle with m = 0 and v < c, then E=0 and p=0. A “non-particle.” No such particle.

But if m = 0 and v = c, then the two equations above are indeterminate. We can’t say one way or the other.

If m = 0 and v = c, we must use 22 2 2 2E = mc +p c .

The energy of such a particle is E = pc. We could detect this particle! It could exist.

Do you know of any massless particles?

photon

neutrino*

*Doubtful. Nobel prize for you if you show mneutrino = 0.

graviton**

**Maybe. Nobel prize for its discoverer. Problem: gravitational fields much, much weaker than E&M fields.

graviton is to gravity as photon is to E&M field

Particles having KE >> E0 (or pc >> mc2) become more photon-like and behave more like waves.

The momentum carried by massless particles is nonzero (E = pc).

Could you stop a freight train with a flashlight?

Could you stop a beam of atoms with a laser beam?

An electron has a rest mass of 9.11x10-31 kg. If you plug that mass into E0 = mc2, you get an energy of 511,000 eV, or 511 keV, or 0.511 MeV.

We sometimes write the electron mass as 0.511 MeV/c2.It is also possible to express momentum in “energy units.” An electron might have a momentum of 0.3 MeV/c.If you are making a calculation with an equation like

22 2 2 2E = mc +p c

and you want to use 0.511 MeV/c2 for the electron mass, please do. It often simplifies the calculation. But watch out…

Because mass and energy are convenient, we sometimes write masses in “energy units.”

What is the total energy of an electron that has a momentum of 1.0 MeV/c?

22 2 2 2E = mc +p c

2 22 2 2

2

0.511 MeV 1.0 MeVE = ×c + c

c c

2 22E = 0.511 MeV + 1.0 MeV

2 2E = 1.26 MeV

E=1.12 MeV

Avoid the common mistake: don’t divide by an extra c2 or multiply by an extra c2 in the 2nd step.

Notice the convenient cancellation of the c’s in the 2nd step.

General Relativity

Something to think about. Is the mass that goes in F = ma (or the relativistic version) the same “thing” as the mass that goes in F = Gm1m2/r2?

Not necessarily!

Experimentally, the two “kinds” of mass are the same to within better than one part in 1012, and most of us believe they are the same anyway, so…

“An observer in a closed laboratory cannot distinguish between the effects of a gravitational field or an acceleration of the lab.”

The principle of equivalence.

The principle of equivalence leads one to conclude that light must be deflected by a gravitational field.

Experimental observation of this effect in 1919 was one of Einstein’s great triumphs.

We investigate more about light and gravity in modern physics classes.

"If A equals success, then the formula is: A=X+Y+Z. X is work. Y is play. Z is keep your mouth shut.“—A. Einstein

More Einstein quotes:“As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.""Relativity teaches us the connection between the different descriptions of one and the same reality"."I sometimes ask myself how it came about that I was the one to develop the theory of relativity. The reason, I think, is that a normal adult never stops to think about problems of space and time. These are things which he has thought about as a child. But my intellectual development was retarded, as a result of which I began to wonder about space and time only when I had already grown up.""The secret to creativity is knowing how to hide your sources.""The important thing is not to stop questioning."Only two things are infinite, the universe and human stupidity, and I'm not sure about the former.""Things should be made as simple as possible, but not any simpler.""Sometimes one pays most for the things one gets for nothing.""Common sense is the collection of prejudices acquired by age 18."Strange is our Situation Here Upon Earth""If you are out to describe the truth, leave elegance to the tailor.""I never think of the future. It comes soon enough.“"Not everything that counts can be counted, and not everything that can be counted counts.""The faster you go, the shorter you are.""The wireless telegraph is not difficult to understand. The ordinary telegraph is like a very long cat. You pull the tail in New York, and it meows in Los Angeles. The wireless is the same, only without the cat. "