Chapter 26Michelson-Morley

ContinuedRelativistic Energy

and Momentum

Hint: Be able to do the homework (both theproblems to turn in AND the recommended ones)you’ll do fine on the exam!

Monday, May 10, 1999 10:30am - 11:20amChs. 20, 26, and 27

You may bring one 3”X5” index card (hand-writtenon both sides), a pencil or pen, and a scientificcalculator with you. Same format!

Hint: Review notes from my review lectures! Tryto do some of the old homework recommendedhomework, and exam problems.

Monday, May 10, 1999 11:30am - 12:30pm

everything we’ve covered

You may bring one 8.5”X11” sheet (hand-writtenon both sides), a pencil or pen, and a scientificcalculator with you.

Monday, December 15, 1997 11:30am - 12:30pm

everything we’ve covered

Format: 5 problems, pick 4.

1 problem on each of the following topics:

Electrostatics Circuits Magnetism

Optics Modern

So, what happens to a clock when it moves?

Well, let’s use the following clock:

Here, the flashbulbgoes off, light bouncesoff the mirror, isdetected at the photocell,triggering a “tick”sound and anotherflash of the bulb.

A simple clock...

photocell flashbulb

mirror

1.5 m

photocell flashbulb

mirror

1.5 m

This clock shouldproduce a tick

every how manyseconds?

td

v

m

m s

2 15

3 108

( . )

/

t = 10-8 sec

So what happens when you put this clockon our boxcar (in a moving reference frame)?

GM R/R

Physics

Rules

u

Flashbulb firstgoes off here

GM R/R

Physics

Rules

u

Photocell receiveslight at this point,causing another

flashHow did thephoton get there?

How far did thisphoton travel?

ut/2

1.5m

d ut/ . ( / )2 15 22 2

d/2 d/2

How long does ittake the photon to

travel this distance?

td

v

m ut

m ss

2 15 2

3 1010

2 2

88( . ) ( / )

/

GM R/R

Physics

Rules

uWhich meansthat the clock onthe train (the movingclock) appearsto be ticking moreslowly than thestationary clock!!!

Which meansthat the clock onthe train (the movingclock) appearsto be ticking moreslowly than thestationary clock!!!

A little geometry tells us the relationshipbetween the two clocks:

ct/2

ut/2

1.5m

Recall, however, that according to Einstein,the speed of light is the same in all frames.

c tm

u t2 2

22

2 2

2215

2

( )( . )

( )

Which means that inside the boxcar, 1.5mis the distance light travels in 10-8/2 s! Callthe time between flashes as measured in theboxcar t0, then

c t c t u t2 2

2

20

2

2

2 2

22 2 2

( ) ( ) ( )

( )( ) ( )c u t c t2 2 2 20

2

tt

v ct

0

2 2 01 /

1

1 2 2v c/

tt

v ct

0

2 2 01 /

What does this mean???

To a stationary observer, moving clocksappear to be ticking more slowly!

And the faster the clock is moving,the more slowly it appears to tick!

Although we used a light clock, the type ofclock used in this experiment is immaterial.It really is TIME itself that is moving atdifferent rates for the two observers!

Although we used a light clock, the type ofclock used in this experiment is immaterial.It really is TIME itself that is moving atdifferent rates for the two observers!

So what happens when we put the Michelson-Morley experiment on our moving boxcar?

Remember, the Michelson-Morleyexperiment provides a null result, whichmeans that interference in never observed!

GM R/R

Physics

Rules

u

GM R/R

Physics

Rules

u

So, whether we sit on the boxcar or standalongside the tracks, light from the two pathswill appear to arrive at the detectorsimultaneously.

But to the observer on the side of the tracks,we know that the time required for light totravel the two, equal-length paths shouldappear to be different.

Again, using our analogy of the boat in thestream, the time difference for light travellingthe two different paths would be:

t t tL c

c

L c

c

||

/

( / )

/

( / )

2

1

2

12 2 2 2u u

But this time difference is NOT observedby the stationary observer! So our equationmust not be correct...

Which is to say, one of the assumptionswe made in writing the equation is incorrect.

WHICH ONE?

tL c

c

L c

c

2

1

2

12 2 2 2

|| /

( / )

/

( / )u u

Well, what if we allowed the two distances to at least appear to be different to the observer alongside the track than to the observer on the boxcar?

L|| is the distance parallel to the motionL| is the distance perpendicular to motion

What relation between L|| and L| willresult in t = 0?

2

1

2

12 2 2 2

L c

c

L c

c

|| /

( / )

/

( / )

u u

L L cL

|| ( / ) 1 2 2u

To the observer alongside the tracks, the distance parallel to the motion appears shorter than the distance perpendicular to the motion!

Of course, to the observer on the boxcarthese two lengths are the same:

L L L|| 0

L0 is called the “proper length” and isdefined to be the length of an object inthe frame in which the object is at rest.

An observer who sees the object in motionwill describe its length in the direction ofmotion as SHORTER than its proper length.

LL

0

L is the apparent length.L0 is the “proper length.”

A train passes a railroad crossing at 100 km/hr. At rest, the train is 1 kmlong. How long does the train appear to a car waiting at the signal?

100 10 1 3600 27 785km hr m hr hr s m s/ ( / ) ( / ) . /

LL

0

( ) ( . / ) / ( / )1000 1 27 78 3 102 8 2m m s m s

L = 999.99999999999 m

m0 rest massL0 proper lengtht0 proper time

m0 rest massL0 proper lengtht0 proper time

In the rest frameof the object:

In the rest frameof the object:

In the lab (wherethe observer is):In the lab (wherethe observer is):

m massL lengtht time

m massL lengtht time

This one’s pretty easy. Our Newtonian definition told us that p = m v

The relativistic formula has exactly the same form, but recall that the mass changes with velocity:

mm

cm

0

2 2 01 v

/

So you will often see the relativistic momentumdefined as:

p m v 0

As was the case in Newtonian mechanics,the relativistic momentum must also beconserved in all collisions. Furthermore,it reduces to the classical expression whenthe velocities are small compared to thespeed of light.

I’m not going to try to derive this one(although you can achieve this resultswith some simple mathematical tricks)...

KE mc m c 20

2KE m c ( ) 1 0

2

Naturally, this expression reduces to ourfamiliar Newtonian expression KE mv 1

22

when v << c.

Here, m0c2 is known as the “rest energy”(E0) of the particle. Which raises thestartling possibility that a particle at reststores an enormous amount of energy.

What is the rest energy of a 0.5 kg baseball?

E m c kg m s J0 02 8 2 160 5 3 10 4 5 10 ( . )( / ) .

Enough energy for 100-W bulb for 10 million years!

KE m c ( ) 1 02

The total energy of a particle consists oftwo parts...

Total Energy = KE + Rest Energy

E m c m c m c ( ) 1 02

02

02

E mc 2

Remember that m is NOT the rest mass inthis expression, and m changes with velocity.

This is the familiar expression you’ve allprobably seen related to Einstein (evenif you had no idea what it meant)

E mc 2

It is useful to derive a relationship betweenrelativistic momentum and energy.

E E m c m c20

20

2 20

2 2 ( ) ( )

202 4

02 4 2

02 41m c m c m c( )

FHG

IKJ

FHG

IKJ

1

11

12 2 02 4

2 2

2 2 02 4

v cm c

v c

v cm c

/

/

/

202 2 2 2 2m v c p c

E E p c20

2 2 2

E p c m c2 2 20

2 2 ( )

Rewritten into its more usual form...

(Total Energy)2 = (momentum)2c2+(rest energy)2(Total Energy)2 = (momentum)2c2+(rest energy)2