Spacetime Invariance Outline

Spacetime Invariance

Spacetime Invariance Relativity and Astrophysics Lecture 05 Terry Herter

Outline  

Simultaneity of events Length measurement  

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Proof of invariance of Spacetime Interval Time Dilation

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Reading

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Lorentz contraction in the direction of motion Invariance of the transverse direction

Spacetime Physics: Chapter 3


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Length measurement 

How do we measure the length of a moving rod – that is, the distance from one end to the other end? 

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We can use the lattice of clocks to mark the location of the two ends at the same time BUT – when the rod lies along the direction of motion someone riding with the rod doesn’t agree that the marking of the positions occurred at the same time! Therefore two different observers will disagree about the validity of the measurement.


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Length-measuring events



Consider a train car moving along for which two lightning bolts strike the ends of the car at the same time according an observer on the ground. 

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Now consider the observer on the train. She measures the front lightning bolt to strike first and the rear bolt later. 

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For him, the char marks on the ground are a valid measurement of the distance.

She concludes that the ground observer made the front mark before the rear and the train has moved forward by the time the rear mark is made. Therefore she concludes that the ground observers measurement of the length of the car too small and concludes that the length is really longer than he (the ground observer) measured. Spacetime Invariance

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Lorentz Contraction 

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The result is that the space separation between the ends of a rod is less in a frame in which the rod is moving than on in which it is at rest. This effect is called Lorentz contraction.  

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Proper length 



A moving rod shrinks in the direction of motion. The length of the rod is largest in the frame in which the rod is at rest. The “rest length” of the rod (measured in a frame in which the rod is at rest).

Note: 

These are measurements that are made – not what we “see” with our eyes.

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Invariance of Transverse Dimension 

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The rod contracts along the direction of motion (longitudinal contraction) There is no change in the transverse dimension 

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The rod does NOT get thinner (or fatter).

Thought experiment – train moving on a track 

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Suppose the ground observer measures the train to get thinner => wheels would fall inside the track The train rider however would see the track shrink => wheels would fall outside the track. But the we can’t have both => no change in width of train Logically inconsistent => can’t change width in transverse direction Hypothetical Earth view

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Hypothetical Train view Spacetime Invariance

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Invariance of Transverse Events 

Consider two equal diameter cylinders moving towards each other.

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There is no change in size in the transverse dimension 

Neither cylinder passes through the other Again, this would be logically inconsistent – which one would “shrink”? Wrong!



Wrong!

Events separated only in the transverse direction that are simultaneous in one frame are simultaneous in both. 

Suppose in the frame of one cylinder a set of simultaneous explosions occur around the rim. These explosions will also be simultaneous in the other frame No preferred direction in the transverse direction, so no explosion can go first (or last)

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Proof of Spacetime Interval Invariance 

Consider a train moving with a velocity, v mirror

v

d′

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Inside the train is light pulse is sent from the bottom of the train car to the top and is reflected back down by a mirror. The photon travels a time, t′ = 2d′/c Spacetime Invariance

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View from outside 

Consider a train moving with a velocity, v mirror

v

L

d′

vt = 2d 

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Outside the train an observer sees the light pulse proceed along the above path. The train moves a distance vt ( = 2d) before the photon returns to where it started. And the photon travels a distance, 2L = ct.

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Invariance 

We want to construct the spacetime interval for each frame

mirror

L

=>

d′

d 2  L2  d 2

And we have:

L  ct / 2

d 

L2  d 2  d 2

Now:

d   ct  / 2 d  vt / 2

Putting this altogether we have

c 2t 2  4d 2  c 2t 2  4 L2  4d 2  c 2t 2  c 2t 2  c 2t  2 “Earth” STI

 c 2t  2 Train STI

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The result is independent of the velocity so we have proved the spacetime interval is constant

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Faster than light? – No! 

Consider a rocket moving a velocity, v, relative to our laboratory frame. 

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The rocket emits two flashes of light separated by time, t′, as measured in the rocket frame.

Using the space time interval, we can figure out the time separation, t, in our lab frame

t 2  x2  t 2  02  t 2  x 2 

But

x  vt 

Substituting this into the spacetime interval and solving for t′ we have



t  t 1  v 2 

1/ 2

Now this expression only makes sense if v < 1, otherwise we have imaginary time, t′ ! 

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

Note – the rocket can travel as close to the speed of light as we wish but can’t exceed it.

The time difference is know as Time Dilation.

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Time Dilation – another proof 

Consider a rocket moving with a velocity, v

d′

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Mirrors

v

Inside the rocket is a “clock” which consists of two mirrors separated by a distance d′ with a light beam bouncing back and forth. Every time the photon hits a mirror, we get a “tick” of the clock. Spacetime Invariance

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... Time Dilation 

To a person sitting on the rocket, the time between ticks is:

t′ = d′/c

d′

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To be clear we are leaving time in units of seconds and distance in meters.

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What does a person outside the rocket (in the lab frame) see?

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To the person in the lab frame, the distance between the mirrors will have changed when a tick occurs.

d

d′

vt 

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If t is the time between ticks as seen by the person in the lab frame, then the mirror will have moved a distance vt.


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... Time Dilation Then d

d 2  d  2  v 2t 2

d′

Or vt

c 2 t 2  c 2 t  2  v 2t 2

d = ct and d = ct

So that



t  t 1  v 2 / c2



1/ 2

Or

t

t 1 v / c 2

Since v < c, the denominator is always less than one and t > t′. Thus, the time between tickets is always longer in the lab frame than the rocket frame.

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So we have: t

t 1 v2 / c2

t = time measured by a person in the lab t = time measured by a person on the rocket 

Again, since (1-v2/c2) < 1, the time between ticks is greater for the person in the lab frame.

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The “clock” on the rocket appears to run slower to the person in the lab frame!!! 

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In fact, it does run slower – decaying particles take longer to decay when they move a high speeds relative to the lab.


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The Tricky Part? 

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On the rocket, a “Timex” must keep the same time as the photon clock. Relativity says that the person on the rocket can’t tell she is moving.  the Timex must slow down too. The same happens for a Rolex, a CD player, her heart beat (!), etc. 

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... Time Dilation

If this didn’t happen then physics would depend on your motion.

Everything slows down!!! Spacetime Invariance

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Worked Examples 

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Problem SP 3-9


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Problem: SP 3-9 

distance moved by sun in one meter of lighttravel time

Aberration of starlight 

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

Suppose a star lies in a direction that is perpendicular to the Earth’s motion around the Sun. Because of the Earth’s motion, the light from the star comes from a slightly different direction than if would to an observer at rest with respect to the Sun.

distance moved by photon in one meter of light-travel time

ct

Sun Frame

This effect is known as aberration

a) Find a trigonometric expression for the aberration angle  shown in the diagram below

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Earth Frame

In view of the Sun frame, Earth moves to the right with a speed vEarth.

 Like running though rain drops 

ct

sin   vEarth

Where the velocity is relative to the speed of light.

  vEarth

Since  is very small, sin  ~ . And  is in radians.


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Problem: SP 3-9 (cont’d) 

b) Evaluate given that vEarth conv = 30 km/sec, giving the answer in radians and arcseconds.  

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For the small angles, sin ~ and tan  Difference so small so that it is not measureable. => Can’t use this as a discriminator between the two theories

Note – This must be an absolute measurement since all sources in the sky are affected in the same way. 

30 km/ sec  10  4 radians 300,000 km/ sec

= 10-4 radians = 20.6 arcseconds

c) The non-relativistic answer gives vEarth = tan . Does this make a difference? 

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vEarth 

Compare observations six months apart when the effect has the opposite sign Must “remember” previous position


vt

vt



 ct

ct

Relativistic

sin    

Non-relativistic

3

  3! 3  tan       3 20

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Problem: SP 3-9 (cont’d) d) Suppose the Earth were moving at v = 0.5 (a larger fraction of the speed of light). How would parts a) and b) change? 

 distance moved by photon in one meter of light-travel time

ct ct

Now we can’t use small angle approximation, so must use

sin   vEarth 

distance moved by sun in one meter of lighttravel time

Sun Frame

So that = 0.52 radians = 30 degrees

Earth Frame

We have that  is now much larger than before.

Reminder 2 radians = 360 degrees => 1 radian = 206265 arcseconds

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