Physics 228 Today: (Special) Relativity: ContinuedMavis, tired of taking the train, acquires a hot...

Physics 228 Today: (Special) Relativity: ContinuedWebsite: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228

First exam - Thursday, Feb 28, 9:40 PMIf you know you will have a conflict - or a last-minute issue - and

cannot make the exam, please tell Prof. Podzorov ASAP.room assignments:

Hill 114: sections 06, 20, 21, 22 (Munson, Podzorov, Xu)ARC 103: sections 02, 03, 04, 05, 14, 15, 16 (Krueger, Tadepalli)

You are allowed 1 formula sheet (both sides may be used, any size font you want), pencils, and a calculator - no devices with networking capability.

Today's lecture is not covered on the exam.HW solutions are posted to the same web page as the lecture notes.

1st postulate (assumption): Physical laws are the same in every inertial reference frame.2nd postulate: speed of light is the same in every inertial reference frame

Monday, February 25, 2013

http://www.physics.rutgers.edu/ugrad/228

http://www.physics.rutgers.edu/ugrad/228

A ClockMavis builds a clock, using a light flash bouncing back and forth between mirrors on the train. If the distance between the mirrors is h, the light bounces from a mirror every h/c seconds.

h = ct

h

x=ut’

Stanley, in the train station, sees Mavis and the train move to the right, as the light bounces between the mirrors. Because the mirrors are moving at constant speed, Stanley sees the light bouncing off at an angle, and going a distance d = √(h2+x2) between the mirrors. With a longer distance, but the same speed c, the time between bounces must be larger. The clocks do not run at the same rate! Stanley sees the moving clock as running slower than his clock.

d=ct’

Algebra: time from one mirror to the other: t’/t = d/h = d/√(d2-x2) = 1/√(1-(x/d)2) = 1/√(1-(u/c)2) or t’ = t/√(1-(u/c)2)


The picture from the book


The β and γ VariablesNotation: in dealing with special relativity, the combination of speed / speed of light, u/c, is common, and is commonly replaced by the symbol β = u/c.

The time dilation factor of 1/√(1-(u/c)2) = 1/√(1-β2) is also common, and is commonly replaced by the symbol γ = 1/√(1-β2). Thus t'/t = γ.Stanley sees more time passing in Stanley's frame than that in Mavis' frame.Of course, Mavis sees the opposite.


iClicker 1 of 3

Whose clock runs faster? Why?Which of the following is true?1) Mavis’ clock is faster because she is in the train moving.2) Stanley’s clock is faster because he is not moving. Mavis is in the train moving.3) Stanley sees his clock as slower because he sees Mavis moving.4) Stanley sees Mavis’ clock as slower because he sees Mavis moving.5) Neither is faster, they are both the same, because there is no preferred inertial reference frame.


Time DilationObservers see moving clocks as being slower than their own stationary clocks. Moving Mavis sees events at a fixed position in her frame as separated by Δt0, while stationary Stanley sees them as separated by Δt = γΔt0.The standard example given is the “twin paradox”. Twin A stays on the earth. Twin B goes off on a really fast spaceship, with v = c/2, to a nearby star and comes back (for example) 40 years later. Twin A has aged 40 years. When twin B returns, the moving clock was slower by a factor of 1/√(1-(v/c)2) = 1/√(3/4) = 1.155, so twin B only aged about 35 years. Twin B is younger!How did we know B is younger? Wouldn’t B think A is younger?Time dilation is well established. While we have not sent people to nearby stars, we regularly measure the lifetimes of moving subatomic particles, and find they are longer by the factor of γ. Standard example: decay of cosmic ray muons. (I do this in my work with beams of muons and pions.)


Clocks in Space

To simplify the discussion, we are going to pretend that Mavis or Stanley see space as filled with clocks that tell the time an event happens at each point in space. Each has their own system of clocks, since if they are moving relative to each other the clocks at the same place read different times.Somehow the clocks in each frame are all aligned with each other, though that is an issue in itself.


Length ContractionIn special relativity, not only do we have the weird effect of time dilation, we have the addition weird effect of length contraction.


Length Contraction


Length Contraction

Mavis is at the center of a train moving at constant velocity. Stanley is at the center of the NB station platform. At the front end of the platform, a strobe light fires when the front of the train goes by. At the rear end of the platform, a strobe light fires when the rear of the train goes by.

The train goes by and Stanley sees the two strobe lights at the same time. He concludes that Ltrain = Lplatform.

Mavis knows the situation as well, and sees the front strobe before the rear strobe. She concludes that the platform is shorter than the train.

So in Mavis’ frame the platform is shorter, while in Stanley’s frame, the length's are the same. The moving object is relatively shorter.


Length Contraction

How much is a moving object shortened?

It is by the same factor that time is dilated:L’ = L/γ

“Proper length”: the length of an object in its rest frame


Length contraction from a ClockMavis turns her clock sideways. The light bounces between front and back of the train in a time t = 2d/c.d = ctStanley, in the train station, sees the light has to catch up with the front of the train, and then goes a shorter distance to return to the back of the train. Assume for the moment the clock length is d', not d.

Forward: d' + ut1' = ct1' ➮ t1' = d'/(c-u)Backward: d' - ut2' = ct2' ➮ t2' = d'/(c+u)Thus t' = t1' + t2' = d'/(c-u) + d'/(c+u) = d' [ (c+u) + (c-u) ] / (c2-u2) = 2cd' / (c2-u2) = [2d'/c] / [1-(u/c)2]So t'/t = [d'/d] / [1-(u/c)2] ➮ t'/t = γ2d'/d.

ct1' = d'+ ut1'

ct2' = d' - ut2'


Length contraction from a Clock

d = ct t'/t = γ2d'/d!

But we found with a vertical clock that t’/t = γ.

This must still be true: Mavis would see the clock running at the same rate in either orientation, so Stanley should see it running at the same rate too.

So γ2d'/d = γ ➮ d' = d/γ.

Thus, moving objects are shorter in the direction of motion by a factor of γ. Length contraction is nonintuitive, maybe even paradoxical, but no more so than time dilation. Both directly result from the speed of light being the same in the two frames.

ct1' = d'+ ut1'

ct2' = d' - ut2'


Length contraction from a Clock

d = ct t'/t = γ2d'/d!

But we found with a vertical clock that t’/t = γ.

This must still be true: Mavis would see the clock running at the same rate in either orientation, so Stanley should see it running at the same rate too.

So γ2d'/d = γ ➮ d' = d/γ.

Thus, moving objects are shorter in the direction of motion by a factor of γ. Length contraction is nonintuitive, maybe even paradoxical, but no more so than time dilation. Both directly result from the speed of light being the same in the two frames.

ct1' = d'+ ut1'

ct2' = d' - ut2'

Special relativity and moving fast: the physics secret to staying thinner and younger!


A Cute Special Relativity Paradox

Relativity leads to a lot of apparent paradoxes, that must be carefully analyzed to figure out what is happening. Here is one on length contraction:

Mavis, tired of taking the train, acquires a hot new sports car, the Ferrari 308vgtc. But she finds that she has a problem when she drives it home - the 308vgtc is 6 m long while her garage is only 5 m long.

Mavis, having studied special relativity, realizes that she can drive the 308vgtc faster than β = 0.306, γ = 1.2, so that it is shorter than the garage. She turns in to the driveway, accelerates hard, and as soon as the back end of the car gets into the garage she breaks hard as Stanley slams the door shut. The vgtc fits into the garage! Or does it?

Have some fun thinking about this or searching for the answer. I will try to get back to it next class.


A Cute Special Relativity ParadoxRelativity leads to a lot of apparent paradoxes, that must be carefully analyzed to figure out what is happening. Here is one on length contraction:

...

Mavis, having studied special relativity, realizes that she can drive the 308vgtc faster than β = 0.306, γ = 1.2, so that it is shorter than the garage. She turns in to the driveway, accelerates hard, and as soon as the back end of the car gets into the garage she breaks hard as Stanley slams the door shut. The vgtc fits into the garage! Or does it?

Or, by the way, does Mavis, once she speeds up, see the garage as shortened from 5 m to 5/1.2 = 4.17 m, so she cannot even hope to fit the 6-m long 308vgtc in the 4.17-m long garage?

Have some fun thinking about this or searching for the answer. I will try to get back to it next class.


Angle Rotation iclicker 2 of 3Mavis in the 308vgtc moving at speed u has a meter stick rotated 45o from the front of the car. What angle does Stanley on the sidewalk measure for the meter stick?1) 45o, both dimensions are shortened the same2) >45o

3) <45o

4) 45o, since only sticks aligned in the direction of motion are shortened.5) It depends on whether Mavis is driving towards or away from Stanley.

45ou


Transverse dimensions unchanged

While the train appears to Stanley shortened in the direction it is moving, it is the same size in the ``transverse’’ directions. When you look at the sideways dimensions of something passing you by, there is the same angle between the light rays reaching your eyes, independent of its speed.


Angle Rotationu

In Stanley’s frame this meter stick is shortened by a factor of γ.

In Stanley’s frame this meter stick is still 1 meter long.

In Stanley’s frame the ``longitudinal” component of this meter stick is shorter, but the “transverse” component is the same, so the angle θ increases.

θ


Distorted shapesMotion of objects at high speeds leads to distorted images of objects, because the light arriving at your eye at the same time is emitted at different time for different distances from you.

For the parallel array of rods, to the top left, at rest, the shape looking at the ends makes sense.

As the rods move faster to the right, you can see the length contraction effect from the red squares becoming thinner and closer together. Notice the vertical lines of bars are curved as well.

Also, the rods appear bent to the left, as you are seeing into their depth at earlier time, and thus when they were further to the left.

Sometimes we can see around objects to things we could not see if they were at rest.


Distorted shapes

The same pictures show you that you see a transverse / vertical stick would be distorted, from being linear to being curved...


iClicker 3 of 3A Star Trek photon torpedo explodes, emitting a spherical light wave, if measured in its rest frame. In your frame, the torpedo’s speed was 0.5c to the right when it exploded. Which picture looks like the “light front” in you frame? (The dark / light circles represent the explosion at earlier / later times.)

a)

b)

c)

d)

e)

The photons expand at the speed of light, with those coming out from

the initial explosion determining the ``light front''.

If we had matter expanding, we would have length contraction and the pattern could be like d, but depends on the speed of the

exploding fragments relative to c/2. Monday, February 25, 2013

Physics 228 Today: (Special) Relativity: ContinuedMavis, tired of taking the train, acquires a hot...

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