# Light, Elevators and Gravity

1. May 12, 2007

### gonegahgah

Hi Pervect

I have done up a couple of diagrams which I was hoping to get your input on.

I have looked into the example that is given where light is sent through an elevator that is accelerating. It describes a curved path and this is used to depict the path of light under gravity due to the principle of equivalence.

I have done up some diagrams of this however mine show light as I believe it should be shown; as an oscillation through a point in space; and not as a particle as is shown in the texts.

The results are interesting. Please have a look and let me know if I can say what I think the results show.

One of the interesting results is that mine depicts the red shifting that occurs under gravity whereas theirs don't.

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2. May 12, 2007

### HallsofIvy

Are you under the impression that say that "light is a wave" means that the light beam itself is moving up and down? That's certainly what you have here implies!

3. May 12, 2007

### pervect

Staff Emeritus
Hmm, well, a falling light beam should blue-shift, not redshift.

I think that what you drew would represent a vibrating particle falling in a Newtonian fashion, at least if I'm interpreting the diagram correctly. What I think you want is to interpret the sine wave as the path traced out by the tip of the E-field vector, i.e

->
---->
------>
---->
->

At least that's the only way that drawing light as a sine wave has a physical interpretation that I can think of - by drawing a line connecting the tips of the E-field for a planar wave.

I've changed the orientation due to the limitations of ascii art. If you trace out the tip of the E-field vectors, they trace out a sinusoidal path, but there is nothing physical actually vibrating.

So to model light falling, you need to transform the E and B fields and not imagine a vibrating particle. As I recall, a component fo either field moving in the parallel direction will not change, while a component in the transverse direction will get multipled by gamma.

What this means to your diagram is that if you replace the dots by E-field vectors, the path of the light will change, but you won't see any graphical depiction of the blueshift. The blueshift effect will be a lot more subtle, it will be related to the B-field vectors that you don't even see in this diagram (they are pointing out of the paper), and their relationship to the E-field vectors.

4. May 12, 2007

### gonegahgah

Hi Pervect

Sorry, I didn't mean for it to depict the light particle itself moving up and down. The dots I have shown are just meant to be points on the light itself at equal time slices from each previous one.

As you say it is meant to depict the em as it passes through a point.

I think you are probably right; it probably should be blue-shift.

I've attached another diagram. You will be pleased to see that even as I have depicted it that the two different frequencies would be following the same path.

Personally I think there is more to it but I will have to think about that.

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5. May 13, 2007

### pervect

Staff Emeritus
OK, I wasn't quite sure what the squiggles represented. If your point is just that light appears to fall in a parabolic path onboard an accelerating spaceship (as long as the fall time is short), I agree.

To re-use an old illustration, if you have a horizontally shooting laser gun and an apple, and the apple and the laser gun shoot "at the same time", (as defined by a lightsource midway between them at the same altitude), the apple and the laser beam will hit the floor at the same time.

Note that you have to be a bit careful about time, if you use the coordinate system associated with an accelerated observer not all clocks will tick at the same rate.

6. May 14, 2007

### gonegahgah

"the apple and the laser beam will hit the floor at the same time."

Cool, I never thought of that. Makes sense. The only problem would be the curvature of the Earth which in a sense, I think, means that when you fire sideways you are always firing up. Is that correct?

One of the problems with my diagram of the em oscillations in the elevator is that the diagram looks like a wave whereas it is really mean to depict a graph of the em oscillations through the centre of the elevator over time due to a light wave passing throught that central line. However it shows an acceptable result.

I will probably have to acede that light bends the same amount regardless of frequency through a gravitational field and is therefore achromatic as science states.

I was thinking that the only thing that distinguishes the accelerating elevator in space from an elevator stationary on Earth is that the blue-shifting on the Earth would be received as a constant shift amount; but in the accelerated elevator the blue shifting would be small at first and keep increasing the amount of shift over time. Is that correct? I'm not saying that it changes the results but is it a difference none-the-less?

Also I've drawn up a diagram. First I would like to get your opinion on what the problems would be for the experiment.

It shows a carriage that is travelling passed a platform. The carriage is as long as the platform. At both ends of the carriage are contacts and at both ends of the platform are contacts. As the carriage passes the platform the contacts briefly complete circuits. When the circuits at the right contact they cause a light at the right of the carriage and a light at the right of the platfrom to both briefly illuminate. The same goes for the left contacts except that they illuminate briefly the lights on the left of the carriage and the platform.

So what are the limitations that will make this experiment useless?

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7. May 14, 2007

### MeJennifer

Parabolic?

8. May 14, 2007

### pervect

Staff Emeritus
For short times, yes, the path of an accelerating object is given by the Newtonian approximation, i.e. it appears parabolic, s=(1/2) a t^2.

I don't think the original poster is necessarily ready for or interested in the equations for the longer term motion of a relativistically accelrating object, if he is, there is at least one other thread on that active right now. The differences between relativistic constant proper acceleration and Newtonian acceleration at large t are not vital to the problem statement, at least not if I am understanding the question.

9. May 14, 2007

### pervect

Staff Emeritus
One thing that you have to beware of is that "at the same time" depends on the observer.

So, when I say that the light beam and the apple hit the floor at the same time, what I really mean is this:

You have a radio transmitter, exactly halfway between the apple and the laser gun. This radio transmitter emits a signal, and when the signal is recived the apple is released, and the laser gun is fired.

You have a clock on the floor which starts ticking when that the apple lands, and another clock on the floor which starts ticking when the laser beam lands. Midway between the two marks on the floor you have another radio transmitter, which sends out another signal on a different frequency, which causes both clocks to stop. You find that the amount of time shown on both clocks is the same.

This is takes quite a bit of space, so that's why I wrote instead that the apple and the laser hit the floor "at the same time".

As far as your other experiment goes, you might want to look at Einstein's "train and platform experiment", for instance in

http://en.wikipedia.org/wiki/Relativity_of_simultaneity

which has some links to Einstein's original text, i.e. http://www.bartleby.com/173/8.html

10. May 16, 2007

### gonegahgah

Hi Pervect

I'm currently reading a book called Einstein's Clocks, Poincare's Maps by Peter Galison which has been offering an interesting insight into the process whereby these principles started to arise.

So I accept the difference between your quick explanation and your longer explanation.

I had a read of Einstein's paper. I will read it further but I understand what he was saying about the passenger in the middle travelling towards the one light source and moving away from the other light source causing the occurance to not appear to be simultaneous to that passenger. I agree with that. I may even be able to show you the example I wanted to show you originally afterall. To me it doesn't contradict; but I will seek your opinion then.

I've drawn up a diagram I think of that train and platform experiment you were refering me to. My illustration is attached. Is it correct?

Correct me if I am wrong but basically it says that you can have two trains pass each other on a single track even if they only have a small section of parallel track that is shorter in length than the length of either train at rest. That this occurs because the trains see the other train as contracted in length when travelling at opposite speeds so that they never collide even though they would not both fit on the parallel tracks together at rest. Is that correct?

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11. May 16, 2007

### pervect

Staff Emeritus
This is a modification of the "barn in the pole" paradox - and yes, I believe you are correct.

In the lab frame, both trains will appear contracted, so they'll both fit on the double track "at the same time".

Suppose that the velocity of both trains is .866 in the lab frame for definiteness.

Then in the lab frame, the trains will appear half their usual length. This is the easiest place to work the problem, and we can see that we expect the trains to be able to pass.

The train frame is trickier. The length of the double-track crossing appears to be halved.

The train we are riding on appears to be its normal length.

But the train coming in the opposite direction is approaching at a higher velocity : 2v/sqrt(1+v^2/c^2) by the velocity addition formula.

I make this out to be .9897c in our example, and the Lorentz contraction factor is 7.

So while the crossing is only 1/2 as long as our train, the oncoming train is 1/7th as long.

It should be able to fit in its entirety on the shorter track, and more importantly it has (or should have) enough rome to move on this track while our train pulls through - our train does not fit on the crossing.

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12. May 16, 2007

### gonegahgah

Hi Pervect

That is one experiment I would like to see in action (though I must admit that myself I believe it would be fun to watch for all the wrong reasons; but that's just me).

I must apologise because I reread the wiki page afterwards and realised you were meaning to refer me to the first thought experiment "The train and platform thought experiment" and not that which I thought.

Instead it refers to a train with an observer on the train in the middle and a light also in the middle and an observer on the platform. The light does a flash as the two observers pass each other. I've done a diagram for this one also below. The experiment is about when they see the flashes of that light hitting the front and back of the train. The observer on the train will of course see the flashes as occuring at the same time; but the observer on the platform would see the flashes as occuring at different times.

I think they need to improve it a little as it uses the following confusing wording: "For the platform observer, light moved more slowly toward the back than towards the front...". This gives the impression that they are saying that light is moving slower or faster than itself when they usually state that the speed of light is constant. I know that they don't mean that but would you agree they are using confusing wordage or is it just me?

It is also confusing because with respect to the observer on the platform they say both "the observer on the platform sees the back of the train moving toward the point at which the flash was given off, and the front of the train moving away from it. This means that the light flash going toward the back of the train will have less distance to cover than the light flash going to the front." and "light moved more slowly toward the back than towards the front". Shouldn't it get to the back faster; not slower (according to the idea) as the first quote says but the second seems to oppose?

Also "... the platform observer sees light cover a greater distance to the front of the train, in less time but at the same speed observed by the train observer." If I were to cover twice the distance in half the time I must be going 4 times the original speed; not the same speed? So how do they cover more distance in less time and still get the same speed?

Does this need to be written more logically/correctly or is it just dumb me?

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