Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm trying to understand the concepts of time dilation and spatial compression. I've been using the classic example of firing a photon of light inside a ship (spaceship, boxcar, etc.) moving at a constant velocity to the second observer. But when I do the Lorentz, I get different values depending upon the direction of travel of the photon relative to the ship.

My understanding is that time dilation for the two observers should be a constant, as should spatial compression, since their relative velocities are constant (at least for the duration of the experiment).

When I fire the photon in the direction of travel of the ship, the measurements all work out as expected. Time on the ship is passing slower. Then when I fire the photon backward in the direction of travel of the ship, all the values that should be constant have different values. So I assumed I'd done the transforms wrong. I redid them and still got different values. hen I simplified the equations by picking values that make the transforms very basic. Ship length = 1 light second for the ship observer. Ship speed = 0.1 light year for the second observer. Even then I get variances in the values that I expect to be constants.

My problem might be in measuring the distance the photon traveled for each observer. In firing forward, the second observer sees the photon travel the compressed length of the ship plus the distance the ship travels. The ship observer see it travel the length of the uncompressed ship. When firing backwards, the second observer sees the photon travel the compressed length of the ship minus the distance the ship travels. Since the ship observer sees the photon travel the same distance and amount of time regardless of direction, the second observer must also see it travel the same time in both directions. But the distance is shortened for the secondary observer.

So either the time dilation or the spatial compression would have to change for the equations to work. But my original premise was those are constants because the relative velocity is assigned to be constant for the experiment! Am I still doing the transforms wrong, or do I have a bad assumption in which values become constants by forcing relative velocity to be constant?

I've reviewed dozens of the time paradox explanations. But those arise from the two observers returning to one frame of reference. I've let my ship sail on beyond the end of the experiment indefinitely, purposely to avoid those problems from interfering in learning how to do the transforms correctly. Now I'm not sure I didn't step into a different paradox. Maybe this should be solved in GR instead?

share|improve this question
1  
Two different observers cannot see the same photon. Photon can be seen only by the person (measuring device) whose eye(s) it comes directly to. The other one sees nothing. That's simply how light can be seen - only directly. –  bright magus May 2 at 18:02

2 Answers 2

"Since the ship observer sees the photon travel the same distance and amount of time regardless of direction, the second observer must also see it travel the same time in both directions". This is a wrong assumption. It's not time alone but the spacetime interval which is the same for both. Try to compute it in all those experiments (between the events 'the photon is emited' to 'its absorbed'), and it will be the same for both observers, although different from exp 1 to 2, if your calculations are correct. Hint: the ground observers sees a longer trayectory in one direction.

Edit in response to the comment:

I'm sorry if that statement mislead you, but it is true in general, although in this particular case, because of the setup of the problem, the interval is the same in both experiments since it is obviously the same for one of the observers.

Having said that lets compute the invariant interval for two photons emitted simultaneously from the center of the train in both directions:

For the first (on the train) observer S: both forward and backward moving photons hit the wall at t = 1/2 so

$ (\Delta \tau_f)^2 = (\Delta \tau_b)^2 = (1/2)^2 - (1/2)^2 = 0$

The invariant interval is zero because light moves along lightlike trajectories.

From the ground observer S': The distance travelled is lenghtened/shortened by an amount $ \beta t $, not the same for each photon:

$ (\Delta \tau'_f)^2 =(t'_f)^2 - (1/2+\beta t'_f)^2 = (\Delta \tau_f)^2 = 0$ $ (\Delta \tau'_b)^2 =(t'_b)^2 - (1/2-\beta t'_b)^2 = (\Delta \tau_b)^2 = 0$

The distances here dont get corrected by the contraction factor because that applies to intervals, not coordinate points. If you solve each equation, you will find that $t'_b \neq t'_f $. This difference doesn't reflect the time dilation effect, but the break of simultaneity. For time dilation, compute the invariant interval for the moving observer S himself:

$ (\Delta \tau_S)^2 = t_S^2 = (\Delta \tau'_S)^2 = (t'_S)^2 - (\beta t'_S)^2 $

$ t_S = t'_S \sqrt{1-\beta^2} $, or $ t'_S = \gamma t_S$

you can see more on this at this wikipedia article

share|improve this answer
    
I can get the transforms to work in both directions. But different from Experiment 1 to 2 is exactly my problem. The direction of travel of the photon seems to change which observer's time runs slower. But without any changes other than the direction of travel of the photon, everything should retain the same computed values. If the ship maintains it's relative velocity, nothing else in the transforms changes. Isn't relative velocity the defining factor from which all the others are computed? If I don't change that, I can't change the rest? –  J-Shipman Apr 1 at 2:54
    
That's what I'd overlooked. Simultaneity. Thank you so much. –  J-Shipman Apr 2 at 11:15

The red (blue) shift of light from gravitational potential is obvious because it is proven by experiments. on the over side the constance of c is part of the concept of Einstein's special theory of relativity. It's the basis for all over considerations. But the gravitational potential doesn't play any role at this moment. Later in his general theory of relativity Einstein unit the space geometry with the gravitational potential and stated out that the light follows this space geometry. The Äther is dead, long live the Äther.

The point is that c is a local constant. It is smaller near big gravitational sources, far away from masses it is higher. The red or the blue shift - this depends from the direction of the experimental measurement in relation to the earth - is very small on the earth because the gravitational potential adds up for all sources and we don't feel this only because the sources well distributed and the gravitational force is the longest acting force.

Saying that the EM radiation travels with constant speed independent of the speed of the source in relation to the space (cosmos, universe) does imply that it travels with constant speed as a function of gravitational potential of the universe in the point he is traveling.

Outgoing from this point of view perhaps you can think again about your question.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.