# Properties of light in 2 moving reference frames

1. Aug 3, 2015

### pixel

In the usual derivation of time dilation in special relativity, we have two frames of reference, A and B, moving relative to each other with velocity v. In A's frame, a light beam is directed vertically upward toward a mirror and reflected vertically back down. In B's frame, the light follows a triangular path due to the relative motion.

If we were instead talking about a ball being thrown up and down vertically in A's frame, B could explain the triangular path by saying that the ball initially had velocity v before it was thrown, there are no horizontal forces acting and therefore due to its inertia it continues to have the velocity v hence a triangular path according to B. (We're assuming we're out in deep space; on earth the ball would follow a parabolic path).

So returning to the case of the light beam, how does B explain the triangular path of the light from a physical point of view given that light doesn't have inertia. It's obvious that a vertical path in from A will be a triangular path in frame B, but what, according to B, is causing it to continue to keep pace with A's frame once it is emitted?

2. Aug 3, 2015

### Staff: Mentor

Light has inertia.
For both A and B, the ball is simply following a straight line, getting deflected according to the usual kinematic laws at the "mirror". I don't see the problem.

3. Aug 3, 2015

### Staff: Mentor

You would use Maxwell's equations if you wanted to get a really complete explanation. In the appropriate limit, Maxwell's equations reduce to ray optics, which is essentially what is typically used in these types of descriptions. But you could always go back to Maxwell's equations and solve it (if you have the patience).

4. Aug 3, 2015

### pixel

The "problem" is that the light source is pointing vertically upward in A's and B's frames, yet once the light is emitted, its motion has a horizontal component in B's frame. I am not doubting that this happens; obviously it must. I am simply asking how B explains this horizontal component. In the case of the ball, that's easy - the ball has an initial horizontal velocity in B's frame, it has inertia and so maintains that horizontal velocity.

You're saying light has inertia. In what sense?

5. Aug 3, 2015

### Staff: Mentor

Using the same laws by which A explains the behavior of the light. QED, Maxwells equations, or one of their simplifications.

6. Aug 3, 2015

### pixel

If my question is so obvious, then it should be capable to give a simple, physical reason for the light's travel in B's frame, analogous to the explanation for the ball, without just saying to solve Maxwell's equations.

7. Aug 3, 2015

### Staff: Mentor

You're thinking of the light source as a little tube pointing up, like the barrel of a laser cannon in first Star Wars movie.

That's not a necessary assumption, and this problem is much easier to understand if you instead think of the source as a horizontal plane like a military phased-array radar antenna (google for this! - it's a fascinating technology and an excellent example of what DaleSpam was saying about how we can calculate the angle at which the beam leaves the source using the laws of classical physics). The direction that the beam travels from such a source depends on when different parts of the antenna are activated - if all of them are activated at the same time the beam will leave vertically, but if they are activated progressively from one side to the other the beam will leave at an angle. Now, when you consider that things that happen "at the same time" in one frame happen at different times in other frames, you'll see how the angle must be frame-dependent.

However, if you do want to think in terms of a little tube pointing up like those laser cannons.... You can. But now you have to ask yourself why you are so sure that if the tube is vertical in one frame it will be vertical in other frames. It turns out that it is not. You can try https://en.wikipedia.org/wiki/Ladder_paradox#Bar_and_ring_paradox for an explanation, or you can consider exactly what it means to say that the tube is vertical: You're saying something about how the various points along the tube line up with the base of the tube all at the same time. But because of the relativity of simultaneity, if they're lined up that way at the same time in one frame, they won't be lined up that way at the same time in another frame.

Last edited: Aug 3, 2015
8. Aug 3, 2015

9. Aug 3, 2015

### Staff: Mentor

I am not trying to be dismissive of your question. I am answering as directly as possible.

Maxwell's equations are the "physical reason" why the light propagates in A's frame to begin with. Those same physical laws govern the behavior in B's frame. The reason it goes diagonally in B's frame is the same reason it goes straight in A's.

10. Aug 4, 2015

### pixel

The basic equations of physics are the reason why everything happens. I can explain the path of the ball in B's frame without saying Lagrange's Equations are the reason it happens. I was looking for a similar explanation for the path of light in B's frame that would illuminate (no pun intended) the physical ideas.

11. Aug 4, 2015

### pixel

Thank you. This is more like what I was looking for. I know that dimensions perpendicular to the relative velocity are measured to be the same in both frames, but didn't consider that there could be a tilt.

12. Aug 4, 2015

### bahamagreen

This is a natural question after reading the original phrasing that the "velocity of light is independent of the source motion". That makes one wonder why the light's path doesn't simply "stay behind", travel vertically, and miss the mirror since velocity has a direction component and in the case of "independent" light it should not be imparted by the motion of a frame. Countless people have been confused by this until discovering that the word "velocity" was a poor choice of English translation of the German word used for speed.

13. Aug 4, 2015

### Staff: Mentor

Both velocity and speed are "Geschwindigkeit" in German (but you can specify "Geschwindigkeitsvektor" to mean velocity vector and "Geschwindigkeitsbetrag" for the magnitude).

You can use a light source that emits light in all directions, and then just look at that part that will hit the mirror. On the other hand, the light source is moving, so it should not be surprising that light can be emitted in a different direction.

14. Aug 4, 2015

### Staff: Mentor

Fair enough.

I don't have a simpler reason in general, but because the light follows Maxwell's equations and because Maxwell's equations are Lorentz invariant you actually can understand the general result easier than analyzing most specific scenarios.

You did not identify a specific mechanism for your light pulse, but the only specific mechanism that I can think of that is easier to understand than just looking at Maxwell's equations would be an incoherent spherical point source, a plate with a hole that is a few times larger than the wavelength of the source/detector, and a mirror of the same size, such that the source/detector, the hole, and the mirror are all lined up. With that setup it should be clear that the only light that passes through the hole in B's frame is the light which was emitted from the point source moving at an angle.

15. Aug 4, 2015

### pixel

I haven't looked at your link yet, but after being initially happy with your explanation some doubts are starting to creep in. For frames moving along the x-direction, events separated in x that are simultaneous in one frame will not be so in the other frame. This follows from the first Lorentz equation t' = gamma (t + vx/c^2). So with the tube in the vertical direction, if the points along the tube are lined up with the base at the same time in A's frame, they will be lined up that way at the same time in B's frame. No?

16. Aug 4, 2015

### pixel

So what if A is using a highly collimated narrow source that is only emitting in one direction (up), say a source with several separated pinholes that guarantee only one direction for the light? He can still bounce the light off a mirror directly above. B will still see the light moving in a triangular path.

17. Aug 4, 2015

### Staff: Mentor

Uhh, that is what I described already. As long as the pinholes are several wavelengths wide (to avoid diffraction) and the last one is many wavelengths from the source the additional pinholes in between don't do much.

18. Aug 5, 2015

### A.T.

The same applies to photons, which have impulse just like the ball

19. Aug 5, 2015

### harrylin

It's not really a tilt, and qualitatively it isn't even specific to SR. If a vertical tube is moving laterally while a light ray propagates inside the tube, the ray cannot go straight up. It's that simple!

20. Aug 5, 2015

### pixel

I am not questioning that the light has a triangular path in B's frame. I am simply looking for how B explains seeing a vertically oriented light source but a triangular path of light, using a high-school or freshman physics level physical explanation.

21. Aug 5, 2015

### pixel

Impulse is integral of force over time. Not sure how that applies here as there is no horizontal force on the photons once they are emitted. Unless you're saying that because the impulse is zero the momentum change is zero.

The atoms emitting the light all have, in B's frame, an initial horizontal velocity. The emitted photons must continue to have that horizontal component, as does the ball. So it's the momentum property of the light, as with the ball, that is key to B's explanation of the triangular path.

If that's the case, then maybe we can put this discussion to rest.

22. Aug 5, 2015

### Janus

Staff Emeritus
Try thinking about it in term of B moving and A standing still. Then what would you expect B to see?

23. Aug 5, 2015

### pixel

I was not questioning what B sees.

24. Aug 5, 2015

### Staff: Mentor

The high-school physics explanation is that there is no such thing as a vertically oriented source of a light beam - those Star Wars laser cannons are science fiction. A real beam source will have a finite width and if all points across that width emit radiation simultaneously, the beam will go straight up, but if you stagger the time of emission across the width of the source the beam will go sideways. Relativity of simultaneity will taje it from there.

It may be easier to analyze the problem using a point source of light which will emit a spherical wavefront instead of a beam.

25. Aug 5, 2015

### pixel

I stated what I think is the explanation I was looking for (above) in terms of the momentum of photons, but I will comment on your last post. Of course there is no such thing as a perfectly collimated beam of light. But let's do this experiment - in all of the photons that are emitted in the various directions, let's just follow the path of the one that happens to go straight up vertically. Then this photon will reflect vertically down in A's frame and will move in a triangular path in B's frame. So the fact that there is a spread in a beam of light is not relevant to the discussion of why any photon that happens to move vertically up will follow a triangular path in B's frame.

Since I started this thread I would suggest that it be put to rest.