Relativity Paradox: What happens to a beam of light that is perpendicular to an observer?

Question

Say that, from my frame of reference, persons A and B are moving perpendicular to the line between them, at the same velocity. Now suppose A shines a light in the direction of B. Since I see A and B moving with the same velocity, they must see each other standing still. Hence, according to A and B, B will see A's light. However, according to me, B is moving, so that by the time the beam reaches where B was when A shone the light, B will have moved and not been able to see it.

How does one resolve this paradox?

Answer 1

No point in repeating (Ω)Mistress Bekki's answer. She nailed it, of course. But I don't think she answered your question. The answer has to do with the finite speed of light. The relativity effect is secondary.

If A intends to shine her light on B, she will see her light pointing at B, but you will see her light pointing off into space to the point where B will be. This is the paradox, and is at the crux of your question.

All of you see A shine a light at where B WILL BE; however, to both A and B, B is already there (no need to 'lead' B), but you see that B is not there yet.

(Seeing this is not as paradoxical as you might think. Imagine two cars coming toward you and the passengers tossing a ball back and forth. To them they are passing it straight to the other car, but to you, they are tossing it to a point where the car will be. But the actual answer in this case, because it involves light, is much cooler, so read on.)

The solution is that, as you are approaching A and B, not only will they appear to get thinner in the perpendicular direction (relativity), but they will look like they have twisted towards you. This twisting is because the light from the edge farthest away from you (in a perpendicular direction) takes longer to get to you, so the light from the farther edge will have come from a 'time' when A was farther from you then the light from the near edge of A.

Instead of being perpendicular to you, A's flashlight will appear to have turned slightly towards you. And you will 'see' it as pointing towards where B will be. (Well not quite pointing at where B will be, but at least initially, A's flashlight will appear directly behind it).

If you ever watch the spaceships go into 'warp drive' on star trek or star wars, you will see how the stars all stretch out as you pass them, this the same effect.

But this is NOT a relativity effect (but relativity will have some effect on how you see it). It has to do with the finite speed of light. If you could see sound waves and were moving towards the source, you would see something similar. (This effect is akin to the 'doppler effect' which occurs with both sound and light)

What relativity does is keep this derivation, as compared with sound waves, point of reference neutral. The speed of light is perceived as the same in all frames of reference.

Let me try to draw a diagram, first with you, A and B are not moving.

A|
. . . . . . . . . ---- You
B|

Now, lets give you a little speed and you will see:

A\
. . . . . . . . . <---- You

B/

Once you pass A and B, they will look like they are pointing in the other direction.

. . . . . . . . . /A
<---you
. . . . . . . . . \B


The bottom line is that A and B will perceive their beam of light as perpendicular to your velocity, but you will not. You will see it as pointing slightly towards you and hitting B at his new position. In theory this is a paradox, but in practice, neither you, nor A, nor B would see anything out of the ordinary.

I'd like to add that at relativistic speeds, you will see some really strange visual distortions. If a string were drawn between A & B, the center, would appear to be "bent" closer to you then A & B, and you could see the light beam follow the string even though the string is bent. But you would also see that the light beam was going straight.

Answer 2

This is only a paradox if you assume that space and time are separately fixed coordinates that are the same to all observers.

The resolution is to postulate that the speed of light IS constant and allow lengths in moving reference frames to contract and times to dilate.

Because A and B are moving toward you perpendicularly, you agree with them on the distance between them. Call it d. Everyone also agrees on the relative velocity, v. They (A and B) say it takes a time t1 = d/c for light to travel between them. But your time, t2, is different. You can figure out how different. During that time t2, A and B (travelling at speed v) cover a distance t2 * v. So according to Pythagorus, the light travels a distance sqrt (d^2 + t2 * v). The time that takes is t2 = sqrt (d^2 + t2*v)/c. Solve that for t2:
t2^2 = t1^2 + (t2 v / c)^2
t2 = t1/sqrt (1 - (v/c)^2)
So the time you ovserve for the light to pass is longer than the time observed by A and B.

Answer 3

That is one of the proofs that time slows down the faster you travel.