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I have just started special theory of relativity. The limiting speed known as speed of light fascinated me most. I asked my teacher:

Consider two massless objects moving in the same direction at the speed of light. What would be their relative velocity w.r.t each other?

My teacher tells me that their relative velocity would be zero. But then the speed of light is universal constant regardless of the motion of its frame of reference, so shouldn't their relative velocity be $c$? What is their relative velocity and how? Do objects moving at the speed of light obey law of addition of velocities?

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  • $\begingroup$ Can you draw a diagram showing the objects you're asking about and what their velocities are. $\endgroup$
    – John Rennie
    Commented Dec 23, 2013 at 14:22
  • $\begingroup$ I am new here,so i don't know enough latex to draw digrams;( $\endgroup$
    – Tom Lynd
    Commented Dec 23, 2013 at 14:25
  • $\begingroup$ The relative velocity is zero, but the relative velocity wrt to YOU is $c$. And you can always use MSPaint or semething like that. $\endgroup$
    – jinawee
    Commented Dec 23, 2013 at 14:36
  • $\begingroup$ @ jinawee,speed of light is said to be independent of the motion of frame of refrence,its always the same,so actually its never relative,am i right? $\endgroup$
    – Tom Lynd
    Commented Dec 23, 2013 at 14:39
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    $\begingroup$ Are you confusing relative speed and speed? $\endgroup$
    – jeremy
    Commented Dec 23, 2013 at 14:42

4 Answers 4

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But then the speed of light is universal constant regardless of the motion of its frame of reference, so shouldn't their relative velocity be $c$? What is their relative velocity and how?

EDIT Upon further review (special thanks to Alfred's comment), I think my original answer is incorrect. It turns out that the question of relative velocities of photons moving in the same direction is a meaningless question. The reason is as follows.

For two objects A and B moving as such,

                       v

               u      -------> A
            -------> B
            ------------------ (ground)

The velocity of B in A's frame is then $$ u'=\frac{u-v}{1-\frac{uv}{c^2}} $$ Notice the denominator? For $u=v=c$, this is zero and we get 0/0 which is an undefined operation, hence the meaningless question.

Do objects moving at the speed of light obey law of addition of velocities?

Not exactly. The Galilean velocity addition, $s=u+v$ does not hold for large-velocity objects. We use the "composition law", $$ s=\frac{u\pm v}{1\pm\frac{uv}{c^2}} $$ where $\pm$ depends on directions/frames. If $uv\ll c^2$, then this does reduce to the Galilean transformation.

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  • $\begingroup$ @ Kyle Kanos So speed of light is relative? Since speed of 1st particle w.r.t other becomes zero? $\endgroup$
    – Tom Lynd
    Commented Dec 23, 2013 at 14:48
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    $\begingroup$ "It is just that, in their co-moving reference frame," There are no reference frames with speed c. $\endgroup$
    – Alfred Centauri
    Commented Dec 23, 2013 at 14:48
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    $\begingroup$ Can someone explain the downvote? $\endgroup$
    – Kyle Kanos
    Commented Dec 23, 2013 at 18:58
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You state "two massless objects moving in the same direction at the speed of light".

This implies the two objects have the same velocity, so the relative velocity is zero.

If two massless objects travel in opposite directions, both at the speed of light relative to a stationary observer, then the relative velocity of the two objects is still the speed of light.

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The answers can be :

  1. You said that they travel in parallel then you know that the relative speed between them is 0.
  2. But suppose that one, or both, intends to measure the relative speed. They send a light beam and wait forever for the returning reflected beam. Obviously they must conclude that the other is at an infinite distance apart. This is the way we measure distances: with light.

IMO both answers are correct.
note: although the speed of light is constant and measured $c$ by any observer it does not invalidate (it is easy to explain) that the speed of light is constant irt to the background, the medium.

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  • $\begingroup$ No time passes in a luminal frame (proper time along the path is zero between any two events) they can't "decide" to do anything and they can't measure a delay. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 26, 2014 at 4:26
  • $\begingroup$ @dmckee possibly it is a question of language, or perspective: In the lab we can send a short pulsed light beam (an object it is), and then split it in two (a decision). If the IOR is the same on both paths then it is impossible to rejoin them. You can time the all operation. I suspect that you know that I know that light is always making a hard decision: What is the shortest path? Take it easy tricky light. 'proper time...is zero..' can imply that proper time is not ontological. I will have to think at GR speed: 43/5571=2*pi*369.2/300000 (in km/s). ;) I feel lucky today starting sexygenary ;) $\endgroup$
    – Helder Velez
    Commented Feb 27, 2014 at 1:29
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The second postulate of special relativity is:

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.(Wikipedia)

Your question is not regarding the second postulate because no inertial observer can observe the constellation of your thought experiment. Thus there is no contradiction.

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