# Weird case for relative velocities and special relativity

This has bothered me for a while. I've been taught that the formula for determining the perceived velocity on an object from another moving object's reference frame under special relativity is:

$$v = \frac{w - u}{1 - wu/c^2}.$$

If two objects travel from earth at $c$ in exactly opposite directions, what is the perceived velocity of one from the other? The formula seems to give a divide by zero error.

Don't forget to use signs. In your example, $w$ is, say, $c$, then $u=-c$. Hence $$v = \frac{w - u}{1 - wu/c^2} = \frac{2c}{2} = c$$

as expected.

• Damn, I looked at the question for a couple of minutes without seeing the sign difference in the velocities :-) – John Rennie Mar 17 '14 at 18:27

If two objects travel from earth at c in exactly opposite directions, what is the perceived velocity of one from the other?

The relativistic velocity addition formula cannot be applied in this case. As you write, if an object has speed $w$ in one frame and you want to know the object's speed in another frame moving with relative speed u, then you would use the relativistic velocity addition formula.

But, there is no reference frame moving with speed c relative to another reference frame.

If there are two objects moving away from Earth with speed $c$ in opposite directions, then each object has speed $c$ in all reference frames.

That is to say, there is no reference frame in which either object is at rest.

Let's look at your question again:

what is the perceived velocity of one from the other?

More precisely, your question is "what is the speed of one of the objects in a reference frame in which the other is at rest".

But there is no reference frame in which either object is at rest.

So, the question is ill posed; it presumes the impossible - that there is a frame of reference in which either object is at rest.