The relativistic velocity addition formula is

$$u = \frac{v+u'}{1+ \frac{vu'}{c^2}}$$


$u$ = velocity of projectile seen by rest observer "A"

$v$ = velocity of moving observer "B" as seen by rest observer "A"

$u'$ = velocity of projectile seen by B

Now the question is this: If $v=c,$ and $u'=-c $

I get an undefined answer. i.e. the relavistic velocity addition formula is undefined.

Whats wrong with setting $v=c,$ and $u'=-c $?

  • 2
    $\begingroup$ possible duplicate of Speed of light travel $\endgroup$
    – Kyle Kanos
    Aug 26, 2014 at 15:34
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    $\begingroup$ An observer can not travel at speed $c$ with respect to anyone. $\endgroup$
    – WillO
    Aug 26, 2014 at 15:38

2 Answers 2


It is important to think about what the velocity addition formula is meant to do. The velocity addition formula should be applied in situations of this type:

When you (person $A$) see someone (person $B$) speeding by you with velocity $v$, and when this second person sees another person (person $C$, one can also substitute objects for people, of course) pass with velocity $u'$, then what is the velocity $u$ at which person $A$ sees person $C$ come by?

As @WillO already pointed out in the comments, if person $B$ is speeding by at $c$, then there is no reference frame in which he/she/it is stationary. Therefore, it is clear that the velocity addition formula should not be applied here. Note that this only rules out situations where both $|v|$ and $|u'|$ equal $c$, since the formula is symmetrical under interchange of the two.

  • $\begingroup$ If I set v = u' = c, I get u = c. It is a definite answer. However since you mentioned that person B cannot be speeding by at c, despite getting a definite u = c, can I also say that the calculation of setting v = u ' = c is wrong? $\endgroup$
    – Candy Man
    Aug 26, 2014 at 16:28
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    $\begingroup$ I would say that it's just not meaningful to apply the velocity addition formula in that situation. $\endgroup$
    – Danu
    Aug 26, 2014 at 16:30
  • $\begingroup$ Alright I see what you mean! $\endgroup$
    – Candy Man
    Aug 26, 2014 at 16:33

This isn't definitive, since the answer is always undefined, but let's be cutesy. Let's let $u' = a*c$ and $v = -a*c$, where $a < 1$


$$\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{ac-ac}{1- \frac{a^{2}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c-c}{-2a}\\ &= 0 \end{align}$$

Now, the reason why this isn't definitive is that you can take different limits, if you want. Say, let $u' = a^{2}c$ and $v = -ac$


$$\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{a^{2}c-ac}{1+ \frac{a^{3}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c\left(2a -1\right)}{3a^{2}}\\ &= \frac{c}{3} \end{align}$$

So, it's clear that, by taking the limit in different ways, you can get an arbitrary answer. It's not valid to choose an observer moving at the speed of light and then take velocities relative to that observer.

  • 1
    $\begingroup$ @lurscher, your edit messed up the MathJax again! $\endgroup$
    – Danu
    Aug 26, 2014 at 15:51
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    $\begingroup$ @lurscher, you broke my latex $\endgroup$ Aug 26, 2014 at 15:51
  • $\begingroup$ oh my. It was right before? It didn't render at all $\endgroup$
    – lurscher
    Aug 26, 2014 at 15:52
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    $\begingroup$ one before last line: \frac{c}{3} (too short edit) $\endgroup$
    – Void
    Aug 26, 2014 at 15:57
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    $\begingroup$ @lurscher: my first draft had broken latex, I fixed it, and then you edited. We must have been concurrently editing it. $\endgroup$ Aug 26, 2014 at 16:01

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