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Isn't it impossible to estimate the velocity of framework through relativistic velocity addition formula when the event moves at speed of the light? $$u=\frac{v-v'}{1-vv'/c^2}$$

if $v=v'=c$ $u=\frac{0}{0}$ undefined,

Note that the $u$ denotes velocity of frame, $v$ denotes velocity of event and $v'$ denotes velocity of event from frame viewpoint.

(because when the event moves at speed of the light result of relativistic velocity addition formula is equal to zero devided by zero which is undefined and indeterminate)


marked as duplicate by John Rennie, Kyle Kanos, Ben Crowell, ACuriousMind, Qmechanic Sep 20 '14 at 19:16

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    $\begingroup$ That should be $u=(v+v')/(1+vv'/c^2)$. $\endgroup$ – lemon Sep 20 '14 at 10:29
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    $\begingroup$ possible duplicate of Speed of light travel $\endgroup$ – John Rennie Sep 20 '14 at 10:41
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    $\begingroup$ Not at all. The frame that moves at the speed of light is always the frame that moves at the speed of light. That's actually the main assumption of special relativity. The expression above isn't undefined, either. Use L'Hospital's rule. $\endgroup$ – CuriousOne Sep 20 '14 at 11:05
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    $\begingroup$ To see a acceptable limit, begin by choose one of the velocities (say $v$) to be $c$, then $u=c$ for all $v' < c$. Now, you may take the limit $v'=c$ $\endgroup$ – Trimok Sep 20 '14 at 11:44
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    $\begingroup$ @Achmed - in physics, when you get into nonsensical math, always ask yourself what is the physical situation you are attempting to describe. Here it seems to be: "while traveling on a train passing a station, if I send a photon in the direction of travel, what would its speed be as seen from the perspective of a photon sent in the same direction from the platform?" The correct answer is that photons lack perspective... $\endgroup$ – Johannes Sep 20 '14 at 15:00