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I know that the velocity of light with respect to anything is constant ($c$) . What, then, is velocity of light with respect to light?

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When you say that an object $A$ has some velocity with respect to another object $B$, you are implying that there is a reference frame where $B$ is at rest, and $A$ moves with the aforementioned velocity in that reference frame. It makes no sense to talk about the movement of a photon with respect to another, because photons have no rest frame.

See: Why can't we make measurements in a photon's rest frame when loop diagrams make measurements possible? and Does a photon in vacuum have a rest frame?

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    $\begingroup$ Hritik Narayan: "When you say that an object $A$ has some velocity with respect to another object $B$, you are implying that there is a reference frame where B is at rest" -- I fully agree. This implies in particular that additional participants can be thought (or even actually be found) who were at rest wrt. $B$ and each other. "and $A$ moves with the aforementioned velocity in that reference frame." -- Or $A$ belonged to that frame, too. And, of course, $A$ might have variable instantaneous speeds. Or instantaneous speed $0$ wrt. $B$ and yet not be at rest wrt. $B$ ... $\endgroup$ – user12262 Sep 12 '15 at 6:45
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When you say that an object $A$ has some velocity with respect to another object $B$, you are implying that there is a reference frame where $B$ is at rest. This means that additional participants ($J$, $K$, ... $P$, $Q$ ...) can be thought of (or can even be actually identified) who were and remained at rest with repect to $B$, and with respect to each other. Then

  • either $A$ had passed some other members of the rest frame to which $B$ belongs, and eventually reached $B$ (and possibly even passed $B$). Let's say that $A$ had passed $K$, $J$, etc., but not $P$, $Q$, or others. For each passage we can evaluate the average speed of $A$'s motion wrt. $B$ and $K$, wrt. $B$ and $J$, and so on. Evaluating the limit of these speed values (if there is one) we obtain the instantaneous speed of $A$ at $B$'s indication of the passage of $A$, and we attribute a direction to $A$'s motion wrt. $B$'s rest frame (say "from $J$ to $B$").

  • Or: $A$ is a member of this rest frame, too; accordingly $A$ and $B$ were at rest to each other, and the value $\vec 0$ is evaluated for $A$'s velocity with respect to $B$.

But it makes no sense to attribute a value of velocity to one photon with respect to another, because photons have no rest frame; photons cannot be said to have been at rest with respect to each other. Arguably it doesn't even make sense to talk about the movement of a photon with respect to another, at all.

See for instance: "Are signal fronts in a beam not at rest to each other?" (PSE/q/104333).

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  • $\begingroup$ Can't understand about the participant particles. Please can you explain it with an easier analogy. $\endgroup$ – AScientist Apr 13 '17 at 14:45
  • $\begingroup$ @ASuckerinMaths: "Can't understand about the participant particles." -- In Einstein's writings we also find them called "observers", or "places", named A, B, C ..., M, M' ...; or "material points". $\endgroup$ – user12262 Apr 18 '17 at 22:38

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