If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is $$ v \oplus w = \frac{v+w}{1+vw} $$

As expected, $v \oplus 1 = 1$, as "nothing can go faster than light".
Similarly, $v \oplus -1 = -1$. (same thing in the other direction)

But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.

So what would the ground observer see? Is this even a meaningful question?

Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.

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    $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Aug 5 '14 at 20:30
  • $\begingroup$ Also of interest: physics.stackexchange.com/q/16018 and the many duplicate and related questions it the sidebar of that one. $\endgroup$ – dmckee --- ex-moderator kitten Aug 5 '14 at 20:50
  • $\begingroup$ I. J. Kennedy: "I'm asking what would happen physically." -- As far as the question concerns only geometric-kinematic relations (and not necessarily "actual emission" etc.) we might simply consider two signal fronts propagating in opposite directions. Of course, the fact (not to say problem) remains that there's no "(inertial) rest frame of a signal front", and hence no "speed with respect to a signal front" defined and to be evaluated. Also related: "Are signal fronts in a beam not at rest to each other?" (PSE/q/104333). $\endgroup$ – user12262 Aug 5 '14 at 22:39
  • $\begingroup$ take a look at my answer for an attempt to provide physical (and mathematical) meaning (comments welcome) $\endgroup$ – Nikos M. Aug 13 '14 at 18:23
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    $\begingroup$ Related: physics.stackexchange.com/q/91149/2451 $\endgroup$ – Qmechanic Aug 26 '14 at 16:00

Physically, SR can't accomodate observers moving at $c$.

Mathematically, the limit is undefined. You have a function $f(v,w)$ defined in the $v-w$ plane, which has a square boundary. You're talking about approaching a corner of the square, but the value of the limit would depend on what path you chose to approach the corner along.

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  • $\begingroup$ Thanks for your answer. Notice I did mention in my question the limit not being well-defined because of dependence on the path. $\endgroup$ – I. J. Kennedy Aug 5 '14 at 20:12

As already noted, Special Relativity cannot account for an observer moving at the speed of light.

It is also instructive to calculate the proper time for the object ${\bf A}$: as you approach the speed of light the proper time becomes zero. So it is even impossible to define in ${\bf A}$'s frame of reference the moment at which ${\bf B}$ will be emitted.

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I'm getting the sense that my fears were correct, it's physically a nonsensical situation.

Applying a formula outside of the context in which it was derived will likely produce nonsensical results.

In the derivation of the relativistic velocity addition formula, and using your notation, there is an object B with uniform velocity $w$ in some inertial reference frame (IFR) A.

Further, A has velocity $v$ in the lab frame of reference.

The velocity of B in the lab frame is then given by

$$\frac{v + w}{1 + \frac{vw}{c^2}}$$

Now, when you stipulate that $v = c$, you're attempting to apply this result outside of the context in which it was derived. In particular, in SR, there are no IFRs with relative speed c.

Put another way, an object with speed c in one IFR has speed c in all IFRs; an object with speed c has no rest frame.

Since, by stipulation, A is an IFR in which B has speed $w$, it must be the case that $v < c$.

Less precisely, only one of the velocities in the relativistic velocity addition formula can validly be set to c - both cannot be c since that would presume an IFR with relative speed c exists.

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Recalling that the relativistic velocity addition formula in 1+1 dimensions in terms of rapidity


is just ordinary addition of rapidities

$$\beta_1+ \beta_2,$$

then it becomes clear that OP is essentially asking

What is $\infty+(-\infty)$ ?

which is not defined mathematically.

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As pointed out by zakk, any proper time of objects moving at light speed would be zero.

That means physically that there does not exist any point of time where you could perform the addition c + (-c). Example: A photon is emitted at A and absorbed at B. There is no point of time A < t < B where the photon is actually moving, the age of the photon at A is the same as at B, and even the hypothetical proper distance between A and B would be zero (even if observers are observing a time lapse between A and B, this observed time lapse does not correspond to any real point of time A < t < B).

This is why it is impossible to subtract (-c).

By the way, this result is confirmed by the second postulate of special relativity: the velocity of light is never (-c), it is always observed at +c.

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    $\begingroup$ I think your last sentence is inaccurate. What about isotropically emitted light - some of it is going in the $-x$ direction (once your coordinate system is specified). $\endgroup$ – Danu Aug 5 '14 at 21:18
  • $\begingroup$ I agree with Danu, simply consider the light cone of an event on a spacetime diagram. $\endgroup$ – Alfred Centauri Aug 5 '14 at 21:28

A more particle-based view of zakk and Moonraker's observation that photons experience zero time between emission and absorption is this:

The postulated event that a massless particle (say a photon) could emit another massless particle (say another photon) in the retrograde direction — or in any direction for that matter — is always zero. A photon from its own perspective is frozen in time throughout its existence, lacking any "moving parts" with which to generate other particles.

The complex time-based behaviors we see in photons as we watch them sail by, things such as frequency, polarization, and even splitting into an electron and positron when passing through an intense electric field gradient, exist only when such photons interact with our sub-c material universe. They are in effect a form of stop-motion photography in which similar photons display their highest probability quantum outcomes over a sequence of overlapping spacetime intervals. We then interpret that sequence as a continuous behavior over space and time, even though in reality it is necessarily composed of innumerable individual photon absorption events.

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Why do i have the sense that this can have a meaningful physical interpretation (even if velocities are both $c$)?

First lets assume that both velocities are $c$ or $1$ in the question's notation. Then the relativistic velocity addition formula simply gives: $1_A \oplus 1_B = 1_B \oplus 1_A = 1_{tot}$, correct? Yes, does the fact that this is mathematicaly well-defined make some difference in the physical interpretation or not? Let's continue.

The situation is completely symmetric to the opposite velocities situation. So by symmetry the answer to the question should be $1_A \oplus -1_B = 1_B \oplus -1_A = 0_{tot}$.

It seems that while the first is mathematicaly well-defined while the second is not, yet the physical problem and interpretation of both (by symmetry) should be similar.

Likewise while one would not measure faster-than-speed of light behavior when a photon emitted another photon (on its own frame), similarly one would not measure a photon emitted in opposite direction (assuming for a minute this experiment could be realised).

Side note: Mathematically, assuming both velocities have same magnitude $0 < a < 1$, the limit at $$1_A \oplus -1_B = \lim_{a \to 1}\frac{a-a}{1-a^2}$$
can be callculated (using L'Hôpital's rule) as: $$1_A \oplus -1_B = \lim_{a \to 1}\frac{a-a}{1-a^2}$$ $$1_A \oplus -1_B = \lim_{a \to 1}\frac{0}{-2a} = 0$$
unambiguously and not depending on path

See also this paper (arxiv, 2007) on The rest mass of a system of two photons in different inertial reference frames


We show that the rest mass of a system consisting of two photons is a relativistic invariant having the same magnitude in all inertial reference frames in relative motion. A scenario which starts with two photons where theirs frequencies are equal to each other the magnitude of the rest mass of the system depends on the angle made by the momentums of the two photons. The behavior of the photons when detected from two inertial reference frames is illustrated using a relativistic diagram which displays in true values the physical quantities (scalar and vector) introduced in order to characterize the dynamical properties of the photon. We consider that the obtained results should be taken in the discussions, some time fierce, between those who ban the concept of relativistic mass and those who consider that its use is harmless.

An interesting (and heated sometimes) discussion of similar question on researchgate

Arguably this is not less "mainstream" than postulating tacyons or FTL (faster-than-light) motion.

Note the above is not inconsistent with the fact that the emited photon will still have velocity $1$ (or $c$) in its own frame (or $-1$ in $A$ photon's frame), but the ground observer will not measure another photon.

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  • $\begingroup$ i understand the downvote (it is mentioned in the answer), yet dont think you have a better answer. $\endgroup$ – Nikos M. Aug 13 '14 at 4:22
  • $\begingroup$ There's no "photon's frame", at least not as anything's frame is ordinarily defined. $\endgroup$ – ACuriousMind Aug 13 '14 at 17:56
  • $\begingroup$ @ACuriousMind, you mean photon's "rest" frame (since photons can have inertial frames), but it doesnt matter, it is not needed (at least as this answer is formulated), the term may be used, but loosely, and does not change the result $\endgroup$ – Nikos M. Aug 13 '14 at 18:07
  • $\begingroup$ @ACuriousMind, if you will, look at it as trying to calculate and provide physical meaning on the edge of SR (per the question) $\endgroup$ – Nikos M. Aug 13 '14 at 18:11

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