Two Photons are Traveling [duplicate]

Two photons, photon A and photon B, are traveling in the same direction at the speed of light, $$c$$. How fast does photon B appear to be traveling from the perspective of photon A?

$$A$$ doesn't have a perspective, there are no coordinate frames moving at $$c$$.

Nevertheless, you can naively use the velocity addition fromual:

$$v_{21} = \frac{v_1+v_2}{1+v_1v_2/c^2}=(c-c)/(1+(c)(-c)/c^2)=0/0$$

well I guess you can't. See 1st paragraph.

• But then the photon would be at rest... Commented Jul 31 at 5:15
• Just think: what should A do to measure the speed of B, and you see the question is meaningless. Commented Jul 31 at 9:39
• @ChristinaDaniel It is not a logical contradiction - it is a domain error. The equations are simply not valid. Commented Jul 31 at 11:44
• Also consider how much time either photon experiences from the perspective of any observer. Photons are always observed to move at $c$ in any reference frame, which means that their Lorentz factor $\frac{\mathrm{d}t}{\mathrm{d}\tau}=\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx\infty$ is infinite. If you invert the differential on the left, you obtain $\frac{\mathrm{d}\tau}{\mathrm{d}t}=\frac{1}{\gamma}=\frac{1}{\infty}=0$, so from any reference frame the photons experience no time. If they experience no time, they can't measure velocity (which is inherently time-based) to begin with. Commented Jul 31 at 16:38
• @controlgroup don't talk about photon reference frame...there are none. It confuses noobs. But you can calculate $\gamma(c)=\infty$ to show it's a domain error. I actually excepted to get $0/1$ from the formula, but $0/0$ was even better.
– JEB
Commented Jul 31 at 17:59