So I just finished learning about the derivation of the addition of velocities...this should imply that I am a newcomer to relativity. So here's my question:(adsbygoogle = window.adsbygoogle || []).push({});

We have a rocket that is traveling away from a lab at the relative velocity of c. (I know this is impossible because it would take infinite energy, but hear me out...). Within the rocket frame, it emits a photon that travels towards the rear of the ship (towards the lab) at velocity -c relative to the rocket. What is the velocity of this photon relative to the lab? My prediction was either it is -c or it is 0...leaning towards the former.

I was looking at this formula: v = (vrel + v') / (1 + vrel*v'). v is the photon's speed relative to the lab. vrel is the rocket's speed relative to the lab (c). v' is the photon's speed relative to the rocket (-c).

Here's where I got:

Let's have the rocket approach the speed of light and let's just have c = 1 and -c = -1. Then the lim as vrel → 1 = (1 + -1) / (1 + 1*-1) = 0 / 0...indeterminate form.

So I tried applying L'hospital's rule (which I don't know if that's even legal). I assumed v' is a constant in the formula since it is a photon that has a constant velocity so the variable to take a derivation is vrel. After taking a derivation of the top and bottom we get: (1 + 0) / (1 + v') = 1 / (1 + v'). Applying the lim vrel → 1, we get 1 / (1 + -1) = 1 / 0 = ∞?

I know my logic is flawed somewhere so any pointers would help.

Thanks.

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# Addition of Velocities: A photon emitted backwards

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