Addition of Velocities: A photon emitted backwards

[wainker1]

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:

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.

 

[Mentz114]

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...)

If you start with an impossibility you can't deduce anything sensible.

 

[wainker1]

If you start with an impossibility you can't deduce anything sensible.

Ok, but let's then look at what happens when the rocket gets really really really really close to c...to do this we take a limit as vrel → 1. Then if you follow through the steps of my post, I eventually get an answer of ∞. So even if we remove the impossibility, my logic still has a flaw somewhere.

 

[yuiop]

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.

Not too surprising to get an indeterminate result when you are calculating something impossible physically.

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.

Using v' = -c = -1 :

The derivative of the numerator (vrel + v') = (vrel -1) wrt vrel is 1.

The derivative of the denominator (1 + vrel*v') = (1 - vrel) wrt vrel is -1.

The result is 1/-1 = -1

So surprisingly we get the right result, probably for all the wrong reasons.

 

[wainker1]

The derivative of the numerator (vrel + v') = (vrel -1) wrt vrel is 1.

The derivative of the denominator (1 + vrel*v') = (1 - vrel) wrt vrel is -1.

The result is 1/-1 = -1

Ah yes, correct mathematics would solve my problem. Thank you for pointing out my mistake and for the help.