Another, more general question about it. If the final derived equation does not generate a divide by zero when v=c, but one of the steps in the derivation does, does that invalidate the final equation when v=c? :scratchin:
Is the relativistic parallel velocity addition formula valid when u=v=c?
So... you mean no, right?Yes, see above.
"At rest" is just as real as "here".But like Pete said, the equation is from a reference point at rest. Which in reality nothing is at rest, and in reality neither photon is at rest.
So... you mean no, right?
You are arguing that it is not valid when both u and v are equal to c.
Nevertheless, the limit is well-defined.
Your sentence does not make much sense, I am saying that $$v=c$$ is not allowed (there is no frame moving at c). That is, $$w(c,c)$$ is not defined. On the other hand, $$w->c$$ when $$u->c$$ and $$v->c$$.Are you saying that v = c is defined at the limit?
$$w(u,v)=\frac{u+v}{1+uv/c^2}$$
$$w(0,2c)=\frac{0+2c}{1+(2*0)/c^2}$$
$$w(0,2c)=\frac{2c}{1}$$
$$w(0,2c)=2c$$
....
Why not? If we assume one photon is at rest wouldn't we assume the other is moving at 2c?
So the jury is still out on the maximum speed observed by two massless particles.
(There is the technical issue that you can't have clocks and rulers at rest with a photon, but that's a side point).
Suffice to say that there exists no frame where an observer with mass sees any massive or massless object moving at a speed higher than c.
The jury says that a massless particle can't measure the speed of anything, because they can't have clocks at rest with them.So the jury is still out on the maximum speed observed by two massless particles.
W=.9944
W=(.8,1)
1+.8/(1+1*.8/c^2)=1