A short question, because the math got me confused. If in my inertial reference frame, I see a dude going away from me at a speed $v$. Then, from his frame of reference (which is also inertial) he also sees me moving with speed $v$. No matter how close he is to the speed of light.
Is that correct ? $$dx/dt = v = dx'/dt'$$ (the gammas cancel out)(I chose $x'$ to be opposite in direction to $x$)
It sounds right, as no dude has a preferred frame of reference. To be sure, let's use the infallible Lorentz transformation (though I cannot, in good conscience, invert $x'$):
Suppose you remain at:
$$ E_1 = (t_1, x_1) = (0, 0),$$
while the dude starts there. After a time $T$, he has moved to $x=D$, placing that event at:
$$ E_2 = (t_2, x_2) = (T, D) $$
while you remain at:
$$ E_3 = (T, 0)$$
for which you measure a speed:
$$ v \equiv \frac{x_2-x_1}{t_2-t_1}=\frac D T$$
Transforming to his frame using:
$$ t' = \gamma(t-\frac{vx}{c^2})$$
$$ x' = \gamma(x-vt) $$
dude sees:
$$ E_1 = (\gamma(0-0), \gamma(0-0))' = (0, 0)'$$
$$ E_2 = (\gamma[T-\frac{vD}{c^2}],\gamma[D-vT])' $$ $$ E_2 = (\gamma[T-\frac{D^2}{Tc^2}],\gamma[D-DT/T])' $$ $$ E_2 = (\gamma[T-\frac{D^2}{Tc^2}], 0)'=(\gamma[T(1-\frac{v^2}{c^2})], 0 )' = (T/\gamma, 0)' $$
meaning he didn't move and his clock ticked slower. Meanwhile, you moved to:
$$ E_3 = (\gamma(T-0), \gamma(0-vT))' = (\gamma T, -\gamma D)'$$
for a velocity of:
$$ v' \equiv \frac{x'_3-x'_0}{t'_3-t'_0}=\frac{-\gamma D}{\gamma T} = -v$$
Note that in your frame, each participant measures the other's speed at the same time, while in dude's frame, you measured his speed before he measured yours.
Yes you are absolutely correct, if you see someone travelling at a speed $v$ (be careful on the distinction between speed and velocity), they also see you travelling at a speed $v$, regardless of how close you are to the speed of light (but not travelling at the speed of light).
