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 $x$)


2 Answers 2


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.

  • $\begingroup$ thanks a lot ! I love your usage of "dude" $\endgroup$
    – Pastudent
    Aug 3, 2020 at 13:55

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).


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