I understand that if I am standing on earth and a space ship takes off and travels at .6c, then I perceive the space traveler receding at .6c relative to me, and the space traveler perceive me as receding at .6c relative to him. If another traveler takes off in the 180-degree opposite direction, then likewise I perceive the other space traveler receding at .6c relative to me, and the other space traveler perceive me as receding at .6c relative to him.
So why don’t they perceive each other as traveling faster than c, the speed of light?
In: 29
The other answers have not told you what speed the spaceships observe each other as traveling at relative to each other. Let me tell you: each spaceship observes the other one as traveling away from it at speed roughly .882c (or exactly (15/17)c), which is still less than the speed of light.
For objects A, B, and C that move along a line, let v*_AB_* be the velocity of A as measured by B, v*_BC_* be the velocity of B as measured by C, and v*_AC_* be the velocity of A as measured by C. In classical physics we have the formulas
> v*_CA_* = -v*_AC_*, v*_BA_* = -v*_AB_*, v*_CB_* = -v*_BC_*
> v*_AC_* = v*_AB_* + v*_BC_*.
In relativity, the first set of formulas remains valid (swapping observers negates the relative velocity), but the formula for combining velocities changes from simple addition to
> v*_AC_* = (v*_AB_* + v*_BC_*)/(1+v*_AB_*v*_BC_*/c^(2)).
To see what this means for your example, let B be Earth, A be a spaceship taking off from (or just passing) Earth at speed .6c, and C be the other spaceship going in the opposite direction from Earth at speed .6c. Call the direction that A travels “positive”, so v*_AB_* = .6c and v*_CB_* = -.6c (note the negative sign). We want to compute v*_AC_*. Since v*_BC_* = -v*_CB_* = .6c,
v*_AC_* = (v*_AB_* + v*_BC_*)/(1+v*_AB_*v*_BC_*/c^(2)) = (.6c + .6c)/(1 + (.6c)(.6c)/c^(2)) = (1.2/1.36)c = (15/17)c
with 15/17 being approximately .882. And v*_CA_* = -v*_AC_* = -(15/17)c. Each spaceship measures the other spaceship as moving away from it at speed (15/17)c, which is less than c.
If numbers v and w are both in the interval (-c,c), then the expression (v+w)/(1+vw/c^(2)) is again in (-c,c), so relative velocities that are below c for one observer will be below c for other observers. Moreover,
> if w = c then (v+w)/(1+vw/c^(2)) = (v+c)/(1+vc/c^(2)) = (v+c)/(1+v/c) = c(v+c)/(c+v) = c.
That captures the idea that if something moves at speed c relative to one observer then it also moves at speed c relative to all other observers. So moving at the speed of light is a phenomenon that’s independent of the observer.
Latest Answers