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
None of those things is an absolute; rather, they are dependent on who or what is doing the observing.
None of those things is an absolute; rather, they are dependent on who or what is doing the observing.
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.
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.
So there are going to be some REALLY clever answers here that probably explain this a lot better than I can. BUT:
They are… from the perspective of an observer at rest at the starting point but importantly neither exceeds C itself from the perspective of the observer.
What happens on each of the ships traveling at .6C is that their own observations of the other are distorted by their own relativistic speeds and they appear to be moving slower, as space behind you is stretched and space infront of you is compressed.
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.
So there are going to be some REALLY clever answers here that probably explain this a lot better than I can. BUT:
They are… from the perspective of an observer at rest at the starting point but importantly neither exceeds C itself from the perspective of the observer.
What happens on each of the ships traveling at .6C is that their own observations of the other are distorted by their own relativistic speeds and they appear to be moving slower, as space behind you is stretched and space infront of you is compressed.
So there are going to be some REALLY clever answers here that probably explain this a lot better than I can. BUT:
They are… from the perspective of an observer at rest at the starting point but importantly neither exceeds C itself from the perspective of the observer.
What happens on each of the ships traveling at .6C is that their own observations of the other are distorted by their own relativistic speeds and they appear to be moving slower, as space behind you is stretched and space infront of you is compressed.
Because that’s simply not how velocities add, even though we’re all used to thinking it is. It’s just that at the sorts of speeds we’re used to, it might as well be. Your teachers didn’t exactly lie to you – but they didn’t tell you the whole truth, either.
The simple total is a rule of thumb that works *really* well at human speeds, basically – you’d need to be involved in something very specific and scientific for it not to be good enough. And it’s what we all get taught at school, because it’s “close enough”. But it isn’t actually correct.
This is all about how the universe *actually* works, and specifically General Relativity (tested many, many times). And at low velocities, it might as well not be. The difference between the number you get by just adding two values, and [the actual result](https://en.wikipedia.org/wiki/Velocity-addition_formula), is *incredibly* small. For most day-to-day purposes it’s undetectable and irrelevant (and a long, long way below the margin of error of anything you’re likely to have available to measure your speed).
But that difference *is* still there. And at high velocities – significant proportions of the speed of light such as this, say – it really starts to show up.
If you add, say, 100mph and 100mph, I make it that the actual result is about 20 *quadrillionths* of one mph less than the simple total of 200mph. You can probably be forgiven for not noticing. But 0.6c plus 0.6c? That adds up to, roughly, 0.88c. That, you’re going to find hard to miss.
Because that’s simply not how velocities add, even though we’re all used to thinking it is. It’s just that at the sorts of speeds we’re used to, it might as well be. Your teachers didn’t exactly lie to you – but they didn’t tell you the whole truth, either.
The simple total is a rule of thumb that works *really* well at human speeds, basically – you’d need to be involved in something very specific and scientific for it not to be good enough. And it’s what we all get taught at school, because it’s “close enough”. But it isn’t actually correct.
This is all about how the universe *actually* works, and specifically General Relativity (tested many, many times). And at low velocities, it might as well not be. The difference between the number you get by just adding two values, and [the actual result](https://en.wikipedia.org/wiki/Velocity-addition_formula), is *incredibly* small. For most day-to-day purposes it’s undetectable and irrelevant (and a long, long way below the margin of error of anything you’re likely to have available to measure your speed).
But that difference *is* still there. And at high velocities – significant proportions of the speed of light such as this, say – it really starts to show up.
If you add, say, 100mph and 100mph, I make it that the actual result is about 20 *quadrillionths* of one mph less than the simple total of 200mph. You can probably be forgiven for not noticing. But 0.6c plus 0.6c? That adds up to, roughly, 0.88c. That, you’re going to find hard to miss.
Latest Answers