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You can't go faster than light, and light can't be additive (if you shine a light from a spaceship, the light is not going $c$+"speed of spaceship", it's just going like it always does).

But what happens when two spaceships, each going 75% the speed of light, are traveling directly towards each other?

Aren't the two spaceships now traveling, in reference to each other, at 150% the speed of light?

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marked as duplicate by Brandon Enright, Kyle Kanos, Nathaniel, John Rennie, Valter Moretti Apr 8 at 7:28

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possible duplicate of Double light speed –  Brandon Enright Apr 8 at 1:20

4 Answers 4

The addition of relative velocities is $$\frac{a+b}{1+ab}$$ so $\frac{.75+.75}{1+.{75}^2}$ = .96 c.

Consider that you are travelling almost the speed of light (1-x) c and you see a space ship pass you at an equal speed. Combining (1-x) c with (1-x)c. This gives $$\frac{2-2 x}{2-2 x+x^2}$$ Since $2-2 x < 2-2 x+x^2$, we have $\frac{2-2 x}{2-2 x+x^2} <1$

The addition of relative velocities is bound by the mathematics of the hyperbolic tangent. So velocity $a$ plus velocity $b$ is the inverse hyperbolic tangent of $a$ plus the inverse hyperbolic tangent of $b$ and then finally take the hyperbolic tangent of the results. The hyperbolic tangent only goes to 1 at the limit of infinity.

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so when I'm in a car driving south on I-95 at 60mph, I'm not passing 60mph north-bound cars at 120mph? When you calculate impact force, it's wrong to add relative velocities? –  ChronoFish Apr 8 at 1:30
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@ChronoFish Technically, yes. You can check, however, that for velocities much smaller than the speed of light, the correct addition formula reduces to a very good approximation to the non-relativistic case. –  EtaZetaTheta Apr 8 at 1:39
    
Um, isn't it $\frac{v_1 + v_2}{1+\frac{v_1 v_2}{c^2}}$ ? en.wikipedia.org/wiki/… –  user80551 Apr 8 at 2:53
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@user80551 I used the term relative velocity, where the speed of light c is set to 1. So $a=v_1/c$ and $b=v_2/c$. –  Daniel Geisler Apr 8 at 4:02
    
Daniel Geisler: "I used the term relative velocity, where the speed of light c is set to 1." -- Yes you did. But then you should be consequent and drop the "c" from your expression "$\frac{.75+.75}{1+.{75}^2}$ = .96 c" as well. –  user12262 Apr 8 at 5:22

No, unfortunately this is one of those "It should be, but it isn't."

Ship A shines a light at ship B. The light leaves ship A at $c$ and it arrives at ship B at $c$. Even if the 2 ships are traveling at 99% $c$, the light still leaves ship A at $c$ and arrives at ship B at $c$. I believe that the reason lies with the time dilation when approaching $c$.

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The Special Theory of Relativity deals with two observers, well... observing a single event, and comparing their observations. So:

You observe two spaceships, S-A and S-B, occupied by Obs-A and Obs-B moving towards you at $0.75c$ from opposite directions.

You observe S-A travelling towards you at $0.75c$; Obs-A observes you moving towards him at $0.75c$. Nothing strange here.

You turn in the opposite direction and observe S-B travelling towards you at $0.75c$; Obs-B observes you moving towards him at $0.75c$. Nothing strange here, either.

You conclude, correctly, that S-A and S-B are approaching each other at $1.5c$. If they started out 30 light-years apart, they'll meet in 20 years, right about where you're standing.

But..

Obs-A will measure Obs-B to be approaching Obs-A at the velocity given in the answer above: $0.96c$. Note: MEASURE

The SPoR lets everyone in these three frames of reference predict what everyone else will observe, what their clocks will read at various events, and why various slow clocks and shortened meter-sticks make everything come out right in the end.

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Thanks this is helpful. It still disturbs me. I think I have a hard time (not) separating observation from experience from "work performed". If after the flight the observers get out and measure the distance covered and the time it took it would work out to 1.5c. So how does this reconcile with the experience of one observer seeing the second spacecraft approach them at .96c? –  ChronoFish Apr 8 at 3:44
    
But in order to "get out and measure the distance covered and the time it took" the observers would have to accelerate to match speeds, and Special Relativity gets a lot messier. –  User58220 Apr 8 at 3:48
    
If obs-A could see obs-C (me) not moving, and over my shoulder saw S-B. Are you saying that the gap between obs-A/Obs-C would be closing at 0.75c and since the gap between obs-A and S-B would be closing at an observed 0.96c it would appear to obs-A that the gap between obs-C and S-B would be 0.21c? Or would the experience change simply because the focal point changed? (That seems hokey to me) –  ChronoFish Apr 8 at 4:01
    
@User58220: I like the emphasis on _measurement_, contrasting with observation (i.e. collecting observational data such as identifying the participants and registering their indications) from which measurements would be derived (e.g. values of pairwise mutual speed, in _pairwise mutual agreement_). It should be emphasized that $A$ and "you" had _measured_ their speed (0.75 c) and that $B$ and "you" had _measured_ their speed (0.75 c) just as $A$ and $B$ _measured_ 0.96 c. "S-A and S-B are approaching each other at 1.5c" -- That's a third party opinion (a.k.a. "improper"). –  user12262 Apr 8 at 5:45

Alice places herself at system C. Now she detect A and B approaching each other from opposite directions, while A and B both have velocity 0.9c. She can safely say, B is approaching A with velocity 1.8c. However, if Bob is located at A and detect the velocity B towards him, he will find that B is approaching with velocity ~0.994c. If Alice and Bob can communicate with no latency(of course, this is against Einstein's theory, since communicating speed cannot be faster than c):

Alice: B is coming to you at velocity 1.8c, now you and B has distance 1.

Bob: No, B is coming to me at velocity 0.994c, and the distance from B to me is only 0.436!

It only depends where you are doing experiments, C or A.

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