# Special relativity

1. May 10, 2015

### TimeRip496

Two spaceships pass by each other. Both are travelling at near the speed of light. How will each spaceship percieve each other speed and time? Besides how do you do it mathematically?

2. May 10, 2015

### phinds

"Traveling at near the speed of light" is an utterly meaningless statement since you gave no reference frame. YOU are traveling at near the speed of light right this very minute.

If you say there is an observer on Earth and he "sees" one ship approaching at near c and another ship approaching at near c from the opposite direction all in his frame of reference, then the the spaceship captains each see themselves as stationary and the Earth approaching them at close to c and the other spaceship approaching at even closer to c.

Look up Lorentz Transform.

3. May 10, 2015

### Ibix

You might also want to check out the relativistic velocity addition formula. You can derive it from the Lorentz transforms, but it's a direct answer to one of your questions.

4. May 10, 2015

### phinds

Thanks for adding this. I should have mentioned it.

5. Jun 22, 2015

### Physicist97

Well, if you are on the spaceship you can simply say that you are the one at rest and the other ship is going at some speed near $c$, (of course there is no way to prove that you are the one who is at rest). You will also see the ship to be time-dilated by the standard formula for time dilation. If you are an observer outside both spaceships and you measure both the spaceships' speeds relative to you, and you wanted to know what speed an observer on one spaceship would measure the other going, you use the relativistic velocity addition formula, as Ibix mentioned. The equation is $w={\frac{u+v}{1+{\frac{uv}{c^{2}}}}}$ , $w$ is the combined speed the observer would measure, $u$ is the speed of one ship, $v$ is the speed of the second ship. By knowing what speed an observer on the ship would measure of the other ship, you can then plug $w$ into the formula for time dilation and find out how that observer would measure time being dilated.

6. Jun 23, 2015

### phinds

There is absolutely no need to "prove" you are at rest in your own frame of reference, you are by definition at rest. And since no one is EVER at rest in any absolute sense, that's the best you can do.

7. Jun 23, 2015

### harrylin

For sure, nobody suggests here that one needs to "prove" that you are at rest in "your own frame of reference". And you are for example by definition not at rest but moving in the ECI frame which we use for GPS.

8. Jun 23, 2015

### phinds

Uh ... I guess I was misled my the words "there is no way to prove that you are the one who is at rest".

9. Jun 23, 2015

### harrylin

That's just one way of stating the PoR.

10. Jun 23, 2015

### phinds

Well, to me it emphatically suggests a belief that SOMEONE is absolutely at rest, you just can't prove you are that one.

11. Jun 23, 2015

### harrylin

It corresponds to the way the PoR was originally formulated, and Einstein's formulation of the PoR is perhaps more neutral.

I agree that the earlier formulation may suggest a certain interpretation, but not emphatically: in SR any standard reference system pretends to be a "rest system" so that other systems are pretended to be "moving", with all implied physical consequences such as relativity of clock synchronization. To get back to my earlier example, in GPS calculations the speed of light is not isotropic relative to you - because you are held to be moving.

Last edited: Jun 23, 2015
12. Jun 23, 2015

### phinds

I'm not getting how any of this is relevant to either the original question or the side issue brought up by my response to physicist97

13. Jun 23, 2015

### harrylin

The side issue that you brought up has probably nothing to do with the original question. I'll therefore end my clarifications as follows: "there is no way to prove that you are the one who is at rest" corresponds to statements like "there is no way to prove that your clock synchronization is right".