# Question on Addition of Velocities

1. Nov 22, 2005

### nemosum

When you calculate the relative velocity of two bodies moving towards eachother at almost c, are you calculating the observed relative velocity, or the actually relative velocity. Because if it's the actual velocity then technically no body could ever move faster the 1/2c due to the fact that there just might be something somewhere else in the galaxy moving toward it at the same speed. But this doesn't make sense because we've already used particle smashers to accelerate particles to near c. How does this all work?:uhh:

Nemo

2. Nov 22, 2005

### Severian596

What's the difference between observed and actual. Who says what's "actual"?

(hint: no one)

Basically, you pick a frame then calculate the relative velocity.

3. Nov 22, 2005

### robphy

The "relative velocity between two inertial observers A and B" is an invariant spacelike 4-vector. All inertial observers measuring that relative velocity will obtain the same result. For this reason, you can choose any frame you find convenient.

4. Nov 22, 2005

### franznietzsche

You're still stuck on the idea of absolute velocity. That and you're adding velocities the wrong way. When you ad 0.5c and 0.5c, you do not get c. You get 0.8c. The formula for adding velocites is not
$$\vec{v} = \vec{u} + \vec{w}$$
but is
$$\vec{v} = \frac{\vec{u} + \vec{w}}{1 + \frac{u w}{c^2}}$$

5. Nov 22, 2005

### EL

I'm not sure the others understood your question correctly, but I'll give it a try the way I interpret it.
Say you are standing still on the ground, and there is one spaceship coming in from the east, and one from the west, both with speed say 0.9c, according to what you measure. Then according to you their relative speed is 1.8c. This is perfectly ok. The crutial thing is only that a single spaceship cannot exceed the speed of light.
However, neither of the captains on the spaceships will measure the speed of the other approaching ship as 1.8c, due to the SR formula for addition of velocities. In no frame any ship will fly faster than c.

Hence I would have to claim that Robphy's answer is not correct.

Last edited: Nov 22, 2005
6. Nov 22, 2005

### nemosum

Thanks EL, you interpreted my question right, I think I understand now. I was about to post an example just like that, but I guess I don't need to.

nemo

7. Nov 22, 2005

### nemosum

OK, here's another question: Using the same example EL did, the spaceship flying from the west would observe the spaceship coming from the east as having more relativistic mass than the observer on the ground would right? And the spaceship from the east would see the same thing with the one coming from the west. So how would they all know whether the observer on the ground was right or if they were, in measuring the relatavistic mass of both spaceships?

nemo

8. Nov 22, 2005

they are all correct =). I know that sounds funny, but with GR you will get differnt sizes, masses, time rates, ect for the same object when viewed from differnt reference frames.

9. Nov 22, 2005

### Severian596

Assuming that one person must be right while the other is wrong assumes that there's an absolute perspective that's "right" all of the time. It's all relative.

Heck, even if you assert that the guy on the ground is "right" because he's at rest during this experiment, how do you know he's not moving relative to someone else who would therefore be "more right." Better yet, how do you determine who's moving more, the guy who's planetside or the "more right" observer...you have to compare their motion to yet another observer to see how they move relative to him or her.

10. Nov 22, 2005

### EL

Well, there's really no need for inwoking GR here. It is perfectly enough with SR.

11. Nov 22, 2005

### EL

Correct.

As Wizardsblade pointed out, observers in different frames will measure different relativistic masses.

Last edited: Nov 22, 2005
12. Nov 22, 2005

### nemosum

lol, It's a little hard to let go of "absolute" perspectives and values. Thanks for the feedback!

nemo