# Relativistic particles

1. Oct 21, 2007

### Niles

1. The problem statement, all variables and given/known data

I have two particles in an accelerator approaching eachother with a relative speed of 0.880c. The particles travel at the same speed as measured in the laboratory. I have to find the velocity of each particle.

3. The attempt at a solution

Ok, I use the transformation of velocities:

u'_x = (2*0.880c)/(1-(-(0.880c)^2/c^2)

- since v_1 = -v_2. I get 0,99c, but apparently it must equal 0,597c?!

2. Oct 21, 2007

### Staff: Mentor

You're using that equation backwards. That equation gives you the relative speed, which you already have, in terms of the lab frame speeds.

3. Oct 21, 2007

### Niles

Can you tell me, what u'_x, v and u_x are? They seem to differ (the meaning of them) every time I encounter a new problem.

v is the speed of object u' relative to u, and u_x is speed of u to stationary observer and u'_x is speed of u' to stat. obs.?

This doesn't add up with our problem? (Btw, I got the real answer, but I still lack the definitions).

4. Oct 21, 2007

### Staff: Mentor

Here's one way of looking at it. In the lab frame, the velocity of the A particle is +u and the B particle is -u. From the frame of the B particle, the lab frame is moving at speed +u. (Call the lab frame S', since it's moving.) So, in the lab frame the speed of the A particle is u' = u.

Transforming to the stationary frame of the B particle, the speed of the A particle with respect to the B particle will be given by:

v = (u' + u)/(1 + u'u/c^2) = 2u/(1 + u^2/c^2)

Make sense?

One way I like to write this is:

$$V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}$$

5. Oct 21, 2007

### Niles

Ahh, I see.

Can you tell me more generally, what they stand for? So I can use them on other examples.

But I liked your explanation! (bookmarked)

6. Oct 21, 2007

### Staff: Mentor

What what stands for? Which variables do you mean?

7. Oct 21, 2007

### Niles

What u_x, u´_x and v stand for in the equations when transforming velocities.

u_x is the speed of the object as seen from a stationary observer?
v is the speed of S' relative to S?
u´_x is the speed of the object in S' as seen from a stationary observer?

8. Oct 21, 2007

### Staff: Mentor

u_x is the speed of object as measured in frame S (stationary)
u'_x is the speed of object as measured in fame S' (moving)
v is the speed of S' as measured in frame S

In that case:
u'_x = (u_x - v)/(1 - v u_x/c^2)

Let me know if that makes sense to you.

Applying this version to your problem:
Speed of A in S = + .88 c
Speed of B in S = - .88 c

If you choose a frame S' moving along with A (say), then v = +.88 c. Then you use the above formula to transform the speed of B in frame S (u_x = -.88 c) to find its speed u'_x with respect to A (which is S').

9. Oct 21, 2007

### Niles

I found an example in my book that I tried to solve, which I hope will clarify things for my, but I'm kinda stuck.

"A planet P between two rockets A and B wants to measure the velocities of A and B. P is watching A and B getting closer to eachother, and the distance between them is getting smaller with the velocity (5/7)c. From A it looks like B is approaching A with the velocity (35/37)c."

Ok, first of all, we notice that A and B are getting approaching eachother with the velocity (5/7)c, so:

(5/7)c = (u_a + u_b) / (1+u_a*u_b/c^2).

Also, from A it looks that B's speed is (35/37)c, so:

(35/37)c = (u_a + u_b) / (1+u_a*u_b/c^2).

Am I way off here?

10. Oct 21, 2007

### Staff: Mentor

This is a tricky one! First note that the highlighted statement is not equivalent to:
Which is good because this equation:

Instead, it means that as seen by P, the distance between A and B is decreasing at a rate of (5/7)c. That's equivalent to saying that: (5/7)c = u_a + u_b

Good.

Now combine the two equations to solve for the speeds.

11. Oct 22, 2007

### Niles

Ahh, I see.. I think I'm getting the hang of this.

Thank you for all your help so far.