Relativistic Relative Velocity Problem

[rtareen]

Homework Statement

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.65c, and the speed of each particle relative to the other is 0.95c. What is the speed of the second particle, as measured in the laboratory?

Relevant Equations

$v_x' = \frac{v_x - u}{1 - uv_x/c^2}$ (1)
$v_x = \frac{v_x' + u}{1 + uv_x'/c^2}$ (2)

Heres how I tried to set up the problem.

I took the laboratory to be S and the frame of the particle whose speed we know to be S', so that the speed of S' relative to S is u = 0.65c.

Also, by convention, S' moves to the right of S, so that S moves to the left of S'.

Next, we know that the speed of particle two as viewed from particle 1's frame is $v_x' = 0.95c$. So since we know $v_x'$ and are trying to find $v_x$, I'd use equation 2. But this gives the wrong answer. If we use equation 1 we end up with the correct answer of 0.784c.

What are the official rules for doing this? How can I know when to use which, if I can't always go by the notation?

 

[PeroK]

Equation (2) is going to give $v_x > v_x'$, so that is clearly wrong. The rule is that you have to set up your reference frames correctly. If you ended up with equation (2) then you got the signs wrong and have the two particles moving in the same direction.

 

[rtareen]

Equation (2) is going to give $v_x > v_x'$, so that is clearly wrong. The rule is that you have to set up your reference frames correctly. If you ended up with equation (2) then you got the signs wrong and have the two particles moving in the same direction.

Sticking with the original notation, would it be correct to say that we must choose S and S' so that $v_x'$ is always moving in the same direction as S' relative to S?

 

[kuruman]

Primes and other such symbols are used differently by different people. Instead of trying to remember where the primes and negative signs go, I find it more straightforward to start with the basic relativistic velocity addition formula is $$v_{\text{AC}}= \frac{v_{\text{AB}}+v_{\text{BC}}}{1+\frac{v_{\text{AB}}*v_{\text{BC}}}{c^2}}.$$ The notation is v_ij = velocity of i relative to j. This is the relativistic analog of a passenger (A) running inside a train (B) which moves relative to the ground (C) with all velocities positive.

In this case I would define A = one particle, and C = other particle in which case v_AC =+ 0.95 c. The leftover subscript, B refers to the lab. What remains is to set one of v_AB or v_BC equal to 0.65 c and solve for the other.

 

[PeroK]

Sticking with the original notation, would it be correct to say that we must choose S and S' so that $v_x'$ is always moving in the same direction as S' relative to S?

I'm not sure I understand what that means. One way to get the numerator is check the non-relativistic case. E.g.
Two particles are created in a low-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is $u = 65 m/s$, and the speed of each particle relative to the other is $v' = 95 m/s$. What is the speed of the second particle, $v$, as measured in the laboratory?

That should be easy $v = v' - u = 95 - 65 m/s$.

Or, more precisely, using the full relativistic formula: $$v = \frac{v' - u}{1 - uv'/c^2}$$
It's probably a good exercise to write out all the frames of reference in full detail, but there can't be any doubt what the answer must be.