# Insights Relativity Variables: Velocity, Doppler-Bondi k, and Rapidity - Comments

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1. Mar 24, 2017

### robphy

2. Mar 24, 2017

### vanhees71

In your story in the very beginning, your terminology is a bit unusual. The relative velocity of two particles is defined frame independently as the velocity of one particle in the rest frame of the other. What you where asking is the particles' velocity in the center-of-mass frame given $\beta_{\text{rel}}=0.5$!

3. Mar 24, 2017

### robphy

Yes, I was probably a little sloppy.
I was thinking that one particle is at rest and the other moves with velocity 0.5c. What is the velocity of the center of momentum frame?

4. Mar 24, 2017

### pervect

Staff Emeritus
I believe that there is a typo in the insight, the proper expression for $\beta$ in terms of k is:

$$\beta = \frac{k^2-1}{k^2+1}$$

not

$$\beta = \sqrt{\frac{k^2-1}{k^2+1}}$$

I checked it a couple of different ways (including algebraic substitution). The second way involves tracing out light signals. If we imagine a signal emitted at t=1 , reflected off an observer moving at some velocity $\beta$, the signal comes back at time k^2. The distance at t=0- is 0. The radar results imply that the distance at time $(k^2+1)/2$ is equal to $(k^2-1)/2$, making the velocity the first expression without the square root.

5. Mar 24, 2017

### robphy

@pervect , yes, that's a stray square root sign. Thanks.

6. Mar 25, 2017

### vanhees71

Yes, that's the right way to state it. It's important that relative velocity is a Lorentz-invariant quantity by defining it in a specific frame, i.e., the velocity of particle 1 in the rest frame of particle 2, and that's used in the definition of the cross-section formula, making the cross section an invariant quantity too.

Related are the somewhat confusing definitions of several invariant quantities like temperature and chemical potential in relativistic statistical/thermal physics or energy and momentum for non-closed systems, where the energy-momentum tensor is not conserved. The latter issue is carefully discussed in Jackson, Classical Electrodynamics.

7. Mar 29, 2017

### Greg Bernhardt

Great job Rob!

8. Mar 29, 2017

### PeroK

You can also do it using the energy momentum transformation. If particle 1 is at rest and particle 2 is moving at velocity $\beta$ (gamma factor $\gamma$) and the COM frame is moving at velocity $\alpha$ then:

$p'_1 = \gamma_{\alpha}m\alpha$

$p'_2 = \gamma_{\alpha}(\gamma m \beta - \gamma m \alpha)$

Equating the two momenta gives:

$\alpha = \frac{\gamma \beta}{1+\gamma} = \sqrt{\frac{\gamma -1}{\gamma +1}} = \frac{\beta}{1+\sqrt{1-\beta^2}}$

This method extends quite well for particles of different masses.

Last edited: Mar 30, 2017
9. Mar 29, 2017

### PeroK

...

$\alpha = \frac{\sqrt{\gamma^2-1}}{\gamma +\frac{m_1}{m_2}} = \frac{\beta}{\frac{m_1}{m_2}\sqrt{1-\beta^2}+1}$