Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 24, 2017 #1

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  2. jcsd
  3. Mar 24, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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##!
     
  4. Mar 24, 2017 #3

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
  5. Mar 24, 2017 #4

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  6. Mar 24, 2017 #5

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @pervect , yes, that's a stray square root sign. Thanks.
     
  7. Mar 25, 2017 #6

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
  8. Mar 29, 2017 #7
    Great job Rob!
     
  9. Mar 29, 2017 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
  10. Mar 29, 2017 #9

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ...

    ##\alpha = \frac{\sqrt{\gamma^2-1}}{\gamma +\frac{m_1}{m_2}} = \frac{\beta}{\frac{m_1}{m_2}\sqrt{1-\beta^2}+1}##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Relativity Variables: Velocity, Doppler-Bondi k, and Rapidity - Comments
Loading...