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

B How do we derive p = γmv ?

  1. Mar 1, 2017 #1
    I thought this mental experiment: consider an inertial frame of reference solidal to a particle moving with velocity v respect to another inertial frame of reference, i gotta find the momentum in the second frame of reference (in the first is 0 since is solidal with the particle)

    p= m* dx/dt

    And then i plug instead of x and t the lorentz tranformation thwt connects them with t' and x'...

    But it doesn't work, where am i wrong? Is that method correct? If not why?
     
  2. jcsd
  3. Mar 1, 2017 #2

    haushofer

    User Avatar
    Science Advisor

    That doesn't work, because the (spatial component of the!) relativistic momentum is p = ymv to start with. You should transform the 4-momentum. This 4-momentum is defined by the relativistic action of a point particle.
     
  4. Mar 1, 2017 #3
    The idea is to find an expression that:
    • is conserved in closed systems (and additive) even when ##v \rightarrow c##,
    • increases in magnitude without bound as ##v \rightarrow c## (because otherwise it would have a maximum limit like ##v## does, and this would contradict the first criterion), and
    • reduces to ##\mathbf{p} \approx m \mathbf{v}## in the classical limit.
    The famous Tolman/Lewis thought experiment demonstrates that if such a quantity exists, then it must be ##\mathbf{p} = \gamma m \mathbf{v}##, which obviously meets the second and third criteria. See here: https://books.google.com/books?id=FrgVDAAAQBAJ&pg=PA76.

    As for the first criterion, experiment verifies that this vector is indeed conserved at all possible values of ##v##. (Think particle accelerators.)
     
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: How do we derive p = γmv ?
Loading...