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

Change in relativistic momentum

  1. Aug 21, 2004 #1
    Is it alright to say that force = rate of change of relativistic momentum

    F = [ m0 v2 / (1 - v2^2/c^2)^1/2 - m0 v1/(1 - v1^2/c^2)^1/2 )] / (t2 - t1)

    and can this relation be used to get sensible results for particles?
    Last edited: Aug 21, 2004
  2. jcsd
  3. Aug 21, 2004 #2


    User Avatar

    Thats close to ordinary force f. To be precise ordinary force f involves the limit of that as t2-t1 becomes infinitesimal dt in a calculus limit. Also, you shouldn't subscript the mass with a zero as it is invariant.
  4. Aug 21, 2004 #3
    DW is right, your notation is wrond. You must let t2-t1 approach 0, it must be alimit.
    Here is the wau you want to write it:

    [tex] F = \frac {d(\frac{mv}{\sqrt{1-v^2 / c^2}})}{dt} [/tex]
  5. Aug 21, 2004 #4


    User Avatar
    Science Advisor

    Yes, but in general, you gain nothing in writing the differential equation in terms of velocity. Just leave it in momentum; equations are far simpler.
  6. Aug 21, 2004 #5
    Yes. Force is defined as

    [tex]\bold F = \frac{d\bold p}{dt}[/tex]
    That is the average force. The instantaneous force is F = dp/dt.

  7. Aug 24, 2004 #6


    User Avatar

    Just to be clear to you, in light of the notation having been used here for a while, it is an expression for ordinary force f, not the four-vector force F.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook