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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

    DW

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    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

    krab

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    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.

    Pete
     
  7. Aug 24, 2004 #6

    DW

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    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.
     
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