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Four momentum expression

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    To write the expression of force in STR

    [tex]\ F=\frac{dp}{dt}=\ m\gamma\ a +\ m\gamma\frac{\ u .\ a}{\ c^2 -\ u^2}\ u[/tex]

    Here a is acceleration

    2. Relevant equations

    I used the equation [tex]\ p=\gamma\ m\ u[/tex]

    I interpreted F as four force,p as four momentum, a as four-acceleration, u as four velocity etc...

    3. The attempt at a solution

    Mere differentitation is giving the answer;But I do not know if the method is correct.Because, [tex]\ p=\gamma\ m\ u[/tex] for 3 velocity---that's for sure.But is it also true for four velocity and four momentum?
     
  2. jcsd
  3. Sep 5, 2009 #2

    gabbagabbahey

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    It occurs to me that you can pretty much answer your own question just by looking up the definitions of "4-momentum", "4-velocity" and "4-force"....surely your text has those definitions?
     
  4. Sep 5, 2009 #3
    Seems my instructor did not formulated the problem in the correct way.The expression is for 3velocity;that for 4 velocity is not that simple.
     
  5. Sep 5, 2009 #4

    gabbagabbahey

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    What exactly is the problem statement?
     
  6. Sep 5, 2009 #5
    To show that 4force can be expressed as

    [tex]
    \ F=\frac{dp}{dt}=\ m\gamma\ a +\ m\gamma\frac{\ u .\ a}{\ c^2 -\ u^2}\ u
    [/tex]
     
  7. Sep 5, 2009 #6

    gabbagabbahey

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    But it can't be expressed that way....for starters, if [itex]\textbf{F}[/itex] is the 4-force, [itex]\textbf{P}[/itex] the 4-momentum and [itex]\tau[/itex] the proper time, then

    [tex]\textbf{F}=\frac{d\textbf{P}}{d\tau}\neq\frac{d\textbf{P}}{dt}[/tex]
     
  8. Sep 5, 2009 #7
    yea,I also suspect that the expression is not meant for 4 force.
     
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