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(gamma)ma=F-[F.(beta)]beta How derive this?

  1. Jul 14, 2009 #1
    (gamma)ma=F-[F.(beta)]beta
    How derive this?
     
  2. jcsd
  3. Jul 14, 2009 #2

    HallsofIvy

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    Re: acceleration

    Would you mind explaining what it is? Until you explain what "gamma", "m", "a", "F", and "beta" mean, it's just a string of symbols.
     
  4. Jul 14, 2009 #3
    Re: acceleration

    gamma is the lorentz factor, m is mass, a acceleration, F force; beta is v/c
    Lev Okun, in his "concepts of mass" has derived it as follows:
    E=ymc^2
    p=ymv
    dp/dt=F
    and next he writes
    a=[F-(F.beta)beta]/ym
     
    Last edited: Jul 14, 2009
  5. Jul 14, 2009 #4
    Re: acceleration

    This is how I feel every time I try to read anything written by a physicist :rolleyes:



    But Halls is right. What is the context of the problem you're working on?

    The left hand side, F-[F.(beta)]beta, looks a lot like you're trying to remove a component from the force vector (it's very close to the force perpendicular to the direction of motion.... is this what you're going for?)
     
  6. Jul 14, 2009 #5

    clem

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    Re: acceleration

    I think the equation you are looking for is
    {\bf F}=\frac{d{\bf p}}{dt}
    \frac{d}{dt}\frac{m{\bf v}}{\sqrt{1-v^2}}
    =m{\bf a}\gamma+m{\bf v}({\bf v}\cdot{\bf a})\gamma^3.
    This is the usual form.
    To get your form, use the above equation to show
    {\bf v}\cdot{\bf F}=m\gamma^3{\bf v}\cdot{\bf a}
    in a few steps.
    I use units with c=1.
    I couldn't find my latex errors. I hope you can read it as is.
     
    Last edited: Jul 14, 2009
  7. Jul 14, 2009 #6

    Fredrik

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    Re: acceleration

    Just add tex tags.

     
  8. Jul 14, 2009 #7

    clem

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    Re: acceleration

    Thank you. I did use [tex]....[/tex], but it kept showing errors.
    I guess it can read your handwriting.
     
  9. Jul 15, 2009 #8
    Re: acceleration

    I got it thanks but using the equation gives F=y^3ma when parallel and F=yma when transverse why is dt so?
     
    Last edited: Jul 15, 2009
  10. Jul 15, 2009 #9

    clem

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    Re: acceleration

    You can't argue with algebra. There is no "why" in algebra.
     
  11. Jul 15, 2009 #10
    Re: acceleration

    Usually it is "y" times mass but here it increases to y^3 times as we move from transverse to longitudinal
     
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