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Four force giving me headaches

  1. Jul 20, 2012 #1
    I attempted to derive a fully expanded expression for the force four vector

    The four velocity being --> (γ,γv) (Only considering one spatial component here)
    The four acceleration is thus, (I've used κ for proper time as I cannot find the tau symbol)

    (∂γ/∂κ,∂γv/∂κ) --> γ(∂γ/∂t,∂γv/∂t)

    Now from here, it turns into a miasma of terms, since both v and γ and functions of the coordinate time (generally), we would have spatial velocity and acceleration mixed together and making it virtually impractical in calculations.
     
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  3. Jul 20, 2012 #2

    Bill_K

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    Heck, it's not all that bad. As you said, the 4-velocity is v = (γc, γv).

    The 4-acceleration is a = dv/dτ = (cγ dγ/dt, γ2a + γ dγ/dt v)
    Use dγ/dt = γ3 (v·a)/c2, so you get a = (γ4(v·a)/c, γ2a + γ4(v·a)/c2 v)

    The 4-force, of course, is m times the 4-acceleration.
     
    Last edited: Jul 20, 2012
  4. Jul 20, 2012 #3

    DrGreg

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    GarageDweller, if you are familiar with hyperbolic functions, how to differentiate them, and the hyperbolic versions of the trig identities, an easy approach (for the one-spatial-dimension case), is to substitute[tex]\begin{align}
    v &= c \tanh \phi \\
    \gamma &= \cosh \phi \\
    \gamma v &= c \sinh \phi
    \end{align}[/tex]
     
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