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Components of the 4-acceleration

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    I read in my textbook that "in the instantanuous reference frame (IRF) [tex][u^{'\gamma}]=(c,\overline{0}) [/tex] imply that the components of the 4-acceleration in the IRF are [tex] [a^{'\gamma}]=(0,\overline{a'}) [/tex]"

    2. Relevant equations

    3. The attempt at a solution
    I don't know why they got a' for the space-like components of the 4-acceleration. Isn't the accelaration = du/d[tex]\tau[/tex] if so then using [tex][u^{'\gamma}=(c,\overline{0}) [/tex] we get zero for both the time-like and space-like components of the 4-acceleration.
    Last edited: Nov 25, 2007
  2. jcsd
  3. Nov 26, 2007 #2

    George Jones

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    Think back to first-year calculus, and suppose [itex]f\left(x\right) = x^3[/itex]. Then, [itex]f\left(2\right) = 8[/itex], and [itex]8[/itex] is a constant, so

    [tex]\frac{d}{dx}8 = 0,[/tex]


    [tex]\frac{df}{dx}\left(2) = 12 \ne 0.[/tex]

    Just because the 4-velocity has a particular value (which could have zero spatial part) in a particular frame at a particular instant doesn't mean that the 4-velocity is constant. If the 4-velocity is not constant, then the 4-acceleration doesn't have to be zero.

    Throw a ball in the air. When the ball reaches its greatest height, its (spatial) velocity is zero, but its (spatial) acceleration still has magnitude [itex]g[/itex]
  4. Nov 26, 2007 #3


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    In addition, note that the 4-acceleration of a particle is orthogonal to its 4-velocity. (Why?)
    Last edited: Nov 26, 2007
  5. Nov 27, 2007 #4
    That example was very helpful. thanks a lot.

    Thanks robphy too.
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