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Magnitude of 4-acceleration

  1. Sep 27, 2013 #1
    I know that if you go an accelerated observer's frame, the 4-acceleration is $$a^\mu=(0,{\bf a})$$ which means the magnitude is just $$-{\bf a}^2$$ which should be invariant. But I'm having a hard time showing this from the general expression for the 4-acceleration. I get to $$a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$ But I don't know where to go from here. Any ideas or am I just wrong somehow?

  2. jcsd
  3. Sep 27, 2013 #2


    Staff: Mentor

    What general expression are you starting from?
  4. Sep 27, 2013 #3
  5. Sep 27, 2013 #4


    Staff: Mentor

    Yes, that will work. How are you getting from that to the second expression you gave in your OP?
  6. Sep 27, 2013 #5
    I don't think the three vector a is the same in the third equation as in the first and second equations. If you want to be working with the same 4 acceleration, you have to apply the Lorentz Transformation to its components.

  7. Sep 27, 2013 #6


    User Avatar
    Science Advisor

    He's right, you know! :wink: In the first and second equation, a is the acceleration 3-vector in the instantaneous rest frame. In the third equation, a is the acceleration 3-vector in an arbitrary rest frame. They agree if you set v = 0 and γ = 1, but not otherwise.
  8. Sep 27, 2013 #7
    I was under the impression [itex]{\bf a}[/itex] was an arbitrary acceleration 3-vector either way? There's nothing in the Wikipedia article that puts any constraints on [itex]{\bf a}[/itex].

    This is the way I went about it:

    The 4-acceleration is: $$a^\mu=\left(\gamma^4\frac{{\bf v}\cdot{\bf a}}c,\gamma^2{\bf a}+\gamma^4\frac{\bf v\cdot a}{c^2}{\bf v}\right).$$

    The magnitude should be given by $$a_\mu a^\mu=\gamma^8\frac{\left|{\bf v\cdot a}\right|^2}{c^2}-\gamma^4|{\bf a}|^2-\gamma^8\frac{|{\bf v\cdot a}|^2}{c^4}|{\bf v}|^2-2\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$

    The first and third terms can be combined to give $$\left(1-\frac{|{\bf v}|^2}{c^2}\right)\gamma^8\frac{\left|{\bf v\cdot a}\right|^2}{c^2}=\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$ So we now have $$a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.$$
  9. Sep 27, 2013 #8
    Oh I think I see what you're saying. So then, is my answer for [itex]a_\mu a^\mu[/itex] the general expression for the norm? This is what I've been trying to find. If so, it's no wonder professors don't spend much time talking about it---it's hideous! :-)
  10. Sep 27, 2013 #9
    You can still improve it as
    [tex]a_\mu a^\mu=-\gamma^4|{\bf a}|^2-\gamma^6\frac{|{\bf v\cdot a}|^2}{c^2}.[/tex]
    [tex]a_\mu a^\mu=-\gamma^6[\gamma^{-2}|{\bf a}|^2-\frac{|{\bf v\cdot a}|^2}{c^2}].[/tex]
    [tex]a_\mu a^\mu=-\gamma^6[(1-\frac{|{\bf v}|^2}{c^2})|{\bf a}|^2-\frac{|{\bf v}|^2|{\bf a}|^2(cos \theta)^2}{c^2}].[/tex]
    [tex]a_\mu a^\mu=-\gamma^6 |{\bf a}|^2[1-\frac{|{\bf v}|^2}{c^2}(1+ (cos \theta)^2)].[/tex]

    That is, if that is an improvement...
    Last edited: Sep 27, 2013
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