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The norm of four-acceleration

  1. Mar 20, 2012 #1
    I saw that the norm of four acceleration is equal to the magnitude of proper frame's acceleration.

    So, if the observer moves in x direction, following equation about norm of it's 4 acceleration is like that

    -(d^2 t / dτ^2) + (d^2 x / dτ^2) = d^2 x / dt^2

    In comoving frame(proper frame), proper time is equals to t, so the first term in the left side vanishes, and the remain term is equal to the right side.

    I tried to proof this relationship generally... but it's very hard for me.

    Now I confused whether the general proof exists, or not...
     
  2. jcsd
  3. Mar 20, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    johnahn, Here's the general relationship. It gets messy! If the 4-velocity is v = (γv, γc) so v·v = c2, then the 4-acceleration is

    a ≡ dv/dτ = (γ2a + γ4(v·a)v/c2, γ4(v·a)/c) where a is the 3-acceleration, a = dv/dt.

    The norm of this is a·a = -γ6a·a + γ6/c2[(v·a)2 - (v·v)(a·a)].

    In the special case that v and a are parallel, it reduces to |a| = γ3|a|.
     
  4. Mar 21, 2012 #3

    Dale

    Staff: Mentor

    The norm of the four-acceleration is invariant. So if you prove it in one frame then you have proved it in all frames.
     
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