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4-velocity -> 4-acceleration?

  1. Oct 8, 2006 #1
    4-velocity --> 4-acceleration?

    Hey all, my first post here.

    Now, Im trying to derive the 4-acceleration from the 4-velocity by explicitly calculating du/d(tau). So from Hartle we have

    u=( gamma, gamma*v)

    where gamma=1/Sqrt(1-v^2)

    and v is the 3 velocity, taking c=1

    So then,

    du/d(tau)=( d(gamma)/d(tau), d(gamma*v)/d(tau) )

    from the book we have

    du/d(tau)=(gamma*(a dot v), gamma*a)

    so then we should have

    d(gamma)/d(tau)=gamma*(a dot v)

    but when i calculate d(gamma)/d(tau) i get....ok im going to show all my steps so people can know what ive done....

    d(gamma)/d(tau)=[d(gamma)/dv)]*[dv/dt]*[dt/d(tau)]

    =[d(1/(1-v^2)^(1/2))/dv]*[a][gamma]

    =[(-1/2)(-2*(v)/(1-v^2)^(3/2)]*[a]*[gamma]

    =[ va/(1-v^2)^(3/2)*[1/gamma]

    =v*a*gamma^4 != v*a*gamma

    does anyone know whats going wrong or what im doing wrong??

    Any help would be appreciated, Thanks in advance.
     
  2. jcsd
  3. Oct 9, 2006 #2

    pervect

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    Staff Emeritus
    Science Advisor

    You should expect that "proper acceleration", which can be defined as the magnitude of your 4-acceleration, is related to coordinate acceleration by a factor of gamma^3, not by a factor of gamma.

    See for instance http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

    (which has this result, but doesn't work through the math, unfortunately).

    But if we work through your example, we get for the two components of the 4-acceleration

    v*a*gamma^4, a*gamma^4

    Note that this makes the 4-acceleration vector Minknowski-perpendicular to the 4-velocity, this is another known result.

    Now when we take the squared magnitude of the above we get

    (1-v^2)*a^2*gamma^8 = a^2*gamma^6

    so the square root of this is a * gamma^3, i.e. the magnitude of the 4-acceleration is gamma^3 times the magnitude of the coordinate acceleration, which is what we expect.
     
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