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Fermi-Walker transport - problem with a minus sign!

  1. Jul 26, 2005 #1
    I am looking at the Fermi-Walker transport of a tetrad transported by an observer in circular motion in Minkowski space. The 0-component of the tetrad should be the 4-velocity of the observer, which should therefore satisfy the FWT DE, but I'm finding that it is equal to the negative of what it should be.... Can anybody find the error in my derivation?

    Assume that in the inertial (Minkowski) lab frame the observer is seen to be orbiting with constant angular velocity [itex]\omega[/itex] at constant radius [itex]r[/itex]. Then the worldline of the observer is given by [itex]\mathcal{P}_0 = \left(\gamma\tau, r\cos(\gamma\omega\tau), r\sin(\gamma\omega\tau), 0)[/itex] where [itex]\tau[/itex] is the proper time of the observer and [itex]\gamma\equiv(1-r^2\omega^2)^{-1/2}[/itex]. I'm working in [itex]c=1[/itex] units with metric signature (+---). This lets us find the 4-vectors

    [tex]u^\mu = \partial_\tau\mathcal{P}_0 = \gamma\left(1, -\omega r\sin(\gamma\omega\tau), \omega r\cos(\gamma\omega\tau), 0\right)[/tex]

    [tex]a^\mu = \partial_\tau u^\mu = -\gamma^2\omega^2r\left(0, \cos(\gamma\omega\tau), \sin(\gamma\omega\tau), 0\right)[/tex]

    Now, we should have that [itex]u^\mu[/itex] satisfies the FWT DE such that [itex]\partial_\tau u^\mu = \left(u^\mu a^\nu - u^\nu a^\mu\right)u_\nu[/itex]. For simplicity, consider 1-component. Then looking at the two sides separately gives us

    [tex]\partial_\tau u^1 = -\gamma^2\omega^2 r\cos(\gamma\omega\tau)[/tex]


    [tex]\left(u^1 a^\nu - u^\nu a^1\right)u_\nu = \left(a^0u^1-a^1u^0\right)u^0 - \left(a^1u^1-a^1u^1\right)u^1 - \left(a^2u^1-a^1u^2\right)u^2 - \left(a^3u^1-a^1u^3\right)u^3[/tex]

    [tex] = \gamma^4\omega^2r\cos(\gamma\omega\tau) - \gamma^4\omega^4r^3\cos(\gamma\omega\tau) = \gamma^2\omega^2r\cos(\gamma\omega\tau) = -\partial_\tau u^1.[/tex]

    It's that final minus sign that shouldn't be there.... I've checked all 16 components of the tetrad and they are all yield the LHS equal to the negative of the RHS of the DE (on occaision because both sides are zero). Anybody have any ideas?

  2. jcsd
  3. Jul 26, 2005 #2

    George Jones

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    There seems to be a sign error here. On the right there is [itex]\partial_\tau u^\mu = a^\mu[/itex], while on the left, [itex]\left(u^\mu a^\nu - u^\nu a^\mu\right)u_\nu = u^\mu a^\nu u_\nu - a^\mu u^\nu u_\nu = u^\mu (0) - a^\mu (1) = - a^\mu[/itex].

    Last edited: Jul 26, 2005
  4. Jul 26, 2005 #3
    True enough... So I guess that solves my immediate problem, thanks! Seems to introduce a potential future one though.... This implies that the DE for Fermi-Walker transport which I've seen many places online, and in MTW, is dependent on the metric of the lab frame, is that right? So if I wanted to do this problem in a Schwarzschild metric for example, or even in Minkowski space with polar coordinates, I'd have to derive a different DE for the tetrad components to satisfy in order to Fermi-Walker transport them?

    Thanks again,
  5. Jul 28, 2005 #4


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    The sign convention strikes again!

    MTW assumes that [itex]u^a u_a=-1[/itex], which is true with a -+++ sign convention. However, the OP used a +--- sign convention.

    I don't know of a clean way of expressing fermi-walker transport in arbitrary sign conventions.
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