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Fermi-Walker transport

  1. Jul 4, 2005 #1

    pervect

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    Imagine a non-rotating particle, one with zero angular momentum. Attach a tetrad of orthonormal vectors to the particle.

    Imagine that we accelerate the particle via a force through it's center of mass, so that it follows some curve through space-time parametreized as [itex]x^i(\tau)[/itex] while continuing to have zero angular momentu.

    In flat space the vectors "attached" to the particle all obey the Fermi-walker transport law.

    [tex]\frac{dv^a}{d\tau} = (u^a \wedge a^b ) v_b[/tex]

    here v is the vector to be transported (one of the orthonormal vectors attached to the particle), u is the 4-velocity of the particle[itex]{dx^i}/{d\tau}[/itex], and a is the 4-acceleration of the particle [itex]{d^2 x^i}/{d\tau^2}[/itex]. The symbol "^" represents the wedge product.

    My question is whether or not this expression works correctly in curved space-times and arbitrary coordinate systems, and if it doesn't, what the correct expression is. I.e. suppose we have a particle with zero intrinsic angular momentum that's orbiting a black hole, given that we know the path of the orbit and the associated 4-acceleration, can we use this formula to transport the particle's coordinate tetrad?
     
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  3. Jul 5, 2005 #2

    George Jones

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    First, some notation. Let [itex] \left\{ \mathbf{e}_\mu \right\} [/itex] be a tetrad along the worldline. I use [itex]\nabla_\mu := \nabla_{\mathbf{e}_\mu}[/itex] for covariant directional dervivatives that do not change the rank of scalar/vector/tensor.

    I think the transition to curved spacetime uses the partial dervivative goes to covariant derivative rule, which I think changes each [itex]d/d\tau[/itex] to the covariant directional derivative [itex]\nabla_{\mathbf{u}}[/itex], including the [itex]d/d\tau[/itex] used in [itex]\mathbf{a} = d \mathbf{u}/d\tau[/itex]. In flat spacetime,

    [tex]\frac{d}{d\tau} = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu} = u^\mu \frac{\partial}{\partial x^\mu}.[/tex]

    Now, for curved spacetime, replace each [itex]\frac{\partial}{\partial x^\mu}[/itex] by [itex]\nabla_\mu[/itex]. By linearity, [itex]u^\mu \nabla_\mu = \nabla_{\mathbf{u}}[/itex].

    Fermi-Walker transport in curved spacetime might be treated somewhere in MTW.

    Regards,
    George
     
  4. Jul 5, 2005 #3

    pervect

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    I finally found this approach in the front of Wald - it makes sense, it's just the prallel transport of a vector along a curve given by the condition

    [tex]
    t^b \nabla_b v^a = 0
    [/tex]

    where v is the vector being transported, and t is the tanget to the curve that's transporting it.

    In a coordinate basis, if the curve is paramaterized by [itex]\tau[/tex] we can re-write this as

    [tex]
    \frac{d v^a}{d\tau} + t^b \Gamma^a{}_{bc} v^c = 0
    [/tex]

    because
    [tex]
    \nabla_b v^a = \partial_b v^a + \Gamma^a{}_{bc} v^c
    [/tex]



    The result I quoted previously was from MTW's coverage of Fermi-Walker transport, which was in the section on accelrated observers. I'll have to study the two results to see why they work out to be the same, but for the problem I'm thinking about, Wald's approach is better.
     
  5. Jul 5, 2005 #4

    George Jones

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    While parallel transport and Fermi-Walker transport agree for geodesics, I don't think they necessarily agree along the worldlines of accelerated observers. Consider an accelerated observer. Because the observer is accelerated, [itex]\mathbf{a} = \nabla_{\mathbf{u}} \mathbf{u} \neq 0[/itex], and the observer's 4-velocity [itex]\mathbf{u}[/itex] is not parallel transported. Let [itex]p[/itex] and [itex]q[/itex] be events on the worldline of the observer such that [itex]\mathbf{u}_q[/itex] is not the parallel transport of [itex]\mathbf{u}_p[/itex]. Let [itex]\mathbf{u}'_q[/itex] be [itex]\mathbf{u}_p[/itex] parallel transported to [itex]q[/itex]. Since parallel transport preserves inner products, parallel transporting a spacelike vector orthogonal to [itex]\mathbf{u}_p[/itex] to [itex]q[/itex] results in a spacelike vector that is orthogonal to [itex]\mathbf{u}'_q[/itex], not (in general) to a vector that is orthogonal to [itex]\mathbf{u}_q[/itex]. Therefore, the tetrad carried along by an accelerated observer is not parallel transported. It is, however, Fermi-Walker transported. Torque is required to maintain parallel transport for accelerated observers.

    Regards,
    George
     
  6. Jul 6, 2005 #5

    pervect

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    Aha - that makes sense, you can parallel transport a vector along an arbitrary path (even a curve representing an accelerating object), but you won't wind up with your time vector as being one of the memebers of your tetrad if you do so.

    The case I'm interested in currently is a geodesic, though - I basically want to find the rate at which the fixed stars appear to rotate for someone with zero intrinsic angular momentum flying by a black hole at an arbitrary velocity.
     
  7. Jul 9, 2005 #6

    George Jones

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    This sounds very interesting. Something related is geodetic precession - the change in the spatial vectors in a tetrad that occurs for an observer that goes once around a circular orbit in Schwarzschild spacetime. Gravity Probe B is measuring this effect, as well as measuring the frame-dragging effect that gets more publicity.

    This thread has got me interested in having a look at Fermi-Walker transport for "hovering" observers in Schwarzschild spacetime.

    Regards,
    George
     
  8. Jul 12, 2005 #7

    pervect

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    I think I've got some results that look like they make sense for rotation, though if they are correct I've got another issue with some other results :-(.

    Anyway, I took a look at the metric (line element) representing a rotating frame of reference

    d[x]^2 + d[y]^2 + d[z]^2 + 2*w*y*d[x]*d[t] - 2*w*x*d[y]*d[t] + (w^2*(x^2+y^2)-1)*d[t]^2

    And I determined that the only non-zero Christoffel symbols were

    [tex]\Gamma^x{}_{ty}=\Gamma^x{}_{yt}=-\Gamma^y{}_{tx} = -\Gamma^y{}_{xt}=w
    [/tex]

    and
    [tex]\Gamma^x{}_{tt}=-w^2x \, \Gamma^y{}_{tt}= -w^2y[/tex]


    These correspond to the coriolis and centrifugal forces in the rotating frame. Since the metric is diagonal (even Minkowskian) at the origin (when x=y=0, the off-diagnoal terms dx*dt and dy*dt are both zero), I conclude that the rotation of a coordinate system can be determined from it's Christoffel symbols alone.

    The first 4 results can be interpreted in terms of parallel transport. They arise from parallel transporting a timelike vector along the x axis, and measuring the y-component of it's acceleration - i.e. from the Coriolis force.

    Furthermore, interestingly enough, all the components of the Riemann tensor vanished for the above rotating metric, at least according to GrtensorII. So apparently information about rotation is not contained in the Riemann (?). This somewhat surprised me.

    Thus it appears that the Christoffel symbols, and the Christoffel symbols alone, hold the information about the rotation of the coordinate system.

    If one has an orthonormal coordinate system, it appears that all one has to do to determine the rotation in the x-y plane is to calculate [itex]\Gamma^x{}_{ty}[/itex] and [itex]-\Gamma^y{}_{tx}[/itex], and if both of them are equal except for sign, this should be the value of the x-y component of rotation.

    Applying this to Newtonian gravity

    ds^2 = (1+2U)(dx^2+dy^2+dz^2) +(-1+2U)dt^2

    gave me the result that the magnitude of the angular frequency of rotation of a particle moving in the +y direction with a velocity of v at was approximately (ignoring small terms of order 1+U).

    [tex] -\frac{2v}{\sqrt{1-v^2}} \frac{dU}{dx}[/tex]

    Of course dU/dx is proportional (modulo some sign issues) to the acceleration of gravity, a_x. This appears at first glance to be reasonable, though I'm not sure how it compares with other calculations of geodetic precession yet.
     
  9. Jul 14, 2005 #8

    pervect

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    Grr - I didn't transcribe my formula for 'w' properly, and I can't edit my old post - the formula actually on my worksheet is

    [tex] w = -\frac{2v}{{1-v^2}} \frac{dU}{dx}[/tex]

    (no square root!), not what I posted.

    v in this formula is the y component of the velocity, which was dervied by setting the x-component of the velocity equal to zero. In this case the result for w, as seen above, is a function of the y-velocity (f(v)=-2v/(1-v^2)) multiplied by the x-acceleration. I would expect that there would be a similar term of opposite sign involving a function of the x-velocity multiplied by the y-acceleration in the more general case where the x-component of the velocity is not zero.
     
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