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Why no absolute derivative in this example of geodesic deviation?

  1. Sep 8, 2014 #1
    On the surface of a unit sphere two cars are on the equator moving north with velocity [itex]v[/itex]. Their initial separation on the equator is [itex]d[/itex]. I've used the equation of geodesic deviation [tex]\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0.[/tex]to find their separation [itex]s=\xi^{\phi}[/itex] after time [itex]t[/itex]. I used [itex]\lambda=t[/itex] to give [tex]\frac{d{}^{2}\xi^{\mu}}{dt{}^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{dt}\frac{dx^{\gamma}}{dt}=0,[/tex]expanded out the Riemann components for a unit sphere to eventually get [tex]\xi^{\phi}=d\cos\left(vt\right),[/tex]which is correct. So far, so good. However, my problem is that in order to do the calculation I assumed the absolute second derivative [tex]\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}[/tex] can be replaced the the ordinary second derivative [tex]\frac{d{}^{2}\xi^{\mu}}{dt{}^{2}}.[/tex] This assumption works, but I don't understand why/how I can get away with it. In other words, why don't I need to calculate the absolute derivative using [tex]\frac{DV^{\alpha}}{D\lambda}=\frac{dV^{\alpha}}{d\lambda}+ V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda}.[/tex]
    Can anyone explain why the ordinary second derivative works in this calculation? Thanks.
     
  2. jcsd
  3. Sep 8, 2014 #2

    WannabeNewton

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    Could you post the details of your calculation? Attach it as a document or image if that's more convenient.
     
  4. Sep 9, 2014 #3
    Thanks, I've attached a pdf showing the calculation.
     

    Attached Files:

  5. Sep 9, 2014 #4
    You are just setting the problem in such a way that allows you to use geodesic normal local coordinates, that gives you the vanishing of the Christoffel symbols at the point you compute the distance, and thus you may use the ordinary derivative.
     
  6. Sep 10, 2014 #5
    What are geodesic normal local coordinates? Are spherical coordinates always geodesic normal local coordinates? I've tried Googling the term, and got something about "applying the exponential map to the tangent space at p". I'm OK with "tangent space" but have no idea what "exponential map" means.
     
  7. Sep 10, 2014 #6
    http://en.wikipedia.org/wiki/Normal_coordinates
    http://en.wikipedia.org/wiki/Fermi_coordinates


    No. It depends on the symmetries of the problem/space.
    http://en.wikipedia.org/wiki/Exponential_map see the Riemannian geometry part, under Properties.

    When you set up the problem so that the distance traveled by the car along the geodesic is a linear function of t, you are introducing a unit time(therefore you can use an orthonormal basis).
     
  8. Sep 10, 2014 #7
    Similarly in GR's geodesic deviation problems it is often possible to simplify the problem by working on linear inertial frames so that the Christoffel symbols vanish at a point or along a geodesic.
     
  9. Sep 10, 2014 #8
    Thanks, I need to think about this.

    I still can't see why the Christoffel symbols vanish along geodesics on the surface of a sphere. Surely the Christoffel symbols will only vanish if we're using a flat metric?
     
  10. Sep 10, 2014 #9
    It's a common misconception. It is the Riemann curvature(second derivatives of the metric) that will only vanish in a flat manifold. The Christoffel symbols(first derivatives of the metric) can be made to vanish at a point by the use of the appropriate coordinates(just like they can appear in a flat space with non-cartesian coordinates), and by the exponential map in the neighborhood of the point along the geodesic.
    It is not always guaranteed at all points of the curve, if this is possible they are called Fermi coordinates .
     
    Last edited: Sep 10, 2014
  11. Sep 10, 2014 #10
    Sorry to labour this, but how do you get from the Christoffel symbols for a unit sphere [tex]\Gamma_{\theta\phi}^{\phi}=\Gamma_{\phi\theta}^{\phi}=\frac{\cos\theta}{\mathbf{\mathbf{\sin\theta}}},\Gamma_{\phi\phi}^{\theta}= \sin\theta\cos\theta[/tex] to Christoffel symbols that vanish?

    I know that the Christoffel symbols are given by:
    [tex]\Gamma_{jk}^{i}=\frac{1}{2}g^{il}\left(\frac{\partial g_{lk}}{\partial x^{j}}+\frac{\partial g_{jl}}{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right).[/tex]How do I get those partial derivatives to vanish? Or, in other words, what is it about being on a geodesic that makes the Christoffel symbols equal zero?
     
  12. Sep 10, 2014 #11
    No, sorry I might have confused you in a previous answer, you need normalized coordinates, spherical coordinates are just orthogonal, that's what you did when you reparametrized the geodesic deviation equation with t.
     
  13. Sep 15, 2014 #12
    Thanks. I've been away for a few days, but still can't see this. In my example how are the new (normalized) coordinates related to the old ([itex]\theta,\phi[/itex]) coordinates? Also, the original line element is [tex]ds{}^{2}=d\theta^{2}+\sin^{2}\theta d\phi^{2}.[/tex]How would I find the new line element using the new coordinates? (I think/hope that if I can see the line element, I might have a chance of seeing how the connections coefficients vanish.)
     
  14. Oct 22, 2014 #13
    But surely (except at the equator) the distance between the geodesics is not measured along geodesics but along circles of constant [itex]\theta[/itex]. Therefore I can't be using geodesic coordinates?
     
  15. Oct 23, 2014 #14

    PeterDonis

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    Yes, it is. At least, the concept of "distance between geodesics" that enters into the equation of geodesic deviation is.
     
  16. Oct 23, 2014 #15

    TSny

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    Here's how I see it. Corrections welcomed!

    I think the confusion is related to the definition of the deviation vector ##\xi^{\mu}##. In a general space, a family of geodesics can be described by some functions ##x^{\mu}(\lambda; p)## where p denotes a particular geodesic and ## \lambda## can be taken as arc length along the geodesic. Two neighboring geodesics would be ##x^{\mu}(\lambda; p)## and ##x^{\mu}(\lambda; p+dp)##.

    For geodesics on the unit sphere that run perpendicular to the equator, you can use the spherical coordinate ##\theta## as arc length ##\lambda## and the spherical coordinate ##\phi## to pick out different geodesics. So, two neighboring geodesics would be ##x^{\mu}(\theta;\phi)## and ##x^{\mu}(\theta;\phi + d\phi)##. The deviation vector between the two neighboring geodesics is defined as $$\xi^{\mu} = \frac{\partial x^{\mu}(\lambda;p)}{\partial p} = \frac{\partial x^{\mu}(\theta;\phi)}{\partial \phi} $$ For example, see http://en.wikipedia.org/wiki/Geodesic_deviation (with modified notation).

    Moreover, if you use spherical coordinates on the unit sphere, then the coordinates ##x^{\mu}## are of course just the spherical coordinates ##\theta## and ##\phi##. Thus, the deviation vector reduces to something fairly trivial: ##\xi^{\theta} = \frac{\partial \theta}{\partial \phi} = 0## and ##\xi^{\phi} = \frac{\partial \phi}{\partial \phi} = 1##.

    This describes the geodesic deviation vector and there is no need to solve the geodesic deviation equation. However, you can check that this deviation vector does satisfy the geodesic deviation equation. But you will need to keep the ##\Gamma## terms in the covariant derivatives.

    Note that the deviation vector does not equal the distance of separation between two neighboring geodesics. The deviation vector is defined so that the coordinate difference between two points on neighboring geodesics (for the same ##\lambda##) is ##\delta x^{\mu} = \xi^{\mu} dp##. This is easily checked for our example where ##\delta x^{\theta} = \xi^{\theta} d\phi = 0 \cdot d\phi = 0## and ##\delta x^{\phi} = \xi^{\phi} d\phi = 1 \cdot d\phi = d\phi##.

    To get the distance ##ds## between two neighboring geodesics, use the metric $$ds^2 = g_{\mu \nu} \delta x^{\mu} \delta x^{\nu} = g_{\phi \phi} d\phi^2 = \sin^2\theta \; d\phi^2$$
     
    Last edited: Oct 23, 2014
  17. Oct 24, 2014 #16
    My physics is struggling here, so I'm going to resort to a classic “argument by authority”. My example of geodesic deviation on a sphere is similar to the CalTech Exercise 24.10 here:

    http://www.pma.caltech.edu/Courses/ph136/yr2004/0424.1.K.pdf

    The solution (Answer 23.10) is given here:

    http://www.pma.caltech.edu/Courses/ph136/yr2011/ps23_09.pdf

    In the CalTech problem the reference geodesic is not along constant ##\phi## geodesics, but along the equator (##\theta=\pi/2##). The solution certainly does involve using the equation of geodesic deviation, but because the connection coefficients vanish along the ##\theta=\pi/2## geodesic the absolute second derivative becomes the ordinary second derivative (CalTech's equation 67 in the solution). The answer to the CalTech problem is ##\xi^{\phi}=0## and ##\xi^{\theta}=b\cos\phi##, where ##b## is the initial separation. Note the similarity to my answer (##\xi^{\theta}=0##, ##\xi^{\phi}=d\sin\theta##) for constant ##\phi## geodesics, where ##d## is my initial separation. Wheeling out the very heavy guns, the same (CalTech) answer is given in Misner, Thorne and Wheeler in the Figure 10 text, page 31.

    The most convincing answer I've had to the question why the absolute second derivative equals the ordinary second derivative in my problem is Void's answer here:

    http://physics.stackexchange.com/qu...oes-the-ordinary-second-derivative-give-the-c

    though I can't honestly say I can confidently follow all of this explanation.
     
  18. Oct 24, 2014 #17

    PeterDonis

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    Applying the term "distance" to this can be misleading, because (except at ##\theta = \pi / 2##) it isn't evaluated along a geodesic curve. If, for example, we were to set up a local Euclidean coordinate chart centered on a point on the reference geodesic, a curve of constant ##\theta##, which is what this is being evaluated along, would appear curved, not straight.
     
  19. Oct 24, 2014 #18
    I forgot about this thread.
    So the part I guess I should have stressed but it seemed obvious is that the sphere is of course a surface of constant curvature, which means that the deviation of the geodesics is also constant. So this fact is what allows you to make the assumption about normalizing the spherical coordinates with a unit time and also computing a correct answer without using the Christoffel symbols.
     
  20. Oct 24, 2014 #19

    TSny

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    In the attachment of post #4, the geodesics are lines of constant ##\phi##. Along these geodesics the ##\Gamma##’s are not zero (with the usual ##g_{\theta \theta} = 1## and ##g_{\phi \phi} = \sin ^2 \theta## ). As you move along one of these geodesics, ##\frac{D^{2}\xi^{\mu}}{D\lambda^{2}} \neq \frac{d^{2}\xi^{\mu}}{d\lambda^{2}}##. Also, as mentioned in my previous post, the deviation vector ##\xi^{\mu}## for this case does not represent the distance between neighboring geodesics.

    In the link given in post #16, the equator is now taken as the “fiducial” geodesic. Along the equator the ##\Gamma##’s are all zero (but the derivatives with respect to ##\theta## of the ##\Gamma##’s evaluated on the equator are not all zero.) In this case you can show that ##\frac{D^{2}\xi^{\mu}}{D\lambda^{2}} = \frac{d^{2}\xi^{\mu}}{d\lambda^{2}}##. Also, for this case, you can show that the deviation vector does represent the distance between neighboring geodesics (at least up to a constant factor).

    The geodesic deviation equation applies to two neighboring geodesics. In the attachment to post #4, the two cars are apparently not infinitesimally close to each other. So, technically, the geodesic equation does not apply directly to the geodesics of the two cars. But since the sphere has the same intrinsic geometry at all points, it is not hard to see how to go from the distance between neighboring geodesics to the distance between finitely separated geodesics.
     
  21. Oct 24, 2014 #20

    TSny

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    Yes. But for two neighboring points whose coordinates differ by an infinitesimal amount, ##\delta x^{\mu}##, the distance between the points can be determined from ##ds^2 = g_{\mu \nu} \delta x^{\mu} \delta x^{\nu}##. You don't need to worry about whether the distance is measured "along a geodesic" here.

    For points separated by a finite distance, then the "distance between the points" needs to be interpreted. If you take this to mean the shortest distance on the sphere between the points, then it would be the distance along a geodesic connecting the points. But, of course, you could also measure distance along any curve between the points.

    For two points widely separated on a curve of constant ##\small \theta##, you could take the distance between the points as measured along the curve of constant ##\small \theta##. This might be a convenient way to define the distance even though it does not correspond to the shortest distance between the points.

    Suppose you have two geodesics such that each geodesic corresponds to a constant value of ##\small \phi## (i.e., two longitude lines) and you select a point on each geodesic corresponding to the same value of ##\small \theta## (i.e., two points on the same latitude). It might be convenient to define "the distance between the two longitude geodesics at that latitude" as the distance between the two points measured along the latitude line (rather than measured along a geodesic). [Of course, for some applications (e.g., airline flights!), this might not be a convenient definition.]
     
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