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Synchronous Coordinates transformation

  1. Jun 13, 2013 #1
    Given a specific metric, is there a easy way to transform it in Synchronous coordinates?

    For example having dsigma2 = (1+z)^2 dt^2 - ds^2 - s^2 dphi^2 - dz^2 ,
    I was able to do some substitutions, but I had to stop at the differential equations presented in the attachement.
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2013 #2

    WannabeNewton

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    Is this for a homework problem? Or a project? There is no easy way in general. If you wish to explicitly construct a set of Synchronous coordinates (which I will hereafter call Gaussian normal coordinates since this is the common name but keep in mind that Synchronous and Gaussian normal coordinates are the same thing) on a neighborhood of a point in space-time, you will end up having to solve differential equations which may or may not be tractable.

    The point of Gaussian Normal coordinates is not the explicit construction but rather the existence theorem (which comes out of what is called Gauss' lemma: http://en.wikipedia.org/wiki/Gauss's_lemma_(Riemannian_geometry)). So given any event in space-time, we can always impose Gaussian normal coordinates on some neighborhood of the event and all we care about is that in these coordinates ##g_{00} = -1## and ##g_{0i} = 0## where ##i=1,2,3##. All the calculation power of Gaussian normal coordinates comes from this property so we really just care about the local existence of said coordinates for general space-time metrics. This is why I asked if you are doing this for a homework problem or a project or not. It really depends on what you want.

    Recall the way Gaussian normal coordinates are geometrically constructed. We start with an initial non-null hypersurface ##\Sigma_{0} ##, and a parameter ##t##. We take a family of geodesics normal to ##\Sigma_{0}## such that in some neighborhood of the event in question, they form a congruence ##\xi^{\mu}## in this neighborhood (meaning they never intersect). ##\xi^{\mu}## will have the property that it is normal to the family of hypersurfaces ##\Sigma_{t}## gotten by evolving the initial hypersurface ##\Sigma_{0}## using the parameter ##t##, where the evolution keeps going until we are out of that chosen neighborhood on which the family of geodesics remain non-intersecting. Using the parameter ##t## as the time coordinate for our metric, and using the fact that ##\xi^{\mu}## is normal to all the hypersufaces in the family (for as long as we don't terminate ##t##), we can put the metric in the form ##ds^{2} = -dt^{2} + h_{ij}dx^{i}dx^{j}## with the same labeling scheme as before. We can also define a quantity called the extrinsic curvature of the hypersurfaces (which measures how they bend in space-time) by ##K_{\mu\nu} = \nabla_{\mu}\xi_{\nu}##. One can show that in Gaussian normal coordinates, we have an evolution equation for the functions ##h_{ij}## given by ##\partial _{t}h_{ij} = 2K_{ij}## so if you can solve this set of differential equations you can determine the explicit form of the functions ##h_{ij}## up to some arbitrary functions but as you can probably tell this may or may not be tractable in practice.
     
    Last edited: Jun 13, 2013
  4. Jun 13, 2013 #3

    Mentz114

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    Excellent answer, again.

    Does ##K_{\mu\nu} = \nabla_{\mu}\xi_{\nu}## mean that in the absence of shear ##K_{\mu\nu}=0## ?
     
    Last edited: Jun 13, 2013
  5. Jun 13, 2013 #4

    WannabeNewton

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    Hi Mentz! Note that ##\xi^{a}## is hypersurface orthogonal, which by Frobenius' theorem implies ##\xi_{[a}\nabla_{b}\xi_{c]} = 0## which one can then use to show that the twist ##\omega_{ab}= 0## (geometrically, the lack of twisting is what allows us to use this foliation). Hence ##B_{ab} = B_{(ab)}## where ##B_{ab} = \nabla_{b}\xi_{a}## is the usual quantity from the kinematical decomposition. If the shear vanishes then ##B_{(ab)} = \frac{1}{3}h_{ab}\theta## where ##\theta = \nabla_{a}\xi^{a}## is the expansion. Since ##K_{ab} = B_{ba} = B_{(ab)}##, ##K_{ab}## vanishes identically iff ##\nabla_{a}\xi^{a} = 0## identically. This need not vanish in general.

    As a simple counter example, consider the flat Friedman universe given in synchronous coordinates by ##ds^{2} = -d\tau^{2} + a^{2}(\tau)\{dx^{2} + dy^{2} + dz^{2}\}## where the function ##a(\tau)## to be solved for is called the scale factor. The geodesic congruence here corresponds to a family of locally inertial, isotropic observers comoving with the hubble flow (the parameter ##\tau## is the proper time as read on a clock carried by any one of these observers) and is given by their 4-velocity field ##u^{a}##. When represented in synchronous coordinates, this is just ##u^{\mu} = \delta^{\mu}_{\tau}## and ##\nabla_{\mu}u^{\mu} = \Gamma ^{\mu}_{\mu\tau} = \partial_{\tau}\ln\sqrt{\left | g \right |}\neq 0## in general.
     
  6. Jun 13, 2013 #5

    Mentz114

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    I know that the twist must be zero if the hypersurfaces are orthogonal to ##\xi^a##, which is why I wanted to know about shear. But in my haste I forgot about the trace part of ##\nabla_{b}\xi_{a}##.

    Question answered, thanks.

    By 'counterexample' I presume you mean a case where ##\nabla_{b}\xi^{a}\ne 0##. I don't understand the last two equations, yet, but I do know this congruence has non-zero expansion.
     
  7. Jun 13, 2013 #6

    WannabeNewton

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    Yes by counter-example I meant a case where ##\nabla_{a}\xi^{a} \neq 0## and I should probably mention that in that last equation, ##g = \text{det}(g_{\mu\nu})## and the bars around ##g## denote absolute value.

    Anyways, if you have further questions about this feel free to pm me; I don't want things to derail too far from the OP's main question :smile:. Cheers!
     
  8. Jun 14, 2013 #7
    Thanks WannabeNewton!

    Im working with D-I junction conditions, and i need the induced metrics of two manifolds to be equal in order to "glue" them together. Having the metrics in gaussian coordinates seems to simplify the problem.
     
  9. Jun 14, 2013 #8

    WannabeNewton

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    Hi nicu! Well Gaussian coordinates do tend to be used sometimes (by which I mean I have seen it used sometimes-my experience with junction conditions is not nearly as extensive as yours) when working on junction conditions (although I'm not sure what D-I stands for here). They work very well with the problem of junction conditions in theory; however, it is true that in general if you wish to transform from some initial set of coordinates to Gaussian coordinates you will end up having to solve complicated PDEs (as mentioned above). You might be interested in this paper: http://cleibovitz.upwize.com/physic...ymmetric matter in co-moving co-ordinates.pdf
     
  10. Jun 14, 2013 #9
    That might work, Thanks!
     
  11. Jun 14, 2013 #10

    George Jones

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    In the original coordinates, what is the hypersurface on which you wish to apply glue?

    D-I = Darmois-Israel?
     
  12. Jun 14, 2013 #11
    I'm trying to connect minkowski ( cylindrical coordinates) to the metric presented above.
     
  13. Jun 14, 2013 #12

    George Jones

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    Along a hypersurface s = constant?
     
  14. Jun 14, 2013 #13
    s is constant, being the radius of a cylinder
     
  15. Jun 16, 2013 #14
    Say we have the situation described above (Minkowski glued with the space time described by the metric above( cylindrical coordinates)) in order define the hypersurface i need to provide intrinsic coordinates that describe it. Usually I would chose {tal , phi, z}. Is there a problem with choosing {s, phi, z} instead?


    tal- proper time
    s- radius of the cylinder
    phi - azimuthal angle
    z - height of the cylinder
     
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