Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Implications of the tensorial form of the EFE

  1. Jul 7, 2015 #1
    The EFE are tensor equations in 4-dimensional spacetime and by virtue of their tensorial form indepedence from the choice of coordinate system is guaranteed, and the same goes for the metric tensor solutions.
    When looking for assumptions that help simplify the process of solving the EFE to find a metric tensor, for instance in cosmology, one usually looks for symmetries like for instance spherical symmetry.
    But can the simplifying assumption for obtaining a solution metric be a coordinate dependent condition?
     
  2. jcsd
  3. Jul 7, 2015 #2

    Mentz114

    User Avatar
    Gold Member

    I think in the case of spherical symmetry there is a topological defintion which is certainly coordinate independent.
     
  4. Jul 7, 2015 #3

    WannabeNewton

    User Avatar
    Science Advisor

    No. The simpliyfying assumptions are always physical. One then chooses a coordinate system which is best adapted to said assumptions i.e. in which the assumptions are manifest. For example spherical symmetry is a statement about the SO(3) Killing fields and is made manifest in a coordinate system in which the metric contains the 2-sphere metric; for the Taub-NUT solution the fact that one has translational symmetry in the plane of the mass distribution can be made manifest in a coordinate system for which the metric is constant on the plane.

    This isn't anything special to Einstein's equations. This is a basic calculation technique used throughout physics, starting from simple classical mechanics problems like a sphere rolling down a moving inclined plane or a rotating spring-mass system: we have a set of simplifying physical assumptions and choose a coordinate system or reference frame that incorporates these assumptions so as to make the problem simpler.
     
  5. Jul 7, 2015 #4
    Hi, WbN.

    Right, so maybe you could help me understand how this works in different scenarios.
    I can see how a simmetry like spherical symmetry is a coordinate independent assumption when it comes together with other assumptions like global staticity or when solving the EFE in vacuum.
    But in the case of cosmological models that use the cosmological principle as assumption for a solution of the EFE when being a static spacetime is not demanded, I can't see how an assumption like the cosmological principle that refers to coordinate spacelike hypersurfaces and a time coordinate that must preserve a certain form dictated by the coordinate hypersurfaces can be a coordinate independent assumption for a 4-dimensional solution.
     
  6. Jul 7, 2015 #5

    WannabeNewton

    User Avatar
    Science Advisor

    That's because a lot of books state the assumptions of homogeneity and isotropy in coordinate-dependent terms, much like how they state that spherical symmetry means even parity under ##\theta \rightarrow -\theta, \phi \rightarrow -\phi##. But isotropy and homogeneity are geometric properties of the space-time relative to some congruence and can be defined in a purely coordinate-independent manner.

    First we assume that on the space-time ##(M,g_{ab})## we have a time-like vector field ##\xi^a## such that ##\xi_{[a}\nabla_b \xi_{c]} = 0## so that there exists a one-parameter family of space-like hypersurfaces ##\Sigma_t## foliated by ##\xi^a##. ##M## is then homogenous and isotropic as measured by the observers in ##\xi^a## if, ##\forall p \in M##, and all unit space-like vectors ##n^a_1, n^a_2 \in T_p M## orthogonal to ##\xi^a## (i.e. direction vectors relative to the observers), there exists some open set ##U \subset M## and an isometry ##\varphi: U \rightarrow U## such that ##\varphi(p) = p##, ##\varphi_* (\xi^a) = \xi^a##, and ##\varphi_*(n^a_1) = n^a_2##.

    From here we can then pick a coordinate system adapted to these symmetries; this will be the usual FRW coordinates. C.f. Wald section 5.1.
     
  7. Jul 7, 2015 #6

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    To add to #5, flat spacetime is a degenerate solution for almost any symmetry assumption because it contains most every symmetry. Thus, the SC metric for M=0 is flat spacetime. The FLRWL metric for total mass of dust=0 and no cosmological constant is flat spacetime (with funny coordinates compared to standard ones). But for the non-flat cases, the SC manifold is completely different from an FLRW manifold.
     
  8. Jul 7, 2015 #7
    This is the specific assumption I am referring to, it assumes a certain congruence just to select a time coordinate that favors certain 3D coordinate hypersurfaces with specific features to obtain a 4D solution. So the only motivation for this assumption seems to be coordinate dependent.
    As previously commented this would not be an issue if the solution was also assumed to be static as the hypersurface orthogonality is in the definition of such spacetimes.
     
  9. Jul 7, 2015 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Hypersurface orthogonality is not a property of a spacetime; it's a property of a congruence of worldlines. Static spacetimes contain such a congruence, by definition; but they are not the only spacetimes that do. The congruence of "comoving" worldlines in FRW spacetime is also hypersurface orthogonal, and that fact is a coordinate-independent feature of the geometry. That's the plain English version of what WannabeNewton said, more precisely, with math.
     
  10. Jul 7, 2015 #9
    Static spacetimes contain that congruence by definition because there
    is no distinction between spacelike hypersurfaces for diffferent time coordinate slices, so the congruence is certainly coordinate independent. This is not the case for non- static spacetimes, in these the congruence must be assumed motivated by the 3D coordinate hypersurfaces desired features of isotropicity and homogeneity.
     
  11. Jul 7, 2015 #10
    The metric solution is, in general, coordinate independent, in any spacetime arrangement. I think you meant to ask, "Can the simplifying assumption for obtaining a coordinate metric solution be a coordinate dependent condition?"
     
  12. Jul 7, 2015 #11

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    That property (all spacelike hypersurfaces orthogonal to the congruence being identical) means the congruence is a Killing congruence (the set of integral curves of a Killing vector field). The fact that a timelike Killing congruence exists and is hypersurface orthogonal is what defines a static spacetime. But the property of being a Killing congruence is not necessary for the congruence to exist or for it to be coordinate independent. Any congruence of worldlines is a coordinate-independent, geometric feature of a spacetime.

    Those features aren't "desired" features; the fact that they are features of particular spacetimes has nothing to do with anyone's desires. Nor does the fact that the congruences, and the hypersurfaces they are orthogonal to, are coordinate-independent, geometric features of the spacetime.
     
  13. Jul 8, 2015 #12
    Sure, but the key point here if you take note of it is that here that geometric feature has to be assumed specifically, it is not something imposed by the EFE. And the reason to impose it is a purely coordinate condition of the 3D slices.


    This sounds like a purely semantic quibble about the word "desire" that misses the bottom of the issue. It is quite clear that the assumption of the congruence is made in order to satisfy certain features of the coordinate 3D hypersurfaces, if we didn't want to have isotropic and homogeneous 3D submanifolds we wouldn't add the assumption of the congruence to restrict the 4D solutions of the EFE as this specific congruence is not a feature of the solutions of the EFE in general. This "desired" requirement that determines the presence of the congruence is obviously not whimsical, it comes dictated by what we observe in the 3D submanifold, but it seems like 4 dimensional manifolds shouldn't be determined by coordinate submanifold properties, should they?

    Again if we wanted a static spacetime as additional assumption for restricting solutions of the EFE, the congruence would come inmediately and requiring isotropicity and homogeneity of the solutions would be indeed a coordinate independent demand as we would already have a congruence by virtue of the static assumption.
    Of course one should maybe ask if the property of staticity is actually coordinate independent. As an example in de Sitter spacetime one can have patches that are either static or expanding depending on the chosen coordinates.
    As another example Minkowski spacetime, depending on the choice of 3D coordinate spacelike hypersurface, euclidean or hyperbolic, we obtain a different time coordinate, the latter an FRW time coordinate that gives an expanding solution of the EFE in vacuum, in the former 3D euclidean hypersurface case gives a purely static solution.
     
    Last edited: Jul 8, 2015
  14. Jul 8, 2015 #13
    Yes, that should be true in general, my question is whether the FRW coordinates are a coordinate dependent metric solution precisely because it uses a coordinate dependent condition.
    That is, it seems like the specific congruence(which is per se a geometric coordinate independent property in 4D) of fundamental observers is imposed on the solutions of the EFE due to the condition of the cosmological principle, which is a purely coordinate hypersurface condition of the 3D submanifolds.
     
    Last edited: Jul 8, 2015
  15. Jul 8, 2015 #14

    Nugatory

    User Avatar

    Staff: Mentor

    Your phrasing is a distinct improvement, I think. In any case, the answer to that question is "No, but the form of the ansatz based on the simplifying assumption will be much more obvious in some coordinate systems".
     
  16. Jul 8, 2015 #15
    The problem is that if you use any coordinates other than the FRW coordinates, not only it is less obvious, you simply have no ansatz,i.e. yo still have a solution of the EFE but you lose the congruence of fundamental observers and of course you can forget about the cosmological principle. How is that assumption not coordinate dependent?

    Compare with the case commented above about Minkoski spacetime, whatever the coordinates you use the congruence is preserved.
    Or look at the (exterior) Schwarzschild solution, no matter what coordinates you use the timelike hypersurface orthogonal congruence is also preserved.
     
  17. Jul 8, 2015 #16

    martinbn

    User Avatar
    Science Advisor

    This doesn't make any sense. You don't lose the congruence, it is a family of curves in the manifold, it is independent of any coordinates you use or not.
     
  18. Jul 8, 2015 #17

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No geometric feature is "imposed by the EFE", except in the obvious sense that solving the EFE tells you the geometry; in that sense every geometric feature is "imposed by the EFE", including homogeneity and isotropy in FRW spacetime.

    The only difference of interest for this discussion between FRW spacetime and Schwarzschild spacetime is that in the latter, the congruence of timelike worldliness that is hypersurface orthogonal is a Killing congruence, whereas in the former it isn't. That is an invariant geometric difference; it's not coordinate-dependent, and it doesn't have to be "assumed specifically" in one case any more than the other.

    The congruence in FRW spacetime isn't assumed; it's discovered in the course of solving the EFE under the assumption of spatial homogeneity and isotropy.

    They aren't "coordinate 3D hypersurfaces"; they are spacelike surfaces that are homogeneous and isotropic. That's a coordinate-independent geometric property. The coordinates get chosen afterwards: once you specify that you are looking for a solution to the EFE that has the geometric feature that it is foliated by a family of spacelike hypersurfaces with a certain coordinate-independent, geometric property, you can then prove (not assume) that you can choose a set of coordinates such that the line element assumes a certain simple form. That makes the solution of the EFE a lot easier. In the course of doing this, you can also prove (not assume) that there is a congruence of timelike worldliness that is hypersurface orthogonal to that family of spacelike hypersurfaces. But none of that makes either the hypersurfaces or the congruence coordinate-dependent. The coordinates depend on the hypersurfaces, not the other way around.

    Only because "static" by definition means a timelike Killing congruence exists. But note the "Killing" property, which is an extra (coordinate-independent, geometric) feature that is not present in FRW spacetime. So, as I said above, the key difference between these two cases is whether or not the congruence is a Killing congruence, not whether or not the congruence is "coordinate-dependent"; neither one is coordinate-dependent.

    This makes no sense at all to me. You appear to be very confused about the role coordinates play in solving the EFE.

    Of course it is; the presence of a timelike Killing congruence which is hypersurface orthogonal is a coordinate-independent, geometric property.

    No, de Sitter spacetime (more precisely, the region inside the cosmological horizon) is static regardless of what coordinates you use. The expression describing the hypersurface orthogonal timelike Killing congruence will be more complicated in some coordinates than in others, but that has nothing to do with whether it's a Killing congruence or not.

    Again, the hypersurface orthogonal timelike Killing congruence is always there, it just looks more complicated in some coordinates than in others. You appear to be very confused about what "coordinate independent" actually means.
     
  19. Jul 8, 2015 #18

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Coordinates aren't the metric. You can describe FRW spacetime in any coordinates you like. The spacetime, including its family of homogeneous, isotropic hypersurfaces with a congruence of timelike worldliness orthogonal to it, remains unchanged. The expression for the line element in coordinates other than FRW coordinates will look more complicated, but that won't change any of the geometric properties.

    For example, a while back we had a thread on Painleve coordinates for FRW spacetime:

    https://www.physicsforums.com/threads/painleve-chart-for-frw-spacetime.556325/#post-3647315

    Note that the "comoving" worldlines are still there (I give the equation for them), and are still orthogonal to a family of spacelike hypersurfaces that are homogeneous and isotropic (I don't give the equation for what those look like in Painleve coordinates in the thread, but it should be easy to work out). The equations describing these things don't look as simple, but they're still there.
     
  20. Jul 8, 2015 #19

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This is not correct. See my example of Painleve coordinates on FRW spacetime; the congruence of "comoving" observers is still there, their worldlines just look more complicated (because they don't stay at the same spatial coordinates). The cosmological principle still applies; the spatial hypersurfaces that are homogeneous and isotropic are still there. They just aren't surfaces of constant coordinate time in Painleve coordinates.
     
  21. Jul 8, 2015 #20
    I wrote any transformation but actually didn't mean any and was late to edit it. Certain coordinate transformations that don't preserve the form of the FRW time coordinate clearly lose the congruence, like those for wich matter is not homogeneous and isotropic for a given spacelike hypersurface.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Implications of the tensorial form of the EFE
  1. EFE simulation (Replies: 10)

  2. Geometry of the EFE (Replies: 8)

  3. Solution to efe (Replies: 3)

  4. Tests of EFE (Replies: 32)

Loading...