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

General relativity mathematics : physically acceptable ?

  1. Apr 24, 2005 #1
    I saw a derivation of the Einstein gravitational field equation based on the following :

    1) gravitational field equivalent to acceleration => curvature (due to local lorentz contraction)

    2) gravitational field is equiv. to energy => curvature has to be linked non-linearly to the energy, which is the 00 element of the stress-energy tensor.

    Einstein then uses the work of Cartan-Riemann about the curvature tensor, and the Ricci tensor R_ij.

    However, the obtention of the curvature tensor is based on the use of closed loops (for the use of the Stokes theorem in the version I saw) :

    the curvature is defined with the parallel displacement of a vector along a closed loop.

    My question is : Is this physically acceptable, since in GR a loop means a loop in space-time, hence coming back in time ???

    This is then not surprising that Goedel found solutions to Einstein's equations in which timelike loops exist.
  2. jcsd
  3. Apr 24, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    A technical note here - acceleration = "curvature" only in a lose sense - the metric coefficients in an accelerated frame cannot be constant. Acceleration does not cause the Riemann curvature tensor to become non-zero in a uniformly accelerated frame - in fact, one can calculate it and show that it is zero.

    This is physically acceptable by interpreting some of the vectors as paths that multiple observers take when travelling through space-time. These paths will naturally involve going forwards in time. Rather than take a single observer through a closed loop, one takes multiple observers through different paths, winding up at the same point, where the multiple-observers compare their clocks and their velocities (the velocity being the direction of their time vector).
  4. Apr 24, 2005 #3
    Right for 1) I should have written : a gravitational field is equivalent to a LOCAL accelerated frame (the acceleration is different in every space-time point).

    For the question : You mean the physical loops have to be seen as a continuous set of observers put along a mathematical loop ?

    Then I'll ask a mathematical question about GR :

    For the derivation of the equation of motion in a curved space-time, it is shown that those are the geodesics of the curved space-time.

    To show this, Einstein uses the following : locally the space-time is flat, hence the local frame is inertial, and Newtons law is valid : the particles move along straight lines locally corresponding to the geodesics of the flat space-time.

    He then says the local coordinates at point p are linked to the global coordinate system via a local coordinate transformation :

    [tex] \xi_p=\xi_p(x^\nu) [/tex]

    where [tex] x^\nu [/tex] are the coordinate in the global frame and the [tex] \xi_p [/tex] the coordinate in the local inertial frame at point p.

    Now : in the global coordinate system, space-time is curved. In the local one, space-time is flat. Note here that the curvature in GR is intrinsic (not like a cylinder which has only an extrinsic curvature, but which is in fact equivalent to a flat space).

    How is then the transformation global<->local possible, since the intrinsic curvature is a quantity invariant under change of coordinates.

    If I have a curved space, for example a sphere, I can change the internal coordinates (theta,phi) as I want, the curvature will remain. If I have a flat space-time, I can change the coordinates as I want, the space-time so parametrized will remain flat.

    How can then he obtain the right equations of geodesics using those special coordinate transformations from the local intertial frame ? (Since those do not exist)
  5. Apr 24, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Because the space-time manifold is locally euclidean (a differentiable manifold)...Let's see,when u mean intrinsic curvature & its "invariance",i think u mean the Ricci scalar,right...?

    Is that so...? What is the necessary & sufficient condition that a space-time be flat...?

    What does not exist?The vierbeins...?

  6. Apr 24, 2005 #5
    Well, I supposed space-time was locally a Minkowski space-time, let us say diag(1,-1,-1,-1) to fix the signs. However, in the Schwarzschild solution, if you travel inside the Schwazschild radius, then the 0-th coordinate transforms from time to space, and the 1st one from space to time....

    I don't know much about the whole GR maths, but let's simplify to a 2-dimensional manifold. In this case, the intrinsic curvatures are equivalent to the principal curvatures. The Gaussian curvature, if I remember well, is the product of both. Then by the Theorema Egregium (from the same guy), it is invariant under reparametrization.

    Just take the Ricci tensor...then I suppose it's obvious that all the invariant under change of basis of the tensor are invariant under coordinate transformation..which is obvious since a change of coordinate induces a local basis change of the tensor. So that there exist invariants of the Ricci tensor under coordinate transformations.

    No, the local coordinate transformation do not exist, since by the Theorema Egregium, the Gaussian curvature is invariant under any coordinate transformation and that GR affirms : in the global frame it's non-zero, whereas in the local one it is zero and that both frames are linke by a change of coordinate (I just speak about the more simple 2d manifold so that we can visiualize it).
  7. Apr 24, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    Well,u started the discussion on mathematics.I'm not really into phenomenology,so this maths stuff is much easier for me.

    Okay.U meant the Ricci scalar.It's a scalar,which means it's invariant to reparametrization (arbitrary change of noniertial reference frames).

    The tensor are invariants (Riemann,Ricci,Weyl),but their components can be changed,because basis in the tangent & cotangent bundle change...

    U're claiming that [tex] R_{mn}\ ^{pq} [/tex] and its contractions are 0...?If so,guess again.I suggest some reading,too.

  8. Apr 24, 2005 #7
    Ok, let's take the Ricci scalar. If I remember well, the Ricci scalar is obtained from some contraction (don't ask me which one now), of the Riemann tensor. (I suppose it is not the only scalar you can build out of the Riemann tensor, that's what I mean with other invariants of the tensor).

    You agree that any change of coordinate, starting from a space-time with a given Ricci scalar for example (or any other invariant), will leave it invariant (by definition in fact).

    So how can we link the local coordinates, in which this scalar is zero, to the global coordinates, in which it's non-zero ? This transformation of coordinates is what Einstein calls "local coordinate transformation", in the sense that at every point, the transformation of coordinate is different.

    For example you have y=cos(x) locally at x=0
    and y=sin(x) locally at x=1,

    which can be summarized in the 1d case as : y=f(x,x).

    In the 4D case, he writes : [tex] \xi_p=\xi_p(x^\nu) [/tex]. Where the index p is very important. If you leave it, you cannot change the value of the Ricci scalar for example....

    Is this correct ?
  9. Apr 24, 2005 #8
    To be precise it is tidal accelerationn that is equivalent to spacetime curvature. And this has nothing to do with local lorentz contraction (whatever that is..).
    No. In fact its quite feasable to have a gravitational field where there is no energy whatsoever. By this I mean that "matter at spatial point A does not imply curvature at spatial point B". What is true is that if, at event P, there is matter (energy-momentum tensor does not vanish) then the Riemann tensor will not be zero

    What does "obtention" mean?

    There are various ways to introduce the Riemann tensor. One is through parallel transport (where you use those loops of yours) and then there is geodesic deviation. E.g.


    That is not clear since you haven't elaborated by what you mean by "physical". In the case you use with your loops it does not require that it be possible to move on such a worldline.

    What is it you mean by "link" anyway???????

  10. Apr 25, 2005 #9


    User Avatar
    Staff Emeritus
    Science Advisor

    This works for the same reason that the a sphere (say the Earth's surface) looks flat if you take a very small piece of it. This isn't unique to a sphere, BTW, it works for any manifold.

    Curvature depends on the higher order derivatives of the metric coefficients. Over a small enough piece of the manifold (space-time for GR), the metric coefficeints do not vary appreciably, and you can calculate distances just as if you were in a flat tangent space to the manifold (a locally Lorentzian space-time for GR).

    For instance, if you calculate the circumference of a circle of radius delta-r, or the area of said circle, you find that the error will go down faster than linearly as you shrink delta-r. Therefore in the limit as delta-r goes to zero, the circumference becomes 2*pi*r for any 2d manifold (you can generalize this appropriately for higher dimensional "circles") and the area becomes pi*r^2.
  11. Apr 25, 2005 #10
    Ok, so why is the tidal acceleration linked to the curvature ?

    (You will see that this can be explained intuitively by local lorentz transfromation. And BTW the schwarzschild metric is obtained in 2 lines with this approach, instead of solving the complete field equation)

    1) If a material point is set in A, it curves spacetime everywhere
    2) The fact that the field equation is non-linear is hiding the following : the gravitational field itself contains energy, but this energy has nothing todo with the stress-energy tensor, so that nontrivial solution in the vacuum of stress-energy is possible

    How do you parallel transport your vector then ? This what I mean is only a mathematical trick that has nothing to do with physics.

    A link between the intrisic coordinate of the curved space (the [tex] \xi_p(x^\nu)[/tex] above.

    Look at a sphere and the tangent plane to it.....Einstein always insisted on the fact that the geometry has to be intrinsic...so there is absolutely no way to transform a delta_x on the sphere as small as you want, to the tangent plane, since this involves the embedding space (in this case R^3), which he asbolutely wanted to avoid. (You transform the coordinates used to localize yourself on the sphere to other ones, so basically you remain on the sphere, so that the curvature cannot vanish by this way).

    But this has nothing to do with as small as you want, the fact is that the Theorema Egregium simply forbids the modification of curvature via intrinsic coordinates transformations.
  12. Apr 25, 2005 #11
    That is incorrect. E.g. consider my favorite example; Its quite possible to have a sheet of matter for which there is a gravitational field outside the sheet and yet the spacetime is not curved, its flat (known as a vacuum domain wall).

  13. Apr 25, 2005 #12
    Well this is because you don't take 1 point (but an infinity of points that cancel curvature mutually (It's like a homogeneous field)).....

    The fact I speak about one point is simply seen in the Schwarzschild metric : you can put the whole mass at r=0, and space-time is curved over all r in [0;infinity]
  14. Apr 25, 2005 #13
    That is not "like" a homogeneous field, it is a homogeneous field. You originally inquired about Einstein's field equations. You didn't state that you were speaking about one single case, i.e. a single point. Note: Regarding the Schwarzschild solution the Ricci tensor and Ricci scalar vanish everywhere except where the material point is.

    I had forgot about the derivation I did which is located here -

    You may find it of use.

  15. Apr 25, 2005 #14
    If the Ricci scalar vanishes, does this mean the space is flat ?
  16. Apr 25, 2005 #15


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    "Flat" requires the entire Riemann curvature tensor is zero in that region.
    Note that Riemann can be decomposed into a Ricci part and a (traceless) Weyl part. When the Ricci scalar is zero, there is still the Weyl part of the Riemann tensor.
  17. Apr 25, 2005 #16
    No. As I indicated above, in the case you mentioned the Riemann tensor does not vanish but the Ricci tensor does as does the Ricci scalar.

    Last edited: Apr 26, 2005
  18. Apr 26, 2005 #17
    Ok, then I have a last question about the consistency of GR :

    1) starting from the equivalence principle : take a spherical grav. field, then by the principle, this is equivalent to a local accelerating frame. If you apply the corresponding local lorentz transformation, then the space-time has to have a time dependent curvature, since the equivalent speed is changing with time (because the local equivalent frame is accelerating). (And it is known experimentally that space-time curvature is time dependent because of the expansion)

    2) However, from Birkhoff theorem, it is known the a spherical grav. field (no charge nor rotation, and in the outer empty space) has to be static (time independent).

    How can you conciliate both ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook