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

The meaning of Weyl curvature caused by gravitational waves

  1. Jul 5, 2012 #1


    User Avatar
    Gold Member

    In his article The Ricci and Weyl Tensors John Baez states that the tidal stretching and squashing caused by gravitational waves would not change the volume as there is 'only' Weyl- but no Ricci-curvature. No additional meaning is mentioned.

    But, beeing not an expert I still have no good understanding, what Weyl-curvature really means and would appreciate any help.

    If I think of spacetime curvature I have effects like Shapiro-Delay/time dilation/the sum of light ray triangles etc. in my mind. But it seems that the Weyl curvature is not responsible for anything else than the tidal effects happening in the x-y-plane, the transverse plane of the wave. Is that right? Perhaps it is sufficient to say, a plane is not curved. The MTW talkes about the plane-wave solution.

    Otherwise gravitational wave measurements should be obscured by Shapiro-Delay/time dilation to a certain extent.

    But on the other side and this puzzles me, gravitational waves are called "ripples of spacetime curvature". From this I would expect the spacetime curvature to oszillate locally between positive and negative values as the wave passes by, which should be measurable by light ray triangles, e.g. However is that right? And if so, in which plane? In the transverse plane or in the plane in which the wave propagates?
  2. jcsd
  3. Jul 5, 2012 #2


    User Avatar
    Gold Member

    For the pp-wave metric
    [tex]ds^2={du}^{2}\,\left( -\left( {z}^{2}+{y}^{2}\right) \,C+2\,y\,z\,B+\left( {y}^{2}-{z}^{2}\right) \,A\right) +{dz}^{2}+{dy}^{2}+2\,du\,dv[/tex]
    where A,B,C are functions of u only, the relevant part of the tidal tensor (in the Y,Z plane, with C=0) is
    \left[ \begin{array}{cc}
    A & B \\\
    B & -A \end{array} \right]
    This is a vacuum solution ( if C=0) so all the curvature is Weyl. Notice that the trace is zero so volume is preserved. The wave is travelling at c in the u-direction, so this is a tranverse wave if A is a wavy function like [itex]a\sin(\omega u)[/itex].

    I think B is a sign of polarization ( rotation around the u-axis ?).
    Last edited: Jul 5, 2012
  4. Jul 5, 2012 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Time dilation is not a curvature effect. Time dilation is an effect you can get in SR, in flat spacetime.

    It may help to think of curvature generically as an effect in which lines that are initially parallel can later on become non-parallel. On a sphere, lines that are initially parallel can later converge. If two particles attract one another gravitationally, their initially parallel trajectories can later converge. If you release a cloud of test particles in a region of space where there is only Weyl curvature, you get divergence in some plane(s) and convergence in some other plane(s), with the net result that the volume stays the same.

    This may be helpful:
    http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.1 [Broken]
    http://www.lightandmatter.com/html_books/genrel/ch09/ch09.html#Section9.2 [Broken] (subsection 9.2.2)
    Last edited by a moderator: May 6, 2017
  5. Jul 5, 2012 #4
    You can say the Riemann tensor has two parts: the Weyl tensor, which describes curvature in regions devoid of matter; and a "source" tensor, based on the stress-energy, which corresponds to curvature from matter at the point in question.

    Imagine you were looking at a small differential four-volume: the Weyl tensor would represent curvature originating from outside this four-volume, while the source tensor (which comes from stress energy and, hence, can be contracted to the Ricci tensor) describes what does come from this volume.

    An analogous case would be finding the electric field at a given point. If you know all the sources in a given volume and assume no sources exist outside, you can calculate the electric field, sure, but there are other valid solutions to the Maxwell equations based on sources from outside the volume of interest.
  6. Jul 6, 2012 #5


    User Avatar
    Gold Member

    Thank you for your comments.
    Mentz114 mentioned the tidal tensor.This tensor vanishes in flat space, as there is no energy exchange (absorbtion, emission). So it is agreed, that there is only Weyl curvature.

    Let us look at the 2 moments where the metric is streched in the x-direction (A) and half a period later in y-direction (B). Now the sum of light ray angles in the xy-plane shall be measured at A and B then, assuming that the change of period of the wave is neglible compared to the time needed to complete the resp. measurement.

    What will these measurements show regarding the sum of the angles?
    Will they prove that the spacetime curvature in the x-y-plane of the gravitational wave oscillates between negative and positive values, or not?
    Last edited: Jul 6, 2012
  7. Jul 6, 2012 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you're getting tangled up here by an incomplete understanding of spacetime curvature. The sum of the interior angles of a triangle measures spatial curvature, but GR isn't concerned with spatial curvature; it deals with spacetime curvature, i.e., 4 dimensions, not 3. It sounds like what you're visualizing is actually curvature of a 2-dimensional space with both dimensions spacelike. In such a space, there is only a single number, the Gaussian curvature, that measures the curvature at any given point. This is not the case in 3+1 dimensions, where you need the whole Riemann tensor to describe the curvature completely. As an example, standard cosmological models these days typically have zero spatial curvature, but they don't have zero curvature. When you talk about constructing triangles out of light rays, I guess you're imagining that as a measurement of the spatial geometry at one instant, but the light rays actually take time to propagate -- they propagate at exactly the same speed as the gravitational wave. Because the spatial geometry is three-dimensional, not two-dimensional, you can't characterize it by a single number that's positive or negative. (The only reason you can get away with this in cosmological spacetimes is that they have rotational symmetry, which is not present in a gravitational wave.)
  8. Jul 6, 2012 #7


    User Avatar
    Gold Member

    Thanks for answering.
    Yes, gravitational waves propagate with lightspeed. My reasoning was that considering a low frequent wave the shape of the distortion of the metric wouldn't change much during the measurement. But nevertheless it isn't infact an instant measurement and it happens within 2 space dimensions.

    I was originally puzzled from what is written
    here, page 11:
    How then should I understand this? Could you kindly comment on that?
    I have in mind that even in the scientific world sometimes space curvature is mixed up with spacetime curvature. Does this happen here?
  9. Jul 8, 2012 #8


    User Avatar
    Gold Member

    I still wonder, if it is correct to say:

    A gravitational wave curves spacetime periodically positive and negative.

    Or is this a priori a wrong statement, because only the stress-energy tensor can be responsible for any curvature of spacetime? In flat space, far away from masses, there is Weyl curvature only.

    However, according to this, page 11:

    there is spacetime curvature.

    Are the triangles shown on page 11 in the plane parallel to the z-direction in which the wave propagates or should they be understood to be in the x-y-plane. I assumed the latter, but may be wrongly. If the former is right, then the extrema of the wave's amplitude could be measured seperately. Any comment to clarify that is very welcome.

    Another approach: Given parallel null-geodesics (perhaps it is better to choose parallel geodesics of test particles). What happens to these geodesics, if a gravitational wave propagates (a) parallel and (b) perpenticular to them? I assume, that only in case (a) the geodesics are bent periodically inwards and outwards. If so, I would conclude that a gravitational wave curves spacetime periodically positive and negative. Please correct, if wrong.

    Supposed correctness, I further assume that geodesics coming closer and moving away from each other, resp. show positive and negative spacetime curvature respectively.
    However testparticles are moving simultaneously in opposite directions along the x- and y-axis as the gravitational wave passes by. So, the conclusion woud be that while the spacetime curvature is positive in x-direction, it is negative in y-direction and vice versa in the next half period.

    I have never heard about this and do not trust this reasoning myself. Any help to improve my understanding is appreciated. Sorry, it must be tiring for you experts.
    Last edited: Jul 8, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook