guenter Posted November 26, 2011 Share Posted November 26, 2011 Hi, my question While Ricci curvature vanishes in the absence of matter (flat space) there is still Weyl curvature propagating with gravitational waves. To my understanding the Weyl tensor is responsible for tidal effects (stretching, shrinking) in the x-y-plane (the wave propagates in z-direction). This is only true for small fields (plan wave solution). But I wonder what else the Weyl curvature is responisble for. Is the x-y-plane flat like Minkowski-spacetime? If not, are triangles in this plane oscillating from concave to convex or whatelse would characterize this curvature? I have seen something like this but wasn't sure about the meaning and will search for it. fits probably better to the sub forum "Relativity". The article I had in mind shows in Fig. 1.1 The propagation of spacetime curvature oscillating light ray triangles. So, according to the phase of the wave the (Weyl?) curvature is negative, zero or positiv. 1. However in which plane? Do these triangles show the curvature in the x-y-plane or parallel to the z-direction? 2. To make the light ray triangles is very fast compared to the period of the gravitational wave. Can one say that the triangles measure (almost) frozen states and not the dynamics of the wave? Supposed there is curvature in the x-y-plane, wouldn't that mean that laser interferometer measurements are disturbed by shapiro delays? I have never heard about that, but perhaps it's marginal if not zero. Any help to better understand these things, especially the meaning of the Weyl curvature in otherwise flat space, is appreciated. Please use layman language and correct my reasoning, if wrong. Link to comment Share on other sites More sharing options...

Mystery111 Posted November 30, 2011 Share Posted November 30, 2011 (edited) Curvature is caused by understanding the Christoffel Symbols. A certain wave equation describing two kinds of waves, [math]\frac{\partial^2 \phi}{\partial t} = c^2 \frac{\partial^2 \phi}{\partial x^2}[/math] It describes basically the kind of waves we attribute to right movers and left movers. In three dimensions this is [math]\frac{\partial^2 \phi}{\partial t^2} = \frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}[/math] This can be written as [math]\eta^{\mu \nu} \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = 0[/math] To make this into a tensor, we will invite now the Christoffel Symbols [math]\nabla_{\mu} g^{\mu \nu} \frac{\partial \phi}{\partial X^{\nu}} + \Gamma^{\mu}_{\mu \alpha} g^{\nu \beta} \frac{\partial \phi}{\partial X^{\beta}} = 0[/math] This equation describes the geodesic of a path taken by the likes a of a photon for an example. Edited November 30, 2011 by Mystery111 Link to comment Share on other sites More sharing options...

guenter Posted December 1, 2011 Author Share Posted December 1, 2011 Where is the path from the null geodesic to my questions? Perhaps these are non trivial. Link to comment Share on other sites More sharing options...

Mystery111 Posted December 7, 2011 Share Posted December 7, 2011 Null, for a photon for instance? I don't quite understand what you mean. Link to comment Share on other sites More sharing options...

guenter Posted December 8, 2011 Author Share Posted December 8, 2011 The term null-geodesic means the light-like path of a photon. Background: The space-time intervall for a photon is zero. You can easily search for that in Wikipedia, to lern more. I will not comment anymore, because its off-topic. MODERATION: Is it possible to contact an expert in GRT/Gravitational waves, who could clarify my questions? I appreciate any help. Thanks, guenter Link to comment Share on other sites More sharing options...

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