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Reading about gravitational lensing here and here, its seems that it can be explained well by assuming a refractive index created by a gravitional potential $$n(\vec{x}) = 1+2\frac{\phi(\vec{x})}{c^2}$$ where $\phi$ is the gravitational potential (negative values, $\phi\to-0$ for $|\vec{x}|\to\infty$).

Looking at this answer or this, this again is the same as the $g_{00}$ term of the metric tensor.

Due to symmetries, the metric tensor has 10 free variables. Now I wonder: if gravitational lensing could indeed be described by just one number (per space point $\vec{x}$), i.e. a scalar field, how come the metric tensor has 10 free variables (per spacetime point)? What are they "good for"? In which situations does the refractive index not tell the whole story?

Harald
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  • How would you describe a gravitational wave using just $T_{00}$? – John Rennie Nov 05 '17 at 17:48
  • Hah, if I would know the answer to this question, I'd probably not have asked my question anyway. So your answer is: gravitational waves cannot be described as a wave-like disturbance of $n(\vec{x})$? Which would would trigger the follow-up question: how exactly does it fail. Ok, I might need to read up a lot on gravitational waves first. – Harald Nov 05 '17 at 20:00
  • I don't think this can be usefully answered without describing all of GR. You are in effect asking why the metric has to be a tensor field and while that has an answer it's not a simple answer. – John Rennie Nov 06 '17 at 05:51
  • I was hoping for an example, real or thought experiment, which shows that just $g_{00}$ is not enough to describe the outcome. Gravitational lensing can be completely described? Gravitational redshift? One of these: https://en.m.wikipedia.org/wiki/Tests_of_general_relativity ? Seeing where just g00 fails to descibe what happens, I hope, lets me better understand what the other parameters effect. – Harald Nov 06 '17 at 21:33

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