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1

The answer to the question in the title is yes, because general relativity is background independent. The gravitational influence of a body is directed toward the body and falls of with distance, and the only reference for direction and distance is the gravitational field (spacetime) itself, so it's impossible for the gravitational field of one body to ...


-1

If a star gravity can change the apparent location of another star on the sky, can it change the gravitational field vector of that star? You are confusing the collective effect of the gravity of a given star on electromagnetic waves, to the gravity of the stars whose images the electromagnetic waves carry when we detect the lensing. The star (or galaxy) ...


4

You may want to look at the interior Schwarzschild solution, which describes a the metric inside a spherically symmetric incompressible mass of constant density and with zero pressure at its surface. It is (in geometric coordinates) $$ ds^2 =\frac{1}{4}\left(3\sqrt{1-(r_s/r_g)} - \sqrt{1-r^2r_s/r_g^3}\right)^2 dt^2 - \frac{dr^2}{1-\frac{r^2r_s}{r_g^3}} - r^2(...


3

Like you said, the Schwarzschild metric applies to a static, spherically symmetric, vacuum spacetime. The Schwarzschild metric applies up to the singularity of the black hole because the region interior to the event horizon is a vacuum: $G_{\mu\nu}=T_{\mu\nu}=0$. The interior of a body such as a star is obviously not a vacuum so the Schwarzschild metric does ...


4

Second, why is it that we describe the black hole until the interior singularity and not the star? Because it can be shown that if the entire mass of an object is inside its Schwarzschild radius then it will collapse to a singularity according to classical general relativity. See Tolman–Oppenheimer–Volkoff limit. The Schwarzschild metric is only valid in ...


2

It looks like there is already an interesting solution available. https://arxiv.org/abs/1701.02098 This assumes an anisotropic fluid in the interior and claims that it satisfies the strong energy condition. The only other solution I have seen involves one that had some unphysical properties that I can't recall. Anyway, here is a link just in case: https://...


-2

You need an app to see the stars in a cloudy sky.


0

After further reading in Kippenhahn, Weigert and Weiss, I try to answer my question 1. This answer seems equivalent to honeste_vivere's answer, but different in the assumptions. From the chain rule for two independent variables, equation (14.5.3): suppose x=g(u,v) and y=h(u,v) are differentiable functions of u and v, and z=f(x,y) is a differentiable function ...


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