I Why Schwarzschild Metric for Deflection of Light & Precession of Perihelia?

davidge
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Why one uses Schwarzschild metric instead of FLRW metric when deriving things such

- deflection of light by the sun
- precession of perihelia of planets

Also, as our solar system is not isotropic nor static, it seems that by using the Schwarzschild metric we would get only an approximation on the results. So why we still get accurate result for the precession of perihelia of Mercury?
 
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davidge said:
Why one uses Schwarzschild metric instead of FLRW metric when deriving things such

- deflection of light by the sun
- precession of perihelia of planets
Because the FLRW metric assumes that spacetime is filled with uniformly distributed matter. A bad approximation on the scale of the solar system.
davidge said:
Also, as our solar system is not isotropic nor static, it seems that by using Schwarzschild metric we get only a approximation on the results. So why we still get accurate result for precession of perihelia of Mercury?
Because it's a very good approximation because the mass of the Sun is such a huge percentage of the total mass of the solar system.
 
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Ah, ok. Thanks.

A bit off topic... but, the name precession of perihelia is due to the fact the we use the nearest distance ##r## of the planet from the sun when computing the angle? Is there a special motivation for non using the aphelion distance instead?
 
No, it's just common use in astronomy.
 
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vanhees71 said:
No, it's just common use in astronomy.
Thanks.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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