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The Einstein Field Equations: $$G_{ab}~=~8\pi T_{ab}.$$ I am familiar with how to obtain the $8\pi$ proportionality factor through correspondence with Newtonian gravity, but am wondering if this factor can be deduced from a quick and dirty geometric argument? This may be a naive question...any other derivations or refutations of my desire for a geometric argument are welcome as well though.

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  • $\begingroup$ What kind of "geometric argument" could work here? you are trying to describe physical phenomena, that is, the actual world! There has to be some input to your model, geometry alone is not enough. Also, in reduced natural units the EFE look like $G=T$, that is, the $8\pi$ is absorbed into the definition of $G$, and then we take $G=1$. This means that the constant of proportionality is kind of abritrary, and you can set it to whatever you want to... $\endgroup$ – AccidentalFourierTransform Apr 11 '16 at 15:43
  • $\begingroup$ Yea the more I think about it the less it seems plausible...I think I'll delete this question $\endgroup$ – ClassicStyle Apr 11 '16 at 15:45
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    $\begingroup$ Its not a bad question though. I'd say let it be. $\endgroup$ – AccidentalFourierTransform Apr 11 '16 at 15:47
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    $\begingroup$ Then again if we think about where that factor comes from, Poisson's equation, its presence can be deduced geometrically there... $\endgroup$ – ClassicStyle Apr 11 '16 at 15:48
  • $\begingroup$ @TylerHG: I would roll back to v3 to have as few implicit assumptions about natural units as possible. $\endgroup$ – Qmechanic Apr 11 '16 at 16:26
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The constant is only $8\pi$ when you're working in units where $G=1$. The number $G$ was originally defined as the constant of proportionality in Newton's

$$F=G\frac{m_1m_2}{r}.$$

So in some sense the only thing we can do is to pass to the Newtonian limit and calculate the force between two point masses.

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