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From my understanding, the gravitational constant $G$ is a proportionality constant used by Newton in his law of universal gravitation (which was based around Kepler's Laws), namely in the equation $F = \frac{G\cdot M\cdot m}{r^2}$. Later, Einstein set forward a different theory for Gravity (based around the equivalence principle), namely General Relativity, which concluded that Newton's law was simply a (rather decent) approximation to a more complex reality. Mathematically speaking, Einstein's Theory was completely different from Newton's Theory and based around his Field equations, which also included $G$ in one of it's terms.

How come two different theories that stemmed from completely different postulates end up having this same constant $G$ with the same numerical value show up in their equations? What exactly does $G$ represent?

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I guess a way to see this would be solving Einstein's equations for a weak field, $g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}$ and you would see that both $G's$ are equal. – jinawee Jun 10 '14 at 21:22
Related: , and links therein. – Qmechanic Jun 10 '14 at 21:23
up vote 12 down vote accepted

Since in the limit of weak gravitational fields, Newtonian gravitation should be recovered, it is not surprising that the constant $G$ appears also in Einstein's equations. Using only the tools of differential geometry we can only determine Einstein's field equations up to an unknown constant $\kappa$: $$G_{\mu\nu} = \kappa T_{\mu\nu}.$$ That this equation should reduce to the Newtonian equation for the potential $\phi$, $$\nabla^2 \phi = 4\pi G\rho \tag{1}$$ with $\rho$ the density fixes the constant $\kappa = \frac{8\pi G}{c^4}. \tag{2}$

In detail, one assumes an almost flat metric, $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where $\eta_{\mu\nu}$ is flat and $h_{\mu\nu}$ is small. Then from writing down the geodesic equation one finds that if $h_{00} = 2\phi/c^2$, one obtains Newton's second law, $$\ddot{x}^i = -\partial^i \phi. \tag{3}$$ Using (3) and taking $T_{\mu\nu} = \rho u_\mu u_\nu$ for a 4-velocity $u_\mu$ with small spatial components, the $00$ component of the field equations (2) is $$2\partial^i \partial_i \phi /c^2 = \kappa \rho c^2.$$ In order to match this with (1), we must have $\kappa = \frac{8\pi G}{c^4}$. (The detailed calculations here are, as is often the case in relativity, rather lengthy and boring, so they are omitted.)

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But when we consider weak gravitational fields we don't get exactly Newton's Law, just a rather good approximation. Does this mean that we get a slightly different (yet entirely usable under experimental verification) $G$, or do we end up getting the exact numerical value for $G$ as in Newton's Laws? – Disousa Jun 10 '14 at 21:31
@Disousa: actually, for the case of orbits with zero angular momentum outside of a spherically symmetric mass distribution, the equations are identical in full relativity and the newtonian case. – Jerry Schirmer Jun 10 '14 at 21:38
@Disousa No. Even without Jerry's comment, the answer would still be no. The situation is wholly analogous to the following problem: determing $\tilde{\kappa}$ so that $\phi\mapsto\sin(\tilde{\kappa}\,\phi)$ and $\phi\mapsto\kappa\,\phi$ have the same slope at $\phi=0$. Only one $\tilde{\kappa}$ will fit the bill: $\tilde{\kappa}=\kappa$. In that sense the "exact numerical values" are the same. We might end up measuring an experimental $G$ more accurately than we did when we only knew Newton's law and thus find that we need to change our value of $G$, but this would also mean we .... – WetSavannaAnimal aka Rod Vance Jun 10 '14 at 22:24
... the constant in Newton's law we should use would need to be updated too. – WetSavannaAnimal aka Rod Vance Jun 10 '14 at 22:25

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