# Gravitational Constant in Newtonian Gravity vs. General Relativity

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: physics.stackexchange.com/q/68067/2451 , physics.stackexchange.com/q/89/2451 and links therein. – Qmechanic Jun 10 '14 at 21:23

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.)
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 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