# Newton's Law of Gravitation, Gauss Law and GR

From One of My Unpublished Papers $$\frac{d^2 x^{\alpha}}{d\tau^2}=-\Gamma^{\alpha}_{\beta \gamma}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau} \tag{1}$$

For radial motion in Schwarzschild’s Geometry we have,

$$\frac{d^2 r}{d \tau^2}=-\frac{M}{r^2}\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2+\frac{M}{r^2}\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2\tag{2}$$

Again from radial motion, we have from Schwarzchild’s metric:

$$d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2\tag{3}$$

Dividing both sides of (3) by $d\tau^2$ we have,

$$1=\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2\tag{4}$$

Using relation (4) in (2), after factoring out $M/r^2$ from the RHS of (2), we obtain:

$$\frac{d^2 r}{d \tau^2}=-\frac{M}{r^2}\tag{5}$$

The inverse square law should hold accurately if proper time is used. Here $r$ represents the coordinate distance along the radius.

One may use the relations:

$$M ~\rightarrow~ GM/c^2\qquad\text{and}\qquad\tau ~\rightarrow~ c\tau,$$

to obtain the exact "form" of the law of Gravitation.

Query: Is equation (5) indicative of the fact that Gauss law may be used in the same classical "form" in GR?

[We may introduce a symbol $F=m\frac{d^2 r}{d\tau^2}$.]

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What do you understand by Gauss law here? The classical field from a source $\rho(r)$ are given by linearly summing over the point fields $F_{tot}(r)=-G\int_Vdy\rho(y)/(r-y)^2$. Gauss law as I understand it here is the divergence form of that. You derived the familiar radial equation (no corrections?), starting from a radial equation from the GR metric for one point mass as input. However, collecting point masses in GR and compute a metric via the nonlinear Einstein equations will not, I assume, give a metric, whose trajectories correspond to divergence free forces outside of some volume. –  NikolajK May 8 '12 at 13:01
@NickKidman this should be the answer, although I would change italics to bold. –  tmac May 8 '12 at 17:50
Gravitational potential for a spherical mass $\phi=-\frac{GM}{r}$. Here we are considering a point P outside the spherical mass in relation to the formula (5) in theoriginal posting. Now inside the spherical body we imagine several smaller distinct[and disjoint] spheres whose centers do not coincide with that of the original one. At the same point P each produces its own potential $\phi_i$. $\phi=\Sigma \phi_i=-\Sigma\frac{Gm_i}{r_i}-\phi_{left-over}$. We have, $\nabla^2 \phi=0$. For each $\phi_i$, $\nabla^2 \phi_i=0$. Therfore for the left over portion, $\nabla^2 \phi_{left-over}=0$. –  Anamitra Palit May 8 '12 at 23:59
Gauss law seems to hold for arbitrary/irregular mass distributions. We have considered those portions where mass density is zero and outside the original large sphere. Example: Point P –  Anamitra Palit May 8 '12 at 23:59
Einstein's Field Equations are $linear$ in the inertial frames of reference since the Christoffel Symbols evaluate to zero value. Incidentally we have considered the $free {\;} fall$ of a body and formula (5) uses proper time.[For trajectories which are not geodesics, inertial agents come into action] –  Anamitra Palit May 9 '12 at 0:17