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From One of My Unpublished Papers $$\frac{d^2 x^{\alpha}}{d\tau^2}=-\Gamma^{\alpha}_{\beta \gamma}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau} \tag{1}$$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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    $\begingroup$ 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. $\endgroup$
    – Nikolaj-K
    May 8 '12 at 13:01
  • $\begingroup$ @NickKidman this should be the answer, although I would change italics to bold. $\endgroup$
    – tmac
    May 8 '12 at 17:50
  • $\begingroup$ 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, $\nabla^2 \phi_{left-over}=0$. $\endgroup$ May 8 '12 at 23:59
  • $\begingroup$ 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 $\endgroup$ May 8 '12 at 23:59
  • $\begingroup$ 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] $\endgroup$ May 9 '12 at 0:17
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Assuming the initial direction of the considered geodesic dr/d{tau}=0 (initial velocity equal to zero), you could calculate the value of d^2r/dt^2 (curvature of geodesic projected on rt-plane) directly from equation (2). Consider however that this (projected geodesic) curvature regards the map coordinates (the acceleration measured by the observer situated very far! i.e. in free, flat space). In equation (5) you have r, which corresponds to the distance in the map (NOT LOCAL), and tau - which is connected to the local time (entities of different spaces!). So, this relation doesn't make sense (in my opinion). However, I like very much the idea to involve Gauss law in GR!!!

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  • $\begingroup$ Metric:c^2dtau^2=c^2g_ttdt^2-g_xxdx^2-g_yydy^2-g_zzdz^2.For Schwarzschid Geometry,setting c=1,dtau^2=[1-2m/r]dt^2-[1-2m/r]^(-1)dr^2-r^2[(d theta)^2+sin^2 theta (dphi)^2] With the above metric dtau is the time interval in the local context while ‘r’ is measured from the origin.This is a concept typical of the metric.It is also to be noted that dr/d tau=0 can be valid in the strong gravity region (example:dropping of a stone) $\endgroup$ Dec 6 '21 at 9:52
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This is all fine, but in local space also the unit of distance (let's call it "rho") is different from "r" of reference coordinates - and not just time. Of course, considering rho as the distance from the origin would not make sense (at r_s it would tend to infinity), but using its first and second derivatives will be correct! (I mean d^2 rho/dtau^2 - for local acceleration, and drho/dtau - for the initial velocity or component of the initial 'geodesic line direction'. Using mixed units, i.e. "dr" - in reference coordinates (in flat-space map) over "dtau" - in local time; is a common but misleading misunderstanding!!! From a logical point of view, it makes no sense!!! It is easier and more natural to consider (and calculate) gravitational acceleration entirely in reference coordinates, i.e. in flat-space, as d^2r/dt^2!!! Unfortunately, this leads to the conclusion that there is still something wrong with the Schwarzschild coefficients... if you calculate the REAL LOCAL acceleration, you will see that it does not coincide with Gauss' law. Please try to do it!!!

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