Weber versus Schwarzschild: GR Much Ado About Nothing

In our last blog we proposed a metric,

$$-c^2d\tau^2=-rc^2dt^2+\frac{1}{2r}\left(dr^2+r^2d\varphi^2\right),$$

in the plane, $\vartheta=\pi/2$, for the Weber force,

$$F_W=\mbox{const.}\times\left(\ddot{r}-\frac{\dot{r}^2}{2r^2}+\frac{c^2H^2}{r^2}\right),$$

whose vanishing coincides with the Euler-Lagrange equations, or, equivalently, the geodesics. 

The Lagrangian describing this geometry is

$$\mathcal{L}=\frac{1}{2}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{nu}=\frac{1}{2}\left\{-rc^2\dot{t}^2+\frac{1}{2r}(\dot{r}^2+r^2\dot{\varphi}^2)\right\},$$

where the dot denotes differentiation with respect to $\tau$. There is no distinction between "coordinate", $t$, and "proper", $\tau$ times here: The introduction of $dt^2$ is just a way of introducing $c$ which appears in Ampere's theory as the ratio of the electromagnetic to electrostatic unit of electricity. The surprise was it could be represented as a velocity. That's Ampere's theory. Now in Weber's theory, which is dynamical, the same constant $c$ is the number of units of statical electricity which are transmitted by a unit electric current in a unity of time. This is Maxwell's definition. We will not get into the discussion of whether the two are equal, that's another story.

Now, it does appear odd that we have claimed that the metric coefficient $g_{00}=-r c^2$ accounts for the Coulomb force, and the only way to unite it with the kinetic energy terms in the above equation is through $c^2$. The potential

$$\Phi=\frac{1}{4\pi\epsilon_0 r}\cdot 4\pi r^2=\frac{r}{\epsilon_0},$$

represents the surface charge density $\sigma$ on a ball of radius $r$, where $\sigma\propto 1/r$ is the surface charge density. The potential satisfies Poisson's equation,

$$\nabla^2\Phi=\frac{1}{r^2}\frac{d}{r^2}\left(r^2\frac{d\Phi}{dr}\right)=\frac{2\sigma}{\epsilon_0},$$

in a spherically symmetric environment. The inverse square dependency of the Coulomb force will be brought in by the constraint that $\dot{t}$ is a cyclic coordinate which leads to an immediate first integral. The Coulomb potential, like its gravitational counterpart, has nothing whatsoever to do with the slowing down of clocks in an electrostatic field! The formal similarity of equations can, at times, be most misleading!

The Euler-Lagrange (EL) equations are

$$\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot{x}^{\mu}}-\frac{\partial\mathcal{L}}{\partial x^{\mu}}=0.$$

The reason for using $\lambda$, and not $\tau$, is that the differential of the latter vanishes on light cones, a condition that determines the speed of light.

Now, $\varphi$ and $t$ are cyclic variables because only their derivatives appear in the Lagrangian. This immediately leads to first integrals of the motion,

$$r\dot{t}=H,  \hspace{20pt}\mbox{and}\hspace{20pt} r\dot{\varphi}=V,$$

where both $H$ and $V$ are arbitrary constants of the motion.

It is easy to check that the geodesic equations

$$\frac{d}{d\tau}\left(g_{\alpha\beta}\frac{dx^{\beta}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{\mu\nu}}{\partial\alpha}\dot{x}^{\mu}\dot{x}^{\nu}=0,$$

coincide with the EL equations. For instance the equation for radial motion is

$$\ddot{r}-\frac{\dot{r}^2}{2r^2}+2c^2 r\dot{t}^2=0.$$

Introducing the first integral of the motion for $\dot{t}$ gives

$$\frac{1}{r^2}\left(r\ddot{r}-\frac{1}{2}\dot{r}^2+(cH)^2\right)=0$$

which is the vanishing of the Weber force, apart from a multiplicative coefficient which cannot be determined. The last term is the Coulomb potential which needs $c^2$ to appear along side the other two terms. Let us now compare this with general relativity's handling of the Schwarzschild field.

The Schwarzschild metric has the form

$$ds^2=Bdt^2-Adr^2-r^2d\Omega^2,$$

where the last term is the element of a two-sphere which will not concern us because we are considering radial motion. The metric coefficients are initially considered functions of both $r$ and $t$, where the "radial coordinate $r$ defines the area of a $2-$sphere and hence it has proper physical meaning." The quote like the others we will use comes from a paper  by Dadhich, "On the Schwarzschild field." But, isn't it a little premature to give $r$ such a "proper" physical meaning when we will find that it can't be extended to the origin of the coordinate system?

Einstein's condition for "emptiness" means that the components of the Ricci tensor vanish. For instance, the time component is

$$R^0_0=-\frac{1}{2AB}\left\{\nabla^2B-\frac{B^{\prime}}{2}\left(\frac{A^{\prime}}{A}+\frac{B^{\prime}}{B}\right)-\ddot{A}+\frac{\dot{A}^2}{2A}+\frac{\dot{A}\dot{B}}{2B}\right\},$$

where the prime and dot denote differentiation w.r.t. $r$ and $t$ respectively. 

We are now told that from a "physical point of view it is obvious that [the] field of a static body cannot depend upon $t$ which is ensured by $R_{01}=-\dot{A}/rA=0\ldots$ which implies $A=A(r)$" But how is this ever to be congruent with the geodesic equations which are an immediate consequence of the variational principle?

The requirement that both $R_0^0$ and

$$R_1^1=R_0^0+\frac{1}{rA}\left(\frac{A^{\prime}}{A}+\frac{B^{\prime}}{B}\right)$$

vanish requires $AB=f(t)=\mbox{const.}$. The same hold true for the Weber metric, but not for the reason that the effect of $A$ in "curving space" exactly cancels out the contribution to the field energy density contained in $B=1+\phi$, where $\phi$ is the Newtonian potential. "In the usual derivation it is here the asymptotic flatness is invoked to set $f(t)=1$." However, wasn't it agreed upon that $A$ and $B$ are independent of $t$ because we are treating a "static" gravitational problem? "Even if $f(t)$ is retained the field will be asymptotically flat and the solution with $f\neq 1$ is physically indistinguishable from the one with $f=1$. So we apply no boundary condition to set $f=1$." What differential equation is the author referring to? to the Laplace equation in the expression for $R_0^0$? And who needs it? when we have the equation for the geodesics readily available?

The claim that $A$ compensates $B$, for otherwise "one does not see how non-linear aspect of gravity is accounted for in the Schwarzschild solution" doesn't hold water since the same condition is found in the Weber metric and there is no talk of "warping" space by the presence of $A$ and the field energy density which appears as a "gravitational charge density." There $A$ and $B$ are definitely functions of time, and $\Phi$, the surface charge density, does not satisfy Laplace's equation. Why should we ever expect it to do so where there is an electrostatic field?

Schrodinger, in his article "The possibility of fulfillment of the relativity requirement in classical mechanics", didn't pick up the fact that "the total interaction energy of two mass point with masses $\mu$ and $\mu'$ with separation $r$,

$$W=\gamma\frac{\mu\mu'\dot{r}^2}{r}-\frac{\mu\mu'}{r},$$

where $\gamma$ is a fitting parameter, does not curve space, $A=1/2r$, but he, moreover, didn't recognize it as Weber's potential with $\gamma=1/2$. (He needed $\gamma$ $6$ times that in order to account for the advance of the perihelion of Mercury.)

Consulting the expression for $R_0^0$, the time derivatives don't even give the time-dependent terms in the Weber force. The condition for the vanishing of this term is

$$\frac{\ddot{r}}{2r}-\frac{\dot{r}^2}{r^2}+2c^2=0,$$

which is not even close to the vanishing of the Weber force given above. If $A$ and $B$ are allowed to depend upon time, the vanishing of $R_0^0$ is

$$\ddot{r}-2\frac{\dot{r}^2}{r}=0,$$

where all effects of the potential $B=1+2\phi$ vanishes, since it solves Laplace's equation. But, then there is the $R_{01}=-\dot{A}/rA$  term which does not vanish. 

Instead of the Schwarzschild metric, it would be much easier (and much more instructive) to begin with

$$\mathcal{L}=\frac{\dot{r}^2-(c\dot{t})^2}{2r}.$$

Then, with the first integral, $r\dot{t}=\surd 2$,  the EL equation is simply

$$\frac{\ddot{r}}{r}-\frac{1}{2}\left(\frac{\dot{r}}{r}\right)^2+\frac{c^2}{r^2}=0,$$

which is the vanishing of the Weber force, whether it be for Coulomb, or Newton.

Dadhich also claims  that even when $B=1$, $A=1/(1-\mathcal{R}/r)$, where $\mathcal{R}=2GM/c^2$ will create a space curvature that will "manifest in tidal acceleration for non-radial motion." This is clearly inaccurate as any tangential acceleration must come from the term $r^2\dot{\varphi}^2$ in the metric. As such there is no linear term, but, we can invent one by dividing the Lagrangian through by $(1-\mathcal{R}/r)^{-1}$ to get

$$ \mathcal{L}=\frac{1}{2}\left(-c^2\dot{t}^2+\frac{\dot{r}^2}{(1-\mathcal{R}/r)^2}+\frac{r^2\dot{\varphi}^2}{(1-\mathcal{R}/r}\right).$$

This was the trick Moller used to derive the geodesic equations---not from the $4$-vector of the velocity---but from the relativistic expression for the orbiting particle, which doesn't exist. Setting the Lagrangian equal to zero, and changing the differentiation variable to $\varphi$ through the first integral,

$$\frac{r^2\dot{\varphi}}{1-\mathcal{R}/r}=L,$$

give

$$\frac{dr}{d\varphi}=\pm\frac{r^2}{L}\sqrt{E-\frac{L^2}{r^2}\left(1-\frac{\mathcal{R}}{r}\right)},$$

where we have used $\dot{t}=H$ [cf. Eqn (7.5.10) in A New Perspective on Relativity].  This identifies the enigmatic $\dot{t}^2$ with the total energy, $E=H^2$.The last term, supposedly, gives the exact expression for the deflection of light by a massive body. The funny thing is that the Newtonian potential is missing; for otherwise it would also explain the advance of the perihelion of Mercury. Why the Coulomb potential was accounted for by the $\dot{t}^2$ term in the Lagrangian above, and why the proper time interval $d\tau^2$ must be included in what was the condition for the vanishing of the Lagrangian appears mysterious. This will be treated in the next blog.

However, there now there is a tangential acceleration,

$$a_{tan}=r\ddot{\varphi}+2\dot{r}\dot{\varphi}=\frac{\mathcal{R}}{r}\frac{\dot{r}\dot{\varphi}}{1-\mathcal{R}/r},$$

which cannot be wished away. In Moller's words, $L$ cannot in general be interpreted as angular momentum, since the notion of a 'radius vector' occurring in the definition of the angular momentum has an unambiguous meaning only in a Euclidean space." This clashes with Dadhich's assertion that "the radial coordinate $r$ defines the area of a $2$-sphere and hence has proper physical meaning", which we quoted above.

The truth of the matter is that by dividing the Lagrangian through by $(1-\mathcal{R}/r)^{-1}$ converts the space part into the Beltrami metric  which is a hyperbolic metric of constant curvature when $\mathcal{R}/r$ is replaced by the mass density $\varrho$ according to $GM/c^2r\rightarrow (4\pi/3)G\varrho/c^2r^2$. This, in fact, applies to the inner solution of Schwarzschild solution [cf. Sec. 9.19.4 in A New Perspective on Relativity].