We have seen, in the last blog, that there is a rather strict analogy between the Schwarzschild metric and Gerber's potential, both of which give the same expression to the perihelion shift of Mercury to order $1/c^2$. It has long been known that to this order the force derived from Gerber's potential is the same as the Weber force. There is a slight difference, however. Whereas the Weber force acts radially between the two masses so that there is no tangential force acting on the orbiting particle, the Schwarzschild metric has a finite tangential force that destroys the conservation of angular momentum.

All accounts of this rely on the misconception that the term that does not allow for the conservation of energy is small so that conservation can be assumed in the first approximation. This is false. The tangential acceleration for the Schwarzschild metric is

$$r\dot{\omega}+2\dot{r}\omega=2gr\frac{\dot{r}\omega}{1-\alpha/r},$$

where $\alpha\equiv 2GM$ and $g=\alpha/2r^2$. This destroys the general relativistic equation for the orbit as

$$\frac{d^2u}{d\varphi^2}+u=\frac{\mathcal{R}}{L^2}+3\mathcal{R}u^2,$$

where $u=1/r$, $\mathcal{R}=2GM/c^2$ is the Schwarzschild radius for the gravitating mass $M$, and the angular momentum is

$$L=\frac{r^2\omega}{1-\alpha/r},$$

which is hardly a constant. The Weber force has no such problems, although the derivation of the advance of the perihelion rests on equating the Weber force with radial acceleration, $\ddot{r}=r\omega^2$, which, we will see, is highly unscrupulous. This is because you are equating a radial acceleration with a radial acceleration.

The Weber force can be derived from the metric

$$-d\tau^2=-rdt^2+\frac{d{r}^2+r^2d\varphi^2}{2rc^2}.$$

Rather than considering the equations of the geodesics, we can consider the Lagrangian

$$\mathcal{L}=\frac{\dot{r}^2+r^2\dot{\varphi}^2}{2r}-c^2r\dot{t}^2,$$

where the dot stands for some parameter which measures the distance along the path, not necessarily $\tau$, which is constant for light rays.

The Euler-Lagrange equations are

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}-\frac{\partial\mathcal{L}}{\partial x}=F,$$

which vanishes along geodesics. $x$, and its derivative $\dot{x}$ stand for any of the three variables in $\mathcal{L}$. It will be appreciated that both $\varphi$ and $t$ are cyclic variables since

$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}=0.$$

Hence, the velocity (and not the angular momentum) $r\dot{\varphi}=V=\mbox{const.}$$, and $r\dot{t}=H=\mbox{const.}$$ Introducing these into the Euler-Lagrange equation for the radial coordinate,

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

gives

$$F_W=\frac{H^2}{r^2}-\frac{1}{c^2}\left(\frac{\dot{r}^2}{r^2}-\frac{\ddot{r}}{r}\right).$$ The terms in $\dot{\varphi}$ dropped out on their own accord.

The first term in the Weber force is the Coulomb potential; it enters in exactly the same way as the Newtonian potential does in general relativity, i.e., through the metric coefficient $dt^2$. This has nothing to do with a hyperbolic metric or the propagation of waves! It is, however, somewhat prophetic that $c^2$ enters in the same way as it does in Ampere's law, as the factor that allows a static force to be added onto Ampere's law.

What allows us to equated $F_W$ with the radial acceleration, $\ddot{r}-r\dot{\varphi}^2$? Nothing! The angular velocity appeared as a conserved quantity in the derivation, so that if we were to equate the Weber force with $\ddot{r}+V\omega$, it would not lead to

$$(1+6\mathcal{R}u)\frac{d^2u}{d\varphi^2}+u=\frac{\mathcal{R}}{L^2}-3\mathcal{R}\left(\frac{du}{d\varphi}\right)^2,$$

since it is not $L$ which is constant. The second term in the coefficient of the second derivative on the left-hand side is what gives the same advance as $3\mathcal{R}u^2$ in the general relativistic equation.

These are remarkable coincidences! Equating acceleration to acceleration seems a little redundant to put it mildly, but to come out with the correct numerical result is prophetic! Then again what has Coulomb's potential to do in the metric coefficient of the $dt^2$ is also unexplainable. Yet, it introduces $c^2$ which is necessary to relate it to the kinetic energy term in the Lagrangian. Moreover, the hyperbolic nature of the metric allows for the propagation of waves at speed $c$. Yet, as we know from the Weber force, these waves will be purely longitudinal.

Finally, it is interesting to remark that the non-conservation of the angular momentum in the Schwarzschild metric comes from a totally extraneous origin in the relativistic mass of the orbiting mass. The three momentum is [cf. Moller, *The Theory of Relativity* Eq. (12) p. 348]

$$m_0\Gamma\left(\frac{\dot{r}}{1-\alpha/r},r^2\dot{\vartheta},r^2\sin^2\vartheta\dot{\varphi}\right),$$

where $m_0$ is the rest mass of the peripheral body and $\Gamma=1/\surd(1-\alpha/r-v^2/c^2)$ is the relativistic factor that depends on the three-velocity,

$$v^2=\frac{\dot{r}^2}{1-\alpha/r}+r^2\dot{\vartheta}^2+r^2\sin^2\vartheta\dot{\varphi}^2.$$

We have set $\vartheta=\pi/2$, but, to quote Rabi. "who ordered this?" Not only is the central mass non-existent in the Schwarzschild metric, insofar as it is there but not there (emptiness!), we now have to do with an orbiting mass that is speed dependent! It is the $\Gamma=1/(1-\alpha/r)$ factor that destroys the conservation of angular momentum since $\Gamma r^2\omega=\mbox{const.}$, and not $r^2\omega$ by itself.

The same result for the expression of angular momentum can be obtained from Fermat's principle of least time [cf. *A New Perspective on Relativity, *Sec. 7.3] so we know it is correct. However, it is amazing to see the degree to which the general relativists go to get what they want.

The nagging questions remain as to why Coulomb's potential should appear in the metric coefficient of the squared time interval when propagation is not even being contemplated, and why the metric needs to be hyperbolic. But these are no stranger than the remarkable coincidences that appear in general relativity.