A "black" hole by definition is black to anyone observing it from the outside. But, is it "black" inside? The same gravitational potential exists whether "inside" or "outside". Then what distinguishes "inside" from "outside"? It was shown by Luneburg [Luneburg, The Mathematical Theory of Optics, p. 172.] that light rays acted upon by a Coulomb potential cannot penetrate into regions greater than a fourth of the inverse of the absolute value of the total energy. We claim that this should be revised to one-half the inverse of the total energy, which is the difference between the minimum and maximum values of radial coordinate of a ray. Here we develop the basic approach, and, in the next blog, we apply it to the Schwarzschild metric in a static gravitational potential. A variant of the phenomenon will give advance of the perihelion of Mercury, but with a completely different explanation. This will illustrate the complete lack of correspondence between the physics that is used to explain phenomena and what is actually happening. In Landau and Lifshitz, Classical Theory of Fields, $\S 101$, the ultra-relativistic Hamilton-Jacobi equation (HJ), $$g^{ij}\frac{\partial S}{\partial x^i}\frac{\partial S}{\partial x^j}-(mc)^2=0,$$ for the action $S$ is used to derive the advance of the perihelion of Mercury. In the HJ equation they introduce the relativistic energy and rest momentum, $mc$, only to claim later on that they use "the non-relativistic energy (without the rest energy)." Certainly, this should have raised eyebrows, but it didn't. It appears that anything goes as long as you come out with the correct numerical result, which, keeps changing as the technology advances. Even Kevin Brown, in his Reflections on Relativity uses numerical equivalence to discredit Paul Gerber's model of the deflection of light by a massive body even though it gives the correct numerical result for the advance of the perihelion of Mercury. This is absurd, and so too, the physics employed to derive the results. Why use the HJ equation, either ultra-relativistic, or non-relativistic, when Einstein's equations can do the job? Note that the HJ equation contains two masses, the central, $M$, and peripheral, $m$ mass---something that general relativity can't do. So what is the role of the peripheral mass when it enters the HJ equation through the rest momentum, $mc$? It serves only to introduce the Newtonian potential, and disappears when discussing the bending of light by a massive body---as it should. But, the physics is completely wrong, and has absolutely nothing to do with the phenomena of the perihelion advance and the bending of light. Even the generalized line element is completely absurd. We are dealing with static phenomena so that there is no distinction between proper and coordinate time. Are we to believe that proper time belongs to Mercury orbiting the sun and coordinate time to us on earth? The influence of a gravitational field on the motion of a massive body or light is liked to a medium with a varying index of refraction. Although the analogy between an inhomogeneous optical medium and the general relativistic formulation has long been known (beginning with Eddington in his Mathematical Theory of Relativity, it nevertheless leads to incorrect interpretations. An example is the red-shift due to the presence of a static potential. Here, one forgets the space part and transforms the proper and coordinate times to frequencies of emission and reception. The metric implies that this be described by a second-order Doppler shift, even though the geometry of observer and emitter has not been specified! When we throw a ball up in the air, the kinetic energy of the ball is being converted into potential energy so that it finally returns to the ground. Light has no mass so its velocity cannot change. What does change, however, is its wavelength; it becomes longer. And since the medium is static, there is no change in the frequency. So we come to the conclusions that gravity cannot cause red-shifts, and the phenomenon involved has nothing whatsoever to do with a Doppler shift. Frequency changes require a source of energy light a moving mobile; yet, here, there is none. The model we are searching for can be found in Luneburg's book, precisely equation (27.31), where he writes the square of the index of refraction as $$ n^2=C+\frac{1}{r}.$$ According to Luneburg, "[t]he light rays in this medium are identical with the paths of particles which move in a Coulomb field of potential $\phi=1/2r$, and with energy $C/2$." Luneburg gives the equation of the trajectory of light rays as $$\varphi-\varphi_0=\int_{r_0}^{r}\frac{Ldr}{r\sqrt{n^2r^2-L^2}},$$ where $L$ is the angular momentum, a constant of the motion. This can easily be derived from Fermat's principle of least time treating the angle variable, $\varphi$ as a cyclic, or ignorable, or kinosthenic, variable. Evaluating the integral he finds $$r=\frac{2L^2}{1+\sqrt{1+4CL^2}\cos\varphi},$$ which are conic sections. For $C>0$, or positive energy, the curves are hyperbolas, for $C=0$, parabolas, and, finally, for $C<0$, ellipses. We are interested in the latter. Extreme values of $r$ are determined by the vanishing of the denominator in the integral of the trajectory, i.e., $(nr)^2=L^2$, or specifically as $$r=\frac{1\pm\sqrt{1-4|C|L^2}}{2C}.$$ When this is equated with the maximum value of $r$ from the equation of the ellipse, $2L^2$, only the negative root is seen to be acceptable. Hence, the conclusion that no light ray can penetrate a region $r>1/2|C|$. This is, effectively, the inverse of a black hole, which we have obtained for a Coulomb potential. A Newtonian potential will work just as well. The equation for the orbit can be derived from the more general equation $$A\dot{r}^2+r^2\varphi^2=n^2,$$ where the index of refraction is $$n^2=B\dot{t}^2+C.$$ The coefficients $A$ and $B$ can be functions of $r$, which make them automatically, functions of $t$. In the case $A=B=1$, the energy integral has been defined as $\dot{t}$, where $t$ is an affine parameter, having absolutely nothing to do with proper time. This is another inaccuracy committed in making the analogy with the mechanical HJ equation. In a previous blog we have defined the Legendre transform with respect to the radial velocity and showed that $B=1/A$. Moreover, $t$ is a cyclic, ignorable, or kinosthenic, variable so that a first integral, $\dot{t}/A=c$, a constant. If $A\neq 1$, the first integral indicates a conservation in contraction with the above analogy with the total energy of the HJ equation. The proof lies in the fact that the radial equation is a geodesic equation and for $n=c/\surd r$, it coincides with the vanishing of the Weber force! Here, $A=1/r$, and the index of refraction given above becomes $$n^2=\frac{c^2}{r}+C,$$ where the constant $C$ still introduces the Coulomb field, $\phi=1/2r$ with energy $C/2$. There is, therefore, no need to transform, say, the Schwarzschild metric into a conformally related metric, $$d\tau^2=Bd\tau^{\prime\;2}$$ $$c^2 d\tau^{\prime\;2}=c^2 dt^2-A^2dr^2-Ad\Omega^2,$$ where $d\Omega$ is an element of a $2$-sphere. Now, since $g_{00}=1$, it is claimed that the energy integral, $\mathcal{E}=dt/ds'$ exists, and $t$, the coordinate time, is an affine parameter. However, this is not the Schwarzschild metric. Obviously not, it is the Beltrami metric for $B=1-(8\pi/3)G\varrho r^2/c^2$, where $1/\surd(G\varrho)$ is the so-called Newtonian free-fall time. Luneburg's choice $A=1/r$ will not give to a systematic (secular) change in the perihelion of the orbit, but Schwarzschild's choice $A=(1-\mathcal{R}/r)^{-1}$, where $\mathcal{R}$ is the Schwarzschild radius, will. It will also give the deflection of light about a massive body, In the former case, the index of refraction become imaginary $n^2=C$, while in the latter case, the constant $C$ disappears, and with it, the gravitational potential. The equation of the trajectory that correctly describes the advance of the perihelion is $$\frac{dr}{d\varphi}=\pm\frac{r^2}{L}\sqrt{-|C|^2+\frac{1}{r}|C|\mathcal{R}-\frac{L^2}{r^2}(1-\mathcal{R}/r)}.$$ The coefficients of the first two terms under the square root have no particularly interesting effect on the orbit (see above). It is "the change in the coefficient of ^1/r^2$", according to Landau and Lifshitz, that "leads to a more fundamental effect---to a systematic (secular) shift in the perihelion of the orbit." Albeit, that "orbit" only applies to Mercury. However, in the derivation of their expression, they have gone from an ultra-relativistic HJ equation to a negative, non-relativistic energy "without the rest energy", although the rest momentum was fundamental in introducing the Newtonian potential! Consistency appears to be lacking. In the bending of light we don't need the constant, $C$, and with an index of refraction given by $$n^2=\frac{c^2}{1-\mathcal{R}/r},$$ the equation of the orbit becomes $$\frac{dr}{d\varphi}=r^2\sqrt{\frac{1}{\Delta^2}-\frac{1}{r^2}(1-\mathcal{R}/r)},$$ where $\Delta=L/c$, the impact parameter. Neglect of the last term under the square root gives the equation $r=\Delta/\cos\varphi$ of a straight line passing at a distance $\Delta$ from the origin. It is the last term that is responsible for the bending of light due to a "fictitious" mass buried in $\mathcal{R}$. It is "fictitious" insofar as the expression for the coefficient $A$ has been derived under Einstein's condition of emptiness. However, that condition results from the assumption that $A$ is time independent. But, how can it be time independent when it depends on the radial coordinate, $r$? As I trust it is clear from the above analysis, there is no need for any of these shenanigans. Maxwell found no need to introduce a Ricci tensor when he derived the index of refraction for his "fish eye" . It turned out that the index of refraction led to a line element of a sphere that allows a mapping of points on the sphere to points in the plane via a stereographic projection. The light rays in such a medium are curves that a stereographic images of geodesics on the unit sphere, i.e. great circles. A great circle intersects the equator at two opposite points. Hence light rays will intersect the unit circle (the image of the equator) at antipodal points. Hence, the optical image, the antipode of the curve, has the same optical image as the original since it has the same path length. This is Maxwell's solution of a perfect optical instrument which images stymatically in $3D$. Maxwell had no idea he was trespassing into the world of non-euclidean elliptic geometry!