Why is a Black Hole Black?

Black holes started off as "frozen" stars until they were renamed black by Wheeler who also introduced the "no hair" theorem which has fallen from prominence together with the misconceived thermodynamics of black holes. The same Newtonian potential is at work in a black hole as in an ordinary star with the exception that its mass has surpassed 10 solar masses so that neither the electron pressure nor the neutron pressure will be sufficient to stop the star's collapse. Both Newtonian gravity and general relativity agree on this, and both predict a collapse to a point of infinite density. The relation that determines the critical radius, $\mathcal{R}=2GM/c^2$, is known as the Schwarzschild radius. We will derive it from elementary considerations of light rays in a static Newtonian field, and show that light rays cannot penetrate into regions greater than $\mathcal{R}$ so that for someone looking in from the outside they will a "black" hole.

The basic equation for the propagation of rays is


where $\Omega$ is a 2-sphere, in an inhomogeneous medium of  radial symmetry, the index of refraction 


where $A$, $B$, and $C$ are coefficients that may depend upon $r$.

We recall that to determine $B$, we observe that $t$ is a kinosthenic variable in that its derivative with respect to another affine parameter, indicated by the dot, appears but $t$ not. Routh, in his Stability of Motion, developed a general procedure for eliminating such variables by defining a new lagrangian


where $T$ is the Legendre transform on  the kinetic energy, $K=A\dot{r}^2/2$,

$$T(\dot{t})=\dot{r}\frac{\partial K}{\partial\dot{r}}-K(\dot{r}),$$

where we define a new variable

$$\dot{t}=\frac{\partial T}{\partial\dot{r}}.$$

Thus the "new" potential is


When $T(\dot{t})$ is subtracted off the original lagrangian


we get a new lagrangian

$$\mathcal{L}^{\prime}=\frac{1}{2}\left(A\dot{r}^2+r^2\dot{\Omega}^2-\frac{\dot{t}^2}{A}\right).$$ This shows that the above index of refraction can be written as


from which we conclude that $B=1/A$. We shall henceforth consider motion in the plane $\vartheta=\pi/2$, so that the 2-sphere reduces to $\dot{\Omega}^2=\dot{\varphi}^2$.

It has been known since the time of Maxwell, that absolute optical instruments, i.e. those that produce perfect and undistorted optical images, have rays of light that are confined to certain regions which include both the point to which the rays coalesce and its conjugate point. Maxwell showed that to any sphere there belongs a conjugate sphere with inverse radius, and to which their corresponds a perfect and undistorted optical image. 

A related medium has an index of refraction


for which light rays in such a medium are "identical with paths of particles which move in a Coulomb potential $\phi=-1/2r$, and with the energy $C/2$" (Luneburg). Depending on the sign of the total energy $C$, three different conic sections are possible. We will concentrate on ellipses for which $C<0$.  The above expression for the index of refraction implies that $A=1/r$, and as we have shown, the Euler-Lagrange (EL) equation for the radial coordinate,

$$\frac{d}{d\tau}\frac{\partial\mathcal{L}^{\prime}}{\partial\dot{r}}-\frac{\partial\mathcal{L}^{\prime}}{\partial r}=0,$$

where $\tau$ is another affine parameter (not proper time!) gives the vanishing of the Weber force. The other two EL equations




give the first integrals




the angular momentum, $L$, and the speed of light in vacuum, $c$. 

Hence, the modified Lagrangian is


This show that the introduction of the speed of light puts a cap on the radial velocity.

Instead of the original Coulomb potential, we will consider the modified gravitational potential introduced by Schwarzschild,


to first-order in $\mathcal{R}$, corresponding to a "weak" gravitational field. But, we are not considering the weak field limit so the second equality will not apply.

Thus, the Schwarzschild index of refraction is


where the first term, taken alone, gives the bending of light in the presence of a massive body, while the second term, taken alone, gives the advance of the perihelion, where $E=-C$. 

Setting the modified modified lagrangian


equal to zero gives the equation of the trajectory as


where $\Delta^2=L^2/E$, a term with dimensions of the square of the distance. It is clear from this expression that we are not considering the "weak" field limit, for, otherwise, the left-hand side would be negative, which is impossible. Writing out the terms explicitly, we have


Einstein, in his 1915 derivation of the advance of the perihelion of Mercury, began with the equation

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

where $u=1/r$, and $\epsilon$ and $\ell$ are the eccentricity and the semi-latus rectum  of the ellipse, respectively.  If we set $L=\ell$, making $L$ have dimensions of length, $1-\epsilon^2=E$, making $E$ dimensionless, then the coefficient of the second expression gives $E=L/\mathcal{R}$, again dimensionless.  Thus, we have an exact identity between our equation and Einstein's: This implies that $\mathcal{R}>r$, and we are definitely not in the "weak" field limit. We shall now see how this comes about.

The vanishing of the equation of the trajectory, $dr/d\varphi=0$, determines where $r(\varphi)$ achieves an extremum value. We may expect  in general that $r$ reaches a maximum and a minimum along a given ray. The last term in the expression is  responsible for the distortion of the ellipse (i.e., the advance of the perihelion) which we will neglect. The two roots are then given by


which indicate the aphelion and the perihelion, respectively. That the two roots be real requires


which can hardly be considered the weak field limit. This has implications on the magnitude of what is considered the perturbation term, $\mathcal{R}/r^3$, which we have discarded. The magnitude of the Schwarzschild radius depends on the magnitude of the central mass, $M$, so we don't expect the inequality to be fulfilled by any old star.

 The distance between the two, $r^{\star}=r_{+}+r_{-}$ is the maximum distance afforded to a light ray. Hence, no light ray can penetrate into a region


Consequently, anyone observing from the outside would see a "black" hole. 

Luneburg shows the possible ellipses with a common focal point at the origin and with the same principal axis $\mathcal{R}/2$, the "classical" Schwarzschild radius, in the diagram below.

Ellipses about a common focal point at the origin within a circle of radius $\mathcal{R}$.

The ellipses have a common focal point at the origin, and all are within the radius $\mathcal{R}$. Luneburg gives the radius "event horizon" as half the Schwarzschild radius. This would necessarily exclude the aphelion of the trajectory. 

Finally, to show consistency, the equation for a ray trajectory in the vicinity of a massive body is obtained from the index of refraction by setting $E=0$. We then obtain


which is again the identical expression obtained from general relativity, with an impact parameter $\Delta=L/c=\ell/\epsilon$, where the last equality follows from the general relativistic expression. Again, the last term is considered a perturbation, but, this time, it gives the same value as the bending of  light in general relativity.

Without it, the equation of the trajectory of the ray, $r=\Delta/\cos\varphi$, is a straight line passing at a distance $\Delta$, the so-called impact parameter, from the origin. There is only a single extremum point, $r_{min}=\Delta$, and the region of all $r>\Delta$ is bright.

Neither the advance of the perihelion nor the bending of light has anything whatsoever to do with a relativistic phenomenon involving proper and coordinate times. It is clear that the same formal appearance of the equations give the impression of a relativistic phenomenon, and this has been the  impediment to further progress in cosmology. From what we have said above, gravity appears as a static phenomenon that affects the propagation of light.

Moreover, our derivation places in doubt the seemingly smallness of the perturbation term, $\mathcal{R}/r^3$, since it is at least the same order of magnitude as the penultimate term in the equation of the trajectory. The two roots $r_{+}$ and $r_{-}$ will certainly be affected by a presence of a third root, invalidating Einstein's and Moller's assumption that the third root can be expressed as a function of the other two.