Why Black Holes Don't Exist in the Schwarzschild Metric, and How to Create One

We are told to imagine that within the Schwarzschild radius, a particle will spiral in to its doom hitting the singularity in an undetermined amount of time. All this comes from extrapolating the Schwarzschild metric beyond its domain of validity, $r<2GM/c^2$. This is undoubtedly why Karl Schwarzschild did not find any such sucking in of material, and why he went beyond his outer solution to create an inner solution for $r<\surd(G\varrho)/c$, where $\varrho$ is the density and $1/\surd(G\varrho)$ is the Newtonian free-fall time.

In our last blog we wrote the generalized Newtonian force as

$$F_N=L^2u^2\left(u+\frac{d^2u}{d\vartheta^2}\right),$$

where $L$ is the angular momentum per unit mass and $u=1/r$. Now the terms in parentheses can be written as a power series

$$\frac{d^2u}{d\vartheta^2}+u=A+Bu+Cu^2+Du^3+Eu^4+Fu^5\cdots$$

where all coefficients are positive. 

The Schwarzschild solution as terms up to $u^2$ with $B=0$, i.e., no in-spiraling. The first term $A=\mu/L^2$, where $\mu=GM$, the gravitational parameter, is the Keplerian ellipse

$$u=\frac{\mu}{L^2}\left(1+\epsilon\cos(\vartheta-\vartheta_0)\right),$$

where $\epsilon<1$ is the eccentricity. The famous square term is responsible for the advance of the perihelion and the deflection of light, where $C=3\mu/(Lc)^2$. Newton's law of inverse square corresponds to an ellipse where the inverse fourth power is a cardioid. The pinch in the cardioid has the effect of slightly rotating the ellipse, as seen below.

limacon

The force at work is therefore a combination of Newton's inverse square and an inverse quartic.

There are then three operative forces in the outer Schwarzschild metric: Newton's inverse square force, the centrifugal repulsion and an attractive inverse power law. However, it is not as Baez claims:

"It is as if on top of Newtonian gravity we had another attractive force obeying an inverse fourth power law. This overwhelms the others at shorter distance, so if you get too close to a black hole, you spiral in to your doom." 

He refers to a Wikipedia article, but the only thing that I found incorrect is the statement that "circular orbits are possible when the effective forces are zero." By "effective" forces it is meant the sum of the above. Moreover, where are the black holes? The coefficient $C=3\mu/c^2$ is so small that the rate of angular change of the line of apsides,

$$\Delta\omega=6\pi\left(\frac{\mu}{cL}\right)^2,$$

per period represents $43"$ per century in the case of Mercury! This hardly can be considered as the source of a black hole!

How then can we create a  black hole? It arises from the squared term above where $B=\mu/L^2$. This is a Cotes spiral, $r=a/\vartheta$. For simplicity, consider all other terms zero. The equation of the trajectory becomes

$$\frac{d^2u}{d\vartheta^2}+\left(1-\frac{\mu}{L^2}\right)u=0.$$

When gravitational attraction overcomes centrifugal repulsion,

$$L^2>\mu$$

the particle will spiral into the center, s shown below.

Cotes spirals

The Archimedes spiral $r=a\vartheta$ is a combination of inverse cube and inverse fifth. Thus, we can expect unstable orbital behavior from odd inverse powers of the force. Since there are no such terms in the Schwarzschild metric, there is no spiraling in of matter. QED

In the next blog we will discuss the centrifugal term, $L^2/r^2$, and its role in eliminating all angular dependencies in the laws of force. We can outline it briefly here. 

Consider the equation for the ellipse,

$$r=(A+B\cos\vartheta)^{-1}.$$

The square of the velocity is

$$v^2=\dot{r}^2+\dot{z}^2,$$

where $\dot{r}=B L\sin\vartheta$ and $\dot{z}=L/r$, having used $L=r^2\dot{\vartheta}$. The force is the negative of one half the derivative with respect to $r$, viz.,

$$F=-\frac{L^2}{r^2}\left(B\cos\vartheta-\frac{1}{r}\right)=\frac{\mu}{r^2},$$

where $\mu=AL^2$. Without the centrifugal force term we would get an angular dependency in Newton's inverse square law. All other forces are derivable in the same way. So instead of adding angle dependencies, the second term in the square of the velocities eliminates them.

This seemingly contradicts what Baez has to say about Newton's propositions 43-45 in Book I. According to Baez, Newton showed that if you add $L^2/r^3$ to the force, "you can find a motion with the same radial dependence but with a different angle dependency." The above example clearly shows that it eliminates the angle dependence!