Your math might check out, but when are you going to consider the Physicality of the system ?

You need to consider the actual mechanism, including the 'bounces that blow away outer layers of the original star, and the fact that density will be greatest at the center, so that a BH's Event Horizon grows from the center outwards.
This is, then, a change in gravity as the event horizon grows. And as changes in gravity propagate at the speed of light, how does that change propagate out of the Event Horizon ?

Theory rather suggests conserved quantities, such as mass, charge and angular momentum, are preserved/encoded on the Event Horizon, and we don't know the ultimate fate of the collapsed matter, as we know of no force that can resist such crushing gravitational pressures.

Nice math presentation, but totally useless without considering the Physics.

6 hours ago, MigL said:

a BH's Event Horizon grows from the center outwards.

Until recently, I believed that the event horizon didn't form until the entire mass was inside its Schwarzschild radius. The rationale behind this belief was that when the mass was almost but not quite inside its Schwarzschild radius, there was no smaller radius for which the mass inside this radius was inside its Schwarzschild radius (though I wasn't considering the increasing density closer to the centre, assuming this would be insufficient). However, I derived a metric of a very simple model of a ball of matter of uniform density. To my surprise, when the entire ball of matter is inside the photon sphere (1.5 Schwarzschild radius), the event horizon forms at the centre and expands as the ball contracts (until the event horizon meets the radius of the ball at the Schwarzschild radius).

The reason the event horizon forms before the ball of matter has collapsed inside its Schwarzschild radius is because the event horizon depends on the gravitational potential which continues to become more negative inside the ball all the way to the centre, unlike the gravitational acceleration which is maximum at the surface in accordance with Birkhoff's theorem (shell theorem in Newtonian gravity). That is, my original belief was based on an erroneous view of what causes the event horizon (gravitational potential vs gravitational acceleration), though the metric makes it very clear.

6 hours ago, MigL said:

This is, then, a change in gravity as the event horizon grows. And as changes in gravity propagate at the speed of light, how does that change propagate out of the Event Horizon ?

For a spherically symmetric collapse there is no change in the gravitation outside of the original radius of the matter (assuming matter isn't ejected). This is also a consequence of Birkhoff's theorem. Thus, there cannot exist spherical gravitational waves.

1 hour ago, KJW said:

the event horizon forms at the centre and expands

I can't remember if it was Oppenheimer who originally worked this out.

1 hour ago, KJW said:

For a spherically symmetric collapse there is no change in the gravitation outside of the original radius

You're right.
The Event Horizon is always spherical.

For those interested, here is the metric I derived of a simple model of a ball of matter of mass [math]M[/math], radius [math]R[/math], and of uniform density:

[math](ds)^2 = f(r^2) (c\ dt)^2 - \dfrac{(dr)^2}{f(r^2)} - r^2 ((d\theta)^{2} + (\sin(\theta)\ d\phi)^2)[/math]

[math]\text{where:}[/math]

[math]f(r^2) = 1 - \dfrac{2GM}{c^2} \bigg(\dfrac{1}{2 \sqrt{R^2}} \Big(3 - \dfrac{r^2}{R^2}\Big)\ H(R^2 - r^2) + \dfrac{1}{\sqrt{r^2}}\ (1 - H(R^2 - r^2))\bigg)[/math]

[math]\text{and}\ \ H(x)\ \ \text{is the Heaviside step function:}[/math]

[math]H(x) = {\begin{cases}1, & x \geqslant 0\\0, & x \lt 0\end{cases}}[/math]

Edited Sunday at 02:35 AM3 days by KJW