Deriving a Schwarzschild radius using relativistic mass

The Newtonian derivation is not a derivation at all. It is a coincidental consequence of the way the Schwarzschild radial coordinate is defined. There is no physical insight to be gained from attempting to derive the Schwarzschild radius this way.

If we start with flat spacetime then the metric is:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

If we now introduce a weak gravitational field, where weak means that the gravitational potential per unit mass $\phi \ll c^2$, then we can use an approximation called the weak field limit to describe the curvature that corresponds to the weak gravitational field. In this approximation the metric becomes:

$$ ds^2 \approx -\left( 1 + \frac{2\phi}{c^2}\right) c^2dt^2 + \frac{1}{1 + 2\phi/c^2}\left(dx^2 + dy^2 + dz^2\right) \tag{1} $$

Remember that this approximation is only valid when $\phi \ll c^2$, but if we ignore this and blunder on regardless we would conclude that there is a coordinate singularity when:

$$ 1 + \frac{2\phi}{c^2} = 0 $$

or:

$$ \phi = -\tfrac{1}{2}c^2 $$

Both sides of this equation are an energy per unit mass, and putting the mass back in produces a possibly more familiar result:

$$ \phi m = -\tfrac{1}{2}mc^2 $$

which is exactly the argument used in the classical approach of calculating when the escape velocity reaches the speed of light.

If we rewrite the weak field equation (1) using polar coordinates:

$$ ds^2 \approx -\left( 1 + \frac{2\phi}{c^2}\right) c^2dt^2 + \frac{dr^2}{1 + 2\phi/c^2} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

then substitute the Newtonian expression for the gravitational potential:

$$ \phi = -\frac{GM}{R} $$

we get something that looks like the Schwarzschild metric:

$$ ds^2 \approx -\left( 1 - \frac{2GM}{c^2R}\right) c^2dt^2 + \frac{dr^2}{1 - 2GM/c^2R} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

But the Newtonian radial coordinate $R$ is not the same as the Schwarzschild radial coordinate $r$. The former is the distance measured from the central point to the position labelled by $R$ while the latter is the circumference of a circle passing through the position labelled by $r$ divided by $2\pi$. However it just so happens that the way the Schwarzschild radial coordinate is defined means that if we replace $R$ by $r$ we get an exact result:

$$ ds^2 = -\left( 1 - \frac{2GM}{c^2r}\right) c^2dt^2 + \frac{dr^2}{1 - 2GM/c^2r} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2 $$

And this is why the Newtonian derivation gives the correct result for $r_s$. It is just a coincidence and shouldn't be regarded as a derivation at all.

This is interesting, however I believe this is a coincidence.

Essentially the relativistic kinetic energy is found with the lorentz gamma factor and momentum:

$E_K[_R] = \int v dp = \int v d(m\gamma v) = ......$ $E_K[_R] = m\gamma c^2 - E_0$

$E_0 = mc^2$

Find gamma using binomial approximation or by taking the first two terms of the Taylor expansion for the reciprocal square root.

$\gamma = 1 + 1/2 v^2/c^2$

Sub $\gamma$ into $E_K[_R]$

$E_K[_R]$ = $mc^2(1 + 1/2 v^2/c^2)$ - $mc^2$

Reduces down to $E_K[_R] = 1/2mv^2$

Thus: $E_K[_R]$ = $E_K$.