# Deriving a Schwarzschild radius using relativistic mass

Introduction

I have shown below two different approaches to deriving the Schwarzschild radius. I know these are less rigorous than the derivation of the Schwarzschild solution however the $\frac{2GM}{c^{2}}$ term still shows up in the metric anyways which in a sense validates the classical method shown below.

Classical method

\begin{equation} \frac{GMm}{r} = \frac{1}{2}mv^{2} \:\:\:\text{and}\:\:\:v =c \:\:\rightarrow \:\: R_{S} = \frac{2GM}{c^{2}} \end{equation}

However the classical method seems less general because it seems to ignore the Lorenz transformations. I offer no judgments rather I am looking to understand why this derivation reconciles with the Schwarzschild metric better.

This next approach is analogous except we set the relativistic potential energy equal to the relativistic kinetic energy. Since we are working in a frame where the gravitating mass $M$ is centered at the origin we only apply the gamma factor to the small mass $m$.

Relativistic method

\begin{equation} \frac{GMm}{r\sqrt{1-\frac{v^{2}}{c^{2}}}} = \frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-mc^{2} \:\:\:\text{and}\:\:\:v =c \rightarrow R_{S} = \frac{GM}{c^{2}} \end{equation}

I feel that the second derivation is more natural but I am somewhat unconvinced it is correct. I was hoping one of you fine stack exchange users could shed some words of wisdom.

• possible duplicate link. Mar 29, 2016 at 14:09
• However, here you find all the proof link with all the details Mar 29, 2016 at 14:12
• Sorry, but I don't think your links are relevant to the question. OP seems to know how to derive the Schwarzschild radius from the Schwarzschild metric. But this might be helpful? Mar 29, 2016 at 15:29
• Do you have any sources for the alleged relativistic potential energy you wrote? Mar 29, 2016 at 22:56
• No I do not, I simply felt this particular inquiry was justified and so I take responsibility for it. I couldn't find any sources to papers which did referenced this particular analysis, so I decided to try PSE instead. If you are dissatisfied with my use of the term relativistic potential energy, try to understand that I wasn't sure what to call it. Mar 30, 2016 at 14:19

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.

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

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.

• Good answer. "No physical insight" might be a bit harsh though - it gets the dependencies on G and M right, but the correct prefactor is coincidence. Mar 29, 2016 at 15:31
• It's no coincidence since the weak field approximation leads to Newton's theory. Mar 29, 2016 at 16:29
• @PeterR: the coincidence is that the expression remains correct when we move outside the weak field approximation. Mar 29, 2016 at 16:52

John Rennie, I think we should clarify that when you go from the second metric to the first, you first perform the transformation $dx^2+dy^2 + dz^2=dr^2 + r^2 d\theta^2+r^2\sin^2\theta d\phi^2$, which gets you to the metric

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

Then in order to get rid of the coefficients on $r^2d\theta^2+r^2\sin^2\theta d\phi^2$, we must replace $r$ with $R$, where $r^2(1+\phi)=R^2$. By a handy coincidence then if $\phi= -C/r$ then it turns out that $dr=dR\left(1+\frac{1}{2}\phi - \frac{1}{2}\phi+O(\phi^2)\right)= dR\left(1+O(\phi^2)\right)$ so to linear order we arrive at your third metric.

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

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

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

$E_0 = mc^2$

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

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

The form of the SC metric has to change if you want to apply relativistically correct conservation of energy as opposed to the classical version. It's not just a matter of redefining the scaling distance. Specifically, you have to replace (1-rs/r) with (1-rs/2r)^2. The resulting metric is not a vacuum solution, but it is approximately so in the weak field limit and the curvature scalar is still zero. I'm no expert in GR, but those I've spoken to object to this formulation for various reasons that I don't fully understand. I find the new metric intriguing because it introduces a "cosmological variable" into the metric tensor so you don't need an ad hoc constant to account for the expansion of the universe. I'm not sure if it fits Hubble's data though. For more on this, check out: https://www.thenakedscientists.com/forum/index.php?topic=69595.0