Say a black hole's Schwarzchild radius is equal to the Planck length then a horizontal formula can be established as
$A = 4\pi \ell^2$
[1]. I've tried to find an analogue of this set up online but can't find any reading material on it. I'll expand further on this at the end and ask a few questions.
We start with the equation for the Schwarzschild radius:
$R_s = \frac{2Gm}{c^2}$
where $G$ is the gravitational constant, $m$ is the mass of the black hole, and $c$ is the speed of light.
If the black hole has a Schwarzschild radius equal to the Planck length, we have:
$\frac{2Gm}{c^2} = \ell_P$
Solving for the mass, we get:
$m = \frac{\ell_P c^2}{2G}$
Next, we use the equation for the radius of the photon sphere:
$R_p = \frac{3}{2}R_s$
Substituting in the expression for $R_s$, we have:
$R_p = \frac{3}{2}\cdot\frac{2Gm}{c^2} = \frac{3}{2}\cdot\frac{2G}{c^2}\cdot\frac{\ell_P c^2}{2G} = \frac{3}{2}\ell_P$
So the radius of the photon sphere in this case is indeed 1.5 times larger than the Planck length, as expected. Then a standard formula would be
$R_p > \frac{3}{2}R_s = \frac{2Gm}{c^2} = \frac{3}{2}R_s$
I have a few questions, this 1.5 difference, can't it simply be calculated, given that you know $R_p$ as
$R_p - \frac{3}{2}R_s$
Well I took a look, its generally said the photon sphere follows
$R_P = \frac{3Gm}{c^2} = \frac{3}{2}R_s$
instead of the usual
$R_s = \frac{2Gm}{c^2}$
Can anyone explain how you obtain these two different results? And why in the formula
$R_P = \frac{3Gm}{c^2} = \frac{3}{2}R_s$
We have a 2 in the denominator on the far RHS but only a 3 in the numerator other middle expression? I would have thought if
$R_s = \frac{2Gm}{c^2}$
Then
$R_s = \frac{(3 \cdot 2)Gm}{2c^2}$
Would lead to...
$R_P = \frac{3Gm}{c^2}$
Right?
Going back now to,
$A = 4\pi \ell^2$
Surely this would be the smallest area that is computable within physics, since physics breaks down below the Planck scales? I've been trying to visualise such a small object and the immense curvature it should possess as posed by general relativity. I took my sights to using the spacetime uncertainty,
$\Delta x\ c\ \Delta t \geq \ell^2_P$
we can state that this equation be taken to the Planck domain as the shortest interval or length:
$ds^2 = g_{tt}\ \Delta x\ c\Delta t \geq \ell^2_P$
And of course, the equation can undergo a curve in the pseudo Reimannian manifold, which is akin to a curve between two Planck regions,
$ds^2 = g\ \Delta x\ c\Delta t \geq \ell^2_P$
Using the spacetime uncertainty, $\Delta x\ c\ \Delta t \geq \ell^2_P$ we can state that this equation be taken to the Planck domain as the shortest interval or length:
$ds^2 = g_{tt}\ \Delta x\ c\Delta t \geq \ell^2_P$
And of course, the equation can undergo a curve in the pseudo Reimannian manifold, which is akin to a curve between two Planck regions,
$ds^2 = g\ \Delta x\ c\Delta t \geq \ell^2_P$
The metric can be rewritten as
$ds^2 = g\ \Delta x\ c\Delta t = g(\Delta \mathbf{q}_1 \Delta \mathbf{q}_2) \geq \ell^2_P $
Where $\mathbf{q}$ is the infinitesimal displacement by generalised coordinates which acts on the integral as $\mathbf{q}(\lambda_1)$ and $\mathbf{q}(\lambda_2)$
$ds = \int_{\lambda_1}^{\lambda_2} d\lambda \sqrt{|ds^2|}$
Can this describe the surface of the black hole in terms of its curvature along a line element?
Any insights as well into more specific and maybe better derivatives into the photon sphere would be much appreciated!
