# Smallest Black Hole and its Photon Sphere

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The Smallest Black Hole

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.

$R_s = \frac{2Gm}{c^2}$

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

$R_p > \frac{3}{2}R_s = \frac{2Gm}{c^2} = \frac{3}{2}R_s$

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 $\mathbf{q}(\lambda_2)$

Which gives the standard curve equation:

$ds = \int_{\lambda_1}^{\lambda_2} d\lambda \sqrt{|ds^2|}$

Considerations:

It is theoretically possible for the photon ring in the micro black hole case to be comprised of virtual particles instead of the usual on-shell photon sphere for macroscopic cases of the same phenomenon.

Edited by TheCosmologist

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