TheCosmologist Posted February 21, 2023 Share Posted February 21, 2023 (edited) 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 [math]A = 4\pi \ell^2[/math] [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: [math]R_s = \frac{2Gm}{c^2}[/math] 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: [math]\frac{2Gm}{c^2} = \ell_P[/math] Solving for the mass, we get: [math]m = \frac{\ell_P c^2}{2G}[/math] Next, we use the equation for the radius of the photon sphere: [math]R_p = \frac{3}{2}R_s[/math] Substituting in the expression for [math]R_s[/math], we have: [math]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[/math] 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 [math]R_p > \frac{3}{2}R_s = \frac{2Gm}{c^2} = \frac{3}{2}R_s[/math] Going back now to, [math]A = 4\pi \ell^2[/math] 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, [math]\Delta x\ c\ \Delta t \geq \ell^2_P[/math] we can state that this equation be taken to the Planck domain as the shortest interval or length: [math]ds^2 = g_{tt}\ \Delta x\ c\Delta t \geq \ell^2_P[/math] And of course, the equation can undergo a curve in the pseudo Reimannian manifold, which is akin to a curve between two Planck regions, [math]ds^2 = g\ \Delta x\ c\Delta t \geq \ell^2_P[/math] Using the spacetime uncertainty, [math]\Delta x\ c\ \Delta t \geq \ell^2_P[/math] we can state that this equation be taken to the Planck domain as the shortest interval or length: [math]ds^2 = g_{tt}\ \Delta x\ c\Delta t \geq \ell^2_P[/math] And of course, the equation can undergo a curve in the pseudo Reimannian manifold, which is akin to a curve between two Planck regions, [math]ds^2 = g\ \Delta x\ c\Delta t \geq \ell^2_P[/math] The metric can be rewritten as [math]ds^2 = g\ \Delta x\ c\Delta t = g(\Delta \mathbf{q}_1 \Delta \mathbf{q}_2) \geq \ell^2_P[/math] Where [math]\mathbf{q}[/math] is the infinitesimal displacement by generalised coordinates which acts on the integral as [math]\mathbf{q}(\lambda_1) [/math] and [math]\mathbf{q}(\lambda_2)[/math] Which gives the standard curve equation: [math]ds = \int_{\lambda_1}^{\lambda_2} d\lambda \sqrt{|ds^2|}[/math] 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 February 21, 2023 by TheCosmologist Link to comment Share on other sites More sharing options...

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