User:JohnOwens/Sandbox

If anyone else takes a look here, note that these are not necessarily correct; there seems to be something wrong with my derivations, which is why I'm poking around here.

Temp links

 * desert (disambiguation)
 * desert (ethics)
 * desert (morality)
 * desert (philosophy)


 * Ampere

My derivations

 * $$R c^2 = 2 G M \; R = {2 G \, M \over c^2} \; M = {c^2 R \over 2 G}$$
 * $$E = M \, c^2 = {c^4 \, R \over 2 G}$$
 * $$S = {k \, A \over 4 \ell_P^2} = {k c^3 \, A \over 4 \hbar G} = {\pi k c^3 \, A \over 2 h G}$$
 * $$T = {E \over S}$$ - no guarantee over this, or that the $$E$$ here is the one equal to $$M c^2$$.
 * $$\sigma = {2 \pi^5 k^4 \over 15 c^2 h^3} = {2 \pi^5 k^4 \over 15 c^2 2^3 \pi^3 \hbar^3} = {\pi^2 k^4 \over 60 c^2 \hbar^3}$$
 * $$q = e_b A\,$$

Old A

 * $$A = 4 \pi \, R^2 = 4 \pi \left({2 G \, M \over c^2}\right)^2 = 4 \pi {4 G^2 \, M^2 \over c^4} = {16 \pi G^2 \, M^2 \over c^4}$$ - but see Weisstein's section below and New A section
 * $$S = {\pi k c^3 \, R^2 \over \hbar G} = {2 \pi^2 k c^3 \, R^2 \over h G}$$
 * $$S = {4 \pi k G \, M^2 \over \hbar c} = {8 \pi^2 k G \, M^2 \over h c}$$
 * $$T = {E \over S} = M \, c^2 {\hbar c \over 4 \pi k G \, M^2} = {c^4 \, R \over 2 G} {\hbar G \over \pi k c^3 \, R^2}$$
 * $$T = {h c^3 \over 8 \pi^2 k G \, M} = {\hbar c^3 \over 4 \pi k G \, M}$$
 * $$T = {h c \over 4 \pi^2 k \, R} = {\hbar c \over 2 \pi k \, R}$$


 * $$e_b = \sigma \, T^4 = {\pi^2 k^4 \over 60 c^2 \hbar^3} \left ({\hbar c^3 \over 4 \pi k G \, M}\right )^4 = {2 \pi^5 k^4 \over 15 c^2 h^3} \left ({h c^3 \over 8 \pi^2 k G \, M}\right )^4$$
 * $$e_b = {\pi^2 k^4 \over 60 c^2 \hbar^3} \frac{\hbar^4 c^{12}}{256 \pi^4 k^4 G^4 \, M^4} = {2 \pi^5 k^4 \over 15 c^2 h^3} \frac{h^4 c^{12}}{4096 \pi^8 k^4 G^4 \, M^4}$$
 * $$e_b = {\hbar c^{10} \over 15,360 \pi^2 G^4 \, M^4} = {h c^{10} \over 30,720 \pi^3 G^4 \, M^4}$$
 * $$q = \sigma \, T^4 A = {\sigma h^4 c^4 \over 64 \pi^7 k^4 \, R^2} = {h c^2 \over 480 \pi^2 \, R^2}$$
 * $$q = {\pi^2 k^4 \over 60 c^2 \hbar^3} \left ({\hbar c^3 \over 4 \pi k G \, M}\right )^4 {16 \pi G^2 \, M^2 \over c^4}$$
 * $$q = {\pi^2 k^4 \over 60 c^2 \hbar^3} \frac{\hbar^4 c^{12}}{256 \pi^4 k^4 G^4 \, M^4} {16 \pi G^2 \, M^2 \over c^4}$$
 * $$q = \frac{c^6 \hbar}{960 \pi G^2 \, M^2}$$

Repeats

 * $$R c^2 = 2 G M \; R = {2 G \, M \over c^2} \; M = {c^2 R \over 2 G}$$
 * $$E = M \, c^2 = {c^4 \, R \over 2 G}$$
 * $$S = {k \, A \over 4 \ell_P^2} = {k c^3 \, A \over 4 \hbar G} = {\pi k c^3 \, A \over 2 h G}$$
 * $$T = {E \over S}$$ - no guarantee over this, or that the $$E$$ here is the one equal to $$M c^2$$.
 * $$\sigma = {2 \pi^5 k^4 \over 15 c^2 h^3} = {2 \pi^5 k^4 \over 15 c^2 2^3 \pi^3 \hbar^3} = {\pi^2 k^4 \over 60 c^2 \hbar^3}$$
 * $$q = e_b A\,$$

Reformulation

 * $$A = {8 \pi G^2 \, M^2 \over c^4} = {8 \pi G^2 \over c^4} \left({c^2 \, R \over 2 G}\right)^2 = {8 \pi G^2 \over c^4} {c^4 \, R^2 \over 4 G^2}$$ - see Weisstein's section below
 * $$A = {2 \pi R^2}\,$$
 * $$S = {k c^3 \over 4 \hbar G} {2 \pi R^2} = {\pi k c^3 \over 2 h G} {2 \pi R^2} = {k c^3 \over 4 \hbar G} {8 \pi G^2 \, M^2 \over c^4} = {\pi k c^3 \over 2 h G} {8 \pi G^2 \, M^2 \over c^4}$$
 * $$S = {\pi k c^3 \, R^2 \over 2 \hbar G} = {\pi^2 k c^3 \, R^2 \over h G} = {2 \pi k G \, M^2 \over \hbar c} = {4 \pi^2 k G \, M^2 \over h c}$$
 * $$T = {E \over S}$$
 * $$T = {c^4 \, R \over 2 G} {2 \hbar G \over \pi k c^3 \, R^2} = {c^4 \, R \over 2 G} {h G \over \pi^2 k c^3 \, R^2} = M \, c^2 {\hbar c \over 2 \pi k G \, M^2} = M \, c^2 {h c \over 4 \pi^2 k G \, M^2}$$
 * $$T = {\hbar c \over \pi k \, R} = {h c \over 2 \pi^2 k \, R} = {\hbar c^3 \over 2 \pi k G \, M} = {h c^3 \over 4 \pi^2 k G \, M}$$

Damn, the different area formula is making things even worse?!? Maybe it's the definition of $$E$$ that's not what I'd expect? It needn't be the mass-energy equivalence of the entire mass of the black hole.


 * $$T = {E \over S} = {E \over {2 \pi k G \, M^2 \over \hbar c}} = {E \hbar c \over 2 \pi k G \, M^2} = {\hbar c^3 \over 8 \pi k G \, M}$$
 * $$E = {\hbar c^3 \over 8 \pi k G \, M} {2 \pi k G \, M^2 \over \hbar c} = {c^2 \, M \over 4}$$ - Why this apparent equivalence? I have no idea. It could also be that $$T = {E \over 4 S}$$.

Let's try a comparison with the gravitational binding energy formula. I doubt it has any relevance, but I'll see what it generates.


 * $$U = {3 G \, M^2 \over 5 \, r} = {3 G \, M^2 \over 5 {2 G \, M \over c^2}} = {3 G c^2 \, M^2 \over 10 G \, M} = {3 c^2 \, M \over 10}$$ - Well, it's in the ballpark, and I'm pretty sure that binding energy formula doesn't allow for relativistic effects. But I still don't think it has anything to do with it.

Using my new S &amp; T

 * $$e_b = \sigma \, T^4 = {2 \pi^5 k^4 \over 15 c^2 h^3} \left ({h c^3 \over 4 \pi^2 k G \, M}\right )^4 = {\pi^2 k^4 \over 60 c^2 \hbar^3} \left ({\hbar c^3 \over 2 \pi k G \, M}\right )^4$$
 * $$e_b = {2 \pi^5 k^4 \over 15 c^2 h^3} \frac{h^4 c^{12}}{256 \pi^8 k^4 G^4 \, M^4} = {\pi^2 k^4 \over 60 c^2 \hbar^3} \frac{\hbar^4 c^{12}}{16 \pi^4 k^4 G^4 \, M^4}$$
 * $$e_b = {h c^{10} \over 1920 \pi^3 G^4 \, M^4} = {\hbar c^{10} \over 960 \pi^2 G^4 \, M^4}$$
 * $$q = {h c^{10} \over 1920 \pi^3 G^4 \, M^4} A = {\hbar c^{10} \over 960 \pi^2 G^4 \, M^4} A$$
 * $$q = {h c^{10} \over 1920 \pi^3 G^4 \, M^4} {8 \pi G^2 \, M^2 \over c^4} = {\hbar c^{10} \over 960 \pi^2 G^4 \, M^4} {8 \pi G^2 \, M^2 \over c^4}$$
 * $$q = {h c^6 \over 240 \pi^2 G^2 \, M^2} = {\hbar c^6 \over 120 \pi G^2 \, M^2}$$

Using their S &amp; T

 * $${\hbar c^3 \over 8 \pi k G \, M} = {h c^3 \over 16 \pi^2 k G \, M}$$
 * $$e_b = \sigma \, T^4 = {2 \pi^5 k^4 \over 15 c^2 h^3} \left({h c^3 \over 16 \pi^2 k G \, M}\right)^4 = {\pi^2 k^4 \over 60 c^2 \hbar^3} \left({\hbar c^3 \over 8 \pi k G \, M}\right)^4$$
 * $$e_b = {2 \pi^5 k^4 \over 15 c^2 h^3} {h^4 c^{12} \over 65,536 \pi^8 k^4 G^4 \, M^4} = {\pi^2 k^4 \over 60 c^2 \hbar^3} {\hbar^4 c^{12} \over 4,096 \pi^4 k^4 G^4 \, M^4}$$
 * $$e_b = {h c^{10} \over 491,520 \pi^3 G^4 \, M^4} = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4}$$
 * $$P = {h c^{10} \over 491,520 \pi^3 G^4 \, M^4} A = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4} A$$
 * $$P = {h c^{10} \over 491,520 \pi^3 G^4 \, M^4} {8 \pi G^2 \, M^2 \over c^4} = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4} {8 \pi G^2 \, M^2 \over c^4}$$
 * $$P = {h c^6 \over 61,440 \pi^2 G^2 \, M^2} = {\hbar c^6 \over 30,720 \pi G^2 \, M^2}$$

as in article

 * $$T_H = {\kappa \over 2 \pi}$$ (natural units)
 * $$T = {\hbar\,c^3 \over 8 \pi G M k}$$
 * $$P = {\hbar\,c^6 \over 15360\,\pi\,G^2 M^2}$$
 * $$t_{\operatorname{ev}} = {5120\,\pi\,G^2 M_0^{\,3} \over \hbar\,c^4}$$

examples in article

 * $$T_{BH_{sol}} = 60 \mathrm{nK} = 6 \times 10^{-8} \mathrm{K}$$
 * $$T_{BH}(4.5 \times 10^{22} \mathrm{kg}) = 2.7 \mathrm{K}$$
 * $$P_{sol} = 10^{-28} \mathrm{W}\,$$
 * $$t_{\mathrm{ev}_{sol}} = 10^{67} \mathrm{year} = 3 \times 10^{74} \mathrm{s}$$
 * $$t_\mathrm{ev}(10^{11} \mathrm{kg}) = 3 \times 10^9 \mathrm{year} = 9 \times 10^{16} \mathrm{s}$$


 * $$P = 3.563\,45 \times 10^{32} \left[\frac{\mathrm{kg}}{M}\right]^2 \mathrm{W}$$
 * $$t_\mathrm{ev} = 8.407\,16 \times 10^{-17} \left[\frac{M_0}{\mathrm{kg}}\right]^3 \mathrm{s} \ \ \approx\ 2.66 \times 10^{-24} \left[\frac{M_0}{\mathrm{kg}}\right]^3 \mathrm{yr}$$
 * $$M_0 = 2.282\,71 \times 10^5 \left[\frac{t_\mathrm{ev}}{\mathrm{s}}\right]^{1/3} \mathrm{kg} \ \ \approx\ 7.2 \times 10^7 \left[\frac{t_\mathrm{ev}}{\mathrm{yr}}\right]^{1/3} \mathrm{kg}$$


 * $$t_\mathrm{ev}(2.28 \times 10^5 \mathrm{kg}) = 1 \mathrm{s}$$
 * $$E(2.28 \times 10^5 \mathrm{kg}) = 2.05 \times 10^{22} \mathrm{J}$$
 * $$P(2.28 \times 10^5 \mathrm{kg}) = 6.84 \times 10^{21} \mathrm{W}$$

derivations

 * $$P = \sigma T^4 A = {\pi^2 k^4 \over 60 c^2 \hbar^3} \left ({\hbar\,c^3 \over 8 \pi G M k}\right )^4 A$$
 * $$P = {\pi^2 k^4 \over 60 c^2 \hbar^3} \frac{\hbar^4 c^{12}}{4096 \pi^4 k^4 G^4 M^4} A$$
 * $$P = {c^{10} \hbar \over 245,760 \pi^2 G^4 M^4} A$$
 * $$A {c^{10} \hbar \over 245,760 \pi^2 G^4 M^4} = {\hbar\,c^6 \over 15360\,\pi\,G^2 M^2}$$
 * $$A = \frac{\frac{\hbar\,c^6}{15,360\,\pi\,G^2 M^2}}{\frac{c^{10} \hbar}{245,760 \pi^2 G^4 M^4}}$$
 * $$A = \frac{\hbar\,c^6}{15,360\,\pi\,G^2 M^2} \frac{245,760 \pi^2 G^4 M^4}{c^{10} \hbar}$$
 * $$A = \frac{16 \pi G^2 M^2}{c^4}$$

Odd, it does come out the same as the usual formula as surface area of a sphere. My mistake must lie elsewhere.

compared to my derivations

 * $$T_{article} = {T_{mine} \over 2}$$
 * $$P_{article} = {q_{mine} \over 16}$$

For some reason, $$T_{mine}$$ is coming out twice as large as $$T_{article}$$. It naturally follows that this would lead to $$q_{mine}$$ being 16 times larger than $$P_{article}$$. I thought event horizons might have a surface area that was not equal to that of a conventional sphere, but that doesn't seem to explain it.

as in article

 * $$S = \frac{A k c^3}{4 \hbar G}$$
 * $$T = \frac{\hbar c^3}{8 \pi k G M}$$

Black hole thermodynamics
(was Black hole entropy)

as in article

 * $$S_{BH} = \frac{k A}{4 l_{\mathrm{P}}^2}$$
 * $$l_{\mathrm{P}}=\sqrt{G \hbar \over c^3}$$

Black hole thermodynamics
(was Laws of black hole mechanics)

as in article

 * $$T_H = \frac{\kappa}{2 \pi}$$ (natural units)
 * $$S_{BH} = \frac{A}{4}$$ (natural units)

as in article
I noticed this in the table of equations with their nondimensionalized equivalents:


 * Thermal energy per particle per degree of freedom
 * $$E = \frac{1}{2} k T$$

This may be the key, if it's half of $$k T$$, much like $$E = \frac{1}{2} m v^2$$ threw me off with that half, when I was very young.

derivations

 * $$2 E = k T\,$$
 * $$E = \frac{k T}{2}$$, $$T = \frac{2 E}{k}$$, $$k = \frac{2 E}{T}$$

as in article

 * $$r_s = \frac{2 G m}{c^2}$$
 * $$r_s = m \times 1.48 \times 10^{-27}$$

as in article

 * $$S = {A k c^3 \over 4 G \hbar}$$
 * $$A = {8 \pi M^2 G^2 \over c^4}$$ - Here's where it's different! That's not the formula for an undistorted sphere.
 * $$S = {2 \pi k G M^2 \over \hbar c}$$

Black Hole Radiation and Volume Statistical Entropy by Mario Rabinowitz

 * Google HTML cache
 * Arxiv PDF

as in article

 * $$P_{SH} = {G \rho \hbar \over 90}$$
 * $$P_{SH} = {\hbar c^6 \over 960 \pi G^2 \, M^2}$$
 * $$\rho = {M \over ({4 \pi \over 3}) R_H^3} = {M \over ({4 \pi \over 3}) ({2 G \, M \over c^2})^3} = {3 c^6 \over 32 \pi G^3 \, M^2}$$
 * $$S_{bh} = {k A c^3 \over 12 \hbar G}$$
 * $$S_{bh} = \frac{1}{3} S_{Bek} = \frac{1}{3} \left[{k A c^3 \over 4 \hbar G}\right]$$

Solar-mass black hole

 * $$M = 1.9889 \times 10^{30} \mathrm{kg}$$
 * $$R = 2953.3 \mathrm{m}\,$$
 * $$E = 1.7876 \times 10^{47} \mathrm{J}$$
 * $$A = 2 \pi R^2 = 5.4800 \times 10^7 \mathrm{m^2}$$ or $$A = 4 \pi R^2 = 1.0960 \times 10^8 \mathrm{m^2}$$
 * $$S = {2 \pi k G \, M^2 \over \hbar c} = 7.2427 \times 10^{53} \mathrm{J/K}$$
 * $$T = {\hbar c^3 \over 8 \pi k G \, M} = 6.1702 \times 10^{-8} \mathrm{K}$$
 * $$P = {\hbar c^6 \over 15,360 \pi G^2 \, M^2} = 9.0081 \times 10^{-29} \mathrm{W}$$


 * $$e_b = {P \over A}$$
 * $$e_b = {P \over 2 \pi R^2} = 1.6438 \times 10^{-36} \mathrm{W/m^2}$$ or $$e_b = {P \over 4 \pi R^2} = 8.2191 \times 10^{-37} \mathrm{W/m^2}$$
 * $$e_b = \sigma T^4 = 5.6705 \times 10^{-8} \mathrm{W \over m^2 K^4} \, (6.1702 \times 10^{-8} \mathrm{K})^4 = 8.2191 \times 10^{-37} \mathrm{W/m^2}$$
 * $$e_b = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4} = 8.2191 \times 10^{-37} \mathrm{W/m^2}$$

1-second black hole

 * $$M = 2.2827 \times 10^5 \mathrm{kg}$$
 * $$R = 3.3895 \times 10^{-22} \mathrm{m}$$
 * $$E = 2.0516 \times 10^{22} \mathrm{J} = 5 \times 10^{12} \mathrm{ton_{TNT}}$$
 * $$A = 2 \pi R^2 = 7.2185 \times 10^{-43} \mathrm{m^2}$$ or $$A = 4 \pi R^2 = 1.4437 \times 10^{-42} \mathrm{m^2}$$
 * $$S = {2 \pi k G \, M^2 \over \hbar c} = 9540.3 \mathrm{J/K}$$
 * $$T = {\hbar c^3 \over 8 \pi k G \, M} = 5.3761 \times 10^{17} \mathrm{K}$$
 * $$P = {\hbar c^6 \over 15,360 \pi G^2 \, M^2} = 6.8386 \times 10^{21} \mathrm{W}$$


 * $$e_b = {P \over 2 \pi R^2} = 9.4738 \times 10^{63} \mathrm{W/m^2}$$ or $$e_b = {P \over 4 \pi R^2} = 4.7369 \times 10^{63} \mathrm{W/m^2}$$
 * $$e_b = \sigma T^4 = 5.6705 \times 10^{-8} \mathrm{W \over m^2 K^4} \, (5.3761 \times 10^{17} \mathrm{K})^4 = 4.7369 \times 10^{63} \mathrm{W/m^2}$$
 * $$e_b = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4} = 4.7369 \times 10^{63} \mathrm{W/m^2}$$

Natural units

 * $$E_\mathrm{P} = {m_\mathrm{P} \ell_\mathrm{P}^2 \over t_\mathrm{P}^2}$$
 * $$P_\mathrm{P} = {m_\mathrm{P} \ell_\mathrm{P}^2 \over t_\mathrm{P}^3}$$
 * $$\sigma = {\pi^2 \over 60} = 0.16449 {E_\mathrm{P} \over t_\mathrm{P} \ell_\mathrm{P}^2 T_\mathrm{P}^4} = 0.16449 {m_\mathrm{P} \over t_\mathrm{P}^3 T_\mathrm{P}^4}$$

Solar-mass black hole

 * $$M = 9.1373 \times 10^{37} m_\mathrm{P}$$
 * $$R = 1.8275 \times 10^{38} \ell_\mathrm{P}$$
 * $$E = 9.1373 \times 10^{37} E_\mathrm{P}$$
 * $$A = 2 \pi \, R^2 = 2.0983 \times 10^{77} \ell_\mathrm{P}^2$$ or $$A = 4 \pi \, R^2 = 4.1966 \times 10^{77} \ell_\mathrm{P}^2$$
 * $$S = {2 \pi k G \, M^2 \over \hbar c} = 5.2458 \times 10^{76} E_\mathrm{P}/T_\mathrm{P}$$
 * $$T = {\hbar c^3 \over 8 \pi k G \, M} = 4.3546 \times 10^{-40} T_\mathrm{P}$$
 * $$P = {\hbar c^6 \over 15,360 \pi G^2 \, M^2} = 2.4821 \times 10^{-81} P_\mathrm{P}$$
 * $$e_b = {P \over 2 \pi \, R^2} = 1.1829 \times 10^{-158} P_\mathrm{P}/\ell_\mathrm{P}^2$$ or $$e_b = {P \over 4 \pi \, R^2} = 5.9146 \times 10^{-159} P_\mathrm{P}/\ell_\mathrm{P}^2$$
 * $$e_b = \sigma T^4 = 5.9146 \times 10^{-159} P_\mathrm{P}/\ell_\mathrm{P}^2 \quad T^4 = 3.5956 \times 10^{-158} T_\mathrm{P}^4$$
 * $$e_b = {\hbar c^{10} \over 245,760 \pi^2 G^4 \, M^4} = 5.9146 \times 10^{-159} P_\mathrm{P}/\ell_\mathrm{P}^2$$