sethna/tex/5.4.tex

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\title{Problem 5.4}
\subtitle{Black hole thermodynamics}
\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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	\maketitle
	
	A black hole of mass $M$ has radius
	\begin{equation}
		R_s = G \frac{2M}{c^2}.
	\end{equation}
	
	Hawking calculated that blackbody radiation emitted from black hole is at temperature:
	\begin{equation}
		T_{bh} = \frac{\hbar c^3}{8 \pi G M \kb}
	\end{equation}
	And use $E = M c^2$ as needed.
	
	\subsubsection*{(a) Specific heat}
	Calculate the specific heat of the black hole.
	
	\subsubsection*{(b)}
	Calculate the entropy of the black hole, by using the definition of temperature $\flatfrac{1}{T} = \pdv*{S}{E}$ and assuming the entropy is zero at mass $M = 0$.
	Express your result in terms of the surface area $A = 4 \pi R_s^2$, measured in units of the Planck length $L^\ast = \sqrt{\flatfrac{\hbar G}{c^3}}$.
	
	\subsubsection*{(c)}
	Calculate the maximum number of bits that can be stored in a sphere of radius one centimeter, in terabytes.
	The Planck length $L^\ast = \sqrt{\flatfrac{\hbar G}{c^3}} \approx \SI{1.6e-35}{\m}$.
	
	\section{Solution} \label{sec:solution}
	
	\subsection*{(a)}

	Ultimately we're going to want $C = \pdv{E}{T}$.
	To start let's get $T_{bh}$ out of massland.
	\begin{align}
		T_{bh} &= \frac{\hbar c^3}{8 \pi G M \kb} \\
		T_{bh} &= \frac{\hbar c^3}{8 \pi G \frac{E}{c^2} \kb} \\
		E &= \frac{\hbar c^5}{8 \pi G T_{bh} \kb} \\
		E &= \frac{\hbar c^5}{8 \pi G \kb} \frac{1}{T_{bh}} \label{eq:tempdepe} \\
		\pdv{E}{T} &= \frac{\hbar c^5}{8 \pi G \kb} \pdv{}{T} \frac{1}{T_{bh}} \\
		\pdv{E}{T} &= -\frac{\hbar c^5}{8 \pi G \kb} \frac{1}{T_{bh}^2}
	\end{align}
	
	This is negative, as we were told to expect.

	\subsection*{(b)}
	Our definiton of temperature is $\flatfrac{1}{T} = \pdv*{S}{E}$, so presumable we want to do some kind of guy like
	\begin{equation}
		S(E) - S(E = 0) = \int_{0}^{E} \dd{E'} \frac{1}{T(E')}
	\end{equation}
	We know that $S = 0$ when $M = 0$, which is the same as $E = 0$, so then

	\begin{equation}
		S(E) = \int_{0}^{E} \dd{E'} \frac{1}{T(E')}
	\end{equation}
	and for $T(E)$ we just plug in an inversion of \cref{eq:tempdepe}.

	\begin{align}
		\frac{1}{T_{bh}} &= \frac{8 \pi G \kb}{\hbar c^5} E \\
		\int_{0}^{E} \dd{E'} \frac{1}{T(E')} &= \int_{0}^{E} \dd{E'} \frac{8 \pi G \kb}{\hbar c^5} E' \\
		S(E) &= \frac{4 \pi G \kb}{\hbar c^5} E^2 \label{eq:entropyunsimple}
	\end{align}
	
	Now we want to write this in better units.
	The Schwarzschild radius is
	\begin{equation}
		R_s = G \frac{2 E}{c^4},
	\end{equation}
	so the surface area $A$ follows
	\begin{align}
		A &= 4 \pi \left(G \frac{2 E}{c^4}, \right)^2 \\
		A &= 16 \pi G^2 \frac{E^2}{c^8}
	\end{align}
	We're told we want this in units of Planck lengths, so
	\begin{align}
		\frac{A}{(L^\ast)^2} &= 16 \pi G^2 \frac{E^2}{c^8} \frac{1}{(L^\ast)^2} \\
		\frac{A}{(L^\ast)^2} &= 16 \pi G^2 \frac{E^2}{c^8} \frac{1}{\flatfrac{\hbar G}{c^3}} \\
		\frac{A}{(L^\ast)^2} &= 16 \pi G^2 \frac{E^2}{c^8} \frac{c^3}{\hbar G} \\
		A_\ast &= 16 \pi G \frac{E^2}{\hbar c^5}
	\end{align}
	Plug this in our $S(E)$ from \cref{eq:entropyunsimple}:
	\begin{align}
		S(E) &= \frac{4 \pi G \kb}{\hbar c^5} E^2 \\
		S(E) &= 16 \pi G \frac{E^2}{\hbar c^5} \frac{\kb}{4} \\
		S(E) &= A_\ast \frac{\kb}{4}
	\end{align}
	and that's a pretty simple result.
	Neato.
	
	\subsection*{(c)}
	A bit has two states, so its entropy is $\kb \log 2$.
	And if we have a sphere of radius $\SI{1}{\cm}$, that can be written in Planck lengths as $\SI{1}{\cm} \approx 6.25 \times 10^{32} L^\ast$.
	So the area is $4 \pi$ times that radius squared, and then we divide by 4, then by $\kb \log 2$, we get our answer:
	\begin{equation}
		\frac{S_{\mathrm{black hole}}}{S_{\mathrm{bit}}} \approx \SI{1.77e66}{\bit} = \SI{2.21e53}{\tera\byte}
	\end{equation}
	\newpage
	\listoftodos

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