From 7f8961aaa5af125e5cb81d9df5e14256fbaaaf7a Mon Sep 17 00:00:00 2001 From: Deepak Mallubhotla Date: Tue, 22 Mar 2022 16:08:48 -0500 Subject: [PATCH] feat: adds 5.4 --- tex/5.4.tex | 137 ++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 137 insertions(+) create mode 100644 tex/5.4.tex diff --git a/tex/5.4.tex b/tex/5.4.tex new file mode 100644 index 0000000..029a4f9 --- /dev/null +++ b/tex/5.4.tex @@ -0,0 +1,137 @@ +\documentclass{article} + +% set up telugu +\usepackage{fontspec} +\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu] +\usepackage{polyglossia} +\setdefaultlanguage{english} +\setotherlanguage{telugu} + +%other packages +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{physics} +\usepackage[binary-units=true]{siunitx} +\usepackage{todonotes} +\usepackage{luacode} +\usepackage{titling} +\usepackage{enumerate} + +% custom deepak packages +\usepackage{luatrivially} +\usepackage{subtitling} + +\usepackage{cleveref} + +\begin{luacode*} + math.randomseed(31415926) +\end{luacode*} + +\newcommand{\kb}{k_{\mathrm{B}}} + +\title{Problem 5.4} +\subtitle{Black hole thermodynamics} +\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}} +% want empty date +\predate{} +\date{} +\postdate{} + +% !TeX spellcheck = en_GB +\begin{document} + \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 + +\end{document}