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fix: typo fixes

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This commit is contained in:
Deepak Mallubhotla 2022-03-16 21:54:55 -05:00
parent e9cee5abf6
commit 63620fcbe5
2 changed files with 3 additions and 3 deletions
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4
tex/3.11.tex
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@ -114,7 +114,7 @@
\subsection{(a)} \subsection{(a)}
Want to show Want to show
\item \begin{equation} \begin{equation}
\left. \pdv{T}{V} \right|_{S, N} = - \left. \pdv{P}{S} \right|_{V, N} \left. \pdv{T}{V} \right|_{S, N} = - \left. \pdv{P}{S} \right|_{V, N}
\end{equation} \end{equation}
@ -133,7 +133,7 @@
\pdv{T}{V_{S, N}} &= - \pdv{P}{S_{V, N}} \pdv{T}{V_{S, N}} &= - \pdv{P}{S_{V, N}}
\end{align} \end{align}
\subection{(b)} \subsection{(b)}
We want: We want:
\begin{equation} \begin{equation}

2
tex/3.8.tex
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@ -107,7 +107,7 @@
The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$. The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$.
So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$. So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$.
Our specific heat per particle satisfies $\N c_v^{(1)}) \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find. Our specific heat per particle satisfies $N c_v^{(1)} = \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find.
Now lets use \cref{eq:3} and solve for $c_v^{(2)}$. Now lets use \cref{eq:3} and solve for $c_v^{(2)}$.
\begin{align} \begin{align}