\section{Supplemental Material}
\label{sec:supplement}
In this supplemental material, we derive the hydrodynamic linear response model, including the thermodynamics and the method of determining the forces arising from the box potential.
\subsection{Hydrodynamic linear response for a normal fluid.}
Our analysis is simplified for a normal  Fermi gas, which is a single component fluid with a mass density is $\rho\equiv n\,m$, where $n$ is the total particle density (we assume a 50-50 mixture of two components) and $m$ is the atom mass. $\rho({\mathbf{r}},t)$ satisfies the continuity equation,
\begin{equation}
\partial_t\rho +\partial_i(\rho\,v_i)=0,
\label{eq:1.3}
\end{equation}
where a sum over $i=x,y,z$ is implied. The mass flux (momentum density) is $\rho\,v_i$, with $v_i({\mathbf{r}},t)$ the velocity field.

The momentum density and corresponding momentum flux $\rho\,v_i v_j$ obey
\begin{equation}
\partial_t (\rho\,v_i) +\partial_j (\rho\,v_i v_j)=-\partial_i p -n\,\partial_i U+\partial_j(\eta\,\sigma_{ij}),
\label{eq:1.5}
\end{equation}
Here, $-\partial_i p-n\,\partial_i U$ is the force per unit volume arising from the pressure $p$ and the externally applied potential $U({\mathbf{r}},t)$. The last term describes the dissipative forces, which arise only from the shear viscosity  $\eta$, as the bulk viscosity vanishes for a unitary gas. The shear stress tensor is $\sigma_{ij}\equiv\partial_iv_j+\partial_jv_i-2\delta_{ij}\nabla\cdot{\mathbf{v}}/3$. Taking the divergence of eq.~\ref{eq:1.5}, and using eq.~\ref{eq:1.3}, we immediately obtain
\begin{equation}
-\partial_t^2\rho+\partial_i\partial_j (\rho\,v_i v_j)=-\partial_i^2 p-\partial_i(n\,\partial_iU)+\partial_i\partial_j(\eta\,\sigma_{ij}).
\label{eq:2.7}
\end{equation}

We are interested in the linear hydrodynamic response to a perturbing external potential $\delta U({\mathbf{r}},t)$, which leads to first order changes in the density $\delta n({\mathbf{r}},t)$ and pressure $\delta p({\mathbf{r}},t)$,
\begin{eqnarray}
n({\mathbf{r}},t)&=&n_0({\mathbf{r}})+\delta n({\mathbf{r}},t)\nonumber\\
p({\mathbf{r}},t)&=&p_0({\mathbf{r}})+\delta p({\mathbf{r}},t)\nonumber\\
U({\mathbf{r}},t)&=&U_0({\mathbf{r}})+\delta U({\mathbf{r}},t).
\label{eq:3.1}
\end{eqnarray}
Here, $n_0({\mathbf{r}})$ and $p_0({\mathbf{r}})$ are the equilibrium (time independent) density and pressure arising from confinement in the box trap potential, $U_0({\mathbf{r}})$. In equilibrium, the velocity field ${\mathbf v}_0({\mathbf{r}},t)=0$ and eq.~\ref{eq:1.5} requires balance of the forces per unit volume arising from the box trap and the pressure,
\begin{equation}
-\nabla p_0({\mathbf{r}})-n_0({\mathbf{r}})\nabla U_0({\mathbf{r}})=0.
\label{eq:1.3b}
\end{equation}
Substituting eq.~\ref{eq:3.1} into  eq.~\ref{eq:2.7} and retaining terms to first order in small quantities, we obtain
\begin{equation}
\partial_t^2\delta n=\frac{1}{m}\nabla^2\,\delta p+\frac{1}{m}\nabla\cdot[n_0({\mathbf{r}})\,\nabla\delta U+\delta n\,\nabla U_0]+\frac{4}{3}\frac{\eta}{n_0 m}\nabla^2\partial_t\delta n.
\label{eq:3.7}
\end{equation}
Here, the second term on the left side of eq.~\ref{eq:2.7} is negligible, as the velocity field is first order in small quantities. To obtain the last term in eq.~\ref{eq:3.7}, we have used the fact that the dissipative force is small compared to the conservative forces and that the density $n_0$ slowly varies in the region of interest. Then we can ignore the spatial derivatives of $\eta$ and $n_0$, taking $\partial_i\partial_j(\eta\,\sigma_{ij})\simeq\eta\,\partial_i\partial_j\sigma_{ij}=4\eta/3\,\nabla^2(\nabla\cdot{\mathbf{v}})$. The velocity field is eliminated using $\nabla\cdot{\mathbf{v}}\simeq -\partial_t\delta n/n_0$, which follows from  eq.~\ref{eq:1.3}.

We retain the spatial dependence of the driving force per unit volume $n_0({\mathbf{r}})\,\nabla\delta U$ in eq.~\ref{eq:3.7}, to assure that it vanishes smoothly at the cloud edges. Here, the perturbing potential $\delta U$ is controlled in the experiments. The force arising from the box potential $-\nabla U_0$ is determined from eq.~\ref{eq:1.3b}, using the equation of state $p_0[n_0({\mathbf{r}}),T]$, where the temperature $T$ is spatially constant in equilibrium. Then,
\begin{equation}
\nabla U_0({\mathbf{r}})=-\left(\frac{\partial p_0}{\partial n_0}\right)_{\!\!T}\frac{\nabla n_0({\mathbf{r}})}{n_0({\mathbf{r}})}.
\label{eq:3.6}
\end{equation}
Eq.~\ref{eq:3.6}, with $(\partial p/\partial n)_T =(c_V/c_P)\,m\,c_0^2$,  yields $\nabla U_0({\mathbf{r}})$, from the equilibrium density profile, where $c_V$ and $c_P$ are the heat capacities per particle at constant volume and pressure, and $c_0$ is the adiabatic sound speed (see Fig.~\ref{fig:soundspeed}).

To proceed further, we need an expression for the first order pressure change, $\delta p$, which is determined from eq.~\ref{eq:6.3T} by setting $n=n_0$ and $T=T_0$,
\begin{equation}
\delta p=m\,c_0^2\,\delta n+mc_0^2\beta\,\frac{T_0}{c_P}\,n_0\,\delta s_1.
\label{eq:6.3}
\end{equation}
As discussed in the thermodynamics section below, the first term in eq.~\ref{eq:6.3} arises from adiabatic compression and in the second term, $\delta s_1$ is the first order change in the entropy per particle, with $\beta$ is the thermal expansivity. Inserting eq.~\ref{eq:6.3} into eq.~\ref{eq:3.7}, we obtain
\begin{equation}
\partial_t^2\delta n-c_0^2\nabla^2\,\delta n-c_0^2\,\beta n_0\frac{T_0}{c_P}\nabla^2\delta s_1-\frac{4}{3}\frac{\eta}{n_0 m}\nabla^2\partial_t\delta n=\frac{1}{m}\nabla\cdot[n_0({\mathbf{r}})\,\nabla\delta U+\delta n\,\nabla U_0].
\label{eq:4.6}
\end{equation}

The evolution equation for $\delta s_1$ is determined by the heating rate per unit volume, $\delta\dot{q}$,
\begin{equation}
T_0(\partial_t+{\mathbf v}\cdot\nabla)\,\delta s_1=\frac{\delta\dot{q}}{n_0},
\label{eq:4.2}
\end{equation}
where $T_0$ is the initial, spatially uniform, temperature.
The heating rate arising from the shear viscosity is second order in $v_i$, which is negligible compared to the heating rate arising from heat conduction. Hence, $\delta\dot{q}\simeq -\nabla\cdot(-\kappa_T\nabla\delta T)\simeq\kappa_T\nabla^2\delta T$, where we neglect the spatial derivatives of $\kappa_T$. Here, $\delta T$ is the first order temperature change, which is determined from eq.~\ref{eq:6.5T} by setting $n=n_0$ and $T=T_0$,
\begin{equation}
\delta T=\frac{T_0}{c_V}\delta s_1 + mc_0^2\,\beta\frac{T_0}{c_P}\frac{\delta n}{n_0}.
\label{eq:4.4}
\end{equation}
Retaining only first order terms in eq.~\ref{eq:4.2} and using eq.~\ref{eq:4.4}, we have
\begin{equation}
\partial_t\delta s_1-\frac{\kappa_T}{n_0 c_V}\nabla^2\delta s_1=\frac{\kappa_Tmc_0^2}{n_0^2c_P}\beta\,\nabla^2\delta n.
\label{eq:4.5}
\end{equation}

We define $\delta \tilde{s}_{1}\equiv\beta T_0\,n_0\delta s_1/c_P$, which has a dimension of density, and take $\nabla^2\rightarrow\partial_z^2$ in eqs.~\ref{eq:4.6}~and~\ref{eq:4.5} to obtain
\begin{equation}
\partial_t^2\delta n-c_0^2(\partial_z^2\,\delta n+\partial_z^2\delta \tilde{s}_{1})-\frac{4}{3}\frac{\eta}{n_0 m}\partial_z^2\partial_t\delta n=\frac{1}{m}\,\partial_z[n_0(z)\,\partial_z\delta U+\delta n\,\partial_z U_0].
\label{eq:1.4}
\end{equation}
and
\begin{equation}
\partial_t\delta\tilde{s}_{1}-\frac{\kappa_T}{n_0 c_V}\,\partial_z^2\delta\tilde{s}_{1}=\frac{\kappa_T}{n_0c_V}\frac{c_P-c_V}{c_P}\,\partial_z^2\delta n,
\label{eq:1.5b}
\end{equation}
where we have used $mc_0^2\beta^2T_0/c_P=(c_P-c_V)/c_V$, as shown in below, eq.~\ref{eq:5.7T}.

For the numerical integration, it is convenient to define the dimensionless variables, $\delta\tilde{n}\equiv\delta n/n_0$ and $\tilde{n}_0(z)\equiv n_0(z)/n_0$, where $n_0$ is the average total density. Similarly $\delta\tilde{s}_{10}=\delta\tilde{s}_1/n_0=\beta T_0\,\delta s_1/c_P$. Further, we define  $\delta\tilde{U}(z,t)\equiv\delta U(z,t)/\epsilon_{F0}$ and $\tilde{U}_0(z)\equiv U_0(z)/\epsilon_{F0}$, where the local Fermi energy, evaluated for the density $n_0$, is $\epsilon_{F0}=mv_{F0}^2/2=\hbar^2(3\pi^2n_0)^{2/3}/(2m)$. In terms of these variables, eqs.~\ref{eq:1.4}~and~\ref{eq:1.5b} become,
\begin{equation}
\partial_t^2\delta\tilde{n}-c_0^2(\partial_z^2\,\delta\tilde{n}+\partial_z^2\delta\tilde{s}_{10})-\frac{4}{3}\frac{\eta}{n_0 m}\partial_z^2\partial_t\delta\tilde{n}=\frac{v_{F0}^2}{2}\,\partial_z[\tilde{n}_0(z)\,\partial_z\delta\tilde{U}+\delta \tilde{n}\,\partial_z\tilde{U}_0]
\label{eq:1.7a}
\end{equation}
and
\begin{equation}
\partial_t\delta\tilde{s}_{10}-\frac{\kappa_T}{n_0 c_V}\,\partial_z^2\delta\tilde{s}_{10}=\frac{\kappa_T}{n_0c_V}\frac{c_P-c_V}{c_P}\,\partial_z^2\delta\tilde{n}.
\label{eq:1.8a}
\end{equation}
We show below how the box force $-\partial_z\tilde{U}_0(z)$ and the box potential $\tilde{U}_0(z)$ are determined from the measured density profile $\tilde{n}_0(z)$.

For the fits, we take the perturbation to have the form,
\begin{equation}
\delta\tilde{U}(z,t)=\delta\tilde{U}_0\,[1-\epsilon\cos(qz-qvt)]\,\frac{1 - \tanh[(z-v t)/w_z]}{2}.
\label{eq:perturb}
\end{equation}
In eq.~\ref{eq:perturb}, $\delta\tilde{U}_0$ is the strength of the perturbation in units of $\epsilon_{F0}$ and $\epsilon$ is modulation depth in the plane of the atoms. The last factor is unity for $z<vt$ and vanishes for $z>vt$, where the width $w_z$ (typically a few microns) assures a smooth transition for the numerical integration.


\subsection{Thermodynamics}
\label{sec:thermo}
\subsubsection{Thermodynamic relations}

For later use in determining the pressure change $\delta p$  and temperature change $\delta T$, we begin by deriving the well-known formula relating the heat capacity at constant pressure $c_P=T(\partial s_1/\partial T)_P$ to the heat capacity at constant volume $c_V=T(\partial s_1/\partial T)_V$. In the following, we define the volume per particle $V_1=1/n$, where $n$ is the local density, and determine the heat capacities per particle from the entropy per particle $s_1(p,T)$.
\begin{equation}
c_V=T\left(\frac{\partial s_1}{\partial T}\right)_{\!\!V_1}=T\left(\frac{\partial s_1}{\partial T}\right)_{\!\!P}+T\left(\frac{\partial s_1}{\partial p}\right)_{\!\!T}\left(\frac{\partial p}{\partial T}\right)_{\!\!V_1}
\label{eq:1.3T}
\end{equation}
Rewriting eq.~\ref{eq:1.3T} using the Maxwell relation $(\partial s_1/\partial p)_T=-(\partial V_1/\partial T)_p$, obtained from the Gibb's free energy, we have
\begin{equation}
c_V=c_P-T\left(\frac{\partial V_1}{\partial T}\right)_{\!\!P}\left(\frac{\partial p}{\partial T}\right)_{\!\!V_1}.
\label{eq:1.7T}
\end{equation}
This result can be written in terms of the thermal expansivity $\beta$,
\begin{equation}
\beta\equiv \frac{1}{V_1}\left(\frac{\partial V_1}{\partial T}\right)_{\!\!P}.
\label{eq:2.1T}
\end{equation}
Using $dp(V_1,T)=0$, we have
\begin{equation}
\left(\frac{\partial p}{\partial T}\right)_{\!\!V_1}=-\left(\frac{\partial p}{\partial V_1}\right)_{\!\!T}\left(\frac{\partial V_1}{\partial T}\right)_{\!\!P}=
\frac{1}{V_1^2}\left(\frac{\partial V_1}{\partial T}\right)_{\!\!P}\left(\frac{\partial p}{\partial n}\right)_{\!\!T}
\label{eq:2.4T}
\end{equation}
In the last step, we have used $n=1/V_1$ so that $(\partial p/\partial V_1)_T=-1/V_1^2\,(\partial p/\partial n)_T$. Inserting eq.~\ref{eq:2.4T}
into eq.~\ref{eq:1.7T} and using eq.~\ref{eq:2.1T}, we obtain the desired relation
\begin{equation}
c_V=c_P-\beta^2T\left(\frac{\partial p}{\partial n}\right)_{\!\!T}.
\label{eq:2.8T}
\end{equation}

For later use, we rewrite the  Maxwell relation $(\partial s_1/\partial p)_T=-(\partial V_1/\partial T)_p$ as
\begin{equation}
\left(\frac{\partial s_1}{\partial p}\right)_{\!T}=-\beta V_1
\label{eq:2.2T}
\end{equation}
and eq.~\ref{eq:2.4T}, using $n=1/V_1$ and $(\partial p/\partial T)_{V_1}=(\partial p/\partial T)_n$, as
\begin{equation}
\left(\frac{\partial p}{\partial T}\right)_{\!\!n}=\beta\, n \left(\frac{\partial p}{\partial n}\right)_{\!\!T}.
\label{eq:3.1T}
\end{equation}

Next, we relate the isothermal sound speed $c_T$ to the adiabatic speed of sound $c_0$,
\begin{equation}
c_0^2\equiv\frac{1}{m}\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}.
\label{eq:4.1aT}
\end{equation}
\begin{equation}
c_T^2\equiv\frac{1}{m}\left(\frac{\partial p}{\partial n}\right)_{\!\!T}.
\label{eq:4.1bT}
\end{equation}

Writing $p(n,s_1)=p[n,T(n,s_1)]$, we have
\begin{equation}
\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}=\left(\frac{\partial p}{\partial n}\right)_{\!\!T}+\left(\frac{\partial p}{\partial T}\right)_{\!\!n}\left(\frac{\partial T}{\partial n}\right)_{\!\!s_1}.
\label{eq:4.2T}
\end{equation}
We rewrite the last factor on the right hand side, using $ds_1(p,T)=0$  to obtain
\begin{eqnarray}
0&=&\left(\frac{\partial s_1}{\partial p}\right)_{\!\!T}\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}+\left(\frac{\partial s_1}{\partial T}\right)_{\!\!P}\left(\frac{\partial T}{\partial n}\right)_{\!\!s_1}\nonumber\\
0&=&-\beta V_1\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}+\frac{c_P}{T}\left(\frac{\partial T}{\partial n}\right)_{\!\!s_1},
\label{eq:4.4T}
\end{eqnarray}
where we have used eq.~\ref{eq:2.2T} and the definition of $c_P$. Hence,
\begin{equation}
\left(\frac{\partial T}{\partial n}\right)_{\!\!s_1}=V_1\frac{\beta T}{c_P}\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}.
\label{eq:4.5T}
\end{equation}
Inserting eq.~\ref{eq:4.5T} and eq.~\ref{eq:3.1T} into eq.~\ref{eq:4.2T} and using $nV_1=1$ then yields
\begin{equation}
\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}=\left(\frac{\partial p}{\partial n}\right)_{\!\!T}+\frac{\beta^2T}{c_P}\left(\frac{\partial p}{\partial n}\right)_{\!\!T}\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}.
\label{eq:5.1T}
\end{equation}
With eq.~\ref{eq:2.8T}, eq.~\ref{eq:5.1T} takes the simple form
\begin{equation}
\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}=\left(\frac{\partial p}{\partial n}\right)_{\!\!T}+\frac{c_P-c_V}{c_P}\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}.
\label{eq:5.2T}
\end{equation}
Using eqs.~\ref{eq:4.1aT}~and~\ref{eq:4.1bT} in eq.~\ref{eq:5.2T}, we obtain the well-known result~\cite{LandauFluids}
\begin{equation}
c_T^2=\frac{c_V}{c_P}\,c_0^2.
\label{eq:5.5T}
\end{equation}
Eqs.~\ref{eq:4.1bT}~and~\ref{eq:5.5T} then determine $\beta^2$ from eq.~\ref{eq:2.8T} in terms of measurable quantities,
\begin{equation}
\beta^2=\frac{c_P}{c_V}\frac{c_P-c_V}{m\,c_0^2\,T}.
\label{eq:5.7T}
\end{equation}

Using the results described above, it is easy to evaluate the pressure change $\delta p$  in terms of the density change $\delta n$ and the change in the entropy per particle $\delta s_1$. Writing the pressure as $p(n,s_1)=p[n,T(n,s_1)]$, we have
\begin{eqnarray}
\delta p&=&\left(\frac{\partial p}{\partial n}\right)_{\!\!s_1}\!\delta n+\left(\frac{\partial p}{\partial s_1}\right)_{\!\!n}\!\delta s_1\nonumber\\
&=&mc_0^2\,\delta n+\left(\frac{\partial p}{\partial T}\right)_{\!\!n}\left(\frac{\partial T}{\partial s_1}\right)_{\!\!n}\!\delta s_1.
\label{eq:6.1T}
\end{eqnarray}
where we have used eq.~\ref{eq:4.1aT}. With eqs.~\ref{eq:3.1T},~\ref{eq:4.1bT},~\ref{eq:5.5T}, and the heat capacity per particle at constant volume  $c_V=T(\partial s_1/\partial T)_n$,  we immediately obtain
\begin{equation}
\delta p=mc_0^2\,\delta n+mc_0^2\,\beta\frac{T}{c_P}\,n\,\delta s_1.
\label{eq:6.3T}
\end{equation}
For the temperature change $\delta T$, we write $T(s_1,n)$, so that
\begin{equation}
\delta T=\left(\frac{\partial T}{\partial s_1}\right)_{\!\!n}\!\delta s_1+\left(\frac{\partial T}{\partial n}\right)_{\!\!s_1}\!\delta n.
\label{eq:6.4T}
\end{equation}
Inserting eqs.~\ref{eq:4.5T}~and~\ref{eq:4.1aT} into eq.~\ref{eq:6.4T} and using $V_1=1/n$, we find
\begin{equation}
\delta T=\frac{T}{c_V}\,\delta s_1+mc_0^2\,\beta\frac{T}{c_P}\frac{\delta n}{n}.
\label{eq:6.5T}
\end{equation}

\subsubsection{Unitary Fermi gas thermodynamics}
For the unitary Fermi gas, universality~\cite{HoUniversalThermo,ThomasUniversal,ThermoLuo} requires that the pressure $p$ and the energy density ${\cal E}$ are functions only of the density and temperature, related by  $p=2\,{\cal E}/3$. Dimensional analysis then shows that the energy density takes the simple form
\begin{equation}
{\cal E}=\frac{3}{5}n\,\epsilon_F(n)\,f_E(\theta),
\label{eq:1.1TP}
\end{equation}
where $\theta\equiv T/T_F$ is the reduced temperature with $T_F$ the local Fermi temperature. For a balanced 50-50 mixture of two spin components of total density $n$, the local Fermi energy is  $k_BT_F=\epsilon_F(n)=mv_F^2/2=\hbar^2(3\pi^2n)^{2/3}/(2m)$. The universal function $f_E(\theta)$ has been measured by Ku et al.,~\cite{KuThermo}, which determines all of the thermodynamic properties. The pressure is then
\begin{equation}
p=\frac{2}{5}n\,\epsilon_F(n)\,f_E(\theta).
\label{eq:1.3TP}
\end{equation}
The entropy density takes a similar form
\begin{equation}
s=nk_B\,f_S(\theta)=n\,s_1(\theta),
\label{eq:1.2TP}
\end{equation}
where $f_S(\theta)$ is determined by  $f_E(\theta)$.

The adiabatic sound speed eq.~\ref{eq:4.1aT} is easily obtained from eq.~\ref{eq:1.3TP}, as eq.~\ref{eq:1.2TP} requires constant $\theta$ for constant $s_1$,
\begin{equation}
c_0^2=\frac{1}{m}\left(\frac{\partial p}{\partial n}\right)_{\!\!\theta}.
\label{eq:1.4TP}
\end{equation}
Eq.~\ref{eq:1.4TP} immediately gives
\begin{equation}
c_0=v_F\sqrt{\frac{f_E(\theta)}{3}}.
\label{eq:1.5TP}
\end{equation}
\begin{figure}[htb]
\begin{center}\
\includegraphics[width=3.0in]{ReducedTempvsC0}
\end{center}
\caption{Reduced temperature $\theta =T/T_F$ versus adiabatic sound speed $c_0/v_{F}$. For $\theta>0.25$, $\theta$ monotonically increases with $c_0/v_F$, showing that the fitted $c_0/v_F$ can be used as a thermometer to determine $\theta$ in the normal fluid region.
\label{fig:soundspeed}}
\end{figure}
Fig.~\ref{fig:soundspeed} shows how the adiabatic sound speed $c_0$ obtained from the measured equation of state varies with $\theta$.

The heat capacity per particle at constant volume takes a simple form. Using eq.~\ref{eq:1.1TP},
\begin{equation}
c_V=\frac{T}{n}\left(\frac{\partial s}{\partial T}\right)_{\!\!n}=\frac{1}{n}\left(\frac{\partial {\cal E}}{\partial T}\right)_{\!\!n}=k_B\frac{3}{5}f_E'(\theta),
\label{eq:2.1TP}
\end{equation}
where $'$ denotes differentiation with respect to $\theta$.

We determine heat capacity per particle at constant pressure from eq.~\ref{eq:2.8T} by finding $\beta$ from eq.~\ref{eq:3.1T}. Using eq.~\ref{eq:4.1bT}~and~\ref{eq:5.5T}, we have generally
\begin{equation}
\beta =\frac{1}{n} \left(\frac{\partial p}{\partial T}\right)_{\!\!n} \frac{1}{m\, c_T^2}=\frac{c_P}{c_V}\frac{1}{mc_0^2}\frac{1}{n}\left(\frac{\partial p}{\partial T}\right)_{\!\!n}.
\label{eq:beta1}
\end{equation}
For the unitary Fermi gas, where $p=2\,{\cal E}/3$, eq.~\ref{eq:2.1TP} shows that $(\partial p/\partial T)_n=2\,n c_V/3$ and
\begin{equation}
\beta_{UG}=\frac{2}{3}\frac{c_P}{m\,c_0^2}.
\label{eq:beta2}
\end{equation}
Then,  eq.~\ref{eq:2.8T} reduces to
\begin{equation}
c_P-c_V=\frac{4}{9}\frac{T}{m\,c_0^2}\,c_V\,c_P.
\label{eq:beta3}
\end{equation}
With $mc_0^2=2\,k_BT_F\,f_E(\theta)/3$ from eq.~\ref{eq:1.5TP}, eq.~\ref{eq:beta3} takes the simple form
\begin{equation}
\frac{c_P-c_V}{c_V\,c_P}=\frac{1}{k_B}\frac{2}{3}\frac{\theta}{f_E(\theta)}.
\label{eq:beta5}
\end{equation}
Solving eq.~\ref{eq:beta5} for $c_P$, and using $c_V$ from eq.~\ref{eq:2.1TP}, then determines
\begin{equation}
\frac{c_P}{c_V}=\frac{f_E(\theta)}{f_E(\theta)-\frac{2}{5}\theta\,f_E'(\theta)}.
\label{eq:beta4}
\end{equation}
\subsubsection{Determination of the box force}
\label{subsec:boxforce}
We find the force arising from the confining potential along one axis $z$, using the measured density profiles $n_0(z)$. We ignore the variation of the density along the line of site and find $n_0(z)$ from the spatially integrated column density, which is obtained from absorption images. Using eq.~\ref{eq:3.6} with $(\partial p/\partial n)_T =(c_V/c_P)\,m\,c_0^2$ from eq.~\ref{eq:5.5T} and eq.~\ref{eq:4.1bT},
\begin{equation}
\partial_z U_0(z)=-m\,c_0^2[\theta(z)]\,\frac{c_V[\theta(z)]}{c_P[\theta(z)]}\frac{\partial_z n_0(z)}{n_0(z)}.
\label{eq:3.6TP}
\end{equation}
 Here, $c_0$ is given by eq.~\ref{eq:1.5TP} and $c_V/c_P$ is given by eq.~\ref{eq:beta4}. As $\partial_z U_0(z)$ is independent of the scale of $n_0(z)$, we find $\partial_z U_0(z)$ from the the scaled density $\tilde{n}_0(z)=n_0(z)/n_0$, where
$n_0$ is the mean central density. Note that the thermodynamic parameters depend only on the reduced temperature $\theta(z)=\theta_0/[\tilde{n}_0(z)]^{2/3}$, where we determine $\theta_0=T_0/T_F(n_0)$ from the fitted adiabatic sound speed $c_0$.

For the numerical integration, we express $U_0$ in units of $\epsilon_{F0}$, $\tilde{U}_0(z)\equiv U_0(z)/\epsilon_{F0}$ . From eq.~\ref{eq:1.5TP}, we have $mc_0^2(\theta)=mv_F^2 f_E(\theta)/3=2\,\epsilon_F f_E(\theta)/3$. With $\epsilon_F/\epsilon_{F0}=[\tilde{n}_0(z)]^{2/3}$ and eq.~\ref{eq:beta4}, we find
\begin{equation}
\partial_z\tilde{U}_0(z)=-\frac{2}{3}\left[f_E(\theta)-\frac{2}{5}\theta f_E'(\theta)\right]\frac{\partial_z\tilde{n}_0(z)}{[\tilde{n}_0(z)]^{1/3}},
\label{eq:2.10TP}
\end{equation}
which is also obtained from eq.~\ref{eq:3.6} by evaluating $(\partial p/\partial n)_T$ directly from eq.~\ref{eq:1.3TP}.
\begin{figure}[htb]
\begin{center}\
\includegraphics[width=3.5in]{BackgroundDensity200}
\end{center}
\caption{Measured background density $n_0(z)$ (blue dots) versus $z$ in $\mu$m, showing the fit of eq.~\ref{eq:fit} (red curve).
\label{fig:backgroundfit}}
\end{figure}
To evaluate the derivative $\partial_z\tilde{n}_0(z)$ on the right hand side of eq.~\ref{eq:2.10TP}, we first fit $\tilde{n}_0(z)=n_0(z)/n_0$ with an analytic function,
\begin{equation}
f(z)= \frac{\tanh[(z - z_{10})/w_1] -
    \tanh[(z - z_{20})/w_2]}{2}\,\sum_n a_n\,z^n.
\label{eq:fit}
\end{equation}
The difference of the $\tanh$ functions produces a top-hat shape of nominal width $z_{20}-z_{10}$ and slopes on each side determined by $w_1$ and $w_2$. The flat top is modulated by the multiplying polynomial.  Fig.~\ref{fig:backgroundfit} shows a typical fit using a fifth order polynomial.

Eq.~\ref{eq:2.10TP} then yields the force profile, $-\partial_z\tilde{U}_0(z)$, Fig.~\ref{fig:force}(a).  The corresponding box potential is then $\tilde{U}_0(z)=\int_0^z dz'\,\partial_{z'}\tilde{U}_0(z')$,  Fig.~\ref{fig:force}(b).
\begin{figure}[htb]
\begin{center}\
\includegraphics[width=3.0in]{BoxForce200.eps}\hspace{0.25in}\includegraphics[width=3.0in]{BoxPotential200}
\end{center}
\caption{Box force profile (a) determined from the measured background density $n_0(z)$ using eq.~\ref{eq:2.10TP}; (b) Corresponding box potential energy in units of $\epsilon_{F0}$ obtained by integrating the force.  Note that the curvature at the bottom of the box potential energy arises from curvature in the bias magnetic field, which produces a confining harmonic potential.\label{fig:force}}
\end{figure}

\subsubsection{Hydrodynamic relaxation time}

The local shear viscosity can be written as $\eta =\tau p$~\cite{BluhmSchaeferModIndep,BluhmSchaeferLocalViscosity}, where $p$ is the pressure and $\tau$ is the hydrodynamic relaxation time. We estimate $\tau$ for the unitary Fermi gas using eq.~\ref{eq:1.3TP} for $p$ and
\begin{equation}
\eta=\hbar\,n\,(\alpha_0\,\theta^{3/2}+\alpha_2),
\label{eq:viscosity1}
\end{equation}
with $n$ the local density, $\alpha_0=2.77$ and $\alpha_2=0.25$~\cite{BluhmSchaeferLocalViscosity}, as discussed in the main paper. Then with $\tau=\eta/p$,
\begin{equation}
\tau=\frac{5}{2}\tau_F\frac{\alpha_0\,\theta^{3/2}+\alpha_2}{f_E(\theta)},
\label{eq:relaxtime}
\end{equation}
where $\tau_F=\hbar/\epsilon_F$ is the Fermi time. For our experiments, where $\epsilon_F\simeq k_B\times 0.15\,\mu$K and $\theta\simeq 0.5$, we find $\tau=0.13$ ms, which is small compared to the period at the highest frequency of 800 Hz that is employed in the experiments.



\end{document}






