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Connection between $\rho _s/\rho $ and the transverse current-current response function

One of the striking properties of superfluids is the ability to flow without friction. This fact allows us to define the normal fluid density $\rho_n$ as the fraction of liquid which is carried along by the walls if they are set in motion. For example [17,18], consider the liquid inside a long tube (see. Fig 2.1), which was at rest at time $t = -\infty$ and was then adiabatically accelerated up to time $t = 0$ (for instance with the exponential law $v(t) = v \exp(\varepsilon t)$ with infinitesimal $\varepsilon >
0$). The normal component $\rho_n$ can be defined through the momentum density $\langle\vec j\rangle$ at $t = 0$

Figure 2.1: Long tube filled with superfluid
\resizebox{7 cm}{3 cm}{\includegraphics{tube.eps}}


$\displaystyle \langle\vec j \rangle = \rho_n \vec v$     (2.25)

and the superfluid density as the difference between the total density $\rho$ and $\rho_n$
$\displaystyle \rho_s = \rho - \rho_n$     (2.26)

The effect of the perturbation caused by the moving walls is given by the energy
$\displaystyle V_{walls} = - \int \vec j(x) \vec v(x,t) d^3x,$     (2.27)

where $\vec j(x)$ is the momentum density and $\vec v(x,t)$ is the external velocity field. The linear response is given by the Kubo formula
$\displaystyle \langle j_i(x,t) \rangle =
\int\limits_{-\infty}^{\infty}ds \int d^3y~\chi_{ij}(x,t;y,s) v_j(y,s),$     (2.28)

where
$\displaystyle \chi_{ij}(x,t;y,s) = i\theta(t-s)\,\langle [j_i(x,t), j_j(y,s)]\rangle$     (2.29)

For uniform systems the response function $\chi_{ij}$ depends only on the difference of its arguments $\chi_{ij} = \chi_{ij}(x-y,t-s)$. The static susceptibility is defined as
$\displaystyle \chi_{ij}(x) =
\int\limits_{-\infty}^{0}\chi_{ij}(x,t) e^{\varepsilon t} dt$     (2.30)

or in terms of Fourier components
$\displaystyle \chi_{ij}(k) =
\int\limits_{-\infty}^{\infty} \frac{\chi_{ij}(k,\omega)}{\omega+i\varepsilon}
\frac{d\omega}{2\pi}$     (2.31)

At time $t = 0$ the linear response function satisfies the equation
$\displaystyle \langle j_i(k)\rangle = \chi_{ij}(k) v_j(k),
\qquad i,j = \{x,y,z\}$     (2.32)

Since $\chi_{ij}(k)$ is a second rank tensor, it can be decomposed into the sum of longitudinal and transverse components
$\displaystyle \chi_{ij}(k) = \frac{k_i k_j}{k^2} \chi_L(k) +
\left(\delta_{ij} - \frac{k_i k_j}{k^2}\right) \chi_T(k),$     (2.33)

Let us consider first the transverse response. Due to the rotational invariance of $\chi_T(k)$ it is enough to examine the response in an arbitrary direction, for example $\chi_{zz}$, that is the momentum response $j_z({\bf k})$ due to an imposed velocity field in the $z$ direction $v_z({\bf k})$. Suppose first that the velocity field is created by dragging the walls of an indefinitely long pipe (see Fig. 2.2) and the cross section of the pipe tends to infinity. This arrangement corresponds to the limiting procedure $k_z \to 0$, followed by $k_x
\to 0$, $k_y \to 0$. Then, the part of the system responding to the shear force is defined as the normal fluid. Carrying out the limiting procedure $k^2 = k_x^2 + k_y^2 + k_z^2 \to 0$ gives
$\displaystyle \rho_n =
\lim\limits_{k_x \to 0\atop k_y \to 0}
\lim\limits_{k_z \to 0}
\chi_{ij}(k)
=\lim\limits_{k \to 0} \chi_T(k)$     (2.34)

Figure 2.2: Illustration of the transverse response. Only the normal component is dragged in $z$ direction
\includegraphics[width=.5\textwidth]{TubeNormal.eps}

Next suppose that the pipe of infinite radius is constrained by two plates, normal to the $z$ axis, with separation between the plates approaching infinity, as illustrated in Fig. 2.3. In this case the entire fluid $\rho = \rho_s +\rho_n$ responds to the external probe. This arrangement corresponds to the limiting procedure $k_x
\to 0$, $k_y \to 0$, followed by $k_z \to 0$. The result of the limiting procedure is
$\displaystyle \rho =
\lim\limits_{k_z \to 0}
\lim\limits_{k_x \to 0\atop k_y \to 0}
\chi_{ij}(k)
=\lim\limits_{k \to 0} \chi_L(k)$     (2.35)

Figure 2.3: Illustration to the longitudinal response. Both superfluid and normal components are pushed in the $z$ direction
\includegraphics[width=.5\textwidth]{TubeTotal.eps}


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Next: Superfluid fraction in the Up: Superfluid density Previous: Superfluid density   Contents