Drag force and energy dissipation

The force with which the impurity acts on the system is

$${\bf F}=-\int \vert\psi \left( {\bf r}\right) \vert^2\,\vec{... ...^3x={g_{i}}(\vec{\nabla}\vert\psi({\bf r})\vert^2)_{{\bf r}=0}$$ (7.21)

Expanding the wave function into the sum of $\phi_0$ and $\delta\psi$ and neglecting terms of order $\delta\psi^2$ we obtain

$${\bf F}={g_{i}}\phi_0\int i{\bf k}\left[ \delta\psi_{{\bf k}}+\le... ...{2m} \left(\frac{\hbar^2k^2}{2m}+2\mu\right)}\frac{d^3k}{\left( 2\pi\right)^3},$$ (7.22)

where we added an infinitesimal positive imaginary part $+i0$ to the frequency ${\bf k\cdot V}$according to the usual Landau causality rule. The drag force is obviously directed along to the velocity ${\bf V}$. The integration (7.22) can be done by using the formula $\frac{1}{x+i0}={\cal P}\frac{1}{x}-i\pi \delta (x)$. Due to the integration between symmetric limits, only the imaginary part contributes to the integral and the final value is real.

$${\bf F}=2({g_{i}}\phi_0)^2 \int \frac{i{\bf k}\frac{\hbar^2 k^2}{... ...frac{\hbar^2 k^2}{2m}\!+\!2\mu\right)}\right) \right]$} \,\frac{d^3k}{(2\pi)^3}$$ (7.23)

It is convenient to do the integration in spherical coordinates with $\vartheta$ being angle between $\bf k$ and $\bf V$. There is no dependence on the angle $\phi$ and it can be immediately integrated out

$$\int f({\bf k})\,d^3k = \int\limits_{0}^{\infty}\!\!\int\limits_0... ...0}^{\infty}\int\limits_{-1}^1 f(k, \cos\vartheta)\,2\pi k^2 dk d(\cos\vartheta)$$ (7.24)

The $\delta$-function can be further developed

$$\delta\left(\hbar{\bf k V}\pm\sqrt{\frac{\hbar^2 k^2}{2m} \left(\... ...V}\sqrt{\frac{\hbar^2 k^2}{2m} \left(\frac{\hbar^2 k^2}{2m}+2\mu\right)}\right)$$ (7.25)

The poles in the integration over $\cos\vartheta$ appear if the square root in the denominator is smaller than one, which leads to the restriction on the values of momentum which contribute

$$\vert k\vert \leq k_{\max} = 2m (V^2-c^2)^{1/2}/\hbar$$ (7.26)

Thus the energy dissipation takes place only if the impurity moves with a speed larger than the speed of sound.

Let us calculate the projection $F_V$ of the force ${\bf F}$ to the direction $\bf V$of the movement of the perturbation. It means that we have to multiply formula (7.23) on ${\bf V}/V$

$$F_V = ({g_{i}}\phi_0)^2~2\!\!\!\!\!\!\!\!\int\limits_0^{2m(V^2-c^... ...rac{({g_{i}}\phi_0)^2m}{2\pi \hbar^4 V^2} \frac{1}{2}\left(2m(V^2-c^2)\right)^2$$

Now we can use that square of the unperturbed wave function gives the density $\phi_0^2 = n$ and the coupling constant can be expressed as

$${g_{i}}= \frac{2\pi\hbar^2 b}{m},$$ (7.27)

which can be obtained from the formula (1.86) recalling that the reduced mass is $\mu = m$ for the scattering on a quenched impurity.

Finally, we obtain following expression for the projection of the force

$$F_{V}=4\pi nb^2mV^2(1-c^2/V^2)^2$$ (7.28)

The energy dissipation, $\dot{E}=-F_{V}V$, can be evaluated by measuring the heating of the gas.

For large $V$ the force is proportional to $V^2$. The energy dissipation per unit time can then be presented as $\dot{E}=-\gamma E$ with the damping rate $\gamma \sim nb^2V.$

Note in conclusion that our perturbative calculations can not describe processes involving dissipation of energy due to creation of quantized vortex rings. Such a creation is possible at $V<c$ but has a small probability for low velocity and for a weak point-like impurity.