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I'm studying superfluid helium 4. I have studied that the superfluid has a velocity : $ \vec{v}_s = \frac{\hbar}{m} \vec{\nabla} \varphi(\vec{r}) $ $\rightarrow$ $ \nabla \times \vec{v}_s = 0 $.

(Where $\varphi$ is the phase of the wave function of the superfluid in the Hartree-Fock approximation)

Now if we consider a simply connected space and if we use Stoke's Theorem we have that $v_s =0 $. But in 1950 experiments with rotating vases (de Osborne 1950) show that the superfluid is not stationary !

To understand this we can consider another experiments in which we have a torus instead of a vase (so we cannot apply stokes's theorem). If we calculate $K= \oint_L \vec{v}_s \cdot d\vec{l} = \frac{\hbar}{m} \delta\varphi_L$ with $ \delta \varphi = 2 n \pi $ with $ n \in Z $.

First problem: also in this case we can have a $v_s = 0 $ when $n=0$

Then my book says that also in a rotating vase we can have a space that is not simply connected because in the vortex core will be a normal fluid due to the high speed.

Second problem: but if we don't have a rotating vase how can be a vortex?

Thanks and have a good day

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Allow me to reformulate. A superfluid is usually described by a complex order parameter $\Psi(x) = \rho(x) e^{i \varphi(x)}$ with $\rho(x) \geq 0$ and $\varphi(x)$ a real field. In the ordered phase, $\rho(x) = \rho_0$, and the phase fluctuates. However, this is not exactly true, because it might happen that $\rho(x_*) = 0 $ for some $x_*$, which is called a vortex, and $x_*$ is the location of the vortex. Note that the decomposition of $\Psi$ into magnitude and phase is only well-defined if the magnitude is non-zero, and thus the phase $\varphi$ is ill-defined at the vortex core!

It is thus dangerous to work with expressions involving $\varphi$ in the presence of vortices, in particular the superfluid velocity is a well-defined concept only away from the vortices. Instead of working with the velocity $v_s$, it is safer to work with the particle current $\vec{j}$. So let's see first what is the connection between the two. In mean-field theory, the superfluid is described by the Lagrangian density

$ L = \frac{i}{2} ( \overline{\Psi} \partial_t \Psi - \Psi \partial_t \overline{\Psi} ) - \frac{1}{2 m} \| \vec{\nabla} \Psi \|^2 - V(|\Psi|) $

from there we can derive the equations of motion

$ i \partial_t \Psi = \frac{1}{2 m} \nabla^2 \Psi + \frac{\partial V}{\partial \overline{\Psi}} $

$ -i \partial_t \overline{\Psi} = \frac{1}{2 m} \nabla^2 \Psi + \frac{\partial V}{\partial \Psi} $

We may now calculate the time-dependence of the particle density $\rho = \sqrt{\overline{\Psi} \Psi}$:

$\partial_t \rho = \frac{1}{2\rho}\left[ \overline{\Psi} \partial_t \Psi + \Psi \partial_t \overline{\Psi} \right] = -\frac{i}{4 m \rho}\left[ \overline{\Psi} \nabla^2 \Psi - \Psi \nabla^2 \overline{\Psi} \right] - \frac{i}{2\rho} \left[ \overline{\Psi} \frac{\partial V}{\partial \overline{\Psi}} - \Psi \frac{\partial V}{\partial \Psi} \right] $

Now in the second term vanishes since $V$ depends only on the magnitude $|\Psi|$, so that we get a current conservation equation:

$ \partial_t \rho = - \vec{\nabla} \cdot \vec{j} \ \ \ , \ \vec{j} = \frac{1}{2m i} \frac{\text{Im}\{ \overline{\Psi} \vec{\nabla} \Psi\}}{\rho} = \frac{\rho(x)}{2m} \vec{\nabla} \varphi$

So this current $\vec{j}$ has a direct physical meaning: it describes the motion of the superfluid density! Now let's assume that $\rho(x) = \rho_0$. Then the expression for the current reduces to

$\vec{j} = \frac{\rho_0}{2 m} \vec{\nabla} \varphi = \rho_0 \vec{v}_s $

So in this limit the current is just the particle density times your superfluid velocity.

If we have now a simply connected region $D$ such that $\rho(x) = \rho_0$ in $D$, then indeed $\vec{\nabla} \times \vec{v}_s = 0$ since it is a gradient. We can use Stokes theorem to infer that all line-integrals inside $D$ of $\vec{v}_s$ vanish: let $\gamma \in D$ be a closed curve bounding an area $A$, then

$ \int_\gamma \vec{v}_s \cdot d \vec{l} = \int_A \vec{\nabla} \times \vec{v}_s \cdot d \vec{\sigma} = 0$

This however does not imply that $\vec{v}_s$ is zero in $D$! for example take the case of $\vec{v}_s$ a constant adnd the path $\gamma$ enclosing a square of side length $L$, where two of the edges are perpendicular to $\vec{v}_s$, so that $\vec{v}_s \cdot d \vec{l} = 0$, and two of the edges are parallel to $\vec{v}_s$, so that $\vec{v}_s \cdot d \vec{l} = \pm \|\vec{v}_s\|$. Then:

$\int_\gamma \vec{v}_s \cdot d \vec{l} = L ( \|\vec{v}_s\| + 0 - \|\vec{v}_s\| + 0) = 0$

as expected, without saying anything about the magnitude of the superfluid velocity.

Now let's consider the case mentioned in your comment, the case that $\gamma$ encloses a circle $C$ of radius $R$ and $\vec{v}_s \cdot d \vec{l} = \|\vec{v}_s\|$ all along $\gamma$, so that

$\int_\gamma \vec{v}_s \cdot d \vec{l} = 2\pi R \|\vec{v}_s\| . $

Now this is impossible if $\vec{v}_s = \frac{\vec{\nabla}\varphi}{2m}$ for a function $\varphi$ defined in the interior of the circle. For then we may calculate the line-integral by evaluating $\varphi$ on the endpoints of $\gamma$, which coincide, and thus give zero. Hence this is only possible if the phase $\varphi$ is not well-defined inside the circle, and this implies that $\Psi(x_*) = 0 $ for some $x_*$ inside the circle, so there has to be a vortex inside! Note that in this case, we should use the well-defined $\vec{j}$ instead of $\vec{v}_s(x) = \frac{\vec{j}(x)}{\rho(x)}$ which is only defined away from the vortex core. Far away from the vortex core, $\rho$ will be constant, so we may write

$\int_\gamma \vec{j} \cdot d \vec{l} = \rho_0 \int_\gamma \vec{v}_s \cdot d \vec{l}$

but if we now use Stokes' theorem, we do not get zero since there is no reason for $\vec{\nabla} \times \vec{j} = 0 $. However, if we insist on using $\vec{v}_s$, which has $\vec{\nabla} \times \vec{v}_s$ everywhere except at the vortex core, we may now longer use Stokes' theorem

$\int_\gamma \vec{v}_s \cdot d \vec{l} \neq \int_C \vec{\nabla} \times \vec{v}_s d \vec{\sigma} $

simply because the right-hand side does not exist! (It is like trying to make sense of the integral $\int_{0}^1 \frac{d x}{x}$.)

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  • $\begingroup$ I have not specified but $\vec{\nabla} \times \vec{v} $ is always true except for the center of the vortex $\endgroup$ – MementoMori Oct 8 '18 at 20:43
  • $\begingroup$ Sorry, i thought that was your question... you wrote $v_s = 0$, but that is not implied by stokes theorem, which just allows to infer $ \text{rot} v_s = 0$ on a simply connected domain the existence of a function $\phi$ s.t. $v_s = \text{grad} \phi$, i.e. the inverse to your opening statement. But by no means does this imply that $v_s = 0$, or am i misinterpreting your statements? $\endgroup$ – Lorenz Mayer Oct 8 '18 at 21:13
  • $\begingroup$ Suppose to have rot($v_s$) =0. Now use stokes theorem $\int rot v \cdot d \sum = \oint \vec{v} \cdot d \vec{l} = 2 \pi R v_s $ so $v_s=0$, where $\sum$ is my surface. My professor does it in this way but as i thought this way is wrong,the last step in general is not true and moreover you have to put strict conditions to make it true. $\endgroup$ – MementoMori Oct 9 '18 at 6:30
  • $\begingroup$ Thank you. One question when you say : "Now let's consider the case mentioned in your comment, the case that $\gamma$ encloses a circle $C$ of radius $R$ and $\vec{v}_s \cdot d\vec{l} =|| \vec{v}_s || $ all along $\gamma$ " $\vec{v}_s \cdot d\vec{l} =|| \vec{v}_s || $ is an hypothesis or something else? $\endgroup$ – MementoMori Oct 9 '18 at 14:10
  • $\begingroup$ It is an assumption on $\vec{v}_s$, namely exactly the assumption necessary to reproduce the equation you commented. $\endgroup$ – Lorenz Mayer Oct 9 '18 at 14:19

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