# Vortex in superfluid?

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

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 $$\int_{0}^1 \frac{d x}{x}$$.)

• I have not specified but $\vec{\nabla} \times \vec{v}$ is always true except for the center of the vortex – MementoMori Oct 8 '18 at 20:43
• 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? – Lorenz Mayer Oct 8 '18 at 21:13
• 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. – MementoMori Oct 9 '18 at 6:30
• 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? – MementoMori Oct 9 '18 at 14:10
• It is an assumption on $\vec{v}_s$, namely exactly the assumption necessary to reproduce the equation you commented. – Lorenz Mayer Oct 9 '18 at 14:19