# 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

• "Now if we consider a simply connected space and if we use Stoke's Theorem we have that $v_s=0$" ...this is not true! Moreover, you can have vortices even in non-rotating vessels, exactly as in water you can have vortices (or vortex loops) in non-rotating containers (the total angular momentum of the fluid will be zero, but the local vorticity can be different from zero). May 8, 2022 at 12:32

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 Oct 8, 2018 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? Oct 8, 2018 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. Oct 9, 2018 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? Oct 9, 2018 at 14:10
• i am not sure what your question is... which calculation in the paper are you refering to? which motion you want to know of whether it is inevitable? Oct 10, 2018 at 15:01