# Conserved quantities of $N$-vortices

In my notes we are given that for $$N$$-vortices with location $${(x_i,y_i)}^N_{i=1}$$ and stegnth $${\Gamma _i}$$.

We have the conserved quanitites of energy, momentum and angular momentum.

How does one prove these to be true. (i am looking for method that doesn't use Hamiltonian if possible)

• Apr 28, 2022 at 17:19

I assume the following is known. Velocity of i-th vortex is given by $$\bar{v}_i=\sum_{j\neq i} \Gamma_j \frac{[\hat{\omega}\times \bar{r}_{ij}]}{r^2_{ij}},\quad (1)$$ where $$\hat{\omega}$$ is a unit vector directed out of the plane and $$\bar{r}_{ij}=\bar{r}_{j}-\bar{r}_{i}$$ is a vector, connecting i-th and j-th votices and $$\Gamma_j$$ is the borticity of the j-th vortex. Interaction energy of i-th and j-th vortices is $$U_{ij}=-\frac{1}{4\pi}\Gamma_i\Gamma_j \ln(r_{ij}).$$ Here are the rough sketches of the proofs of the conservation laws.
Energy conservation. Taking time derivative from the sum of all interaction energies, we obtain $$\dot{U}=\sum_{ij}\dot{U}_{ij}=-\frac{1}{4\pi}\sum_{ij}\Gamma_i\Gamma_j \frac{\bar{v}_j-\bar{v}_i}{r_{ij}^2}$$ We can notice that $$\dot{U}_{ij}=-\dot{U}_{ji}$$, which ensures that the above sum is zero, and therefore the energy is conserved.
Momentum conservation. Momentum of the system is $$P=\sum_i \Gamma_i \bar{v}_i$$ Substituting here Eq. (1) and accounting for $$\bar{r}_{ij}=-\bar{r}_{ji}$$, yields $$P=0$$.
Angular momentum conservation. Angular momentum of the system is $$M=\sum_{i} M_i=\sum_{i} [\bar{r}_{i}\times\Gamma_i\bar{v}_i ].$$ Substituting Eq. (1), expanding cross-product and accounting for $$\hat{\omega}$$ being perpendicular to the plain, we obtain $$M=\sum_{ij,i\neq j} M_{ij}=\sum_{ij,i\neq j}\Gamma_i \Gamma_j \frac{\hat{\omega}(\bar{r_i},\bar{r_j}-\bar{r_i})}{r_{ij}^2}$$ One can notice that $$M_{ij}+M_{ji}=\Gamma_i\Gamma_i\hat{\omega}$$, which makes the whole sum equal to $$M=\sum_{ij,i\neq j}\Gamma_i \Gamma_j \hat{\omega},$$ which is clearly a conserved quantity.