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)
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.
