# Symmetry, conservation laws and coordinates

Consider the following Lagrangian $$L_1 = \iint \left(u_x^2+u_y^2+v_x^2+v_y^2\right)\mathrm d x\, \mathrm dy \tag{01}$$ where subscripts denote (partial) derivatives.

The following transformation preserves $$L$$:

$$u_x\to u_x\cos\phi-u_y\sin\phi; \quad u_y\to u_y \cos\phi+u_x\sin\phi; \tag{02a}$$ $$v_x\to v_x\cos\phi-v_y\sin\phi;\quad v_y\to v_y \cos\phi+v_x\sin\phi\tag{02b}$$

for some parameter $$\phi$$. You can find this by looking for the generators of the symmetries associated with Laplace's eqtn.

A natural question is, what is the conservation law associated with this symmetry? To this end, for an infinitesimal angle $$\delta \phi$$, we find $$\delta u_x =-u_y\delta \phi; \quad \delta u_y =u_x\delta \phi\tag{03a}$$ $$\delta v_x =-v_y\delta \phi; \quad \delta v_y =v_x\delta \phi\tag{03b}$$

Applying this to variations in $$L_1$$, we have $$\delta L = 0 \implies \iint \left(-u_xu_y+u_yu_x+ -v_xv_y+v_yv_x\right)\mathrm d x\, \mathrm dy = 0.\tag{04}$$ We confirm the Lagrangian stays fixed, but I am having a hard time interpreting this conservation law (does this just say mixed partials commute?).

Next, consider a mapping $$u=\sqrt{\rho}\cos\theta\,; \quad v=\sqrt{\rho}\sin \theta \tag{05}$$

The Lagrangian becomes

$$L_2=\int \int \left[\frac{\rho_x^2+\rho_y^2}{\rho}+4\rho(\theta_x^2+\theta_y^2)\right] \mathrm d x\, \mathrm dy\tag{06}$$

Is there any intuitive way, starting with this Lagrangian, to find the symmetry that was apparent in Cartesian coordinates? The fact that $$L_1$$ just depends on gradients, while $$L_2$$ does not, seems to make the symmetry analysis yield different results (which is consistent with what I've read: eg p 181 of Olver's book) and in particular it's not obvious how to find a relationship for how $$\rho$$ maps, as it depends on nonlocal (ie integral) quantities.

I) Hints for the first part:

1. Concentrate on $$u$$ and forget about $$v$$ as they enter in similar fashion.

2. Let's also put in a conventional $$\frac{1}{2}$$ factor, i.e. the Lagrangian density becomes \begin{align}{\cal L}~=~&\frac{1}{2}u_{\mu}u^{\mu}, \qquad u_{\mu} ~:=~ d_{\mu} u,\cr d_{\mu}~:=~&\frac{d}{dx^{\mu}}, \qquad \mu~\in~\{1,2\}.\end{align}\tag{A}

3. Instead of letting OP's symmetry act on the derivatives $$u_{\mu}$$, it can be viewed as originating from an infinitesimal (so-called horizontal) rotation $$\delta x^{\mu}~=~\epsilon \varepsilon^{\mu\nu}x_{\nu}\tag{B}$$ in the worldsheet. It in turn induces a so-called vertical infinitesimal transformation $$\delta_0 u~=~-u_{\mu}\delta x^{\mu} ~\stackrel{(B)}{=}~-\epsilon u_{\mu}\varepsilon^{\mu\nu}x_{\nu},\tag{C}$$ so that the total infinitesimal variation $$\delta u~=~\delta_0 u+u_{\mu}\delta x^{\mu}~\stackrel{(C)}{=}~0\tag{D}$$ is zero.

4. The corresponding Noether current \begin{align} j^{\mu} ~=~~~~~&\frac{\partial {\cal L}}{\partial u_{\mu}}\frac{\delta_0 u}{\epsilon}+{\cal L}\frac{\delta x^{\mu}}{\epsilon}\cr ~\stackrel{(A)+(B)+(C)}{=}& -u^{\mu}u_{\lambda} \varepsilon^{\lambda\nu}x_{\nu}+\frac{1}{2}u_{\lambda}u^{\lambda} \varepsilon^{\mu\nu}x_{\nu}\end{align}\tag{E} satisfies a continuity equation $$d_{\mu}j^{\mu}~\approx~0\tag{F}$$ on-shell.

II) Hints for the second part:

1. Since the transformation from rectangular/Cartesian to polar coordinates takes place in the target space while the symmetry acts in the worldsheet, they seem to be separate issues.
• Thank you for this helpful response! Can you explain where (C) comes from? 1. Is it the chain rule? 2. Or does it follow directly from the total variation of L vanishing? Commented Jul 12, 2023 at 6:11
• I updated the answer. Commented Jul 12, 2023 at 6:27