# Noether's theorem

Euler-Langrange Equation

$\frac{\partial L}{\partial \varphi} - \partial_\mu \frac{\partial L}{\partial \varphi'} = 0$

Action

$\mathcal S = \int \mathcal L(\varphi, \partial_\mu\varphi, x^\mu)\,d^4x$

transition

$\varphi \mapsto \varphi + \delta$

$\mathcal L \mapsto \mathcal L + \partial_\mu J$

L difference

$d\mathcal L = \frac{\partial\mathcal L}{\partial\varphi}\partial\varphi + \frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi'$

$= \frac{\partial\mathcal L}{\partial\varphi}\partial\varphi - \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'})\partial\varphi + \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'})\partial\varphi + \frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi'$

$= \frac{\partial\mathcal L}{\partial\varphi}\partial\varphi - \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'})\partial\varphi + \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi)$

$= (\frac{\partial\mathcal L}{\partial\varphi} - \partial_\mu\frac{\partial\mathcal L}{\partial\varphi'})\partial\varphi + \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi)$

$= \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi)$

$\partial_\mu J = \partial_\mu(\frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi)$

$\partial_\mu (\frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi - J) = 0$

Noether's current

$j = \frac{\partial\mathcal L}{\partial\varphi'}\partial\varphi - J \text{ is const}$

## Examples

### Euclidean translations

The Langrangian is

$\mathcal L = \sum_a \frac{1}{2}m_a\dot{\hat x}_a^2 - V(\hat x_1, \hat x_2, ...)$

Transition is

$\hat x \mapsto \hat x + \hat \delta,\quad \dot {\hat x} \mapsto \dot x, \quad t \mapsto t$

Because potential depends on the relative positions of each particles,

$\mathcal L \mapsto \mathcal{L}$

Noether's current is

$j = \frac{\partial\mathcal L}{\partial \dot {\hat x}}\partial \hat x = \sum_a m_a\dot {\hat x}_a\hat\delta=\text{const}$

because of $\hat \delta$ is arbitary direction,

$\sum_a m_a\dot {\hat x}_a=\text{const}$

### Euclidean rotation

The Langrangian is

$\mathcal L = \sum_a \frac{1}{2}m_a\dot{\hat x}_a^2 - V(\hat x_1, \hat x_2, ...)$

Transition is

$\hat x \mapsto \hat x + \hat \delta \times \hat x,\quad \dot {\hat x} \mapsto \dot {\hat x} + \hat \delta \times \dot {\hat x}, \quad t \mapsto t$

Because potential depends on the relative positions of each particles,

$\mathcal L \mapsto \mathcal{L}$

Noether's current is

$j = \frac{\partial\mathcal L}{\partial \dot {\hat x}}\partial \hat x = \sum_a m_a\dot {\hat x}_a\cdot(\hat\delta \times {\hat x})=\sum_a\hat\delta \cdot({\hat x} \times m_a\dot {\hat x}_a )=\text{const}$

because of $\hat \delta$ is arbitary direction,

$\sum_a({\hat x} \times m_a\dot {\hat x}_a )$

### time translations

The Langrangian is

$\mathcal L = \sum_a \frac{1}{2}m_a\dot{\hat x}_a^2 - V(\hat x_1, \hat x_2, ...)$

Transition is

$\hat x \mapsto \hat x + \frac{\partial\hat x}{\partial t}\epsilon ,\quad \dot {\hat x} \mapsto \dot {\hat x} + \frac{\partial\dot {\hat x}}{\partial t}\epsilon, \quad t \mapsto t + \epsilon$

Because potential depends on the relative positions of each particles,

$\mathcal L \mapsto \mathcal{L} + \frac{\partial\mathcal{L}}{\partial t}\epsilon$

Noether's current is

$j =\frac{\partial\mathcal{L}}{\partial \dot {\hat x}} \frac{\partial\hat x}{\partial t}\epsilon - \mathcal L\epsilon = (\frac{\partial\mathcal{L}}{\partial \dot {\hat x}}\dot {\hat x}- \mathcal L)\epsilon = \text{const}$

because of $\hat \delta$ is arbitary direction,

$\sum_a \frac{\partial\mathcal{L}}{\partial \dot {\hat x}}\dot {\hat x}- \mathcal L=\text{const}$

called Hamiltonian is conserved