# Conserved Noether charge and gravity

1. Dec 5, 2011

### haushofer

If one considers the Lagrangian of a non-relativistic particle in a gravitational field,

$$L = \frac{m}{2}(\delta_{ij}\dot{x}^i \dot{x}^j + 2 \phi(x^k) )$$

it transforms under

$$\delta x^i = \xi^i (t), \ \ \ \ \delta \phi = \ddot{\xi}^i x_i$$

as a total derivative:

$$\delta L = \frac{d}{dt}(m \dot{\xi}^i x_i)$$

My question is: why is the corresponding Noether charge

$$Q = p_i \xi^i - m \dot{\xi}^i x_i$$

not conserved if one uses the equations of motion? I'm staring at the problem now for quite some time, missing something obvious, but I can't see it :)

2. Dec 5, 2011

### haushofer

It's confusing, because the EOM for phi is the Poisson equation, which is not dynamical; it could be obtained by adding

$$\lambda[\partial_i \partial^i \phi - 4 \pi G \rho]$$

to the Lagrangian, where lambda is a Lagrange multiplier. But this doesn't change the situation.

3. Dec 5, 2011

### haushofer

So I more or less have the answer to my question; the variation $\delta L$ also contains a

$$\frac{\partial L}{\partial \phi}\delta \phi = m \ddot{\xi}^i x_i$$

term, which I don't want. So how to get the corresponding Noether charges?

4. Dec 6, 2011

### haushofer

Does somebody know where this problem is treated?

5. Dec 8, 2011

### haushofer

I think I have the answer, so nevermind.