# Conserved charge

1. Mar 18, 2007

### kakarukeys

Is there an inconsistency between the definition of conserved charge and conserved current in Hamiltonian and Lagrangian formulation?

For example, $$H = \int T^{00} d^3x$$ is a conserved charge,
$$\frac{dH}{dt} = \{H, H\} = 0$$

But we have $$\partial_\mu T^{\mu\nu} = 0$$ implies
$$\int (\partial_\mu T^{\mu 0}) d^3x = \int (\partial_0 T^{00} + \partial_i T^{i0}) d^3x = 0$$ so it seems
$$\frac{d}{dt}\int T^{00}d^3x = - \int \partial_i T^{i0} d^3x \neq 0$$

I'm very puzzled.

Last edited: Mar 18, 2007
2. Mar 18, 2007

### Crosson

Why is the last integral non-zero? If the charge is constant, shouldn't the current be zero?

3. Mar 18, 2007

### kakarukeys

So there's a contradiction. In general if there is a boundary, the last integral is not zero.

4. Mar 19, 2007

### dextercioby

Yes, of course. It's the case on curved manifolds which occur in GR, for example. The boundary terms are very important. However, as it's usually presented in field theory in Minkowski space, the hypersurface integrals are always chosen to be 0.

5. Mar 19, 2007

### kakarukeys

So is there a condition
$$T^{i0} = 0$$ at boundary?

6. Mar 19, 2007

### da_willem

If you take a volume in which the charge is conserved that will mean there is no net charge flowing in or out of the boundary, wich is the last condition you mention (integrated over the surface).