Noether current

1. Oct 26, 2013

spookyfish

Hi,

I read about Noether's theorem, which states that if, under a continuous transformation, the Lagrangian is changed by a total derivative
$\delta \cal L = \partial_\mu F^\mu$

then there is a conserved current
$$j^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi - F^\mu$$

However, I have seen in a different place the formulation that if the action is invariant, then the conserved quantity is:

$$\frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi - T^{\mu \nu}\delta x_\nu$$
where $T^{\mu \nu}$ is the energy-momentum tensor.

Is the second formulation equivalent to the first? or is it a particular case

Last edited: Oct 26, 2013
2. Oct 26, 2013

vanhees71

Neither of both formulae is the most general case of a symmetry and Noether's theorem but special cases.

The first case is the symmetry of the action under a variation of the field and unvaried space-time coordinates, where the Lagrangian changes by a total four-gradient, which means that the action is invariant.

The second case is a symmetry under a more general transformation, where the space-time coordinates and fields are changed under the transformation and the Lagrangian is invariant. Of course a symmetry is still present also under such transformations, if the Lagrangian changes by a total four-gradient. Then the Noether current is
$$\frac{\partial}{\partial (\partial_{\mu} \phi)} \delta \phi - T^{\mu \nu} \delta x_{\nu}-F^{\mu}.$$
You find this derived in some detail in my quantum-field theory manuscript:
http://fias.uni-frankfurt.de/~hees/publ/lect.pdf