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Noether currentby spookyfish
Tags: classical action, classical fields, current, noether, noether currents, noether theorem 
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#1
Oct2613, 02:18 PM

P: 48

Hi,
I read about Noether's theorem, which states that if, under a continuous transformation, the Lagrangian is changed by a total derivative [itex] \delta \cal L = \partial_\mu F^\mu [/itex] then there is a conserved current [tex] j^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi  F^\mu [/tex] However, I have seen in a different place the formulation that if the action is invariant, then the conserved quantity is: [tex] \frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi  T^{\mu \nu}\delta x_\nu [/tex] where [itex] T^{\mu \nu} [/itex] is the energymomentum tensor. Is the second formulation equivalent to the first? or is it a particular case 


#2
Oct2613, 03:30 PM

Sci Advisor
Thanks
P: 2,351

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 spacetime coordinates, where the Lagrangian changes by a total fourgradient, which means that the action is invariant. The second case is a symmetry under a more general transformation, where the spacetime 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 fourgradient. Then the Noether current is [tex]\frac{\partial}{\partial (\partial_{\mu} \phi)} \delta \phi  T^{\mu \nu} \delta x_{\nu}F^{\mu}.[/tex] You find this derived in some detail in my quantumfield theory manuscript: http://fias.unifrankfurt.de/~hees/publ/lect.pdf 


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