
#1
Nov1412, 11:53 AM

P: 7

Hello,
Where can I find a good explanation (book) of the derivation via Noether's theorem of the three momentum and angular momentum operators of the usual maxwell lagrangian ? Thank you! 



#2
Nov1412, 12:34 PM

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P: 11,863

This is standard QFT (actually QED) material, any thorough book should have it. Check out a nice treatment in Chapter 2 of F. Gross' "Relativistic Quantum Mechanics and Field Theory", Wiley, 1999.
In purely classical context (no operators), advanced electrodynamics books should also have this. 



#3
Nov1412, 01:03 PM

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#4
Nov1412, 01:29 PM

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P: 11,863

Maxwell Lagrangian
Can you calculate [itex] T^{\mu\nu} [/itex] and [itex] M^{\lambda}_{~~\mu\nu} [/itex] from the Lagrangian and the general Noether formula which for the energymomentum 4 tensor reads:
[itex] T^{\mu}_{~~\nu} [/itex] = ([itex] \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\rho})}[/itex] [itex] \mathcal{L}\delta^{\mu}_{\lambda} [/itex]) X [itex] \frac{\partial x'^{\lambda}}{\partial\epsilon^{\nu}} [/itex], where [tex] x'^{\mu} = x^{\mu} + \epsilon^{\mu} [/tex] 



#5
Nov1412, 01:57 PM

P: 7





#6
Nov1412, 04:34 PM

P: 7

And now? How I relate this to the momentum and total angular momentum operators ? 



#7
Nov1412, 04:41 PM

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The momentum should be [itex] T^{0i} [/itex], just like energy is [itex] T^{00} [/itex]. For angular momentum, you should derive the general formula using the linearized version of a general Lorentz transformation (i.e. a linearized spacetime rotation):
x'^{μ}=x^{μ}+ϵ^{μ} _{ν} x^{ν}, where ϵ^{μν} =  ϵ^{νμ} A minor change T^{μν}=−F^{μρ}∂^{ν}A_{ρ}+1/4 F^{2}g^{μν} 



#8
Nov2012, 12:29 PM

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Every place, I have looked seems to use the result in some form without actually deriving it. 



#9
Nov2512, 11:17 AM

P: 146




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