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Maxwell Lagrangian |
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| Nov14-12, 11:53 AM | #1 |
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Maxwell Lagrangian
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! |
| Nov14-12, 12:34 PM | #2 |
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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. |
| Nov14-12, 01:03 PM | #3 |
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| Nov14-12, 01:29 PM | #4 |
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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 energy-momentum 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] |
| Nov14-12, 01:57 PM | #5 |
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| Nov14-12, 04:34 PM | #6 |
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And now? How I relate this to the momentum and total angular momentum operators ? |
| Nov14-12, 04:41 PM | #7 |
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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 space-time rotation):
x'μ=xμ+ϵμ ν xν, where ϵμν = - ϵνμ A minor change Tμν=−Fμρ∂νAρ+1/4 F2gμν |
| Nov20-12, 12:29 PM | #8 |
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Recognitions:
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Every place, I have looked seems to use the result in some form without actually deriving it. |
| Nov25-12, 11:17 AM | #9 |
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