Maxwell Lagrangian


by Petraa
Tags: lagrangian, maxwell
Petraa
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#1
Nov14-12, 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!
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dextercioby
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#2
Nov14-12, 12:34 PM
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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.
Petraa
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#3
Nov14-12, 01:03 PM
P: 7
Quote Quote by dextercioby View Post
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.
I've been watching the book and yes, the book treats it but don't deduce them. He just announces and perform some calculations with them

dextercioby
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#4
Nov14-12, 01:29 PM
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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]
Petraa
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#5
Nov14-12, 01:57 PM
P: 7
Quote Quote by dextercioby View Post
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]
I'll try it.
Petraa
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#6
Nov14-12, 04:34 PM
P: 7
Quote Quote by dextercioby View Post
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]
[tex] T^{\mu\nu}=-F^{\mu\nu}\partial^{\nu}A_{\rho}+\frac{1}{4}F^{2}g^{\mu\nu}[/tex]

And now? How I relate this to the momentum and total angular momentum operators ?
dextercioby
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#7
Nov14-12, 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 space-time rotation):

x'μ=xμμ ν xν, where

ϵμν = - ϵνμ

A minor change

Tμν=−FμρνAρ+1/4 F2gμν
Meir Achuz
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#8
Nov20-12, 12:29 PM
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Quote Quote by Petraa View Post
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!
I too would be interested in seeing this for EM angular momentum.
Every place, I have looked seems to use the result in some form without actually deriving it.
LayMuon
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#9
Nov25-12, 11:17 AM
P: 146
Quote Quote by Petraa View Post
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!
Maggiore "Modern introduction in QFT"


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