electromagnetic energy is not Gauge invariant?by JustinLevy Tags: electromagnetic, energy, gauge, invariant 

#1
Feb809, 02:35 AM

P: 886

I assume I am making a mistake here. Can you please help me learn how to fix them?
In electrodynamics, the gauge transformations are: [tex]\vec{A} \rightarrow \vec{A} + \vec{\nabla}\lambda[/tex] [tex]V \rightarrow V  \frac{\partial}{\partial t}\lambda[/tex] These leave the electric and magnetic fields unchanged. The electromagnetic energy density is proportional to: [tex] u \propto (E^2 + B^2)[/tex] So the energy density should be gauge invariant. A guage transformation shouldn't even shift it by a constant. However, the energy density can also be written in terms of the potentials and charge distributions as: [tex] u = \frac{1}{2}(\rho V + \vec{j} \cdot \vec{A})[/tex] If I let [itex]\lambda[/itex] = e^(r^2), then V is unchanged, and A is just changed by a radial field which vanishes at infinity. Here, the energy density IS changed, and not by a constant amount either ... it changes by an amount depending on j. What gives? Even weirder, is some textbooks start with the potentials form to derive the fields form. I don't see any place they fix any gauge in doing such derivation. Please help. 



#2
Feb809, 03:23 PM

Sci Advisor
P: 815

[QUOTE]
[tex]A_{\mu}J^{\mu} \rightarrow A_{\mu}J^{\mu} + J^{\mu}\partial_{\mu} \lambda[/tex] or [tex]\int d^{3} x \ \delta (A_{\mu}J^{\mu}) = \int d^{3} x \ \partial_{\mu} (\lambda J^{\mu}) = 0[/tex] regards sam 



#3
Feb809, 05:24 PM

P: 886

I understand that [tex]\partial_{\mu} J^{\mu}=0[/tex] because it is a statement of conservation of charge. But I don't understand why the following is zero for every possible J or lambda [tex]\int d^{3} x \ \delta (A_{\mu}J^{\mu}) = \int d^{3} x \ J^{\mu} \partial_{\mu} \lambda = 0[/tex]. Let's try just a simple circulating current, and a lambda polynomial in x and y: [tex]\vec{j} = y\hat{x} + x\hat{y}[/tex] [tex]\lambda = xy[/tex] noting that for this choice [tex]\partial_{\mu} \lambda = y\hat{x} + x\hat{y}[/tex] Looking at the result: [tex]\int d^{3} x \ J^{\mu} \partial_{\mu} \lambda = \int d^{3} x \ (y\hat{x} + x\hat{y}) \cdot (y\hat{x} + x\hat{y}) \neq 0 [/tex] So that doesn't seem to be gauge invariant either!? 



#4
Feb1109, 04:57 PM

Sci Advisor
P: 815

electromagnetic energy is not Gauge invariant?
[QUOTE]
I was trying to tell you that the EM interaction is gauge invariant because its change reduces to a hepersurface integral [tex]\int d^{4} x \ \delta (A_{\mu}J^{\mu}) = \int d^{4} x \ J^{\mu} \partial_{\mu} \lambda = \int d^{4} x \ \partial_{\mu}(\lambda J^{\mu}) = 0[/tex] Assuming there is no current at infinity, the last integral vanishes since it can be changed to a surface integral at infinity. If [itex]J^{\mu}[/itex] is treated as externally given source, then the canonical energymomentum tensor [tex]T^{\mu\nu} = \frac{1}{4} \eta^{\mu\nu} F_{\sigma \rho}F^{\sigma \rho}  F^{\mu \sigma} \partial^{\nu}A_{\sigma} + \eta^{\mu\nu}A_{\rho}J^{\rho}[/tex] will have the following undesirable properties 1) It is not conserved: [tex]\partial_{\mu}T^{\mu\nu} = (\partial^{\nu}J_{\sigma})A^{\sigma} \ \ (1)[/tex] Notice that T is conserved in the absence a current,i.e.,free electromagnetic field. 2) It is not gauge invariant. Indeed, it changes like [tex] \delta T^{\mu\nu} = \partial_{\sigma} (F^{\sigma \mu} \partial^{\nu} \lambda + \eta^{\mu\nu} J^{\sigma} \lambda )  J^{\mu}\partial^{\nu} \lambda \ \ (2) [/tex] This "problem" stays with us even in the absence of sources; [tex] \delta T^{\mu\nu} = \partial_{\sigma}\left( F^{\sigma \mu} \partial^{\nu} \lambda \right) [/tex] However, in this case, the apparent violation of gauge invariance is no reason for concern because the change in T is a total divergence, which upon integration leads to a surface term, making no contribution to the total (measurable) energymomentum 4vector of the EM field; [tex] \delta P^{\nu} = \int d^{3} x \delta T^{0 \nu} = \int d^{3} x \partial_{\sigma} ( F^{\sigma 0} \partial^{\nu}\lambda}) = \int d^{3} x \partial_{j} (F^{j 0} \partial^{\nu}\lambda) = 0 [/tex] Thus, even though the energy density [itex]T^{00}[/itex] and the momentum density [itex]T^{0j}[/itex] ARE NOT gauge invariant, the total energy [itex]P^{0}[/itex] and the total momentum [itex]P^{j}[/itex] are gaugeinvariant quantities. To resolve the above two problems when [itex]J_{\mu} \neq 0[/itex], the sources must be included in the dynamical description, i.e., our complete theory must include the dynamical fields which cause the currents. For example, Dirac's fields in the current [itex]J^{\mu} = \bar{\psi} \gamma^{\mu} \psi[/itex] . These matter fields will contribute to the total energymomentum tensor on the LHS of eq(1) & (2), whereas the RHS of eq(1) will vanish and the RHS of eq(2) will again be a total divergence leading to a gaugeinvariant energymomentum 4vector. It is a good exercise to do the calculations on the QED Lagrangian [tex]\mathcal{L} = i\bar{\psi} \gamma^{\mu}\partial_{\mu}\psi (1/2) F^{2}  J_{\mu}A^{\mu}[/tex] Try it. regards sam 



#5
Feb1409, 06:09 PM

P: 886

The rest of what you wrote makes sense as well. Thanks for your help. 


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