electromagnetic energy is not Gauge invariant?


by JustinLevy
Tags: electromagnetic, energy, gauge, invariant
JustinLevy
JustinLevy is offline
#1
Feb8-09, 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.
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samalkhaiat
samalkhaiat is offline
#2
Feb8-09, 03:23 PM
Sci Advisor
P: 815
[QUOTE]
Quote Quote by JustinLevy View Post
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]
In any gauge field thory, It is a very important to understand that gauge fields couple to CONSERVED current [itex]\partial_{\mu}J^{\mu} = 0[/itex], if you use this you find

[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
JustinLevy
JustinLevy is offline
#3
Feb8-09, 05:24 PM
P: 886
Quote Quote by samalkhaiat View Post
In any gauge field thory, It is a very important to understand that gauge fields couple to CONSERVED current [itex]\partial_{\mu}J^{\mu} = 0[/itex], if you use this you find

[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
I wanted to talk about the electromagnetic energy density (proportional to [itex](E^2+B^2)[/itex]), but you have instead talked about a relativistic scalar density (proportional to [itex](E^2-B^2)[/itex]). Therefore I am not sure how to relate your response back to the original question. Are you saying [itex]A_{\mu}J^{\mu}[/itex] is gauge invariant but the electromagnetic energy density is NOT?


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!?

samalkhaiat
samalkhaiat is offline
#4
Feb11-09, 04:57 PM
Sci Advisor
P: 815

electromagnetic energy is not Gauge invariant?


[QUOTE]
Quote Quote by JustinLevy View Post
I wanted to talk about the electromagnetic energy density (proportional to [itex](E^2+B^2)[/itex]), but you have instead talked about a relativistic scalar density (proportional to [itex](E^2-B^2)[/itex]). Therefore I am not sure how to relate your response back to the original question. Are you saying [itex]A_{\mu}J^{\mu}[/itex] is gauge invariant but the electromagnetic energy density is NOT?
[itex]A_{\mu}J^{\mu}[/itex] is the interaction Lagrangian. It is not proportional to the free electromagnetic Lagrangian [itex]E^{2} - B^{2}[/itex]. Why do you need to worry about the gauge invariance of energy density or even the momentum density? There are no fundamental reasons for requiring them to be gauge invariant! The quantities that we measure are ENERGY and MOMENTUM. Therefore, they must be gauge invariant. I will show you this below.
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 energy-momentum 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) energy-momentum 4-vector 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 gauge-invariant 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 energy-momentum 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 gauge-invariant energy-momentum 4-vector. 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
JustinLevy
JustinLevy is offline
#5
Feb14-09, 06:09 PM
P: 886
Quote Quote by samalkhaiat View Post
Assuming there is no current at infinity...
Ah, okay. That is why my "counter-example" fails.
The rest of what you wrote makes sense as well.

Thanks for your help.


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