Electromagnetic tensor and energy

[itssilva]

From introductory courses on EM, I was given 'sketchy' proofs that, in a EM field in vacuum, magnetic energy density is B² and electric energy density is E² (bar annoying multiplication factors; they just get under my skin, I'll skip them all in the following). Other facts of life: -F^μνF_μν, the free EM Lagrangian (density), goes about like B² - E², which is an invariant, and δ(j^μA_μ - F^μνF_μν) = 0, or something, = Maxwell's equations (ME). Nice; however, Lagrangians are supposed to be energy - for a nonrelativistic test particle, L is just T - V -, and this is mysterious: why did someone put tensor F in the definition of the free EM Lagrangian, if it doesn't give the total energy density of the EM field in vacuum B²+ E² ? (If one can't find some other tensor F' and correct the full Lagrangian to give the ME, that can't be done, but in that case I want to know what F means, since j^u is supposed to be a Dirac invariant that couples to the field, which vanishes if the matter field does not interact with the EM field; if that sign in the E² were right, I'd be unashamed to call F's contraction the kinetic energy of the free field).

 

[andresB]

T-V is no the energy.

To actually answer your question, the lagrangian have to be gauge and relativistic invariant. There are only two combination of the Field tensor that fulfill that requirement, but only the one that you wrote is a scalar, the other is not invariant under parity transformation.

 

[stedwards]

T-V is no the energy.

To actually answer your question, the lagrangian have to be gauge and relativistic invariant. There are only two combination of the Field tensor that fulfill that requirement, but only the one that you wrote is a scalar, the other is not invariant under parity transformation.

I'm not sure we want invariance of the lagrangian density under a single parity transformation, but should, under CPT.

 

[andresB]

Not in general, but that lagrangian should give us Maxwell's equation and then it have to be invariant under parity. Note that the QED corrections to the classical lagrangian are also invariant under parity.

 

[stedwards]

Not in general, but that lagrangian should give us Maxwell's equation and then it have to be invariant under parity. Note that the QED corrections to the classical lagrangian are also invariant under parity.

I'm not familiar with QED, However we can examine $\nabla \times B - \partial_t E = J$. If we pick E and J to be vectors, B is a pseudovector. Picking B as vector, then E and J are pseudo vectors.

Either way, under a parity transformation of the equation, the vectors pick up a sign change and the axial vectors do not, or I'm was to sleepy for this and should stop posting.

 

[andresB]

I don't understand what you are trying to say.

But I suppose you are right that in general any combination of the scalar and pseudoscalar, to build the lagrangian, but the idea is to connect it with the known facts about electromagnetism, and for that you need the lagrangian to be invariant under parity transformation.

 

[my2cts]

Given a Lagrangian, which better be a scalar density, the Noether theorem delivers the energy momentum and angular momentum tensor densites.
However there are snakes in the grass if F*F is used as a Lagrangian. The resulting Noether densities are unacceptable.
One way out is to use the Belinfante trick. This is the standard procedure.
You can also use the Fermi Lagrangian, which does give completely acceptable Noether densities.
I have a reference on this if you are interested.

 

[stedwards]

I don't understand what you are trying to say.

But I suppose you are right that in general any combination of the scalar and pseudoscalar, to build the lagrangian, but the idea is to connect it with the known facts about electromagnetism, and for that you need the lagrangian to be invariant under parity transformation.

I'm sorry. I misunderstood your statement. Let me try again, if I may.

Why must the classical electromagnetic lagrangian density, under variation of the 4-vector potential, be invariant under a parity transformation? I hope this long description is the animal we're talking about.

 

[andresB]

Not sure,

The easy answer is that it leads to parity invariant equation of motion that fit with the experiments. Not sure if there is a deeper reason.

 

[stedwards]

Not sure,

The easy answer is that it leads to parity invariant equation of motion that fit with the experiments. Not sure if there is a deeper reason.

Thank you, Andres. You must be thinking along the lines of quantum field theory. In this, I haven't a clue. I'd very much like to ask you about the QED motivation, but we would diverge far from original question.

 

[andresB]

Well, I suppose you are free to create a new thread, But I'm not an expert in QED (I'm workin on the modified lagrangian that take into account some QED correction, but my work is on the classical level) so not sure If I could help you there