Energy location of Electromagnetic and Charge-current Fields.

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Main Question or Discussion Point

What practically obtained experimental evidence could be used to distinguish between the location of electromagnetic energy density given by these contending terms,

[tex]E^2+B^2[/tex]

vs.

[tex]\bar{J}\cdot \bar{A}+\rho \phi \ ,[/tex]

where A is the magnetic potential, [itex]\rho[/itex] is charge density and [itex]\phi[/itex] is the electric potential?
 
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Answers and Replies

  • #2
Meir Achuz
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There is no experimental evidence to distinguish which is energy density. There is no rigorous theoretical reason to choose one over the other. Poynting's theorem shows how EM energy passes through a closed surface, but not which choice is energy density. Surface integrals that are too often ignored lead to the claim that one or the other is energy density. If surface integrals are included properly, then the energy EM within a closed surface is the same for each. You might want to look at
<http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf> [Broken]
which is a controversial look at this, suggesting \rho\phi +j.A. is EM density. It really shows that either will work for the energy within a closed surface (including surface integrals), but there is no meaning to EM energy density at a point.
 
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  • #3
Bill_K
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Sorry, the energy density is a component of the stress-energy tensor, which is what gravity couples to. Consequently it must be unique and well-defined at every point. For electromagnetism E^2 + B^2 is the correct form. The second form would imply an electromagnetic energy density of zero in any charge-free, current-free region, which is obviously not the case.
 
  • #4
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Bill_K,

When you lift an apple or a stone to a height of 2 meters in the gravity field of the earth surface,
where do you feel the potential energy is "obviously" stored?

A poll would for sure come with a majority of people voting for the stone as the energy storage.
The stone or the apple falling on their head would bias their opinion very strongly.

A few theoreticians would pretend energy is stored in the field.
They know fields better than apples and stones.

I think that none of these answers are correct.
Potential energy is not stored.
You can only store apples, stones or charges.
Charges owns their fields.

Michel
 
  • #5
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Bill_K,

The "gravity couples" to the stress-energy tensor, not only the energy density.
Are you sure that accounting for all the terms would support the idea that energy is "obviously" stored in the fields?

How can you distinguish the two statements in calculating this coupling?
I lack experience in using the Einstein's field equations.
Without more information, my best guess is that the Einstein's equation would not favor one or the other point of view.
After all, the stress-energy tensor can be used to formulate the whole physics, not only general relativity, and I do not remember having seen any argument to decide where "potential energy" is stored.
My feeling is that this is not a good question, at least as long as I could not see an impact on a result.
But this is an interresting question.

Michel
 
  • #6
Meir Achuz
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Sorry, the energy density is a component of the stress-energy tensor, which is what gravity couples to.
I am glad you have succeeded in a rigorous unification of EM and GR.
Please let Einstein know.
 
  • #7
Bill_K
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Yes indeed I am sure. Physics questions are not decided by popular poll.

Are you seriously claiming that an electromagnetic wave in vacuum has zero energy density? That the energy in a magnetic field is not really in the field, it's stored down inside the magnet? Note that the second expression involves A and φ and is not even gauge invariant, therefore does not represent a genuine physical quantity. It's sometimes a handy thing to use when calculating the total energy as an integral, but it does not correctly represent the energy density.

Regarding your example of the gravitational potential of a stone, well you've chosen a very lopsided example where it sounds silly to mention the gravitational field of a stone, nevertheless that's where the potential energy is located. How exactly could it be stored in the stone? A stone on the ground is identical, atom for atom, to a stone raised in the air. It's just for convenience that we think of the "potential energy of the stone", but that's not where it's really located. What about the Earth-Moon system - where is the gravitational energy stored in that case? In the Moon? In the Earth? Or maybe some of it in the Moon and some of it in the Earth? Of course not, it's stored in the field.
 
  • #8
Bill_K
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I am glad you have succeeded in a rigorous unification of EM and GR.
Please let Einstein know.
Thanks, but the interaction between gravity and electromagnetism has been understood for about a century now.
 
  • #9
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There is no experimental evidence to distinguish which is energy density. There is no rigorous theoretical reason to choose one over the other. Poynting's theorem shows how EM energy passes through a closed surface, but not which choice is energy density. Surface integrals that are too often ignored lead to the claim that one or the other is energy density. If surface integrals are included properly, then the energy EM within a closed surface is the same for each. You might want to look at
<http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.3421v3.pdf> [Broken]
which is a controversial look at this, suggesting \rho\phi +j.A. is EM density. It really shows that either will work for the energy within a closed surface (including surface integrals), but there is no meaning to EM energy density at a point.
Thanks. That's just what the doctor ordered.

Jerrold Franklin certainly puts his reputation on the line with some very strong statements such as
The fairly common belief that there can be electrostatic field energy in otherwise empty space is wrong.
and
The fairly common belief that there can be magnetostatic
field energy in regions without current is wrong.

I've very tempted to recast his equations in differential forms, where possible, and look for an expression of electromagnetic energy-momentum.
 
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  • #10
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Note that the second expression involves A and φ and is not even gauge invariant, therefore does not represent a genuine physical quantity.
Hello Bill_K. How do you think gauge invariance counts against expression two?

By the way, thanks for pointing out the dichotomy; that
[tex]
E^2+B^2
[/tex]
is an absolute energy density, and
[tex]
\bar{J}\cdot \bar{A}+\rho \phi \ ,
[/tex]
is not.
 
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  • #11
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My feeling is that this is not a good question, at least as long as I could not see an impact on a result.
But this is an interresting question.

Michel
Thank you. And I would be very interested to know why you think this might not be a good question if you still think so. If a question I've made doen't seem to make sense, I what to know why! Most, if not all, of physics consists in answering ill posed questions in my opinion, and upon such time where a question can be made that is not ill posed, physics would become uninteresting.
 
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  • #12
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Phrak,

For me, saying that it is not a good question is absolutely not pejorative.
On the contrary!
And I agree totally with statement: physics is more about learning to ask the good question than learning to answer any question.

To go back to the thread, what is then the meaning of "storing energy"?
For fuels, for electrical batteries, this has an every-day meaning. (good for a poll, and obvious)
It is essentialy about solid things from which we can get some work being done.
This simple picture breaks down if we look at these example on a microscopic level.
But does that mean that gasoline is a field carried in a liquid?

Really, I like your question, but I don't see how to chose between the two point of views.
And again, it seems that I learned from this question.
But it is not a "good question" because, I think, nature doesn't speak this language.
How could that be answer experimentally?
Experiment could only validate the two points of view, the two formulas.

Michel
 
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  • #13
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Yes indeed I am sure. Physics questions are not decided by popular poll.
But can we use the word "obvious" in physics, then?
 
  • #14
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Phrak,

For me, saying that it is not a good question is absolutely not pejorative.
On the contrary!
And I agree totally with statement: physics is more about learning to ask the good question than learning to answer any question.

To go back to the thread, what is then the meaning of "storing energy"?
For fuels, for electrical batteries, this has an every-day meaning. (good for a poll, and obvious)
It is essentialy about solid things from which we can get some work being done.
This simple picture breaks down if we look at these example on a microscopic level.
But does that mean that gasoline is a field carried in a liquid?

Really, I like your question, but I don't see how to chose between the two point of views.
And again, it seems that I learned from this question.
But it is not a "good question" because, I think, nature doesn't speak this language.
How could that be answer experimentally?
Experiment could only validate the two points of view, the two formulas.

Michel
As Bill_K brought up, one of the hidden assumptions is that energy is localizable. Assuming this, [itex]E^2+B^2[/itex] and [itex]\bar{J}\cdot \bar{A}+\rho \phi[/itex] give different locations for the energy density. (I tend to favor the later because it is manifestly generally covariant, where the former is not--but that's not enough.) Obviously, an experiment measuring the curvature of spacetime would sort it out, but this seems surely beyond reach with any means currently available, involving extraordinary energy. I'm curious to know if there is any experiment in reach that could sort this out.

Really, if this is science and not philosophy, then experimental evidence should decide it.
 
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  • #15
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2
Phrak,

Your comment on invariance is right on the spot.
As I mentioned, the whole stress-energy tensor should be considered,
and doing so, I am not sure about who would bend the space: charges or fields.

The invariance needs the field point of view to include the Poynting vector.
The paper mentioned by Meir Achuz makes that crystal clear.

Finally, Feynmann and Wheeler have shown us how electrodynamics can be build totally without any field. They even gave the Lagrangian for that, by the way, quite obvious.
What could we conclude from that?
 
  • #16
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I'm not sure what you're saying about the Poynting vector. But I don't think it should be taken too seriously. Insisting on both Lorentz invariance and Poynting's theorem requires electric and magnetic charge to appear out of nowhere, even for a beam of light in vacuum, as I recall from the last time I looked.
 
  • #17
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I didn't intend to hijack the OP, but the invocation by lalbatros of the Poynting vector has cause me to review some old notes.

Poynting's theorem seems to be the origin of placing energy in propagating electromagnetic fields in the form (1/2)(E^2+B^2) yet the theorem and it's usage have errors.

The Poynting equation is

[tex]\nabla \cdot \left( E \times B \right) + \frac{1}{2} \partial_t \left( E^2+B^2 \right) = -E \cdot J[/tex]

It's further assumed it can be applied in a vacuum, setting J to zero.

[tex]\nabla \cdot \left( E \times B \right) + \frac{1}{2} \partial_t \left( E^2+B^2 \right) = 0[/tex]

Now, there is a definite value this would obtain on the right hand side, in charge and current density. One is not free to choose what the right hand side contains, nor can one arbitrarily add or subtract terms. Arbitrarily setting [itex]-E \cdot J[/itex] to zero, expecting an energy continuity equation for light, is a mathematical error.

My next assignment in this distraction is find out what's really on the right hand side by mathematical identity. I'm fairly sure it isn't zero, and I'm not so sure of -E \cdot J.

Next, in the language of differential forms, Poynting's equation is the temporal component of a pseudo vector. The spacial components of the pseudo vector must be as equally valid as the temporal component if flat space is Minkowskian. A Lorentz boost of the left hand side of Poynting's equation obtains the terms

[tex][B \times \partial_t E - \frac{1}{2}\nabla B^2 + (\nabla \cdot B) B + E(\nabla \cdot E)] + [-E \times \partial_t B - \frac{1}{2}\nabla E^2 + (\nabla \cdot E) E + B(\nabla \cdot B)] [/tex]

This is a big mess, but clearly obtains nonzero charge densities in the forms [itex]\nabla \cdot E[/itex] and [itex]\nabla \cdot B[/itex], the second of which is magnetic charge density. This states that a beam of light has electric and magnetic charge in one frame of reference and not another. Nevermind the fact that they arise. The fact that there is a preferred frame where there are none is an issue in relativity.
 
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  • #18
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I'm going to look at this a little more closely.

The Lorentz invariant quantity that contains the Poynting equation (and as an added bonus, is manifestly covariant) is

[itex]*(G ^ *dG)-*(F ^ *dF)=[/itex] the stuff on the right hand side.

F is the Farday tensor, * is the Hodge duality operator, G=*F, d is the exterior derivative.

Automatically there seems something wrong with it, as dF should be set to zero for no magnetic charge (dF=d^2A=0; all exact forms are closed), yet Poynting's theorum requires it for contributing terms.

However the term [itex]*(G \wedge *dG)[/itex] has all the right properties. It is directly derivable from Maxwell's 4 equations, is generally covariant, has units of energy density and is magnetic charge and current free and should represent at least one form of electromagnetic energy (There could be higher order terms in terms of F and G), and can be generalized to include magnetic charge/current if preferred.
 
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  • #19
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[itex]*(G ^ *dG)-*(F ^ *dF)=[/itex] the stuff on the right hand side.
This was a condensed notation. The wedge operator (^) did not show in LaTex, and could lead to confusion.

[tex]*(G \wedge *dG)-*(F \wedge *dF)= *(G \wedge *J)[/itex]

is what I should have written.

dF is identically zero for no magnetic charges.

[tex]*(G \wedge *dG) = *(G \wedge *J)[/itex]

In the vacuum where the conditions may exist without charges and currents, J=0. So for a region in spacetime with nothing but electromagnetic fields,

[tex]*(G \wedge *dG) = 0 \ .[/itex]

This would be the correct form for the energy continuity equation, in the absence of electric charge and current, where spacetime could be curved, and ignoring any higher order terms in G of energy flux density if there are any.

See Wikipedia, Maxwell's Equations, Differential Geometric Formulations.
http://en.wikipedia.org/wiki/Maxwell's_equations#Differential_geometric_formulations
 
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