I Momentum from static E and H fields? (So this video claims)

Summary
This video states that a static E field and a static H field at right angles to each other, have momentum. Is this true?
Is it true, what he's saying from 04:28 to 04:56 ? I have my doubts, but I thought I'd better ask here.

 

DrClaude

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D. Babson, S. P. Reynolds, R. Bjorkquist and D. J. Griffiths
Hidden momentum, field momentum, and electromagnetic impulse
American Journal of Physics 77, 826 (2009); https://doi.org/10.1119/1.3152712
Abstract said:
Electromagnetic fields carry energy, momentum, and angular momentum. The momentum density, ##\epsilon_0 (\mathbf{E} \times \mathbf{B})##, accounts (among other things) for the pressure of light. But even static fields can carry momentum, and this would appear to contradict a general theorem that the total momentum of a closed system is zero if its center of energy is at rest. In such cases, there must be some other (nonelectromagnetic) momenta that cancel the field momentum. What is the nature of this “hidden momentum” and what happens to it when the electromagnetic fields are turned off?
 
Thanks, I'll take a look.
 

vanhees71

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What the guy in the video forgot to mention is that you only get a complete and consistent treatment of energy and momentum in electromagnetism, if you use complete relativistic equations for both the em. field and matter.

Another great free online resource are the essays by Kirk McDonald:

http://puhep1.princeton.edu/~mcdonald/examples/

The ones about "hidden momentum" are gems!
 

clem

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There is electromagnet momentum in the crossed static electric and magnetic fields, but there is no "hidden momentum" as Babson, et al propose. McDonald's papers on "hidden momentum are wrong.
Perhaps that is why he doesn't publish them.
 

vanhees71

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Where are these papers wrong? What's called "hidden momentum" (though in my opinion a misnomer, because it's simply properly relativistically defined momentum) is a well-known relativistic phenomenon, known since the very early papers (particularly von Laue's) about SRT.
 

clem

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Every model in Babson et al is wrong. In their most obvious error, they write

"The cleanest example of hidden momentum is the following: Imagine a rectangular loop of wire carrying a steady current. Picture the current as a stream of noninteracting positive charges that move freely within the wire. When a uniform electric field E is applied (see Fig. 9), the charges will accelerate up the left segment and decelerate down the right one."

Figure 9 shows that the E field in the two vertical segments of the loop is different than the E field in the two horizontal segments of the "wire carrying a steady current." This violates Ohm's law, j=\sigma E.

List the site of one of McDonald's papers, and I will point out his error.
 

vanhees71

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You understand that the relativistically correct Ohm's Law reads
$$\vec{j}=\gamma \sigma (\vec{E}+\vec{v}/c \times \vec{B})$$
or in covariant form
$$j^{\mu} = \sigma F^{\mu \nu} u_{\nu}?$$
Most mistakes about electrodynamics is using non-relativistic approximations for the motion of the charge carriers and then using field momenta, which are of a higher order in the relativistic expansion (formally in powers of ##|v|/c=|\beta|##. That's the case in the current-loop example shown in said Fig. 9 of the paper by Babson et al.

I don't see any obvious errors in the nice paper by Babson quoted in #2. I'm also not aware of any errors in McDonald's manuscript quoted in #4.
 

clem

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Are you saying that \gamma E is constant in their example?
 

clem

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Which of McDonald's papers should I look at?
 

vanhees71

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Concerning Babson's paper, what's unclear with Eq. (14), which is the correct equation to use if you want to take into account relativistic effects as the electromagnetic field momentum. If you make the non-relativistic approximation cf. Eq. (13) you also have to neglect the field momentum. Otherwise you come to the wrong conclusion that there's a non-zero net-momentum though the field-wire system as a whole doesn't move.

The here discussed example is also discussed by McDonald with some more very nice elaborations starting from it:

/home/arch/hees2/paper/textbooks/class-edyn/electrodynamics-of-moving-media-Penfield.djvu

A more puzzling one is


Another good explanation is found in

 

clem

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I am not concerned with Eqs. 13 and 14. Your relativistic Ohm's law equation shows that, if j is constant (as B says), \gamma E can't vary.
 

clem

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The long McD paper doesn't really derive hidden momentum.
Its mistake is assuming the center of energy theorem.
The Griffith's Hnizdo paper just repeats Babson.
 

vanhees71

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Are you claiming the center-of-momentum (not energy!) theorem is wrong? That's ridiculous, because that would deny SRT as a whole!
 

clem

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I think that we can agree that "ridiculous" is not a proof.
A proof the "center of energy" theorem (quote from Grifffiths) in an EM field is illusive.
McD refers to a paper by Griffiths, but Griffith's just refers to a paper by Coleman and Van Vleck.
 

vanhees71

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Where is this quote from? How can this in a Poincare invariant theory be true? If a theory is Poincare covariant the center-of-momentum theorem automatically holds thanks to Emmy Noether. See, e.g.,

https://arxiv.org/abs/physics/0501134v1
 
Last edited:

clem

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"Where is this quote from? How can this in a Poincare invariant theory be true? "
What quote do you mean. My post only quoted you and two words from a Griffiths paper.
 

vanhees71

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Sigh. If you quote something, you should give the source. Which "Griffiths paper" are you talking about?
 

clem

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The reference is in the McD paper you listed:
D.J. Griffiths, Dipoles at rest, Am. J. Phys. 60, 979 (1992) .
"Center of Energy Theorem" is always used by Griffiths , but why does the name matter when \gamma E is not constant for Babson?
 

vanhees71

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How do you come to the claim that the proof of the "center-of-energy theorem" (it seems to be the usual expression indeed, and it's not that wrong on 2nd thought, because the corresponding weight is indeed the energy density) is elusive? It's a very fundamental consequence of the symmetry of physics under proper orthochronous Lorentz boosts (or more precisely the corresponding one-parameter subgroup of the proper orthochronous Poincare group describing boosts in an arbitrary fixed direction), saying that for a closed dynamical system there always exists an inertial reference frame where the total momentum vanishes. It's just part of the 10 fundamental conservation laws following from the space-time symmetry of SRT. The proof for electromagnetism is given in the quoted paper by Coleman and van Vleck and in the important and famous paper by Belinfante (of 1940!) cited there.

I also don't see anything in the quoted paper by Griffiths which contradicts standard relativistic theory; to the contrary Griffiths discusses the resolution of apparent paradoxes, which always occur when using non-relativistic approximations under circumstances where a relativistic treatment or more careful systematic non-relativistic approximations are due.

He nicely discusses even the case of electrodynamics with magnetic monopoles and corresponding nonstandard magnetic dipoles. So far all empirical evidence rules out (elementary) magnetic monopoles and nonstandard magnetic dipoles (as is also explained at the end of the paper).
 

clem

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Your lengthy post about things I hadn't brought up indicates that we are talking at cross purposes.
For now, I just want to repeat again that keeping ##{\bf j}## constant while ##{\bf\gamma E}## varies contradicts
##{\bf j=\gamma\sigma E.}##
 
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I just want to repeat again
Well all you do is repeating your enigmatic claims that basicly "everyone is wrong", and give us no justification whatsoever. It does not look good.
 

vanhees71

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Your lengthy post about things I hadn't brought up indicates that we are talking at cross purposes.
For now, I just want to repeat again that keeping ##{\bf j}## constant while ##{\bf\gamma E}## varies contradicts
##{\bf j=\gamma\sigma E.}##
Obviously we have some misunderstanding here. The correct Ohm's Law is
$$\vec{j}=\gamma \sigma (\vec{E}+\vec{v}/c \times \vec{E}),$$
where ##\vec{v}## is the velocity of the charge carriers making up the current. In covariant form it reads
$$j^{\mu}=\sigma F^{\mu \nu} u_{\nu},$$
where
$$u^{\mu}=\gamma \begin{pmatrix} 1 \\ \vec{v} \end{pmatrix},$$
is the four-velocity of the charge carriers making up the current.

BTW. All this is well-established physics, already understood by Minkowski in 1910 (posthumously published by Born, who completed the work). It's also all contained in the famous first textbook on SRT by Minkowski. I don't know, why you question this well-established physics all the time!

As I said, it's somewhat unfortunate to call the here addressed phenomenon "hidden momentum", because in fact it's simply momentum, which is not hidden at all, but somehow this terminology came up in the later literature of the mid 1960ies for some reason.
 
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clem

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"why you question this well-established physics all the time!"
What did I question there? I know textbook physics, and regret Minkowski's premature death.
I meant to write your Ohm's law equation with ##\bf v\times B=0##. B external is zero isn't it?
 

vanhees71

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As I emphasized before, if you want to treat this problem using macroscopic electrodynamics and not the simplified version with the free charge carriers as in Griffiths's textbook and papers, you have to use the correct relativistic formulae, and there to take into account the ##\vec{v} \times \vec{B}## term is important, because otherwise it's not correct to all relativistic orders in ##\beta##, and since we discuss "hidden momentum" here, we must be relativistically accurate.

By chance, I've just written an Insights article about the much simpler problem of a straight wire. Maybe that helps to understand the importance of the relativistically complete version of Ohm's Law when discussing relativistic covariant electrodynamics:


Another example, where it is important, is the homopolar generator, which also usually gives rise to a lot of confusion, which is just due to the fact that Ohm's Law is not written down in its full relativistic form:

 

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