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

In summary, the guy in the video said that there is "hidden momentum" in electromagnetic fields, but the paper by Babson, et al. says that there is no hidden momentum. McDonald's papers on "hidden momentum" are wrong.
  • #1
Swamp Thing
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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.

 
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  • #2
See
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?
 
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  • #3
Thanks, I'll take a look.
 
  • #4
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!
 
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  • #5
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.
 
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  • #6
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.
 
  • #7
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.
 
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  • #8
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.
 
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  • #9
Are you saying that \gamma E is constant in their example?
 
  • #10
Which of McDonald's papers should I look at?
 
  • #11
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

http://puhep1.princeton.edu/~mcdonald/examples/mansuripur.pdf
Another good explanation is found in

https://doi.org/10.1119/1.4812445http://arxiv.org/abs/1303.0732
 
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  • #12
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.
 
  • #13
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.
 
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  • #14
Are you claiming the center-of-momentum (not energy!) theorem is wrong? That's ridiculous, because that would deny SRT as a whole!
 
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  • #15
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.
 
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  • #16
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
 
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  • #17
"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.
 
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  • #18
Sigh. If you quote something, you should give the source. Which "Griffiths paper" are you talking about?
 
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  • #19
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?
 
  • #20
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).
 
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  • #21
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.}##
 
  • #22
clem said:
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.
 
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  • #23
clem said:
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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  • #24
"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?
 
  • #25
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:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/
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:

https://www.physicsforums.com/insights/homopolar-generator-analytical-example/
 
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  • #26
Are you saying that including ##{\bf v\times B}## makes Babson correct?
 
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  • #27
You never specified, why you think that Babson eta al are incorrect. I don't see anything incorrect there. It's over 100 years old standard relativistic electrodynamics! Babson uses another model for the current, namely freely moving charge carriers and not the one with friction.

If you want to treat the problem using the more realistic model of charge carriers with friction, from which Ohm's Law derives, you have to use the correct relativistic version of Ohm's Law, which reads
$$\vec{j}_{\text{cond}}=\gamma \sigma (\vec{E} +\vec{v}/c \times \vec{B}).$$
 
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  • #28
Is there a practical experiment that one can do with, say, a cylindrical polystyrene or electrolytic capacitor and some neodymium magnets to demonstrate something like this:
200px-Poynting-Paradoxon.svg.png

Wikipedia: the circulating flow of energy contains an angular momentum.[14] This is the cause of the magnetic component of the Lorentz force which occurs when the capacitor is discharged. During discharge, the angular momentum contained in the energy flow is depleted as it is transferred to the charges of the discharge current crossing the magnetic field.

I'm thinking of something like, mounting the capacitor and magnets on a fidget spinner. You can have a piece of string holding a switch open, and you burn the string to close the switch and short the capacitor without nudging the rotor.
Theoretically and practically, can one do this at home? And would the result depend on how we run the wire from the capacitor to the switch, or would it be one of those line integrals that are path-independent?

Edit: You can't have both the magnets and the capacitor on the rotor, of course. The magnets would have to be fixed.
 
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  • #29
To understand this well-known example, again you need to consider the complete system consisting of electromagnetic fields and "mechanical stresses" (which are again on the microscopic level electromagnetic of course) to analyze this situation correctly. Again you have to treat it relativistically. It's very similar to the discussion of the Trouton-Noble experiment a la von Laue.

Another possibility is the more modern approach to just define the electromagnetic total energy-momentum vector and angular-momentum tensor in a manifestly covariant way by defining it in the rest frame of the capacitor and then writing the corresponding three-dimensional hypersurface in a manifestly covariant way a la Rohrlich, Butler et al. For details, see Jackson's, Classical electrodynamics, 2nd or 3rd edition.
 
  • #30
vanhees71 said:
To understand this well-known example ... you have to treat it relativistically.

By "understand", do we mean "merely" :wink: knowing how to balance the books when we compare the picture in two different reference frames? Or do we also mean being able to predict observable experimental results in the frame of the macroscopic objects?

I will need to go through a long and steep learning curve before I can benefit directly from the books and papers that have been referenced in this thread, but I do have a couple of questions by way of "tl; dr".

Talking of observable consequences of hidden momentum, I think the takeaway from the various discussions seems to be that there are no observable consequences of the static field when it's just sitting there -- but I'm not sure, so just thought I'd confirm that.

However, I get a sense that a changing crossed E x H field would produce mechanical impulses that could perhaps be measured in principle. Is this true? And if it is true, do we need to do calculations in the moving charges' reference frame in order to calculate the effect? If so, it means that there is a demonstrable difference between a magnetic field produced by a few electrons moving really fast, versus lots of electrons "drifting" along lazily e.g. in metal, even when the two cases have the same field strength. This is something I had no clue about prior to following this thread.
 
  • #31
Sure, a changing ##\vec{E} \times \vec{B}## means there's change in energy flow/momentum density with time and this means some net force is acting on the system. For a closed system the total momentum doesn't change due to spatial translation invariance which is a fundamental symmetry of Minkowski space and thus this must hold true for any properly defined special-relativistic theory.
 
  • #32
So when the E x B changes, the system becomes temporarily "open" to that extent -- it that the EM radiating away as a wave or EM pulse?
 
  • #33
If you take into account all charges and fields, it's a closed system.

That's the deeper reason why we need the electromagnetic field to describe nature! Given the well-established fact that space and time have to be described relativistically (special relativity is sufficient for the argument of course) and that this still implies that for any inertial observer space is a Euclidean affine manifold which is homogeneous and isotropic, also the momentum-conservation law must hold (because of homogeneity of space due to Noether's theorems).

On the other hand there cannot be causal instantaneous interactions, i.e., in a closed system of interacting charges the acceleration of one charge, i.e., the change of its momentum due to the forces from the other charges, cannot instantaneously be compensated by a change of the other charges momentum. So you need some local entity that takes up the momentum to keep momentum conservation right at all times, and that's the electromagnetic field.
 
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  • #34
vanhees71 said:
Sure, a changing ##\vec{E} \times \vec{B}## means there's change in energy flow/momentum density with time and this means some net force is acting on the system.

Re. my post #28 with the coaxial capacitor and axial magnetic field, and considering that changes in hidden momentum should result in mechanical impulses...
200px-poynting-paradoxon-svg-png.png


Let the magnetic field be provided by a pair of solenoids in series, one at each end of the coaxial capacitor.

Let's excite the capacitor and coil with frequencies in the MHz range, such that the reactive power is a couple of hundred watts. But let the E and H frequencies differ by, say 1 KHz. Then the hidden angular momentum will have a 1 KHz oscillation from clockwise to anticlockwise around the capacitor's axis as per the phase variation between the E and the H. Can the resultant torque on the capacitor and/or coil be detected with a suitable acoustic transducer (maybe fiber optic based to avoid EMI) ?

Better still, we can use a toroidal coil to make sure that there is no energy radiating away, and that E and H don't overlap anywhere except in the capacitor.
1571325615338.png
 
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  • #35
Of course you also have to take into account the "mechanical (Poincare) stresses" to keep the entire system in balance. Doing so you have a closed system, where the center-momentum theorem (aka center-of-energy theorem) is fulfilled.

For more simple examples of this kind, look for "Trouton-Noble experiment" and "Tolman's paradox".
 

1. What is momentum from static E and H fields?

Momentum from static E and H fields is a concept in physics that describes the transfer of momentum from electromagnetic fields that are not changing over time. This means that the fields are not producing any waves or radiation, but are instead stationary.

2. How is momentum from static E and H fields different from other forms of momentum?

Momentum from static E and H fields is different from other forms of momentum because it is not associated with the motion of particles or objects. Instead, it is related to the energy and momentum of electromagnetic fields themselves.

3. What is the significance of momentum from static E and H fields?

Momentum from static E and H fields plays a crucial role in understanding the behavior of electromagnetic fields and their interactions with matter. It also has practical applications in technologies such as particle accelerators and electromagnetic propulsion systems.

4. How is momentum from static E and H fields calculated?

Momentum from static E and H fields can be calculated using the Poynting vector, which describes the direction and magnitude of energy flow in an electromagnetic field. The momentum is then calculated by dividing the energy by the speed of light.

5. Can momentum from static E and H fields be observed in real life?

Yes, momentum from static E and H fields can be observed in real life through experiments and observations of electromagnetic fields. It is also used in practical applications such as the manipulation of particles in particle accelerators and the propulsion of spacecraft using electromagnetic fields.

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