I understand from basic conservation of energy and momentum considerations, it is clear in classical electrodynamics that the fields should be able to have energy and momentum. This leads to the usual Poynting vector and energy density relations for electromagnetic fields.

However, I do not know how to interpret situations where there is a net linear momentum in a static electromagnetic field. The fields aren't propagating. It doesn't make sense to me that momentum can be divorced from motion.

As a concrete example to discuss: Consider a massless string of length $L,$ with a spherical shell on each end with a magnetic dipole moment m and positive charge q. The radius of the sphere $R\ll L,$ or alternatively, consider the dipoles to be perfect "point" dipoles. Let the string be along the $y$ axis, with one dipole at the origin and the other at $y=+L.$ If the magnetic dipole at the origin is oriented in the $-z$ direction, and the other dipole in the $+z$ direction, if you calculate the total linear momentum in the fields, the answer is:

$$p_\text{em} = m q \frac{\mu_0}{2 \pi L^2}\, \hat{x}$$

While this is an unstable equilibrium, it is an equilibrium. So classically the state can remain static with no need to evoke other external entities, interactions, etcetera. So there doesn't seem to be any of the usual potential pitfalls to save us here.

Please, can someone explain how a static field can have momentum?

  • $\begingroup$ A static field does not propagate? Consider a new particle was created now (a pair). Do you think that its gravitational and electrostatic field influences the whole universe in no time or, as I think, they will propagate at c speed? $\endgroup$ Commented Mar 19, 2011 at 12:34
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    $\begingroup$ I mean static in the usual sense. static = not changing in time. Your example is not a static field, as a field spreading out is clearly changing in time. $\endgroup$
    – Edward
    Commented Mar 19, 2011 at 12:56

5 Answers 5


This is a fairly subtle question! Griffiths recently published a paper on this.

Hidden momentum, field momentum, and electromagnetic impulse:

Electromagnetic fields carry energy, momentum, and angular momentum. The momentum density, $ϵ_{0}(E\times B)$, 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?

Free version of the above link.

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    $\begingroup$ This is link only. $\endgroup$
    – akrasia
    Commented Aug 25, 2014 at 21:50
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    $\begingroup$ It sure is! Tough to do better than Griffiths. $\endgroup$
    – Andrew
    Commented Aug 26, 2014 at 0:10
  • $\begingroup$ Perhaps you could provide a summary of the paper in your answer? $\endgroup$ Commented Jun 18, 2023 at 21:15

It is possible to show that the total momentum of any static system is zero in an inertial frame where nothing is moving. This does not mean that the momenta associated with various components of that system are individually zero. As you point out, there can be finite electromagnetic momentum associated with static charge distributions. Even though there is no obvious motion in the system, the momentum associated with the matter distribution is actually nonzero. It is equal and opposite to the electromagnetic momentum.

This is often referred to as the hidden mechanical momentum. It is a special case of a much more general result that the net momentum of an extended object need not be parallel to its center of mass velocity.

Electrodynamics books like Griffiths or Jackson have a nice microscopic interpretation for this effect in the simple case of a magnetic dipole placed near a charge. Internally, the dipole may be thought of as containing a current loop. The charges in this current loop accelerate and decelerate in response to the external electric field. One may show that this gives them a net momentum that is exactly equal and opposite to the electromagnetic momentum. Note that this is an intrinsically relativistic effect. It does not arise if Lorentz factors are neglected when computing the momenta of the circulating charges.

  • $\begingroup$ Your answer tells me that the total momentum of the system is zero, because there is mechanical momentum to balance the field momentum, and then give details on the mechanical momentum. This is not what I was asking though. The issue is that the field has momentum, but is not propagating at all. Please focus on the field. $\endgroup$
    – Edward
    Commented Mar 19, 2011 at 13:59
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    $\begingroup$ I was trying to point out that it is only the total momentum that needs to be zero. There is no requirement that the various interacting components of a static system individually have zero momenta. If the field did not have any momentum in your example, the system's total momentum would be nonzero. I think you'd agree that this situation would be worse. Additionally, one might want to say that there is some ambiguity in splitting up the momentum between various parts of an interacting system. It is only the total that matters. $\endgroup$
    – Stingray
    Commented Mar 19, 2011 at 14:09

Nonzero Momentum in static field configuration is actually a good thing. Consider a coaxial cable carrying DC current and voltage. Internally it has a constant E and H field. The Poynting flux is nonzero and shows that there is energy transport. The energy flow is indeed in the direction ExH.

The fact that the field configuration shows no movement is irrelevant. Momentum is moving energy, not something else moving!

-- Jos


I think the problem here is that the "massless string" actually requires electromagnetic forces to hold the two charges together. My preliminary calculations (which i need to refine but think are fundamentally correct) show that if you replace the string with a charge of -(1/4)q halfway between the two charges of q, the integral of ExB is zero. This extra charge would result in electrostatic equilibrium and would supply the electrostatic force not supplied mathematically by the massless string.

  • $\begingroup$ This is a really interesting interpretation. Has anyone verified the calculation? $\endgroup$
    – JJMalone
    Commented Oct 6, 2017 at 20:48

Momentum is a conserved quantity that belongs to a system. When particles are in motion the usual linear and angular moment

In the image of the dipole of the water molecule here Electric_dipole_moment WP you see that the molecule is subject permanently to a tension (red in the image).

The definition of the Electric Dipole moment

is a measure of the separation of positive and negative electrical charges in a system of charges, that is, a measure of the charge system's overall polarity.

In the simple case of two point charges, one with charge +q and one with charge −q, the electric dipole moment p is:

dipole moment ~ charge * distance (of charges)

compare this with Angular moment

angular moment ~ linear moment * distance (to a point)

where linear moment ~ mass * velocity as both charge and l.m. are states of the particle in a given moment that represent a potential for action, both formulas of angular and dipole moments are quite similar.

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    $\begingroup$ How does this answer the question? $\endgroup$
    – Marek
    Commented Mar 19, 2011 at 16:05

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