How is it possible to have momentum in a situation of perfectly static electric and magnetic fields?

I´m not the specialist. But in my memory was this paradoxon due to relativistic maxwell equations. I´ve found a text in WIKI, which I attend:

I hope for discussion.

Hanno is correct in using a WIKI answer. Wikipedia may not be trustworthy for many things, but is quite useful for established science and mathematics. The reference given is

Feynman Lectures on Physics, Sections 17-4 and Volume 2, Chapter 17, section 4 and the end of Chapter 27, Section 6.

I had just reread those sections before I saw Hanno's answer. I was not sure if I should give my own answer or simply give the reference. Now all I have to do is give Hanno an up vote.

Dear Hanno and Joseph,

Thank you very much for your references. My example is not a cylindrical capacitor having a radial field, but a parallel plates capacitor having field in say X-direction and the magnetic field is in y-direction, then EXH, the field momentum lies in z-direction. Here no angular momentum comes to the rescue. We have simply to explain the field momentum in the z-direction, given that there is no motion anywhere in the system.

Thanks and regards,

Rajat

Dear Joseph,

do you have a copy of the cited Feynman chapter? I would like to read it again, but have no access to this volume.

Dear Joseph, no efforts please. It´s incredible, but I found an old german version of the Feynman lectures. You are right, he discusses parallel plate condensator in Chapter 27-5 of my edition as a typical example for the Poynting vector S.

Sorry, but it took some time. Feynmans example deals with charging a capacitor, no with completely static fields. Rajats contradiction holds.

Dear Rajat,

did you try to define the situation during the positioning of the condensator and the magnetic static field. During the positioning you will induce currents in the condensator which influence the magnetic field. Could it be, that our regarded system is to small, we forget the surroundings?

Lynn B Wilson

The Poynting vector is an energy flux density. The simple continuity-like equation form of Poynting's theorem shows that the time rate of change of the energy density of the electromagnetic fields [dW/dt] + the rate of electromagnetic energy flux flowing out of a surface [Div(S)] = the energy lost due to momentum transfer between particles and fields [-J.E]. In the case you describe, I do not think there would be a finite divergence no a finite dW/dt. Thus, since J.E = 0 the equation holds.

Another thing to note is that time-variation of E or H is not required for the individual photons carrying these fields to have momentum. Photons have momentum/energy regardless of whether their flux has a time-variation. However, unless the capacitor you speak of discharges, I do not think there will be any energy flux flowing into/out of the system.

Hanno, Rajat, Read Feynman carefully, including the magnet and charge later in 27-6. Angular momentum is imparted to the capacitor and must be conserved. There is no energy flowing into or out of the system, but the system has energy and momentum.

Thanks Hanno for all your replies. My question is after the positioning etc. has been done and perfectly static electric and magnetic fields have been set up. Even then we have ExH non-zero!

Dear Lynn Wilson,

Since My question is about perfectly static fields , no photons could be there, i suupose. No energy transfer is there between particles and fields. It is perfectly static situation.

Regards,

rajat

Dear Rajat,

I´m still opposing. I think the completely static situation is a construction. In the past, this case had to be built up. And we all know, your integrals and field have extension to infinity. How will a completely static field take this development into account?

To be honest, I´m overwhelmed.

Static fields carry momentum. The momentum is circulating in the space. The momentum was acquired when the field was built up. The momentum will go somewhere when the field is turned off.

It is not possible to charge a capacitor in a magnetic field or place a field across a charged capacitor without introducing momentum. The momentum will go somewhere when the field is removed or the capacitor discharged. The momentum will circulate in the field during the time the field and charges are static. The angular momentum will be conserved.

The Poynting vector P = E X H is a mathematical model of a physical effect. This model successfully describes energy transfer in an electromagnetic field. It appears to be correct in all situations where it is applied. Nevertheless, it is a model. We are not constrained to believe that a model applies outside the bounds where it was developed. The explanation of circulating momentum (in a static charged capacitor with a magnetic field applied perpendicular to the electric field) may be an ad hoc explanation for a place where the model fails. A capacitor with a magnetic field may be static at a particular time. It is not possible to create the capacitor with a magnetic field without moving charges to the capacitor and imposing the field.

You may choose to redefine fields and momentum or restrict the use of the Poynting vector to certain bounds. Any justified change to the current model and its application will be an interesting and rewarding endeavor.

Many thanks Joseph and Hanno.

It seems momentum can remain in some form other than (~~ Evergy/speed). When we really build up the static situation described above, momentum flows into the system. Yes. And then it stays as the Poynting vector EXH as long as the system remains in place. Just as energy has many forms, it seems we have to admit that momentum also has this poynting vector form.

Regards,

Rajat

Dear Rajat,

I´m still thinking. And my impression is, our "local system static C and M" is to small. The poynting vector is ok, but the regarded system is to small.

Please correct me, if possible.

Daniel Baldomir

Dear Rajat,

It is very curious that you repeat the same question that Eric Lord, who I have tried to convince of the zero value of the electromagnetic momentum in the case of static fields. I recommend you to go to his question for seeing only the aspect of energy current for the Poynting vector. I will try to answer you in a more general case.

1. Notice that the Poynting vector P that you have used is only vectorial product of the electric and magnetic fields E and H. But this is a general vectorial field too and therefore extended in all the space, not only in a point. Therefore a derivative doesn't gives all the physical information for this mathematical tool. If you want to go much deeper on the geometrical and topological information for the electromagnetic fields you can read the book

D.Baldomir and P.Hammond, Geometry of Electromagnetic Fields, Clarendon Press, 1996.

2. The physical interpretation of the Poynting vector is in energy-momentum tensor time-space components (energy currents) or its the space-time components (momentum density). The motion equation is

daTab= -FacJc

where usually the a,b,c indices are greek letters for taking the values 0,1,2,3. Your equation is only for b=0. That is to say, you have lost the other 3 equations of conservation in Electrodynamics.

3. The electromagnetic momentum p definition includes an integral on a closed surface

p=IntS (P) dS

and as the Poynting P of the static fields depends of r^(-4) obviously it is zero for all the large surfaces. Therefore you are not going to obtain any physical linear or angular momenta for static electromagnetic fields.

Eugene F Kislyakov

Rajat,

since magnetic charges are absent, any magnetic field (static also) is moveing electric charges. It is their momentum. Momentum is conserved globally and locally (due to flows inside system). But, also electrodynamic is internally contradictory on very small distances (self action). See Landau and Lifshiz, second volume ("Theory of fields") and also Feynman.

Regards,

Eugene.

@Eugene,

I´ve learnt that magnetic fields can just accellerate charges, if they have a velocity (Lorentz equation). If they don´t move, there is no acceleration.

Daniel Baldomir

Eugene,

Electrodynamics is not contradictory at all, in fact it is the best theory of Physics. What happens is that there is a transition between classical and quantum theories ( which stick perfectly well ) at certain distances. But the same that happens with mechanics or other physical theory.

Respect to the self-interaction this is not possible because it is always an abelian theory working perfectly.

The net momentum flow and net energy flow is equal to zero in a static electromagnetic field. There is no question in this regard. The net flow must be zero or the fields would not be static. Net flow requires that flow out must equal flow in. That raises the question is there any flow across the surface. Net flow does not negate flow through an element of the surface smaller than the total surface.

The question still remains, "What is the significance of the Poynting vector in a static electromagnetic field?" Is there angular momentum in a static electromagnetic field? Is there energy and momentum flow in a static electromagnetic field?

The Poynting vector P = E x H represents the electromagnetic field energy flow for a unit area. The energy balance requires a full surface integral of N. A superposition of static electric and magnetic fields in the absence of currents might have non-zero values for the Poynting vector at various spatial locations. The divergence vanishes over a closed surface.

The mathematics of electrostatic and magnetostatics were developed by considering fixed charges and fields. The extension to moving charges and current flow requires further consideration. Moving charges and currents require electromotive force and resulting radiation and resistive heating losses. Conservation of energy and momentum must consider that fields carry energy and momentum.

Consider a surface enclosing a charging capacitor with an imposed magnetic field. The electromotive force is supplied by a battery that is exterior to the surface. The Poynting vector indicates that the energy and momentum are balanced by the field. The field supplies the energy and momentum to the capacitor. The battery field is not required.

Is this interpretation valid? Is field momentum qualitatively the same as particle momentum. In some contexts this is called hidden momentum. There are interpretations that propose an impulse to the capacitor if it is discharged. Is this impulse measurable for a light-weight capacitor fitted with strain gauges?

I have attached several reports that deal with field momentum. Should we accept the analyses or attempt to confirm experimentally?

Eugene F Kislyakov

Joseph,

now the difference between particle and field is unclear as a whole. In quantum field theory it disappears. Static electron has magnetic moment. But the question is: Who has seen static electron? To unswer on a question has coil magnetic moment or not, we must put it into external magnetic field. The answer is known.

Do You know the problem of energy of two capacitors after they are connected?

Many thanks Joseph for your references.

The question is not about whether the total flux through a closed surface vanishes or not. It is more about local "flows" of energy, if at all there are any. Momentum means there must be such local flows. Now what is it that flows, when the fields are static?

Thank you Eugene for joining the discussion.

Regards,

Rajat

Rajat

The references provided were concerned with momentum of an electric field. The development of the concept of momentum balance or conservation leads to the Poynting vector. Logically and physically we must conserve the momentum.

Use a mechanical example. Compress a spring and place it between two fixed plates. The spring exerts a force on the plates and the plates an equal and opposite force on the spring. Apply F=ma to the forces. They are static so F=ma does not apply to the equal and opposite forces, because nothing accelerates. There is no change in momentum.

One could argue otherwise. If the spring is compressed between the plates for a very long time, the spring will loose its tension and its shape will change. The plates will be distorted at the position of contact with the spring.

Compression of the spring was purely mechanical. We did not require fields to explain the total momentum. We conclude for the capacitor that fields must carry momentum. A charged, isolated capacitor is like a compressed spring, in that, it contains stored energy. Unlike the spring, fields carried momentum into the capacitor and fields are present for the charged, isolated capacitor.

The Poynting vector indicates that field momentum is added to a charging capacitor. The Poynting vector implies that a charged, isolated capacitor has momentum in its field and it is moving in the direction of the Poynting vector.

The usual approach for textbooks is to integrate over a closed surface and conclude that the net flow is zero, then state that using the Poynting vector at any location will give anomalous results. Feynmann concludes at the end of chapter 27, volume II that angular momentum has been added to a capacitor so the momentum must flow to conserve angular momentum.

Momentum must be conserved, linear and angular. There is no reason to require flow only for angular momentum. A spring has momentum during compression. It does not require momentum flow while compressed, it is stored. Momentum is conserved on release or into releasing the tension if compression continues.

Momentum in a capacitor might be stored as tension in the materials and released upon discharge. There is no need to require flow.

As you stated, Rajat, we may have to think of it differently. There should be a way to measure it.

Thank you very much Remi and Joseph for your efforts at making things clearer.

I want to make the following rather pedestrian observation:

We know that the Electric field can accelerate a charge. Now, what can possibly accelerate something? A force i.e. a quantity that can change momentum dp/dt. How can a force lead to a change in momentum unless it itself is endowed with the ability to suck in or impart momentum? Thus the force or the field must be a storehouse of momentum that can act as a source as well as a sink of momentum. It means that the static electric field also (not just the Poynting vector) must carry momentum. For that matter any force field must be such a storehouse of momentum, since it can give only momentum to another and that is what it has and that is what it does i.e. change momentum.

Regards,

Rajat

Rogier Brussee

You have to redefine momentum (this is well known in Quantum physics where the impulse operator changes from

i\hbar\nabla

i\hbar \nabla + \vec A

There is also a purely classical argument simply form the lagrangian.

But think about it this. Say, the momentum component p_x is the generator of translations in the x-direction (in geometric/symplectic terms: \frac{\partial}{\partial x} is the Hamiltonian vector field of the function p_x on the co-tangent bundle). With a an electric and magnetic field present translation invariance ((or rotation invariance) is broken, and so there is no reason for a naive version of momentum preservation to hold.

Well said Rogier.

Thank you.

What about the suggestion that a purely static electric field may also be a potential reservoir of momentum, since it can impart momentum or take it away ?

Should we allow the static fields themselves, not just the cross product ExH, to carry momentum? Can we support it with a fiber bundle argument? p -> p +eA in Lagrangian dynamics of a charged particle in external fields will break down for a static electric field, since to generate static electric field we require time varying vector potential (E = - \del \phi -- dA/dt).

Similarly, for the magneto-static field which also can change momentum by changing direction. an we say static B-field has momentum?

Regards,

Rajat

I am not sure what is well-understood at this state of the conversation. If you don't mind, I'll just state my point of view and see if anyone finds it mysterious:

- Confusions that I have seen in the literature (recently a field angular momentum problem by Mansuripur) seem to lie in the general unawareness of what is called "hidden momentum". This is the momentum that is present in a physical system but does not manifest in observable motion. How can it do that? I'll give an example.

- Consider a static electric field Eo, along the x-axis, produced by a parallel-plate capacitor. Nothing is moving. I am not interested in how we got to this, I want to assume it as my starting point, state 0.

- Inside the capacitor, along the y-axis, I arrange a wire coil with square cross section: think of a Slinky not wound round in circles, but in squares. I want to make a rectangular parallelepiped space with a homogeneous magnetic field, in the y-direction. So the axis of the "squinky" will be along y, and the squares that it traverses will lie in the z-x plane. This coil is made of superconducting wire to eliminate resistive losses.

- Now I connect the coil to a battery + resistor, and it gets current running through the coil. When the current is big enough, we provide a by-pass to the battery and then switch it out. Voila, the coil is running in the coil without any further resistive losses.

- And we have a magnetic field Bo, essentially constant within the squinky, and nearly zero outside it. Since the squinky is contained within the region of electric field Eo, the interior contains not only the static B field, but also the Poynting vector of magnitude Eo*Bo/μo along the z-axis.

- So if we accept the standard interpretation, the spatial integral of the (Poynting vector)/(c^2) gives the field momentum. So if the dimensions of the squinky were: length L, sides 2S, that means we have about 4(LS^2)(Eo*Bo)/(μo)*c^2) z^ of momentum in the field that we didn't have before. So where did it come from?

[UPDATE: I realized later that I actually visualized each square as having side 2S, not S. I have fixed this here and a few points following. It does not affect the argument.]

- The answer lies in the electrons that were previously at rest: They are now circulating the squinky coils, with most of their motion in the z-x plane. I claim that this circulating motion, when time-averaged and summed over all electrons, will give this magnitude of momentum along the z-axis. Thus the field momentum is necessary to balance the so-called "hidden momentum" of the electrons.

- Why doesn't the momentum of a circulating electron average to 0? There are two factors:

a) In circulating through the squinky, the electrons traverse each square coil: from (z,x) = (-S,S) = A, to (S,S) = B, to (S,-S) = C, to (-S,-S) = D; and then back to (-S,S) = A. But notice that during the leg AB, the electron is at the highest potential energy, while during the leg CD it is at the lowest potential energy. That means that all during AB the electron is traveling relatively slowly, but during CD it is traveling fast. During BC it is accelerating and during DA it is decelerating, so these two would likely cancel out (although we ought to give it a second thought), But the fact that we have a decided difference between the two legs of travel AB and CD gives us the starting point for a non-zero result.

b) So let's calculate the average momentum over one cycle:

AB: momentum = mV(ab) z^, duration = 2S/V(ab);

BC: momentum = -mV(bc) x^, duration to be calculated

CD: momentum = -mV(cd) z^, duration = 2S/V(cd)

DA: momentum = mV(da) x^, duration to be calculated

To calculate the time-average of the momentum, we multiply the momentum by the duration for each leg, sum over legs, and divide by the total time. For AB and CD, it's easy:

AB: mV(ab) z^ * (2S/V(ab) = 2mS z^

CD: -mV(cd) z^ * (2S/V(cd) = -2mS z^

for BC and DA, we can integrate instead of multiply; but since the integral of the velocity is just the displacement, the integral is still S; the time will be the same for the two (time-reversal invariance), so BC gives -2mS/T x^, DA gives 2mS/T x^

So we add these four terms up and we get a big fat 0 ! Average momentum = 0. Does that mean there's no hidden momentum after all?

No, it means that we forgot that hidden momentum is a relativistic notion, so to calculate it we need to use the relativistic formula for momentum:

P_relativistic = mγv

where γ = 1/sqrt(1 - (v/c)^2)

When we use these:

AB: mγ(ab)v(ab) * 2S/v(ab) z^ = 2mSγ(ab) z^

CD: -mγ(cd)v(cd) * 2S/v(cd) z^ = -2mSγ(cd) z^

BC and DA are more complicated, but fortunately will cancel each other from time-reversal invariance. The only problem is we don't know exactly how long it would take to traverse them; but we can estimate it as the average of the other two, so that the total time would be twice the sum of the times for AB and CD:

total time =~2*(2S/v(ab) + 2S/v(cd)) = 8S(v(ab) + v(cd)/(v(ab)*v(cd))

average momentum ~ z^ S(mγ(ab) - mγ(cd)/[2S(v(ab) + v(cd)/(v(ab)*v(cd))]

By conservation of energy, mγ(ab) - mγ(cd) = - q*Eo*2S

=> -q*Eo*v(ab)*v(cd)/(v(ab) + v(cd))

Now we need to estimate the number of electrons; however, although we're definitely heading for a non-zero number here, I need to backtrack: It is not quite right to donate all the potential energy change to the velocity of the electrons, because this is a "drift velocity", and there could be a lot more kinetic energy tied up in random-walking that this could go to.

Fortunately, we do have information on the drift velocity of the electrons, because it is exactly this that gives rise to the current that produces Bo. But I need to think a little bit more to figure out how to incorporate it properly.

I'm also not thrilled about having to estimate the cycle time this way. It may be better to do an instant average over all electrons than a time-average of one electron's cycle.

I'll be back.

OK, I've thought about it:

- It's definitely going to be better to do a summation/integration over all electrons at one time.

- It's OK to assume that all the potential energy is going into "center-of-mass" drift velocity: Otherwise the electron gas would not only be heating up instead of drifting faster, but also cooling down instead of slowing down. That sounds like a good candidate for violation of the 2nd law of thermodynamics.

Further, since the Lorentz transform converts the energy-momentum 4-vector between frames, I use one transformation between the electron gas and the lab frame, and the total energy-momentum of it transforms like that of a single particle. So it's legitimate to treat a localized set of electrons as a single particle, with mass equal to the center-of-momentum total energy.

- So let's go back the summation over electrons: At one instant of time, I want to add the momenta of all the electrons. The easiest way to o it is to do a line integral over the entire circuit, squinky and all. If we consider the linear number density of electrons along the wire (units = # electrons/cm), n(s) ds is the number of electrons in a segment of wire of length ds, located at distance s from the beginning point. The momentum of that segment's electrons is thus:

n(s) mγ v ds ; and this is to be integrated over the length of the circuit:

- We can add up the segments in any order, but

- We must of course respect the vector of the integrand; and

- We keep in mind that γ and v are also functions of s.

So let's do the integrals for all the BC and DA segments first. Now it is obvious that each BC segment will have an opposite integral to the nearest DA segment, because their conditions will be identical, but in reverse order by s. So the net momentum from all these segments together gives zero.

For the segments we did before, the analysis works out very much the same:

AB: integral [n(s) m γ(ab) v(ab) ds = mγ(ab) integral [n(s) v(ab)] ds

CD: integral [n(s) m γ(cd) v(cd) ds = mγ(cd) integral [n(s) v(cd)] ds

How do we finally evaluate those integrals? Easy: The current running through each point of the wire is: I = (electron charge) * (linear density of electrons) * (speed)

= q n(s) |v|

So each AB segment has integral:

mγ(ab) integral [n(s) v(ab)] ds = mγ(ab) 2S [n(s) v(ab)]

= mγ(ab) 2S [I/q] z^

and likewise CD gives:

- mγ(cd) 2S [I/q] z^

So together they give:

total integral due to one square coil = m(γ(ab) - γ(cd)) 2S *I/q z^

As we remarked before:

mγ(ab) - mγ(cd) = - q*Eo*2S/(c^2)

[Previously I neglected the factors of c !]

So each coil gives a contribution to the total momentum of:

-q * Eo * 2S * 2S * I/(q*(c^2)) z^

= -4 * S^2 * Eo * I/(c^2) z^

The total for the whole squinky we get by multiplying by the number of coils, N; this gives the total total as:

P_electrons = - 4N * S^2 * Eo * I/(c^2) z^

As we know from studying coils:

L * Bo = μo * N * I , so:

I = Bo * L/(μo * N) and hence:

P_electrons = - 4N * S^2 * Eo * Bo * L/(μo * N*(c^2)) z^

= - 4 L*S^2*Eo*Bo/(μo*(c^2)) z^

As mentioned above, the momentum we customarily assign to the Poynting vector in this situation would be:

4 L*S^2)(Eo*Bo)/(μo*(c^2) z^

So they are exactly the same, except in sign.

Conclusion: In this very simple case we can see that the generation of the crossed fields (giving rise to the Poynting vector distribution) required production of a state in which there was hidden momentum: ordinary mechanical momentum that, due to relativistic effects, does not result in net motion. To create this condition, there must be somewhere an equal amount of momentum to cancel it out. As shown in much more general calculations, it's the Poynting vector.

Similar observations could be made concerning field energy density and field angular momentum.

Rajat Pradhan:

To return to your question about whether state E-only or B-only fields could hold field momentum or angular momentum (we already know they can hold energy): It boils down to the question of whether or not you need to produce electrons with enduring momentum or angular momentum to create that specific form of the field.

In the example just worked out, if there had been no E-field, there would be no hidden momentum, because there would be no difference in the magnitude of the momentum contributed by AB and CD. But it seems that there would be some angular momentum, even if not hidden.

Eugene F Kislyakov

Neal,

E-only or B-only fields can be only in one inertial system, in all the others there are both E and B fields.

Regards,

Eugene.

Eugene,

that would be analogous to the rest-frame of a particle, in which it has energy but no momentum. A change of frame would naturally entail picking up some momentum.

What is intriguing to me at the moment is the fact that a static magnetic field (without E fields present) has no angular momentum, even though the magnetic coil certainly has angular momentum. Evidently the angular momentum that was added to counterbalance the electrons' AM was carried away by radiation.

But in that case, why is it that the field momentum counterbalancing the hidden momentum isn't carried away by radiation likewise, instead of hanging around as the configuration of the Poynting vector? It seems odd.