The essence of this question is simplicity itself: There is an electron in a curl-free $\vec{A}$ field. The electron is stationary so its m$\vec{v}$ momentum is 0. However, it has "momentum" from the term q$\vec{A}$ of the equation $\vec{p}$ = m$\vec{v}$ + q$\vec{A}$. Note that the usual notion that $\vec{A}$ is arbitrary/nonphysical seems to be contradicted by the empirical validation of the existence of the Aharonov-Bohm Effect. Also there is, a time-domain Ahronov-Bohm Effect reported by Nikolova, N. K. et al in "Detection of the Time-dependent Electromagnetic Potential at 1.3 GHz" equation (20). Also consider Carron, N.J. "On the Fields of a Torus and the Observability of the Vector Potential" (7.13) formulating Maxwell's Equations without gauge transformations.

What is the intuitive description of this "momentum"?

Note further that the aforelinked paper by Carron specifically states that for the case in question (a single reference frame involving a torus, electron and observer at rest with respect to each other) Maxwell's Equations may be formulated without gauge transformations. This gauge invariant formulation, while not being Lorentz invariant, is, nevertheless, convenient for analysis in our single inertial frame. Can anyone now provide the answer to my question without invoking the "arbitrary nature of gauge choice"? Here is the relevant passage from Carron:

enter image description here enter image description here enter image description here

Any further complication of this question is made necessary only because some answers attempt to attribute this q$\vec{A}$ "momentum" to the electron's m$\vec{v}$ momentum being somehow converted to or stored as "potential" momentum in the $\vec{A}$ field. That is equivalent to saying it is not possible to position the electron in the $\vec{A}$ field without contributing to its source current. Really? Please do not allow the following thought experiment (intending to demonstrate the positioning of an electron in an $\vec{A}$ field without contributing to its source current) to cloud the simplicity of the original question. If you are asserting that it impossible to position an electron in an $\vec{A}$ field without contributing to the $\vec{A}$ field's source current, please explain how the following process does so:

An electron is approaching along the torus's axis of symmetry. There are exactly 2 directions in which it can make this approach (there being only 2 directions of travel on a line). The electron is slowed to a stop. The way the electron is slowed to a stop is to place a static charge on the torus that goes to zero when the electron stops. One may posit that this, somehow, increases the current sheet $\vec{J}$ but one may not make that argument for both directions of approach. Hence there is at least one scenario in which the electron can come to a rest in the $\vec{A}$ field without contributing to $\vec{J}$ (hence without somehow "storing" its m$\vec{v}$ as q$\vec{A}$ "potential" momentum).

Now, back to the original, simple, question:

One way of intuitively explaining what q$\vec{A}$ means would be to describe the system in two states:

1) The poloidal current sheet, $\vec{J}$, is zero hence $\vec{J}$=$\vec{A}$=q$\vec{A}$=0

2) The poloidal current sheet, $\vec{J}$, is non-zero hence q$\vec{A}$ is non-zero.

How does the system's behavior differ in these two cases since its property given by q$\vec{A}$ differs in these two cases? If this momentum is not conserved then in why may we say this system embodies conservation of momentum?


6 Answers 6


Nice question. This is connected to gauge invariance.

The interaction term in the Lagrangian (interaction between charge and field) is not gauge invariant. Thus, a curl-free vector potential, which corresponds to zero magnetic field, appears to have a non-zero momentum. Nevertheless the equation of motion is not changed in a new gauge, i.e., you will get exactly the same rate of change of the particle's momentum, which in this case is zero.

Another way of stating the answer is that $\vec A$ has no independent meaning, and thus neither does the term $q\vec A$. The field $\vec A$ is not physical, only $\vec E$ and $\vec B$ are real.

P.S. The diagram you have is wonderful. How did you make it (what software did you use)?

  • $\begingroup$ Mmm … real in which sense? The potentials are in the Lagrangian, not the fields. Some sensors measure the intensity of the field (number of particles) never the E or B fields themselves. Other sensors might measure phase differences (canonical conjugate to the number of particles, almost) but not the fields themselves. How are they real? If you put a test particle attached to a spring in an electric field, you are measuring how much energy-momentum is transferred to the probe and then stored in the spring. $\endgroup$
    – Hector
    Commented Feb 18, 2015 at 17:30
  • $\begingroup$ In the sense that physical effects depend on E, B alone, not on potentials (at least in classical theory). Physical effects, and these fields, are gauge-invariant, while the potentials are not. The qA term does depend on gauge, but the equation of motion (e.g., for the charge you mention attached to a spring) comes out the same in all gauges. So trying to measure a gauge-dependent term is not useful. The simplest example is the corresponding term for the scalar potential: q$\Phi$. We can add a million volts worldwide to $\Phi$, and while q$\Phi$ increases, nothing really changes in the lab. $\endgroup$ Commented Feb 19, 2015 at 22:25
  • $\begingroup$ I concur with you in that the equations of motion are independent of the gauge choice. I never said that $(φ,A)$ were real. They are "more real" in the sense that those quantities are in the Lagrangian. My point is that classical sensors measure neither $(E,B)$ nor $(φ,A)$. The physical effects depend on the number of photons and their relative phases. Take for example, $E=E(t,x)$ and superimpose $E=-E(t,x)$. The classical amplitude vanishes (at least at an instant). On the other hand, if $E=E(t,x)$ has 20 photons and $E=-E(t,x)$ has 20 photons, you get 40 photons discarding non-linear effects $\endgroup$
    – Hector
    Commented Feb 20, 2015 at 0:43

There are two kinds of momentum.

One kind is simply a frame dependant portion of a larger tensor. It is exemplified in the total stress-energy tensor, which is a symmetric rank two tensor that is divergence free. The divergence free part means that is is conserved locally in in the sense that the momentum (or energy) in a region at one time, is equal to the momentum (or energy) at a later time plus the net flow of momentum (or energy) into the region.. Each of the four scalars, $E$, $\hat{x}\cdot \vec{p}$, $\hat{y}\cdot \vec{p}$, $\hat{z}\cdot \vec{p}$ is itself conserved separately via the exact same process. This is a useful thing precisely because it has a local conservation law. And that doesn't mean that the total momentum (or energy) at one time is equal to the total momentum at a later time, momentum (or energy) could flow in from infinity and change the total while still being locally conserved. Also, some expressions for momentum (and energy) explicitly store momentum (or energy) at infinity in the sense that surface terms have nonzero contribution to total energy even when the region in question is unbounded.

A completely different and second kind of momentum is the canonical momentum conjugate to a generalized coordinate in a Lagrangian. A Lagrangian isn't just a scalar and a Lagrangian density isn't just a scalar field. It is a function. The exact same physical setup cold have a Lagrangian that is a function of $r,\theta,\dot{r},\dot{\theta}$ or a Lagrangian that is a function of $x,y,\dot{x},\dot{y}$ and they are completely different Lagrangians (since they are different functions). A Lagrangian has a canonical momentum associated with every single generalized co-ordinate. These canonical momentum may or may not be conserved. They are only conserved if the Lagrangian doesn't depend on the generalized coordinate in question. For a charged particle in an external field $(\Phi,\vec{A})$, the external field $(\Phi,\vec{A})$ just is, and has the dynamics it has. The Lagrangian just depends on the particle, but the canonical momentum conjugate to whatever generalised coordinates you feel like using for your personally favorite Lagrangian isn't physically relevant. If you pick different generalized coordinates, you get different Lagrangians, and hence different momentum, that may or may not be conserved. The physical dynamics of the actual configuration will be the same, but you simply can't read too much in to the canonical momentum associated with your particular choice of generalized coordinates. If you choose different generalized coordinates, you get different canonical momentum conjugate to those coordinates, and often they won't be conserved in any sense of the word.

If you have a potential that doesn't depend on a generalized position, then you can make a Lagrangian that doesn't depend on that position, and the corresponding momentum conjugate to that momentum will be conserved, and it will then be useful because it is conserved. But when you pick a generalized coordinate where the Lagrangian does depend on it, then the canonical momentum conjugate to your arbitrary choice of a coordinate doesn't mean much, if anything, physically. It's really only given the word momentum, since for $x,y$ and the Lagrangian $L(x,\dot{x},y,\dot{y})=\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\dot{y}^2-U(x,y)$ the canonical momentum conjugate to $x$ is $\hat{x}\cdot \vec{p}$ (the linear mechanical momentum) and for $r,\theta$ the Lagrangian $L(r,\dot{r},\theta,\dot{\theta})=$


has for the canonical momentum conjugate to $\theta$, angular momentum $mr^2\dot{\theta}$.

What is the intuitive description of this "momentum"?

It is not intuitive because it is not even physical. You can imagine a value that depends on your actual physical objects (and their derivatives, and time). But then you make a mathematical function (a Lagrangian) that is a function of variables that you arbitrarily get to choose. The numerical value is that value that depended on your objects, but the Lagrangian is a mathematical function, specifically a Lagrangian for cartesian and polar coordinates are completely different Lagrangians, even for the same physical potential energy. And they have different momentum. To do a Legendre transformation, you need a variable conjugate to your variables, these conjugate variables are called momentum since in particular situations they reduce to actual physical linear and angular momentum, and they are useful if/when the Lagrangian is invariant with respect to the variable (since in that case the conjugate momentum is conserved). These conjugate variables are mathematical, not physical, since they depended on an arbitrary mathematical (and not physical) choice. Specifically the choice of mathematical parametrization. If the numerical Lagrangian had a physical symmetry, then a choice of coordinates that respects that is natural, and in that case then the canonical momentum conjugate to that coordinate is both natural and conserved.

If you want an intuitive understanding of Legendre transformations and the conjugate variables involved, you can consider the graphs of the functions in a product space of the domain and the range, and look for conjugate variables with derivatives that match up. But that is a mathematics question about functions and their parametrizations.

How does the system's behavior differ in these two cases [$J=0$, and $J\neq 0$] since its property given by $qA$ differs in these two cases?

I like that this time you say the canonical momentum is a property of the system (since it isn't meaningfully a property of the electron). But it still isn't a physical property of the system of "interacting particle and field," it is a property of your choice of a mathematical parametrization of the physical system. The actual physical system evolves with the charge moving according to the Lorentz Force Law, this follows from the change in canonical momentum listed in my other answer. The field evolves according to Maxwell's Equation, though this does not follow from the change in canonical momentum or even from the Lagrangian in question. As far as this Lagrangian is concerned the $\vec{A}$ just is what it is, and evolves how it evolves. It is an external field, and it's spatial and temporal distribution is taken as a given input to this problem. It is also an external field in that it satisfies Maxwell's equations without the electron being considered a source term. (Though the boundary conditions is a completely different issue.)

But we don't even know that $q\vec{A}$ differs in these two cases. If $\vec{r}(t)$ is the position of the electron and $q$ is the charge of the electron, then $q\vec{A}=q\vec{A}(\vec{r}(t),t)$. If you had an $\vec{A}_1(x,y,z,t)=\vec{A}_1(x,y,z,0)$ as shown above for $\vec{J}\neq 0$, then we could define another $\vec{A}_2$ such that $\vec{A}_2(x,y,z,t)=\vec{A}_1(\vec{r}(0),0)$, and if in this case $\vec{r}(t)=\vec{r}(0)$, then two canonical momentum are actually both constant and equal. And this field $\vec{A}_2$ is one of the many many vector potentials that fit with $J=0$. This is a separate issue from the canonical momentum depedning on your coordinates, it is about gauge freedom, that specifying the charge and current doesn't tell us what $\vec{A}$ is. So it's not obvious that the two canonical momentum are unequal. They might be equal. This isn't surprising when you have a gauge dependant quantity ($\vec{A}$) inside a quantity (the canonical momentum) that depended on your coordinates.

If this momentum is not conserved then in why may we say this system embodies conservation of momentum?

The physical system embodies conservation of the sum of field and mechanical momentum. The Lagrangian in question doesn't have any representation of the field momentum, so there simply isn't going to be any conservation of canonical momentum, except in trivial cases, such as if the particle has always been at rest and the field as always been the same as well. I can't actually figure out the grammar enough to answer any better.

For all I know you wanted to talk about the canonical stress energy tensor, a rank two tensor, which comes from a completely different variation. It's much nicer in that it has a zero divergence and so it has a local conservation law for the total momentum. Unfortunately, it often is derived in a way where the canonical stress energy tensor is not a symmetric rank two tensor (and so physically it would be like a system whose parts have more and more angular velocity when you look at smaller and smaller region).

Canonical does not mean best.


You are correct that Aharanov and Bohm's effect means that the potential $A^\mu$ cannot be blithely disregarded.

However, that doesn't mean it's physically meaningful or well defined in the way you seem to imagine. It's still a gauge variant quantity meaning it can be locally set to zero by a gauge transformation. Its physical significance (independently of $F^{\mu\nu}$, that is) is only evident when considering large spatial regions where you may measure a geometric phase of some sort; but the canonical momentum $\vec{p} = m \vec{v} + q\vec{A}$ is a local operator so it cannot "see" those phases.

Geometrically, what's happening here is that the connection $A^\mu$ tells you how to perform parallel transport along a section of the vector bundle, or, intuitively, how the phase changes when you take a particle along a path. If that path is closed and infinitesimally small, the total shift corresponds to the curvature $F^{\mu\nu}$. When the path is large, you get a geometrical phase such as the Aharanov-Bohm effect. Another way to look at this is to think about a lattice formulation of QED: the plaquette goes to $F^{\mu\nu}$ in the continuum limit while larger Wilson loops retain a nontrivial gauge invariant phase. It is the same because the fundamental geometrical ideas are the same.

The gist of all of this is that while the connection is an important geometrical and physical object (it is not true that only $\mathbf{E}$ and $\mathbf{B}$ are "real"), the choice of gauge with which to represent it is definitely unphysical. If you're interested in observables that depend on $A^\mu$ you are limited to gauge invariant quantities such as $F^{\mu\nu}$ or large Wilson loops. The momentum $\vec{p} = m \vec{v} + q\vec{A}$ is not such a quantity.

An excellent reference here is M. Atiyah's text "The Geometry of Yang-Mills fields". All he says about nonabelian theories is also mutatis mutandis valid for electromagnetism.


The diagram for A resembles that of a field due to a dipole that varies inversely as r cube which is not the case here though directions of A are OK. However the situation here is that of a current sheet of a cylinder that has been deformed so that the ends meet. There is no B outside. By redefining A all over the space you end up with zero A outside and same B inside. This shows that qA here does not has any physical significance in this problem as much as a constant potential does not have.


UPDATE Thank you for posting back your thought experiment. My calculations apply again.

I think I found the source of your confusion. Your analysis is missing the displacement current. You seem to think that the linear momentum of the electron could only go back to the current in the torus:

Hence there is at least one scenario in which the electron can come to a rest in the $A$ field without contributing to $J$ (hence without somehow "storing" its [linear momentum] as … "potential" momentum).

But, you want to stop the electron by changing the electric field:

The way the electron is slowed to a stop is to place a static charge on the torus that goes to zero when the electron stops.

I quote Wikipedia:

In electromagnetism, displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

The linear momentum of the electron did not go the current $J$ but it was transferred to the component of the electromagnetic field due to the "displacement current" equivalent to the time-varying electric field. The electromagnetic field then carries that linear momentum and probably will radiate it away.

Finally, remember how picking the zero level for gravitational potential energy is irrelevant. You can solve a projectile problem using $z=0$ at sea level or at the floor of the classroom. The same is true about the canonical momentum with the additional complication that you can pick that zero level of canonical momentum to have a different value at every point (as long as that choice of zero level is just a gauge transformation). While you can change the value of the canonical momentum almost at will, the total changes in linear momentum still satisfy conservation laws.

UPDATE First I will rant about your flip-flopping. I did not see you original question but Timaeus's original post seemed to address a somewhat different question than what I read. That would have been the first version. I answered to the second(?) incarnation in which the electron was moving:

What we are left with is a stationary electron in an unchanging A field that has not been increased.

My calculations showed you a way to properly compute the transfer of momentum. Now, you changed the question to a more basic one. The calculations I did no longer apply. In your old question, you contrasted the non-conservation of linear momentum of the electron with the apparent constancy of the $A$ field.

The third version now reads

There is an electron in a curl-free A field. The electron is stationary so …

This is a major re-write that asks a more basic question

What is the intuitive description of this "momentum"?

Now, I will try to answer it.

The canonical momentum is the derivative of the Lagrangian respect to the velocity. It has no intuitive description in general. There are theorems about the conservation of canonical momentum that link it to underlying symmetries, but those theorems are not easy to hand wave.

The equation you found $p=m v + q A$ links the canonical momentum and the linear momentum in Cartesian coordinates. If you were using other coordinate system, say spherical, the canonical momentum would not necessarily be computed in terms of the linear momentum and the $A$ field.

Your new emphasis in the Aharonov-Bohm effect versus "the usual notion that $A$ is arbitrary/nonphysical" has a very simple explanation. At the quantum level, $A$ is not non-physical, period. Did you expect classical Physics to be valid at all regimes? If you want an explanation of what $A$ represents at the quantum level, well … that is yet another question (version 4.0).

Answer to the old question As posed, your question is impossible.

Let us assume there is an incoming electron … that it is being slowed to a halt by … an $E$ field that goes to zero …

Short answer. If the $E$ field is changing in time, then the curl of the $B$ field cannot be zero. Since the curl cannot vanish, the $B$ field itself cannot vanish. Hence the $B=0$ marks outside the torus are incorrect. While your graph might have been done by an observer at rest respect to the torus, the previous reasoning is valid for any frame in which the charge is slowed down by a time-varying electric field: $B\neq 0$ at the point where the charge is at the moment when it is being slowed down.

Long answer. Let us concentrate on a neighborhood centered on the particle. To first order in displacements from the position of the particle, we can assume that the $B$ field is

enter image description here

where all the quantities are time dependent, except for $x$, $y$, and $z$ that measure displacements from the particle position. The $z$ axis is the axis of symmetry.

Cylindrical symmetry reduces the above to

enter image description here

No magnetic monopoles reduces that to

enter image description here

If we require that $\partial_t E$ lies along the axis of symmetry (*), we use one of Maxwell's equations to get

enter image description here

I left out several constants and the weird eDot is $\partial_t E$.

The only way the Poynting vector $S=E\times B \propto (E \cdot\partial_t E ) r \hat r$ would vanish is if $E$ is constant. Since there is transfer of energy, transfer of momentum should follow.

So far, the existence of the electric field that stops the charge implies that there is energy exchange between the neighborhood containing the particle and the rest of the universe.

What we are left with is a stationary electron in an unchanging $A$ field that has not been increased. … That means this stationary electron has momentum and we cannot claim that its momentum has somehow been transferred to the $A$ field as "potential momentum".

The key of your argument seems to be that the constancy of $A$ at the location of the electron imply that the mechanical momentum of the electron cannot have been transferred to the electromagnetic field. Moreover, the value of $A$ you refer to is the value of the $A$ produced by the torus.

I think one must use the total field when we compute the Poynting vector. Since we cannot compute the total field at the location of a point particle, we make a series expansion and analyze the behavior in a neighborhood of the particle.

I am aware of a non-algebraic mistake in (*). I just do not have time to hand wave properly.

  • $\begingroup$ Please see the first paragraph of the revised wording of the question. $\endgroup$ Commented Feb 14, 2015 at 19:54

It sounds like you want an electron, a statically charged torus, and then to consider the two cases of the torus with and without a current.

You also bring up the canonical momentum conjugate to position: $\vec{p}+e\vec{A}$, which evolves according to:

$$ \frac{d}{dt}(\vec{p}+e\vec{A})=e\left[\vec{E}+\vec{v}\times\vec{B}\right]+e\frac{d\vec{A}}{dt}=e\left[-\vec{\nabla}\phi -\frac{\partial \vec{A}}{\partial t} +\vec{v}\times(\vec{\nabla}\times\vec{A})\right]+e\left[(\vec{v}\cdot\vec{\nabla})\vec{A}+\frac{\partial \vec{A}}{\partial t}\right].$$

The first three terms tell you how the mechanical momentum changes, and the last two tell you how the canonical minus the mechanical evolves. The canonical momentum is not a conserved quantity (if you want a conserved momentum, use the mechanical and field momentum, the canonical momentum is simply the momentum conjugate to the position of the particle). This momentum even just takes as an $\vec{A}$, the $\vec{A}$ external to the particle. The vector potential in the above formula is not the total vector potential, just the external one. And as such it most definitely not part of the momentum of the particle, it is merely the momentum of the Lagrangian.

Let's look at the issues that still exist in your setup. Firstly, if you allow the charge to pull the torus, then (if the torus has current), it will produce a magnetic field outside the torus. But the title of your question sait curl free, so let's just apply a force to the torus to keep it still. Since this Lagrangian didn't have a mass for the torus and just takes the $\vec{A}$ as both external and given, that's fine. (Note that had the torus had current, it would have weighed more from the additional mechanical momentum of the mobile changes and the nonzero $\vec{B}$ field inside, so your initial electron would have to be closer to come to rest at the same place.)

So let's consider the torus as mechanically stationary, statically charged, and with constant uniform current (either zero or nonzero depending on the case). Still we have choices, because their is a gauge freedom.

Because of the charged torus, there is a constant electrostatic field $\vec{E}$, which we can get as the negative of a gradient of a scalar function $V$. Because of the constant $\vec{B}$ field inside, which we can get as the curl of a vector function $\vec{A}_0$. Now because of our gauge freedom, there are many possible potentials. Here's one pair:

$$ \Phi_A=V, \qquad \vec{A}_A=\vec{A}_0.$$

Clearly they produce the correct constant fields. Here is another pair: $$ \Phi_B=0, \qquad \vec{A}_B=\vec{A}_0+t\vec{\nabla}V.$$

These different choices have the same $\vec{E}$ and $\vec{B}$ and the same dynamics for the electron. But they have different canonical momentum, so that's how unphysically-meaningful the canonical momentum is.

There are tons more choices because we can make the scalar potential do virtually whatever we want by just taking its gradient and integrating that from $t=0$ to the current time and adding that to the vector potential. And we get a different $\vec{A}$ when we do that.

That said, I still can't tell what you want to know. Fix one of the vast number of $\vec{A}=\vec{A}(x,y,z,t)$ available, then you can either compute the mechanical momentum like normal, or you can use the canonical momentum and your knowledge of $\vec{A}=\vec{A}(x,y,z,t)$ to find the mechanical momentum.

In either case, the electron is pulled by the electrostatic force from the electric field due to the charge on the torus. In either case the electron doesn't feel a force due to the steady current of the stationary torus. The torus itself is more massive with the current and the $\vec{B}$ field so it resists the pull of the electron better by itself, but you have to pull it just as hard to undo the electron pull. If you change the current on the torus, the electron will feel additional forces. If you change the charge on the torus, the electron will feel additional forces, and part of them will be due to the currents required to change the charge.

If you want a conserved momentum that doesn't change, consider the full stress energy tensor (with both the mechanical momentum density and the field momentum density) and just avoid having net momentum flow in or out from spatial infinity (the full stress energy tensor is locally conserved, so that's the only way to change the total momentum). Then you can see momentum flow into or out of the fields as the charges lose or gain mechanical momentum.

  • $\begingroup$ @JamesBowery I'm not saying it is impossible, I just that you wanted to conserve the canonical momentum. I've edited my answer, please comment if I'm still misunderstanding the question. $\endgroup$
    – Timaeus
    Commented Feb 10, 2015 at 6:22
  • $\begingroup$ You said: "It sounds like you want an electron, a statically charged torus, and then to consider the two cases of the torus with and without a current." No. The only reason I introduced the static charge to the torus was to construct the desired end state: No electric or magnetic fields outside the torus's minor radius, except the electric field of an electron at rest in that space. The two conditions are then 1) with an A field (which is neither electric nor magnetic) in that space and 2) without an A field in that space. Please don't say the electron contributed to the A. $\endgroup$ Commented Feb 13, 2015 at 23:33
  • $\begingroup$ Timaeus, please read the first paragraph of the revised question -- which dramatically simplifies and clarifies the question in a way that should allow you to similarly dramatically simplify and clarify your answer. I'm leaning toward accepting your answer but I would like this question and answer to be simple and clear enough that it can be more relevant. $\endgroup$ Commented Feb 14, 2015 at 20:03
  • $\begingroup$ @JamesBowery I read the new question, and wrote a completely different response. I still can't tell what you really want to know. $\endgroup$
    – Timaeus
    Commented Feb 15, 2015 at 5:39
  • $\begingroup$ Let me take a whack at it: q$\vec{A}$ is a measure of the momentum that would be imparted to the electron if $\vec{J}$, hence its associated $\vec{A}$, went to zero. The way this momentum is imparted to the electron is by the potential formulation of Faraday's law: $\vec{E} = -\vec{\nabla}\phi - \frac{\partial \vec{A}}{\partial t}$ The time-changing $\vec{A}$ produces a radiating $\vec{E}$ field that contains energy which may, if it encounters the electron, impart energy, hence momentum, to the electron. The energy in the radiating $\vec{E}$ is reduced subsequent to this interaction. $\endgroup$ Commented Feb 15, 2015 at 7:04

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