# Do the integral forms of Maxwell's Equations have limited applicability because of retardation?

In the usual bookwork treatment, it is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. I have always thought that neither version is more fundamental than the other and each has its place in solving problems. (See also Which form of Maxwell's equations is fundamental, in integral form or differential form? )

But: I have a conceptual problem with applying the integral forms of these equations in cases where there is time-dependence and the "size" of the loop or area means there is a significant light travel time across the regions considered compared with the timescale on which fields vary.

e.g. Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire. My first instinct is that this ought to be solved using the inhomogeneous wave equations to give A- and V-fields that are dependent on the retarded time - hence leading to the E- and B-fields.

But what about using Ampere's law in integral form? What is the limit of its validity? If we write $$\oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \int \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$$ then presumably the $t$ that is defined on each side of the equation cannot be the same, since a change in $I$ at time $t$, presumably leads to a change in $B(r)$ at a time $t + r/c$? I suppose one does not care about this so long as the timescale for a current change is $\gg r/c$.

So my question is: Are the integral forms of Maxwell's equations inherently limited by this approximation, or is there a way of formulating them so that they take account of the finite size of a region in cases where the fields are time-variable?

• you might want to make use of the integral form to get to fields when you can exploit a certain symmetry. In general I don't see how to use the integral forms to determine $\mathbf E$ and $\mathbf B$. So in this respect the differential eqs are what you really want to use to find the fields. – Phoenix87 Feb 1 '15 at 11:26
• @Phoenix87 Not really what I'm asking; obviously it only makes sense to use integral forms in cases with high symmetry. I'm asking if that is the only consideration, or is there a more fundamental problem in the case of time-varying fields? If you like, consider a magnet passing axially through a large circular loop (a problem that inspired this question). In the integral form of Faradays's law, at what time do you work out the changing flux on the RHS in order to get E(t) on the LHS? Or is it just hopeless and you can only use the differential forms? – Rob Jeffries Feb 1 '15 at 11:47
• My previous comment still applies then. If you want to be able to compute those integrals, knowledge of the integrand is required. Or, if you want to use that as an ""integral"" equation, some symmetry is required. – Phoenix87 Feb 1 '15 at 12:00
• I am guessing here, but I think you might need to do the integral for each time step because the integral is independent of that. So if each value changes at each time, you need to resolve that equation for each time step. I only see the integral form in static cases, though I work with plasmas which can have time-varying volumes/areas (to add an extra complication). So I do not see or use the integral form of Maxwell's equations very often. – honeste_vivere Feb 1 '15 at 15:28
• @RobJeffries : of course the integral equations are not usable in the case you ask of. Take just the integral on a closed curve. To travel along the curve takes time, all the more so if the curve diameter is big. The current on the RHS is given at a time $t$, while travelling along the curve may take a significant time $\Delta t$ s.t. if one starts the trip on the curve at the time $t$, by the time $t + \Delta t$ the current is completely different. So, the RHS and LHS won't be equal. – Sofia Feb 1 '15 at 20:04

I'll assume that we are neglecting any curvature of space (no GR) and any quantum effects (no QM).

The differential and integral forms are entirely equivalent, but the integral forms are not as physically intuitive as you might hope in cases that aren't static or quasistatic.

The other factor is that their utility could be questionable. For instance, in a highly symmetric situation in statics you might use Gauss or Ampere to find an Electric or Magnetic field. The same laws hold when not in statics, but they might not be as useful.

Also, some relationships between terms and more practical matters that only hold in statics might not hold anymore.

Let's look at an example. You have a very large conducting wire, and a magnetic field that is changing on time scales faster than the radius of the circuit divided by $c$. Since we are outside of statics or quasistatics there is no more equality between the electric potential difference across a battery terminal and the total emf through the circuit at a fixed time. But there is still an equality between the flux through the circuit of the time rate of change the magnetic field and the part of the emf through the circuit due to the electric force because that result doesn't depend on any static or quasistatic result. But it does have to be interpreted more carefully.

The integral equations hold for a time-slice in a fixed frame, so fix a frame. The $\vec{B}$ flux through the circuit is a scalar quantity that has a value at all times, and it changes for two reasons: from the instantaneous rate of change of the $\vec{B}$ field located along some fixed surface through the instantaneous locations of the circuit, and the instantaneous motion of the charges in the circuit through reacting to the instantaneous $\vec{B}$ field at the instantaneous locations of the circuit.

The time rate of change of $\vec{B}$, can be integrated over the some fixed surface through the instantaneous locations of the circuit, and that will (by Faraday's Law) give the integral $-\oint \vec{E}\cdot d\vec{\ell}$ along the instantaneous locations of the circuit. It's not the case that the $\vec{B}$ field out there caused the electric field to have this circulation, in fact the circulation of the electric field causes the magnetic field to change, so the causality is completely the other way. It's best to think that electric and magnetic fields don't have independently specifiable time rates of change. Particles can have a velocity, and then forces determine particle acceleration, but the curl of the $\vec{E}$ and $\vec{B}$ (and the sources) force the fields to have the time rate of change that they have. So each field has the value it does because of the previous value and the field time derivative, and the field time derivative is determined. It's almost a first order system (except that the current depends on the source so the electric field has some second order characteristics in that its time change depends on particle velocities). But the magnetic field straight up must evolve according to what the circulation of the electric field dictates (since there are no magnetic monopole currents). So the electric force per unit charge integrated along the instantaneous circuit locations is (as always) numerically equal to the instantaneous flux of $-\partial \vec{B}/\partial t$ through the loop. However the causality is that the circulation of the $\vec{E}$ field is causing the flux of $-\partial \vec{B}/\partial t$ to be what it is. Specifically, it the instantaneous $\vec{E}$ everywhere along that instantaneous surface that is making the $-\partial \vec{B}/\partial t$ flux be what it is along that instantaneous surface.

Now, if instead you looked at the rate of change of the instantaneous total magnetic flux you get that contribution, plus another do to the moving wire. In quasistatics you get that the other contribution equals the magnetic force per unit charge integrated along the instantaneous location of the wire. And you got it from the non-monopole law. So in statics all together you get the integral of the Lorentz force per unit charge equals $-d\Phi/dt$. The no-monopole law still holds, but you don't get the $-d\Phi/dt$ result because the velocity of the charges is no longer equal to the velocity of the circuit piece plus a velocity parallel to the circuit piece.

And even if you had the whole emf it is is no longer equal to the potential difference across the part of the circuit with a battery.

However, every integral form of the equations holds. I described Faraday's law, the part that still hold (the instantaneous flux of $-\partial \vec{B}/\partial t$ equals the line integral of $E$ around the loop).

The no-monopole law still holds, but it doesn't get you the results it used to (such as about magnetic emf), but it still does give you a vector potential. It still gives you that field lines entering a region leave the region.

Gauss' law still holds, so for any instantaneous volume, the instantaneous flux through the surface is proportional to the charge instantaneously inside. And it still gives you that field lines start and stop on electric charges.

The continuity equation still holds, in integral form. The charge is the instantaneous charge within, the current is the instantaneous flux of charge through the instantaneous surface.

Ampere's law says you can pick a loop, instantaneous in a frame, and the flux of current through it (the instantaneous $I$) plus the flux of the instantaneous displacement current through the instantaneous surface numerically equals the circulation of the $\vec{B}$ field through the instantaneous loop. But again the causality is that the instantaneous circulation of $\vec{B}$ in a region minus the instantaneous current flux $I$ through the region is proportional to rate of change of the orthogonal component of the $\vec{E}$ field and actually makes the $\vec{E}$ change in that way. So the current in that direction and the circulation around that direction tell you how that component of the $\vec{E}$ field changes, and it is the circulation and current right there (and right then), that determines that (well in the neighborhood). And again the $\vec{E}$ evolves based on what the $\vec{E}$ was before plus the time change based on the $\vec{B}$ nearby and the $\vec{J}$ nearby. So Ampere holds just as well in integral form.

All of Maxwell's equations hold just as well in integral form. You can even get the version with the total time derivative of the flux versions of the laws if the loop in question is a fixed loop in space. And we can still see the causality clearly, so it is known what makes each field be what it is.

It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems.

Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every point then the integral versions hold for any surface/loop. (This also assumes you write the integral versions in the complete and correct form with the flux of the time partials of the fields and/or with stationary loops.)

Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire.

Ampere's Law is correct, but it is not also helpful. If you know the circulation of $\vec B$, you can use it to find the total current (charge and displacement). If you know the total current (charge and displacement), then you can find the circulation of $\vec B$. But solving for $\vec B$ itself is hard unless you have symmetry.

What about using Ampere's law in integral form? What is the limit of its validity?

It is completely valid, but it might not be helpful. When you write: $$\oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{a}$$ then the $t$ that is used on each side of the equation is exactly the same.

When $I(t)$ changes, then $\vec B$ field nearby changes quickly, and when there is a changing $\vec B$ field there is a circulating electric field, so as the region of changing $\vec B$ field expands, so does the region of newly circulating electric fields. Both expand together. Eventually the expanding sphere of changing $\vec B$ field and changing circulating electric fields finally starts to reach the Amperian loop (together), and only then does the circulation of $\vec B$ on the far away Amperian loop change. If there was just one change in $I(t)$, then the expanding shell of changing electric fields continues expanding, and you are stuck with the new value of the circulating $\vec B$ field, based on the current that changed a while back.

So, to solve for $\vec B$, you'd need both $I$ and $\partial \vec E /\partial t$ and the latter you need for all the empty space on a surface through the Amperian Loop. Maxwell's equations don't have limited validity and do not need to be modified. They just aren't always as useful as you might want them to be.

• Is this an example of what you are saying: consider a current that starts at zero, ramps up quickly, and settles on a constant value. There will be an $E$ field existing in an expanding shell, but no $E$ field inside or outside of the shell. There is a $B$ field inside the shell but not outside. Before the shell reaches the Amperian loop, the integrated displacement current exactly cancels the real current, and Ampere's Law correctly predicts no $B$ at the loop. After the shell of $E$ passes, we have only the real current crossing the surface, and the $B$ field exists at the loop. – garyp Apr 23 '15 at 14:26
• @garyp Yes, however since you can't have isolated current your shell has some thickness. – Timaeus Apr 23 '15 at 14:29
• Thank you. This issue has bothered me for some time. I know my shell has some thickness because the current must ramp up over some non-zero time interval (I had to observe the character limit in comments!). But I don't understand what this has to do with isolated currents. – garyp Apr 23 '15 at 14:39
• @garyp Each piece of current has an expanding shell, but current has to come from somewhere. So you need a bunch of pieces of current to make a complete circuit, the size of the circuit also increases the thickness of your shell because the total shell is the sum of the shells of each piece of current. For an infinitely long wire of current, the shell had infinite thickness since the wire is infinitely long so parts of it are arbitrarily far away, luckily their contributions are small so it starts to settle down. – Timaeus Apr 24 '15 at 13:11

I don't really see what the problem is. If $B$ is zero on the boundary, your equation shows $0= \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$, so $$\iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}=-I(t)$$

Why is it that you expect that this equation isn't satisfied? I can tell very little about the quantity $\frac{\partial \vec{E}(r,t)}{\partial t}$

The differential versions of the Maxwell equations imply the integral versions, and the integral versions imply the differential versions, so you can't break one without breaking the other.

Not all systems of PDEs are local wave equations, so the fact that Maxwell's equations can be phrased in terms of PDEs (local laws) doesn't mean that effects are local. Conversely, the fact that Maxwell's equations can be phrased in terms of integrals ("global" laws) doesn't mean that the effects are global.

• The surface integral will reduce to the area covered by magnetic field propagation at time t. If you use the same formula for that new reduced surface you will conclude that the line integral is zero at the perimeter of the reduced boundary, because from your first usage of the formula you conclude that the R.H.S is zero. – MOON Feb 10 '15 at 12:22
• @yashar I don't know what you're saying. You're not saying that the $\iint\dot{\bf E}\cdot d\vec{\bf A}$ integral will be zero, correct? If you reduce the surface to be too small then the $\iint\dot{\bf E}\cdot d\vec{\bf A}$ integral bounds will change and the RHS will no longer be zero, so it's not OK to bring the loop inside the nonzero B field and have the RHS quantity be zero. – user12029 Feb 10 '15 at 17:36
• @yashar You have a surface $S$ and its boundary $\partial S$. One integral is $\oint_{\partial S} {\bf B}\cdot d{\bf \ell}$ and the other is $\iint_{S} \dot{\bf E}\cdot d{\bf A}$. You can't change the bounds of one without changing the bounds of the other. – user12029 Feb 10 '15 at 17:38

The answer depends on whether by "applicability," you mean "validity" or "usefulness." The integral and differential forms of Maxwell's equations are completely equivalent, so the integral forms remain completely valid even in a highly relativistic context. However, they're rarely as useful as the differential forms in this context, because they're nonlocal, so the relevant integrals can't be performed using only locally accessible data. And also, relativistic problems rarely have any symmetries that can be used to simplify them enough to where the integral forms are useful.

I couldn't put a picture in the comments and I am too lazy to do any drawing without my tablet.

I think the problem or at least part of it, is because the magnetic field has discontinuity in space so its curl at those points is not well defined and as a result someone cannot change the surface integral of the curl into a line integral over the perimeter of that surface.

If you consider a circle of radius 10c, c speed of light, then at time t = 1s and distance 10c magnetic field is zero because it didn't have time to reach there. So the line integral is zero. However the equivalent of the line integral is the surface integral which is not zero. In this case I think at the border of propagation of magnetic filed there is discontinuity, before it it is non zero but after it it is zero, due to the fact that nothing can travel faster than light. Thus, you cannot use stokes theorem to get the line integral from the surface integral. Therefore the line integral is valid for r/t<=c. • The integral and differential forms are equivalent. As time progresses the region of nonzero $\vec{B}$ field expands, and in order to change the $\vec{B}$ field, a rotational electric field develops. This expanding region of rotational electric field contributes equally and oppositely to the current in the $\vec{\nabla}\times\vec{B}$ integral. Once the expanding electric field surpasses the 10c circle, then you stop including it and just see the contribution due just to the current, and at that point your LHS is finally nonzero. – Timaeus Feb 26 '15 at 20:40