Vector Potential and Elementary Gauge Theory

Having established Maxwell’s equations we can now take note of some interesting properties. For one, there are no magnetic charges, hence ∇ ⋅ B = 0∇⋅B=0. This means that instead of using BB we can define another entity AA by using the fact that for functions that are smooth, all derivatives exists and are continuous, we have the vector identity ∇ ⋅ (∇ × A) = 0∇⋅∇×A=0. This has important implications. If

then equation (4.4) is automatically fulfilled. Substituting this into (4.3) we get

∇×E=−∂B∂t=−∂∇×A∂t=−∇×∂A∂t

We now use the vector identity: ∇ ×  ∇ φ ≡ 0∇×∇φ≡0, where φ(x)φxt is any smooth function of coordinates xx and time t, to integrate EE and we find

E=−∇φ−∂A∂t(4.8)

The vector field AA is commonly referred to as the vector potential field and scalar φφ is known as the potential field, or voltage field. Together they are called gauge potentials in physics literature.

The equation for the EE-field should be familiar to most readers with the possible exception of the last term. Equation (4.8) is simply the normal elementary textbook definition of the electric field as a gradient of a voltage but with an additional time-derivative (dynamic) term. It means we can have an electric field without a voltage drop in a dynamic situation. With the potential fields we can write Maxwell’s equations to be a set of equations for φ  and  AφandA. Let us rewrite equation (4.1):

∇×H=∇×∇×Aμ=∂D∂t+J=∂∂tϵ−∇φ−∂A∂t+J

or

∇×∇×Aμ=∂∂tϵ−∇φ−∂A∂t+J(4.9)

and from equations (4.2) and (4.5) we find

∇⋅ϵ−∇φ−∂A∂t=ρ.(4.10)

We now have another version of Maxwell’s equations through equations (4.7)–(4.10). As the reader will have noticed, the potentials, A,   φA,φ are not uniquely defined. One can, for instance, add a term ~ ∇ f∼∇f where ff is some function to AA and the BB field is not affected (∇ ×  ∇ f ≡ 0)∇×∇f≡0. This freedom in the choice of the potentials is known as gauge invariance. Let us look at this in some more detail (compare with [15] for a similar argument).

Let γ(x, t)γxt be an arbitrary scalar field. Then let us change the gauge potentials according to the following transformation

φ→φ−∂∂tγA→A+∇γ.(4.11)

Now

and

E→−∇φ−∂∂tγ−∂∂tA+∇γ=−∇φ−∂A∂t.

Both fields are unchanged! The transformation in equation (4.11) is known as a gauge transformation, and since the fields are unchanged under this transformation we speak of gauge symmetry. We see that a physical system is described by a whole family of gauge potentials that differ by a gauge transformation. By picking a particular set of gauge potentials we are making a gauge choice. The above might seem a trivial matter but it has profound importance in theoretical physics. The interested reader is highly encouraged to refer to the literature on this matter.

We will now show that we can always find a solution that satisfies

∇⋅A+μϵ∂∂tφ=0(4.12)

by making an appropriate gauge choice. Assume we have a particular solution A^′,   φ^′A′,φ′. We look for a particular gauge transformation that will satisfy (4.12). We have using (4.11)

∇⋅A′+∇γ+μϵ∂∂tφ′−∂∂tγ=∇⋅A′+μϵ∂∂tφ′+Δγ−∂2∂t2γ=0

or

Δγ−∂2∂t2γ=−∇⋅A′+μϵ∂∂tφ′.

The right-hand side is the known solution which acts as a source term for a wave equation which we can solve for γγ. This way we can always find gauge potentials that satisfy (4.12). We do not have to find γγ explicitly, but we can use this calculation as a motivation to use (4.12) as an additional requirement on A,   φA,φ. Equation (4.12) is known as the Lorenz gauge. Another typical choice is the Coulomb gauge

The Lorenz gauge is typically used for situations where the wavelength is comparable to the physical sizes. It is standard in microwave theory and antenna theory for obvious reasons. For integrated circuits one can often get by with the Coulomb gauge which corresponds to the Lorenz gauge in the long-wavelength approximation.

Maxwell’s Equations in Terms of External Sources

In electrical engineering it is natural to think of currents and charges as being impressed on an electric system through voltage or current sources. These impressed entities will then give rise to electromagnetic fields. We will now write Maxwell’s equation in terms of such impressed currents and charges.

The current can be divided into two parts. A conduction component,

and an impressed current, J_iJi. The charge can be divided the same way. By taking the divergence of equation (4.1) we recover the continuity equation:

∂ρ∂t=−∇⋅J(4.15)

which relates the charge to the current. The addition of the time derivative of the DD-field to Ampere’s law (∇ × H = J)∇×H=J was Maxwell’s famous generalization that created a self-consistent description of the field equations. For the current we have

J=Jc+Ji=σE+Jiρ=ρc+ρi∇⋅σE=−∂ρc∂tcontinuityequationfortheconductioncomponent.

Putting this together we get

∇×H=∂ϵE∂t+J=∂ϵE∂t+Jc+Ji=∂ϵE∂t+σE+Ji∇×H−∂ϵE∂t−σE=Ji(4.16)

∇⋅ϵE=ρ=ρc+ρi∇⋅ϵE−ρc=ρi.(4.17)

These equations show the fields as a result of external source currents and charges.

Full-Wave Approximation – Single Frequency Tone Formulation

In this book we will generally look at these equations not as a function of time but as a function of frequency. We get there by simply assuming the time dependence scales as e^jωtejωt where we follow the convention in most engineering books. By doing this we get

∇×H=jωϵE+σE+Ji=jωϵ1+σjωϵE+Ji.(4.18)

From the continuity equation for the conduction component we get

∇⋅σE=−∂ρc∂t=−jωρc(4.19)

This together with the charge equation (4.17) gives

∇⋅ϵE=ρ=ρc+ρi=∇⋅σE−jω+ρi→∇⋅ϵE+∇⋅σEjω=ρi(4.20)

We can now define an effective permittivity

ϵ′=ϵ1+σjωϵ(4.21)

and we have

∇ × H = jω ϵ^′ E + J_i

∇×H=jωϵ′E+Ji(4.22)

∇ ⋅ (ϵ^′E ) = ρ_i.

∇⋅ϵ′E=ρi.(4.23)

Using equations (4.6)–(4.8) we get

∇⋅B=∇⋅∇×A=0∇×E=∇×−∇φ−jωA=−jω∇×A=−jωB∇×H=∇×Bμ=1μ∇×∇×A=1μ∇∇⋅A−∇2A=jωϵ′−∇φ−jωA+Ji.

We use the Lorenz gauge which in frequency domain looks like

∇ ⋅ A + jϵ^′μω φ = 0

∇⋅A+jϵ′μωφ=0(4.24)

and get

∇(∇ ⋅ A ) − ∇²A =  ∇ (−jϵ^′μω φ) − ∇²A = jωμ ϵ^′ (− ∇ φ − jωA ) + μJ_i

∇∇⋅A−∇2A=∇−jϵ′μωφ−∇2A=jωμϵ′−∇φ−jωA+μJi

After rewriting we find

∇²A + ω² μ ϵ^′A =  − μJ_i

∇2A+ω2μϵ′A=−μJi(4.25)

We also have from equation (4.23)

∇⋅ϵ′E=ϵ′∇⋅E=ϵ′∇⋅−∇φ−jωA=−ϵ′Δφ+jωϵ′jϵ′μωφ=−ϵ′Δφ−ω2ϵ′ϵ′μφ=ρi

After rewriting we find

Δφ+ω2ϵ′μφ=−ρiϵ′(4.26)

Equations (4.25) and (4.26) are Maxwell’s equation in yet another form. Knowing A,   φA,φ will give us B   and  EBandE through equations (4.7) and (4.8). We will use these in numerous examples in this and the following chapters.

Long Wavelength Approximation

In the long wavelength approximation, λ ≫ lλ≫l, where λ = 2πc/ωλ=2πc/ω is the wavelength and ll is a length scale of the model, we find the second term on the left-hand side of (4.25), (4.26) disappears and we are left with:

∇²A =  − μJ_i.

∇2A=−μJi.(4.27)

We can also write this directly from the fields as

∇ × H = μJ_i  (Ampere ‘ s  law),

∇×H=μJiAmpere’slaw,(4.28)

where (4.27) follows if we use gauge ∇ ⋅ A = 0∇⋅A=0 and

Δφ=−ρiϵ′.Thisfollowsalsofrom∇⋅ϵ′E=ρiifweusegauge∇⋅A=0(4.29)

For integrated circuits the long wavelength approximation is often appropriate since the dimensions are much smaller than any wavelength.

Solutions to Maxwell’s Equations

Note the equations are of the same form where the only difference is the vector form for the vector potential and scalar from for the voltage field equation. The general solutions are known for certain sources. Here we will almost exclusively look at cases where the sources are Dirac delta functions of one sort or other. The specific solution will depend on the boundary conditions and most of the time will be spent establishing those.

For completeness, in this section we will discuss a common approach to solve wave style equations such as Maxwell’s. The first subsection covers the general solutions and we will discuss how to handle the all-important boundary conditions in the following subsection. We follow the presentation given in [1].

General Solution

We will start discussing the general solution of the one-dimensional case and follow with the two- and three-dimensional versions in the following subsections. There is a rich literature describing these methods and also a list of references at the end of the chapter.

1D – Solution

Let us consider an equation similar to (4.26) in free space

d2φxdx2+k2φx=−δx−x0.(4.30)

This is known as Helmholz’s equation in one-dimensional free space subject to the boundary condition at infinity, φ(±∞) = 0φ±∞=0. The response at xx is due to the delta source at x₀x0. Let us consider the homogeneous equation

d2φxdx2+k2φx=0

This is the same as (4.30) when x ≠ x₀x≠x0. The solution to this equation that satisfies the boundary conditions at infinity is

φx=Aejkxx>x0Be−jkxx<x0.

The unknown constants AA, BB can be determined by the boundary condition at x = x₀ ± Δx=x0±Δ where ΔΔ denotes an infinitesimally small interval. Integrating (4.30) from x = x₀ − Δx=x0−Δ to x = x₀ + Δx=x0+Δ we find

dφdxx=x0−Δx=x0+Δ+∫x=x0−Δx=x0+Δk2φxdx=−1.

Since φ(x)φx is continuous the last term on the left-hand side disappears when Δ → 0Δ→0.

We get

dφdxx=x0−Δx=x0+Δ=−1φx+Δ=φx−ΔikAejkx+Be−jkx=−1Aejkx=Be−jkx.

Solving for AA and BB gives

φx=j2kejkx−x0x>x0j2ke−jkx−x0x<x0=i2kejkx−x0.(4.31)

Long Wavelength Approximation

When kx₀ ≪ 1kx0≪1 (long wavelength approximation) we find

φlwx=j2k1+jkx−x0x>x0j2k1−jkx−x0x<x0=j2k1+jkx−x0=j2k−x−x02.

Since in the long wavelength approximation Helmholz equation reduces to Poisson equation where φ_lwφlw is defined with an arbitrary constant factor, we can simply relabel the constant term for φ_lwφlw and end up with

φlwx=−x−x02+C.(4.32)

Let us verify by examining the Poisson equation

d2φlwxdx2=−δx−x0.

When x ≠ x₀x≠x0 we see ∂2φlwx∂x2≡0. Let us integrate around the singularity as we did earlier

∫x0−Δx0+Δ∂2φlwx∂x2dx=−∫x0−Δx0+Δδx−x0dxdφlwxdxx0−Δx0+Δ=−12−12=−1=r.h.s.

Indeed, in one dimension and the long wavelength approximation (4.32) solves the Poisson equation.

2D – Solution

In two dimensions equation (4.30) becomes

∂2φxy∂x2+∂2φxy∂y2+k2φxy=−δy−y0δx−x0(4.33)

We can now use the Fourier transform

φxy=∫−∞∞φ˜ejβx−x0dβ.

This gives

∂2φβy∂y2−β2φ˜βy+k2φ˜βy=−δy−y0

The solution to this equation is the one-dimensional free-space Green’s function

φ˜βy=j2κejκy−y0

where κ=k2−β2. We find

φxy=∫−∞∞j2κejκy−y0ejβx−x0dβ=j4H01kx−x02+y−y02

where the last equality relates the Hankel function of order 0, H01 to the integral. We see the solution is simply a composition of a continuous spectrum of plane waves.

Long Wavelength Approximation

The two-dimensional solution in the long wavelength approximation can be found using similar techniques as in the one-dimensional case we noted earlier. Here we will show the solution for the special case of cylindrical symmetry which we will take advantage of later in this chapter when we discuss inductance and current elements.

Helmholz equation in cylindrical symmetry becomes, with a delta function at x = 0x=0,

∇ ⋅  ∇ φ = Cδ(x).

∇⋅∇φ=Cδx.(4.34)

Outside of x = 0x=0 we have

∇⋅∇φ=1r∂∂rr∂φrθ∂r+1r2∂2φrθ∂θ2=0.

To further simplify we will assume there is no θθ-dependence and we find

1rddrrdφrdr=d2φrdr2+1rdφrdr=0.

This has the general solution

φ(r) = D ln r + B.

φr=Dlnr+B.

To find out the constants we need to integrate (4.34) around x = 0x=0. Let us choose a sphere centered at x = 0x=0 with radius ΔΔ as an integration volume. We find for the left-hand side using the divergence theorem (see Appendix B)

∫∇⋅∇φdV=∫∇φ⋅rrda=∫dφdrrdθ=dφdrr2π=2πD.

The right-hand side of (4.35) becomes, as before, CC. Putting all this together we have

2πD=C→D=C2π.

We have for the long wavelength solution to the Helmholz equation in two dimensions:

φr=C2πlnr+B.(4.35)

3D – Solution

We finally present the 3D solution. We will use it in Chapter 6. Equation (4.30) becomes with a source at r = r₀r=r0

∂2φxyz∂x2+∂2φxyz∂y2+∂2φxyz∂z2+k2φxyz=−δr−r0.

Let us first do a change of variables ρ = r − r₀ρ=r−r0. We find we have created a spherically symmetric model. By realizing

∫δ(r − r₀)dV =  ∫ δ(ρ)4πρ²dρ,

∫δr−r0dV=∫δρ4πρ2dρ,

we find

δr−r0=δρ4πρ2.

By substituting

φρ=uρρ,

we find for the Helmholz equation

∇2+k2uρ=−δρ4πρ.

The solution for u(ρ) = Ae^jkρuρ=Aejkρ and we find

φρ=Aejkρρ.

The boundary condition at ρ = 0ρ=0 must be used to determine AA. The Helmholz equation is

∇2+k2φρ=−δρ4πρ2.

After integrating over a small volume we find, as with the one-dimensional case,

φ0+=φ0−,∫∇2φρdV=−1.

After applying the divergence theorem we get

∫∇2φρdV=∮∇φρ⋅ds=Ajkejkρρ−ejkρρ24πρ2ρ→0=−4πA.

So

A=14π.

The general solution is

φxyz=14πr−r0ejκ⋅r−r0+14πr−r0e−jκ⋅r−r0,κ2=κx2+κy2+κz2=ω2ϵ′μ.

Similarly, for the vector potential we get

Ar=μ4π∫Jir′r−r′ejκ⋅r−r′+e−jκ⋅r−r′dr′.

Long Wavelength Approximation

The solution to the long wavelength equations (4.27) and (4.29) in three dimensions are well known. We see the equations are essentially identical and the general solution that vanishes at infinity is

Ar=μ4π∫Jir′r−r′dr′.(4.36)

The solution for φφ is similarly

φr=14πϵ′∫ρir′r−r′dr′.(4.37)

For a point charge at the origin, ρ_i(r^′) = qδ(r^′)ρir′=qδr′ we find

φr=q4πϵ′r.

which is the familiar electrostatic potential from a point charge.

Boundary Conditions

Identifying the relevant equations and how to solve them is a helpful exercise. It is often the easiest part of any investigation. The real problem comes when taking into account what happens at the boundaries, either in time and/or in space. To the novice, the opposite often appears true. To help us there are many great examples in the literature on how to handle the boundaries, and here we will go through the basic methods and leave it to the reader to explore more if needed.

Fundamentally, what one does is to put a “pill box” at the boundary that extends a little into each material with a large surface area. In Figure 4.1 it extends ε/2ε/2 into each region. The volume is thus infinitesimal, while the area is macroscopic. They key thing to note here is that the equations are still valid in this volume and to find out what the boundary conditions are one simply integrates the equations over the small volume. For some entities, portions of the equation will be proportional to the volume and thus small while other entities will be proportional to the area and thus large. We will go through Maxwell’s equations with this method to show explicitly how the conditions work out.

Figure 4.1 Boundary conditions pill box.

Let us look at Ampere’s law (4.28)

∇ × H = J.

∇×H=J.

At the boundary between two media we put a small pill box of height εε which is much smaller than any other dimension in the problem. We integrate Ampere’s law over this volume

∫ ∇  × HdV =  ∫ JdV.

∫∇×HdV=∫JdV.

We can here use Stoke’s theorem for the left-hand side and denoting by A_rea = A_reanArea=Arean where A_reaArea is the area and nn is the outward normal to the area segment

∫ ∇  × HdV =  ∮ H × A_readA_rea = A_rea(H_{t, +} − H_{t, −}) + εH_n → A_rea(H_{t, +} − H_{t, −}),    ε → 0.

∫∇×HdV=∮H×AreadArea=AreaHt,+−Ht,−+εHn→AreaHt,+−Ht,−,ε→0.

Volume Current

For the right-hand side without delta functions, which we call volume current, we get

∫JdV~ε → 0    when   ε → 0.

∫JdV∼ε→0whenε→0.

Putting it all together we get at the boundary

∫ ∇  × HdV = A_rea(H_{t, +} − H_{t, −}) =  ∫ JdV = 0.

∫∇×HdV=AreaHt,+−Ht,−=∫JdV=0.

or

H_{t, +} = H_{t, −}.

Ht,+=Ht,−.(4.38)

Surface Current

We also see when we have a delta function on the right-hand side, a surface current,

∫JdV =  ∫ J_sδ(y)dV = J_sA_rea    when   ε → 0.

∫JdV=∫JsδydV=JsAreawhenε→0.

Putting this together with the left-hand side we get

A_rea(H_{t, +} − H_{t, −}) = J_sA_rea

AreaHt,+−Ht,−=JsArea

or

(H_{t, +} − H_{t, −}) = J_s

Ht,+−Ht,−=Js(4.39)

Similarly, we can use the charge equation (4.23)

∇ ⋅ (ϵ^′E ) = ρ_i.

∇⋅ϵ′E=ρi.

We integrate over volume

∫ ∇  ⋅ (ϵ^′E )dV =  ∫ ρ_idV

∫∇⋅ϵ′EdV=∫ρidV

For the left-hand side we use Gauss’ law

∫∇⋅ϵ′EdV=∮ϵ′E⋅ndArea=Areaϵ+′En+−ϵ−′En−+ε…→Areaϵ+′En+−ϵ−′En−,ε→0

Volume Charge

For the case of volume charge, the right-hand side is now treated the same way as for the volume current earlier so we end up with

ϵ+′En+=ϵ−′En−(4.40)

Finally, for the case of surface charges we have

Surface Charge

ϵ+′En+−ϵ−′En−=ρsArea(4.41)

This way of treating the boundary is standard and can be found in many textbooks.

Field Energy Definitions

Let us look at the concept of energy in these fields. It is outside the scope of this book to fully derive this here but we will make some plausible arguments as to their validity: see [4] for details. Let us start with a static electric field where there are no currents.