Posts Tagged ‘Electricity & Magnetism’

Physics Friday 137

September 24, 2010

Part 17: Momentum Transport and Radiation Pressure

Let us consider a linearly-polarized plane wave propagating in the z direction, with plane of polarization in the x direction; so we then have
electric and magnetic fields

,
with .
Let us now compute the Maxwell stress tensor . As you may recall from here, the components of this symmetric (Cartesian) tensor are given by
.
First, we recognise that . Secondly, , and . Thus, for the electric parts, we see except for when i=j=1, so the electrical component only gives diagonal terms; we have
,
,
and
;
Next, for magnetic components, we have . Secondly, , and ,
so again, we only have diagonal terms,


,
so our tensor’s off-diagonal terms are zero. As for the diagonal terms, we note that for T11,
,
where we used the facts that , and that .
Similarly,
.
Lastly,
.
Recalling that the flux of the ith component of momentum in the j direction is given by , we see that the only momentum flow is of the z component of momentum, pz, flowing in the z direction; the wave carries momentum, aligned with its direction, in the direction of its motion. If we time-average over the cycle of the wave, we obtain .
(This last statement, in terms of the root-mean-square average, is valid for elliptically-polarized light as well).

Now, suppose we have a flat surface, perpendicular to our wave (and thus it is normal to the z axis in our coordinates); and that this surface is a perfect absorber. As the wave arrives at this surface, we see that the average linear momentum per unit area flowing into the surface in unit time is the vector p with components
;
as previously established, this is zero except for the z component:
.
Now, since the wave is being absorbed, this much momentum is lost from the field per unit time per unit of area of the surface. By conservation of the momentum, the surface must be gaining this momentum per unit time, and since change in linear momentum with time is a force (Newton’s second law); we see that the above gives the force on the surface per unit area; that is to say, a pressure equal to
; this is radiation pressure.
Note also that we demonstrated here that the average energy density in the wave is
;
thus, we see that the radiation pressure on a perfect absorber due to a plane electromagnetic wave incident normally to the surface is equal to the energy density of that wave.

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Physics Friday 136

September 17, 2010

Part 16: Energy in Electromagnetic Waves

As we saw here, the flow of energy in electromagnetic fields is given by the Poynting vector , the power flowing per unit of cross-sectional area. Now, let us consider electromagnetic plane waves in vacuum. As we showed here, they are transverse waves with the electric and magnetic fields both perpendicular to the direction of propagation, and that the electric and magnetic fields are perpendicular to each other as well, and proportional, with . Since the fields are perpendicular, the magnitude of their cross product is simply the product of their magnitudes:
, and we can see by the right-hand rule that S will be in the direction of propagation , so electromagnetic waves transport energy in their direction of propagation; an obvious result. We also see that since S is proportional to the square of a field magnitude, we recover the important optical property that the intensity of an eletromagnetic wave is proportional to the square of its amplitude.
More specifically, for a linearly-polarized wave, we have sinusoidally varying fields. With z the direction of propagation and polarization in the x direction, we then have

,
and so
.
The cosine-squared term oscillates between zero and one, and has a time average of 1/2. Taking the time average, then, we have
,
where Erms is the root-mean-square amplitude of the electric field, and which is equal to the peak amplitude over the square root of two.

For circularly-polarized waves, the magnitudes of the electric and magnetic fields are constant, so the Poynting vector is a constant; with propagation again in the z direction,
.
For general elliptical polarization, we can again express in terms of the root-mean-square amplitude (with mean over time) as
,
which reduces to the linear or circular case in each of those limits.

Also, since the energy density for electric fields is , and the energy density for magnetic fields is . Applying these to our wave, and taking the time average over a cycle, the average energy density is

And dividing the energy flux by the energy density confirms that the speed of the energy flow is
,
the speed of light in vacuum, as expected.

Physics Friday 135

September 10, 2010

Part 15: Polarization of Electromagnetic Waves

Last week, we began discussing electromagnetic waves in a vacuum, obtaining the fact that light and other electromagnetic waves are transverse waves, with the electric and magnetic fields perpendicular to the direction of propagation. We also established that the fields are perpendicular to each other, and proportional: for waves in vacuum, we have .
So, since the magnetic field is determined by the electric field and the propagation direction, we will focus on the electric field E for the following description, with the magnetic field behaving similarly due to the above.
Given our unit vector in the direction , we may choose a pair of orthogonal unit vectors and in the plane with normal , so that the three form an orthonormal basis for space. However, to simplify the discussion, let the wave propagate in the x direction, so that becomes , while and become and . Now, we can express our electric field in terms of x and y components. Recall that we have electric field , which for our wave propagating in the z direction is
,
or, breaking into components,
,
where Ex and Ey are complex numbers, and we take the real component of the right-hand side above to get the physical field. Specifically, that Ex and Ey are complex means each carries a phase, not necessarily the same.
Suppose, though, that Ex and Ey do have the same phase, so that is real. Then, taking the real part, we have
,
where φ is the phase. We see then that the electric field vector maintains a constant angle θ with the x axis, , and oscillates sinusoidally in this line. Light in this mode is a linearly polarized wave; and we have retrieved another classic optical phenomenon from Maxwell’s laws.
Now, suppose our phases are different. In particular, let us start with the simple case that the two components have the same magnitude, but differ in phase by π/2 (90°); then , and we have
, or letting Ex set our zero phase,
.
Note that at a particular fixed point in space, then, the electric field vector is of constant magnitude, but rotates in direction with angular frequency ω, sweeping out a circle. Thus, we dub such a wave circularly polarized. Specifically, as we look into the wave, we note that for the upper (+) sign, the vector rotates counter-clockwise, and in the lower (-) sign, the vector rotates clockwise. We say that the former is left circularly polarized, or has positive helicity, while the latter is right circularly polarized, or has negative helicity.
In the case where the phases of Ex and Ey differ by a value other than zero or π/2, or they differ in both phase and amplitude, then the tip of the electric field vector traces an ellipse in the xy-plane, and we call the light elliptically polarized.
Now, note that if we add two circularly polarized waves of equal magnitude and opposite helicity, the result is a linearly polarized wave, with the angle of polarization depending on the phase difference between the circularly polarized waves. More generally, defining the complex-valued vectors ,
we see that left circularly polarized light is given by
,
and right circularly polarized light is given by
,
where E+ and E are complex amplitudes.
Further, any complex vector may be expressed as a complex linear combination of , and so the general wave may be expressed as
,
and the circular polarizations form a basis: any electromagnetic wave may be decomposed into left and right circularly polarized components. [This is key to the phenomenon of optical rotation.]
For the above basis, let us set , so that r is the ratio of the magnitudes of the components, and φ the phase difference. Then, with a little math, one can show that the ellipse traced by the electric field vector has semi-minor axis to semi-major axis ratio of
, and thus eccentricity of , and that the major and minor axes of this ellipse are rotated from the x and y axes by an angle φ/2.
If r=1, we get, as previously noted, a linearly polarized wave, at an angle θ=φ/2.

Physics Friday 134

September 3, 2010

Part 14: Electromagnetic Waves in Vacuum

Writing Maxwell’s Laws so that all fields are on the left hand sides of the equations, we have



.
The second and third equations, Gauss’ Law of Magnetism and Faraday’s Law, are homogeneous equations; while the first and fourth, Gauss’ Law and Ampère’s Law, are inhomogeneous; they depend on the charge and current densities.
However, let us consider fields in empty space, so that ρ and J are zero as well; all four equations are then homogeneous, and so solutions may be superimposed (added).
Let us first examine solutions to these with harmonic time dependence, since Fourier superposition allows us to construct arbitrary solutions from these. So, we start with

,
where and are complex numbers, and we take real components of the results to get the physical fields.
Then our vacuum Maxwell’s Laws equations



,
become



,
or, eliminating the common complex exponentials (which are always nonzero), we obtain equations for the (complex) amplitudes



.
Note that if we take the divergence of the third equation, we obtain the second, and if we take the divergence of the fourth equation, we obtain the first, as the divergence of a curl is always zero. Thus, the first two equations are not independent, but are implicit in the latter two, so we just need

and
.
Taking the curl of the first equation, we have
.
Now,
,
since E has zero divergence. Thus
.
but our second equation tells us that
,
so we can plug this in to get
.

(a Helmholtz equation).
Similarly, taking the curl of the second equation
,
so the zero divergence of the magnetic field gives
,
and substituting the first equation into this gives
, the same differential equation which E must obey.
Letting our fields vary in one direction only, say z, the field equation reduces from

to
,
which has solutions of the form , where . We see then that k is the angular wavenumber of a wave with frequency ω and phase velocity c. Adding back the time dependence, we see the basic one-dimensional solution is
,
or using the dispersion relation ,
,
plane waves propagating in ±z.
Now, solving the Helmholtz equations for E and B and Maxwell’s equations, as well, we begin with plane waves with vector (angular) wavenumber k

,
where E0 and B0 are constant (complex) vectors, and, as before, physical quantities are obtained by taking the real part of complex quantities.
Plugging these into the Helmholtz equation, we use the fact that

to find that

and so

remains our dispersion relation.
Now, we need only find what the constant vectors E0, B0, and k must obey to satisfy Maxwell’s laws. Given a vector field that is the product of a constant vector and a scalar function, the divergence is thus the dot product of the constant vector and the gradient of the scalar ( for v constant), so the zero divergence equation for the electric field gives





.
Similarly, the divergence equation for B gives
.
Thus, both the electric and magnetic fields are perpendicular to the direction of propagation: electromagnetic waves are transverse waves.
For a scalar field φ and a constant vector v, one has
,
so the first curl equation

becomes




,
or, since , this is

,
where is the unit vector in the direction of propagation; hence, E0 and B0are perpendicular; the electric field, magnetic field, and direction of propagation are all mutually orthogonal for an electromagnetic wave, and for a wave in vacuum, the electric and magnetic field amplitudes are proportional with .

Physics Friday 133

August 27, 2010

Part 13: Electromagnetic FIelds and Angular Momentum

Last week, we considered momentum conservation in electromagnetism, finding that the momentum density is equal to , where is the Poynting vector; and that the momentum flux density tensor is , where is the Maxwell stress tensor, a symmetric rank-two tensor with components . Today, we consider angular momentum.
Recalling that for a particle of momentum p at displacement x from the origin, the angular momentum about the origin is
, and the force on a particle of charge q is , so the torque about the origin on the particle is
.
Converting to charge and current densities, if is the total (mechanical) angular momentum of the charges in a volume V, then
.
However, we showed last week that
,
where is the divergence of the Maxwell stress tensor , which is a vector with ith component .
Thus, we plug this in to get:
.
Note that since , ,
and the order can be interchanged between time derivatives and volume integrals, so the above becomes:
,
and so we see is the angular momentum of the fields in the volume, and so is the angular momentum density of the electromagnetic fields.
Now, let us consider the flux term. The ith component of the cross product of vectors a and b is , where is the Levi-Civita symbol. Thus, the ith component of is
.
Now, using the product rule, , and , so
,
and so
,
since due to the symmetry of and the antisymmetry of the Levi-Civita symbol (exchange the labels on the dummy indices j and k, then reverse orders on , with no sign change, and on , with sign change, to find that the sum is its own opposite, and therefore zero).
Note that the cross product of vectors a and b is the vector with ith component . Similarly, the cross product of the vector a with the tensor is the tensor with components
.
In this vein, we see then that , and therefore, that
,
the ith component of the divergence of the tensor (or, more accurately, pseudotensor) . Thus, since the corresponding components are equal,
,
and so
,
giving us our integral continuity equation. Just as is the linear momentum flux density tensor of electromagnetic fields, so we see that is the angular momentum flux density tensor of electromagnetic fields. Thus, the differential form for our continuity equation is
.
And just as one may find the total force the fields exert on the charges in a volume by integrating the total force via , one can do the same with to find the total torque.

Physics Friday 132

August 20, 2010

Part 12: Electromagnetic Fields and Momentum

Last week, we considered energy conservation of electromagnetic fields, and found the continuity equation, which told us that the Poynting vector gives the flow of energy in electromagnetic fields.
Now, let us consider linear momentum. Since electric and magnetic fields exert forces on charged particles, they change the momentum of these particles; therefore, momentum conservation means that the fields must themselves carry momentum. The total momentum in a volume is then the sum of the total mechanical momentum of the charged particles in that volume, , and the total momentum of the electromagnetic fields in the volume, . This sum, , must then obey a continuity equation; as momentum is conserved, the time change in this quantity is entirely due to the flux out of the volume.
The standard (differential form) continuity equation is
,
where v is the flux of the quantity φ. When φ is a scalar, v is a vector quantity, as the flow has both magnitude and direction. However, momentum is a vector quantity. So, letting Ptot be the momentum density, the flux is not a vector, but a rank-two tensor. The component Gij of this tensor represents the flux of the ith component of electromagnetic momentum in the j direction. For reasons similar to those for the mechanical stress tensor, this tensor must be symmetric: Gij=Gji, so we have six independent components.

Thus, for a volume V bounded by surface S with outward normal , we have
,
where is the vector whose ith component is (the rank-two tensor acts as a function mapping vectors to vectors, here applied to ). Splitting the momentum into mechanical and electromagnetic components,
.
Now, let us consider the change in mechanical momentum. The force on a particle with charge q is . The total change in momentum for the particles in a volume is the sum of the forces on the particles; for discrete charges qi, this is
;
converting to a continuous charge distribution, we have , , and the sum becomes a volume integral:
.
Now, we can use Maxwell’s laws to eliminate the charge and current densities in favor of the fields. From Gauss’ Law, ; and from Ampère’s Law, .
Thus,
.
Now, the product rule for the partial derivative of the cross product tells us
,
so
,
and we have
.
Now, Faraday’s Law tells us , so we have
.
Or, using , then
,
so
.
Note the near symmetry between the electric term and the magnetic term . Note, however, the term to make them symmetric would be , but since , this term is zero, and may be freely added:
so
. Thus, we have
.

Now, the product rule for the gradient of a dot product is
.
Letting a=b,
,
so
.
Thus,
.
Examining the components,

and
.
Adding these,

.
And since
So, using the Kronecker delta , we have
.
This is the ith component of the divergence of the tensor with components . Thus, we define the tensor with components
. This is called the Maxwell stress tensor. Then, we have
.
Note that has units of momentum flow per area, or momentum/(area)(time), which is equivalent to force/area, the units of stress; hence the name Maxwell stress tensor.

So, we then identify the term as the time derivative of the electromagnetic momentum in the volume. We then see the electromagnetic momentum density is ; the momentum density is parallel and proportional to the energy flux density with proportionality constant . We thus see that the momentum flux density tensor of electromagnetic fields is
. Thus, if is the mechanical momentum density, then the differential continuity equation for electromagnetic momentum is
.

Or, using the integral form
,
and applying the divergence theorem,
.
Note that the flux per unit area of momentum across the surface S, is the force per unit area transmitted across the surface S and acting on the fields and particles within V, and is , which has ith component
. One can thus consider the force on a material object in electromgnetic fields by considering a boundary surface S enclosing the object, and integrating up the total force via .

[Within dielectric media, the issue of momentum of the electromagnetic fields is more complicated; see the Abraham–Minkowski controversy.]

Physics Friday 131

August 13, 2010

Part 11: Electromagnetic Energy and Poynting’s Theorem

We previously noted that the rate of work done per unit times by electric fields (magnetic fields do no work) in a volume V with current density J is
.
This represents the power being transfered out of the electromagnetic fields and into mechanical or thermal energy. Now, let us examine further the conservation of energy.

First, we can use Ampère’s law to eliminate J in the above, working only in terms of the fields instead. In SI units, we have
,
so
,
and so

Now, ,
so
,
and we recall from here that the term in the parentheses above is the energy density of the electric field UE,
so
.

Next, for the dot product with the curl, we use the product rule for divergence of a cross product (similar to our usage here, albeit with E instead of A):
.
Now, Faraday’s Law gives us the curl of the electric field:
; thus
,
and .
Plugging these in to our work equation,
;
and
,
where is the energy density of the magnetic field, which we found here. Thus, we have
;
assuming that the total electromagnetic energy density U is the sum of the electric and magnetic energy densities, this is
;
defining , this is written
.
This has the form of a general continuity equation for electromagnetic energy: represents the rate of removal of energy from the fields. The vector S, then, describes the flux of electromagnetic energy from one location to another. This equation of energy conservation is called Poynting’s theorem, and S is called the Poynting vector, and has units of intensity (or power per unit area). Since only the divergence of the Poynting vector appears in the theorem, one might think that it is arbitrary in that adding the curl of any vector field will not change that divergence; however, relativistic considerations confine the Poynting vector to the specific value .

Physics Friday 130

August 6, 2010

Part 10: Electrodynamic Potentials and Gauge Symmetry

As noted here, the introduction of Faraday’s Law means that in electrodynamics, the electric field is not irrotational, and we no longer have the scalar electric potential φ with . But we still have no magnetic monopoles, so we can still use a magnetic vector potential A, with .
Next, we examine Faraday’s Law:
;
moving the right-hand term over, we can put this as
.
Rewriting in terms of A,
.
The spatial and time derivatives commute, so we have

.
So, while the electric field is no longer irrotational, the quantity is, and so can be written as the gradient of a scalar function:
,
or, solving for the electric field,
.
This and thus satisfy the two of Maxwell’s equations that are homogeneous: Faraday’s Law and the absence of magnetic monopoles.
Note that when the magnetic field is static, , and the above reduces to our electrostatic potential.

Recall that when we first discussed the vector potential, it was noted that the vector potential is not unique, and that adding the gradient of any scalar field to A does not change the resulting field. We called such a transformation a gauge transformation, and noted that we could choose a potential such that , calling that choice the Coulomb gauge.
Now, for electrodynamics, we can again add the gradient of a scalar function Λ to A without changing the magnetic field. However, if we perform the transformation from A to , we see that
. With , we then see that to keep the electric field unchainged, we must simulateously change the scalar field from φ to (so that we have cancelling terms).
This is the gauge transformation for electrodynamics, and the invariance of the electric and magnetic fields under these is called gauge invariance.
As before, the Coulomb gauge (also known as the “transverse gauge”) is that where the potentials are chosen so that . In this case, we plug our expression for the electric field due to potentials into Gauss’ Law:

.
.
Since the time and space terms commute, and noting that the divergence of the gradient is the Laplacian,
.
In the Coulomb gauge, the second term on the left is zero, and so our scalar potential obeys the Poisson’s equation
,
the same equation the electric potential obeys in electrostatics; though, we can no longer compute the electric field from this potential alone, but must find the magnetic vector potential. As in electrostatics, the solution to this is given by
, the instantaneous Coulomb potential; hence the name.

Another convienient choice of gauge is done by expressing both Gauss’ Law and Ampère’s law in terms of the potentials. For Gauss’ Law, we have
,
as noted above. Rearranging Ampère’s law so that the field terms are on the same side, we have
.;
using the fact that ,
.
Plugging in and ,
.
Now, the formula for curl of a curl tells us that
; and our time and space derivatives commute, so
, and so we have

.
.
Now, to remove φ from the second equation, we use the freedom of gauge transformations to choose potentials that obey
,
this is known as the Lorenz gauge condition, and the resulting gauge the Lorenz gauge. It is named for Ludvig Lorenz (not to be confused with Hendrik Lorentz).
If we make this choice, the second equation (Ampère’s Law) becomes
,
an inhomogeneous (vector) wave equation with wave velocity c. For the first equation (Gauss’ law) , the Lorenz condition means
; making this substitution,

,
and we have an inhomogeneous wave equation for the scalar potential; this choice has uncoupled the equations for φ and A.
Note that if we have φ and A obeying the Lorenz condition, and we make the gauge transform

,
with a scalar field Λ, then
,
so if Λ obeys the homogeneous wave equation , then the transformed fields φ‘ and A‘ also obey the Lorenz condition. Thus, the Lorenz gauge is not a single choice of potentials, as is the Coulomb gauge, but instead a restricted class of choices, with all potentials in it belonging to the Lorenz gauge.
The Lorenz gauge places φ and A on more equal and symmetric footing, and is also independent of the choice of coordinate system, making it the natural choice for dealing with electromagnetism in special relativity.

Physics Friday 129

July 30, 2010

Part 9: Maxwell’s Laws

Summing up so far, electrostatics uses Gauss’ law, which in differential form states
.
For magnetostatics, we had (Gauss’ law of magnetism, equivalent to saying that magnetic monopoles do not exist), and Ampère’s Law, which in differential form states
.
Lastly, we began electrodynamics with Faraday’s Law, which in differential form is

(we have used SI units for all of these).
However, these four equations, as written here, are inconsistent; one cannot expect the static equations to hold for dynamic situations. In recognizing this, and providing the correction, is where James Clerk Maxwell made his great achievement.
Specifically; the problem is in Ampère’s law. Taking the divergence of both sides,
.
but the divergence of a curl is always zero, so the left-hand side must be zero, and Ampère’s Law as formulated for magnetostatics requires . But by the continuity equation, ; so this condition only holds when the charge density is fixed.

According to J.D. Jackson in his Classical Electrodynamics textbook (my textbook for E&M at Caltech), Maxwell’s repair can be reasoned as follows: One begins with the continuity equation . One then notes that Gauss’ law says , and so, taking its time derivative, one sees that
; plugging this into the continuity equation, the result is

,
so the quantity must always be solenoidal (divergence-free); thus, one replaces J in the magnetostatic form of Ampère’s law with this quantity; then both sides are always divergence free:
.
Maxwell dubbed this added term the displacement current. While it has units of current density, it is not an actual current of flowing charges, but instead indicates that, just as a current generates a magnetic field, so does a time-varying electric field.
The four equations,
.



are collectively known as Maxwell’s equations, and are the collective basis of all classical electrodynamics.
From the last two, we see that a time-varying electric field generates a magnetic field (Ampère’s Law), and a time-varying magnetic field generates an electric field (Faraday’s Law); the combination of these two makes possible electromagnetic waves.

Physics Friday 128

July 23, 2010

Part 8: Work, Induction, and Magnetic Energy Density

Previously, we derived the energy density of the electric field by considering the work done in assembling a charge distribution from infinite separation, with the assembly done slowly enough as to be a quasi-static system. While we can find the energy density of the magnetic field by considering the work done in forming a particular current density, we cannot do so using magnetostatics; we must use Faraday’s Law of induction.

First, consider a circuit with a constant current I. If the flux Φ through the circuit changes, then an electromotive force will be generated around the circuit. This will change the current in the circuit; to oppose this, and keep the current constant, the current source must do work. As we found here, the power delivered to a current (work per unit time) by a voltage V is P=IV. Thus, the emf does work per unit time of , and so the work per unit time needed to oppose this and keep the current constant is . Now, Faraday’s Law tells us that (using SI units) . Thus, our current source delivers power ; or, in terms of differentials, a small change in flux δΦ is countered by work .

Next, consider a system of n circuits, with respective currents Ii, i=1,2,…,n. Then the flux in the ith circuit is

where dSi is the vector surface element for a surface bounded by the ith circuit (I’ve used dS rather than the usual dA to avoid confusion with the magnetic vector potential A).
Now, as we noted here, the definition of the vector potential combined with the Kelvin-Stokes theorem tell us that
,
where dsi is a vector line element of the ith circuit.
Thus, by the above single-circuit case, when there is a change in the magnetic field, and thus the fluxes, the current source of current Ii to maintain the current must deliver power equal to
.

So, then the total work necessary to take these n circuits from zero current to some final values a time T later, is
.
Now, the result should be independent of the particular “path” through intermediate values, so to simplify, we ramp up the currents proportionally, so that there is some increasing function of time f(t), with f(0)=0 and , with some constants of proportionality ci, for all i=1,2,…,n. The key then, is to note that the magnetic field generated by a current has magnitude proportional to that current, and so B will be linearly proportional to f(t), and thus all the Φi will be proportional to f(t). Thus, dubbing the constants of this latter proportionality by ki,

and so
. Since the final values of the current and flux are , and , respectively, this says that for proportional ramp-up,
, and so the work in setting up these currents is
,
and thus, for n circuits with currents Ii and fluxes Φi, the energy stored is
,
and using our expression for flux in terms of the line integral of the vector potential,
.

Now, let us instead consider a continuous current distribution, with current density J. As we did in our argument here, we break up the distribution into elemental current loops. An elemental loop will have a path C with line element ds parallel to the local current density, so that Jds=Jds; and we have a small perpendicular cross-section Δσ, so that the current in the loop is I=JΔσ. Thus, the contribution to the total stored energy by this element is
;
but, as we noted in our argument here,
JΔσds=JΔσds=Jd3r, and so the sum over all of these elemental loops becomes a volume integral:
.
Now, recall that Ampère’s Law states that
, and so, using this to replace the current density in the above, we see
.
Now, the product rule for the divergence of a cross product states that for two vector fields v and w,
.
Letting v be B and w be A, and solving for the first term on the right-hand side, we see
,
and so
.
This second term is the volume integral of a divergence; thus, by the divergence theorem,
, where S is the surface bounding our volume of integration. Now, a realistic current distribution can be expected to be of finite spatial extent; thus, as we expand our volume, the surface will eventually come to be far from the current distribution. As we noted here, for distant fields, the dipole term dominates, and the vector potential goes as r-2, and the field goes as r-3; thus, their cross product will have a magnitude that goes as r-5, while the surface of integration has area that goes as r2; thus, as the volume is expanded to all space, this surface integral will go to zero, and we get
. But from the definition of the vector potential, , so the above is
, and we identify the quantity being integrated over all space as the energy density of the magnetic field: .
Compare this to the energy density of the electric field, .