classical field theory: maxwell equations

Classical Field Theory: Maxwell Equations

April 28, 20101

1J.D.Jackson, ”Classical Electrodynamics”, 2nd Edition, Section 6Classical Field Theory: Maxwell Equations

Introduction

I Electrostatics and Magnetostatics deal with steady-state problemsin electricity and in magnetism.

I The almost independent nature of electric and magnetic fieldsphenomena disappears when we consider time-dependentproblems.

I Time varying magnetic fields give rise to electric fields andvice-versa. We then must speak of electromagnetic fields ratherthan electric or magnetic fields.

I The interconnection between electric and magnetic fields and theiressential sameness becomes clear only within the framework ofSpecial Theory of Relativity


Faraday’s Law of Induction

Faraday (1831), observed that a transient current is induced in a circuit if:

1. A steady current flowing in an adjacent circuit is turned on or off

2. The adjacent circuit with a steady current flowing is moved relativeto the first circuit

3. A permanent magnet is thrust into or out of the circuit

The changing magnetic flux induces an electric field around the circuit,the line integral of which is called the electromotive force E and causes acurrent flow according to Ohm’s law: ~J = σ~E ( σ is the conductivity).

The magnetic induction in theneighboring of the circuit is ~B and themagnetic flux linking the circuit isdefined by

F =

∫S

~B · ~n da (1)


The electromotive force around the circuit is defined by

E =

∮C

~E ′ · d ~ (2)

where ~E ′ is the electric field at the element d ~ of the circuit C .Thus Faraday’s observation is summed up in the mathematical law:

E = −kdF

dt(3)

That is, the induced electromotive force around the circuit isproportional to the time rate of change of magnetic flux linking thecircuit.

The sign is specified by Lenz’s law, which states that the induced

current (and accompanying magnetic flux) is in such a direction to

oppose the change of flux through the circuit. For SI units k = 1,

Gaussian units k = 1/c .


Faraday’s law for a moving circuit

∮C

~E ′ · d ~= −kd

dt

∫S

~B · ~n da (4)

This is eqn (3) in terms of integrals. We can observe that :

I The induced electromotive force isproportional to the total time derivativeof the flux.

I The flux can be changed by changing:the magnetic induction or the shape or theorientation or the position of the circuit.

If the circuit C is moving with a velocity ~v in some direction the totaltime derivative in eqn (4) must take into account the motion e.g.convective derivative

d

dt=

∂

∂t+ ~v · ~∇ and

d~B

dt=∂~B

∂t+(~v · ~∇

)~B =

∂~B

∂t+ ~∇×

(~B × ~v

)+ ~v

(~∇ · ~B

)Classical Field Theory: Maxwell Equations

The flux through the circuit may change because:

I the flux changes with time at a point

I the translation of the circuit changes the location of the boundary.

Eqn (4) can be written as (why?):∮C

[~E ′ − k

(~v × ~B

)]· d ~= −k

∫S

∂~B

∂t· ~n da . (5)

In a comoving coordinate system∮C

~E · d ~= −k

∫S

∂~B

∂t· ~n da (6)

where ~E is the electric field in the comoving frame. Thus

~E ′ = ~E + k(~v × ~B

)(7)

With the present choice of units for charge and current, Galileancovariance requires that k = 1/c (why?).


Finally, the transformation of the electromotive force integral into asurface integral leads to (how?)∫

S

(~∇× ~E +

1

c

∂~B

∂t

)· ~n da = 0 (8)

Since the circuit C and the bounding surface S are abritrary, theintegrant must vanish at all points in space. Thus the differential formof Faraday’s law is:

~∇× ~E +1

c

∂~B

∂t= 0 . (9)

Note that for time-independent electrostatic fields : ~∇× ~E = 0.

KELVIN-STOKES THEOREM :relates the surface integral of the curl of avector field over a surface S in Euclidean3-space to the line integral of the vectorfield over its boundary∫

S

∇× ~F · d~S =

∮∂S

~F · d ~ (10)

z

x y

SC

0

n

da


Energy in the Magnetic Field

I Typically in magnetostatics, the creation of a steady-stateconfiguration of currents and associated magnetic fields involves aninitial transient period during which the currents and fields arebrought from zero to the final values.

I For such time-varying fields there are induced electromotive forceswhich cause the sources of current to do work.

I Since the energy in the field is by definition the total work done toestablish it, we must consider these contributions.


Energy in the Magnetic Field

• Suppose that we have a circuit with a constant current I flowing in it.• If the flux through the circuit changes an electromotive force E isinduced around it.• In order to keep the current constant, the sources of current mustdo work.• To determine the rate we note that the time rate of change of energyof a particle with velocity ~v acted on by a force ~F is

dE

dt= ~v · ~F

With a changing flux the added field ~E ′ on each conducting electron ofcharge q and drift velocity ~v gives rise to a change in energy per unittime of q~v · ~E ′ per electron.Summing over all the electrons in the circuit, we find that the sources dowork to maintain the current at the rate

dW

dt= −IE =

1

cIdF

dt


Thus, if the flux change through a circuit carrying a current I is δF , thework done by the source is :

δW =1

cI δF

We consider the problem of the work done inestablishing a general steady-statedistribution of currents and fields.

I We can imagine that the build-up processoccurs at an infinitesimal rate so that~∇ · ~J = 0 (~J is the current).

I The current distribution can be broken upinto a network of elementary loops, thetypical ones of which is an elemental tubeof current of cross-sectional area ∆σfollowing a close path C and spanned by asurface S with normal ~n

Figure: Distribution ofcurrent density broken upinto elementary currentloops


We can express the increment of work done against the inducedelectromotive force in terms of the change of the magnetic inductionthrough the loop:

∆(δW ) =J∆σ

c

∫S

~n · δ~Bda (11)

the extra ∆ is because we consider only one elementary circuit.• If we express ~B in terms of the vector potential ~A i.e. ~B = ~∇× ~A thenwe have

∆(δW ) =J∆σ

c

∫S

(~∇× δ~A

)· ~n da (12)

With application of Stokes’s theorem this can be written

∆(δW ) =J∆σ

c

∮C

δ~A · d ~ (13)

but J∆σd ~≡ ~Jd3x since ~l is parallel to ~J. Evidently the sum over allsuch elemental loops will be the volume integral.• Hence the total increment of work done by an external source due to achange δ~A(~x) in the vector potential is

δW =1

c

∫δ~A · ~J d3x (14)


By using Ampere’s law

~∇× ~H =4π

c~J

we can get an expression in terms of the magnetic fields. Then

δW =1

4π

∫δ~A · (~∇× ~H) d3x (15)

which transforms to (how?)

δW =1

4π

∫ [~H · (~∇× δ~A) + ~∇ · (~H × δ~A)

]d3x (16)

If the field distribution is assumed to be localized, the 2nd integrantvanishes (why?) and we get

δW =1

4π

∫~H · δ~Bd3x (17)

which is the analog of the electrostatic equation for the energy change

δW =1

4π

∫~E · δ~Dd3x (18)

where ~D = ~E + 4π~P is the electric displacement and ~P the electric

polarization (dipole moment per unit volume).


If we bring the fields from zero to the final values the total magneticenergy will be (why?)

W =1

8π

∫~H · ~B d3x (19)

which is the magnetic analog of the total electrostatic energy

W =1

8π

∫~E · ~D d3x (20)

The energy of a system of charges in free space i.e. electrostatic energyis:

W =1

2

∫ρ(~x)Φ(~x) d3x (21)

The magnetic equivalent for this expression i.e. the magnetic energy is

W =1

2c

∫~J · ~A d3x (22)


Maxwell Equations

Maxwell’s equations are based on the following empirical facts:

1. The electric charges are sources of the vector field of the dielectricdisplacement density ~D. Hence, for the flux of the dielectricdisplacement through a surface enclosing the charge we have

1

4π

∮S

~D · ~nda = Q =

∫V

ρd3x (23)

This relation can be derived from Coulomb’s force law.

2. Faraday’s induction law:

E =

∮C

~E · d~l = −1

c

∂

∂t

∫S

~B · ~nda (24)

3. The fact that there are no isolated monopoles implies∮S

~B · ~nda = 0 (25)

4. Ampere’s law: ∮C

~H · d~l =4π

c

∫S

~J · ~nda (26)


Maxwell Equations

The basic laws of electricity and magnetism can be summarized indifferential form:

Coulomb’s law ~∇ · ~D = 4πρ (27)

Ampere’s law (~∇ · ~J = 0) ~∇× ~H =4π

c~J +

1

c

∂~D

∂t(28)

Faraday’s law ~∇× ~E +1

c

∂~B

∂t= 0 (29)

no free magnetic poles ~∇ · ~B = 0 (30)

where ~E and ~B are the averaged ~E and ~B of the microscopic or vacuumMaxwell equations. The two extra field quantities ~D and ~H usually calledthe electric displacement and magnetic field (~B is then called the

magnetic induction and ~M is the macroscopic magnetization)

Da = Ea + 4π

(Pa −

∑b

∂Q ′ab∂xb

+ . . .

)→ ~D = ~E + 4π~P + . . .(31)

Ha = Ba − 4π(Ma + . . . ) → ~H = ~B − 4π ~M + . . .(32)


I The quantities ~P, ~M, Qab represent the macroscopically averagedelectric dipole, magnetic dipole and electric quadrupolemoment densities of the material medium in the presence of ofapplied fields.

I Similarly, the charge and current densities ρ and ~J are macroscopicaverages of the free charge and current densities in the medium.

I The macroscopic Maxwell equations are a set of 8 eqns involvingthe components of 4 fields ~E , ~B, ~D and ~H.

I The 4 homogeneous eqns can be solved formally by expressing Eand B in terms of the scalar potential Φ and the vector potential ~A

I The inhomogeneous eqns cannot be solved until the derived fields ~Dand ~H are known in terms of ~E and ~B. These connections whichare implicit in (32) are known as constitute relations, e.g. ~D = ε~E

and ~H = ~E/µ (ε: electric permittivity & µ magnetic permeability).

I All but Faraday’s law were derived from steady-state observationsand there is no a priori reason to expect that the static equationshold unchanged for time-dependent fields.


The above equations without the red term in Ampere’s law areinconsistent. While Ampere’s law ( ~∇ · ~J = 0) is valid for steady - stateproblems, the complete relation is given by the continuity equation forcharge and current

~∇ · ~J +∂ρ

∂t= 0 . (33)

Maxwell replaced ~J in Ampere’s law by its generalization

~J → ~J +1

4π

∂~D

∂t(34)

Maxwell called the added term displacement current, without it therewould be no electromagnetic radiation (Can you repeat his steps?).Maxwell’s equations, form the basis of all classical electromagneticphenomena. When combined with the Lorentz force equation

~F = q

(~E +

~v

c× ~B

)(35)

and Newton’s 2nd law of motion, provide a complete description of the

classical dynamics of interacting charged particles and EM fields.


Vector and Scalar Potentials

In electrostatics and magnetostatics we have used the scalar potential Φand the vector potential ~A to simplify certain equations.Since ~∇ · ~B = 0, we can define ~B in terms of a vector potential ~A :

~B = ~∇× ~A (36)

Then Faraday’s law ~∇× ~E + 1c∂~B∂t = 0 can be written

~∇×

(~E +

1

c

∂~A

∂t

)= 0 (37)

Thus the quantity with vanishing curl can be written as the gradient of ascalar potential Φ:

~E +1

c

∂~A

∂t= −~∇Φ or ~E = −~∇Φ− 1

c

∂~A

∂t(38)

The definition of ~B and ~E in terms of the potentials ~A and Φ satisfies

indentically the 2 homogeneous Maxwell equations. While ~A and Φ are

determined by the 2 inhomogeneous Maxwell equations.


If we restrict our considerations to the vacuum form of the Maxwellequations the inhomogeneous form of Maxwell equations can be write interms of the potentials as:

∇2Φ +1

c

∂

∂t

(~∇ · ~A

)= −4πρ (39)

∇2~A− 1

c2

∂2~A

∂t2− ~∇

(~∇ · ~A +

1

c

∂Φ

∂t

)= −4π

c~J (40)

These eqns are equivalent to Maxwell eqns but they are still coupled.Thanks to the abritrariness in the definition of the potentials we canchoose transformations of the form

~A→ ~A′ = ~A + ~∇Λ (41)

Φ→ Φ′ = Φ− 1

c

∂Λ

∂t(42)

Thus we can choose a set of potentials (~A,Φ) such that

~∇ · ~A +1

c

∂Φ

∂t= 0 (43)


This way we uncouple the pair of equations (39) and (40) and leave two

inhomogeneous wave equations, one for Φ and one for ~A

∇2Φ− 1

c2

∂2Φ

∂t2= −4πρ (44)

∇2~A− 1

c2

∂2~A

∂t2= −4π

c~J (45)

This set of equations is equivalent in all respects to the Maxwell

equations.


Gauge Transformations : Lorenz Gauge

The transformations

~A→ ~A′ = ~A + ~∇Λ (46)

Φ→ Φ′ = Φ− 1

c

∂Λ

∂t(47)

are called gauge transformation and the invariance of the fields underthese transformations is called gauge invariance.The relation between ~A and Φ :

~∇ · ~A +1

c

∂Φ

∂t= 0 (48)

is called Lorenz condition 2. [Prove that there will always exist potentialssatisfying the Lorentz condition].The Lorenz gauge is commonly used because:

I It leads to the wave equations (44) and (45) which treat Φ

and ~A on equal footings

I It is a coordinate independent concept and fits naturally intothe considerations of special relativity.

2The condition is from the Danish mathematician and physicist Ludvig Valentin Lorenz (1829-1891) and not

from the Dutch physicist Hendrik Lorentz (1853-1928)Classical Field Theory: Maxwell Equations

Gauge Transformations : Coulomb Gauge

In this gauge~∇ · ~A = 0 (49)

From eqn (39) we see that the scalar potential satisfies the Poisson eqn :

∇2Φ = −4πρ (50)

with solution

Φ(~x , t) =

∫ρ(~x ′, t)

|~x − ~x ′|d3x ′ (51)

The scalar potential is just the instantaneous Coulomb potential due tothe charge density ρ(~x , t). This is the origin of the name Coulombgauge.From eqn (40) we find that the vector potential satisfies:

∇2~A− 1

c2

∂2~A

∂t2= −4π

c~J +

1

c~∇∂Φ

∂t(52)


Finally, if we define the transverse (or solenoidal) current:

~Jt =1

4π~∇× ~∇×

∫ ~J

|~x − ~x ′|d3x ′ (53)

and longitudinal (or irrotational) current ~Jl for which ~∇× ~Jl = 0 whichmay cancel the contribution of the term with the potential Φ.Then, the wave equation for ~A can be expressed entirely in terms of thetransverse current ~Jt [Can you explain it?]

∇2~A− 1

c2

∂2~A

∂t2= −4π

c~Jt (54)

The Coulomb or transverse gauge is often used when no sourcesare present. Then Φ = 0, and ~A satisfies the homogeneous waveequation. The fields are given by

~E = −1

c

∂~A

∂t, and ~B = ~∇× ~A (55)


Green Functions for the Wave Equations

The wave eqns (44), (45) and (54) all have the same structure,

∇2Ψ− 1

c2

∂2Ψ

∂t2= −4πf (~x , t) (56)

• To solve (56) it is useful to find a Green function (as in electrostatics)• In order to remove the time dependence we introduce a Fouriertransform with respect to frequency• We suppose that both Ψ(~x , t) and f (~x , t) have a Fourier integralrepresentation

Ψ(~x , t) =1

2π

∫ ∞−∞

Ψ(~x , ω)e−iωtdω , f (~x , t) =1

2π

∫ ∞−∞

f (~x , ω)e−iωtdω

(57)with the inverse transformations,

Ψ(~x , ω) =

∫ ∞−∞

Ψ(~x , t)e iωtdt , f (~x , ω) =

∫ ∞−∞

f (~x , t)e iωtdt (58)


When the representation (57) are inserted into (56) it is found that theFourier transform Ψ(~x , ω) satisfies the inhomogeneous Helmholtzwave equation for each value of ω and k = ω/c(

∇2 + k2)

Ψ(~x , ω) = −4πf (~x , ω) (59)

Equation (59) is an elliptic PDE similar to Poisson eqn to which itreduces for k = 0.The Green function G (~x ,~x ′) appropriate to (59) satisfies theinhomogeneous equation(

∇2 + k2)Gk(~x ,~x ′) = −4πδ(~x − ~x ′) (60)

If there are no boundary surfaces, the Green function can only depend on~R = ~x − ~x ′, and must be spherically symmetric, that is depend only onR = |~R|. This means that in spherical coordinates Gk(R) satisfies

1

R

d2

dR2(RGk) + k2Gk = 4πδ(~R) (61)


In other words, everywhere except R = 0 the quantity RGk(R) satisfiesthe homogeneous equation

d2

dR2(RGk) + k2 (RGk) = 0

with solution :RGk(R) = Ae ikR + Be−ikR

The general solution for the Green function is :

Gk(R) = AG(+)k (R) + BG

(−)k (R) (62)

where

G(±)k (R) =

e±ikR

R(63)

with A + B = 1 and the correct normalization condition at R → 0

limkR→0

Gk(R) =1

R(64)

The first term of (62) represents a diverging spherical wavepropagating from the origin, while the second term represents aconverging spherical wave.


To understand the different time behaviors associated with G(+)k and

G(−)k we need to construct the corresponding time-dependent Green

functions that satisfy(∇2 − 1

c2

∂2

∂t2

)G

(±)k (~x , t;~x ′, t ′) = −4πδ(~x − ~x ′)δ(t − t ′) (65)

note that the source term for (59) is −4πδ(~x − ~x ′)e iωt′

Using the Fourier transforms (57) the time-dependent Green functionsbecome (how?)

G (±)(R, τ) =1

2π

∫ ∞−∞

e±ikR

Re−iωτdω (66)

where τ = t − t ′ is the relative time.The integral actually is a δ function and the Green function becomes:

G (±)(R, τ) =1

Rδ

(τ ∓ R

c

)(67)

or

G (±)(~x , t;~x ′, t ′) =1

|~x − ~x ′|δ

(t ′ −

[t ∓ |

~x − ~x ′|c

])(68)


? The infinite space Green function is thus a function only of therelative distance R and the relative time τ between the source and theobservation point.The Green function G (+) is called the retarded Green function and G (−)

is called the advanced Green functionParticular integrals of the inhomogeneous wave equation (56) are

Ψ(±)(~x , t) =

∫ ∫G (±)(~x , t;~x ′, t ′)f (~x ′, t ′)d3x ′dt ′

to either of these maybe added solutions of the homogeneous equation inorder to specify a definite physical problem.EXAMPLES

Φ(~x , t) =

∫ ∫1

|~x − ~x ′|δ

(t ′ −

[t ∓ |

~x − ~x ′|c

])ρ(~x ′, t ′)d3x ′dt ′

~A(~x , t) =

∫ ∫1

|~x − ~x ′|δ

(t ′ −

[t ∓ |

~x − ~x ′|c

]) ~J(~x ′, t ′)

cd3x ′dt ′


Poynting’s Theorem : Conservation of Energy

For a continuous distribution of charge and current, the total rate ofdoing work by the fields in a finite volume V is∫

V

~J · ~Ed3x (69)

This represents a conversion of EM energy into mechanical or thermalenergy and it must be balanced by a corresponding rate of decrease ofenergy in the EM field within the volume V .In order to derive this conservation law explicitly we use the Maxwell eqnsto express (69) in other terms. We use Ampere law to eliminate ~J∫

V

~J · ~Ed3x =1

4π

∫V

[c~E ·

(~∇× ~H

)− ~E · ∂

~D

∂t

]d3x (70)

If we use the vector identity,

~∇ · (~E × ~H) = ~H · (~∇× ~E )− ~E · (~∇× ~H)


together with the Faraday’s law, we get∫V

~J · ~Ed3x =−1

4π

∫V

[c ~∇ ·

(~E × ~H

)+ ~E · ∂

~D

∂t+ ~B · ∂

~H

∂t

]d3x (71)

The terms with the time derivatives can be interpreted as the timederivatives of the electrostatic and magnetic energy densities.If we also remember that the sum/integrals

WE =1

8π

∫~E · ~Dd3x and WB =

1

8π

∫~H · ~Bd3x (72)

represents the total EM energy (even for time varying fields). Then thetotal energy density is denoted by

u =1

8π

(~E · ~D + ~B · ~H

)≡ 1

2

(ε0~E

2 +1

µ0

~B2

)(73)

Then equation (71) can be written (how?):

−∫

V

~J · ~Ed3x =

∫V

[∂u

∂t+

c

4π~∇ ·(~E × ~H

)]d3x (74)


Since the volume is abritrary, this can be cast into the form of adifferential continuity equation or conservation law

∂u

∂t+ ~∇ · ~S = −~J · ~E (75)

The vector ~S represents the energy flow and is called Poynting vector

~S =c

4π

(~E × ~H

)≡(~E × ~H

)(76)

• Poynting’s theorem (Conservation of energy) : The physical meaningof the above relations is that the time rate of change of EM energywithin a certain volume, plus the energy flowing out through theboundary surfaces of the volume per unit time, is equal to the negativeof the total work done by the fields on the sources within the volume.

• In other words Poynting’s theorem for microscopic field (~E , ~B) is a

statement of conservation of energy of the combined system of

particles and fields.


• If we denote the total energy of the particles within the volume V asEmech and assume that no particles move out of the volume we have

dEmech

dt=

∫V

~J · ~Ed3x (77)

and the total field energy within V as

Efield =

∫V

ud3x =1

8π

∫V

(~E 2 + ~B2

)d3x (78)

Poynting’s theorem expresses the conservation of energy for thecombined system as:

dE

dt=

d

dt(Emech + Efield) = −

∮S

~n · ~Sda (79)


Poynting’s Theorem : Conservation of Momentum

The conservation of linear momentum can be similarly considered.The total EM force on a charged particle is

~F = q

(~E +

~v

c× ~B

)(80)

If we denote as ~Pmech the sum of all the momenta of all the particles in avolume V we can write the Newton’s 2nd law (how?),

~Ftot ≡d~Pmech

dt=

∫V

(ρ~E +

1

c~J × ~B

)d3x (81)

where (~J ≡ ρ~v) we converted the sum over particles to an integral overcharge & current densities.We can use Maxwell equations to eliminate ρ and J from (81) by using

ρ =1

4π~∇ · ~E , ~J =

c

4π

(~∇× ~B − 1

c

∂~E

∂t

)(82)


After some manipulations we can show (how?) that the rate of changeof mechanical momentum of eqn (81) can be written

d~Pmech

dt+

d

dt

∫V

1

4πc

(~E × ~B

)d3x (83)

=1

4π

∫V

[~E (~∇ · ~E )− ~E × (~∇× ~E ) + ~B(~∇ · ~B)− ~B × (~∇× ~B)

]d3x

We can identify the volume integral on the left as the total EMmomentum ~Pfield in the volume V

~Pfield =1

4πc

∫V

(~E × ~B

)d3x (84)

The integrant can be interpreted as the density of EM momentum

~g =1

4πc

(~E × ~B

)(85)

The EM momentum density ~g is proportional to the energy-flux density ~Swith proportionality constant c−2.


Maxwell Stress Tensor

In order to establish that (84) is a conservation law for momentum, wemust convert the volume integral on the right into surface integral ofsomething that will be identified as momentum flow.By defining the Maxwell stress tensor Tab as

Tab =1

4π

[EaEb + BaBb −

1

2δab

(~E · ~E + ~B · ~B

)](86)

where the a-th component can be written as[~E (~∇ · ~E )− ~E × (~∇× ~E )

]a

=∑

b

∂

∂xb

(EaEb −

1

2δab~E · ~E

)(87)

also (~∇ · ~T

)a

=∑

b

∂

∂xbTab (88)

Then (84) can be written as:

d

dt(Pmech + Pfield)a =

∑b

∫V

∂

∂xbTabd

3x (89)


Then via the divergence theorem we get

d

dt(Pmech + Pfield)a =

∮S

∑b

Tabnbda (90)

If (90) represents a statement of conservation of momentum,∑b

Tabnb

is the a-th component of the flow per unit area of momentum across thesurface S into the volume V .

In other words it is the force per unit area transmitted across the surface

S and acting on the combined system of particles and fields inside V .


Conservation of Angular Momentum

The derivation of the EM angular momentum shares the same tacticalapproach as that of the linear momentumLet us define the mechanical angular momentum of the system as

~Lmech = ~r × ~pmech (91)

where ~pmech is the mechanical momentum density. Then

d~Lmech

dt= ~r ×

(ρE +

1

c~J × ~B

), (92)

and substitution of ρ and ~J from Maxwell’s eqns leads to

d

dt

[~Lmech +

1

4πc~r ×

(~E × ~B

)](93)

=1

4π~r ×

[~E (~∇ · ~E )− ~E × (~∇× ~E ) + ~B(~∇ · ~B)− ~B × (~∇× ~B)

]


By using the definition of the Maxwell stress tensor (86), we can simplifyeqn (94) considerably

d

dt

(~Lmech + ~Lfield

)= ~r × ~∇ · ~T (94)

where~Lfield = ~r × ~g (95)

has the interpretation of being the EM field angular momentumdensity In integral form since

~r × ~∇ · ~T = ~∇ ·(~r × ~T

)and ~∇×~r = 0 (96)

we getd

dt

(~Lmech +

∫V

~Lfieldd3x

)=

∫S

(~r × ~T

)· ~nda (97)

The right-hand side of this equation represents the integrated torque

density due to the fields over the boundary surface S .


classical field theory: maxwell equations

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