Lecture 1 The Maxwell Equations

The Maxwell Equations

ames Clerk Maxwell June 13th 1831 – November 5th 1879)

1

… and God said:

1 ∫SE ⋅ dA = ε0 Qenclosed

∫ B ⋅ dl = µ 0 ∫ J ⋅ dA + µ 0ε0 C

S

Gauss’s law for E

d E ⋅ dA Ampere’s law ∫ dt S

displacement current, ID d Faraday’s law ∫CE ⋅ dl = − dt ∫SB ⋅ dA

∫ B ⋅ dA = 0 S

Gauss’s law for B … and there was light! 2

… and God said:

1 ∇⋅E = ρ ε0 ∇ × B = µ 0 J + µ 0ε 0 ∇×E = −

∂ B ∂t

∇⋅B = 0

Gauss’s law for E ∂ E ∂t

Ampere’s law Faraday’s law Gauss’s law for B

3

… and there was light!

1 ∇⋅E = ρ ε0

Gauss’s law for E

– electric field diverges from charges ∂ ∇×E = − B ∂t

Faraday’s law

– E curls around the rate of change of B if B varies in time ∇⋅B = 0

Gauss’s law for B

– there are no magnetic monopoles, i.e., lines of B are closed loops ∇ × B = µ 0 J + µ 0ε 0

∂ E ∂t

Ampere’s law

displacement current density, JD

4

The Maxwell equations in vacuum and the displacement current

5

The Displacement Current The Maxwell equations ∇⋅B = 0

∇×E = −

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t Consider

∂ B ∂t

1 ∇⋅E = ρ ε0

∂ ∇ ⋅ ( ∇ × B ) = µ 0∇ ⋅ J + µ 0 ε 0∇ ⋅  E   ∂t 

0 = µ 0∇ ⋅ J + µ 0

∂ ( ε 0∇ ⋅ E ) ∂t

= µ 0∇ ⋅ J + µ 0

∂ ρ ∂t

∂ ρ + ∇ ⋅ J = 0 continuity equation ∂t This expresses the local conservation of electric charge – the charge density ρ and current density J are all functions of position x and time t: electric charge is 6 conserved throughout space. ∴

Example 1 Charging a capacitor, Part 1. Consider a parallel plate capacitor being charged by a current I that flows in wires along the z axis: Q( t )

− Q( t )

a

a

By Gauss’s law, the electric field between the plates, i.e., for r < a, isσ( t ) Q( t ) ˆ ∂ Q ( t ) ˆ ˆ E( t ) = k= k J D ( t ) = ε 0 E( t ) = k 2 2 ε0 ε 0 πa ∂t ε 0 πa By Ampere’s law,

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

7

Example 1 Charging a capacitor, Part 1. (cont’) B = Bφ ( r ) φˆ The curl of B is in the z direction, and B is azimuthal: 1 ∂ ∂ ∂ ( ∇ × B) r = Bz − Bφ = − Bφ = 0 r ∂φ ∂z ∂z ∂ ∂ ( ∇ × B ) φ = Br − Bz = 0 ∂z ∂r 1∂ 1 ∂ 1∂ ( ) ( rBφ ) ( ∇ × B) z = rBφ − Br = r ∂r r ∂φ r ∂r Consider the Amperian loop C1,

∂   ∫S1 ( ∇ × B ) ⋅ dA = µ 0 ∫SJ1 ⋅ dA + µ 0ε0 ∫S1  ∂t E  ⋅ dA ∂  ˆ  ˆ ∫CB1 ⋅ dl = µ 0 ∫SJ1 ⋅ kdA + µ0ε0 ∫S1  ∂t E  ⋅ kdA ∂  ˆ  ˆ ∫C1 ( B ⋅ φ)rdφ = µ 0 I + µ 0ε0 ∫S1  ∂t E  ⋅ kdA

8

Example 1 Charging a capacitor, Part 1. (cont’) In the quasistatic approximation, µ I Bφ ( r ) × 2πr = µ 0 I Bφ ( r ) = 0 2πr Consider the Amperian loop C2,

∂   ∫S2 ( ∇ × B ) ⋅ dA = µ0 ∫SJ2 ⋅ dA + µ0ε0 ∫S2  ∂t E  ⋅ dA µ0 I  Q ( t )  2 Bφ ( r ) = Bφ ( r ) × 2πr = µ 0 ε 0  πa = µ 0 I 2 2πr  ε 0 πa 

Consider the Amperian loop C3,

∫

S3

( ∇ × B ) ⋅ dA = µ 0 ∫SJ ⋅ dA + µ 0ε 0 ∫S  ∂ E  ⋅ dA 3

 Q ( t )  2 Bφ ( r ) × 2πr = µ 0 ε 0  πr 2  ε 0 πa 

3

 ∂t 

Bφ ( r ) =

µ 0 Ir 2πa 2 9

Example 1 Charging a capacitor, Part 1. (cont’)

Consider the Amperian loop C,

∫ B ⋅ dl = ∫ ( ∇ × B ) ⋅ dA = µ ∫ ( J + J ) ⋅ dA = µ ∫ J ⋅ dA = µ I C

0

Sa

D

Sa

0

∫ B ⋅ dl = ∫ ( ∇ × B ) ⋅ dA = µ ∫ ( J + J ) ⋅ dA = µ ∫ C

Sb

0

Sb

D

0

Sb

Sa

0

J D ⋅ dA = µ 0 I10

Scalar and vector potentials

11

Potential Functions The Maxwell equations ∇⋅B = 0

∇×E = −

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

∂ B ∂t

1 ∇⋅E = ρ ε0

Any vector function whose divergence is 0 can be written as the curl of another vector: ∇ ⋅ B = 0 ⇒ B = ∇ × A( x, t )

vector potential

∂ ∂ ∂   ∇ × E = − B ⇒ ∇ × E = − ( ∇ × A ) = −∇ ×  A  ∂t ∂t  ∂t  Any vector function whose curl is 0 can be written as the gradient of a scalar function: ∂  ∂  ∇ ×  E + A  = 0 ⇒ E + A = −∇ V ( x, t ) scalar12 potential ∂t  ∂t 

Potential Functions (cont’) The Maxwell equations B =∇∇⋅ B ×A = (0x, t )

∂∂ E = −∇ ∇V × (Ex,=t )−− BA( x, t ) ∂∂t t

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

1 ∇⋅E = ρ ε0

∂ ∂ 1 ∇ ⋅ E = −∇ 2V − ∇ ⋅  A  = − ( ∇ ⋅ A ) − ∇ 2V = ρ ∂t ε0  ∂t  ∂ ∂  ∇ × B = ∇ × ∇ × A = ∇ ( ∇ ⋅ A ) − ∇ A = µ 0 J + µ 0 ε 0  − ∇V − A  ∂t  ∂t  2

2 ∂ ∂ ∇ µ 0 ε 0 V + ∇ ⋅ A  + µ 0 ε 0 2 A − ∇ 2 A = µ 0 J ∂t ∂t  

The electric and magnetic fields are physical quantities, that could be measured, are uniquely determined if the 13 sources and boundary conditions are established.

Gauge Transformations and Gauge Invariance The potentials A(x, t) and V (x, t) are not uniquely determined by the charge and current sources in a system. Let f (x, t) be an arbitrary scalar function. Define ∂f A ′ ≡ A + ∇f V′ ≡V − ∂t Then,

∇ × A ′ = ∇ × A + ∇ × ∇f = ∇ × A = B

∂ ∂ ∂ ∂f ∂ − A ′ − ∇V ′ = − A − ∇f − ∇V + ∇ = − A − ∇V = E ∂t ∂t ∂t ∂t ∂t B and E are invariant with respect to the gauge transformation ∂f A → A ′ ≡ A + ∇f V →V′ ≡V − ∂t 14

Gauge Choices and Equations for V and A The Coulomb gauge

∇⋅A = 0

This gauge choice makes A unique: A ′ ≡ A + ∇f ∇ ⋅ A′ = 0 Demanding

∇ ⋅ A′ = ∇ ⋅ A + ∇ 2 f = ∇ 2 f ∇2 f = 0 would mean requiring

∇f or = constant

A′( ∞ ) = 0, A( ∞ ) = 0 ⇒ ∇f = 0 ∂ 1 2 − ( ∇ ⋅ A) − ∇ V = ρ Imposing the Coulomb gauge, ∂t ε0 1 2 − ∇ V = ρ reduces to Poisson’s equation ε0 Recall the Green’s Function 1 of G ( x − x′) = 2 –∇ : 4π x − x′

It follows that 1 ρ( x′, t ) d 3 x′ 15 V ( x, t ) = ∫ 4π ε0 x − x′

Gauge Choices and Equations for V and A (cont’) The scalar potential is thus also uniquely determined, except for an arbitrary additive constant. Imposing the Coulomb gauge, ∂ ∂2   ∇ µ 0 ε 0 V + ∇ ⋅ A  + µ 0 ε 0 2 A − ∇ 2 A = µ 0 J ∂t ∂t   reduces to

∂  ∂2  ∇ µ 0 ε 0 V  + µ 0 ε 0 2 A − ∇ 2 A = µ 0 J ∂t  ∂t 

The vector potential A(x, t) will in general be complicated if the sources vary in time. The Coulomb gauge is thus not a convenient gauge choice for problems involving the 3 generation of radiation. ′ 1 ρ ( x , t ) d x′ Remark – The scalar V ( x, t ) = 4π ε0 ∫ x − x′ potential has an unphysical aspect: It depends on the charge density ρ at all source points x’ at the very same time t. 16

Gauge Choices and Equations for V and A (cont’) ∂ The Lorentz gauge µ 0 ε 0 V + ∇ ⋅ A = 0 ∂t Imposing the Lorentz gauge, ∂ 1 2 − ( ∇ ⋅ A) − ∇ V = ρ ∂t ε0 2 ∂ ∂ ∇ µ 0 ε 0 V + ∇ ⋅ A  + µ 0 ε 0 2 A − ∇ 2 A = µ 0 J ∂t ∂t  

reduce to ∂2 1 2 µ 0ε 0 2 V − ∇ V = ρ ∂t ε0

∂2 µ 0ε 0 2 A − ∇ 2 A = µ 0 J ∂t

17

Example 2 – V and A for a point charge at rest @ the origin q V ( x, t ) = A ( x, t ) = 0 4π ε0 r ∂ q rˆ It follows that E = A( x, t ) − ∇V ( x, t ) = 2 ∂t 4π ε0 r B = ∇ × A( x, t ) = 0 Consider the gauge transformation with

qt f ( x, t ) = 4π ε0 r

V ′( x, t ) = V ( x, t ) −

∂ f ( x, t ) = 0 ∂t

A′( x, t ) = A( x, t ) + ∇f ( x, t ) = −

qt rˆ 2 4π ε0 r 18

Energy and momentum of electromagnetic fields

ohn Henry Poynting September 9th 1852 – March 30th 1914)

19

Poynting’s Theorem The Maxwell equations ∇⋅B = 0

∇×E = −

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

∂ B ∂t

1 ∇⋅E = ρ ε0

Consider

  ∂   ∂ ∂ ( εijk Ei Bk ) − Bk  εijk Ei  E ⋅ ∇ × B = Ei  ε ijk Bk  =  ∂x j  ∂x j  ∂x j  ∂ ( ε jik Ei Bk ) + Bk ε kji ∂ Ei =− ∂x j ∂x j = −∇ ⋅ ( E × B ) + B ⋅ ∇ × E = −∇ ⋅ ( E × B ) − B ⋅ = µ 0E ⋅ J + µ 0ε 0E ⋅

∂ E ∂t

∂ B ∂t 20

Poynting’s Theorem (cont’) ∂ ∂ Therefore, − ∇ ⋅ ( E × B ) − B ⋅ B = µ 0 E ⋅ J + µ 0 ε 0 E ⋅ E ∂t ∂t 1  ∂ 1 ∂ ε 0 E ⋅ E + B ⋅ B + ∇ ⋅  E × B  = −E ⋅ J ∂t µ 0 ∂t  µ0  1  ∂ 1 2 1 2  B + ε 0 E  + ∇ ⋅  E × B  = −E ⋅ J Poynting’s theorem ∂t  2µ 0 2   µ0  This expresses the local conservation of energy – the energy density u and Poynting vector (energy flux) S: 1 2 1 1 2 u= B + ε0 E S = E× B 2µ 0 2 µ0 are all functions of position x and time t.

21

Poynting’s Theorem (cont’) If a particle with charge q moves, then the change of its kinetic energy K is equal to the work done on q:workdK = F ⋅ vdt = q( E + v × B ) ⋅ vdt = qE ⋅ vdt dK = qE ⋅ v dt

kinetic energy theorem

For a general continuous distribution of charge, with charge density ρ (x, t) and kinetic energy density uK(x, t), the total kinetic energy in an infinitesimal volume d 3x at x satisfies ∂u K 3 ∂u K 3 ( ) d x = ρd x E ⋅ v = E ⋅ ( ρv ) = E ⋅ J ∂t ∂t

22

Example 3 Charging a capacitor, Part 2. Q( t )

− Q( t )

a

a

On the cylindrical surface of radius a that encloses the volume between the plates Q ˆ µ 0 I ˆ quasistatic E= k B = φ 2 ε 0 πa 2πa approximation The energy flux density on the surface is 1 QI ˆ ˆ QI S = E × B = 2 3 k × φ = − 2 3 rˆ µ0 2π ε 0 a 2π ε 0 a

23

Example 3 Charging a capacitor, Part 2. (cont’) Field energy flows radially into the cylinder as Q increases, i.e., when I > 0. The total rate at which field energy flows into the cylinder QI QId P = 2 3 × 2πad = 2π ε 0 a π ε0 a 2 2   d Q d  d 1  Q  2  =  ε 0   × πa d  =  2  2 dt  2π ε0 a  dt  2  ε 0 πa   d 1 d =  ε 0 E 2 × πa 2 d  ≡ U E dt  2  dt 2

Field energy flowing in through the surface builds up in the field between the plates. The energy stored in a capacitor resides in the electric field.

24

Example 3 Charging a capacitor, Part 2. (cont’) Magnetic field energy between the plates is negligible compared to UE: a 1 UM ≡ ∫ Bφ2 ( 2πrd ) dr 0 2µ 0 2

µ0 I 2d a 3 µ0 I 2d 1  µ 0 Ir  = ( 2πrd ) dr = r dr =   2 4 ∫ ∫ 2µ 0 0  2πa  4πa 0 16π a

It follows that U M µ 0 I 2 d 2π ε0 a 2 I 2a 2 1 I 2a 2 = × = µ 0ε 0 = 2 2 2 UE 16π Q d 8Q c 8Q 2 So, unless the characteristic time of the charging process is comparable to the time for light to cross the radius of the capacitor, UM << UE. Remark – We must neglect UM to be consistent with the quasistatic approximation. Homework: Work through Example 4 @ Page 420

25

Field Momentum The Maxwell equations ∇⋅B = 0

∂ ∇×E = − B ∂t

∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

1 ∇⋅E = ρ ε0

Consider F = q( E + v × B ) = ∫ ρ( E + v × B ) d 3 x = ∫ ( ρE + J × B ) d 3 x V

or

V

1 ∂  f = ρE + J × B = ε 0 ( ∇ ⋅ E ) E +  ∇ × B − ε 0 E  × B ∂t   µ0 1 ∂ ∂ = ε0 ( ∇ ⋅ E) E − B × ∇ × B − ε0 ( E × B ) + ε0E × B µ0 ∂t ∂t 1 ∂ = ε0 ( ∇ ⋅ E) E − B × ∇ × B − ε0 ( E × B ) − ε0E × ∇ × E µ0 ∂t

26

Field Momentum (cont’) 1 ∂ f = ε0 ( ∇ ⋅ E) E − B × ∇ × B − ε0 ( E × B ) − ε0E × ∇ × E µ0 ∂t 1 = ε 0 [ ( ∇ ⋅ E ) E + ( E ⋅ ∇ ) E] + [ ( ∇ ⋅ B ) B + ( B ⋅ ∇ ) B ] µ0 1 1 2 ∂ 2 − ∇ ε 0 E + B  − ε 0 ( E× B ) 2µ 0  ∂t 2 since

  ∂ ∂ ( B × ∇ × B ) i = εijk B j ( ∇ × B ) k = εijk B j  ε klm Bm  = ε kij ε klm B j Bm ∂xl ∂xl   ∂ ∂ ∂ ( ) = δil δ jm − δim δ jl B j Bm = B j Bj − Bj Bi ∂xl ∂xi ∂x j ∴ ( B × ∇ × B) i =

∂ 1 2   B  − ( B ⋅ ∇ ) Bi ∂xi  2 

27

Field Momentum (cont’) ∂ ∂ f + ( µ 0 ε 0S ) = f + ε 0 ( E × B ) ∂t ∂t 1 1 2 1 2 = −∇  ε 0 E + B  + ε 0 [ ( ∇ ⋅ E ) E + ( E ⋅ ∇ ) E] + [ ( ∇ ⋅ B ) B + ( B ⋅ ∇ ) B] 2µ 0  µ0 2 The momentum density of the electromagnetic field is dPem = µ 0 ε 0S dV The angular momentum density of the electromagnetic field is dL em dPem = r× = µ 0 ε 0r × S dV dV

Homework: Read Example 5 @ Page 422

28

Electromagnetic waves in vacuum

Heinrich Rudolf Hertz February 22nd 1857 – January 1st 1894)

29

The Maxwell Equations in Vacuum The Maxwell equations ∇⋅B = 0 ∂ ∇ × B = µ 0 J + µ 0ε 0 E ∂t

∇×E = −

∂ B ∂t

1 ∇⋅E = ρ ε0

In vacuum, i.e., without any matter at all: ρ = 0 and J = 0; these reduce to ∂ ∇⋅B = 0 ∇×E = − B ∂t ∂ ∇ × B = µ 0ε 0 E ∂t

∇⋅E = 0

Note 1: There may be static fields in a region where ρ = 0 and JNote = 0,2: produced by static charges … electromagnetic waves and … currents outside the region. 30

Derivation of the Wave Equation The Maxwell equations in vacuum ∇⋅B = 0 ∇ × B = µ 0ε 0 Consider

∇×E = − ∂ E ∂t ∇×∇×E = −

∂ B ∂t

∇⋅E = 0 ∂ ∇×B ∂t

2 ∂ ∇( ∇ ⋅ E ) − ∇ 2 E = −µ 0 ε 0 2 E ∂t

wave equation

∂2 ∴ −µ 0 ε 0 2 E + ∇ 2 E = 0 ∂t

One can similarly show that ∂2 − µ 0ε 0 2 B + ∇ 2B = 0 wave equation ∂t

c=

1 31 µ 0ε 0

The General Plane Wave Solution A harmonic plane wave has definite values of wavelength λ and frequency ν , related by λ ν = c. The wave fronts of a plane wave are infinite planes, perpendicular to the direction of propagation determined by the wave vector: k = k x ˆi + k y ˆj + k z kˆ The wave number 2π ω satisfies the dispersion k =k = = λ ν= c relation: λ k with

ω = 2π ν

phase velocity

The most general linearly polarized harmonic plane wave: B( x, t ) = B 0 exp[ i ( k ⋅ x − ωt ) ]

E( x, t ) = E 0 exp[ i ( k ⋅ x − ωt ) ]

where B0 and E0 are constant vectors. Note: The real parts are understood to be the physical32 fields.

The General Plane Wave Solution (cont’) E satisfies the wave equation:

∂2 − µ 0ε 0 2 E + ∇ 2E = 0 ∂t

∂2 2 E ( x , t ) = − ω E 2 ∂t 2 2 2   ∂ ∂ ∂ 2 ∇ E( x, t ) =  2 + 2 + 2 E( x, t ) = −( k x2 + k y2 + k z2 ) E ≡ − k 2 E  ∂x ∂y ∂z 

if the k = ω /c. An Example of a Plane Wave Solution B( x, t ) = B0 exp[ i ( kz − ωt ) ] ˆj where

E( x, t ) = E0 exp[ i ( kz − ωt ) ] ˆi

E0 B0 = c 33

The General Plane Wave Solution (cont’) The Maxwell equations ∇⋅B = 0 ∇ × B = µ 0ε 0 require

∇⋅E = 0 ∂ E ∂t

k ⋅ B0 = 0

transverse wave

k × B 0 = −µ 0 ε 0 ω E 0

∇×E = −

∂ B ∂t

k ⋅ E0 = 0 k × E 0 = ωB 0

E0 and B0 are perpendicular to each other, as well as perpendicular to k, with E0 x B0 parallel to k. The energy flux is in the direction of k: Poynting vector 1 1 S = E × B = E 0 × B 0 cos 2 ( k ⋅ x − ωt ) µ0 µ0 34

The General Plane Wave Solution (cont’)

k × B 0 = −µ 0 ε 0 ω E 0

k × E 0 = ωB 0

ω 1 kB0 = k × B 0 = − µ 0 ε 0 ωE 0 = 2 E0 ⇒ B0 = E0 c c Note: How to treat complex waves As long as we consider only expressions linear in the fields we may delay taking the real part. But before doing any calculation that is nonlinear in E and B, we must first take the real part, reducing the fields 35 to real numbers. Homework: Work through Examples 6 & 7 @ Page 429

The Maxwell equations in matter

36

… and God said:

11 -free enclosed ∫SE∫SE⋅ d⋅ AdA==ε εQ0 Qenclosed

Gauss’s law for E

d ∫CB ⋅ dl = µ 0∫S∫SJJfree⋅ d⋅AdA+ µ+0µε0 εdt ∫SE ⋅ dA Ampere’s law displacement current, ID d Faraday’s law ∫CE ⋅ dl = − dt ∫SB ⋅ dA

∫ B ⋅ dA = 0 S

Gauss’s law for B … and there was light! 37

… and God said:

∇ ⋅ D ≡ ∇ ⋅ ( εE ) = ρfree

Gauss’s law for E

∂ ∂ 1  ∇ × H ≡ ∇ ×  B  = J free + ( εE ) = J free + D ∂t ∂t µ  Ampere’s law ∂ ∇×E = − B ∂t ∇⋅B = 0

Faraday’s law Gauss’s law for B 38

… and there was light!

Gauss’s Law When dielectrics are present the source of E will include bound charge ρ b as well as free charge ρ f: 1 1 ∇ ⋅ E = ρ = ( ρf + ρ b ) Gauss’s law ε0 ε0 The bound charge density ρ b describes the effect of (atomic) electric dipoles as sources of electric field: ρ b = −∇ ⋅ P Here, P(x, t) is the polarization field or electric dipole moment density in matter – the origin of the electric dipoles is the separation of positive and negative charges within atoms. ∇ ⋅ ( ε 0 E + P ) ≡ ∇ ⋅ D = ρf Gauss’s law Here, D(x, t) is the displacement field.

39

Polarization Current The polarization current JP is the current associated with change of electric polarization, which occurs when positive charge shifts one way and negative charge another:

Since matter is electrically neutral, ρ ∂ ρb + ∇ ⋅ J P = 0 ∂t It follows that

b

and JP satisfy

∂ ∂  ( − ∇ ⋅ P ) + ∇ ⋅ J P = ∇ ⋅  − P + J P  = 0 ∂t  ∂t 

∂ JP = P ∂40 t

Ampere’s Law When matter is present the source of B will include bound current Jb, polarization current JP and free current Jf; in addition to ∂E/∂t: Ampere’s law ∂ ∂ ∇ × B = µ 0 J + µ 0ε 0 E = µ 0 ( J f + J b + J P ) + µ 0ε 0 E ∂t ∂t The bound current density is an effective Jb = ∇ × M current density equivalent to the (atomic) magnetic dipoles from the magnetization: Here, M(x, t) is the magnetization field or magnetic dipole moment density of the matter – the origin of the magnetic dipoles is the current associated with electron orbital motion and spin. Ampere’s law 1  ∂ ∂   ∇ × H ≡ ∇ ×  B − M  = J f + ( ε0E + P ) = J f + D 41 ∂t ∂t  µ0 

Linear Constitutive Equations In many applications of electromagnetism in matter, the polarization P is proportional to E P = χeε 0E

χ e = electric susceptibility

and the magnetization M is proportional to H M = χmH

χ m = magnetic susceptibility

For such linear materials, D and E (B and H) satisfy the following constitutive equations: D = ε 0 E + P = (1 + χ e ) ε 0 E ≡ εE 1 H ≡ B − M ⇒ B = µ 0 H + M = (1 + χ m ) µ 0 H ≡ µH µ0 where the constant ε is the permittivity of the material, 42 and µ the permeability.

Boundary Conditions of Fields The Maxwell equations in matter ∇ ⋅ D = ρf

∇⋅B = 0

Consider the surface S separating two linear media, with permittivity ε 1 and permeability µ 1 on one side of S, and ε 2 and µ 2 on the other side. Dn 2 − Dn1 = σ f

δ

2 1

Bn 2 − Bn1 = 0 G

In the limit δ → 0, 3 3 ( D − D ) dA = D ⋅ d A = ∇ ⋅ D d x = ρ d ∫ n 2 n1 ∫S ∫V ∫V f x = Qenclosed-free = ∫ σf dA 3 ( B − B ) dA = B ⋅ d A = ∇ ⋅ B d x=0 ∫ n 2 n1 ∫S ∫V

43

Boundary Conditions of Fields (cont’) The Maxwell equations in matter ∂ ∂ ∇ × H = Jf + D ∇×E = − B ∂t ∂t 2

E t 2 − E t1 = 0

1

C

δ

H t 2 − H t1 = K f × nˆ

In the limit δ → 0, d 0 = − ∫ B ⋅ dA = ∫ ( ∇ × E ) ⋅ dA = ∫ E ⋅ dl = ∫ ( Et 2 − Et1 ) ⋅ dl S C dt S d ∫ ( K f × nˆ ) ⋅ dl + 0 = ∫S J f ⋅ dA + dt ∫SD ⋅ dA = ∫ ( ∇ × H ) ⋅ dA = ∫ H ⋅ dl = ∫ ( H t 2 − H t1 ) ⋅ dl S

C

44