Chapter 8

Conservation Laws

8.1 Conservation of Charge and Energy

As was already shown, the continuity equation

∂

∂tρ+∇ · J = 0, (8.1)

can be derived by considering the flow of charges from a given volume,

∂Q

∂t=

∂

∂t

∫

ρ(r)d3r = −∫

J(r) · da = −∫

∇ · J(r)d3r (8.2)

The continuity of charges expresses the fact that the total charge is conserved.Another conservation law can be derived from the Faraday’s and Am-

pere’s laws

∇× E+∂B

∂t= 0 (8.3)

∇×B−1

c2∂E

∂t= µ0J. (8.4)

By differentiating the energy density (7.90) with respect to time we get that

∂u

∂t=

∂

∂t

1

2(ϵ0E

2+1

µ0B2) = ϵ0E ·

(

1

ϵ0µ0∇×B−

1

ϵ0µ0µ0J

)

+1

µ0B ·(−∇× E)

(8.5)or

∂

∂t

1

2(ϵ0E

2 +1

µ0B2) = −

1

µ0∇ · (E×B)− J · E. (8.6)

This is the Poynting’s theorem which expresses the fact that energy is con-served.

80

CHAPTER 8. CONSERVATION LAWS 81

Indeed the energy required to assemble a static distribution of chargesis given by (7.89) and the the energy required to assemble a distribution ofcurrents is given by (7.90). Then from the Poynting’s theorem (8.6), the rateof change of energy

∂u

∂t=

∫

∂

∂t

(

1

2(ϵ0E

2 +1

µ0B2)

)

d3r

= −∫

1

µ0(E×B) · da−

∫

J · Ed3r. (8.7)

Note that the second term is the rate at which the total external work isdone on all charges

∂W

∂t=

∂

∂t

∫

ρ (E+ v ×B) · vdt d3r ·

=

∫

ρ (E+ v×B) · v d3r

=

∫

E · (ρv) d3r

=

∫

E · J d3r. (8.8)

If the system is closed (i.e. there are no external work), then the Poynting’stheorem (8.6) can be written as

∂u

∂t= −∇ · S (8.9)

where

u =1

2(ϵ0E

2 +1

µ0B2) (8.10)

is the energy density and

S =1

µ0E×B (8.11)

is the Poynting vector. What it describes is the energy flux similarly to howthe current density describes the flux of charges.

For example, consider a current I flowing down the wire which heats theenvironment with a power that we are to calculate. The electric field in thewire is

E =V

Lz (8.12)

and the magnetic field at the surface is

B =µ0I

2πaφ (8.13)

CHAPTER 8. CONSERVATION LAWS 82

and therefore the Poynting vector is pointing towards the center of the wire

S =1

µ0E×B. (8.14)

Then we can use the Poynting’s theorem to find the power∫

∂u

∂td3r = −

∫

∇ · Sd3r = −∫

S · da =V I

2πaL2πaL = V I (8.15)

in agreement with (7.85).

8.2 Conservation of Momentum

When one considers motion of charged particles in electromagnetic field veryoften one encounters situations when the linear or angular momentum ofparticles is not conserved. The problem is that in addition to momentumcarried by particle the momentum can also be carried by fields such as electricand magnetic fields.

Consider the total electromagnetic force in some volume

F =

∫

Vρ (E+ v×B) d3r

=

∫

V(ρE+ J×B) d3r (8.16)

Using two Maxwell’s equations (i.e. ϵ0∇ · E = ρ and 1µ0∇×B − ϵ0

∂B∂t = J)

the force density can be rewritten as

f = ρE+ J×B

= ϵ0 (∇ · E)E+

(

1

µ0∇×B− ϵ0

∂

∂tE

)

×B. (8.17)

and the third Maxwell equation (i.e. ∇× E + ∂B∂t = 0) can used to rewrite

the time derivative of Poynting vector as

µ0∂S

∂t=

∂

∂t(E×B)

=∂E

∂t×B+ E×

∂B

∂t

=∂E

∂t×B− E× (∇× E) (8.18)

CHAPTER 8. CONSERVATION LAWS 83

or

ϵ0∂E

∂t×B = µ0ϵ0

∂S

∂t+ ϵ0E× (∇×E) . (8.19)

By substituting (8.19) into (8.17) and using the fourth Maxwell equation (i.e.∇ ·B = 0) we obtain

f = ϵ0 (∇ · E)E−1

µ0B× (∇×B)− ϵ0

∂E

∂t×B

= ϵ0 [(∇ · E)E− E× (∇× E)] +1

µ0[(∇ ·B)B−B×∇×B]− µ0ϵ0

∂S

∂t.(8.20)

This can be modified further using

∇E2 = 2 (E ·∇)E+ 2E× (∇× E) (8.21)

∇B2 = 2 (B ·∇)B+ 2B× (∇×B) (8.22)

to obtain

f = ϵ0

[

(∇ · E)E+ (E ·∇)E−1

2∇E2

]

+1

µ0

[

(∇ ·B)B+ (B ·∇)B−1

2∇B2

]

−µ0ϵ0∂S

∂t.

(8.23)Now if we define the Maxwell stress tensor (spatial components of the socalled energy-momentum or stress-energy tensors)

Tij ≡ ϵ0

(

EiEj −1

2δijE

2

)

+1

µ0

(

BiBj −1

2δijB

2

)

= ϵ0

⎛

⎝

12 (E

21 − E2

2 − E23) E1E2 E1E3

E2E1 −E21 + E2

2 − E23 E2E3

E3E1 E3E2 −E21 −E2

2 + E23

⎞

⎠+

1

µ0

⎛

⎝

12 (B

21 −B2

2 −B23) B1B2 B1B3

B2B1 −B21 +B2

2 − B23 B2B3

B3B1 B3B2 −B21 − B2

2 +B23

⎞

⎠(8.24)

then the force density can be expressed as

fi =∑

j=1,2,3

∂

∂xjTji − µ0ϵ0

∂Si

∂t(8.25)

or using the vector notation

f = ∇ ·T− µ0ϵ0∂S

∂t(8.26)

CHAPTER 8. CONSERVATION LAWS 84

where the divergence is taken with respect to either first or second index inTij would not make a difference due to symmetry of the stress tensor

Tij = Tji. (8.27)

This equation represent the flow of momentum that can be exchanged be-tween charged particles and electric and magnetic field without violating themomentum conservation. We can now go back to the volume integral toexpress the external force as

F =

∫

Vf d3r =

∫

ST · da− µ0ϵ0

∂

∂t

∫

VSd3r. (8.28)

and in the static case as

F =

∫

ST · da. (8.29)

Thus, T is nothing but a force per unit area, or stress tensor whose diagonalterms describe pressure and off-diagonal terms describe shear.

For example, consider a hemisphere of a uniformly charged sphere ofradius R and charge Q. Since this is a static case the stress tensor (a localquantity) calculated over surface should be sufficient to calculate the netforce on the hemisphere,

F =

∫

disk

T · da+

∫

bowl

T · da

= −∫ 2π

0

dφ

∫ R

0

rdrT · z+∫ 2π

0

dφ

∫ π/2

0

sin θdθR2T · r (8.30)

In Cartesian coordinates the radial unit vector is

r = sin θ cosφx+ sin θ sinφy + cos θz (8.31)

and the relevant components of the stress tensor on the “bowl” are given by

Tzx = ϵ0EzEx = ϵ0(ρR)2 sin θ cos θ cosφ (8.32)

Tzy = ϵ0EzEy = ϵ0(ρR)2 sin θ cos θ sinφ (8.33)

Tzz =ϵ02

(

E2z − E2

x −E2y

)

=ϵ02(ρR)2

(

cos2 θ − sin2 θ)

(8.34)

where

ρ =Q

4πϵ0R3(8.35)

CHAPTER 8. CONSERVATION LAWS 85

or

(T · r)z = Tzx · dax + Tzy · day + Tzz · daz

= ϵ0(ρR)2(sin2 θ cosφ2 cos θ + sin2 θ sin2 φ cos θ +1

2cos3 θ −

1

2sin2 θ cos θ)

=ϵ02(ρR)2 cos θ. (8.36)

Similarly on the “disk”

(T · z)z = Tzz =ϵ02

(

E2z − E2

x − E2y

)

= −ϵ02(ρr)2 (8.37)

and thus,

Fz =

∫ 2π

0

dφ

∫ R

0

rdrϵ02(ρr)2 +

∫ 2π

0

dφ

∫ π/2

0

sin θdθR2 ϵ02(ρR)2 cos θ

= ϵ0ρ2π

∫ R

0

r3dr + ϵ0ρ2πR4

∫ π/2

0

sin θ cos θdθ

= ϵ0ρ2πR4

(

1

4+

1

4

∫ π

0

sin(2θ)d(2θ)

)

=3

4ϵ0

(

Q

4πϵ0R3

)2

πR4

=1

4πϵ0

3Q2

16R2. (8.38)

Note that the stress tensor can be generalized to the stress-energy (orenergy momentum) tensor

Tµν =

⎛

⎜

⎜

⎝

u µ0ϵ0S1 µ0ϵ0S2 µ0ϵ0S3

µ0ϵ0S1 −T11 −T12 −T13

µ0ϵ0S2 −T21 −T22 −T23

µ0ϵ0S3 −T31 −T32 −T33

⎞

⎟

⎟

⎠

(8.39)

and then the conservations of momentum and energy can be combined in asingle covariant equation

fµ +∇µTµν = 0 (8.40)

where fµ is the external four-force of the change of external (let say mechan-ical) momentum per unit time. The momentum stored in electromagneticfield or in the momentum density

g = µ0ϵ0S = ϵ0 (E×B) . (8.41)

CHAPTER 8. CONSERVATION LAWS 86

In the absence of the external momentum (i.e. f = 0) the conservation ofmomentum implies

∫

ST · da = µ0ϵ0

∂

∂t

∫

VSd3r (8.42)

or in a differential form∂g

∂t= ∇ ·T. (8.43)

Moreover, one can also define the angular momentum density

l = r× g = ϵ0 [r× (E×B)] (8.44)

that must also be conserved. For example consider a long solenoid withcurrent I, radius R and n turns per unit length and two coaxial cylindricalconducting shells of length l and radii a < R and b > R with charges Qand −Q respectively. When the current is switched off the cylindrical shellsstart to rotate due to flow of angular momentum that was initially stored inelectromagnetic fields. Initially the electric field between shells was

E =Q

2πϵ0l

1

ss (8.45)

and the magnetic field inside solenoid was

B = µ0nI z. (8.46)

This corresponds to momentum density

g = ϵ0 (E×B) = −µ0nIQ

2πlsφ (8.47)

and angular momentum density

l =µ0nIQ

2πlss× φ = −

µ0nIQ

2πlz (8.48)

in the region a < s < R or the total angular momentum

L = −µ0nIQ

2πlz

∫ R

a

sds

∫ 2π

0

dφ

∫ l

0

dzz

= −1

2µ0nIQ

(

R2 − a2)

z. (8.49)

This angular momentum went into the angular momentum of charged cylin-ders due to induced electric field (Faraday’s law),

E =

{

−12µ0n

dIdt

R2

s φ for s > R

−12µ0n

dIdt sφ for s < R.

(8.50)

CHAPTER 8. CONSERVATION LAWS 87

Then the total torque on the cylinders is

Nb = r× (−QE) =1

2µ0nQ

dI

dt

R2

bb× φ =

1

2µ0nQR2dI

dtz (8.51)

Na = r× (QE) = −1

2µ0nQ

dI

dta a× φ = −

1

2µ0nQa2

dI

dtz (8.52)

and the total angular momentum

L =

∫ 0

I

1

2µ0nQ

(

R2 − a2) dI

dtz = −

1

2µ0nIQ

(

R2 − a2)

z. (8.53)

The angular momentum that was initially stored in the electric and magneticfield (8.49) is now in the angular momentum of the charges on the cylinders(8.53).