Chapter 8 Conservation Laws
We study the conservation of energy, momentum, and angular momentum, in
electrodynamics.
8.1 Charge and Energy
J
8.1.1 The Continuity Equation
Q
Review the conservation of charge.
r
The charge in the volume v is Q(t ) = ∫ ρ (r , t )dτ .
v
v
S
r
The current flow out through the boundary S is ∫ J ⋅ nˆ da .
S
r r
dQ
= − ∫ J ⋅ da .
dt
Surface
r
r
r
r
∂ρ
∂ρ
d
τ
=
−
J
⋅
d
a
=
−
∇
⋅
J
d
τ
Æ
=
−∇
⋅
J
∫ ∂t
∫
∫
∂t
Surface
The local conservation of charge is I =
1.
2.
Conservation of charge can be derived from Maxwell’s equations.
Conservation of charge is a consequence of the laws of electrodynamics.
8.1.2 Poynting’s Theorem (Conservation of energy)
The work necessary to assemble a static charge and a static current is:
1
1
ε
We = 0 ∫ E 2 dτ ( W = ∑ ∑ q jVi = ∫ ρVdτ =
E 2 dτ , Ch2.4.3),
∫
2 All _ Space
2
2ε 0
i =1− n j =1, j > i
1
1 2 1 r r
1 r r
1
2
and Wm =
B
d
τ
(
W
=
LI
=
A
⋅
I
dl
=
A ⋅ Jdτ =
B 2 dτ ,
∫
∫
∫
∫
2µ0 All _ Space
2
2
2
2 µ0
Ch7.2.4).
This suggests that the total energy stored in electromagnetic fields is:
U em =
⎛
1
1 2⎞
⎜⎜ ε 0 E 2 +
B ⎟dτ
∫
µ 0 ⎟⎠
2 All _ Spalce ⎝
According to the Lorentz force law, the work done on a moving charge q is
r r
r
r r
r r
r r
r r r
dW = F ⋅ dl = qE + qv × B ⋅ dl = qE ⋅ dl = qE ⋅ v dt = E ⋅ (qv ) dt .
(
)
Change the view of point charges to volume charge, we have
(
)
r
r r
r
dW r r
= E ⋅ qv = ∫ E ⋅ (ρv )dτ = ∫ E ⋅ Jdτ .
dt
V
V
r r
E ⋅ (qv ) is the work done per unit time.
P = IV = ( JA)(El ) = ( JE )( Al ) is the work done per unit time.
r r
r r
E ⋅ J is the work done per unit time, per unit volume. That means E ⋅ J may be
ε E 2 B2
related to the volume energy stored in electromagnetic fields 0 +
.
2
2µ0
r
r
r
r
r
r 1
r
∂E
∂E
Since ∇ × B = µ0 J + µ0ε 0
, we can replace J with J = ∇ × B − ε 0
.
∂t
∂t
µ0
r
r r 1 r
r
r ∂E
E⋅J =
E ⋅ ∇ × B − ε0E ⋅
∂t
µ0
r r
r
r r
r
r
r r
r
r r
∇⋅ E × B = B ⋅∇× E − E ⋅∇× B Æ E ⋅∇× B = B ⋅∇× E − ∇⋅ E × B
r
r r 1 r
r
r r
r ∂E
E⋅J =
B ⋅ ∇ × E − ∇ ⋅ E × B − ε0E ⋅
∂t
µ0
r
r
r r 1 ⎛ r ⎛ ∂B ⎞
r r⎞
r ∂E
⎟ − ∇ ⋅ E × B ⎟ − ε0E ⋅
E ⋅ J = ⎜ B ⋅ ⎜⎜ −
⎟
µ0 ⎜⎝ ⎝ ∂t ⎟⎠
∂t
⎠
(
)
(
(
(
(
))
(
)
)
)
r r
r r ∂ ⎛ B2 ε 0 2 ⎞
1
E ⋅ J = − ∇ ⋅ E × B − ⎜⎜
+ E ⎟⎟
µ0
∂t ⎝ 2µ0 2
⎠
(
)
r r
dW
= P = IV = ∫ E ⋅ J dτ
dt
V
r r ∂ ⎛ B 2 ε 0 2 ⎞⎤
⎡ 1
dW
+ E ⎟⎟⎥ dτ
= ∫ ⎢− ∇ ⋅ E × B − ⎜⎜
∂t ⎝ 2µ0 2
dt V ⎣ µ0
⎠⎦
(
dW
d ⎛ B2 ε 0 2 ⎞
1
= − ∫ ⎜⎜
+ E ⎟⎟dτ −
dt
dt V ⎝ 2µ0 2
µ0
⎠
Energy stored in the system:
)
r
r
r
∫ E × B ⋅ da
(Poynting’s theorem)
Surface
B2 ε 0 2
+ E , the decrease of the system energy means
2µ0 2
r r
d
the power consumption E ⋅ J = − (u B + u E ) , if the integral volume is extending to
dt
infinity. If we have a finite volume with a bounding surface, we may have energy
transported through the surface.
The energy per unit time, per unit area, transported by the fields is called the Poynting
r 1 r r
E × B (energy flux density)
vector: S =
µ0
(
)
r r
dW
dU
= − em − ∫ S ⋅ da
dt
dt
S
The work done on charges will increase their mechanical energy.
r r
dW
d
dW
d
= ∫ umech dτ &
= − ∫ uem dτ − ∫ S ⋅ da
dt
dt V
dt
dt V
S
r
r
d
(umech + uem )dτ = − ∫ S ⋅ dar = − ∫ ∇ ⋅ S dτ
∫
dt V
S
V
(
)
r
∂
(umech + uem ) = −∇ ⋅ S -- conservation of energy
∂t
r
∂ρ
= −∇ ⋅ J -- conservation of charge
∂t
(
)
r 1 r r
S=
E × B flow of energy
µ0
r
J flow of charge
Example: When current flows down a wire, work is done, which shows up as Joule
heating of the wire. Though there are certainly easier ways to do it, the energy per unit
time delivered to the wire can be calculated using the Poynting vector.
r
V
µI
1
1 ⎛ µ IV ⎞
IV
, B = 0 Æ S = − rˆ EB = − ⎜ 0 ⎟rˆ = −
rˆ
µ0
µ0 ⎝ 2πaL ⎠
2πaL
L
2πa
r r
r r
dW
S
⋅
d
a
=
−
IV
,
=
−
S
∫S
∫S ⋅ da = IV
dt
E=
r 1 r r
E × B being used for electron’s scattering
Energy transported into the wire: S =
µ0
with phonons and impurities (resistance)
The direction of Poynting vector represents the way that energy is transported
through.
Exercise: 8.1, 8.2