Jackson Notes part 2 January 2017
1
Energy in the Electromagnetic field
1.1
Electrostatic case
Then the
We put a charge in a region where there is an electrostatic field
force acting on the charge is = and the work done by the fields on the
charge as it moves from to along path is
Z
·
=
= −
= −
Z
Z
·
∇Φ
Φ = (Φ − Φ )
independent of the path Since the work done by the fields must decrease
the stored energy, we may interpret the term Φ as the potential energy of the
system comprising the charge and the fields
= − = Φ − Φ
For a system of point charges at positions we may write the potential
as
Φ () =
1 X
40 =1 | − |
When we compute the potential at the position of a charge we do not include
charge (charges do not exert forces on themselves, or equivalently, we ignore
the infinite self-energy of the charge). Thus the energy of charge in the
presence of all the other charges is
= Φ ( ) =
1
40
X
=1 6=
=
| − |
X
(1)
=1 6=
Thus we see that the energy is the sum of the energies of all the pairs of
charges we can form with charge The energy of the total distribution appears
to be
X
X
X
=
=
=1
=1 =1 6=
However, we have to be careful not to double count pairs of charges. ( =
but we only want one of them.) (See Lea and Burke Ch 25 for more on this).
So the correct expression is
=
1 X X
40
=1 6=
1 1 X X
=
| − |
2 40 =1
=1 6=
1
| − |
(2)
Next we look at what happens when the charge distribution is continuous.
We replace with the charge element = () 3 and with (0 ) 3 0 and
replace each sum with an integral to obtain
Z Z
Z
1 1
() (0 ) 3 3 0 1
=
=
() Φ () 3
(3)
2 40
| − 0 |
2 all space
where we used eqn (24) in Notes 1 for Φ () Notice, though, that we have no
way to express the constraint 6= and the self-energy is now included. Using
Poisson’s equation (Notes 1 eqn 27), we may eliminate the charge density and
obtain the energy in terms of the fields alone:
Z
¢
1 ¡
=
−0 ∇2 Φ Φ () 3
2
∙Z
¸
Z
³
´
0
· Φ∇Φ
· ∇Φ
= −
∇
− ∇Φ
2
¸
∙Z
Z
0
· ̂ −
·
Φ∇Φ
= −
2
∞
If the sources of the fields are localized, then the potential decreases at least as
fast as 1 as → ∞ and so the surface integral is zero. (It goes as 1 12 2 = 1
as → ∞) Thus
Z
0
2
=
2 all space
and the energy density is
0 2
(4)
2
Extension to fields in media (see Jackson §4.7— we’ll do more with this
with
later). We replace 0
=
=
1
·
2
(5)
and this result is true in media with 6= 0 as well.
{Digression that shows the equivalence of (5) and (3).
Z
Z
Z h
³
´
i
1
·
= − 1 ∇Φ
·
= − 1
· Φ
− Φ∇
·
=
∇
2
2
2
⎤
⎡
Z
I
Z
1
· ̂ − Φ ⎦ = 1 () Φ ()
= − ⎣ Φ
2
2
∞
}
2
Example: energy of a pair of point charges.
Two charges, 1 and 2 are separated by the displacement 2 To simplify
the calculation we can put the origin at the position of one of the charges, 1
Then the electric field at point is
() =
1 +
2 = 1 + 2 − 2
||3
| − 2 |3
and the energy density (4) is
! Ã
!
Ã
− 2
− 2
2 0
· 1 3 + 2
() =
1 3 + 2
3
3
2
||
| − 2 |
||
| − 2 |
#
"
22
( − 2 )
12
2 0
=
+
+ 21 2 3 ·
2 ||4
| − 2 |4
|| | − 2 |3
The first two terms are the self-energy terms, and the volume integral of each
is infinite.1 The third term is the interaction energy. It is the only term that
changes as the charges move, and thus (classically) it is the only energy we can
extract from the system.
Z
− 2 3
0
int = 2
21 2 3 ·
2
|| | − 2 |3
Z
1
2
1 ·∇
= 0 1 2 ∇
3
||
| − 2 |
µ
¶
¸
Z ∙
1
1
·
1 −
·∇
1 3
∇
∇
∇
= 2 0 1 2
| − 2 | ||
| − 2 |
||
Z
Z
1
1
1
1 · ̂ −
= 2 0 1 2
∇
∇2 3
|
−
|
|
|
|
−
|
|
|
2
2
∞
1 Note
that the infinity arises at = 0 :
∞
0
3
1
42
4
∞
= − 4
0
The surface integral is zero (integrand goes as 13 while goes as 2 as
→ ∞), so we have
Z
1
2
int = 40 1 2
() 3
| − 2 |
1 2
=
|2 |
This is the expected result. Compare with eqn (1).
1.2
Magnetic energy
The work done by the fields on moving charges (currents), per unit time, is
Z
=
= ·
When the currents are confined to wires, the expression simplifies
XZ
·
=
=
loops
We may use Stokes theorem to re-express the line integral:
´
X Z ³
×
· ̂
∇
=
loops
where is a surface spanning loop and then from Faraday’s law
Ã
!
X Z
−
=
· ̂
loops
So the work done by the fields is
= −
= −
X
loops
X
Z
· ̂
Φ
loops
where Φ is the magnetic flux through surface Work done by the fields
decreases the stored energy, = − Thus
X
=
Φ
loops
³
´
Now we let the charge density build up infinitely slowly
' 0 and thus ∇ · ' 0
This means the "field lines" of are closed loops to which we can apply the result
4
we have just derived. The current = ∆ where ∆ is a small area element
perpendicular to Thus
=
X
∆
loops
=
X
∆
loops
=
X
loops
=
Z
∆
Z
· ̂
Z ³
´
×
· ̂
∇
Z
loop
· =
X Z
loops
loop
∆
·
·
where we used the fact that is parallel to Next we use Ampere’s law to
write in terms of the fields. Remember: we are changing things infinitely
slowly so ' 0
Z ³
´
×
·
=
∇
Now, from the cover,
´
³
´
³
´
³
· × = · ∇
× − · ∇
×
∇
and =
we have
Thus, with =
Z h
³
´
³
´i
·
×
+
· ∇
×
=
∇
We use the usual trick of converting the first integral to a surface integral, and
→ 0 at least as fast as 12 for localized
arguing that it is zero. (Note that
while goes as 13 .) Then
Z
·
=
∝
and so
For a linear medium,
1
=
2
Z
5
·
and thus
1
=
2
and the energy density is
Z
=
In vacuum
=
·
1
·
2
1 2 0 2
=
20
2
Combining with the electric energy, we get
¶
µ
1
2
em =
0 2 +
2
0
1.3
(6)
Conservation of energy, general case
the fields do work per
When there exists a current distribution and field
unit volume to maintain the current at a rate
= ·
Using Ampere’s law, we may eliminate This time we allow arbitrary time
variation.
!
Z
Z Ã
=
×
−
·
∇
·
³
´
· ∇
×
using
Substitute for
³
´
³
´
³
´
·
×
=
· ∇
×
−
· ∇
×
∇
and use Faraday’s law and the divergence theorem to get:
#
Z
Z "
³
´
³
´
· =
· ∇× −∇· × −
·
Ã
!
#
Z "
I ³
´
×
· ̂
=
· −
−
· −
In a linear medium, we can simplify the term in square brackets:
Z
Z
I
´
³
= − 1
· ̂
·
−
·
·+
2
or
Z
Z
I
´
1 ³
· ̂ −
·
= −
·
·+
2
6
(7)
is the Poynting vector
×
Remembering that ̂ is the outward
where
normal, we may interpret this result as:
rate of change of stored energy in the volume = flow of energy into the volume − work done by fields
Applying the divergence theorem to the first term on the RHS, we may write
the energy conservation equation in differential form:
em
·
= − ·
+∇
(8)
Compare with the equation of charge conservation (Notes 1 eqn 2). Here the
non-zero term on the right hand side shows that energy may be converted to
non-electromagnetic forms by the currents.
1.4
Systems of conductors
With the usual convention that Φ → 0 at infinity, a spherical conductor of
radius carrying charge has potential = Similarly, the surface of
any conductor is an equipotential and the value of the potential is proportional
to the charge the conductor carries. The constant of proportionality depends
on the size and shape of the conductor:
=
(9)
where is the capacitance. Note that equals the charge on the conductor
when it is raised to unit potential.
Now suppose we have a collection of charged conductors. The fields produced by all of the conductors affect the potential on each of them. Thus we
may write
X
=
(10)
=1
The coefficients depend on the geometry. We can invert the matrix to
obtain the relation
X
(11)
=
=1
If all the conductors except one are grounded, and the remaining one has = 1
then
= =
Compare this relation with equation (9). The coefficients are called capacitances while the with 6= are called coefficients of induction.
The capacitance is the charge required to raise conductor
to unit potential with all the others grounded.
7
Note that this definition is not quite the same as the capacitance of a system
of two conductors defined in elementary treatments (e.g. Lea and Burke Ch 27
and example below). Since the name is the same, you must use context to
decide which is meant.
We may also express the energy of the system in terms of the Since
charge exists only on the surface of the conductors (J Problem 1.1), and each
surface is an equipotential, equation (3) becomes:
Z
Z
X
1
=
() Φ () 3 =
() Φ () 3
2 all space
=1
Z
Z
X
1X
=
() =
()
2 =1
=1
=
1X
2 =1
=
1 XX
1 XX
=
2 =1 =1
2 =1 =1
(12)
As an example of finding coefficents of capacitance, consider a simple system
composed of two concentric, spherical conducting shells of radii and
Let the potential be zero at infinity, and let the charges on the two conductors
be and respectively. Then we can use the spherical symmetry and Gauss’
law to find the electric fields, and hence the potentials on the two surfaces.
=0
For
For
= ̂
2
= + ̂
2
To find the potentials, use the line integral:
Z 2
·
Φ1 − Φ2 =
For
1
We start from infinity where we chose Φ = 0 Then:
¯∞
Z ∞
+ ¯¯
+
= −
Φ − Φ∞ = Φ =
¯
2
+
=
and then:
¯
µ
¶
Z
¯¯
1 1
=
−
=
−
Φ − Φ =
2
¯
8
(13)
and so
Φ =
+
+
µ
1 1
−
¶
µ
=
+
¶
Using superposition, we can think of this as the constant potential inside the
charged sphere of radius due to its charge plus the potential at the surface
of the charged sphere of radius due to its own charge
Now we want to put all this into the language of capacitances. First:
X
=
or, equivalently:
µ
¶
=
µ
11
21
¶µ
12
22
¶
For our example, we can see that
¶
µ
¶
µ
1 1
11 12
=
1 1
21 22
Then we can also write:
=
X
or
where
µ
11
21
Thus:
µ
µ
12
22
¶
¶
=
µ
11
21
1
=
−
¶
12
22
µ
=
µ
¶−1
−
−
11
21
=
1
12
22
µ
¶µ
¶µ
¶−1
=
1
1
=
−
µ
( − )
− +
1 1
1 1
¶
¶
µ
−
¶µ
−
−
(14)
¶
Now put in the expressions for the potentials and double check:
¶
µ
µ + + ¶
¶
µ
1
1
−
−
=
=
+
− +
−
−
−
+ +
¶
µ
¶
µ
−
=
=
−
−
as expected.
The capacitances of the two conductors (from (14)) are
11 =
1
−
and
9
22 =
1 2
−
¶
The “capacitance” of this system, as defined in elementary treatments for
= − = is (equation 13)
pair =
40
=
=
= 11
4
( − )
( − )
Is this relation true in general? No.
In general, if = − we have:
µ
¶ µ
¶µ
¶ µ
¶
11 12
11 − 12
=
=
−
21 − 22
21 22
and thus
¯
¯
pair = ¯¯
∆
¯ ¯
¯
¯ ¯
¯
1
¯=¯
¯
¯ ¯ 11 − 12 − 21 + 22 ¯
Using Green’s reciprocation theorem (J problem 1.12), we can show that 12 =
21 and thus
¯
¯
¯
¯
1
¯
pair = ¯¯
11 − 212 + 22 ¯
whereas
µ
11
21
12
22
¶
So
11 =
=
µ
11
21
12
22
¶−1
=
1
11 22 − 12 21
µ
22
−21
−12
11
¶
22
22
1
=
=
2
11 22 − 12 21
11 22 − 12
11 − 212 22
which is not equal to pair in general.
For this system of two spheres, 22 = 12 (you might want to investigate the
conditions necessary for this relation to hold) and we have:
11 =
1
= pair
11 − 22
The energy of the system is:
=
=
=
1
· ·
2
¶
µ
¶µ
¢
1¡
1 1
1 1
2
µ 2
¶
2
+ 2 +
2
In the special case that = − = we get
µ
¶
µ
¶
2
2 2
2 1 1
1 2
=
−2
+
=
−
=
2
2
2 pair
as expected.
10
(15)
1.5
Poynting’s theorem for harmonic fields
The above discussion does not include any dissipative effects that convert energy to non-electromagnetic forms. The easiest way to include these effects is
by working with the Fourier transforms of the Maxwell equations. This immediately gives us the results for AC circuits, for example, by considering a
single frequency component. The general results for time dependent fields with
dissipation are in Jackson §6.8.
For a single frequency component
³
´
´
³
( ) = Re
( ) − = 1
∗ ( )
( ) − +
2
When multiplying two such terms, we have to be careful. For example, the rate
at which the fields do work is
i
h
i
h
( )
( ) = Re ( ) · Re
´1³
´
1 ³
∗ ( )
( ) − +
=
( ) − + ∗ ( )
2
2
1 h
( ) + ( ) ·
( ) −2
∗ ( ) + ∗ ( ) ·
=
( ) ·
4
i
∗ ( ) 2
+ ∗ ( ) ·
´
³
1
() + () ·
() −2
=
Re ∗ () ·
2
Taking the time average, the second term vanishes, and we get
i
h
1
()
() = Re ∗ () ·
2
In what follows, the "Re" will be implicit. Then with these conventions, the
transformed Maxwell’s equations become
·
∇
·
∇
×
∇
∇ × +
= f
= 0
=
=
and the time-averaged rate at which the fields do work in a volume is
Z
1 ∗
()
() ·
=
2
Z h
i
1
∗ ·
()
×
∗ −
=
∇
2
But
³
´
³
´
´
³
×
−
· ∇
×
∗
∗· ∇
·
×
∗ =
∇
11
So
Z
³
´
i
´
1 h ³
×
−
∗ ·
()
∗· ∇
∗ +
=
−∇ · ×
2
Z
³
´
i
´
1 h ³
−
∗ ·
()
∗ ·
∗ +
=
−∇ · ×
2
The time averaged rate of energy flow is
´
´
³
³
=1
×
∗ = 1
∗ ×
2
2
making
1
2
Z
I
Z
i
h∗
∗·
= − · −
·−
2
()
∗ () ·
I
Z
= − · ̂ − 2 ( − )
(16)
Note that the last term has a real part only if
=
1 ³ ∗ ´ ∗ () ¯¯ ¯¯2
· =
¯ ¯
4
4
has an imaginary part, that is, has an imaginary part, indicating losses.
Re ( ) is the time-averaged electric energy density at frequency and simlarly
for Equation (16) is the analog of equation (7) for harmonic fields.
12
Now consider a system with all the fields confined inside a volume as
is zero on the bounding surface (This means
shown. The Poynting flux
we have no radiation resistance. See J §6.9 for a fuller discussion that includes
radiation resistance). Instead power is provided to the system by an input
current . is the potential difference across the system, input terminal to
output terminal, as shown in the diagram. Then we modify equation (16) as
follows:
Z
Z
1
1 ∗
∗
() · () + 2 ( − )
=
2
2
We can interpret this in terms of the impedance of the system (see, eg, Lea Ch
2 Pr 14, Jackson page 266):
= = ( − )
where is the resistance and is the reactance. (The minus sign is a result
of choosing the time dependence to be − .) Then, if the conductivity is real,
Z
Z
() + 4 ( − )
()∗ ·
| |2 =
¾
½Z ¯ ¯
Z
1
¯ ¯2
− =
¯ ¯ + 4 ( − )
2
| |
Thus, equating real and imaginary parts, we have
Z ¯ ¯2
1
¯¯
=
¯
¯
2
| |
and
=−
4
| |2
Z
( − )
(17)
At low frequencies, the stored energies may be expressed in terms of capacitance
(eqn 15) and inductance as
=
where =
1 ||2
1
1 | |2
2
=
and = | |
4
4 2
4
= − so that (17) reduces to
µ
¶
1
1
= − 2
= −
as expected.
13
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