Download 4.3.2 The multipole expansion

The principle of superposition (ie fields add vectorially) means that
P
~ r) =
E(~
1 X qi (~r − ~ri )
.
4πǫ0 i |~r − ~ri |3
~r
i
~ri
O
In the limit of (infinitely) many charges this becomes
P
~ r) = 1
E(~
4πǫ0
Z
V
~r − ~r′
(~r − ~r′ )
.
dV ′ ρ(~r)
|~r − ~r′ |3
~r
~r′
O
To return to a single charge at position ~r1 set ρ(~r) = q1 δ(~r − ~r1 ) which gives the
original result.
Finally as
∂φ
ρ(~r′ )
dV
≡
−
,
|~r − ~r′ |
∂ri
V
~ = −∇φ.
~
where we have defined a potential, φ by E
So
Z
1
ρ(~r′ )
φ(~r) =
.
dV ′
4πǫ0 V
|~r − ~r′ |
∂
−
E(~r)i =
∂ri
4.3.2
1
4πǫ0
Z
′
The multipole expansion
First expand
1
|~
r +~a|
for r ≫ a. We have
∞
X
1
1
~ r )n 1
=
(~a · ∇
|~r + ~a|
n!
r
n=0
1
1
1
1
1
+ ai ∂i + (ai ∂i aj ∂j ) + · · ·
r 1!
r 2!
r 2
2 2
1
1 ~a · ~r 3(~a · ~r) − a r
.
− 3 +
+O
=
5
r
r
2r
r4
=
Thus
Z
1
ρ(~r′ )
φ(~r) =
dV ′
4πǫ0 V
|~r − ~r′ |
Z
1 ~r′ · ~r 3(~r′ · ~r)2 − r 2 r ′2
1
′
′
dV ρ(~r )
+ 3 +
+ ... ,
=
4πǫ0 V
r
r
2r 5
45
(upon using the Taylor expansion).
This gives
φ(~r) =
1 ri pi
1 ri rj Qij
1 Q
+
+
+ ... .
3
4πǫ0 r
4πǫ0 r
4πǫ0 2r 5
where
Q =
pi =
Qij =
Z
dV ′ ρ(~r′ )
ZV
ZV
(total) charge or monopole term
dV ′ ri′ ρ(~r′ )
dipole moment
dV ′ (3ri′ rj′ − r ′2 δij )ρ(~r′ )
quadrupole moment
V
This expansion is valid in the far zone, ie r ≫ r0 .
• If Q 6= 0 the monopole term dominates
φ(~r) =
1 Q
.
4πǫ0 r
~ field emanates from a point charge at the origin.
The E
• If Q = 0 the dipole term dominates
φ(~r) =
1 p~ · ~r
.
4πǫ0 r 3
If the charge density is given by two equal opposite charged particles close
~ − δ(~r′ )] then
together, ie ρ(~r′ ) = q[δ(~r′ − d)
Z
~ − δ(~r′ )] = q d~ ,
~p =
dV ′ ~r′ q[δ(~r′ − d)
this justifies the name.
• If Q = 0, p~ = 0, the quadrupole term dominates
φ(~r) =
1 ri rj Qij
.
4πǫ0 2r 5
Qij is symmetric (Qij = Qji ) and traceless (Tr Q = 0). A simple example is
given by the charge distribution of two opposite dipoles ρ(~r) = q[δ(~r − ~a) −
δ(~r)−δ(~r −~a−~b)+δ(~r −~b)] which gives p~ = 0 and Qij = 2q[~a·~bδij − 23 (ai bj +aj bi )].
Similar considerations also hold for the gravitational potential. Simply replace
1
→ G,
4πǫ0
q → m.
46
Chapter 5
Some Physical Examples of
Tensors: I
We now consider some examples
5.1
Polarisation of Media
• Electric
~ = ǫ0 E
~ + P~ ,
D
~ is the (total) electric field, D
~ is the external electric field (from free
where E
charges) called ‘electric displacement field’ and P~ is the polarisation vector
~ · P~ . We have
from (averaged) bound charges, ρb = −∇
~ = χij Ej + . . . ,
Pi = Pi (E)
for a linear, anisotropic field.
• Magnetic
~ = µ0 H
~ + µ0 M
~
B
~ is the external magnetic field (due to external currents), B
~ is (total)
where H
~ is the magnetic polarmagnetic field, called ‘magnetic induction field’ and M
~ ×M
~
isation vector from (averaged) bound charges/currents ~jb = ∂ P~ /∂t + ∇
~ = χij Hj + . . . ,
Mi = Mi (H)
again for a linear, anisotropic field.
For both cases χ is the susceptibility tensor (tensor by the quotient theorem).
47
5.2
Ohm’s Law
Current density:
~j def
= electrical charge transported through a unit area ⊥ to ~j per unit time,
~j
and so
~j = j~n ,
~n2 = 1 ,
where j = total charge to cross unit area in direction ~n per unit time.
The current is
def
i =
Z
~,
~j · dS
(or alternatively I for the current). The units are Cs−1 = A (amps).
Ohm’s Law:
~ applied to the material,
Current density is proportional to the electric field E
~,
~j = σ E
where σ is the conductivity.
~ = ~0 and σ = 0 is a (perfect) insulator
σ = ∞ is a (perfect) conductor where E
(for example a vacuum). We shall usually consider good conductors, such as copper
wire.
~
E
φ1
A
φ2
The potential difference is V = φ1 − φ2 (measured in volts, V ). Now E = V /l
(potential gradient) so j = σV /l and hence
Z
~ = jA = σ V A ,
i = ~j · dS
l
A
48
or
i=
V
R
with R =
l
.
σA
R is the resistance, measured in V /A or Ohms, Ω.
We can generalise Ohm’s Law by writing
ji = σij Ej ,
where σ is the conductivity tensor (by the quotient theorem).
• Layered material of conductor and insulator where the current can only flow
in the layers.
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0000000000000
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0000000000000
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0000000000000
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0000000000000
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z
y
x
1111111111111
0000000000000
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0000000000000
1111111111111


σ0 0 0
σ =  0 σ0 0 
0 0 0

 

j1
σ0 E1
 j2  =  σ0 E2  .
j3
0
giving
~
• Uniform material, no prefered direction, isototropic, so σij = σ0 δij and ~j = σ0 E
~ are always parallel to each other.
(the original Ohm’s Law when ~j and E
5.3
5.3.1
The Inertia Tensor
Angular momentum and kinetic energy
Suppose a rigid body of arbitrary shape rotates about a fixed axis with angular
velocity ~ω = ω~n. Consider a blob of matter dm at point P , with position vector
ω
P
O
~r
49
dm
~r relative to O. The velocity of a little element dm = ρdV where ρ ≡ ρ(~r) is the
(mass) density is
~v = ~ω × ~r .
Repeating previous results:
• Simple proof δr = δθr sin φ = |δθ~n × ~r| and so ~v = δ~r/δt = ω
~ × ~r where
~ω
δθ
~r
φ
~ω
~ω = δθ/δt~n.
• More complicated proof:
Use previous result for rotation matrix R(θ, ~n),
δri = Rij (δθ, ~n)rj − ri
= (−ǫijk nk δθ + O((δθ)2 ))rj ,
δθ
δri
= (~n × ~r)i ,
δt
δt
or ~v = ~ω × ~r again.
Angular momentum
~ of an element of mass dm = ρdV at ~r is d~h =
The angular momentum d~h (or dL)
ρ(~r)dV ~r × ~v giving
Z
~h =
ρ~r × (~ω × ~r)dV ,
body
giving
hi =
Z
ρǫijk rj ǫklm ωl rm dV
Z
ρ(δil δjm − δim δjl )rj ωl rm dV .
body
=
body
Thus
hi = Iij ωj ,
Iij =
Z
ρ(r 2 δij − ri rj )dV .
body
50
The geometric factor I(O) (where O is the origin) is the moment of Inertia tensor.
It is a tensor because ~h is pseudovector, ~ω is a pseudovector and hence from the
quotient theorem I is a tensor.
[Furthermore note that Iij is symmetric and independent of the ~n axis chosen.]
Kinetic energy
For the kinetic energy, T , we have dT = 21 (ρdV )(~ω × ~r)2 or
Z
1
ρǫijk ωj rk ǫilm ωl rm dV
T =
2 body
Z
1
=
ρ(δjl δkm − δjm δkl )ωj rk ωl rm dV
2 body
Z
1
=
ρ(ω 2 r 2 − (~r · ~ω )2 )dV
2 body
Z
1
ρ(r 2 δij − ri rj )dV ωi ωj ,
=
2 body
or
1
1
T = Iij ωi ωj ≡ ~ω I~ω .
2
2
Alternative (more familiar) forms
Often write
L =
T =
I (n) ω
1 (n) 2
ω
2I
with L = ~h · ~n ,
I (n) = Iij ni nj .
where I (n) is now the moment of inertia with respect to ~n,
Z
Z
(n)
2
2
2
I = Iij ni nj =
ρ(r − (~r · ~n) ) dV ≡
ρr⊥
dV ,
body
body
with r⊥ being the perpendicular distance from the ~n-axis.
Simpler if only interested in one ~n, but for a different ~n must re-compute; Iij contains
all the information.
Sometimes write I (n) = Mkn2 , where kn is called the radius of gyration (ie distance
of a point mass giving the same result).
51
An example
For example, a cube of side a of constant density, ρ with M = ρa3 ,
z
y
a
x
O
Iij (O) =
Z
ρ(r 2 δij − ri rj )dV .
This gives
So by symmetry
a
dxdydz (x2 + y 2 + z 2 ) − x2
10 3
a
= ρ 3 y xz + 13 z 3 xy 0 = 23 ρa5 = 23 Ma2 ,
Z a
= ρ
dxdydz(−xy)
0
a
= −ρ 21 x2 21 y 2 z 0 = − 41 ρa5 = − 14 Ma2 .
I11 = ρ
I12
Z

I(O) = Ma2 
5.3.2
2
3
− 14
− 14

− 41 − 14
2
1 
.
3 −4
1
2
−4
3
Parallel Axes Theorem
Often simpler to find the moment of inertia tensor about the centre of gravity, G,
rather than an arbitrary point, O. There is, however, a simple relationship between
~ =R
~ we have ~r = R
~ + ~r′ ,
them. Taking O to be the origin, and OG
ω
P
dm
~r′
O
~r
~
R
G
52