4C
Vector Calculus
State the value of ∂xi /∂xj and find ∂r/∂xj , where r = |x|.
[1]
∂xi
= δij
∂xj
[1]
xj
∂
1
∂r
=
(xi xi )1/2 = 2xj (xi xi )−1/2 =
∂xj
∂xj
2
r
Vector fields u and v in R3 are given by u = rα x and v = k × u, where α is a
constant and k is a constant vector. Calculate the second-rank tensor dij = ∂ui /∂xj , and
deduce that ∇ × u = 0 and ∇ · v = 0.
∂
∂ui
=
(rα xi ) = αxj rα−2 xi + rα δij
∂xj
∂xj
[2]
(∇ × u)i = ijk
[2]
as ijk is antisymmetric in j and k while the term in brackets is
symmetric.
∇·v =
[2]
∂xk
= ijk αxj xk rα−2 + rα δjk = 0
∂xj
∂
(jkl kk ul ) = jkl kk αrα−2 xl xj + rα δjl = 0
∂xj
as jkl is antisymmetric in j and l while the term in brackets is
symmetric.
When α = −3, show that ∇ · u = 0 and
∇×v =
3(k · x)x − kr2
.
r5
∇ · u = αrα + rα δii = −3r−3 + 3r−3 = 0 if α = −3
(∇ × v)i =ijk
∂
(klm kl um )
∂xj
= (δil δjm − δim δjl ) kl −3r−5 xm xj + r−3 δmj
= − 3ki r−5 r2 + 3r−5 kj xi xj + 3r−3 ki − kj r−3 δij
= 3r−5 (k · x)x − kr−3 i
[2]
[10]
9C
Vector Calculus
Write down the most general isotropic tensors of rank 2 and 3. Use the tensor
transformation law to show that they are, indeed, isotropic.
[2]
λδij and µijk .
Check that Rli Rmj δij = Rli Rmi = (RT R)lm = δmn
[4]
and Rli Rmj Rnk ijk = lmn det R = lmn
Let V be the sphere 0 6 r 6 a. Explain briefly why
Z
xi1 . . . xin dV
Ti1 ...in =
V
is an isotropic tensor for any n.
A rotation of the basis is equivalent to a backward rotation of the sphere.
Spheres have no preferred direction, so the integral will still be T .
[2]
Z
Hence show that
Z
xi xj dV = αδij ,
xi xj xk dV = 0
V
V
Z
and
xi xj xk xl dV = β(δij δkl + δik δjl + δil δjk )
V
for some scalars α and β, which should be determined using suitable contractions of the
indices or otherwise.
[You may assume that the most general isotropic tensor of rank 4 is
λδij δkl + µδik δjl + νδil δjk ,
where λ, µ and ν are scalars.]
[1]
R
These are all isotropic and symmetric
R in the indicies.
R 4 So V xi xj dV
must be α δij . Contract i = j: 3α = V xi xi dV = V r sin θ drdθdφ =
4πa5 /5 so α = 4πa5 /15
R
By isotropy,
R V xi xj xk dV = µijk . But the RHS is antisymmetric, so
µ = 0 and V xi xj xk dV = 0.
[2]
By isotropy and hint, the integral must be λδij δkl + µδik δjl + νδil δjk .
Check symmetry
i ↔ j swaps µ ↔ ν so must have µ = ν and similarly
R
µ = λ. So V xi xj xk xl dV = β(δij δkl + δik δjl + δil δjk ) for some β.
[3]
Contract i = j and k = l: xi xi xk xk = r4
Z
r4 r2 sin θdrdθdφ =
V
[4]
4πa7
7
and
β(δii δkk + δik δik + δik δik ) = β(9 + 3 + 3)
so β = 4πa7 /105.
Deduce the value of
Z
x × (Ω × x) dV ,
V
where Ω is a constant vector.
4πa5
ijk xj klm Ωl xm dV =(δil δjm − δim δjl )Ωl
δjm
15
4πa5
Ωi 8πa5
=(3δil − δil )Ωl
=
15
15
Z
V
[2]
so
R
V
x × (Ω × x) dV = 8πa5 Ω/15
[20]
11C Vector Calculus
The electric field E(x) due to a static charge distribution with density ρ(x) satisfies
E = −∇φ ,
∇·E=
ρ
,
ε0
(1)
where φ(x) is the corresponding electrostatic potential and ε0 is a constant.
(a) Show that the total charge Q contained within a closed surface S is given by
Gauss’ law
Z
Q = ε0 E · dS .
S
Assuming spherical symmetry, deduce the electric field and potential due to a point charge
q at the origin i.e. for ρ(x) = q δ(x).
Z
[2]
Q=
Z
V
Z
∇ · E dV = 0
ρ dV = 0
V
n · E dS by divergence theorem.
S
ρ(x) = qδ(x) and E = E(r)n. Hence by Gauss’ law q = 4πr2 0 E(r).
[4]
So E(r) = q/4π0 r2 and φ = q/4π0 r.
(b) Let E1 and E2 , with potentials φ1 and φ2 respectively, be the solutions to (1)
arising from two different charge distributions with densities ρ1 and ρ2 . Show that
Z
Z
Z
Z
1
1
φ1 ρ2 dV +
φ1 ∇φ2 · dS =
φ2 ρ1 dV +
φ2 ∇φ1 · dS
(2)
ε0 V
ε0 V
∂V
∂V
for any region V with boundary ∂V , where dS points out of V .
[5]
∇ · (φ2 ∇φ1 − φ1 ∇φ2 ) = ∇φ2 · ∇φ1 + φ2 ∇2 φ1 − ∇φ1 · ∇φ2 − φ1 ∇2 φ2
= φ2 ∇2 φ1 − φ1 ∇2 φ2 and (1) gives ∇2 φ = −ρ/ε0 . Hence
Z
Z
ρ2
ρ1
(φ2 ∇φ1 − φ1 ∇φ2 ) · n dS =
−φ2 + φ1 dV
⇒ answer
ε
ε0
0
S
V
(c) Suppose that ρ1 (x) = 0 for |x| 6 a and that φ1 (x) = Φ, a constant, on |x| = a.
Use the results of (a) and (b) to show that
Z
ρ1 (x)
1
Φ=
dV .
4πε0 r>a r
[You may assume that φ1 → 0 as |x| → ∞ sufficiently rapidly that any integrals over the
‘sphere at infinity’ in (2) are zero.]
[9]
[20]
Take V = {r > a} and ρ2 = qδ(x), so that φ2 = q/4πε0 r. Now part
(b) gives
Z
Z
Z
Z
q
1
q
1
q
φ1 ρ2 dV +
Φ
dS =
ρ1 dV +
∇φ1 · ndS
2
ε0 V
4πε
a
ε
4πε
r
4π
0
0 V
0
0a
r=a
r=a
R
But ρ2 = 0 in V , and r=a ∇φ1 · n dS = Q1 /ε0 = 0 by part (a) so
Z
Φq
q
1
=
ρ1 dV
⇒ answer
2
ε0
4πε0 V r