Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 6

For a Cartesian coordinate change

x_{i}^{'} = R_{ij} x_{j}

with

R

an orthogonal matrix, a rank-

n

tensor obeys

T_{i_{1} \dots i_{n}}^{'} = R_{i_{1} j_{1}} \dots R_{i_{n} j_{n}} T_{j_{1} \dots j_{n}},

(1)

with repeated indices summed according to Einstein notation. A tensor is isotropic if

T_{i_{1} \dots i_{n}}^{'} = T_{i_{1} \dots i_{n}}

for every proper rotation matrix

R

. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.

b

 0  0

Solution

 0  0

Under a proper rotation matrix

R

, the product of coordinates transforms as a rank-

n

tensor. The unit ball is unchanged by the substitution

y = R x

, whose Jacobian determinant is one. Hence

R_{i_{1} j_{1}} \dots R_{i_{n} j_{n}} S_{j_{1} \dots j_{n}} = \int_{B} (R x)_{i_{1}} \dots (R x)_{i_{n}} d V_{x} = \int_{B} y_{i_{1}} \dots y_{i_{n}} d V_{y} = S_{i_{1} \dots i_{n}} .

(1)

Thus

S

is an isotropic tensor. An isotropic second-rank tensor has the form

S_{ij} = A δ_{ij}

. Taking the trace and using spherical coordinates,

3 A = \int_{B} ∣ x ∣^{2} d V = 4 π \int_{0}^{1} r^{4} d r = \frac{4 π}{5} .

(2)

Therefore the isotropic second moment of a ball is

S_{ij} = \frac{4 π}{15} δ_{ij} .

(3)

An isotropic third-rank tensor has the form

B ϵ_{ijk}

, but

S_{ijk}

is symmetric in every pair of indices, forcing

B = 0

. Alternatively,

x \mapsto - x

makes its integrand odd on the ball. Thus

S_{ijk} = 0.

(4)

c

 0  0

Solution

 0  0

Choose an orthonormal basis with

n = e_{3}

. A half-turn about

n

kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, is

σ_{ij} = a (δ_{ij} - n_{i} n_{j}) + d n_{i} n_{j} + h ϵ_{ikj} n_{k},

(1)

with real constants

a, d, h

. The Kronecker delta, the outer product

n_{i} n_{j}

, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about

n

, so this form is also sufficient. In the chosen basis,

σ = a h 0 - h a 0 00 d .

(2)

The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as

E = E_{⊥} + E_{∥}

, where

E_{∥} = (n \cdot E) n

. The electric current density is

J = a E_{⊥} + d E_{∥} + h n \times E_{⊥} .

(3)

For a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

If $E_{⊥} = 0$ , then $J = d E$ , with no additional symmetry required.
If $E_{∥} = 0$ , then $J ∥ E ⟺ h = 0$ . Within this family $h = 0$ is exactly the condition that the conductivity tensor is symmetric; then $J = a E$ . An additional half-turn symmetry about a transverse axis is one way to force $h = 0$ .
If both components are nonzero, then $J ∥ E ⟺ h = 0 and a = d$ . These conditions give the isotropic second-rank tensor $σ = a I$ , and then $J = a E$ for every direction.