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 Incoming links: Cartesian moments of a three-dimensional ball

Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 9C / Solution Created 2026-09-24 Updated 2026-10-14

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Here isotropic tensors are invariant under proper rotation matrices,

R^{T} R = I

and

det R = 1

. The isotropic second-rank tensor and isotropic third-rank tensor forms in three dimensions are

T_{ij} = A δ_{ij}, T_{ijk} = B ϵ_{ijk},

(1)

with scalar coefficients

A, B

. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two tensor, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of

(1, 2, 3)

survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the parity of a permutation. This gives a multiple of the Levi-Civita symbol.

The tensor transformation law verifies invariance of these forms:

T_{ij}^{'} = A R_{i p} R_{j q} δ_{pq} = A δ_{ij}, T_{ijk}^{'} = B R_{i p} R_{j q} R_{k r} ϵ_{pq r} = B (det R) ϵ_{ijk} = B ϵ_{ijk} .

(2)

The first equality uses orthogonality and the second uses the determinant formula. If reflections were also required while using the ordinary tensor transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which

B ϵ_{ijk}

is an isotropic tensor.

The ball is unchanged by any rotation matrix, and rotating coordinates has Jacobian determinant one. Thus changing variables

y = R x

gives

R_{i_{1} j_{1}} \dots R_{i_{n} j_{n}} \int_{V} x_{j_{1}} \dots x_{j_{n}} d V = \int_{V} y_{i_{1}} \dots y_{i_{n}} d V = T_{i_{1} \dots i_{n}} .

(3)

These moments are therefore isotropic tensor integrals. For the Cartesian moments of a three-dimensional ball, the second moment is

α δ_{ij}

. Taking its tensor contraction gives

3 α = \int_{V} r^{2} d V = 4 π \int_{0}^{a} r^{4} d r = \frac{4 π a ^{5}}{5}, α = \frac{4 π a ^{5}}{15} .

(4)

The third moment vanishes:

x \mapsto - x

preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric Levi-Civita symbol.

The fourth moment is symmetric under every permutation of its four indices. The general rank-four isotropic tensor form consequently has equal coefficients, yielding