Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 6

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

Use three components of the proposed isotropic tensor form. Since $T_{1122}=0$, one gets $\alpha =0$. Since $T_{1212}={ϵ}_{123}^2=1$, one gets $\beta =1$. Finally $T_{1221}={ϵ}_{123}{ϵ}_{213}=-1$ gives $\gamma =-1$. Hence contraction of two Levi-Civita symbols gives

The integration domain is invariant under every rotation matrix, and its volume element is unchanged. The substitution $\mathbf{y}=R\mathbf{x}$ therefore provesThus $I$ is an isotropic tensor. It is also completely symmetric in its four indices, because the coordinate factors commute. In the general isotropic form, exchanging indices $j,k$ equates $\alpha$ and $\beta$, and exchanging $j,l$ equates $\alpha$ and $\gamma$. Therefore

The contraction in the PDF is $I_{iijj}$. By Einstein notation, its integrand is $({\sum }_i{x}_i^2)({\sum }_j{x}_j^2)=r^4$. In spherical coordinates,

Contract the isotropic expression from (i). Since ${\delta }_{ii}=3$ and ${\delta }_{ij}{\delta }_{ij}=3$,The value in (ii) gives $A=4\pi /105$, so the isotropic fourth moment of a ball isFor example ${\int }_B{x}_1^{4}dV=4\pi /35$ and ${\int }_B{x}_1^2{x}_2^{2}dV=4\pi /105$, illustrating the factor of three between these two fourth moments.

Write the inverse tensor asInsert this into $({\delta }_{ik}+\lambda X_{ik})Y_{kj}={\delta }_{ij}$ and equate powers of $\lambda$. One obtains successivelyThus the Neumann series givesEvery contraction is an ordinary ordered matrix product. No transpose or symmetry assumption is used. Multiplying the truncated series on either side by $I+\lambda X$ leaves $I-{\lambda }^4{X}^4$, which verifies the error order. For a fixed finite matrix this expansion converges whenever $|\lambda |\Vert X\Vert <1$ in a submultiplicative matrix norm.