Paper 3, Section II, B

Vector Calculus | Part IA, 2009

A second-rank tensor T(y)T(\mathbf{y}) is defined by

Tij(y)=S(yixi)(yjxj)yx2n2 dA(x)T_{i j}(\mathbf{y})=\int_{S}\left(y_{i}-x_{i}\right)\left(y_{j}-x_{j}\right)|\mathbf{y}-\mathbf{x}|^{2 n-2} \mathrm{~d} A(\mathbf{x})

where y\mathbf{y} is a fixed vector with y=a,n>1|\mathbf{y}|=a, n>-1, and the integration is over all points x\mathbf{x} lying on the surface SS of the sphere of radius aa, centred on the origin. Explain briefly why TT might be expected to have the form

Tij=αδij+βyiyjT_{i j}=\alpha \delta_{i j}+\beta y_{i} y_{j}

where α\alpha and β\beta are scalar constants.

Show that y(yx)=a2(1cosθ)\mathbf{y} \cdot(\mathbf{y}-\mathbf{x})=a^{2}(1-\cos \theta), where θ\theta is the angle between y\mathbf{y} and x\mathbf{x}, and find a similar expression for yx2|\mathbf{y}-\mathbf{x}|^{2}. Using suitably chosen spherical polar coordinates, show that

yiTijyj=πa2(2a)2n+2n+2y_{i} T_{i j} y_{j}=\frac{\pi a^{2}(2 a)^{2 n+2}}{n+2}

Hence, by evaluating another scalar integral, determine α\alpha and β\beta, and find the value of nn for which TT is isotropic.

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