Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 12 / a

Using the Einstein summation convention, reorder the scalar components and sum over the two dummy indices separately:

a_{i} b_{k} b_{i} a_{k} = (a_{i} b_{i}) (a_{k} b_{k}) = (a \cdot b)^{2} .

The repeated suffixes each represent the three-dimensional component sum.

ii

Contract the Kronecker deltas in the first term to obtain

a_{i} b_{i}

. In the second term,

δ_{ij}

identifies the two

a

components, while the two

b_{k}

components already contract. Therefore

a_{i} b_{k} δ_{km} δ_{im} + a_{i} b_{k} a_{j} b_{k} δ_{ij} = a \cdot b + (a \cdot a) (b \cdot b) .

In three dimensions each contracted delta is the trace of the identity,

δ_{jj} = δ_{kk} = 3

. Thus the exponent is six and the base is

a \cdot a

. Hence

(a_{i} a_{i})^{δ_{jj} + δ_{kk}} = (a \cdot a)^{6} = ∣ a ∣^{12} .

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