Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 2 / c

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- Solution c

Solution

The Lie derivative of the spatial covector

j_{α}

along

n^{μ}

is

L_{n} j_{α} = n^{μ} \nabla_{μ} j_{α} + j_{μ} \nabla_{α} n^{μ} .

To show that it is spatial, contract with

n^{α}

. Differentiating

n^{α} j_{α} = 0

along

n^{μ}

gives

n^{α} n^{μ} \nabla_{μ} j_{α} = - a^{α} j_{α},

whereas

n^{α} j_{μ} \nabla_{α} n^{μ} = j_{μ} a^{μ} .

The terms cancel, so

n^{α} L_{n} j_{α} = 0.

A spatial projector therefore acts trivially:

⊥^{α}_{β} L_{n} j_{α} = L_{n} j_{β} .

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