Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 3

Decompose each index with ${\delta }^{\rho }{}_{\mu }={\perp }^{\rho }{}_{\mu }-n^{\rho }{n}_{\mu }$. Since $n^{\nu }{n}_{\nu }=-1$, one has $n^{\nu }{\nabla }_{\mu }{n}_{\nu }=0$, so the second index of ${\nabla }_{\mu }{n}_{\nu }$ is spatial. Its spatial projection in the first index is $-K_{\mu \nu }$ by definition, while its normal projection isHence

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Use $n_{\mu }=-\alpha {\nabla }_{\mu }t$. Differentiating, contracting with $n^{\rho }$, and using $n^{\rho }{\nabla }_{\rho }t=1/\alpha$ gives a gradient of $\log \alpha$ plus a component parallel to $n_{\mu }$. The acceleration is orthogonal to $n^{\mu }$, so spatial projection removes the parallel part:

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Because $n^{\mu }{\partial }_{\mu }\varphi =-2\Pi$,Writing $\varphi ={\varphi }_R+i{\varphi }_I$ and $\Pi ={\Pi }_R+i{\Pi }_I$ givesEvery quantity on the right is real, so the Noether charge density is explicitly real valued.

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By definition,Thereforewith $n_{\alpha }{Q}^{\alpha }=0$.

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Taking the divergence of the Noether current,because the two gradient-product terms cancel. The field equation and its conjugate are $\square \varphi =V^{\prime }\varphi$ and $\square \bar{\varphi }=V^{\prime }\bar{\varphi }$, with real $V^{\prime }(|\varphi {|}^2)$. Hence

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Insert $J^{\mu }=Q{n}^{\mu }+Q^{\mu }$:The extrinsic-curvature convention gives ${\nabla }_{\mu }{n}^{\mu }=-K$. For a spatial vector,ThereforeThe constants are

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In adapted coordinates,Multiplying the conservation law by $\alpha$ givesEquivalently, combining the last two terms,This is the coordinate form of the 3+1 Noether-current conservation law.

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