Past exam of the relativity course of the University of Cambridge / 2017 / 1

Use the Christoffel symbol formula with its first index raised:The diagonal metric tensor has $g_{rr}=e^{2\chi }$ and $g_{\theta \theta }=r^2{e}^{2\chi }$, with inverse components $g^{rr}=e^{-2\chi }$ and $g^{\theta \theta }=r^{-2}{e}^{-2\chi }$. Since ${\partial }_r\chi ={\chi }^{\prime }$ and all angular derivatives vanish,For example, ${\Gamma }^r{}_{\theta \theta }=-\frac{1}{2}{e}^{-2\chi }{\partial }_r(r^2{e}^{2\chi })$. The other components vanish. The formula supplied in the paper gives the first-kind coefficients ${\Gamma }_{ijk}$; multiplication by the inverse metric tensor is needed for the requested second-kind coefficients. These polar coordinates are understood away from $r=0$.

For a scalar field, ${\nabla }_{\mu }\varphi ={\partial }_{\mu }\varphi$. Apply the product rule to the stress-energy tensor, using metric compatibility to pass the covariant derivative through $g^{\mu \nu }$:The middle terms cancel because the Levi-Civita connection gives a symmetric Hessian of a scalar field. Thus, with the sign conventions of this question, the scalar stress-energy divergence identity becomesThe last equality is the Klein-Gordon equation. This stress-energy conservation follows from the field equation and metric compatibility; it does not require imposing the Einstein field equations.

Take metric signature $(+,-,-,-)$ and the supplied covariant electromagnetic field tensor, so $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. Raising both indices reverses the time-space components but leaves the spatial components unchanged. Its complete contraction is thereforeDefine the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${\epsilon }^{0123}=+1$. Direct contraction givesThese electromagnetic field invariants are complete tensor contractions. The second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under spatial inversion.

For $T^{0i}$ the metric trace term vanishes. Since $F^0{}_j=E_j$,Consequently the Poynting vector is exactly the time-space row in the stated $c=1$ units: