Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 34A (Electrodynamics)

Use $x^0=ct$. The stress-energy tensor component $T^{00}$ is energy per unit volume; $c{T}^{i0}$ is energy flux in direction $i$; $T^{0j}/c$ is momentum density in direction $j$; and $T^{ij}$ is the flux of momentum component $j$ through a surface normal to direction $i$. In a symmetric tensor, energy flux equals $c^2$ times momentum density. Thus ${\partial }_{\mu }{T}^{\mu 0}=0$ is energy conservation and ${\partial }_{\mu }{T}^{\mu j}=0$ is momentum conservation.

Differentiate the electromagnetic stress-energy tensor. The current term isThe other two terms cancel by the homogeneous Maxwell equations. To see the contraction explicitly, multiply ${\partial }_{\mu }{F}_{\beta \alpha }+{\partial }_{\beta }{F}_{\alpha \mu }+{\partial }_{\alpha }{F}_{\mu \beta }=0$ by $F^{\mu \alpha }$. The first and third contracted terms are equal after swapping indices and using antisymmetry; the middle is $-\frac{1}{2}{\partial }_{\beta }(F^{\mu \alpha }{F}_{\mu \alpha })$. ThereforeIt cancels differentiation of the trace term, leavingThe sign is tied to the printed convention ${\partial }_{\mu }{F}^{\mu \nu }=-{\mu }_0{J}^{\nu }$ and the displayed field-strength tensor.

In the local rest frame the four-velocity is $(c,0,0,0)$, the current is $(q{n}_{0}c,0,0,0)$, and the pressureless particle fluid has only $T^{00}=m{n}_0{c}^2$ nonzero. Covariance therefore givesThe absence of random rest-frame velocities is what makes this a dust stress-energy tensor rather than a fluid with pressure. The Lorentz four-force on a particle and its equation of motion are $f^{\nu }=q{F}^{\nu }{}_{\mu }{u}^{\mu }$ and $m{u}^{\mu }{\partial }_{\mu }{u}^{\nu }=q{F}^{\nu }{}_{\mu }{u}^{\mu }$.

Particle-number conservation gives ${\partial }_{\mu }(n_0{u}^{\mu })=0$. It also follows from charge conservation ${\partial }_{\mu }{J}^{\mu }=0$ when $q\ne 0$, obtained by antisymmetric double differentiation of Maxwell's equations. ConsequentlyAdding part (i) proves ${\partial }_{\mu }(T_{\mathrm{em}}^{\mu \nu }+T_{\mathrm{fluid}}^{\mu \nu })=0$. Energy and momentum exchanged with the field cancel in the total tensor.