Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 4

Under a gauge transformation, the electromagnetic field tensor changes byThe partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

A Lorentz transformation preserves the quadratic form of the Minkowski metric for every four-vector. Substituting $x^{\prime \sigma }={\Lambda }^{\sigma }{}_{\mu }{x}^{\mu }$ givesThe coefficient difference is a symmetric matrix, and a symmetric quadratic form that vanishes on every vector is zero: evaluating it on basis vectors and their sums recovers each coefficient. ConsequentlyThis is the defining metric-preservation relation of the Lorentz group.

Insert ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+\alpha {\omega }^{\mu }{}_{\nu }$ into the Lorentz group condition and keep the coefficient of $\alpha$. It givesDefine ${\omega }_{\mu \nu }={\eta }_{\mu \rho }{\omega }^{\rho }{}_{\nu }$. The displayed equation is ${\omega }_{\nu \mu }+{\omega }_{\mu \nu }=0$. Raising both indices with the symmetric Minkowski metric preserves this antisymmetry:These are the generators of the Lie algebra of the Lorentz group. In particular ${\omega }^{\mu }{}_{\mu }=0$. The mixed-index matrix need not be antisymmetric in the ordinary Euclidean transpose; it is antisymmetric after lowering an index with $\eta$.

For the active Lorentz transformation of a vector field, coordinates are held fixed and $A^{\prime \mu }(x)={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }({\Lambda }^{-1}x)$. Its field strength transforms as a two-index Lorentz tensor. The index contractions in the Maxwell Lagrangian are invariant by ${\Lambda }^T\eta \Lambda =\eta$, so the density transforms as a Lorentz scalar:For $\Lambda =I+\alpha \omega$, its inverse is $I-\alpha \omega +O({\alpha }^2)$. Expanding the coordinate argument therefore givesSet $v^{\mu }(x)=\alpha {\omega }^{\mu }{}_{\nu }{x}^{\nu }$. Part (iii) implies ${\partial }_{\mu }{v}^{\mu }=\alpha {\omega }^{\mu }{}_{\mu }=0$. The product rule then converts the variation into a total derivative:Its integral changes the action only by a boundary term, which is the condition needed for Noether's theorem.

For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_{\rho }=\beta {ϵ}^{\nu }{\partial }_{\nu }{A}_{\rho }$, and $\delta \mathcal{L}={\partial }_{\mu }(\beta {ϵ}^{\mu }\mathcal{L})$. Differentiating the Maxwell Lagrangian with respect to the field gradient givesThus Noether's theorem yields $j^{\mu }={ϵ}_{\nu }{T}^{\mu \nu }$, where the canonical stress-energy tensor of the Maxwell field isIt is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,need not vanish. For example, take $A_0=a{x}^1$, $A_2=b{x}^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.

Nor is it gauge-invariant: while $F$ and $\mathcal{L}$ are unchanged, a gauge transformation giveswhich is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the canonical stress-energy tensor.

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are