Past exam of the symmetry and relativity course of the University of Oxford / 2022 / 1

Write the Lorentz transformation as $x^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{x}^{\nu }$. Acting on a scalar field, the chain rule gives the four-gradientFor the specified Lorentz boost, the inverse is obtained by changing $\beta$ to $-\beta$, soThese are precisely the covector component rules for the lowered coordinates. Raising the index with the Minkowski metric ${\eta }^{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ reverses the time component, yieldingIndeed ${\partial }^{\prime 0}=\gamma ({\partial }^0-\beta {\partial }^1)$ and ${\partial }^{\prime 1}=\gamma ({\partial }^1-\beta {\partial }^0)$, the contravariant four-vector rules. Thus the index on a partial derivative with respect to an upper coordinate is lower.

The four-gradient rules from a giveThe Lorentz transformation coefficients are constant, so the partial derivatives do not differentiate them. This proves the Lorentz invariance of the d'Alembert operator:For example, the two longitudinal terms transform as $-{\gamma }^2({\partial }_0+\beta {\partial }_1{)}^2+{\gamma }^2({\partial }_1+\beta {\partial }_0{)}^2=-{\partial }_0^2+{\partial }_1^2$, using ${\gamma }^2(1-{\beta }^2)=1$ and commuting partial derivatives. The transverse terms are unchanged. Applied to a scalar field, $\square$ therefore gives a Lorentz scalar.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.

Let $s=k^{\mu }{x}_{\mu }$. The Lorenz gauge condition on the plane electromagnetic wave givesThe vacuum wave equation gives $k_{\mu }{k}^{\mu }=0$ for a nonzero amplitude; equivalently $k^0=\omega /c$ and $|\mathbf{k}|=\omega /c$. Use the gauge transformation $A^{\prime \mu }=A^{\mu }+{\partial }^{\mu }\chi$ withBecause $\square \chi =0$, this is a residual Lorenz gauge symmetry and $k_{\mu }{ϵ}^{\prime \mu }=0$. For a propagating plane electromagnetic wave, $k^0\ne 0$, so take $\lambda =-{ϵ}^0/k^0$. ThenThe second equality follows from $-k^0{ϵ}^{\prime 0}+\mathbf{k}\cdot {\mathbf{ϵ}}^{\prime }=0$. The electric field and magnetic field are unchanged by this gauge transformation. Thus Lorenz gauge permits a representative in radiation gauge; Lorenz gauge alone does not require the spatial amplitude of every representative to be transverse.