Past exam of the theoretical physics I course of the University of Cambridge / 2019 / 1

Use Minkowski metric $g_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ and units with the speed of light equal to one. Thus $u^{\mu }{u}_{\mu }=-1$, and $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }{u}_{\nu }$ is the spatial projector of a timelike congruence. Keep the medium velocity fixed when varying the electromagnetic four-potential. Since the electromagnetic field tensor is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, its variation givesBoth tensors are antisymmetric. Integration by parts, with variations vanishing at the boundary, therefore givesThe Euler-Lagrange field equation is ${\partial }_{\mu }(F^{\mu \nu }-M^{\mu \nu })=0$. Renaming the free index and using antisymmetry givesThese are the inhomogeneous Maxwell equations with the bound charge and bound current encoded by the polarization-magnetization tensor.

There is a sign inconsistency in the original PDF: varying its action gives a plus sign for the spatially projected magnetic term, whereas its displayed target equation has a minus sign. The following Lagrange multiplier for antisymmetry calculation derives the result of the action as printed.

Temporarily vary all sixteen entries $M^{\mu \nu }$ independently and addVariation of the symmetric Lagrange multiplier imposes $M^{\mu \nu }+M^{\nu \mu }=0$. DefineOn this constraint surface, $C_{\nu \mu }=-C_{\mu \nu }$. Differentiating the original contractions before imposing the constraint givesandIn particular, the reversed order $\sigma \rho$ in the last factor of the action matters. The unrestricted Euler-Lagrange field equation, evaluated on the antisymmetry constraint, isSubtract its index-transposed equation to eliminate the symmetric multiplier:A direct check is to restrict the action to a rest-frame magnetic component $M^{12}=m$, $M^{21}=-m$: its quadratic contribution is $+m^2/(2\chi )$, whose derivative has a positive sign. To obtain the PDF's target equation instead, one must reverse the sign of that quadratic term in the action, or replace its last $M^{\sigma \rho }$ by $M^{\rho \sigma }$. These are alternative repairs; changing conventions cannot make the printed action and printed equation agree.

Choose $A^{\mu }=(\Phi ,\mathbf{A})$, so that $F_{i0}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$, with ${ϵ}_{123}=1$. The definitions of the polarization-magnetization tensor imply $M^{0i}=-P_i$ and $M^{ij}={ϵ}_{ijk}{M}_k$. Separating temporal and spatial indices,The Levi-Civita symbol contraction equals $2{\delta }_{k\ell }$, givingThus $\mathbf{P}$ is the electric dipole moment per volume, or electric polarization, and $\mathbf{M}$ is the magnetic dipole moment per volume, or magnetization. The signs are those of dipole interactions in a Lagrangian; the associated fixed-dipole potential energies have the opposite signs. Consistently, the equation in part (a) has componentsso ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\mathbf{J}}_b={\partial }_t\mathbf{P}+\nabla \times \mathbf{M}$ are the bound charge and bound current.

In the rest frame $u^{\mu }=(1,0,0,0)$ and $h_{\mu \nu }=\mathrm{diag}(0,1,1,1)$. The two material contractions in the printed action become $-{\mathbf{P}}^2$ and $-2{\mathbf{M}}^2$. Consequently its material Lagrangian density isThe Euler-Lagrange field equations are algebraic, since no derivatives of the material variables occur. They giveThese also follow from the $i0$ and $ij$ components of the corrected equation in part (b). If instead the intended starting point was the PDF's displayed equation in part (b), its predictions would beThe two magnetic answers express the documented sign error, rather than different choices of frame. Here the inverse coefficients in the action require nonzero $\kappa ,\chi$; zero response is obtained as a limiting constitutive law.

The constants specify isotropic, nondispersive linear response: $\kappa$ controls electric polarization, and $\chi$ controls magnetization. Their numerical relation to conventional electric susceptibility and magnetic susceptibility depends on the signs and field normalization in the action. In the rationalized units used here, the printed action gives the constitutive relationsThus its relative permittivity and magnetic permeability areHere conventional ${\chi }_m$ is defined by $\mathbf{M}={\chi }_m\mathbf{H}$, not by $\mathbf{M}={\chi }_m\mathbf{B}$. With the alternative repaired action that yields the PDF's target equation, ${\epsilon }_r$ is unchanged but ${\mu }_r=(1-\chi {)}^{-1}$ and ${\chi }_m=\chi /(1-\chi )$. In particular, a positive conventional electric susceptibility corresponds to negative $\kappa$ in the printed conventions; the problem does not specify signs for these constants.