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With the momentum-flux sign convention used for the Maxwell stress tensor, Maxwell's equations implywhere $\mathbf{g}$ is the electromagnetic momentum density and ${\mathbf{f}}_{\mathrm{L}}$ is the Lorentz force density exerted by the field on matter. Thus total field-plus-matter momentum is locally conserved.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

Use $\dot{M}>0$ for inward mass accretion rate, and $T_m>0$ for magnetic torque adding angular momentum to the disk. Combining the continuity equation with conservation of angular momentum identifies the inward mass flux as $(dh/dr{)}^{-1}d\mathcal{G}/dr$. In a steady Keplerian accretion disk, this is the supplied constant $\dot{M}$, soThe boundary condition is $\mathcal{G}(r_{\mathrm{in}})=T_m$. Since $d\Omega /dr=-3\Omega /(2r)$, the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Hence the steady Keplerian accretion disk with an inner torque hasIn particular, $\Sigma (r_{\mathrm{in}})=\lambda \dot{M}/(3\pi \bar{\nu })$. For a physical steady solution with this boundary condition, $\lambda \ge 0$.

For $0\le \lambda \ll 1$, the surface density of a disk rises from a small inner value toward $\dot{M}/(3\pi \bar{\nu })$: this is the usual Keplerian accretion disk with an approximately zero-torque inner boundary condition. For $\lambda \gg 1$, the inner surface density of a disk is large and decreases approximately as $r^{-1/2}$, giving a torque-dominated accretion disk. This approximation applies where $r/r_{\mathrm{in}}\ll (\lambda -1{)}^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

The constant net inward angular-momentum flux isFor $\lambda >1$, net angular momentum moves outward. The strong-torque profile is therefore decretion disk-like in its stress-dominated structure, but the specified mass flux remains inward: a genuine net decretion disk requires an outward mass flux, not merely large $\lambda$.

At $r_{\mathrm{in}}=r_m$, the Maxwell stress tensor estimate givesUsing the magnetospheric truncation radius scaling in part (b), all dimensional parameters cancel:If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi )$; retaining $\Sigma \simeq 2H\rho$ and exact free-fall speed throughout gives $\sqrt{2}/(4\pi )$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda \gg 1$.

Let $\mathbf{\sigma }$ be the stated Maxwell stress tensor. Taking its divergence and usinggivesThe Maxwell equations in vacuum with sources areSubstitution, together with antisymmetry of the cross product, yieldsTherefore the required vector is the electromagnetic momentum densityandThis is local conservation of electromagnetic momentum. The tensor component ${\sigma }_{ij}$ is the flux of $i$-momentum in the $j$ direction in the convention of the question, $\mathbf{g}$ is field momentum per unit volume, and $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ is the Lorentz force density transferred to matter.

Differentiate $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, use the vacuum Maxwell equations, and use $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$. Componentwise vector identities then givewhereThis is the negative of the conventional Maxwell stress tensor; with the sign used here, its divergence is the outward momentum flux.

Differentiating $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation lawMoving the divergence to the left identifies