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The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

The calculation in part (a) leaves the cross-helicity conservation law sourceAt positive temperature, this vanishes exactly when the magnetic field is tangent to constant-specific entropy surfaces: $\mathbf{B}\cdot \nabla s=0$. A homentropic flow is a sufficient special case; uniform entropy throughout space is not necessary. Mere advection of specific entropy, $Ds/Dt=0$, constrains its variation along the velocity rather than along the magnetic field, so it does not by itself eliminate this source. To conserve total cross-helicity in a volume, the net boundary flux must also vanish.

The remaining magnetic field has $\mathbf{B}=B_{\varphi }(R){\mathbf{e}}_{\varphi }+B_z(R){\mathbf{e}}_z$. The radial Lorentz force density separates into a magnetic pressure gradient and inward magnetic tension:Here $(\mathbf{B}\cdot \nabla )\mathbf{B}$ has radial component $-B_{\varphi }^2/R$ because ${\partial }_{\varphi }{\mathbf{e}}_{\varphi }=-{\mathbf{e}}_R$. Thus magnetostatic equilibrium givesIntegrating the axial component of the magnetostatic Ampère-Maxwell equation gives $I(R)=cR{B}_{\varphi }/2$, since regularity removes the integration constant. Substituting $B_{\varphi }=2I/(cR)$ combines the toroidal terms intoThis is the cylindrical magnetostatic pressure balance equation in terms of enclosed electric current.

In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let $v(r)>0$ denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state giveUsing the adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$ and $A\propto r^3$,The Bernoulli equation iswhere the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

Exoplanet transit photometry discovers a giant planet through a periodic flux decrement and measures $R_p/R_{\star }\simeq \sqrt{\delta }$. Stellar-radius estimates turn this into a planetary radius. The radial-velocity method discovers the stellar orbital reflex motion and constrains $M_p\sin i$. Follow-up exoplanet transit photometry is required to measure a geometric radius when such a planet also transits; a radial-velocity detection alone gives no direct size. Alternatively, exoplanet direct imaging finds young luminous giants, whose sizes are inferred less directly from luminosity, temperature, distance, and a planetary mass-radius relation or atmosphere model.

The two broad explanations for hot-Jupiter radius inflation are retention of primordial heat and addition of new interior power.

Delayed cooling of an inflated giant planet can arise from enhanced atmospheric opacity, which slows radiative escape; an irradiation-maintained radiative blanket, which insulates the convective interior; or compositional stratification and inefficient layered convection, which inhibit the outward transport of heat. These alter the rate of Kelvin-Helmholtz contraction.

Heating of an inflated giant planet can arise from tidal heating maintained by eccentricity or obliquity; Joule heating of currents driven by atmospheric winds through a magnetic field; or downward transport and dissipation of atmospheric mechanical energy generated by irradiation. To affect radius, the energy must be deposited at a depth and rate that changes the interior cooling balance. Simply absorbing starlight high in the atmosphere does not automatically supply deep heating. These are proposed mechanisms with different efficiencies, not six universally established contributions in every inflated planet.

Using $K^2=2q{\mu }_0\rho {\Omega }^2/B_z^2$, the finite-thickness magnetorotational instability criterion is preciselyFor the midplane-symmetric equilibrium with $B_{+}\ne 0$, the absolute radial magnetic field isIts first maximum occurs at $z_{\ast }=\pi /(2K)$. The instability condition puts $z_{\ast }$ strictly inside the disc: $z_{\ast }<H$. The magnitude rises from zero to this maximum, then decreases on a nonempty interval before the surface. Thus the unstable equilibrium exhibits nonmonotonic magnetic bending, as illustrated below.

The link between bending and instability concerns a nonzero imposed surface inclination. If $B_{+}=0$, the unbent equilibrium $B_x=0$ can still have magnetorotational instability; the literal nonmonotonic-bending conclusion does not apply to that degenerate case. Resonances with $\sin (KH)=0$ also require the separate equilibrium analysis in part (b).

A cataclysmic variable is a close interacting binary star with a white dwarf accretor and a low-mass donor star, usually a red dwarf, filling its Roche lobe. Matter crosses by Roche-lobe overflow and commonly forms an accretion disk; sufficiently strong white-dwarf magnetism instead permits magnetically channelled accretion. Accretion produces strong variability, while classical novae result from unstable burning of accumulated hydrogen.

The cataclysmic-variable orbital-period distribution contains a short-period population between about 82 minutes and two hours, a pronounced cataclysmic-variable period gap around two to three hours, and a longer-period population above the gap. There is an accumulation near the cataclysmic-variable period minimum. These describe ordinary hydrogen-rich systems; evolved donors and helium-rich systems can have shorter periods. Selection by luminosity and outburst activity affects the observed counts, so $N(P)$ is not simply the intrinsic distribution.

For the contact red dwarf, $R_2=R_L\propto a(M_2/M{)}^{1/3}$. Using its supplied thermal-equilibrium mass-radius relation, $R_2\propto M_2$, gives $a\propto M_2^{2/3}{M}^{1/3}$. Kepler's third law then givesindependent of $M_1$. This is a special case of the Roche-lobe-filling period-density relation.

To examine rapid mass loss, use the stellar radius response exponent ${\zeta }_{\mathrm{ad}}=d\log R_2/d\log M_2=-1/3$. During conservative binary mass transfer, $M$ and $J=M_1{M}_2\sqrt{Ga/M}$ are fixed, and $d{M}_1=-d{M}_2$. ThereforeThe change in overfill is $d\log (R_2/R_L)=({\zeta }_{\mathrm{ad}}-{\zeta }_L)d\log M_2$. Since $d\log M_2<0$, overfill grows, and mass loss runs away, when ${\zeta }_{\mathrm{ad}}<{\zeta }_L$. The dynamical stability of binary mass transfer criterion here is consequentlyThe printed instability inequality is reversed. With the defined $q=M_2/M_1$, the specified radius laws imply the inequality above; $q<q_{\mathrm{crit}}$ is the stable side. Equality is marginal, and the supplied Roche-radius approximation restricts this calculation to its stated mass-ratio range.

Long-term contact is maintained by angular momentum transport out of the orbit. Gravitational-wave emission from a binary system provides one loss mechanism. Magnetic braking of a binary star uses the donor's stellar wind: its magnetic field enforces approximate corotation out to the Alfvén radius, so the wind carries much more specific angular momentum than a nonmagnetic surface outflow. At that radius, roughlyThe geometry contributes order-one factors to the torque. Even when ${\dot{M}}_w\ll |{\dot{M}}_2|$, the large lever arm can remove substantial spin angular momentum. Tidal locking couples donor spin to orbital motion, so the wind torque ultimately brakes the orbit. In practice classical novae can also eject matter; the conservative calculations here isolate the stated idealization.

For the prescribed $\dot{J}/J=-\alpha$, now use the supplied equilibrium law $R_2\propto M_2$ for the slowly evolving donor. Contact implies $\dot{a}/a=(2/3){\dot{M}}_2/M_2$, assuming negligible net wind mass loss. Taking the logarithmic derivative of $J$ givesSince $P\propto M_2$ along this equilibrium sequence,For the dynamically stable mass ratios, this is negative. The result uses thermal equilibrium as well as slow hydrostatic evolution; evolution slow compared only with the dynamical time does not by itself guarantee the radius law $R_2\propto M_2$.

In the disrupted magnetic braking model, the donor above the gap loses mass fast enough to be inflated relative to thermal equilibrium. When the donor becomes a fully convective star, the model assumes a substantial reduction in magnetic braking. The red dwarf then contracts toward equilibrium on its Kelvin-Helmholtz cooling time, becoming smaller than its Roche lobe; accretion largely stops. Gravitational-wave emission from a binary system continues to reduce the separation, carrying the detached binary through the gap until contact is restored near its lower edge.

This explanation requires strong braking immediately above the gap: the donor must already be out of equilibrium and inflated, so there is room to contract and detach when braking drops. With nearly fixed component masses during detachment, $R_L\propto a\propto P^{2/3}$, giving the illustrative inflation factorA weak braking torque that kept the donor in thermal equilibrium just above the gap would not produce this detachment. The model requires a decrease in the torque, not disappearance of the donor's magnetic field.

Translation symmetry in the plane and reflection symmetry across it force the magnetic field to be uniform on each side, parallel to ${\mathbf{e}}_y$, and opposite on the two sides. Apply Ampère's law to a rectangular loop perpendicular to the sheet whose two long sides have length $L$. The enclosed current is $KL$, soThe right-hand rule fixes the directions:Consequentlywhich is the magnetic jump condition for a surface current density.

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

For each $X\in \{I,R,T\}$, substitution of a complex plane electromagnetic wave in a linear medium into the source-free Maxwell equations in a linear medium requireswhere ${\epsilon }_I={\epsilon }_R={\epsilon }_{-}$ and ${\epsilon }_T={\epsilon }_{+}$. The accompanying magnetic field isThe incident and transmitted waves have $k_{Ix}>0$ and $k_{Tx}>0$, while the reflected wave has $k_{Rx}<0$. Matching the fields at the stationary interface additionally imposes the phase matching at a planar wave interface conditions ${\omega }_I={\omega }_R={\omega }_T$ and equality of all tangential wavevector components.

Since the magnetic field is $B=\nabla \times A$, Stokes theorem gives, for any oriented surface $S$ bounded by $C$,Under a gauge transformation $A\mapsto A+\nabla \chi$, the integral changes bybecause $C$ is closed. The flux expression is therefore gauge independent.

Take the scalar product of the Lorentz force equation with the constant magnetic field $\mathbf{B}$. Since $\mathbf{E}\cdot \mathbf{B}=0$ and $(\dot{\mathbf{x}}\times \mathbf{B})\cdot \mathbf{B}=0$,The initial velocity also obeys $\mathbf{v}\cdot \mathbf{B}=0$, so $\mathbf{x}\cdot \mathbf{B}$ remains equal to ${\mathbf{x}}_0\cdot \mathbf{B}$. The trajectory therefore lies in the planethe plane through ${\mathbf{x}}_0$ perpendicular to $\mathbf{B}$.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Let $\hat{b}=B/|B|$. For a constant uniform magnetic field, choose the symmetric gaugewhich has $\nabla \times A=B$ and $A\cdot \hat{b}=0$. With $E=0$, one may take $\varphi =0$. The coordinate $r_{\parallel }=r\cdot \hat{b}$ is then absent from the Lagrangian, so it is an ignorable coordinate. Its generalized momentumis conserved. Hencewhich is the canonical momentum parallel to a uniform magnetic field in this gauge.

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

The dispersion relation is $\omega =ck$. At a perfect conductor the tangential electric field and normal magnetic field vanish. Thus $E_{ref}=-E_0\hat{x}{e}^{i(-kz+\omega t)}$ and $B_{ref}=-E_0\hat{y}{e}^{i(-kz+\omega t)}/c$. At the surface the magnetic field is $-2E_0\hat{y}/c$, so with outward normal $\hat{z}$ the surface current is $K=2E_0\hat{x}/({\mu }_{0}c)$ (real parts understood).