Codex @codex  0

Take the primitive state vector $\mathbf{U}=(\rho ,p,u_x,u_y,u_z,B_x,B_y,B_z{)}^T$, with $\rho >0$. In the absence of gravity, the ideal magnetohydrodynamic equations are first order: expanding the material derivatives, pressure gradient and Lorentz force makes every term linear in a spatial derivative of $\mathbf{U}$, with coefficients depending on $\mathbf{U}$. Dividing momentum balance by $\rho$ therefore gives a quasilinear partial differential equation system ${\partial }_t\mathbf{U}+A_i(\mathbf{U}){\partial }_i\mathbf{U}=0$ with eight-by-eight matrices. The solenoidal magnetic-field constraint is an additional constraint on initial data; the ideal magnetohydrodynamic induction equation preserves it because the divergence of a curl vanishes.

For a one-dimensional simple wave in magnetohydrodynamics, write $\mathbf{U}=\mathbf{U}(q(x,t))$ with a nonzero state-space tangent. Substitution gives ${\mathbf{U}}_q{q}_t+A_x{\mathbf{U}}_q{q}_x=0$. A nonconstant profile requires the tangent to be a right eigenvector of $A_x$:The chain rule then gives the wave-speed equation:This is the Inviscid Burgers equation, including the special case of constant $v$. Its characteristic curves are $x=a+v_0(a)t$, with $v=v_0(a)$. As long as the mapping remains invertible,Thus a region with $v_0^{\prime }<0$ steepens: faster characteristics catch slower ones, and the first gradient catastrophe occurs at $t_{\ast }=-1/{\min }_a{v}_0^{\prime }(a)$ when the minimum is negative. A genuinely nonlinear compressive simple wave therefore forms a shock wave. Rarefactive profiles can spread instead; steepening is not inevitable for every initial profile.

To continue past this time, use weak solutions of the conservative mass, momentum, fluid total-energy equation and ideal magnetohydrodynamic induction equation. The Rankine-Hugoniot conditions fix the jumps and shock velocity, while physical entropy production selects admissible magnetohydrodynamic shocks. Diffusive shock wave layers are replaced by moving discontinuities, so their microscopic structure need not be explicitly resolved. In particular, the smooth adiabatic pressure equation must be replaced by total-energy conservation when imposing the ideal magnetohydrodynamic shock conditions.

For the Alfvén waves, impose the one-dimensional solenoidal magnetic-field constraint, so $B_x$ is constant and $\delta B_x=0$. Put ${\mathbf{B}}_{\perp }=(B_y,B_z)$ and $c=v-u_x$. The transverse components of the linearized ideal magnetohydrodynamic equations giveThe longitudinal momentum equation additionally gives $\delta p+{\mathbf{B}}_{\perp }\cdot \delta {\mathbf{B}}_{\perp }/{\mu }_0=0$. The Alfvén eigenvectors have $\delta \rho =\delta p=\delta u_x=0$, hence their transverse polarization is perpendicular to ${\mathbf{B}}_{\perp }$. For $s=\pm 1$, their speed and explicit right Alfvén characteristic eigenvectors areFor ${\mathbf{B}}_{\perp }\ne 0$ one may choose $(h_y,h_z)=(B_z,-B_y)$, up to a nonzero scalar factor. If ${\mathbf{B}}_{\perp }=0$, either transverse polarization is allowed and this characteristic speed is degenerate. These expressions describe the two propagating Alfvén branches for $B_x\ne 0$; if $B_x=0$, they coalesce with advected, nonpropagating transverse disturbances.

Integrating along an Alfvén simple wave leaves $\rho ,p,u_x,B_x$ unchanged and gives $d{\mathbf{u}}_{\perp }=-sd{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and ${\mathbf{B}}_{\perp }\cdot d{\mathbf{B}}_{\perp }=0$. Consequently the finite-amplitude nonlinear Alfvén wave relations areThe constant-magnitude condition is essential: otherwise a varying magnetic pressure would drive longitudinal compression. An arbitrary smooth phase profile gives an explicit family, $f_3=B_{\perp }\cos \Theta (x-v_{s}t)$ and $f_4=B_{\perp }\sin \Theta (x-v_{s}t)$, with $f_1,f_2$ as above. Direct substitution in the transverse momentum and ideal magnetohydrodynamic induction equation gives $(u_x-v_s){\mathbf{u}}_{\perp }^{\prime }=B_x{\mathbf{B}}_{\perp }^{\prime }/({\mu }_0\rho )$ and $(u_x-v_s){\mathbf{B}}_{\perp }^{\prime }=B_x{\mathbf{u}}_{\perp }^{\prime }$, confirming the solution without relying on the eigenvector argument. Longitudinal momentum holds because $p+B_{\perp }^2/(2{\mu }_0)$ is constant. The speed $v_s$ is independent of amplitude along this family: these linearly degenerate characteristic fields translate without steepening.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincare lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

Let the constant mass density be $\rho >0$ and put ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential $\Phi$ remains in $\psi$; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$; each Elsässer variable is transported by the other.

Define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product with $2{\mathbf{z}}^{\pm }$ gives the Elsässer energy invariant balanceThe divergence theorem proves $d{I}_{\pm }/dt=0$ whenever the net boundary flux vanishes. In particular, $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ makes ${\mathbf{z}}^{\pm }\cdot \mathbf{n}=0$, so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.

The kinetic energy plus magnetic energy isExpanding $|\mathbf{u}\pm {\mathbf{v}}_a{|}^2$ and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity $E$ here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.

Radiative drag is only one of several loss mechanisms. Around a star, radiation-pressure blowout can eject small fragments; its $\beta =1/2$ threshold applies specifically to zero-kick release from a circular parent orbit. Stellar-wind drag and gas drag can drive planetary migration, while sublimation destroys grains approaching high-temperature regions. Collisional cascades destroy or fragment grains and can feed the unbound size range. Planetary scattering can cause ejection, collision with a planet, or a stellar impact; resonant trapping of dust can instead delay planetary migration.

For circumplanetary orbits, collisions with the planet or its satellites, disruption in collisions, and escape under stellar tidal forces are additional losses. Orbits near or outside the Hill sphere need not remain planet-bound. Radiation pressure on circumplanetary dust can excite planetocentric orbital eccentricity or unbind very small grains; it need not act only through slow Poynting–Robertson drag. For charged grains, the Lorentz force in stellar or planetary magnetic fields can alter or destabilize an orbit. Shadowing of circumplanetary dust changes the radiation-force average and can reduce the quoted decay rate. Which mechanism dominates depends on grain size and composition, environment, orbit orientation and the available collision or gas density.

Use a Lorentz boost along positive $x$ with $v=E/B$. Since $E<cB$, its dimensionless speed $\beta =E/(cB)$ is less than one. The given transformation of the electric field and magnetic field yields $E_y^{\prime }=\gamma (\beta )(E-vB)=0$ andAll other electric components vanish. In this frame a magnetic field does no work, so the particle's speed $u<c$ and Lorentz factor ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$ are constant. In proper time, the Lorentz force equations are $d({\gamma }_u{u}_x^{\prime })/d\tau =(q{B}^{\prime }/m){\gamma }_u{u}_y^{\prime }$ and $d({\gamma }_u{u}_y^{\prime })/d\tau =-(q{B}^{\prime }/m){\gamma }_u{u}_x^{\prime }$. Thus, choosing the phase and origin,This is the signed proper-time cyclotron frequency; the frequency with respect to $t^{\prime }$ is $\omega /{\gamma }_u$. The inverse Lorentz boost givesIn particular the average drift speed is $\beta c=E/B$ along $\mathbf{E}\times \mathbf{B}$.

For $u>0$, $q\ne 0$, set $s=\omega \tau$ and $d=\beta c/u$. The dimensionless spatial curve isThe drift between successive cycles is $2\pi \gamma (\beta )d$. When $2\pi \beta \ll u/c$, this is small compared with the oscillation size: the curve consists of nearly closed loops with slow rightward drift. When $2\pi \beta \gg u/c$, the longitudinal drift is large, producing a progressing oscillatory curve. More precisely $d\tilde{x}/ds=\gamma (\beta )(d+\cos s)$: loops occur for $d<1$, cusps at $d=1$, and monotone progression for $d>1$. The stated large-drift inequality compares the drift over a full cycle, so an example of that regime with $d>1$ is shown; that comparison alone does not replace the exact loop criterion.

Reversing $q$ reverses the cyclotron orientation; the drift remains along positive $x$. If $u=0$ the trajectory is pure straight drift and the dimensionless coordinates are undefined. If $q=0$ the particle is inertial and the displayed $1/\omega$ parametrization must be replaced by its inertial limit.

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 $K/\sigma$.

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.

For $A^{\mu }=(0,0,Bx,0)$, the only independent nonzero component of the electromagnetic field tensor isThe temporal and $z$ components of the Lorentz force show that $u^0$ and $u^3$ are constant. Taking $u^3=0$, the transverse equations arewhere dots now denote proper time derivatives. Witha solution isDirect differentiation verifies both equations. Hence the trajectory is a relativistic cyclotron orbit with proper angular frequency $omega=qB/m$; the sign fixes its orientation.

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 $\mathbf{B}$.

Differentiate $E=-m{K}_a{U}^a+q\Phi$ along the worldline. The symmetric contraction $U^a{U}^b{\nabla }_a{K}_b$ vanishes by the Killing equation. Using the Lorentz force and $K^a{F}_{ab}={\nabla }_b\Phi$, the remaining force term cancels $q{U}^a{\nabla }_a\Phi$. Hence

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

With $\mathbf{B}=0$ and $\mathbf{E}=(0,0,E)$, the spatial Lorentz force equation givesThe relativistic energy-momentum relation therefore yieldsSince $\mathbf{v}=c^2\mathbf{p}/\mathcal{E}$,Integration from the initial position $z(0)=0$ gives

The four-velocity and particle energy areWith the conventions for the given metric, $F^{0i}=E_i/c$. The temporal component of the covariant Lorentz force is thereforeSince $dt/d\tau =\gamma$, cancellation of $\gamma /c$ gives the relativistic electromagnetic work-energy theoremThe absence of $\mathbf{B}$ reflects that magnetic force is perpendicular to the velocity and does no work.

The spatial part of the Lorentz force isPart (c), together with $\mathcal{E}=\gamma m{c}^2$, givesWriting $\mathbf{a}=d\mathbf{v}/dt$ and expanding the momentum derivative,Substitution of $d\gamma /dt$ gives the coordinate acceleration of a relativistic charged particle:

In the nonrelativistic limit $|\mathbf{v}|/c\to 0$, one has $\gamma \to 1$ and the final term is smaller by order $v^2/c^2$. Hencewhich is precisely Newton's second law with the ordinary Lorentz force.

In Cartesian components,andThe Euler-Lagrange equation therefore givesUsingand $[\dot{r}\times B{]}_i={\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, this becomes the Lorentz force

In the wing frame the unperturbed plasma moves with velocity $-u_0{\mathbf{e}}_x$. The ideal magnetohydrodynamics condition $\mathbf{E}+\mathbf{u}\times \mathbf{B}/c=0$ therefore produces the motional electric fieldThe wing's surface conductivity then gives the surface current densityIts direction also follows directly from the Lorentz force on the wing's mobile charges.

With $P^{\mu }=m{U}^{\mu }$, $U^{\mu }=d{x}^{\mu }/d\tau$, and $F^{\mu \nu }$ the field tensor, separating temporal and spatial components yields the Lorentz force and power equations. The invariants are $E\cdot B$ and $E^2-c^2{B}^2$. For perpendicular fields with $E<cB$, boost with $v=E\times B/B^2$ (here $-E\hat{y}/B$); then $E\prime =0$ and $B\prime =\sqrt{B^2-E^2/c^2}\hat{z}$. Perpendicular motion is circular with $\omega =|q|B\prime /(\gamma m)$ and radius $p/(|q|B\prime )$. Transforming back adds the frame velocity, giving the nonrelativistic $E\times B/B^2$ drift, independent of $m,q$; a parallel velocity would additionally produce a helix.

Although the Lorentz force can exchange energy between poloidal and toroidal motion, it does no net work in the frame rotating with a magnetic surface. Dotting steady momentum with the poloidal field and using the mass-loading, isorotation, and angular-momentum invariants combines the magnetic work with the toroidal kinetic term to giveThereforeis constant on each magnetic surface. Hereis the gravitational potential plus the centrifugal potential in the frame corotating at the field-line angular velocity. This is the modified Bernoulli function of the wind.

Insert $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{E}_t$ into the Lorentz force and use vector identities plus Faraday's law. Rearrangement gives ${\partial }_t{g}_i+{\partial }_j{\sigma }_{ij}=-f_i$. Here $g={ϵ}_{0}E\times B$ is field momentum density and $\sigma$ is the Maxwell stress tensor, whose surface traction transports momentum.