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Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

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.

Resolve the uniform upstream field into components normal and tangential to the spherical shock:The ideal magnetohydrodynamic shock conditions leave the normal field continuous and multiply the tangential field by the gas compression ratio $\chi$. Therefore

Resolve the upstream uniform field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$ into components normal and tangential to the spherical shock:The ideal magnetohydrodynamic shock conditions preserve the normal magnetic component and multiply the tangential component of a weak passive field by the gas compression ratio. Part (b) therefore gives