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For smooth inviscid barotropic fluid with potential body forces, barotropic vorticity transport and $\nabla \cdot \mathbf{\omega }=0$ give $D\mathbf{\omega }/Dt=(\mathbf{\omega }\cdot \nabla )\mathbf{u}-\mathbf{\omega }\nabla \cdot \mathbf{u}$. Combine this with material surface element evolution to obtain $D(\mathbf{\omega }\cdot d\mathbf{S})/Dt=0$. Integrating over a material surface proves the flux law. Its boundary must move with the fluid; flux through a fixed open surface is generally not conserved. By Stokes theorem the same statement is Kelvin's circulation theorem around the material boundary.

Parameterize a material surface by fixed particle labels $a,b$, with tangents $\mathbf{a}={\partial }_a\mathbf{x}$, $\mathbf{b}={\partial }_b\mathbf{x}$. The material derivative gives $D\mathbf{a}/Dt=(\nabla \mathbf{u})\mathbf{a}$ and similarly for $\mathbf{b}$. Differentiate $d\mathbf{S}=(\mathbf{a}\times \mathbf{b})dadb$ and use the cross product identityThis proves the displayed oriented-area evolution law, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. It allows both area change and rotation; even incompressible flow need not preserve the magnitude of a material area element.

This is magnetic flux freezing in ideal magnetohydrodynamics. In addition to the supplied ideal magnetohydrodynamic induction equation, a physical magnetic field obeys Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$.

Parameterize the material surface by two particle labels $a,b$, so $d\mathbf{S}=({\partial }_a\mathbf{x}\times {\partial }_b\mathbf{x})dadb$ and $D\mathbf{x}/Dt=\mathbf{u}$. Differentiating these tangents gives the material surface element evolution lawwhere $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Thus the material magnetic-flux transport theorem isExpand the curl in the ideal magnetohydrodynamic induction equation:Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ is the material derivative, the first three terms cancel the corresponding terms in the surface transport formula. The remaining integrand is $\mathbf{u}\nabla \cdot \mathbf{B}=0$. ThereforeNo incompressible flow assumption is needed: the changing surface area cancels the compression term. This proof assumes smooth velocity and magnetic field and a surface carried with the fluid; the ideal magnetohydrodynamic induction equation by itself, for an arbitrary non-solenoidal vector field, would leave the final divergence term.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.