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Advection transports a quantity with a velocity field. For a scalar field $f$, its local advective rate is $\mathbf{u}\cdot \nabla f$, the spatial contribution to the material derivative. The same term transports the components of a vector field.

A Fourier mode advected with a homologous spherical flow has phase $\mathbf{q}\cdot \mathbf{r}/R(t)$. Its material derivative in the background vanishes because $\dot{\mathbf{k}}=-(\dot{R}/R)\mathbf{k}$. Thus contraction shortens its physical wavelength while preserving its phase on each fluid element. Linearizing the continuity equation and pressureless momentum equation gives $\dot{\delta }=-i\mathbf{k}\cdot {\mathbf{u}}_1$ and ${\dot{\mathbf{u}}}_1+(\dot{R}/R){\mathbf{u}}_1=-i\mathbf{k}{\Phi }_1$. Combining these with the Poisson equation for Newtonian gravity yields $\ddot{\delta }+2(\dot{R}/R)\dot{\delta }-4\pi G\rho \delta =0$ for nonzero compressive wavenumber. It is a local bulk calculation, without the moving cloud-boundary conditions.

For concentration $C$ per unit surface area, conservation on a moving material interface givesThe material derivative follows the surface particles. The two dilution terms are tangential expansion and curvature-induced area change from normal velocity $u_n$; the last term is surface diffusion.

Convective acceleration is the spatial part $(\mathbf{u}\cdot \nabla )\mathbf{u}$ of the material derivative of fluid velocity. It can remain nonzero in steady flow, because a moving particle passes through positions with different velocities. The identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ relates it to vorticity.

With velocity gradient $A_{ij}={\partial }_i{v}_j$, set $\Omega =(A-A^T)/2$. A vector rigidly rotating with the material has material derivative $D_t\mathbf{p}=-\Omega \mathbf{p}$, so its corotational derivative vanishes. Its unit rotational coefficient is fixed by an objective time derivative; adding $-\xi D\mathbf{p}$ describes the material's flow alignment of a polar order parameter.

Let $h_c=\mathbf{u}\cdot \mathbf{B}$ be the cross-helicity density. The ideal magnetohydrodynamic induction equation implies $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}-\mathbf{B}\nabla \cdot \mathbf{u}$. Dotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density, since it is perpendicular to the magnetic field. The material derivative of $h_c$ is thereforeUsing $dh=Tds+dp/\rho$ for specific enthalpy $h$ and specific entropy $s$, together with Gauss's law for magnetism, converts this toThe source vanishes when magnetic field lines lie in constant-specific entropy surfaces, in particular in a homentropic flow. Integrating over a volume conserves its total cross-helicity only if the boundary flux also vanishes. This local law is derived in Ogilvie's astrophysical fluid dynamics notes.

For $F=(F_1,F_2,F_3)$ in Cartesian coordinates,

For a continuously differentiable vector field, the divergence is the limiting outward flux per unit volume through a small closed surface. Opposite faces of a small rectangular box almost cancel; the remaining differences in the three components give the sum of partial derivatives above. The divergence theorem extends this local balance to a whole region.

For $F=(x,y,0)$, the divergence is $1+1+0=2$. In a planar unit square, the right edge contributes outward flux $1$ and the top contributes $1$, while the left and bottom contribute zero. The total $2$ equals the integral of the planar divergence over area $1$.

If $F$ is a velocity field, positive divergence means local volume expansion, not automatically creation of mass. The continuity equation gives $D\rho /Dt=-\rho \nabla \cdot F$, where $D/Dt$ is the material derivative. For this field, a uniform initial density evolves as $\rho (t)=\rho (0)e^{-2t}$ while material volumes grow as $e^{2t}$. Their product stays constant.

The Eulerian time derivative measures a continuum field's rate of change at a fixed Eulerian coordinate. For a velocity field $\mathbf{u}$, the material derivative instead follows a fluid element and satisfies $Df/Dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$. A nonuniform steady flow has zero Eulerian time derivative of velocity but can have nonzero material acceleration.

With real $k,U$ and $\mu >k^2$, substitution into the nonlinear Ginzburg-Landau equation cancels the material derivative and gives $Q^2=\mu -k^2$. A complex $k$ makes the modulus spatially varying, so it generally does not admit this constant-amplitude nonlinear Ginzburg-Landau equation ansatz.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

Take a static uniform fluid of mass density $\rho >0$, constant gas pressure, and background magnetic field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$ with $B_0>0$. Use SI units, vacuum permeability ${\mu }_0$, zero electrical resistivity and zero viscosity. Write $\mathbf{B}={\mathbf{B}}_0+{\mathbf{b}}_{\perp }(z,t)$ and $\mathbf{u}={\mathbf{u}}_{\perp }(z,t)$, with both disturbances perpendicular to ${\mathbf{e}}_z$. Retain only linear order in their amplitudes. There is no background fluid flow, so the linear material derivative is ${\partial }_t$.

The continuity equation has $\nabla \cdot {\mathbf{u}}_{\perp }=0$, hence ${\partial }_t\delta \rho =0$; choosing no initial mass density fluctuation gives $\delta \rho =0$. For this shear mode the gas pressure fluctuation is also zero. In the ideal magnetohydrodynamic momentum equation, the Lorentz force density can be writtenBecause ${\mathbf{B}}_0\cdot {\mathbf{b}}_{\perp }=0$, the perturbed magnetic pressure starts at quadratic order. The surviving transverse magnetic tension gives the first equation below. Expanding the ideal magnetohydrodynamic induction equation gives the second:Differentiate the first in time and substitute the second. Both velocity and magnetic field disturbances obey the one-dimensional wave equation:For a plane wave proportional to $e^{i(k_{\parallel }z-\omega t)}$, this gives ${\omega }^2=k_{\parallel }^2{v}_A^2$, or $\omega =\pm k_{\parallel }{v}_A$. The magnetic and velocity amplitudes satisfy ${\mathbf{b}}_{\perp }=-(B_0{k}_{\parallel }/\omega ){\mathbf{u}}_{\perp }$: they have opposite signs for propagation along $+{\mathbf{e}}_z$ and equal signs for propagation along $-{\mathbf{e}}_z$. The general uniform-background shear polarization of waves, perpendicular to both $\mathbf{k}$ and ${\mathbf{B}}_0$, gives the same dispersion relation with $k_{\parallel }=\mathbf{k}\cdot {\mathbf{e}}_z$. Its group velocity is $\pm v_A{\mathbf{e}}_z$, whereas its phase velocity normal to an oblique wavefront has magnitude $v_A|k_{\parallel }|/|\mathbf{k}|$.

Let $D_n={\partial }_t+{\mathbf{v}}_n\cdot \nabla$ be the material derivative of the normal fluid, and $\zeta >0$. The displayed phenomenological density damping preserves the condensate continuity equation because its coefficient is real. Under a Galilean boost, both velocities shift by the boost velocity and the field acquires the Schrödinger phase; $D_n|\psi {|}^2$ is invariant. Thus the normal-fluid derivative, rather than a derivative along the condensate flow, is essential to the model.

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.

Let $g=\dot{\gamma }$ and use $G_{yx}=g$ for the source's velocity gradient convention. Seek the homogeneous stationary stress reached under steady simple shear flow. The material derivative of that stress tensor vanishes. In the $xy$ plane,The $xx$, $yy$ and $zz$ components of the constitutive equation give, respectively,Thus, putting $D=1+\alpha (2\tau -\alpha )g^2$ and using the shear stress calculated below,The out-of-plane shear components vanish on the homogeneous branch. The two normal-stress differences areThe ratio is defined at nonzero shear when $N_1\ne 0$; at $g=0$ both differences vanish, and the displayed constant is their limiting ratio. An arbitrary isotropic pressure in the total Cauchy stress tensor cancels from both differences; it is not an arbitrary addition to the constitutively defined extra stress $\mathsf{T}$ itself.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

The magnetic second-moment tensor is $M_{ij}=B_i{B}_j$. This is the magnetic dyadic called the Maxwell tensor in the paper; the isotropic part of the usual Maxwell stress tensor is already contained in the magnetohydrodynamic total pressure. The solenoidal magnetic-field constraint givesConsequently the magnetohydrodynamic momentum equation in part a becomes $D_t{u}_i=-{\partial }_{i}p+{\partial }_j{M}_{ij}$, where $D_t={\partial }_t+u_n{\partial }_n$ is the material derivative. Applying its product rule and the ideal magnetohydrodynamic induction equation givesThus the evolution uses only $\mathbf{u},M,p$, together with ${\partial }_i{u}_i=0$. To determine $p$, take the divergence of the magnetohydrodynamic momentum equation. Since ${\partial }_t{\partial }_i{u}_i=0$ and $u_j{\partial }_j{\partial }_i{u}_i=0$,This is the pressure Poisson equation for incompressible flow, with boundary conditions supplied by the imposed normal acceleration and normal stress, or by periodicity or decay at infinity as appropriate. For example, the normal component of the magnetohydrodynamic momentum equation requires $\mathbf{n}\cdot \nabla p=\mathbf{n}\cdot (\nabla \cdot M-D_t\mathbf{u})$ at a prescribed boundary. The pressure is determined up to a spatially constant function of time.

For equivalence to one resolved magnetic field, the initial $M$ must actually be $\mathbf{B}{\mathbf{B}}^T$ for a divergence-free vector field. In particular it is a symmetric positive semidefinite matrix of matrix rank at most one. The tensor equations also admit more general stresses; the averaged $M$ in part d need not have matrix rank one.

To first order in fluid velocity divided by the speed of light, the Lorentz transformation of electromagnetic fields gives ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{u}\times \mathbf{B}$. Applying Ohm's law in the local fluid frame gives $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\eta \mathbf{J}$. Neglecting the displacement current in Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. Substitution in Faraday's law givesHere ${\eta }_m$ is the magnetic diffusivity; the paper's $\eta$ is electrical resistivity, rather than magnetic diffusivity. Expanding the curl with the solenoidal magnetic-field constraint gives the resistive induction equation in its compressible form:The left side is the material derivative of the magnetic field. The three right-hand terms respectively describe field-line stretching and reorientation, amplification or dilution by compression or expansion, and magnetic diffusion. In particular, compression has $\nabla \cdot \mathbf{u}<0$ and amplifies a field component perpendicular to the compression.

The assumptions are nonrelativistic bulk motion, the low-frequency approximation permitting omission of the displacement current, a conducting-fluid continuum, isotropic local Ohm's law with spatially constant scalar electrical resistivity, constant magnetic permeability ${\mu }_0$, and $\nabla \cdot \mathbf{B}=0$. Hall, electron-pressure and other non-Ohmic corrections to the generalized Ohm's law are excluded. Incompressible flow is not required: imposing it would remove the compression term.

Follow a fluid element with position $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives the material derivativeApply Newton's second law to a material element of volume $dV$ and mass $\rho dV$. The net pressure force is $-\nabla pdV$, and gravity gives $-\rho \nabla \varphi dV$. Dividing by the element mass, neglecting viscosity and using the chain rule componentwise gives the Euler equation

For a static spherical isothermal gas, $p=c_s^2\rho$ and $dp/dr=-GM(r)\rho /r^2$. Trial density $\rho =A/r^2$ gives enclosed mass $M(r)=4\pi Ar$, soThis is the singular isothermal sphere. Its central density diverges, although the enclosed mass tends to zero linearly and its total mass inside a finite radius is finite. It is not the finite-central-density regular isothermal solution.

Truncate at $r_0$ with boundary pressure $p_0$. Then $M=4\pi A{r}_0=2c_s^2{r}_0/G$ and $p_0=c_s^2\rho (r_0)$, givingThe external medium is being treated as a prescribed pressure boundary; its own gravitational field is excluded from the cloud self-energy calculation.

For a monatomic perfect gas, the equilibrium is isothermal but the internal-energy relation still uses its adiabatic index $\gamma =5/3$:The gravitational self-energy follows by assembling shells:Both integrals remain finite at the singular center. ConsequentlyThe pressure-confined singular isothermal gas sphere therefore satisfies the virial theorem including the work of the confining surface pressure. One can see its origin directly by integrating $\mathbf{r}\cdot \nabla p$: integration by parts gives $\oint p\mathbf{r}\cdot d\mathbf{S}-3\int pdV$, whose surface term is $4\pi r_0^3{p}_0$. An isolated vacuum-boundary version would omit that term and could not truncate this same positive-temperature isothermal profile at finite radius. The equality expresses equilibrium force and energy balance, rather than proving dynamical stability.

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.