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Differentiate the advection equation componentwise: $D_t({\partial }_i\theta )=-({\partial }_i{u}_j){\partial }_j\theta$. Along a Lagrangian trajectory, the scalar gradient therefore evolves by the negative transpose of the velocity gradient tensor. If $F$ is the deformation gradient, this says $\mathbf{k}(t)=F(t{)}^{-T}\mathbf{k}(0)$, as expected for a covector. Fluid material vectors and scalar gradients transform oppositely under stretching.

A fluid element is an infinitesimal material parcel followed along a Lagrangian trajectory.

The Lagrangian integral time is the area under a normalized velocity autocorrelation along a Lagrangian trajectory. If this correlation is integrable, Taylor turbulent dispersion has asymptotic mean-square displacement $2R_L(0)t_{L}t$. The integral time need not equal every short-time decorrelation scale, particularly when the autocorrelation oscillates.

For stationary statistics along a Lagrangian trajectory $\mathbf{X}$, the velocity autocorrelation depends on the lag $s$. Integrating velocity twice gives the mean-square displacement $2{\int }_0^t(t-s)R_L(s)ds$. If $R_L$ is integrable, the long-time displacement is diffusive. Its normalized one-sided area is the Lagrangian integral time.

A fluid particle follows a Lagrangian trajectory, so its position satisfies $\dot{x}=u(x,t)$ rather than the velocity evaluated permanently at its initial position. Assume $\omega \ne 0$, and writeSet $\alpha =k{X}_0$ and $\theta =\alpha -\omega t$. A Taylor expansion of the velocity about $X_0$ givesThus $\dot{a}=\sin \theta$, andAt the next order,so the required second-order trajectory in a sinusoidal velocity field isBoth corrections vanish at $t=0$. All retained terms have units of length. The secular contribution gives the Stokes driftalthough the Eulerian time average of the given velocity is zero. The particle samples the wave at its displaced position, producing this second-order drift. The expansion is in the dimensionless parameter $|k{u}_0/\omega |\ll 1$, for fixed times; it ceases to be uniform when the accumulated drift changes the wave phase by order unity.

For a scalar fluid property $f(\mathbf{x},t)$, the Eulerian time derivative ${\partial }_{t}f$ measures change at a fixed Eulerian coordinate. The material derivative, also called the Lagrangian time derivative, measures change while following one fluid element. If its Lagrangian trajectory is $\mathbf{X}(\mathbf{a},t)$, where $\mathbf{a}$ is its fixed Lagrangian coordinate, then ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$.

The chain rule therefore givesThe last term accounts for entering a region where the property differs. For example, a spatially varying but time-independent velocity field has zero Eulerian time derivative, while a moving fluid element can have nonzero material acceleration. The same formula applies componentwise to a vector field.

At a fixed point beside a river, the Eulerian coordinate description measures $\mathbf{u}(\mathbf{x},t)$. Its local acceleration ${\partial }_t\mathbf{u}$ measures change at that fixed point. Following a floating parcel gives a Lagrangian trajectory $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives its physical material acceleration:The extra term is the convective acceleration. For a river that narrows steadily, an observer at one fixed position measures a constant velocity, yet a parcel speeds up as it reaches the narrowing. A parcel following a steady bend also experiences a centripetal acceleration even when its speed is constant. An Eulerian calculation of the physical acceleration must include the same convective acceleration; the two descriptions do not predict different forces.

Put $\theta =x-t$. Taking the gradient of the velocity potential givesThe motion is not generally incompressible; a velocity potential ensures irrotationality, not zero divergence. Both components have zero time average at each fixed point because their sine and cosine averages vanish.

The fluid acceleration is the material derivative $D\mathbf{u}/Dt={\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}$. Direct differentiation givesAveraging over one period at fixed $(x,y)$ therefore yieldsThis need not be the derivative of the mean velocity: the nonlinear advective term remains after averaging.

The dyed particle obeys the Lagrangian trajectory equationsExpand $x=ϵx_1+{ϵ}^2{x}_2+O({ϵ}^3)$ and $y=ϵy_1+{ϵ}^2{y}_2+O({ϵ}^3)$, for bounded times. The first-order equations give $x_1=0$, ${\dot{y}}_1=\cos t$, so $y_1=\sin t$. At second order,Integrating with zero initial values verifiesThe velocity averaged along this particle over $0\le t\le 2\pi$ is its displacement divided by $2\pi$. HenceThis Stokes drift is a Lagrangian mean; it does not contradict the zero Eulerian mean velocity at a fixed point. The secular term is a small-parameter expansion, not an assertion of uniform accuracy as $t\to \infty$.

Differentiate the material derivative of the passive scalar component by component. With ${\sigma }_{ij}={\partial }_j{u}_i$,The chain rule along a Lagrangian trajectory $\dot{X}=u(X,t)$ therefore gives the covector evolution of a material scalar gradientThe transpose matters: a gradient is transported as a covector, rather than as a material separation vector. When $u_3=0$, the third row of $\sigma$ vanishes, andLet $k_h=(k_1,k_2{)}^T$ and setFor $c>0$, the matrix exponential gives the exact horizontal solutionThe two projections are onto the eigenspaces with eigenvalues $-c$ and $+c$. If $p_{-}\ne 0$, write $p_{-}=k_0(\cos \varphi ,\sin \varphi {)}^T$ with $k_0=|p_{-}|>0$. ThenThe stated growing asymptotic requires this nonzero projection. An initial gradient entirely in the other eigenspace decays, and when $c=0$ the horizontal gradient is constant.

For constant vertical shear, write $h=({\sigma }_{13},{\sigma }_{23}{)}^T$. Integrating the third component using the exact horizontal solution yieldsConsequently, if $D=h\cdot (\cos \varphi ,\sin \varphi )$ is nonzero, $k_3\sim -k_{0}D{e}^{ct}/c$, andIf $D=0$, exponential growth of $k_3$ is absent, but the ratio still tends to zero on the growing horizontal branch. Thus exponential vertical growth is generic, not unconditional. The scalar-gradient inclination in a steady planar strain is the tangent of the angle between the scalar's normal and the horizontal plane. Equivalently, it is the cotangent of the inclination of an isoconcentration surface to that plane. Large $\alpha$ corresponds to nearly horizontal scalar surfaces; small $\alpha$ corresponds to nearly vertical ones.

For the probability calculation, take $\Gamma ,\Lambda >0$. The independent random variables $a,b$ with normal distributions of variance ${\Gamma }^2$ make $c$ distributed according to the Rayleigh distribution, withThe advertised Gaussian-over-Rayleigh inclination distribution also needs $D$ to be independent of $(a,b)$; its Gaussian marginal alone does not imply this. Under that intended assumption, for $A>0$,The cumulative distribution function is zero for $A\le 0$. Differentiating givesand the probability density function is zero for negative $A$. Its integral is one because the cumulative distribution function tends to one at infinity. Independence is a real hypothesis: $D=(\Lambda /\Gamma )a$ has exactly the stated Gaussian marginal but makes $\alpha \le \Lambda /\Gamma$ almost surely, contradicting the unbounded support of this density. A sufficient physical model is isotropic Gaussian vertical shear, independent of $a,b$, whose projection in any horizontal direction has variance ${\Lambda }^2$.

If the strain varies slowly, the horizontal gradient must align with the current growing direction, and the vertical-to-horizontal ratio must adjust before that strain changes. Both adjustment times are of order $c^{-1}$. Thus the local condition is $c{\tau }_{\sigma }\gg 1$, and the typical condition for the specified statistics isThis is an adiabatic approximation, not a uniform statement about arbitrarily weak instantaneous strains: samples with $c{\tau }_{\sigma }\lesssim 1$ retain memory of earlier strain. The instantaneous eigenvector must likewise turn slowly compared with $c$.

For rapidly fluctuating vertical shear, insert the large-time horizontal solution into its stochastic differential equation:The linear combination $W=\cos \varphi W^{(1)}+\sin \varphi W^{(2)}$ is a Brownian motion: it has independent Gaussian increments and increment variance $({\cos }^2\varphi +{\sin }^2\varphi )dt=dt$. For the signed inclination $\beta =k_3{e}^{-ct}/k_0$, the Itô product rule introduces no extra cross term because the exponential factor is deterministic. It follows thatThis is an Ornstein-Uhlenbeck process, explaining why white-noise vertical shear gives an Ornstein-Uhlenbeck inclination. Its Fokker-Planck equation isIn a stationary state the Fokker-Planck probability current $j=-cBp-(g^2/2)p^{\prime }$ is constant. A nonzero constant current cannot give an integrable, nonnegative density at both ends of the real line, so $j=0$. Integrating $p^{\prime }/p=-2cB/g^2$ and normalizing with the Gaussian integral givesThe stationary variance is $g^2/(2c)$. More explicitly the explicit Ornstein-Uhlenbeck solution has mean $\beta (0)e^{-ct}$ and variance $g^2(1-e^{-2ct})/(2c)$, confirming convergence to this density. Here $c>0$ and $g\ne 0$ are needed for the stated smooth stationary density; at $c=0$ there is no confinement, while $g=0$ gives a point mass at zero.

A large molecular Schmidt number $\mathrm{Sc}=\nu /\kappa$ separates the velocity's smallest smooth length scale from the scalar's smaller diffusive length scale. Follow a Lagrangian trajectory $X(t)$ and measure displacement $x$ from it. Within a sufficiently small scalar packet, Taylor expansion of the smooth velocity field givesThe moving frame removes uniform sweeping, leaving a linear random velocity gradient tensor. Thus the local scalar equation is the advection-diffusion equation with velocity $Cx$, and averaging over realizations of $C$ gives the corresponding ensemble average. The approximation requires the packet to lie within the velocity's locally smooth region. Its thin, diffusive direction can satisfy this condition even when its long direction eventually becomes too large for one global linearization; applications to a spatially varying flow then use local packet descriptions. The statistical conclusions below refer to the linear-flow model and the usual mixing assumptions, not to every conceivable trace-free random matrix process.

Write $Q=\int \chi dA$ and assume sufficiently rapid decay to discard boundary terms. Integration by parts givesFor $J_{ij}=\int x_i{x}_j\chi dA$, integrate the advective term once and the diffusive term twice. Since ${\partial }_k(x_i{x}_j)={\delta }_{ki}{x}_j+{\delta }_{kj}{x}_i$ and $\Delta (x_i{x}_j)=2{\delta }_{ij}$, one obtainsThese are mass conservation and the evolution of the unnormalized second moment tensor. The first moment obeys $\dot{M}=CM$, so a centered packet stays centered.

For a positive Gaussian scalar packet in a linear incompressible flow, $B$ must be a positive-definite matrix. Diagonalizing its symmetric matrix and applying the Gaussian integral givesThe factor $Q^2$ in the last identity matters because $J$ has not been divided by the scalar mass. Substitution into the advection-diffusion equation also givesThese identities agree with the second moment equation and prove that the Gaussian family is preserved. For an initially negative packet the same analysis applies after replacing $\chi$ by $-\chi$ and taking its positive mass.

To see the roles of stretching and diffusion explicitly, let the deformation gradient $F$ solve $\dot{F}=CF$, $F(0)=I$. Incompressibility gives $\det F=1$. Differentiating $F^{-1}J{F}^{-T}$ integrates the moment equation:Without diffusion, $J=FJ(0)F^T$ and $\det J$ is constant. In a statistically stationary, mixing flow with positive Lyapunov exponent $\lambda$, the two stretching exponents are $\lambda$ and $-\lambda$. Writing the ordered eigenvalues of $J$ as $e^{2{\rho }_1},e^{2{\rho }_2}$, their typical initially nondiffusive evolution isThe fluctuation scale presumes the usual central limit theorem for accumulated strain; it is not implied by incompressibility alone. During this stage the packet becomes long and thin but keeps its area and peak concentration.

In the instantaneous principal-axis basis, write $\tilde{C}$ for the rotated velocity gradient tensor. Differentiation of an eigenvalue givesThe formulas hold through a repeated eigenvalue by continuity of their sum. The compressed physical variance is $e^{2{\rho }_2}/Q$. Its contraction is arrested when it is of order $\kappa /\lambda$, the square of the Batchelor scalar microscale ${\ell }_B\sim \sqrt{\kappa /\lambda }$. If its initial length is $L\gg {\ell }_B$, the crossover occurs atThereafter ${\rho }_2$ has stationary fluctuations associated with the strain and the diffusive floor, while ${\rho }_1$ continues to grow typically at rate $\lambda$. Under sufficiently fast mixing, the central part of the distribution of ${\rho }_1$ is approximately Gaussian with mean $\lambda t+O(1)$ and variance $D_{\rho }t+o(t)$ for a strain-dependent coefficient $D_{\rho }$. This central approximation does not determine far tails, which control high concentration moments; those require a full large deviation principle. The two logarithmic widths need not be independent. The area now increases and $f$ decreases typically as $e^{-\lambda t}$. This is diffusion arrest of Gaussian packet compression.

For example, a constant principal strain $C=\mathrm{diag}(\lambda ,-\lambda )$ gives the exact physical variancesThe thin variance approaches its diffusive floor, the long variance grows exponentially, and the mass-to-area relation gives the quoted typical amplitude decay. A random strain additionally makes rare histories important for concentration moments.

For a precise statistical formulation define the logarithmic area increaseThe Gaussian identities give the exact relation $f(t)=f(0)e^{-H_t}$; henceAssume the long-time strain statistics give a large deviation principle for $H_t/t$ on $s\ge 0$, with convex good rate function $I(s)$, $I(\lambda )=0$. At exponential accuracy the probability of $H_t/t$ near $s$ is $e^{-tI(s)}$. Combining that cost with the amplitude factor and applying the Laplace principle for probability measures gives the large-deviation decay rates of a Gaussian scalar packetThe initial amplitude only changes a time-independent prefactor. One can see the minimization directly by dividing the possible area-growth rates into small intervals: each contributes its probability cost times $e^{-\mu ts}$, and the interval with least total exponential cost dominates. An interval around a minimizing $s$ gives the matching lower bound. The nonnegative area increase bounds the exponential weight by one, so unusually large positive growth cannot produce a divergent weighted tail. Using the area rather than only ${\rho }_1$ also avoids discarding possibly important fluctuations of the compressed direction.

Suppose $I$ is strictly convex and differentiable, with its minimum at $\lambda >0$ and finite negative right derivative at zero. DefineFor $0<\mu <{\mu }_{\ast }$ the minimizing rate $s_{\mu }$ is positive and uniquely solvesFor $\mu \ge {\mu }_{\ast }$, convexity puts the minimum at the endpoint $s=0$, givingHigh moments are dominated by rare histories with little net area growth, in which a packet keeps a substantial concentration. Their probability cost is $I(0)$, independent of the moment order. A finite threshold is conditional: if $I^{\prime }(0+)=-\infty$, the minimizer can approach zero without reaching it at any finite $\mu$.

Below the threshold, differentiation of the minimizing expression cancels the terms involving $s_{\mu }^{\prime }$ because $I^{\prime }(s_{\mu })+\mu =0$. Thus ${\gamma }_{\mu }^{\prime }=s_{\mu }$, andStrict negativity uses a nondegenerate rate function with a unique zero at $\lambda$: for positive $\mu$, $s_{\mu }<\lambda$ and $I(s_{\mu })>0$. For deterministic strain the ratio can instead be constant. Near the typical rate, if $I(s)\simeq (s-\lambda {)}^2/(2D)$ with $D>0$, then ${\gamma }_{\mu }\simeq \lambda \mu -D{\mu }^2/2$ for small $\mu$. If this quadratic expression holds all the way to zero, it gives the illustrative full resultThe quadratic form is an example, not a universal law of turbulent strain.

A thin scalar packet is stretched until molecular diffusion broadens its compressed direction; conservation of its scalar mass then converts area growth into dilution. Ensemble averages sample different stretching histories, producing different decay rates for different moments and saturation when weakly stretched histories dominate. The present moment calculation is specifically for the centered Gaussian packet and averaging over the flow; signed random initial fields with overlapping packets can also have cancellations and require their initial-field statistics.

Finally, positive exponential decay is a statistical conclusion requiring the strain assumptions above, not a consequence of the local differential equation by itself. For $C\equiv 0$ and initially isotropic physical variance $L^2$, the exact solution has $J=Q(L^2+2\kappa t)I$ and $f=Q/[2\pi (L^2+2\kappa t)]$. Its moments decay algebraically as $t^{-\mu }$. Thus the general equations for the exponential rates are the rate-function minimization above; numerical rates and a finite saturation threshold require the long-time statistics of the flow.

Take a positive initial separation $0<l_0\ll \eta$. In the dissipation range, the velocity field is smooth, so the difference of the two Lagrangian trajectories obeys the linearized separation equation $\dot{\mathbf{l}}\simeq (\nabla \mathbf{u})\mathbf{l}$. Chaotic stretching gives a positive Lyapunov exponent of order ${\tau }_{\eta }^{-1}$ and a typical separationThis needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.

Once $l$ lies in the inertial range, the Kolmogorov 1941 theory gives a relative speed of order $(ϵl{)}^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim (ϵl{)}^{1/3}$ integrates toAfter loss of the entry-scale memory, the statistical Richardson pair dispersion law is $\mathbb{E}[l^2]\sim g_Rϵ(t-t_1{)}^3$. Thus the characteristic separation grows as $(t-t_1{)}^{3/2}$ while $\eta \ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.

For $l\gg L$, the particle velocities become approximately independent. If their Lagrangian velocity autocorrelations have a finite Lagrangian integral time $T_L\sim L/U$, the diffusive large-scale pair dispersion has effective relative diffusivity of order $UL$. After a further velocity-decorrelation transient,This is the long-time mechanism of Taylor turbulent dispersion applied to the difference of two decorrelated trajectories. The root mean square separation eventually loses memory of the $L^2$ term.

Reaching $l=L$ at the end of the inertial range takesOrder-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer eddy turnover time.

The three-regime turbulent pair-separation model in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.

Since the particle starts at the origin, its Lagrangian trajectory satisfies $\mathbf{X}(t)={\int }_0^t\mathbf{v}(s)ds$. Differentiating the mean-square displacement givesHere we set $\tau =t-s$ and assume finite second moments so that averaging and integration commute. Under Lagrangian stationarity write the autocorrelation as $R_L(\tau )$. Uniform fluid-tracer sampling in homogeneous incompressible turbulence gives $R_L(0)=⟨|\mathbf{u}{|}^2⟩$. A second integration yields the exact Taylor turbulent dispersion formulaFor times short compared with velocity decorrelation, continuity at zero gives $R_L(\tau )\simeq R_L(0)$ throughout the integral. ThusThe particle retains nearly its initial velocity: this is ballistic transport.

If $R_L$ is integrable, then at long times the displacement derivative approaches $2{\int }_0^{\infty }{R}_L(\tau )d\tau =2R_L(0)t_L$, using the Lagrangian integral time. ThereforeThis is diffusive transport: successive long-time displacements lose memory of one another. In an isotropic three-dimensional diffusion equation, $⟨X^2⟩=6D_{T}t$, so $D_T=⟨|\mathbf{u}{|}^2⟩t_L/3$. The shorthand $t\ll t_L$ and $t\gg t_L$ presumes that $t_L$ is representative of the decorrelation time. With strongly oscillatory correlations, the area-based integral time need not characterize every crossover; the exact integral formula remains the appropriate starting point.

The three questions concern different statistics. Taylor turbulent dispersion follows the absolute displacement of one marked fluid element from its release position, or the spreading of the centre of a cloud. Its input is the velocity autocorrelation along a Lagrangian trajectory. It relates short-time ballistic motion to long-time effective diffusion. This is not the shear-flow mechanism ordinarily called Taylor dispersion.

Richardson pair dispersion instead follows the separation of two fluid elements, or the growth of a cloud relative to its moving centre. Common sweeping by an eddy much larger than the cloud contributes to absolute displacement but largely cancels from relative displacement. The relative diffusivity depends on separation; the classical inertial-range prediction is a $t^3$ mean-square separation regime after memory of the release separation is lost.

Obukhov-Corrsin theory, also called Kolmogorov-Obukhov-Corrsin theory, concerns spatial fluctuations of a passive scalar, their transfer across scales and their destruction by molecular diffusion. Its objects are scalar structure functions and scalar spectra, with an inertial-convective prediction controlled by scalar-variance flux and the turbulent energy supply. Thus Taylor describes one-particle transport, Richardson describes relative spreading, and Obukhov-Corrsin describes scale-by-scale scalar mixing. Following a cloud's location is not the same as predicting its internal concentration fluctuations.

Let the undisturbed water surface be $z=0$, the oil's upper surface $z=\eta (x,t)$ and its lower surface $z=b(x,t)$, so $h=\eta -b$. Leading hydrostatic pressure in the nearly inviscid water is $p_w=-{\rho }_{w}gz$ relative to atmospheric pressure; in the oil it is $p_H=\rho g(\eta -z)$. Equality at the interface gives $\rho gh=-{\rho }_{w}gb$, henceThus the horizontal hydrostatic driving gradient isThis reduced gravity accounts for most of the oil thickness being below the undisturbed water level.

Because the oil is much more viscous than the water and air, tangential tractions on both faces are negligible. In a thin layer, the leading horizontal velocity is consequently independent of depth: the floating extensional viscous gravity current is a plug, rather than the parabolic profile of a no-slip gravity current. Integrating incompressibility $u_x+w_z=0$ between the moving material boundaries, and using their kinematic boundary condition, gives conservation of oil volume,To derive the factor $4$ in the extensional balance, incompressibility gives $w_z=-u_x$. The Newtonian fluid stress tensor has ${\sigma }_{zz}=-p-2\mu u_x$. Normal-stress balance relative to the hydrostatic part therefore requires $p=p_H-2\mu u_x$. The horizontal stress then satisfiesThus the longitudinal tensile force in an oil slice, relative to its hydrostatic pressure, is $4\mu h{u}_x$ per unit width. Balancing its difference across the slice with the horizontal driving force $-h{p}_{H,x}dx$ yieldsThe coefficient is the planar extensional Trouton ratio, arising from both the direct horizontal strain and the pressure correction required by the transverse normal-stress condition.

Using the pressure gradient above, integration in $x$ givesThe stated nose stress condition sets $C(t)=0$, not an arbitrary imposed inlet tension. ConsequentlyFor a material slice injected at time $t_0$, let $\tau =t-t_0$ be its age. The thickness equation along its Lagrangian trajectory is $dh/dt=-h{u}_x=-k{h}^2$, with $h=h_0$ at injection. DefineThen $h=h_0/(1+\tau /T)$. The slice injected during $d{t}_0$ has volume $h_0{u}_{0}d{t}_0$ per unit width, so at fixed observation time $t$,Since a newly injected slice has $x(t,t)=0$, integration and subsequent differentiation along a material slice give the constant-flux floating extensional gravity current:The oldest slice forms the nose. Measuring time from the start of injection, $t_0=0$ there, soIts speed ${\dot{x}}_N=u_0(1+t/T)$ is exactly the material velocity at the nose, as required.

For a fixed position $x$ behind the nose, the age needed to reach it is determined by $x/u_0=\tau +{\tau }^2/(2T)$, independently of the observation time. Hence both thickness and velocity become stationary there once the nose has passed. Eliminating $\tau$ givesTheir product is $h_0{u}_0$, so the steady profile carries precisely the inlet volume flux; integrating it to $x_N(t)$ gives the total injected volume $h_0{u}_{0}t$.

The lengthis the horizontal distance over which reduced-gravity driving causes an order-one change in inlet speed and thickness by viscous extension. The associated age is $T=8\ell /u_0$; the exact profile contains $x/(4\ell )$. It is an extensional adjustment length, not a shear-diffusion scale. The result applies while the stated thin-layer and negligible-inertia approximations remain valid.

The MHD induction equation can be written in material form asIts $y$ component immediately givesFor the Cartesian magnetic flux function, the $x,z$ components of the MHD induction equation, or the $y$ component of the magnetic vector potential equation, giveThe additive function of time in $\psi$ has no effect on $\mathbf{B}$. Choose this gauge so that $C=0$. Then the flux label is materially conserved:This is magnetic flux freezing in the two-dimensional geometry.

To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each Lagrangian trajectory, $\psi$ is fixed, and the assumed divergence givesThus $B_y$ remains a function of $\psi$ and time alone. Its gradient stays parallel to $\nabla \psi$, so the $y$ component of the Lorentz force remains zero. The axial momentum equation isbecause $(B_x{\partial }_x+B_z{\partial }_z)\psi =0$. With $u_y=0$ initially, it remains zero. Hence neither axial force nor axial motion is generated. This flux-surface preservation during magnetic-tube expansion argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.

Parametrize a streamline by $\mathbf{r}(s)$ with $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r}(s))$. The chain rule givesTherefore $H$ is constant along every streamline. A different regular parametrization merely multiplies this derivative by a scalar and gives the same conclusion. For time-dependent flow, the claim concerns streamlines at a fixed time, rather than Lagrangian trajectories.

The dyed particle satisfies the Lagrangian trajectory equationsFor the proposed approximation, $x=O({\epsilon }^2)$ andAlsoThe initial conditions hold, verifyingthrough order ${\epsilon }^2$.

Over one period, the periodic terms return to their initial values while the secular term changes $x$ by ${\epsilon }^2\pi$. Hence the dyed particle has Stokes driftto this order, despite the zero Eulerian mean velocity.

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.