Set $t=0$ just inside the entrance, $z=0$, $\rho ={\rho }_0$ and $\theta ={\theta }_0$. The stated continuity gives $\dot{\rho }(0)=0$. Until the first axis crossing,For the full orbit, including any crossings, use the transverse complex coordinate $u=x+iy$:This helical solenoid trajectory with zero incident angular momentum is a circle of radius ${\rho }_0/2$ about transverse guiding centre $u_0/2$, combined with uniform axial motion. Its cyclotron angular frequency is $|qB|/(\gamma m)=2|\Omega |$, and its axial pitch per turn is $\pi w/|\Omega |$. The speed check is $|\dot{u}{|}^2+w^2={\Omega }^2{\rho }_0^2+w^2=v^2$.

The physical radius is $|u|={\rho }_0|\cos \Omega t|$. When the cosine changes sign, a nonnegative-radius coordinate adds $\pi$ to $\theta$; this is a coordinate change, not a discontinuous turn of the particle. For $q=0$ or $B=0$, the limit is simply $u=u_0$, $z=vt$.

In a uniform-field region, $z$ is cyclic and $\gamma$ is constant. Its equation is $d(\gamma m\dot{z})/dt=0$. The radial Euler-Lagrange equation isso with $\dot{\theta }=-\Omega$ and $qB/(\gamma m)=2\Omega$,The radial equation applies on intervals with a regular signed cylindrical-coordinate description; the Cartesian solution handles axis crossings.

In a field-free region, the equations areFor this particle $p_{\theta }=0$, so $\dot{\theta }=0$ and $\ddot{\rho }=0$ outside, until a coordinate change is needed at an axis crossing.

Axial velocity is constant within each uniform region, but not across the ends. In a smooth fringe,The entrance develops an azimuthal speed $|\Omega |{\rho }_0$ while radial velocity remains zero. Conservation of total speed givesForward passage presumes the radicand is positive. The exit fringe changes axial velocity again. Neglecting these small changes is a paraxial approximation, requiring $|\Omega |{\rho }_0\ll v$, rather than an exact consequence of cylindrical symmetry.

The azimuth is a cyclic coordinate, so its canonical momentum is conserved:Initially the velocity is parallel to the axis in a field-free region, so $p_{\theta }=0$. Conservation of canonical angular momentum across a solenoid fringe therefore gives, inside the uniform field,This holds away from the coordinate singularity $\rho =0$; the Cartesian orbit will be smooth there.

Time independence also conserves the energy Hamiltonian,The velocity-linear magnetic term cancels in this Legendre transform. Thus $\gamma ={\gamma }_v=(1-v^2/c^2{)}^{-1/2}$ is constant, as expected because a static magnetic Lorentz force does no work. Mechanical angular momentum can change even though $p_{\theta }$ and the total speed are conserved.

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.

Position-dependent density. The density ${\mathcal{L}}_1$ remains independent of $\psi$ itself, so the shift Noether charge survives with its altered canonical momentum:A prescribed generic function $\rho (z)$ supplies a fixed spatial scale and breaks scale invariance. For the dilation-current candidate with the corresponding ${\mathcal{H}}_1$ and ${\Pi }_1$, the scale-breaking sources in derivative field theories giveThus $Q_2$ is generally not conserved. Constant density recovers the earlier case; specially scale-invariant profiles and suitably invariant domains would require separate treatment.

Quartic strain. The extra term $-\alpha ({\psi }^{\prime }{)}^4$ in ${\mathcal{L}}_2$ again preserves a uniform field shift. Its time momentum is unchanged, while its spatial current becomes $-AK{\psi }^{\prime }-4\alpha ({\psi }^{\prime }{)}^3$. ThereforeThe new Hamiltonian density is ${\mathcal{H}}_2={\mathcal{H}}_0+\alpha ({\psi }^{\prime }{)}^4$. Under waveform dilation, the quartic term scales with $b^4$ rather than $b^2$, andHence there is no corresponding conserved $Q_2$ for nonzero generic $\alpha$, even if the energy in its candidate expression is updated.

Field-dependent quadratic stiffness. In ${\mathcal{L}}_3$, the term $-\beta {\psi }^2({\psi }^{\prime }{)}^2$ breaks the uniform field shift:when spatial flux vanishes. Thus $Q_1$ is not generally conserved. However all terms still contain exactly two derivatives in total, with a field of zero dilation weight. The factor ${\psi }^2$ gains no extra power of $b$, so waveform dilation remains an action symmetry. Its charge isHere $\Pi =A\rho \dot{\psi }$ as before. Relative to the original charge, the energy term changes; the symmetry survives despite the nonlinear field dependence. These statements concern symmetry conservation, separately from any restrictions on parameters needed for positive elastic energy.

For a rightgoing wave packet, put $\xi =z-ct$ and assume $f^{\prime }$ decays sufficiently rapidly. ThenThe shift charge becomesIt is time independent. It vanishes for a localized displacement packet that returns to the same displacement at both ends, but need not vanish for a localized strain pulse joining different asymptotic displacements.

For the dilation charge, substitute $z=\xi +ct$:The change of variables removes all time dependence. This charge also need not vanish; choosing an origin at the packet's energy centroid can make it zero. Conservation assumes the integral exists and there is no boundary flux, rather than merely assuming that the energy is finite.

Both Noether charges require boundary conditions that admit the stated transformations and no net boundary-current flux. Take a sufficiently localized field on the whole rod, with the weighted integrals below finite. Fixed endpoints at finite positions generally spoil the waveform-dilation symmetry.

Write $A=\pi a^2$ for the cross-sectional area, and use $z$ consistently for the longitudinal coordinate: the paper's later $x$ is the same coordinate. The canonical momentum is a density per reference length,The Legendre transform gives the Hamiltonian density in its proper canonical variables:The Euler-Lagrange equation isFor a twice-differentiable wave packet $\psi =f(z-ct)$, the derivatives are $\ddot{\psi }=c^2{f}^{\prime \prime }$ and ${\psi }^{\prime \prime }=f^{\prime \prime }$. Thus the nontrivial packet solves the wave equation whenThe negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $\mathcal{H}$ has units $\mathrm{J}{\mathrm{m}}^{-1}$, and $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. The result follows from the supplied axial stiffness in the Lagrangian. For a freely contracting thin elastic rod the relevant longitudinal modulus is normally Young's modulus, rather than the bulk modulus; the question's given Lagrangian fixes which coefficient is being used.

For dissolved, nonadsorbing polyethylene glycol, the usual new interaction is a depletion force. Polymer centres are excluded from a layer of thickness of order their radius of gyration $R_g$ around each particle. When the particles approach, their exclusion layers overlap. This increases the volume accessible to polymer and its translational entropy, lowering the colloid pair free energy. Equivalently the surrounding polymer osmotic pressure pushes the particles together.

In the dilute-polymer Asakura-Oosawa model, let $h=R-2a$ be the surface gap and $R_e=a+R_g$. Thenfor $2a\le R<2R_e$, and $U_{\mathrm{dep}}=0$ for $R\ge 2R_e$. Here ${\Pi }_p\simeq n_p^{(\mathrm{polymer})}{k}_{B}T$ is the reservoir polymer osmotic pressure. The overlap-volume formula follows by adding two equal spherical cap volumes with height $R_e-R/2$. Thus the added attraction has range about $2R_g$ and grows with polymer concentration. The total interaction is approximately hard exclusion plus screened electrostatic repulsion plus Van der Waals force and depletion attractions.

Increasing polymer concentration therefore creates a family of phase diagrams, rather than a unique shift determined by polymer size alone. Where the attraction overcomes repulsion, colloid-poor fluid can coexist with a colloid-rich phase, and fluid–crystal coexistence generally broadens. For sufficiently long-ranged attraction, stable fluid–fluid separation can occur. For the stipulated small polymer, the attraction is short-ranged: fluid–fluid separation can be metastable to crystallization, and a rapid attraction increase can instead produce an arrested colloidal gel. Residual long-ranged repulsion can oppose bulk separation. Simulations and free-volume calculations confirm that longer-ranged electrostatic repulsion can counteract depletion-driven separation.^[ref]

The nonadsorption assumption matters. If the polyethylene glycol adsorbs to the particles, a dense attached layer can give steric stabilization, whereas chains attaching to two particles can produce polymer bridging. The paper does not specify surface affinity or polymer concentration, so depletion is the standard conditional answer, not a prediction of an inevitable particular final phase.

First hold the surface electric charge fixed and increase the bulk salt concentration. Since ${\kappa }^2\propto c$, the Debye–Hückel screening length decreases as $c^{-1/2}$. For surface gap $h=R-2a\ge 0$,Thus both the range and the strength of repulsion decrease. In the repulsive phase diagram, the sample moves downwards at fixed volume fraction and can melt from a crystal into a fluid. Because screening length is already a diagram axis, changing salt moves the state through that diagram rather than independently changing its fixed-charge coexistence curves. If Van der Waals forces become dominant, however, adding salt may instead destabilize the dispersion and cause aggregation. These are distinct physical regimes.

Second increase $\sigma$ at fixed salt concentration and particle size. The electric charge is $Q=4\pi a^2\sigma$, so the pair repulsion increases approximately as ${\sigma }^2$ within the Debye–Hückel approximation, while the reservoir screening length remains unchanged. Freezing then becomes possible at lower volume fraction or shorter screening length. At fixed ${\lambda }_D$, the coexistence curves shift towards lower $\eta$; a previously fluid sample may crystallize.

These statements assume fixed, uniform charge and a salt reservoir dominating the screening. Strong charging requires nonlinear screening and possibly charge regulation. At very low added salt or appreciable particle concentration, particle-released counterions also contribute to ionic strength, so the screening length is no longer independently fixed by the added salt alone.

Write the particle volume fraction as $\eta =(4\pi a^3/3)n_p$. In a stable, repulsion-dominated colloid, low $\eta$ gives a disordered fluid. Increasing $\eta$ makes the interaction between neighboring particles important and can produce an ordered crystal, separated from the fluid by a phase coexistence curve. Increasing the Debye–Hückel screening length at fixed charge strengthens the repulsion at a given separation and allows freezing at smaller geometrical volume fraction.

The geometric intuition is that particles behave as though they occupy more than their bare diameter: roughly, an effective exclusion diameter $d_{\mathrm{eff}}$ extends out to the separation where $U_{\mathrm{el}}$ is of order $k_{B}T$. A hard sphere comparison then uses ${\eta }_{\mathrm{eff}}\sim \eta [d_{\mathrm{eff}}/(2a){]}^3$. This explains the trend; it is not an exact freezing criterion.

At short Debye–Hückel screening length and negligible attraction, the diagram approaches hard sphere fluid–crystal coexistence. For longer-ranged soft repulsion, body-centered cubic lattice and face-centered cubic lattice crystals can both be stable, with the former often appearing towards the softer, lower-density freezing boundary. Screened-Coulomb simulations exhibit both these crystal structures.^[ref]

No universal quantitative diagram follows from $(\eta ,{\lambda }_D)$ alone: the charge, Hamaker constant and surface conditions must also be fixed. If salt suppresses the stabilizing barrier enough for Van der Waals force attraction to dominate, a colloid-poor/colloid-rich separation or aggregation can replace part of this repulsive reference diagram. Dense colloidal glasses or colloidal gels may also be observed after kinetic arrest; they should not be mistaken for additional thermodynamic equilibrium thermodynamic phases.

Let $\epsilon ={\epsilon }_0{\epsilon }_r$ be the permittivity, $\beta =(k_{B}T{)}^{-1}$, $e>0$ the elementary charge, and $c$ the bulk salt formula-unit number density. The Boltzmann distribution givesConsequently the exterior Poisson-Boltzmann equation isFor small $Q$, meaning $2\beta e|\varphi (a)|\ll 1$, linearize the exponentials. The Debye–Hückel approximation givesThis is also ${\kappa }^2=\beta e^2{\sum }_i{z}_i^2{n}_{i,\infty }/\epsilon$: the divalent cation contributes $4c$ and the two monovalent anions contribute $2c$. In terms of molar salt concentration $C$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, replace $c$ by $N_{A}C$; if $C$ is in $\mathrm{mol}{\mathrm{L}}^{-1}$, use $c=10^3{N}_{A}C$. The molar ionic strength is $3C$ in the same concentration units.

The boundary conditions are $\varphi (\infty )=0$ and, by Gauss's law,Here the surface charge is uniform and the spherically symmetric interior field is zero. Set $u=r\varphi$; the radial equation reduces to $u^{\prime \prime }={\kappa }^{2}u$. Decay excludes the growing exponential, so write $\varphi =B{e}^{-\kappa (r-a)}/r$. Then ${\varphi }^{\prime }(a)=-B(1+\kappa a)/a^2$, givingIt solves the linearized Poisson-Boltzmann equation, not the full nonlinear equation at finite $Q$. The radial ion profiles, consistently to first order, areThe exponential Boltzmann distribution formulas above express the full mean-field relation, but substituting a linearized potential into them is accurate only to the retained order. A useful electroneutrality check isThe result is the Debye-Hückel potential of a charged sphere.

The particle carries positive surface charge density $\sigma =Q/(4\pi a^2)$. Its surrounding electric double layer has an excess of negative chloride counterions and a deficit of positive magnesium coions. Thus $\varphi (r)>0$ decreases to zero, $n_{\mathrm{Cl}}(r)$ decreases towards its bulk value, and $n_{\mathrm{Mg}}(r)$ increases towards its bulk value. The diffuse charge density is negative and tends to zero.

If $c$ denotes the bulk number of dissolved salt formula units per unit volume, the far-field number densities are $n_{\mathrm{Mg},\infty }=c$ and $n_{\mathrm{Cl},\infty }=2c$. The bulk is neutral even though these number densities differ. In the dilute mean-field picture the total diffuse electric charge is $-Q$, ensuring overall electroneutrality. A molecularly bound compact layer would require additional surface-chemistry information and is not assumed in the supplied potential.

In a magnesium chloride chemical solution, the principal pair interactions are hard sphere exclusion, repulsion of overlapping electric double layers, and attractive Van der Waals forces. The positive surface attracts chloride counterions and repels magnesium coions; mobile ions screen the bare Coulomb's law interaction. The competition between screened electrostatic repulsion and Van der Waals force attraction is described by DLVO theory.

For well-separated identical weakly charged spheres, the approximate repulsive pair potential energy, at centre separation $R\ge 2a$, isIt is a screened Coulomb pair potential. An attractive Hamaker constant adds a negative short-range contribution, while overlap is forbidden. Very near contact, solvation and molecular surface structure can also matter. No polymer-induced interaction should be assumed before polymer is added in part (f).

Each addition starts a new statistically identical waiting episode at zero gap. Thus the addition rate is the reciprocal of a mean first-passage time, not the reciprocal of the mean gap:This can also be derived without the steady density. If $T(x)$ is the mean time to reach $\delta$ starting at gap $x$, its backward Smoluchowski equation isThe first boundary is reflecting and the second absorbing. An integrating factor gives $T^{\prime }(x)=[1-e^{\beta Fx}]/(D\beta F)$, whenceAt weak force, $\tau \simeq {\delta }^2/(2D)$ is the time for a freely diffusing wall to move a monomer length, with reflection at the tip. At strong force, opening the gap requires work $F\delta$, with Boltzmann distribution suppression $e^{-\beta F\delta }$. The drift establishes a near-tip thermal length ${\ell }_F=k_{B}T/F$ and a diffusion time ${\ell }_F^2/D$; consequentlyThe exponentially long wait explains why the growth slows. There is no additional chemical waiting time in the imposed absorbing-threshold model; such a time would be needed to describe the separate reaction-limited polymerization Brownian ratchet regime.

For a small opposing force, the relevant parameter is $q=\beta F\delta \ll 1$. A Taylor series givessoIn particular the zero-force rate is finite, despite the numerator and denominator of the original expression both vanishing.

For $q\gg 1$, the exponential dominates the denominator:Opening a monomer-sized gap is exponentially rare at high opposing force. The algebraic $F^2$ prefactor belongs to this first-passage model and must not be discarded when quoting the leading asymptotic rate.

Put $k=\beta F$ and $q=k\delta$. The absorbing boundary condition for a diffusion gives $A=j_0{e}^q/(Dk)$, henceThis is nonnegative throughout the allowed interval. Normalizing the probability density givesThereforeEquivalently the normalized density isThe denominator is positive for $q>0$, since $e^q>1+q$. The dimensions are consistent: $j_0$ has units of inverse time, $A$ and $p$ of inverse length, and $q$ is dimensionless.

In this polymerization Brownian ratchet model, a gap of width $\delta$ immediately accommodates one monomer. That event advances the tip by $\delta$ and resets the gap to zero. The threshold is therefore an absorbing boundary condition for a diffusion,All waiting-time probability lies within $0\le x<\delta$, so the steady occupation probability density satisfiesThere is a reinjection flux $j_0$ at $x=0$ equal to the absorption flux at $x=\delta$. The nonpenetration constraint at zero is a reflecting boundary condition for a diffusion during each individual waiting episode. For the stationary reset process it must not be confused with setting the net current $J(0^{+})$ to zero: the source at reset supplies $J(0^{+})=j_0$. This is how a normalized steady density can coexist with a nonzero exit flux.

In the steady state, conservation of probability requires $J_x=0$, so $J=j_0$ is constant. The sign is positive because increasing gap reaches the incorporation boundary, whereas the deterministic drift points towards smaller gap. ThusAn integrating factor $e^{\beta Fx}$ givesSubstitution makes the check particularly transparent:so ${\partial }_{t}p=-J_x=0$. The constant offset is necessary for nonzero transport; a pure Boltzmann distribution would have $J=0$ and could not describe a steady succession of addition events. The $F=0$ limit is taken after applying the boundary conditions, avoiding the apparent singularity of the displayed parametrization.

The gap is not a fixed mechanical separation. Thermal fluctuations of the wall compete with drift towards the microtubule under the opposing force. A monomer can be incorporated only after the gap reaches its length. The Smoluchowski equation describes this overdamped drift and diffusion of gap probability density.

For gap potential energy $U(x)=Fx$, the equation and its probability current areThe drift mobility is $D\beta$, the overdamped consequence of the fluctuation-dissipation relation for a Langevin particle. Equivalently, this coefficient is fixed by requiring the zero-current solution to be the Boltzmann distribution $p\propto e^{-\beta U}$. Solve this equation with the appropriate exclusion, incorporation and probability-normalization conditions. The steady flux through the incorporation threshold gives the addition rate $j_0$; the elongation speed is $v=\delta j_0$.

There is a regime distinction in the printed problem. The absorbing boundary imposed in part (c) means instantaneous incorporation once a gap opens, and leads to the diffusion-limited Brownian ratchet with constant load. A genuinely reaction-limited polymerization Brownian ratchet instead assumes fast gap equilibration followed by a finite chemical addition rate, giving an exponential gating factor multiplying that chemical rate. The equations requested in the subsequent parts specify the former model, which is the one solved here.