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Use the same resistive-force theory convention as before: the force density is the fluid's force on the filament. Let $\phi$ be the usual cylindrical azimuth. Choose the orientation of a left-handed helix so that, as $z$ increases, $\phi$ decreases. With $s=\sin \theta$, $c=\cos \theta$,Reversing the orientation of the tangent leaves the local drag tensor unchanged. Define $k={\xi }_{\perp }-{\xi }_{\parallel }$. From resistive-force theory, the relevant force-density components areThese scalar axial and azimuthal components are uniform along the helix. Integration over its contour length $\ell$, and use of the axial torque density $a{f}_{\phi }$, give the axial resistance matrix of a slender helix:Thus $B=C$, as also required by the Lorentz reciprocal theorem. The signs here belong to $(F,L{)}^T=-\left(\begin{array}{cc}A& B\\ B& D\end{array}\right)(U,\Omega {)}^T$, with $F,L$ forces and moments on the helix. The handedness reversal of helical hydrodynamic resistance reverses $B,C$ but leaves $A,D$ unchanged. If the positive rotation direction or handedness convention is reversed, the coupling sign reverses with it.

As a check on physical admissibility, the determinant of helical resistance in resistive-force theory isTogether with $A,D>0$, this makes the hydrodynamic resistance matrix positive definite and the viscous power loss $A{U}^2+2BU\Omega +D{\Omega }^2$ positive. Isotropic local drag would have $B=0$, so rotating a helix would not propel it in this approximation.

For the pair, use a local additive resistive-force theory model: both helices have the same drag coefficients, and inter-helix hydrodynamic interactions and unresolved end effects are omitted. The right-handed helix has resistance entries $(nA,-nB,-nB,nD)$. Define the signed motor motion byThe printed relative rotation specifies a magnitude rather than which helix rotates relative to which. Choosing ${\Omega }_2-{\Omega }_1=\omega$ instead reverses all signed speeds below.

The common translation speed $U$ and zero total force and torque obeyInternal motor forces and torques cancel in these totals. IntroduceSolving the three linear equations yields the opposite-handed counterrotating helical swimmer:For $n>0$, ${\xi }_{\perp }>{\xi }_{\parallel }$ and $\omega >0$, $B>0$, so $U<0$, ${\Omega }_1>0$ and ${\Omega }_2<0$. The opposite-handed helices counterrotate but contribute thrust in the same axial direction. The formula satisfies the specified relative rotation without identifying either laboratory rotation rate with the motor rate.

At $n=0$, the second helix supplies no hydrodynamic resistance. The remaining helix must have $F=L=0$, and invertibility of its hydrodynamic resistance matrix forces $U={\Omega }_1=0$. Formally ${\Omega }_2=-\omega$ is the rotation of a zero-resistance motor shaft or vanishing second rotor; there is no physical finite second helix to propel or to provide a reaction torque. This is the vanishing reaction rotor in a helical swimmer.

At $n=1$, equal-length opposite-handed helices haveThe translation-generated axial torques cancel between the two helices, and equal counterrotation balances the rotational torques. Their propulsive forces add, rather than cancel, because handedness and rotation both reverse. These are equal-length opposite-handed helices.

For $n\to \infty$, the large reaction helix limit isThe increasingly long second helix nearly anchors the whole assembly: its very small translation and rotation suffice to balance the finite force and torque generated by the first helix. The first helix rotates at almost the full motor rate, but moving the large resistive second helix makes the common translation tend to zero. These are idealized limits of the additive local model, rather than a resolution of short-helix end effects or infinitely extended interacting filaments.

Take $\mathbf{F},\mathbf{G}$ to be the force and couple exerted by the body on the fluid, equivalently the external force and couple needed to maintain its motion. This fixes the sign for a positive hydrodynamic resistance matrix. If one uses the fluid's force on the body, both resultants have the opposite sign.

For two Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\tilde{\mathbf{u}},\tilde{\mathbf{\sigma }})$ in the same fluid domain $\mathcal{D}$, with zero body force, the Lorentz reciprocal theorem for Stokes flow isHere $\mathbf{n}$ points out of the fluid. To see the identity, take the divergence of the difference of the two cross-work fluxes. The stress divergences vanish, while incompressibility and symmetry of the Newtonian fluid stress tensor reduce the remaining terms to $2\mu (\mathbf{e}:\tilde{\mathbf{e}}-\tilde{\mathbf{e}}:\mathbf{e})=0$. The divergence theorem proves the result.

On the body, the no-slip boundary condition is $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{r}$, and the far-field contribution vanishes for decaying exterior flows. Reciprocity becomesWith $q=(\mathbf{U},\mathbf{\Omega }{)}^T$ and $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}q$, this is $q^T\mathsf{R}\tilde{q}={\tilde{q}}^T\mathsf{R}q$ for all pairs, so ${\mathsf{R}}^T=\mathsf{R}$. The power identity givesEquality would force zero strain throughout the connected exterior domain, hence a rigid fluid motion. Decay at infinity makes that motion zero, and the no-slip boundary condition then gives $q=0$. This proves the symmetry and positivity of a rigid-body resistance matrix:

For the helix, assume $0<\varphi <\pi /2$ and use cylindrical unit vectors. Its arclength element and tangent areIntegrating to $\theta =L\cos \varphi /b$ gives $\boxed{\text{total arc length}=L}$. In a combined axial translation and rotation, $\mathbf{V}=U{\mathbf{e}}_z+b\Omega {\mathbf{e}}_{\theta }$. The slender-body force density, or local resistive-force theory, givesWriting $s_{\varphi }=\sin \varphi$, $c_{\varphi }=\cos \varphi$, its axial and azimuthal components areThe axial couple density is $b{f}_{\theta }$. Integration over arclength gives the axial resistance matrix of a slender helix:Thus pure translation gives $(F_z,G_z)=(AU,BU)$, and pure rotation gives $(F_z,G_z)=(B\Omega ,D\Omega )$. The mixed coefficients agree, as reciprocity requires, and $AD-B^2=C^2{L}^2{b}^2/2>0$. The sign of $B$ follows the handedness in the parametrization; reversing handedness reverses propulsion.

For the helical microswimmer with a spherical head, the diagram's rotation relation is ${\Omega }_0=\Omega -\omega$. Neutral buoyancy and the absence of external forcing make the whole swimmer force-free and torque-free. Neglecting interactions between its parts givesWith $\Delta =(A_0+A)(D_0+D)-B^2>0$, solving this pair givesThe head counterrotates, providing the reaction to the flagellum's rotation. The figure fixes this relative rotation convention; it is duplicated and corrupted in the local TeX.

For a very large head, the coefficient regime is $A_0\gg A$ and $D_0\gg D$. ConsequentlyThe large translational resistance $A_0$ makes the microswimmer slow even though the flagellum rotates almost at the motor rate.

For a very small head, $A_0\ll A$ and $D_0\ll D$. The appropriate denominator retains the translation-rotation coupling:Both tend to zero with $D_0\propto a^3$, while ${\Omega }_0\simeq -\omega$. The motor mainly rotates the low-resistance head, and produces little flagellar motion relative to the fluid.

In the intermediate coefficient regime $A_0\ll A$ and $D_0\gg D$, the head supplies a strong rotational reaction with little translational penalty. ThenThe head radius has dropped out. To find the optimal pitch of a helical microswimmer, put $z=\tan \varphi$. The dimensionless speed becomes $z/(z^2+2)$ and has derivative $(2-z^2)/(z^2+2{)}^2$. Hence, for $\omega >0$ with the chosen handedness,The angle is measured from the horizontal plane. An intermediate pitch combines axial and transverse tangent directions, allowing anisotropic drag to convert rotation into translation; either extreme removes that coupling.

The geometric head-size regimes quoted in the paper hold with the logarithmic slenderness factor treated as fixed. More precisely, $A_0\ll A$ requires $a\ll L(1+{\cos }^2\varphi )/(3|\log ϵ|)$, while $D_0\gg D$ requires $a^3\gg L{b}^2(1+{\sin }^2\varphi )/(4|\log ϵ|)$. These coefficient inequalities state the validity of the intermediate approximation when the logarithm is quantitatively important. The many-turn and local-drag assumptions of slender-body theory also remain in force.

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.

On the circular cylinder, $\rho =R$ is constant, so the given line element reduces to $d{s}^2=R^{2}d{\varphi }^2+d{z}^2$. Unroll the circular cylinder with $u=R\varphi$. The line element is then $d{s}^2=d{u}^2+d{z}^2$, so the usual Euclidean geometry formula measures arc length in the $(u,z)$ plane.

The endpoint has infinitely many lifts $(R({\varphi }_1+2\pi k),z_1)$, with $k\in \mathbb{Z}$. For each lift, the straight line segment is shortest; its length is $\sqrt{R^2({\varphi }_1+2\pi k{)}^2+z_1^2}$. Comparing the lifts gives the unrestricted shortest path on a circular cylinder:If $\delta$ is a minimizing angular increment, a shortest geodesic is parametrized by $(\rho ,\varphi ,z)=(R,\delta t,z_{1}t)$, $0\le t\le 1$. This is a helix when both increments are nonzero, and a circular arc or vertical generator in the limiting cases. A nonminimal lift also gives a geodesic, but it is only shortest within its fixed winding class.

For $0\le {\varphi }_1\le 2\pi$, put $m=\min ({\varphi }_1,2\pi -{\varphi }_1)$. Then $\boxed{L_{\min }=\sqrt{R^2{m}^2+z_1^2}}$. At ${\varphi }_1=\pi$, both winding directions give distinct shortest paths of equal length.

Put $L=5\pi /2$. The curve tangent is ${\mathbf{r}}^{\prime }=(-b\sin \theta ,b\cos \theta ,c)$. For the first line integral, the scalar triple product identity gives $(\mathbf{a}\times \mathbf{r})\cdot {\mathbf{r}}^{\prime }=\mathbf{a}\cdot (\mathbf{r}\times {\mathbf{r}}^{\prime })$. Its cross product isThe needed definite integrals areand ${\int }_0^L(-\theta \sin \theta -\cos \theta )d\theta =[\theta \cos \theta -2\sin \theta {]}_0^L=-2$. HenceFor the second line integral, substitute the curve into ${\mathbf{v}}_2\cdot {\mathbf{r}}^{\prime }$. The cubic terms combine to $b^3\cos \theta$, leavingThe fundamental theorem of calculus therefore givesBoth calculations retain the full path from $\theta =0$ to $5\pi /2$, rather than inadvertently closing the helix after one turn.

Integrating ${\varphi }_x=2xy$ gives $\varphi =x^{2}y+h(y,z)$. Matching ${\varphi }_y=x^2+3y^2+z^2$ then gives $h=y^3+y{z}^2+k(z)$. Finally ${\varphi }_z=2yz$ forces $k$ to be constant. A scalar potential isThe initial and final points of the helix are $(b,0,0)$ and $(0,b,cL)$, where $L=5\pi /2$. The fundamental theorem for line integrals givesagreeing with the directly evaluated $I_2$.

Substituting the endpoint parameter values givesThe curve lies on the unit circular cylinder $x^2+y^2=1$. Its azimuth increases from zero to $\pi /2$ while its height increases linearly, so it is one quarter-turn of a rising helix. The arrow in the sketch indicates increasing parameter.

Vary $r$ by $r+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation of the functional isThe boundary term from integration by parts is zero. The fundamental lemma of the calculus of variations therefore gives $\boxed{F_{1,r}-d{F}_{1,r^{\prime }}/dz=0}$. For $I_2$, independently vary $r$ and $\theta$ with zero endpoint variations; the two Euler-Lagrange equations areFor the optical path, Fermat principle makes $\int \mu ds$ stationary. In cylindrical coordinates its integrand is $L=\mu (r,\theta )s$, with $s=\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$. There is no explicit $z$ dependence. The Beltrami identity for several dependent variables givesThe conserved quantity is the axial component of the optical momentum of a ray. As $\mu >0$, squaring and inverting yieldsNow suppose $\mu =\mu (r)$ and seek a path of radius $R>0$. The angle is a cyclic coordinate, so $L_{{\theta }^{\prime }}=\mu R^2{\theta }^{\prime }/\sqrt{1+R^2{\theta }^{\prime 2}}$ is conserved. At fixed radius this is strictly increasing in ${\theta }^{\prime }$, forcing ${\theta }^{\prime }=\omega$ constant. The radial Euler-Lagrange equation reduces toConsequently the helical rays in a radially graded refractive index haveA real nonzero helix exists at such a radius when $0<-R{\mu }^{\prime }(R)/\mu (R)<1$, and its helical pitch is $2\pi /|\omega |$. If ${\mu }^{\prime }(R)=0$, the solution is instead a straight axial light ray. Thus radial dependence permits constant-radius rays for suitable profiles; an arbitrary radial profile need not support a nonzero real helix. This qualification is necessary, for example when the refractive index increases with radius.

On the cylindrical surface $r=R$, the line element is $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. For a single-valued path parametrized by increasing $\theta$,The integrand has no explicit $z$, so its Euler-Lagrange equation gives the constant conjugate quantity $z^{\prime }/\sqrt{R^2+z^{\prime 2}}$. Its derivative with respect to $z^{\prime }$ is strictly positive, hence $z^{\prime }$ itself is constant. The two endpoints determineThis is a constant-pitch helix on the cylinder. To verify global minimality, unroll the cylinder with planar coordinates $(R\theta ,z)$. The path becomes a straight line between points separated by $\pi R$ and $z_2-z_1$, so the triangle inequality for arc length givesThis is the standard description of geodesics on a circular cylinder.

For steady currents in vacuum, Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enclosed}}$. A rectangular loop with one long side inside an ideal long solenoid and the other outside gives $B\ell ={\mu }_0(n\ell )I$, because the exterior field is negligible and the perpendicular sides contribute zero. Thus $B={\mu }_{0}nI$, approximately uniform away from the ends.

Write the electron electric charge as $q_e=-e$, with $e>0$. The Lorentz force is $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$; it does no work. For perpendicular motion, $m{v}^2/r_L=evB$ givesHere ${\omega }_c$ is the positive angular speed; the rotation sense is determined by the negative electron electric charge.

The parallel velocity is unaffected, while the perpendicular component rotates. Hence the path is a helix with $v_{\parallel }=v\cos \theta$ and $r_L=mv\sin \theta /(eB)$. For on-axis entry into a solenoidal tube, the transverse circle passes through that axis: its guiding centre is displaced by $r_L$, and its maximum distance from the tube axis is $2r_L$, not $r_L$.

Using the nonrelativistic approximation appropriate to $1\mathrm{keV}$,The cyclotron period is $2\pi /{\omega }_c=0.893\mathrm{ns}$, so the flight lasts $4.783$ revolutions. The electron therefore crosses the axis again four times before leaving. At $3^{\circ }$, $2r_L=0.279\mathrm{mm}<a$. Wall clearance requires $2mv\sin \theta /(eB)\le a$, givingThe transit is longer than a full orbit, so the maximum transverse excursion really is sampled before exit.

For the solenoidal time-of-flight energy window, the prescribed time fixes $v_{\parallel }=l/t_0=v_0=\sqrt{2E_0/m}$, while $0\le v_{\perp }\le eBa/(2m)$. ThereforeThese are total kinetic energies. The upper value is the ideal grazing limit; if touching the wall counts as rejection, it is a supremum approached from below.

Write the bead position in Cartesian coordinates as ${\mathbf{x}}_B=(a\cos {\theta }_B,a\sin {\theta }_B,b{\theta }_B)$. Differentiation gives $|{\dot{\mathbf{x}}}_B{|}^2=(a^2+b^2){\dot{\theta }}_B^2$, while the free particle has $|\dot{\mathbf{x}}{|}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. Thus the kinetic energy is the sum of these squared speeds weighted by half the respective mass. The squared spring length isA spring obeying Hooke's law with zero natural length has potential energy $k{\ell }^2/2$. Consequently $L=T-k{\ell }^2/2$, and deleting the additive constant $-k{a}^2/2$ givesThe spring length is the straight separation of its endpoints, not distance measured along the helix. The fixed frictionless constraint is already incorporated by using the generalized coordinate ${\theta }_B$; it supplies no extra potential energy.

The continuous transformation $\theta \mapsto \theta +ϵ$, ${\theta }_B\mapsto {\theta }_B+ϵ$, $z\mapsto z+bϵ$, $r\mapsto r$ moves both endpoints by the same screw motion and preserves the fixed helix. Both $\Delta$ and $u$ are unchanged, so $\delta L=0$. Noether theorem therefore givesThis is the Noether conserved quantity for a mechanical point symmetry applied to helical symmetry. In physical Cartesian terms it equals total $z$ angular momentum plus $b$ times total $z$ linear momentum: the bead contributes $M{a}^2{\dot{\theta }}_B+b(Mb{\dot{\theta }}_B)$. Neither pure axial translation nor pure axial rotation preserves the fixed coil separately. Their conserved combination is a consequence of the helical symmetry, not two independently conserved momenta. Substitution of the four equations of motion verifies $\dot{\mathcal{Q}}=-kar\sin \Delta +k(ar\sin \Delta +bu)-bku=0$.^[ref]

The pitch is the axial displacement per complete turn. For a circular helix with $z=a\varphi +b$, the signed pitch for increasing angle is $2\pi a$, and its magnitude is $2\pi |a|$. Pitch is independent of the cylinder radius.

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].

The solenoid turn density $n$ counts turns per unit axial length in a uniformly wound solenoid. The total turn count $N$ is dimensionless, whereas $n$ has units ${\mathrm{m}}^{-1}$. A $1000$-turn coil of length $0.50\mathrm{m}$ has $n=2000{\mathrm{m}}^{-1}$; an axial interval of length $s=0.010\mathrm{m}$ then contains $ns=20$ turns. The effective circumferential surface current density in the closely spaced circular-turn model is $K=nI$, with units $A/m$. ^[ref].

For a single-layer helix, the axial advance per turn is approximately $1/n$. The circular-turn model neglects this advance and the associated axial component of electric current, as well as the detailed magnetic field close to individual wires. The number of turns around the whole coil is always $N=n\ell$ here; it must not be substituted for $n$ in $B={\mu }_{0}nI$.

If the second rotor has zero drag, it cannot supply reaction torque. The single remaining helix must be both force-free and torque-free; a positive-definite resistance matrix forces its translation and rotation to vanish. The relative motor motion is absorbed by the zero-resistance rotor.