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Stereoscopy supplies depth because different cameras see different projections of the same track. A functional reconstruction first matches the same interaction vertex and track segments across views. A calibrated lens and window model maps each measured image coordinate back to a ray inside the liquid. Refraction at the windows must be included. Intersecting the rays, or finding their closest agreement when measurements have errors, estimates three-dimensional coordinates. Several points then determine a trajectory, rather than only a projected arc.

An original ideal-camera model makes the geometry explicit. Take two parallel pinhole cameras with centers $(0,0,0)$ and $(b,0,0)$, focal length $f$, and optical axes along $z$. Use virtual image planes in front of the pinholes so that image coordinates have the same orientation as the scene. A point $(X,Y,Z)$ then givesSubtracting the horizontal image coordinates gives disparity $d=u_L-u_R=fb/Z$, henceFor stipulated $b=0.20\mathrm{m}$, $f=0.050\mathrm{m}$, $u_L=0.010\mathrm{m}$, $u_R=0.005\mathrm{m}$ and $v_L=0.002\mathrm{m}$, the reconstructed point is $(0.40,0.08,2.0)\mathrm{m}$. These are ideal optical inputs, not dimensions of Alvarez's apparatus. This model omits refraction and lens distortion; actual calibration replaces its simple ray geometry. Since $|dZ/dd|=fb/d^2$, the same disparity error gives worse depth precision for distant points.

The Lorentz force then supplies a momentum scale. With magnetic field $B$ uniform and negligible stopping power over a fitted segment, bubble-chamber track readout gives $p_{\perp }=|q|Br$. The local three-dimensional helix tangent supplies the missing direction: if $\alpha$ is its angle to the field, $p=p_{\perp }/\sin \alpha$. In practical units,Here $e$ is the elementary charge and $c$ the speed of light. A short, almost straight arc poorly constrains its radius of curvature; a visually sharp track can therefore have poor momentum precision.

Stereoscopy enables three-dimensional track reconstruction.^[ref] Calibrated camera positions and optical geometry are needed before a photographed arc becomes a physical radius of curvature. Neutral particles do not directly leave the same continuous ionization trail as charged particles; an interaction or decay can instead reveal them through charged secondary tracks. A photon's inferred path between interactions is not a photographed luminous line.

In a uniform magnetic field, with negligible stopping power and elastic scattering over the fitted segment, the transverse projection is a circle. The Lorentz force turns transverse momentum at angular rate $v_{\perp }/r$, givingHere $r$ is measured in the plane perpendicular to the magnetic field. The relation remains valid with relativistic physical momentum; it does not require $p=mv$. Thus curvature determines magnetic rigidity, not mass or kinetic energy alone. For chosen $|q|=e$, $B=1.5\mathrm{T}$ and $r=0.40\mathrm{m}$, $p_{\perp }\simeq 0.180\mathrm{GeV}/c$. This is an invented example, not a fit to a historical photograph. Reconstructing a three-dimensional helix also determines the component parallel to the magnetic field; the transverse radius of curvature alone does not. Determining the sign of electric charge additionally requires the field direction and the direction of travel, as in the charge-sign ambiguity of a curved track.

In a uniform magnetic field, a nonrelativistic particle of mass $m$ and nonzero charge $q$ completes one circular transverse orbit in time $T_c=2\pi /{\omega }_c$, where ${\omega }_c=|q|B/m$. Parallel velocity changes the helix pitch but not this period. In relativistic motion the constant Lorentz factor multiplies the period: $T_c=2\pi \gamma m/(|q|B)$. The direction of rotation depends on the charge sign; the period is positive.

Local resistive-force theory for a helix gives $AD-B^2={\ell }^2{a}^2{\xi }_{\parallel }{\xi }_{\perp }$. Positive local drag therefore guarantees a positive-definite axial resistance matrix and positive viscous power loss. The identity also keeps the denominator in counterrotating two-helix balances positive.

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.

The gyroradius is the radius of the transverse circular motion of a charged particle in a uniform magnetic field. In nonrelativistic classical mechanics, with zero electric field and no other forces, it is $m{v}_{\perp }/(|q|B)$, where $v_{\perp }$ is the speed perpendicular to the magnetic field. The parallel component of velocity changes the advance of the helix, rather than its transverse radius.

Mirroring the handedness of a helix reverses its axial translation-rotation coupling coefficients while preserving its pure translation and rotation resistances. With $\mathbf{t}=\cos \theta {\mathbf{e}}_z-\sin \theta {\mathbf{e}}_{\phi }$ for a left-handed helix and the force-on-body convention, $B=C=\ell a({\xi }_{\perp }-{\xi }_{\parallel })\sin \theta \cos \theta$.

In a uniform magnetic field, the Lorentz force leaves the component of velocity parallel to the field constant and rotates the perpendicular component at the nonrelativistic cyclotron frequency $|q|B/m$. The resulting path is a helix around a field line.

For a positive refractive index $\mu (r)$ in cylindrical coordinates, a light ray with constant radius $R>0$ and varying axial coordinate has the form $\theta (z)={\theta }_0+\omega z$ when it exists. Fermat principle makes $L=\mu (r)\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$ stationary. Conservation of the momentum conjugate to the cyclic angle makes ${\theta }^{\prime }$ constant along a constant-radius path. The radial Euler-Lagrange equation is thengiving the displayed formula. A nonzero real helix requires $0<-R{\mu }^{\prime }(R)/\mu (R)<1$. If ${\mu }^{\prime }(R)=0$, $\omega =0$ gives a straight axial light ray; a generic radial profile need not support a constant-radius helix. Its helical pitch is $2\pi /|\omega |$. This ray model is useful for a graded-index optical fiber when the refractive index varies smoothly and geometrical optics applies; it does not by itself impose wave-mode quantization or boundary confinement.

A helical symmetry combines an axial rotation and a proportional axial translation. A helix with $z=b\theta$ is unchanged as a set by this transformation. If a mechanical Lagrangian is invariant, the Noether conserved quantity for a mechanical point symmetry is $L_z+b{P}_z$, where the two momenta include all moving components. Pure axial rotation and pure axial translation need not separately be symmetries. The helical symmetry of a potential is a related special case in which the constraint is replaced by an invariant potential.^[ref]

As the length ratio of an opposite-handed reaction helix tends to infinity, its translation and rotation tend to zero as $1/n$, while the finite first helix rotates at almost the full motor rate. Small motions of the long rotor balance finite force and torque. The common propulsion vanishes in this limit of the additive model.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

A nucleic acid double helix consists of two nucleic acid strands winding around a common axis. In the familiar DNA arrangement, sugar–phosphate backbones lie outside and paired nucleobases lie inside, stacked along the axis. The strands are antiparallel nucleic acid strands. The flattened base pairing ladder above displays partners and direction; winding that ladder into a helix adds spatial organization. RNA can also form double-helical paired regions, so the geometry is not exclusive to DNA. See Alberts et al., Molecular Biology of the Cell: DNA structure.

The useful surprise is that a structure explains a copying principle. Complementary base pairs allow each DNA strand to specify the other's nucleic acid sequence; separate strands can each act as templates in DNA replication. This is a structural constraint, while the enzymes and stages of DNA replication supply the actual process. A nucleic acid sequence specifies letter order without, by itself, giving every detail of the three-dimensional helix.

On 25 April 1953, James Watson and Francis Crick published a model with two oppositely directed chains, outside phosphate groups, and specific base pairs. Their short paper explicitly connected this pairing to a possible copying mechanism, while describing the model as requiring further experimental tests. See the original ^[ref], with an accessible ^[ref].

Rosalind Franklin and Raymond Gosling's paper in the same issue had a different role: it presented X-ray diffraction evidence from DNA fibres and its structural interpretation, including the helical character of the wetter B form and evidence for phosphate groups on the outside. They distinguished forms produced at different water contents and assessed compatibility with the Watson–Crick model. Read their original ^[ref]. The model and the experimental analysis are distinct contributions; a diffraction pattern is evidence constraining a molecular geometry, rather than a direct picture of letter-by-letter base pairing.

Two coaxial opposite-handed helices linked by an internal motor share a translation speed but counterrotate. In an additive local-drag model, a length ratio $n$ multiplies the second helix's diagonal resistances by $n$ and reverses its coupling. With ${\Omega }_1-{\Omega }_2=\omega$, force and torque balance give $U=-2nBD\omega /[AD(1+n{)}^2-B^2(1-n{)}^2]$.

For a regular differentiable curve $\mathbf{x}(t)$, its speed is $ds/dt=|{\mathbf{x}}^{\prime }(t)|>0$, where $s$ is arc length. The oriented unit tangent vector is $\hat{\mathbf{T}}={\mathbf{x}}^{\prime }/|{\mathbf{x}}^{\prime }|$. The curvature of a space curve is $\kappa =|d\hat{\mathbf{T}}/ds|$. Equivalently, for a twice differentiable regular curve,For the helix, differentiation gives ${\mathbf{x}}^{\prime }=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. ConsequentlyIndeed $d\hat{\mathbf{T}}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne (0,0)$; the excluded constant curve has no well-defined unit tangent vector. For $a=0,b\ne 0$ the curve is a straight line with zero curvature of a space curve.

For a force $\mathbf{F}\delta (\mathbf{x})$ acting on an unbounded incompressible fluid, the Stokeslet and its pressure areThey solve $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}\delta =0$ and $\nabla \cdot \mathbf{u}=0$. Approximate a slender body by a line distribution of these point forces on its centreline. At a point labelled by arclength $s$,To extract the large logarithm, write $s^{\prime }=s+\ell$ and approximate $\mathbf{X}(s+\ell )-\mathbf{X}(s)\simeq \ell \mathbf{t}$, $\mathbf{f}(s+\ell )\simeq \mathbf{f}(s)$, where $\mathbf{t}={\mathbf{X}}^{\prime }$ is the unit tangent vector. The near-line contribution has kernel $(I+\mathbf{tt})/(8\pi \mu |\ell |)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line giveThe far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf{tt}{)}^{-1}=I-\mathbf{tt}/2$, inversion proves the leading slender-body force densityThis is force on the fluid; the fluid's force on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.

The stated arclength describes the circular helix $\mathbf{X}=(a\cos \theta ,a\sin \theta ,b\theta )$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi \mu /\ln (L/R)$. ThenRotation about $z$ has velocity $\mathbf{v}=\Omega (-a\sin \theta ,a\cos \theta ,0)$, so $\mathbf{t}\cdot \mathbf{v}=\Omega a^2/q$. The axial force density is $f_z=-c\Omega a^{2}b/q^2$. Integrating over one turn givesFor translation, $\mathbf{v}=W\hat{\mathbf{z}}$ and $\mathbf{f}=cW(2\hat{\mathbf{z}}-(b/q)\mathbf{t})$. Only the tangential term contributes to axial couple. Since $(\mathbf{X}\times \mathbf{t}{)}_z=a^2/q$, its density is $-cW{a}^{2}b/q^2$, givingThese are the off-diagonal coefficients of the axial resistance matrix of a slender helix. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf{tt})$, the rotation-induced axial force is $\int \hat{\mathbf{z}}\cdot D(\hat{\mathbf{z}}\times \mathbf{X})ds$, whereas the translation-induced axial torque is $\int (\hat{\mathbf{z}}\times \mathbf{X})\cdot D\hat{\mathbf{z}}ds$. They agree pointwise by symmetry of $D$. The full Stokes-flow statement is the Lorentz reciprocal theorem.

For completeness, retaining the literal elliptic helix instead gives $s^{\prime }=\sqrt{a^2{\sin }^2\theta +{\cos }^2\theta +b^2}$ and the axial coupling of an elliptic helixThus reciprocity holds for the literal curve as well; only the geometric coefficient and arclength change. In all these formulas the equal and opposite force or couple on the wire reverses the displayed sign.

With no electric field, the Lorentz force is $F=q\dot{r}\times B\hat{z}$. The equation from Newton's second law and its first integral areThe cyclotron frequency is $|\omega |$; the sign of $\omega$ specifies the sense of rotation. Since $\hat{x}\times \hat{z}=-\hat{y}$, the initial data giveThe component equations are $\dot{x}=\omega y$, $\dot{y}+\omega x=u+a\omega$ and $\dot{z}=v$. For $\omega \ne 0$, differentiating the first givesSolving this linear differential equation and then using $y=\dot{x}/\omega$ givesThe perpendicular projection is a circle of radius $|u/\omega |$ centred at $(a+u/\omega ,0)$, the guiding centre. The velocity parallel to the magnetic field stays constant, producing helical motion in a uniform magnetic field.

When $a\omega +u=0$, this simplifies toThus the trajectory is a helix about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega |$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega >0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.

For $\omega =0$, the Lorentz force vanishes and the solution is instead $r(t)=a\hat{x}+ut\hat{y}+vt\hat{z}$, also the limit of the general formula. The special condition then forces $u=0$.

The specified vector potential gives $\mathbf{B}=(0,0,B)$. Put $\omega =qB/m$, retaining its sign. The Lorentz force law yields $\ddot{x}=\omega \dot{y}$, $\ddot{y}=-\omega \dot{x}$ and $\ddot{z}=0$. For initial velocities $(u,v,w)$ and $\omega \ne 0$, integration givesThe transverse motion is a circle centred at $(x_0+v/\omega ,y_0-u/\omega )$ with radius $\sqrt{u^2+v^2}/|\omega |$. Thus the full orbit is a helix around a line parallel to the magnetic field, with period $2\pi /|\omega |$ and signed axial advance $2\pi w/|\omega |$ per turn. The charge sign determines the sense of rotation. If $w=0$ it is a circle, if $u=v=0$ it is a straight line parallel to the field, and if $qB=0$ all motion is uniform rectilinear motion.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

Lift the angular coordinate continuously to a real coordinate and write $\Delta z=z_B-z_A$ and $\Delta {\varphi }_m={\varphi }_B-{\varphi }_A+2\pi m$ for a prescribed winding number $m\in \mathbb{Z}$. Assume $z_A<z_B$ for the calculation; reversing the endpoints changes no length. The arc length functional on the circular cylinder isThe Euler-Lagrange equation says $R^2{\varphi }^{\prime }/\sqrt{1+R^2{\varphi }^{\prime 2}}$ is constant. This expression is strictly increasing in ${\varphi }^{\prime }$, so ${\varphi }^{\prime }$ is constant. The candidate is the helixIt is a global minimum within the chosen winding class: the integrand is strictly convex, and Jensen inequality bounds the functional below by $(z_B-z_A)\sqrt{1+R^2(\Delta {\varphi }_m/\Delta z{)}^2}$, with equality only for constant slope. Equivalently, unrolling the cylinder makes the minimizing path a straight segment.

The signed pitch of a helix, measured for an increase of $2\pi$ in angle, isIts geometric magnitude is $|P_m|$. If only the physical endpoints are fixed, choose the integer $m$ minimizing $|\Delta {\varphi }_m|$; opposite points may give two equally short helices. If the lifted endpoint angles are prescribed, use that winding class. A zero angular increment gives a straight generator, the limiting infinite-pitch case.