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A magnetic field quantizes closed semiclassical orbits into Landau levels. Their passage through the Fermi energy modulates the density of states and scattering, producing Shubnikov-de Haas oscillations of resistivity. The fundamental frequency for a regular extremal orbit perpendicular to the field isThus two distinct fundamentals ordinarily identify two inequivalent extremal areas, rather than two arbitrary points or two spin directions automatically giving two different geometric areas.

The particle-hole equality of tetragonal extremal orbit areas is a special constraint of the exact half-filled nearest-neighbor band stated here. For $B\parallel z$, its regular closed endpoint sections are the electron contour at $k_{z}c=\pi$ and the hole contour at $k_{z}c=0$ described in b. Translation $(k_{x}a,k_{y}a)\mapsto (k_{x}a+\pi ,k_{y}a+\pi )$ changes the sign of both cosines and maps one enclosed pocket onto the other. Their areas are equal. Direct integration givesConsequently this bare model predictsnot two distinct regular fundamental frequencies. The intermediate $k_{z}c=\pi /2$ contour is a separatrix with a divergent cyclotron period, not another ordinary smooth extremal pocket to insert into the Onsager formula. The larger occupied-area slice at $k_z=0$ is the complement of a hole pocket on the Brillouin torus; counting that complement as a second enclosed planar orbit would give a spurious ratio.

For comparison, an additional small-$k_x,k_y$ parabolic approximation would replace the exact surface by $(k_{x}a{)}^2+(k_{y}a{)}^2\simeq 4+\cos (k_{z}c)$ and suggest a rough belly-to-neck ratio $5/3$. That approximation is uncontrolled at half filling and erases the actual topology; it cannot be asserted as an exact prediction of the supplied cosine band. The observation of genuinely different frequencies in a real material therefore needs further information, such as longer-range hopping, a departure from half filling, extra bands, symmetry breaking or a resolved spin-dependent effect. The universal area-ratio inference is valid, but a nonunit numerical ratio is not determined consistently by the complete stated band assumptions alone.^[ref]

The electric field does work $qEd$ on the initially stationary particle. Thus the entry speed into the magnetic region isIn the source geometry the magnetic field is perpendicular to the page, and the particle enters perpendicular to the straight interface. Its return to that interface occurs after a semicircle. The separation between exit and return points is its orbit diameter, so the required separation $d$ gives $r=d/2$. ConsequentlyFor the drawn field into the page and a positive charge entering upwards, the trajectory bends left. Its speed is constant while in the magnetic field, so the time there is the semicircle length divided by $v_0$:The last form is half the cyclotron period and independently checks the geometric calculation.

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.

When uniform $\mathbf{E}$ and $\mathbf{B}$ are parallel, the relativistic Lorentz force splits into independent hyperbolic motion along the fields and circular motion in the perpendicular plane. The circle is periodic in proper time, while electric acceleration makes the helix's longitudinal pitch grow exponentially from one proper-time cyclotron period to the next.