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Use two collinear quantum point contacts in a clean two-dimensional electron gas. Surface split gates deplete the electron gas beneath them, leaving two short constrictions facing one another across an open region. The first injects a directional beam from a biased reservoir; the second collects carriers into a downstream reservoir, with electric current flowing in the same spatial direction through both constrictions. Additional surrounding contacts can absorb carriers that miss the collector. A direct collector signal at zero magnetic field, sensitive to alignment and to obstruction of the intervening path, demonstrates ballistic electron transport across the separation.

An illustrative GaAs device uses gate gaps of a few hundred nanometres and electrically tunable constriction widths of tens to a few hundred nanometres, with injector–collector separation of a few micrometres. These are typical design scales, not dimensions supplied for a particular chip in the question. The electrical width can be much smaller than the lithographic gap because depletion extends laterally. A one-mode opening has width comparable to roughly half a Fermi wavelength; it is not simply any opening below one micrometre. Low-temperature operation suppresses inelastic scattering and resolves transverse modes. The direct trajectory length should be comfortably below the relevant transport mean free path.

Collimation is provided by the smooth exit from a constriction. In adiabatic electron collimation, transverse mode action is approximately conserved, giving $p_{\perp }W\simeq \mathrm{constant}$. As the channel opens from width $W_0$ to $W_{\mathrm{out}}$, $p_{\perp ,\mathrm{out}}\simeq p_{\perp ,0}{W}_0/W_{\mathrm{out}}$. At the same time the bottom of the electrostatic potential falls, increasing longitudinal kinetic energy. Thusand the angular distribution narrows. The reciprocal collector selects this forward-directed set of trajectories. A narrow abrupt aperture by itself may diffract into a broad fan; the smooth funnel and its potential profile are important. Here “focusing” refers primarily to collimation and efficient injector–collector coupling, not a claim that every emitted trajectory converges to one exact geometric point.

In scanning gate microscopy, a conducting tip biased sufficiently negatively creates a local repulsive potential and, when strong enough, a small depleted region in the electron gas beneath it. Raster the tip and record the collector current or differential conductance. Placing the perturbation on a strongly contributing path deflects or backscatters electrons and changes collection, producing spatial contrast. A primary scanning-probe experiment records conductance changes as a negatively charged tip depletes and backscatters the two-dimensional gas.^[ref]

The resulting image is a map of transport sensitivity to the tip, not a literal nonperturbing photograph of an electron trajectory. Broad electrostatic gating of either constriction should be distinguished from local beam disruption. Weak smooth disorder can create branches even in a high-mobility sample. Where electron phase coherence is preserved, interference between alternative paths can add fringes, often with a scale comparable to half a Fermi wavelength for a round-trip path-length change. Ballistic motion and phase-coherent interference are related but distinct requirements.

A small perpendicular magnetic field curves the imaged paths through the Lorentz force, $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$. Reversing $B$ reverses the bending, and a fixed aligned collector generally loses the initially straight beam unless a curved path connects it. For a circular orbit at the Fermi velocity,For one valley with two nearly equally occupied spin species at small field, state counting in the Fermi disc gives $n_s=2\pi k_F^2/(2\pi {)}^2=k_F^2/(2\pi )$. HenceSpin polarization or extra valley degeneracy changes the density-to-wavevector relation; the stated low-field GaAs estimate uses the ordinary spin-degenerate case.

The electron mobility gives the momentum relaxation time in a parabolic-band Drude model, $\mu =e{\tau }_{\mathrm{tr}}/m^{\ast }$. ThereforeThe effective mass cancels from both ${\ell }_{\mathrm{tr}}$ and $r_c$ when expressed in terms of density and mobility. For the supplied density,With $m^{\ast }=0.067m_0$, $m_0\simeq 9.1093837139\times 10^{-31}\mathrm{kg}$, one also obtains $v_F\simeq 1.937\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$ and ${\tau }_{\mathrm{tr}}\simeq 38.1\mathrm{ps}$. The requested lengths areA useful independent ratio is ${\ell }_{\mathrm{tr}}/r_c=\mu |B|=3$. The standard constants used here are explicitly tabulated, rather than inferred from the sample.^[ref]

Mobility measures momentum relaxation. Predominantly small-angle elastic scattering can occur more frequently than that transport time, and the phase-coherence time is different again. Thus the calculation supports a several-micrometre transport scale, without proving that every electron experiences literally no perturbation over that distance.

As temperature rises, the directed ballistic collector signal decreases. Increased electron-phonon scattering and other inelastic processes reduce survival of the directed beam; thermal averaging broadens the mode and angular selection, and interference contrast also diminishes. A diffuse background current can remain, so disappearance of beam contrast need not mean all current through the device vanishes.

In vacuum, the source-free Maxwell equations arePut $\psi =\omega (t-z/c)$ and denote the constant electric amplitude by $E_0$. A corresponding traveling-wave magnetic field isBoth divergences vanish because both the electric field and the magnetic field depend only on $z,t$ and have zero $z$ component. Direct differentiation gives $\nabla \times \mathbf{E}=(\omega E_0/c)(\cos \psi ,\sin \psi ,0)=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{B}=(\omega E_0/c^2)(-\sin \psi ,\cos \psi ,0)=c^{-2}{\partial }_t\mathbf{E}$. A time-independent background magnetic field satisfying the homogeneous equations could be added; the formula is the wave component.

Introduce the magnetic vector potential and electric potential by $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. These identities automatically give the divergence equation for $\mathbf{B}$ and Faraday's law. For this wave choose $\varphi =0$ and $\mathbf{A}=(c/\omega )\mathbf{B}$; the derivatives just calculated verify both potential equations.

The electromagnetic force is the Lorentz force, $Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. On the smooth constraint plane $z=z_0$, the constraint force is normal to the plane. Both $\mathbf{v}$ and $\mathbf{B}$ are tangential, so $\mathbf{v}\times \mathbf{B}$ is normal and is balanced by that constraint force; it does not affect the tangential motion. ThusA particular motion is ${\mathbf{r}}_{\parallel }={\mathbf{r}}_0-[Q{E}_0/(m{\omega }^2)](\cos \psi ,\sin \psi )$, a circle of radiusThe general planar motion also includes a constant drift velocity; the particle moves on a circle when that drift is zero. The constraint force can do no work because it is perpendicular to the velocity.

In a Hall effect measurement, a current flows along $x$ and a magnetic field is applied along $z$. The transverse Lorentz force initially deflects the mobile charges, building a transverse electric field until the net transverse current vanishes. Defining $R_H=E_y/(j_x{B}_z)$, the one-carrier Drude model gives $R_H=1/(nq)$. Thus electron-like carriers give $R_H=-1/(ne)$ and hole-like carriers give $R_H=+1/(pe)$.

A nearly filled band can conduct through holes: its missing electrons behave as positive mobile charges. Metals can have both electron-like and hole-like Fermi surface sheets. In the weak-field two-carrier approximation the Hall coefficient isThe sign depends on the carrier densities and squared mobilities; it need not identify the most numerous carrier. Experimentally the Hall voltage changes sign when the field or current reverses, allowing subtraction of unwanted longitudinal-voltage pickup. Anisotropic bands and different scattering rates complicate a literal conversion of $R_H$ into a carrier count.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

Relaxation-time model. The Drude model treats mobile charge carriers as independent particles accelerated by the Lorentz force between scattering events. Collisions randomize the drift momentum toward local equilibrium and are represented by a uniform, field-independent momentum relaxation time. For the present calculation the carrier density, isotropic mass and relaxation time are fixed, the fields are uniform and stationary, and quantum orbital quantization and intercarrier correlations are neglected. The steady drift velocity obeysHere $n$ is the number of carriers of that species per unit volume, $q$ their signed charge, $m>0$ their inertial or isotropic band mass, $\tau$ the momentum relaxation time, and ${\sigma }_0$ their single-species electrical conductivity at zero field.

One-carrier tensor. With $\mathbf{B}=B\hat{\mathbf{z}}$ and $\mathbf{J}=nq\mathbf{v}$, solving for the electric field givesSince $(\mathbf{J}\times \mathbf{B}{)}_x=B{J}_y$ and $(\mathbf{J}\times \mathbf{B}{)}_y=-B{J}_x$,Thus the resistivity tensor, defined by $\mathbf{E}=\mathbf{\rho }\mathbf{J}$, isIts off-diagonal signs follow from the stated axes and signed $q$. For zero transverse electric current, $E_y/J_x=B/(nq)$ is the Hall electrical resistivity; reversing charge reverses it. The single-species longitudinal electrical resistivity is independent of $B$ in this ideal isotropic model.

Adding electrons and electron holes. Let both species have density $n$ and charges $q=\pm e$, with $e>0$. The stipulated equal zero-field electrical conductivity means equal positive charge carrier mobility $\mu ={\sigma }_0/(ne)$, even if their masses and momentum relaxation times separately differ. Put $r=1/{\sigma }_0$ and $h=B/(ne)$. Inverting each electrical resistivity matrix givesBoth species experience the same electric field, so their electric current densities add. Therefore it is their conductivity tensors, rather than their resistivity tensors, that must be summed. The Hall parts cancel andThis is compensated two-carrier magnetoresistance. Although the net Hall field vanishes, each carrier species is bent in its own direction; maintaining the total longitudinal electric current requires an increasing longitudinal field. Equal carrier numbers alone do not ensure zero Hall response if charge carrier mobilities differ; both equalities given here are used.

Large-field limit and quantum oscillations. Expanding the exact diagonal entry,For $\mu B\gg 1$,Here the last equality applies to either species with its own mass and transport time; the stipulated charge carrier mobilities give the same threshold. This is the classical strong-field condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$: carriers turn substantially before their drift momentum relaxes.

Observable quantum oscillations additionally require sufficiently sharp Landau levels and a sufficiently cold Fermi-Dirac distribution, approximatelyWhen ${\tau }_q\simeq {\tau }_{\mathrm{tr}}$, the disorder field scale is comparable to the onset scale for large classical magnetoresistance. The thermal condition can demand a much larger field, and when ${\tau }_q\ll {\tau }_{\mathrm{tr}}$ even the disorder criterion is stronger. The quantitative damping factors in Q2(c) show that these are crossover scales rather than sharp visibility boundaries. A large smooth $B^2$ magnetoresistance therefore does not itself demonstrate Landau level; it can exist while thermal smearing suppresses oscillations. Conversely, the asymptotic condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$ is stronger than merely reaching unity, so there is no universal ordering of the two experimental onset fields without temperatures and lifetime information.

Scaling function. The material's actual zero-field longitudinal electrical resistivity is ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$. Dividing the exact answer by this value givesThe assumption that $n$ stays constant is essential to eliminating further temperature-dependent parameters. $B{\sigma }_0$ is a dimensionful scaling variable; $B{\sigma }_0/(ne)=\mu B$ is its dimensionless version. The corresponding fractional increase is $[\rho (B)-\rho (0)]/\rho (0)=(\mu B{)}^2$.

What the supplied data test. In the original PDF's left panel, the field dependence is strongly temperature dependent. In the right panel, plotting the fractional magnetoresistance against the squared rescaled field makes the displayed 125, 150, 175, 200 and 225 K curves collapse approximately onto one straight line, over the field intervals where they overlap. This supports the proposed one-parameter scaling and, on that range, the quadratic field law. The 100 K curve does not collapse: it bends away and has a different slope in the rescaled plot. The indicated scaling is therefore supported for the displayed 125–225 K data, not for the entire displayed temperature range. The graph alone does not locate a sharp transition between 100 and 125 K or establish behavior above 225 K.

This empirical collapse is Kohler's rule. It is consistent with a common temperature-dependent charge carrier mobility or relaxation scale and a fixed carrier population; it is not a unique proof of exact compensation or of one microscopic scattering mechanism. A change in relative scattering times, carrier population or Fermi-surface structure could spoil the collapse, and this figure alone does not select the cause of the 100 K departure.

There is a factor-of-two normalization convention in the printed text. The ${\sigma }_0$ introduced for each species gives ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$, whereas the final paragraph calls $1/{\sigma }_0$ “${\rho }_0$.” If ${\rho }_0$ denotes the experimentally measured zero-field electrical resistivity, the consistent slope isIf one instead keeps the literal single-species definition ${\rho }_s=1/{\sigma }_0$, the same model has $[\rho (B)-{\rho }_s]/{\rho }_s=-1/2+[B/{\rho }_s{]}^2/(2n^2{e}^2)$. The choice rescales and offsets the axes by temperature-independent constants, so it does not change which temperature curves collapse, but it must be resolved before inferring a carrier density from the slope.

A magnetic field quantizes the transverse cyclotron motion into Landau levels. For a parabolic band, this is a harmonic oscillator with $E_n=\hslash {\omega }_c(n+1/2)$, $n=0,1,\dots$, where ${\omega }_c=eB/m_c$ and the half arises from its zero-point energy. Equivalently, semiclassical orbit quantization gives the stated flux sequence in increments of $h/e$.

To relate real and momentum space, take $\mathbf{B}=B\hat{\mathbf{z}}$ and electron charge $-e$. The semiclassical Lorentz force isConsequently $dx=\hslash d{k}_y/(eB)$ and $dy=-\hslash d{k}_x/(eB)$. The real-space orbit is a rotated copy of the transverse wavevector orbit, scaled by $\hslash /(eB)$. Their areas satisfy $S_r=[\hslash /(eB){]}^2{S}_k$. Since $B{S}_r=(n+1/2)h/e$,This is Onsager quantization for the ordinary half-integer phase. For a circular parabolic orbit, $S_k=\pi k^2$ reproduces the Landau energy spacing. The flux increment $h/e$ here concerns a single electron; it is distinct from the superconducting pair flux unit.

A current-carrying sample in a transverse magnetic field develops a transverse electric field because the Lorentz force deflects its charge carriers. In steady state this Hall voltage balances the transverse deflection. With Hall coefficient $R_H=E_y/(j_x{B}_z)$, one isotropic electron band gives $R_H=-1/(ne)$ and one hole band gives $R_H=+1/(pe)$.

When electrons and electron holes both conduct, the weak-field two-carrier result isThe two-carrier Hall coefficient weights each species by mobility squared. Its sign changes whenTemperature changes dopant ionization, intrinsic electron-hole generation, and the two charge carrier mobilities, so it can move the system across this condition. For example, a p-type semiconductor can be hole-dominated in its extrinsic regime but become negative in its intrinsic regime if $n\simeq p$ and ${\mu }_e>{\mu }_h$. A sign change is possible, not mandatory: in many n-type semiconductors electrons dominate at both temperatures. The density majority alone does not determine the Hall sign in a two-carrier material.

In a sufficiently tight binary star, the main sequence companion can fill its Roche lobe. Roche-lobe overflow then transfers gas through the inner Lagrange point towards the neutron star. The gas carries angular momentum, so it generally forms an accretion disk rather than falling radially onto the surface. Internal stresses carry angular momentum outwards and permit accretion inwards. Capture of a stellar wind is another possible supply mechanism if the companion does not fill its Roche lobe.

Near the neutron star, the rapidly increasing magnetic dipole field, with $B\propto r^{-3}$, becomes dynamically important when its magnetic pressure $B^2/(2{\mu }_0)$ is comparable to the inflow stresses. The accretion disk is truncated around a magnetospheric radius. Ionized material attaches to magnetic field lines and streams along them in magnetically channelled accretion, forming an accretion column above one or both magnetic poles. The Lorentz force impedes motion across the magnetic field, but does not forbid motion parallel to it. Gas can therefore reach the polar surface, with its gravitational potential energy released in a hot shock wave and radiation.

This description presumes the neutron star rotates slowly enough for material to reach it: sufficiently rapid rotation can centrifugally expel the inflow instead. A strong magnetic field by itself does not guarantee surface accretion.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

Normalize the five-dimensional tangent as in part (c), and let $s=\sqrt{1-{\xi }^2}>0$. Since $u=c\tau$, one has $d\tau /d\lambda =s/c$, soSubstitution into the corrected equation in part (d) givesThis is the Lorentz force form in four-dimensional Minkowski spacetime. The effective charge-to-mass coefficient is proportional to the conserved fifth-direction momentum: in $c=1$ units and a potential normalization identifying $A$ with the electromagnetic four-potential, $q/m=\xi /\sqrt{1-{\xi }^2}$. The factor converting the geometrical $A$ to an SI potential fixes the physical charge units. Opposite signs of $\xi$ give opposite effective charges, and $\xi =0$ gives a neutral inertial projection. Antisymmetry gives $d({\eta }_{ab}{u}^a{u}^b)/d\tau =0$.

This is a Kaluza-Klein theory mechanism: a higher-dimensional geodesic appears as forced motion after projection. With the explicitly supplied $+---$ base signature, the positive fibre term gives a second positive direction in the five-dimensional metric. The derivation is a valid geometrical reduction of that metric; a conventional single-time spacetime would require a spacelike extra direction and corresponding sign changes. If one instead retains the sign printed in part (d), the inferred effective charge reverses, but that version is not the geodesic equation of the metric supplied here.

A Stern-Gerlach experiment uses a nonuniform magnetic field to exert force $F_z\simeq {\mu }_z{\partial }_{z}B$ on a magnetic dipole. The two resolved beams reveal discrete projections of spin, rather than a continuum of classical dipole orientations. For the ground-state silver chemical atom, filled shells contribute no net angular momentum and the outer $5s$ electron has orbital angular momentum zero, leaving an effective spin-$1/2$ magnetic moment. This provides evidence for the electron's two-valued intrinsic spin.

Neutral silver chemical atoms can pass through the apparatus without the large charge-dependent Lorentz force that would deflect and broaden a free-electron beam. Neutrality isolates the small magnetic-gradient force far more cleanly. The later parts use an ideal spin filter model; a literal free-electron apparatus requires additional control of charge forces.

The canonical momentum for the stated symmetric-gauge Lagrangian isSince $\mathbf{B}$ is uniform and time independent, its Euler-Lagrange equation becomesThis is the magnetic Lorentz force, which is unchanged by constant spatial translations. The Lagrangian itself changes byA total time derivative changes the action only by $ϵ[F_{\mathbf{n}}(t_f)-F_{\mathbf{n}}(t_i)]$. For variations with fixed endpoints this boundary contribution has zero variation, so the equations of motion are unchanged. The translation is therefore a symmetry of the dynamics even though this particular gauge makes the Lagrangian only quasi-invariant.

Take $\mathbf{B}=B\hat{\mathbf{z}}$. The magnetic Lorentz force givesThus $v_x+i{v}_y=(v_x(0)+i{v}_y(0))e^{-i\Omega t}$, while $v_z$ is constant. Integrating the transverse velocity gives a circle of radius $r_L=v_{\perp }/|\Omega |$ about a fixed guiding centre; adding $z(t)=z_0+v_{\parallel }t$ gives a helical motion in a uniform magnetic field. The magnetic force does no work, so the speed and kinetic energy remain constant. The handedness reverses with the sign of $q$. If $v_{\parallel }=0$ the helix degenerates to a circle; if $v_{\perp }=0$ it is a straight line parallel to the field.

Seek a constant transverse guiding-centre drift ${\mathbf{v}}_d$ whose magnetic Lorentz force balances the applied force. DefineSince $\mathbf{F}\cdot \mathbf{B}=0$, the vector triple-product identity gives $q{\mathbf{v}}_d\times \mathbf{B}=-\mathbf{F}$. Writing $\mathbf{v}={\mathbf{v}}_d+\mathbf{u}$ therefore reduces the full equation to $m\dot{\mathbf{u}}=q\mathbf{u}\times \mathbf{B}$, the ordinary gyromotion equation. Its transverse orbit is circular in the drifting frame, soAveraging over a gyroperiod removes the transverse oscillatory velocity and leaves this drift, together with any constant parallel velocity. For an electric force $\mathbf{F}=q\mathbf{E}$, this becomes the charge-independent $\mathbf{E}\times \mathbf{B}/B^2$ drift; a charge-independent force instead gives opposite drift directions for opposite charge signs.

Let $\mathcal{E}=\sqrt{m^2{c}^4+c^2{\pi }^2}$. Hamilton's equations giveSubtracting $qd{A}_i/dt$ from the second equation gives the familiar Lorentz force, $\dot{\pi }=q(E+\dot{r}\times B)$, with $E=-\nabla \varphi -{\partial }_{t}A$ and $B=\nabla \times A$. This also checks the signs of both potential terms.

The beam energy is a total pion energy, so the momentum follows the relativistic relation $p^2{c}^2=E_{\pi }^2-m_{\pi }^2{c}^4$. Taking the standard charged-pion rest energy $m_{\pi }{c}^2=139.57$ MeV,^[ref]For motion perpendicular to a uniform magnetic field, the Lorentz force supplies the transverse momentum change and gives $r=p/(qB)$. With the stipulated $qB=1$ eV s m${}^{-2}$,The unit cancellation leaves length after dividing by $c$ in metres per second. Equivalently, the usual magnetic-rigidity conversion is $p[\mathrm{GeV}/c]\simeq 0.2998B[\mathrm{T}]r[\mathrm{m}]$ for unit charge. The sign of the pion charge chooses the bending direction, while the radius uses its magnitude. Treating the pion as massless would give about $0.831$ m and is not a good approximation at this modest total energy. The production target does not determine the downstream orbit once the pion momentum and magnetic field are specified.

An Abrikosov vortex combines a core of suppressed superconducting order parameter, circulating electric current and a $2\pi$ winding of the quantum phase. In the bulk limit a singly wound vortex carries the superconducting flux quantum ${\Phi }_0=h/(2e)$. More precisely it is the fluxoid that is quantized; appreciable current on the integration contour prevents identifying enclosed flux alone with the integer.

For a type-II superconductor, the Ginzburg-Landau parameter is $\kappa =\lambda /\xi >1/\sqrt{2}$. The superconducting mixed state allows vortices between the lower critical magnetic field and upper critical magnetic field. A short superconducting coherence length permits a high orbital upper field,Thus surviving a large applied field generally requires accommodating many vortices, not excluding flux everywhere. The Abrikosov vortex lattice provides their equilibrium arrangement in the simplest isotropic model. Material anisotropy, disorder and thermal fluctuations can change that arrangement and its stability. Spin-polarization limits can also constrain an actual upper field, beyond this orbital estimate.

Carrying a large transport current density is a separate problem. For a straight vortex with tangent $\hat{\mathbf{t}}$, the driving Lorentz force per unit length is ${\mathbf{f}}_L=\mathbf{J}\times {\Phi }_0\hat{\mathbf{t}}$. Vortex motion produces a mean electric field $\mathbf{E}=\mathbf{B}\times {\mathbf{v}}_L$ and hence dissipative power density $\mathbf{J}\cdot \mathbf{E}$. A large $B_{c2}$ alone therefore does not imply a large usable superconducting critical current.

Flux pinning holds vortices at favorable defects: suppressing the order parameter costs less where superconductivity is already weakened. Precipitates, dislocations, grain boundaries and deliberately introduced damage can provide pinning, subject to material-specific trade-offs. A defect scale comparable with a core size can be effective; a defect that destroys connected superconducting paths is harmful. In a coarse force balance, a maximum pinning force density $F_p$ gives $J_{c}B\simeq F_p$ for transverse transport. Vortex depinning and thermally activated flux creep can produce a voltage below an ideal static threshold.