Relativistic electromagnetism

Relativistic electromagnetism combines electric and magnetic fields into Lorentz tensors.

An infinite uniform line with proper charge per unit length ${\lambda }_0$ moving axially at speed $u$ has Lorentz-contracted density $\lambda =\gamma {\lambda }_0$ and current $I=\lambda u$. Transverse distance $r$ is unchanged by the boost. The rest electric field ${\lambda }_0\hat{\mathbf{r}}/(2\pi {ϵ}_{0}r)$ transforms into $\mathbf{E}=\lambda \hat{\mathbf{r}}/(2\pi {ϵ}_{0}r)$ and $\mathbf{B}=\mathbf{u}\times \mathbf{E}/c^2$. This equals the magnetic field ${\mu }_{0}I\hat{\mathbf{\phi }}/(2\pi r)$ from Ampère's law. The formulas concern a uniform stationary current pattern, not an accelerating finite filament's radiation.

For an ideal infinite filament with zero line charge density and current $I$ in a laboratory frame, the line-integrated four-current is $(0,I)$ in temporal/axial components. An axial boost of speed $u$ gives the title density and current. The laboratory magnetic field ${\mu }_{0}I\hat{\mathbf{\phi }}/(2\pi r)$ therefore becomes both an electric field from ${\lambda }^{\prime }$ and a magnetic field from $I^{\prime }$. Charge neutrality is frame-dependent for a current-carrying system. A model built from oppositely charged moving and resting populations makes this result explicit; the charge of each fixed material particle set remains invariant.

An antisymmetric tensor combines the electric polarization and magnetization of a medium. For Minkowski metric $(-,+,+,+)$ and units with light speed one, choose $M^{i0}=P_i$ and $M^{ij}={ϵ}_{ijk}{M}_k$, together with $F_{i0}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$. Then the Lorentz scalar interaction is $M^{\mu \nu }{F}_{\mu \nu }/2=\mathbf{P}\cdot \mathbf{E}+\mathbf{M}\cdot \mathbf{B}$. The bound charge and bound current have four-current $j_b^{\mu }={\partial }_{\nu }{M}^{\mu \nu }$, giving ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\mathbf{J}}_b={\partial }_t\mathbf{P}+\nabla \times \mathbf{M}$. This current is identically conserved because partial derivatives commute and $M$ is antisymmetric.

For metric signature $(-,+,+,+)$, a particle of mass $m$ and charge $q$ coupled to a four-potential has the reparametrization-invariant action

Electromagnetic minimal coupling replaces the free kinetic momentum in an action or Hamiltonian by a combination involving the electromagnetic potential. For $L=-m\sqrt{{\dot{x}}^a{\dot{x}}_a}-q{A}_a{\dot{x}}^a$, the canonical covector is $P_a=-m{\dot{x}}_a/\sqrt{{\dot{x}}^2}-q{A}_a$, so $P_a+q{A}_a$ recovers the potential-independent covector. Under $A_a\mapsto A_a+{\partial }_a\chi$, the action changes by an endpoint term. The fixed-endpoint equations are the relativistic Lorentz force equation and are gauge invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the interaction changes byFixed-endpoint equations of motion are therefore gauge invariant.

Varying the relativistic charged-particle action with fixed endpoints givesor the equivalent raised-index equation.

The four-potential combines the electric potential and magnetic vector potential as $A^{\mu }=(\varphi /c,\mathbf{A})$. The electromagnetic four-potential $A_{\mu }$ transforms underwhile the antisymmetric field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is unchanged.

An electromagnetic multipole expansion expresses the potentials or fields of a localized source as successively smaller monopole, dipole, quadrupole, and higher moments in the ratio of source size to observation distance.

A retarded potential evaluates each source at the event on the source's past light cone. This accounts for propagation at the speed of light and enforces causality.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.

The causal magnetic vector potential generated by a current density in the Lorenz gauge, with no additional incoming homogeneous field. Each source contribution is evaluated at its retarded time. The integral is a convolution with the retarded wave-equation Green function; its validity includes the stated boundary and source conditions.

For a spatially uniform line current $I\Theta (t)$, the causal potential outside the line is $A_z=({\mu }_{0}I/2\pi )\mathrm{arcosh}(ct/r)$ for $ct>r$, and zero before arrival. The finite causal integration length makes it finite at finite time, unlike an absolute magnetostatic potential for an infinite wire.

The induced electric field opposes the increasing axial current, while the magnetic field circles it. Their Poynting vector points radially outward and transports energy into the growing magnetic-field region.

The ideal fields after instantaneous switching behave as $E_z,H_{\varphi }\propto (t-r/c{)}^{-1/2}$ near arrival, and the radial Poynting vector as $(t-r/c{)}^{-1}$. A finite switching rise time smooths the ideal front; the singular solution does not represent a finite-energy real pulse.

For a source point $y(s)$ and field event $(x,t)$, the retarded time solves

For a stationary distant detector, the arrival time of a light signal emitted at $(t,\mathbf{x}(t))$ is $t_a=t+|{\mathbf{x}}_D-\mathbf{x}(t)|/c$. Its derivative is $d{t}_a/dt=1-\mathbf{n}\cdot \mathbf{u}/c$, where $\mathbf{n}$ points from the source to the detector. This exact relation follows by differentiating the distance; a constant viewing direction is a far-field approximation. For a source approaching near the speed of light, a long emission interval can correspond to a short arrival interval. Combining it with the $1/\gamma$ cone of relativistic beaming yields synchrotron pulse compression of order $r/(c{\gamma }^3)$.

For $R(s)=|x-y(s)|$ and $|v|<c$, the function $F(s)=s+R(s)/c$ hasIt is strictly increasing, so the retarded-time equation has at most one solution.

The potentials of a point charge $q$ arewith every source quantity evaluated at the retarded time.

The Liénard-Wiechert field is the electric field and magnetic field obtained from the retarded potentials of a moving point charge. In the nonrelativistic radiation zone, the electric radiation field is ${\mathbf{E}}_{\mathrm{rad}}=q\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})/(4\pi {ϵ}_0{c}^{2}R)$ at retarded time. Its outgoing Poynting vector integrates to the Larmor formula. Coulomb and induction fields must also be retained when a near-field calculation is required.

Let $\mathbf{R}$ join the instantaneous position of charge q to the field point, with constant $\mathbf{\beta }=\mathbf{v}/c$. Transforming its rest-frame Coulomb field givesThe electric field is radial from the current position, reduced along the velocity and enhanced transversely. This present-position form is compatible with the causal retarded solution specifically because motion is uniform. There is no acceleration-dependent radiation field, although the moving charge has a magnetic field and transports electromagnetic energy.

The Lorenz gauge condition isor ${\partial }_{\mu }{A}^{\mu }=0$ in covariant notation.

For a transverse monochromatic electromagnetic wave with angular frequency $\omega \ne 0$, ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$. The displayed electric potential and magnetic vector potential therefore give $-\nabla \Phi -{\partial }_t\mathbf{A}=\mathbf{E}$. Transversality gives $\nabla \cdot \mathbf{A}=0$, so the Lorenz gauge holds. For $\mathbf{E}=E_0(\cos \theta ,\sin \theta ,0)$ and $\theta =\omega t-Kz$, taking a curl yields $\mathbf{B}=(K/\omega )\hat{\mathbf{z}}\times \mathbf{E}$. The vacuum Maxwell equations require $K^2={\omega }^2/c^2$. The field has circular polarization, with $|\mathbf{B}|=|\mathbf{E}|/c$ and perpendicular electric and magnetic directions.

With convention ${\nabla }^{\mu }{F}_{\mu \nu }=-{\mu }_0{J}_{\nu }$, write the electromagnetic field tensor as $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$. Commuting the contracted covariant derivatives gives ${\nabla }^{\mu }{\nabla }_{\nu }{A}_{\mu }={\nabla }_{\nu }({\nabla }^{\mu }{A}_{\mu })+R^{\alpha }{}_{\nu }{A}_{\alpha }$. The Lorenz gauge condition removes the gradient-of-divergence term, leaving the displayed wave equation for a covector field. A locally inertial coordinate system does not eliminate the Ricci tensor term.

An Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$ changes $\partial \cdot A$ by $\square \lambda$. Consequently gauge fixing by a quadratic function of $\partial \cdot A$ retains transformations satisfying $\square \lambda =0$, subject to the same boundary conditions. Eliminating those residual transformations is part of specifying an inverse with a particular Green function prescription.

The electromagnetic field tensor isIt packages the electric and magnetic fields into an antisymmetric rank-two Lorentz tensor and transforms as

Choose coordinates $x^{\mu }=(ct,\mathbf{x})$ and Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. With the convention $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, lowering both tensor indices gives $F_{0i}=+E_i/c$ while leaving $F_{ij}=F^{ij}$. An array with all upper indices cannot be used unchanged as the corresponding array with all lower indices. The propagation Lorentz four-vector $n^{\mu }=(1,\mathbf{s})$ for unit spatial direction $\mathbf{s}$ similarly has $n_{\mu }=(1,-\mathbf{s})$. These signs matter in contractions that encode Maxwell equations.

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

In oriented Minkowski coordinates with $(+---)$ signature and ${ϵ}^{0123}=+1$, the conventional dual is ${\tilde{F}}^{ab}=\frac{1}{2}{ϵ}^{abcd}{F}_{cd}$. For $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$ in $c=1$ units, ${\tilde{F}}^{ab}{F}_{ab}=-4\mathbf{E}\cdot \mathbf{B}$. A definition without the half produces twice this dual and contraction $-8\mathbf{E}\cdot \mathbf{B}$. The factor accounts for summing both orders of each antisymmetric index pair.

The source-free electromagnetic Lagrangian density is quadratic in the electromagnetic field tensor. Its Euler-Lagrange field equation is ${\partial }_{\mu }{F}^{\mu \nu }=0$. Antisymmetry of $F$ makes the density invariant under gauge transformations, and contraction of its indices makes it a Lorentz scalar.

The source-free Maxwell Lagrangian gives $S=-\frac{1}{4}\int F_{\mu \nu }{F}^{\mu \nu }{d}^{4}x$ in mostly-minus Lorentzian signature. Its Euclidean action is the displayed positive quadratic functional of the electromagnetic field tensor. Variation, with boundary terms vanishing, yields ${\partial }_{\mu }{F}_{\mu \nu }=0$.

Vary the antisymmetric auxiliary electromagnetic field tensor and the electromagnetic four-potential independently. The first variation sets $F_{ab}={\nabla }_a{A}_b-{\nabla }_b{A}_a$ algebraically, and variation of the potential gives ${\nabla }_b{F}^{ab}=0$. Substitution of the algebraic equation reduces the first-order action to the usual negative quadratic Maxwell action. Metric variation gives the same stress tensor after eliminating the auxiliary field, because the chain-rule term multiplying its variational equation then vanishes.

Because the electromagnetic field tensor is the exterior derivative $F=dA$ of a potential, $dF=d^{2}A=0$. For a torsion-free connection this iswhich is the covariant form of the two homogeneous Maxwell equations.

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 nonrelativistic charged particle entering a uniform magnetic region normally at one plane follows a circle of radius $r=mv/(|q|B)$. If the field extends through a normal thickness $b$, the particle enters the strict region beyond the far plane only for $r>b$. Equality gives tangency, not crossing. For transmission, its deflection angle obeys $\sin \theta =b/r$. The result assumes a sharp field boundary, fixed speed in the magnetic region and no electric field there.

Rapid circular particle motion perpendicular to a locally uniform magnetic field, superposed on parallel streaming when the parallel velocity is nonzero. In Gaussian units, its signed cyclotron frequency is given in the title and its gyroradius is $v_{\perp }/|{\Omega }_s|$. Slowly varying fields permit a guiding center description.

For a constant perpendicular force in a uniform magnetic field, write the particle velocity as $\mathbf{v}={\mathbf{v}}_d+\mathbf{u}$. Choosing $q{\mathbf{v}}_d\times \mathbf{B}=-\mathbf{F}$ leaves the ordinary gyromotion equation for $\mathbf{u}$. The gyroperiod-averaged transverse velocity is therefore the displayed drift. An electric force gives the same drift for both charge signs; a charge-independent applied force gives opposite drifts.

For a curved field direction $\mathbf{b}=\mathbf{B}/B$, define the curvature vector $\mathbf{\kappa }=(\mathbf{b}\cdot \nabla )\mathbf{b}$. Motion along curved field lines produces a gyroaveraged drift perpendicular to both field and curvature. In a purely toroidal field this and magnetic gradient drift cause opposite vertical drifts for opposite charge signs.

A slowly varying field magnitude exerts the gyroaveraged force $-\mu \nabla B$ on a particle with magnetic moment $\mu =m{v}_{\perp }^2/(2B)$. Applying the perpendicular-force guiding-centre drift formula gives the displayed velocity. Its direction depends on the sign of the particle charge.

The magnetic contribution to the Lorentz force on a point charge. It is perpendicular to that charge's instantaneous velocity, so its direct work on the charge is zero. Summing contributions in a conductor leads to the magnetic force on a current-carrying wire; mechanical work in moving circuits must also account for the electric fields and constraints that maintain the current.

For a square spanning $0\le x,y\le a$, with current counterclockwise as seen from positive $z$, an applied field $B_z=\alpha xy$ vanishes on two edges. Integrating $Id\mathbf{\ell }\times \mathbf{B}$ on the other two gives the displayed force. Its direction reverses with the current. The formula concerns the loop, not unspecified supply leads, and cannot be replaced by a small-loop approximation without checking the field variation.

The radius of a charged particle orbit perpendicular to a uniform magnetic field. In a parabolic band $r_c=m^{\ast }{v}_{\perp }/(|q|B)$, or $r_c=\hslash k_F/(e|B|)$ for a Fermi-level electron. Charge sign sets the curvature direction, while the radius uses magnitudes.

Magnetic rigidity measures the resistance of a charged particle's momentum to bending by a magnetic field. The relativistic cyclotron orbit gives $p_{\perp }=|q|Br$, where $r$ is the transverse gyroradius. For motion perpendicular to a uniform magnetic field, $p_{\perp }=p$ is the whole momentum. A measured magnetic rigidity fixes momentum per magnitude of electric charge; it does not by itself fix mass, kinetic energy, or the sign of the electric charge. This distinction makes the Anderson positron track experiment an especially revealing application of the Lorentz force.

The Anderson positron track experiment identified a light positively charged particle, now called the Positron, by combining magnetic bending, loss of kinetic energy in lead, and particle range. Each observation answered a different question: bending constrained momentum, the lead supplied a direction of travel, and particle range tested the proposed mass.

Carl David Anderson's The Positive Electron (1933), pp. 491–493, describes the 1932 discovery in a vertical cloud chamber with a $15,000\mathrm{gauss}$ magnetic field. A $6\mathrm{mm}$ lead plate separated two photographed track segments. The effective viewing depth was $1\mathrm{cm}$ along the magnetic field, with $14\mathrm{cm}$ transverse diameter: this restricted out-of-plane motion. The photographic readout supplied curvature on both sides. Comparisons with electron tracks under the same conditions constrained ionization; specific ionization was not directly measured for these events. He tested accidental alignment of unrelated tracks and rejected a reversed electron interpretation requiring an increase of kinetic energy in lead. Assuming unit electric charge, the mass bound was about twenty electron masses, rather than an exact mass determination. The paper credits Robert Andrews Millikan with the design and Seth Henry Neddermeyer with assistance.^[ref]

Preparation used rapid piston expansion and a water–alcohol mixture in the cloud chamber; cooling permitted droplets to form along ionization trails. A camera photographed these droplets. Initially, expansions occurred at random, so many photographs contained no tracks. Later counter triggering improved the useful yield.^[ref]

In The Production and Properties of Positrons, his 1936 lecture, Carl David Anderson describes the chamber inside an electromagnet and explains why the lead plate was decisive: after losing kinetic energy, a particle bent more sharply. He describes a proton hypothesis predicting only $5\mathrm{mm}$ of particle range, against an observed path longer than $50\mathrm{mm}$. These are historical values from the lecture, not the stipulated example below. The point was the incompatibility of particle range with the momentum inferred from curvature, rather than positive electric charge alone.^[ref]

A photograph records a path but supplies no arrow of time. The Lorentz force depends on $q\mathbf{v}\times \mathbf{B}$: reversing both electric charge and direction of travel leaves the spatial bending compatible with the same path. Thus a particular arc can represent a Positron travelling one way or an electron travelling the other. Knowing the magnetic field's polarity is necessary but insufficient. Gravity does not justify assuming that every cosmic ray track travels downward.

The lead plate adds a physical ordering. For a single particle crossing a passive absorber without an accelerating interaction, the outgoing momentum is smaller, so the outgoing gyroradius is smaller in the same magnetic field. The less curved segment comes first. Once this direction is assigned, the cross product in the Lorentz force determines the sign of electric charge. For the convention below, $\mathbf{B}=B\hat{\mathbf{z}}$ points out of the page; at upward velocity $\mathbf{v}=v\hat{\mathbf{y}}$, $\mathbf{v}\times \mathbf{B}=vB\hat{\mathbf{x}}$. Bending to the right therefore means positive electric charge.

Magnetic forces do no work: the change of kinetic energy occurs through interaction with the material, not through the bending magnetic field. The ideal construction suppresses multiple scattering, kinetic energy fluctuations, and loss of kinetic energy in the gas. In an actual analysis one compares arcs away from the plate and allows a change of tangent from multiple scattering; one cannot interpret every kink as a change of electric charge. Short visible arcs, uncertain projection and small bending also limit a determination of magnetic rigidity.

For a concrete independent example, stipulate perpendicular motion, $|q|=e$ with $e$ the elementary charge, uniform magnetic field $B=1.00\mathrm{T}$, incoming gyroradius $r_{\mathrm{in}}=0.100\mathrm{m}$ and outgoing gyroradius $r_{\mathrm{out}}=0.0500\mathrm{m}$. These are invented teaching inputs. Converting SI momentum to electronvolt units givesso $p_{\mathrm{in}}=29.9792458\mathrm{MeV}/c$ and $p_{\mathrm{out}}=14.9896229\mathrm{MeV}/c$. Both a Positron and a proton with these momenta trace the same ideal arcs. Their kinetic energies differ because the relativistic energy-momentum relation givesThe second expression avoids subtracting nearly equal quantities when $pc\ll m{c}^2$. For this calculation stipulate rounded rest energies $m_e{c}^2=0.511\mathrm{MeV}$ and $m_p{c}^2=938.272\mathrm{MeV}$. The Positron interpretation gives $K_{\mathrm{in}}\simeq 29.473\mathrm{MeV}$ and $K_{\mathrm{out}}\simeq 14.487\mathrm{MeV}$; the proton interpretation gives only $0.479\mathrm{MeV}$ and $0.120\mathrm{MeV}$. The corresponding proposed losses of kinetic energy are about $14.985\mathrm{MeV}$ and $0.359\mathrm{MeV}$.

The small proton kinetic energy follows from $K\simeq p^2/(2m_p)$, whereas here the Positron has $pc\gg m_e{c}^2$ and $K\simeq pc-m_e{c}^2$. Consequently magnetic bending alone cannot reject the proton. One needs an independent material constraint: could a proton at the proposed kinetic energy penetrate the plate and then travel at least the observed visible distance in the specified gas? A particle range shorter than that visible distance contradicts the hypothesis. The visible track need not include the stopping point: it already supplies a lower bound on particle range. No numerical particle range is assigned to this teaching example; that would require stopping data for the actual material and gas conditions. The experimental insight is to combine a geometric momentum measurement with an independent test of penetration.

The charge-to-mass ratio is a body's electric charge $q$ divided by its mass $m$. Its SI unit is the coulomb per kilogram, $\mathrm{C}\mathrm{k}{\mathrm{g}}^{-1}$. In a prescribed electric field and magnetic field, nonrelativistic acceleration depends on $q/m$ through the Lorentz force. Two bodies with the same initial position, velocity, and charge-to-mass ratio therefore have the same trajectory in this ideal model. For an electron, write $q=-e$ with $e>0$ the elementary charge: $q/m=-e/m$, whereas the positive number $e/m$ is the magnitude of the charge-to-mass ratio.

A velocity selector uses perpendicular electric field and magnetic field to transmit a narrow range of speeds along a chosen beam direction. Assume uniform static electric field $\mathbf{E}=E\hat{\mathbf{y}}$ and magnetic field ${\mathbf{B}}_s=B_s\hat{\mathbf{z}}$, with $E,B_s>0$, and an incident velocity $\mathbf{v}=v\hat{\mathbf{x}}$. The cross product gives $\mathbf{v}\times {\mathbf{B}}_s=-v{B}_s\hat{\mathbf{y}}$, so the transverse Lorentz force isFor an electron, the electric force points downward and the magnetic force upward. A faster electron initially bends upward; a slower one initially bends downward. At the selected speed, both forces cancel along the whole straight trajectory. The cancellation is independent of mass and nonzero electric charge; it selects velocity, not a particular charge-to-mass ratio. A component of velocity parallel to the magnetic field is not constrained by the cancellation, so the incident beam must also have its direction restricted.

A real exit slit accepts a finite range of speeds and incident directions. A zero displacement at one screen alone does not establish zero force throughout the path: a curved trajectory could return to that screen position. The ideal velocity selector interpretation requires a straight aligned beam, known overlapping electric field and magnetic field, and sufficiently narrow entrance and exit slits. Its useful role in the Thomson charge-to-mass experiment teaching model is to supply speed independently of the subsequent magnetic bending.

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

The Lenard cathode-ray window experiment lets cathode rays leave a gas-discharge tube through a thin aluminium foil while preserving the tube's vacuum. This separates production from propagation.

Philipp Lenard's lecture, delivered on 28 May 1906, describes sealing a small hole in a metal plate with aluminium foil. Luminous phosphor grains first rested on the window, then glowed above it in air. His improved tube separated production and observation chambers; coated paper served as a fluorescent screen. An aperture revealed spreading beyond a narrow straight beam. Evacuating the observation chamber reduced spreading and attenuation without interrupting production. Interposed layers and stepped foil stacks gave shadow comparisons; approximate scaling with mass had exceptions.^[ref]

The experimental advantage is unusually clean: reducing the gas content in the observation chamber tests whether gas is needed to carry the cathode rays, while retaining the conditions needed to generate them. The window is a boundary that transmits some electrons while confining the source environment. A spot outside establishes a transmitted effect; the aperture and changed observation environment provide additional ways to distinguish propagation from the source's behavior. This logic is useful whenever a complicated source would otherwise be changed by the very intervention intended to test the emitted beam.

Several inferences require care. A darker fluorescent screen can mean fewer arriving electrons, less deposited energy per electron, or redistribution by electron scattering beyond the observed area. A visible endpoint depends on the screen's sensitivity and background. It is not an intrinsic electron range independent of initial kinetic energy, gas density, and observation geometry. For a modern comparison, measure a reference at fixed source conditions, use background subtraction, and compare the whole luminous patch as well as a small central region. A wider collection area tests redistribution; changing source-to-screen separation also changes collection geometry. These are proposed controls for interpreting a readout, not additional historical observations attributed to Lenard.

Nor is fluorescence a unique signature of cathode rays: X-rays can also excite it. Lenard notes that a later tube could transmit both, and that slow cathode rays failed to pass its window.^[ref] In modern terms, cathode rays are electrons carrying negative electric charge, whereas X-rays are photons with zero electric charge. Magnetic bending of the incident electrons, interpreted through the Lorentz force, supplies a discrimination that brightness alone cannot. Passage through foil alone determines neither the sign of electric charge nor its magnitude, and it does not measure charge-to-mass ratio. The separate Thomson charge-to-mass experiment explains that inference. Even a measured charge-to-mass ratio needs an independent electric charge measurement to determine mass.

An original toy model separates two ways to lose a central cathode ray signal. Let $x$ be forward depth into a uniform gas. Normalize the entering electron population to one. Write $U(x)$ for the fraction still in a narrow forward channel, and $D(x)$ for the fraction diverted into a wider collection channel. Assume constant removal rate $a$ per unit depth in both channels, and constant transfer rate $b$ from $U$ to $D$, with no return. Removal means that an electron stops or is trapped before reaching the collecting plane; it does not mean that its electric charge disappears. The wide ideal collector accepts all surviving members of both channels. This is a stipulated two-population differential equation model, not a general law of electron transport:The first equation gives exponential decay. Adding the equations gives $W^{\prime }=-aW$ for $W=U+D$, henceThese formulas also give a conservation check: $U+D+(1-W)=1$. The fraction $1-W$ has been removed before the plane, while $D$ remains present but misses the narrow collector. An exponential fit to the narrow signal therefore measures $a+b$, not the removal rate $a$ by itself. This is an identifiability issue: one declining signal cannot distinguish the two unknown rates.

For a worked example, choose $a=0.15\mathrm{c}{\mathrm{m}}^{-1}$ and $b=0.35\mathrm{c}{\mathrm{m}}^{-1}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\mathrm{cm}$,Thus about $63.2\%$ misses the narrow collector, although only about $25.9\%$ has been removed in this model. Treating the whole narrow-signal deficit as stopping would confuse redistribution with removal.

Suppose a calibrated readout is proportional to arriving electron number with the same response for either channel. Choose an empty-path signal $R_0=1020$ arbitrary units, of which $B=20$ is background; the two collection geometries have the same calibrated reference. The calculated readings at $2\mathrm{cm}$ are $R_{\mathrm{n}}=20+1000e^{-1}\simeq 387.88$ and $R_{\mathrm{w}}=20+1000e^{-0.3}\simeq 760.82$. After background subtraction, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original inverse problem has the solutionA single narrow reading would instead allow every nonnegative pair with $a+b=0.50\mathrm{c}{\mathrm{m}}^{-1}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\mathrm{c}{\mathrm{m}}^{-1}$. In a real fluorescent screen, the assumed response must be tested: deposited energy and conversion to light need not remain constant as electrons slow. Repeating the comparison at several depths tests the constant slopes, although straight slopes alone do not establish the microscopic mechanism.