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A radiative lifetime depends on the available electromagnetic modes as well as on the atomic transition. In the ideal parallel-perfect conductor geometry of polarization-selective spontaneous emission between conducting plates, an in-plane transition dipole has no coupled propagating electric dipole radiation mode below the relevant waveguide cutoff frequency.

If that idealized electric dipole transition is the only decay channel, its rate is zero and the corresponding lifetime diverges. This is a statement about a restricted channel and geometry, not a general prediction of an immortal excited atom. Finite plates, absorbing surfaces, dipole misalignment and other allowed transitions can restore decay. A normal transition dipole can couple to the surviving transverse electromagnetic mode.

A coaxial cable has an inner conductor surrounded by a concentric outer conductor, with dielectric between them. In the ideal homogeneous lossless Transverse electromagnetic mode, electric field is radial and magnetic field circles the axis; the return current flows on the outer conductor. For inner radius $a$ and inner shield radius $b$, its characteristic impedance isFor air dielectric this is approximately $60\log (b/a)$ ohms. A cable's characteristic impedance describes travelling waves, not its DC wire resistance. Finite conductivity, imperfect shielding and higher modes limit the ideal description.

An electromagnetic waveguide confines an electromagnetic wave across its axis while allowing it to travel along the axis. Conducting walls or changes of refractive index select transverse normal modes: reproducible cross-sectional field patterns with their own longitudinal wavenumbers. A hollow metal pipe and an optical fiber are different realizations of this principle. Their usefulness is to deliver electromagnetic energy along a chosen route instead of letting it spread freely.

For a uniform vacuum-filled conducting guide, write a normal mode's axial and time dependence as $e^{i(\beta z-\omega t)}$. Here $\omega >0$ is angular frequency, $\beta$ is the longitudinal wavenumber, and $k_{\perp }^2$ is the transverse eigenvalue selected by the wall boundary conditions. The electromagnetic wave equation separates intoThe geometry consumes part of the available squared wavenumber: when $\omega /c<k_{\perp }$, this branch is an evanescent wave rather than a propagating normal mode. The waveguide cutoff frequency is therefore a property of a specified branch and geometry. A transverse electromagnetic mode in a suitable two-conductor geometry can have $k_{\perp }=0$ and no positive geometric cutoff. An optical fiber needs its own dielectric boundary conditions, rather than the metal-wall formulas below.

The rectangular electromagnetic waveguide makes these ideas explicit with a single sine profile. The scalar-potential conducting waveguide modes give a broader construction. At an open aperture, a confined field can radiate into the electromagnetic aperture angular spectrum; it need not remain one guided normal mode.

Two perfectly conducting wide strips of width $a$ and separation $d\ll a$, filled with a homogeneous nonmagnetic dielectric of absolute permittivity $\epsilon$, support a transverse electromagnetic mode. Ignoring edge fields, their distributed capacitance is $C^{\prime }=\epsilon a/d$ and distributed external inductance is $L^{\prime }={\mu }_{0}d/a$. The magnetic-energy integral ${\mu }_0(I/a{)}^{2}ad/2=L^{\prime }{I}^2/2$ gives the latter. Their characteristic impedance is $Z_0=(d/a)\sqrt{{\mu }_0/\epsilon }$ and propagation speed is $1/\sqrt{{\mu }_0\epsilon }$. These parameters are per unit length, distinct from a lumped capacitor and inductor.

Define $\hslash {\omega }_0=E_2-E_1$ and use radiation energy density $u(\omega )$ per unit angular frequency, excluding the zero-point contribution. The Angular-frequency Einstein coefficients give rates per atom: $B_{12}u({\omega }_0)$ for absorption, $B_{21}u({\omega }_0)$ for stimulated emission, and $A_{21}$ for spontaneous emission. Stimulated emission adds a photon to the stimulating mode; spontaneous emission persists with no incident photons.

In thermal equilibrium the Boltzmann distribution gives $N_2/N_1=e^{-\hslash {\omega }_0/(k_{\mathrm{B}}T)}$. Balancing up and down rates yieldsMatching the Planck law for every temperature fixesNo degeneracy factor is needed for these nondegenerate levels.^[ref]

For plates of large area $A$ and spacing $d$, the surviving low-frequency transverse electromagnetic mode has one polarization, $\mathbf{E}\parallel \hat{\mathbf{z}}$, and in-plane wavevector. Periodic in-plane counting givesMultiplying this single-polarization two-dimensional photon density of states by $\hslash \omega$ and the Bose-Einstein distribution produces the thermal radiation between parallel conducting plates:The same rate balance therefore givesThese $B$ coefficients refer to the radiation polarization in their respective environments: isotropic free-space $B$ must not automatically be equated to the coefficient for a specified $z$-polarized field.

For hydrogen, take the quantization axis along $z$, neglect spin-orbit coupling, and assume perfect plates, a wide in-plane region and an unmodified atomic wavefunction: $a_{\mathrm{B}}\ll d\ll c/{\omega }_0$, with negligible surface-induced mixing or shifts. The electric-dipole selection rules for hydrogen require $\Delta l=\pm 1$; the allowed $z$ field has $\Delta m_{\ell }=0$. Hence the polarization-selective spontaneous emission between conducting plates givesThe thin-cavity enhancement of a normal electric dipole applies: the $2p_0\to 1s$ dipole points along $z$ and couples to the enhanced mode density. For $m_{\ell }=\pm 1$, the transition dipoles lie in the $x$–$y$ plane and couple to none of these low-frequency modes: the electric-dipole decay rate is zero in the ideal model, so this channel's lifetime is infinite. Real finite, imperfect plates and additional decay or mixing channels limit the lifetime. The order-one orientation factor for the allowed transition does not change the requested lifetime scaling.

An empty perfectly conducting rectangular electromagnetic waveguide hasThe lowest transverse electric mode has indices $(1,0)$ or $(0,1)$. A simply connected hollow metallic guide has no transverse electromagnetic mode. With $a=b=10^{-2}\mathrm{m}$, the two lowest modes are degenerate and their cutoff frequency isPropagation with nonzero axial energy transport requires $f>f_c$; the equality is the lower limiting frequency, with zero axial group velocity. A filling medium would replace $c$ by $1/\sqrt{\mu ϵ}$.

The fundamental mode of an ideal coaxial cable is a transverse electromagnetic mode: the electric field is radial between the conductors, the magnetic field is azimuthal, and energy flows along the cable through the dielectric. There is no cutoff for this mode; higher modes, conductor losses and dielectric dispersion can matter outside the ideal transmission-line regime.

Ignoring edge fields in the parallel-strip geometry, $E_x=V/a$ and $H_y=I/b$. Per unit length the charge is $ϵbV/a$ and the linked magnetic flux is $\mu aI/b$. Thus the parallel-plate transmission line hasFor a segment $dz$, charge conservation, equivalently the appropriate integral of the Ampère-Maxwell equation, gives $I(z)-I(z+dz)=Cdz{\partial }_{t}V$. Applying Faraday's law around a longitudinal rectangle between the plates gives $V(z)-V(z+dz)=Ldz{\partial }_{t}I$. These are the lossless telegrapher's equations,Differentiate the second with respect to $z$ and use the first, and conversely for $I$:The signal speed is $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, equal to $c$ for a vacuum-filled line. It is the speed of the coupled electromagnetic disturbance, rather than the electron drift velocity. Electrons locally adjust their motion as the field arrives; one electron need not travel the cable's length to transmit a signal. The Poynting vector in the dielectric integrates to power $VI$.

For a hollow, air-filled rectangular electromagnetic waveguide with conducting walls, transverse confinement givesA transverse electric mode of a rectangular waveguide permits one zero index, but not two; a transverse magnetic mode of a rectangular waveguide requires both indices positive. Take $w=20\mathrm{mm}$ and $h=10\mathrm{mm}$. The dominant mode is ${\mathrm{TE}}_{10}$, withThe waveguide cutoff frequency is the limiting value: positive axial group velocity and propagating power require $f>f_c$. There is no transverse electromagnetic mode in this single hollow conducting pipe; otherwise a zero cutoff would incorrectly follow.

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

Distributed line parameters. The geometry is the wide parallel-plate transmission line described in the question, with absolute permittivity $\epsilon =ϵ{\epsilon }_0$. A length $dx$ has charge $CVdx$ and field $E=V/d$, givingThe opposed plate currents have sheet density $I/a$ and confine the leading magnetic field $H=I/a$ between the plates. The magnetic energy per unit length is ${\mu }_0{H}^{2}ad/2=L{I}^2/2$, soHere $C$ and $L$ are per unit length, in $\mathrm{F}{\mathrm{m}}^{-1}$ and $\mathrm{H}{\mathrm{m}}^{-1}$, not lumped component values. Internal conductor fields and edge corrections are neglected.

Voltage and current evolution. The current entering a short length minus that leaving it is the rate of increase of its plate charge:The voltage drop across that length is its inductive drop, $V(x)-V(x+dx)=Ldx{\partial }_{t}I$. Dividing by $dx$ gives the lossless Telegrapher's equations,

Propagation speed. Differentiate these equations once more and eliminate the other variable:The wave equation speed is

Travelling-wave impedance. For a forward wave $V=F(t-x/v)$, take $I=V/Z_0$. Substitution into ${\partial }_{x}I+C{\partial }_{t}V=0$ gives $1/(v{Z}_0)=C$, hence the characteristic impedanceThe backward-wave current has the opposite sign relative to its voltage; the fixed ratio above applies to one travelling direction.

Matching the antenna. The voltage reflection coefficient is $\Gamma =(Z_{\mathrm{load}}-Z_0)/(Z_{\mathrm{load}}+Z_0)$. No reflection requires $Z_0=5\Omega$. With $a=0.020\mathrm{m}$, $ϵ=1.5$, and $\sqrt{{\mu }_0/{\epsilon }_0}\simeq 376.7\Omega$,The resulting $d/a\simeq 0.016$ is consistent with the wide-plate approximation.

Both transverse forces. The matched transverse electromagnetic mode has electric attraction and magnetic repulsion. For time-averaged power $P=100\mathrm{kW}$,The force on one plate excludes its own field. Thus the average electric attraction and magnetic repulsion per unit length are respectivelyThe electric and magnetic pressure cancellation in a matched TEM line follows directly from $V_{\mathrm{rms}}=Z_0{I}_{\mathrm{rms}}$:The equality also holds instantaneously for a single travelling wave. Quoting either $0.628\mathrm{N}{\mathrm{m}}^{-1}$ alone would omit the opposing contribution. This is the ideal electromagnetic force in the stated lossless uniform-dielectric model; a standing wave would not generally have the same cancellation.

Assume a hollow, perfectly conducting rectangular electromagnetic waveguide filled with air approximated by vacuum. Let its wider side be $a=0.040\mathrm{m}$ and its narrower side $b=0.020\mathrm{m}$. The longitudinal propagation constant of a mode isso the waveguide cutoff frequency isFor a mode with transverse electric polarization, one of $m,n$ may vanish, but they cannot both vanish. For a mode with transverse magnetic polarization, both must be positive. A single hollow conducting guide has no transverse electromagnetic mode with zero cutoff.

The lowest cutoff therefore belongs to the ${\mathrm{TE}}_{10}$ mode, varying across the wider side:Strictly, a traveling mode transporting energy down the guide requires $f>f_c$. At exactly $f_c$, $k_z=0$ and the axial group velocity vanishes. Below cutoff the field is evanescent. A different dielectric filling would replace $c$ by the wave speed in that material.

Wide perfectly conducting plates with $d\ll c/{\omega }_0$ admit only a normal-electric-field transverse electromagnetic mode at the transition frequency. The electric-dipole interaction therefore couples to the normal component of the atomic transition dipole. A normal dipole has enhanced available mode density; an in-plane dipole has no allowed electric-dipole radiation channel below the higher-mode cutoffs. The conclusion assumes negligible changes to the atom and negligible absorption, leakage and symmetry-breaking mixing.

Two wide perfect-conductor plates separated by $d$ support a gapless transverse electromagnetic mode whose electric field is normal to the plates. At $\omega \ll c/d$, multiply the single-polarization two-dimensional photon density of states by $\hslash \omega$ and the Bose-Einstein distribution to get the title's spectrum. This is a low-frequency statement; it need not describe the full thermal spectrum if thermally populated higher branches cross their cutoffs.

A guided electromagnetic wave whose electric field has no component along the propagation axis. Its magnetic field can have a longitudinal component. In an empty perfectly conducting rectangular electromagnetic waveguide, transverse wave numbers determine a nonzero cutoff frequency. This differs from a transverse electromagnetic mode, in which both fields are transverse.

A transverse electric mode of a rectangular waveguide has $E_z=0$ while its axial magnetic field is generally nonzero. For time dependence $e^{-i\omega t}$ and axial dependence $e^{i\beta z}$, set $\Delta =(\omega /c{)}^2-{\beta }^2$. The Maxwell equations reconstruct the transverse components from $B_z$:The axial complex amplitude satisfies a transverse Helmholtz equation, $({\partial }_x^2+{\partial }_y^2+\Delta )B_z=0$. Zero tangential electric field at each perfect conductor wall becomes a Neumann boundary condition on $B_z$. Separation of variables givesThe index $m$ counts half-periods across $a$, and $n$ across $b$. Either index can vanish, allowing ${\mathrm{TE}}_{10}$; both cannot vanish. Indeed a spatially constant axial oscillating magnetic field would give nonzero changing magnetic flux through the cross-section, whereas the line integral of tangential electric field around its metal boundary is zero. The integral form of Faraday's law excludes this ${\mathrm{TE}}_{00}$ candidate. Dividing the reconstruction formulas by $\Delta =0$ would be invalid and cannot produce the separate transverse electromagnetic mode.

For $m=1,n=0$, choose $B_{\ast }=-i(\pi /a)E_0/\omega$. The formulas recover exactly the $E_y$, $B_x$ and $B_z$ in rectangular electromagnetic waveguide. In particular, the transverse magnetic field accompanies $B_z$ and ensures zero divergence; an oscillating longitudinal magnetic field on its own is not the guided solution.

A transverse electromagnetic mode has both electric field and magnetic field transverse to its propagation axis. An ideal two-conductor guide can support such a mode without a geometric waveguide cutoff frequency; an empty simply connected perfect conductor pipe cannot.

The waveguide cutoff frequency separates oscillatory axial propagation from exponential decay for a particular normal mode. For the vacuum-filled rectangular electromagnetic waveguide, the dispersion relation isFor a transverse electric mode of a rectangular waveguide, $m,n$ are nonnegative integers but not both zero. For a mode with $B_z=0$ and nonzero $E_z$, both indices must be positive. The lowest cutoff for $a>b$ is consequently ${\omega }_{10}=\pi c/a$, or ordinary frequency $f_c=c/(2a)$.

A square cross-section instead gives equal ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$ cutoffs. The absence of a transverse electromagnetic mode is specific to this hollow pipe, rather than all electromagnetic waveguides.^[ref]

For $\omega >{\omega }_{mn}$, choose real $\beta >0$ for propagation toward $+z$. For $0<\omega <{\omega }_{mn}$, a field decaying into a semi-infinite guide hasA finite section can contain both growing and decaying axial solutions set by its two ends; the single decaying solution is not a complete finite-section transmission calculation. At exact cutoff $\beta =0$, the two axial solutions become a constant and a linear function of $z$: there is no finite axial wavelength for a traveling sinusoid.

In the chosen $a=30\mathrm{mm}$ guide at $f=0.8f_c=3.99723277\mathrm{GHz}$, the ${\mathrm{TE}}_{10}$ branch has $\alpha =20\pi {\mathrm{m}}^{-1}$. Its amplitude decay length is $1/\alpha \simeq 15.9155\mathrm{mm}$; after $50\mathrm{mm}$, its normalized wave amplitude is $e^{-\pi }\simeq 0.0432139$. This is geometric evanescence even with ideal walls, not Joule heating. A single decaying branch has zero cycle-averaged axial Poynting vector: substituting $\beta =i\alpha$ makes $B_x$ a quarter-cycle out of phase with $E_y$. Coupling across a finite section requires the complete boundary-value solution.