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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.

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.

Below the lowest waveguide cutoff frequency, the axial wave number of the ${\mathrm{TE}}_{10}$ mode is $i\kappa$, with $\kappa >0$. For a real peak scale $E_0$ and time convention $e^{-i\omega t}$, $E_y=E_0\sin (\pi x/a)e^{-\kappa z}$, $B_x=-i\kappa E_y/\omega$ and $B_z=-i{E}_0(\pi /a)\cos (\pi x/a)e^{-\kappa z}/\omega$. The electric field and magnetic field are in quadrature. A single decaying mode in a lossless semi-infinite guide consequently has zero time-averaged Poynting vector, although its instantaneous field stores energy.

A useful stipulated dispersive medium obeys the Klein-Gordon equation $u_{tt}-c^2{u}_{xx}+{\omega }_0^{2}u=0$, with $c>0$ and ${\omega }_0>0$. Its positive-angular frequency dispersion relation isHere ${\omega }_0$ is the gap at zero wavenumber, not the central angular frequency ${\omega }_c=\omega (k_0)$ of a wave packet; the parameter in the preceding formula is $A={\omega }_0/c$. For $k>0$, the phase velocity satisfies $v_p>c$, the group velocity satisfies $0<v_g<c$, and $v_p{v}_g=c^2$. In this stipulated medium $c$ is the limiting propagation speed of the model; it need not mean the vacuum speed of light. The same square-root dispersion relation appears in waveguide phase and group velocities, with a waveguide cutoff frequency playing the role of ${\omega }_0$.^[ref]

For the original dimensionless choice $c={\omega }_0=k_0=1$, a carrier wave has phase velocity $v_p=\sqrt{2}$, while a narrow wave packet has leading group velocity $v_g=1/\sqrt{2}$. After time $t=8$, the carrier's zero-wave phase reference has moved to $x=8\sqrt{2}\simeq 11.31$, whereas the leading wave envelope centre is at $x=8/\sqrt{2}\simeq 5.66$. A carrier wave crest therefore moves through the wave envelope; it does not carry an identifiable piece of the envelope with it.

The origin-to-curve secant on the dispersion diagram gives phase velocity, while its local tangent gives group velocity. As $k\to 0^{+}$, $v_p\to \infty$ and $v_g\to 0$; at $k=0$ the phase velocity itself is undefined. As $k\to \infty$, both approach $c$, from opposite sides. On the same positive-angular frequency branch with $k<0$, both signed velocities are negative and their product remains $c^2$. These conventions distinguish direction from phase speed. Reversing both $k$ and $\omega$ represents the same real plane wave; changing just one sign changes its direction.

For an original finite-bandwidth check, take equal-amplitude plane waves at $k_1=0.8$ and $k_2=1.2$, still with $c={\omega }_0=1$. The exact two-tone wave envelope speed iswhile its carrier wave has phase velocity $(\sqrt{1+0.8^2}+\sqrt{1+1.2^2})/2\simeq 1.421337$. Their product is exactly $1$: rationalizing the difference gives $v_{\mathrm{beat}}=2/({\omega }_1+{\omega }_2)$. Nevertheless $v_{\mathrm{beat}}\ne {\omega }^{\prime }(1)$; the derivative describes the limit of vanishing separation, not an arbitrary pair of tones. The two-tone wave envelope repeats indefinitely, so it has no isolated first arrival and cannot measure front velocity.

The acoustic pressure satisfies $p_{tt}=c_0^2\Delta p$. Rigid walls impose vanishing normal velocity, equivalently the Neumann boundary condition on pressure. Separation of variables gives transverse modes $\cos (m\pi x/a)\cos (n\pi y/a)$ with $m,n\ge 0$. With axial factor $e^{ikz-i\omega t}$, their dispersion relation isThe $(0,0)$ mode is nondispersive. Every other mode propagates only above its waveguide cutoff frequency ${\omega }_{mn}=\pi c_0\sqrt{m^2+n^2}/a$. Thus ${\omega }_{\min }=\pi c_0/a$.

Expand the prescribed entrance pressure in this cosine basis, with coefficients $A_{mn}$. Impose outgoing propagation and decay at infinity, with no incoming disturbance. At $\omega ={\omega }_{\min }/2$, only the constant transverse mode propagates, with $k_{00}=\pi /(2a)$. All other modes have $k=i{\chi }_{mn}$, where ${\chi }_{mn}=\pi \sqrt{m^2+n^2-1/4}/a$. The exact modal field is the outgoing constant mode plus ${\sum }_{m+n>0}{A}_{mn}\cos (m\pi x/a)\cos (n\pi y/a)e^{-{\chi }_{mn}z-i\omega t}$.

Since the slowest evanescent decay rate is $\sqrt{3}\pi /(2a)$, the far field isIf that area average is zero, there is no propagating far-field component, and the field decays exponentially instead. For sufficiently regular entrance data the omitted terms are $O(e^{-\sqrt{3}\pi z/(2a)})$.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.

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.

Guided electromagnetic waves propagate along a structure whose boundaries select transverse field patterns. At a perfect conductor the tangential electric field vanishes. In a hollow rectangular guide, TE modes have $E_z=0$ and TM modes have $H_z=0$. Their longitudinal wavenumber obeysBelow the waveguide cutoff frequency ${\omega }_c=v{k}_{\perp }$, that mode is evanescent rather than freely propagating. TM modes require both indices positive; TE modes may have one zero but not both. For $a>b$, the lowest cutoff is the TE${}_{10}$ mode, with $f_c=v/(2a)$. A hollow single-conductor guide has no TEM mode, whereas a two-conductor transmission line can support one without this geometrical cutoff.^[ref]

For a propagating TE or TM mode, $v_p=v/\sqrt{1-{\omega }_c^2/{\omega }^2}$ and $v_g=v\sqrt{1-{\omega }_c^2/{\omega }^2}$. The phase velocity can exceed $v$ without superluminal information transfer. Frequency-dependent group delay can broaden pulses; excitation of several modes adds intermodal dispersion.

Choose the guide axis as $z$, with $0<x<a$ and $0<y<b$. The electric field is along $y$, parallel to the walls' $b$ dimension. Take a homogeneous lossless nonmagnetic dielectric with relative permittivity $ϵ$, ideal conducting walls, and time dependence $e^{-i\omega t}$. The lowest transverse electric mode of a rectangular waveguide isAt $x=0,a$, this tangential electric field vanishes. At $y=0,b$ the electric field is normal to the wall; a nonzero value is supported by surface electric charge and violates no conducting boundary condition. The field has zero divergence. The associated normal magnetic field also vanishes at each wall, as verified below. The fundamental sine across the wider dimension minimizes the nonzero transverse wave number; ${\mathrm{TE}}_{00}$ is not a physical propagating mode of a hollow simply connected guide.

Below the waveguide cutoff frequency $f_c=c/(2a\sqrt{ϵ})$, put $\beta =i\kappa$ with $\kappa >0$ to select decay for $z>0$. The evanescent transverse electric mode in a rectangular waveguide then hasFor the fixed frequency in the question,The decay length increases monotonically and diverges as $ϵ\to 4^{-}$; its formal limit at $ϵ\to 0^{+}$ is $a/\pi$. At $ϵ=4$ the mode is at cutoff, with zero axial wave number. For $ϵ>4$, $\beta =(\pi /a)\sqrt{ϵ/4-1}$ is real and the mode propagates without axial attenuation in this ideal lossless model. There is no real finite evanescent decay length to continue beyond cutoff. The paper's graph notation $l$ denotes the same $\ell$ as its preceding formula.

From Faraday's law of induction, $\nabla \times \tilde{\mathbf{E}}=i\omega \tilde{\mathbf{B}}$. Writing $q=\pi /a$ givesThus $B_x$ vanishes on the $x$ walls and $B_y$ vanishes on the $y$ walls, meeting the normal magnetic boundary conditions. The fields also satisfy $\nabla \cdot \mathbf{B}=0$. For real $E_0$ the physical fields are $E_y=E_0\sin (qx)e^{-\kappa z}\cos \omega t$, $B_x=-(\kappa /\omega )E_0\sin (qx)e^{-\kappa z}\sin \omega t$, and $B_z=-(q/\omega )E_0\cos (qx)e^{-\kappa z}\sin \omega t$.

The time-averaged Poynting vector isAll products are imaginary because the electric field and magnetic field are in quadrature. A single decaying evanescent wave stores reactive energy without transmitting average power into a semi-infinite lossless guide. A finite below-cutoff section can transmit power when both independent evanescent solutions are present; that is a different boundary-value problem.

The skin effect also gives exponential field decay, but in a good metal $k$ has both real and imaginary parts and there is a real inward mean Poynting vector. That flux supplies Joule heating $\sigma |\tilde{E}{|}^2/2$. Here attenuation comes from geometric cutoff in a lossless region, with no conductivity and no dissipative power sink; exponential decay alone does not establish absorption.

Assume a hollow square electromagnetic waveguide with perfectly conducting walls, filled with vacuum or air. For side $a$, the waveguide cutoff frequency isA transverse electric mode can have one zero index, but not both. A transverse magnetic mode requires both indices positive. Thus the lowest cutoff belongs to the two degenerate transverse electric modes ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$, withThe corresponding cutoff vacuum wavelength is $2a=2.0\mathrm{cm}$. Strictly, a travelling mode carrying power along the guide needs $f>f_c$: at the cutoff itself, the axial wavenumber and group velocity are zero. The number above is therefore the lower frequency threshold, not a propagating mode with nonzero longitudinal energy flux exactly at that frequency. A dielectric filling would replace $c$ by the wave speed in that medium.

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.

For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency $f_c=10\mathrm{GHz}$ and carrier frequency $f=30\mathrm{GHz}$, $\sqrt{1-(f_c/f{)}^2}=\sqrt{8}/3$. ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check $v_p{v}_g=c^2$ follows directly from the dispersion relation. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.

MIT's documented slotted-line waveguide measurement shorts one end of a rectangular electromagnetic waveguide and excites it through a probe coupled by capacitance. A second such probe moves through a longitudinal slot, aligned with the wall electric current to reduce disturbance. Its output locates axial standing wave nulls. The book's standing-wave setup chooses a frequency above the first waveguide cutoff frequency and below higher-mode cutoffs. Lowering it below the first cutoff instead reveals spatial decay away from the source. The text supplies apparatus and an observable comparison, rather than a table of measured null positions.^[ref]

The inference can be checked without assuming that detector output is proportional to electric field amplitude. At fixed measured frequency, record several minima at coordinates $z_0,\dots ,z_N$; then estimate ${\lambda }_g=2(z_N-z_0)/N$ when the minima are consecutive. This uses positions alone. Compare the result with the guided wavelength prediction using independently measured $a$ and $f$. The $40\mathrm{mm}$ spacing above is a synthetic example, not a reading from the MIT apparatus.

Additional proposed controls are to keep probe insertion fixed, repeat the scan in both directions, recheck source frequency, and compare shallow probe settings for shifts in the minima. These test probe loading, mechanical backlash and drift. A deliberately reflecting termination supplies the standing wave; varying its position should translate the pattern without changing adjacent-null spacing in a uniform single-mode region. Source-adjacent evanescent waves, unintended reflections, additional propagating normal modes, detector noise and finite positional resolution can spoil that inference. A disappearing remote signal alone does not distinguish below-cutoff decay from loss or poor excitation, and a scan at one frequency does not measure a pulse's group velocity.

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.