Hankel Functions and Asymptotic Behavior of Bessel Functions

The Bessel equation $x^2 y''+x y'+(x^2-\nu^2)y=0$ has an irregular singular point at infinity, so far from the origin its solutions do not decay as powers or grow as exponentials — they oscillate. Physics on unbounded domains needs to know exactly how: with what amplitude, at what wavelength, and with which direction of energy flow. The real pair $(J_\nu,Y_\nu)$ settles amplitude and wavelength but not direction, because each is a standing wave — the cylindrical analogue of $\cos$ and $\sin$ . This article changes basis to the two complex combinations that carry a definite direction, the Hankel functions (Bessel functions of the third kind), and extracts the single most-used fact in the whole Bessel toolkit: the large-argument asymptotics. The move is exactly the passage from $(\cos\theta,\sin\theta)$ to $(e^{+i\theta},e^{-i\theta})$ . The Hankel functions $H^{(1)}_\nu=J_\nu+iY_\nu$ and $H^{(2)}_\nu=J_\nu-iY_\nu$ diagonalize the asymptotic behaviour: each collapses to a single travelling exponential $\sqrt{2/\pi x}\;e^{\pm i(x-\nu\pi/2-\pi/4)}$ at large $x$ . Under the $e^{-i\omega t}$ time convention $H^{(1)}_\nu(kr)$ is outgoing and $H^{(2)}_\nu(kr)$ incoming, so choosing $H^{(1)}$ imposes the radiation condition automatically. We derive the leading term three equivalent ways — Sommerfeld's contour integral by steepest descent, the WKB (Liouville–Green) reduction of the Bessel equation, and real and imaginary parts recovering $J_\nu$ and $Y_\nu$ — and mark where the fixed-order formula fails: the turning-point region $x\approx\nu$ , whose uniform Airy description is a separate story.

Key takeaways

The Hankel functions $H^{(1)}_\nu(x)=J_\nu(x)+iY_\nu(x)$ and $H^{(2)}_\nu(x)=J_\nu(x)-iY_\nu(x)$ are the complex combinations of Bessel functions that behave as outgoing and incoming cylindrical waves.
At large argument each collapses to one exponential, $H^{(1,2)}_\nu(x)\sim\sqrt{2/\pi x}\;e^{\pm i(x-\nu\pi/2-\pi/4)}$ — the cylindrical analogue of the spherical $e^{\pm ikr}/r$ .
Real and imaginary parts give the universal asymptotics of $J_\nu,Y_\nu$ : amplitude decaying as $x^{-1/2}$ with phase $x-\nu\pi/2-\pi/4$ , so consecutive zeros settle to spacing $\pi$ .
The leading form follows from steepest descent on Sommerfeld's contour integral, or the WKB form of the Bessel equation; both fix the amplitude $\sqrt{2/\pi}$ and the universal $-\pi/4$ Maslov phase.
Hankel functions are the natural radiation-condition solutions of 2D scattering, cylindrical Green's functions, waveguides, and antenna theory.
Key idea: Switch from the real basis $(J_\nu,Y_\nu)$ to the complex basis $(H^{(1)}_\nu,H^{(2)}_\nu)$ : each Hankel function is asymptotically a single travelling exponential $\sqrt{2/(\pi x)}\,e^{\pm i(x-\nu\pi/2-\pi/4)}$ , so it imposes the radiation condition automatically.

Where this lives today

Hankel functions are the working language wherever a wave must be outgoing in a cylindrically symmetric world. In computational electromagnetics they are the radiation boundary condition: finite-element and boundary-element solvers truncate an infinite domain by matching the outer boundary field to $H^{(1)}_m(kr)$ , and the method of moments for antenna and radar-cross-section work rests on the 2D Green's function $G=\tfrac{i}{4}H^{(1)}_0(kr)$ . Optical-fibre design uses the pair directly: guided core modes are $J_\nu$ , the evanescent cladding field is $K_\nu$ (imaginary-argument Hankel), and mode-cutoff conditions come from matching their asymptotics at the core boundary. In acoustics the same functions describe sound radiating from or scattering off a cylinder. At the frontier, the asymptotic distribution of Bessel zeros feeds regularized vacuum-energy sums — Casimir forces between cylinders, heat-kernel expansions — where McMahon's spacing and the Airy corrections near $x\approx\nu$ must both be handled for finite answers.

The physical picture

Think of $(J_\nu,Y_\nu)$ as the cylindrical $(\cos,\sin)$ and of $(H^{(1)}_\nu,H^{(2)}_\nu)$ as the cylindrical $(e^{+i\theta},e^{-i\theta})$ : the real pair builds standing waves, the complex pair travelling waves with a definite radial direction of energy flow. The whole content of "far from the source" is two numbers — how fast the amplitude fades and what the phase is. The first figure shows the amplitude answer on the graph of $J_0$ : the oscillation is trapped between the envelopes $\pm\sqrt{2/\pi x}$ , and the zeros march out toward a fixed spacing.

Figure

The universal large-x oscillation of J₀(x): the curve is trapped inside the envelope ±√(2/πx), so at x = 10 the amplitude bound is 0.2523 and J₀(10) = −0.2459, while successive zeros settle to a spacing Δx → π = 3.14

The phase answer is where the direction of travel lives. Reduce the Bessel equation to Helmholtz form: $y=u/\sqrt{x}$ kills the first-derivative term and leaves $u''+(1-(\nu^2-\tfrac14)/x^2)u=0$ , so far out $u''+u\approx0$ and $u\sim e^{\pm ix}$ . The amplitude decays as $x^{-1/2}$ because in two dimensions the energy on a ring of circumference $2\pi x$ is conserved — spread the same power over a longer ring and $|y|^2\propto1/x$ . The constant offset $-\nu\pi/2-\pi/4$ is accumulated WKB phase: the $-\pi/4$ is the universal turning-point (Maslov) shift, and $-\nu\pi/2$ records the centrifugal-barrier strength. The second figure makes the basis change concrete — the real pair as projections, the complex pair as a rotating phasor that marches outward.

Figure

Real basis versus complex basis: (Jν, Yν) are the cylindrical cosine and sine that stand still, while H⁽¹⁾ν = Jν + iYν and its conjugate H⁽²⁾ν rotate on the Argand circle of radius √(2/πx) and march outward as e^{i(kr − ωt)} — the outgoing cylindrical wave

Because $H^{(1)}_\nu$ is purely outgoing, multiplying by the physicist's $e^{-i\omega t}$ gives $e^{i(kr-\omega t)}$ — crests moving to larger $r$ . That is why Green's functions, scattered fields, and antenna patterns are written with $H^{(1)}$ : the choice of Hankel function is the choice of boundary condition at infinity.

The core equations

The definition is a pure change of basis, chosen so the two members are complex conjugates and each singles out one direction of travel. Equations (1) and (2) are exact identities for all $x$ and all orders $\nu$ .

H^{(1)}_\nu(x)=J_\nu(x)+i\,Y_\nu(x),\qquad H^{(2)}_\nu(x)=J_\nu(x)-i\,Y_\nu(x) \qquad (1)

where: $J_\nu$ — Bessel function of the first kind, order $\nu$ (regular at the origin for $\nu\ge0$ ); $Y_\nu$ — Bessel function of the second kind, the Neumann function (singular at the origin); $\nu$ — order, generally complex; $x$ — argument, physically the radial phase $kr$ . Reading: the Hankel functions are to $(J_\nu,Y_\nu)$ what $e^{\pm i\theta}$ are to $(\cos\theta,\sin\theta)$ . The map is linear and invertible, so the real Bessel functions are the real and imaginary parts of $H^{(1)}_\nu$ (for real $\nu,x$ , since then $H^{(2)}_\nu=\overline{H^{(1)}_\nu}$ ).

J_\nu=\tfrac{1}{2}\!\left(H^{(1)}_\nu+H^{(2)}_\nu\right),\qquad Y_\nu=\tfrac{1}{2i}\!\left(H^{(1)}_\nu-H^{(2)}_\nu\right) \qquad (2)

where: $H^{(1,2)}_\nu$ — Hankel functions of the first and second kind. Reading: standing waves are equal-weight superpositions of the two travelling waves, just as $\cos\theta=\tfrac12(e^{i\theta}+e^{-i\theta})$ . The central quantitative fact is the large-argument form. For fixed order and $x\to\infty$ each Hankel function is one travelling exponential, with a universal $x^{-1/2}$ amplitude and a phase that is linear in $x$ up to the constant offset.

H^{(1,2)}_\nu(x)\sim\sqrt{\frac{2}{\pi x}}\,\exp\!\left[\pm i\!\left(x-\frac{\nu\pi}{2}-\frac{\pi}{4}\right)\right],\qquad x\gg|\nu^2-\tfrac14| \qquad (3)

where: $\sqrt{2/\pi x}$ — the $x^{-1/2}$ amplitude fixed by 2D energy conservation; $-\nu\pi/2-\pi/4$ — the WKB phase, containing the order-independent $-\pi/4$ turning-point shift. Reading: $H^{(1)}$ (upper sign) is outgoing, $H^{(2)}$ (lower sign) incoming; the two are complex conjugates for real arguments. Taking the real and imaginary parts of (3) gives the oscillatory asymptotics of the real Bessel functions — the forms that make every Bessel function look like a damped cosine at large $x$ .

J_\nu(x)\sim\sqrt{\frac{2}{\pi x}}\cos\!\left(x-\frac{\nu\pi}{2}-\frac{\pi}{4}\right),\quad Y_\nu(x)\sim\sqrt{\frac{2}{\pi x}}\sin\!\left(x-\frac{\nu\pi}{2}-\frac{\pi}{4}\right) \qquad (4)

where: $\cos,\sin$ — the real and imaginary parts of the Hankel exponential. Reading: same amplitude and phase for both, with $Y_\nu$ leading $J_\nu$ by a quarter cycle, so their zeros interlace and consecutive zeros of either approach spacing $\pi$ . The leading term is only the first of an asymptotic (divergent, but optimally truncatable) series in $1/x$ . Its coefficients are set by the order alone.

H^{(1)}_\nu(x)\sim\sqrt{\frac{2}{\pi x}}\,e^{i\chi}\sum_{k=0}^{\infty}\frac{i^{k}\,a_k(\nu)}{x^{k}},\quad a_k(\nu)=\frac{\prod_{j=1}^{k}\big(4\nu^2-(2j-1)^2\big)}{k!\,8^{k}} \qquad (5)

where: $\chi\equiv x-\nu\pi/2-\pi/4$ ; $a_k(\nu)$ — the asymptotic coefficients, with $a_0=1$ and $a_1=(4\nu^2-1)/8$ . Reading: for half-integer $\nu$ the product $4\nu^2-(2j-1)^2$ eventually vanishes and the series terminates — that is exactly why the spherical Bessel functions are elementary. That the two Hankel functions form a genuine basis is certified by their Wronskian, nonzero everywhere and fixed (being $\propto 1/x$ ) by Abel's identity applied to the Bessel equation.

W\{H^{(1)}_\nu,H^{(2)}_\nu\}=H^{(1)}_\nu\,{H^{(2)}_\nu}'-{H^{(1)}_\nu}'\,H^{(2)}_\nu=-\frac{4i}{\pi x} \qquad (6)

where: $W$ — the Wronskian. Reading: it never vanishes, so $H^{(1)}_\nu$ and $H^{(2)}_\nu$ are linearly independent for every $x>0$ ; equivalently $W\{J_\nu,Y_\nu\}=2/\pi x$ . The engine behind the asymptotics is Sommerfeld's contour-integral representation, whose large- $x$ evaluation by steepest descent gives the whole series of (5).

H^{(1)}_\nu(x)=\frac{1}{i\pi}\int_{C}\exp\!\big(x\sinh t-\nu t\big)\,\mathrm{d}t \qquad (7)

where: $C$ — a contour from $-\infty$ to $+\infty+i\pi$ in the complex $t$ -plane. Reading: the large $x$ multiplies $\sinh t$ , so the integral is dominated by the saddle of $\sinh t$ and everything follows from the local geometry there. Finally, half-integer-order cylindrical Hankel functions collapse to the elementary spherical Hankel functions, the 3D radiation solutions — the bridge to the next article.

h^{(1)}_\ell(x)=\sqrt{\frac{\pi}{2x}}\,H^{(1)}_{\ell+1/2}(x)\;\xrightarrow{\;x\to\infty\;}\;(-i)^{\ell+1}\frac{e^{ix}}{x} \qquad (8)

where: $h^{(1)}_\ell$ — spherical Hankel function of the first kind, order $\ell$ . Reading: in 3D the amplitude decays as $1/x$ (energy on a sphere of area $4\pi x^2$ ), one power faster than the cylindrical $x^{-1/2}$ .

Derivation AThe large-argument asymptotics by steepest descent

The goal is to turn Sommerfeld's exact integral representation (7) into the leading travelling-wave form (3) and to see where each piece is born — the exponential $e^{ix}$ from the saddle value, the $x^{-1/2}$ amplitude from the saddle width, the $-\pi/4$ from the descent direction, and the $-\nu\pi/2$ from the order term at the saddle. This is the canonical example of Debye's saddle-point method, and it fixes the constant $\sqrt{2/\pi}$ that the WKB argument alone cannot supply.

Step 1. Write $H^{(1)}_\nu(x)=\tfrac{1}{i\pi}\int_C e^{x f(t)}\,\mathrm{d}t$ with $f(t)=\sinh t-(\nu/x)t$ . For fixed $\nu$ and $x\to\infty$ the dominant exponent is $x\sinh t$ ; the $\nu t$ piece is a slowly varying correction. Why: isolating the large parameter multiplying a fixed function is the prerequisite for any saddle-point estimate.

Step 2. Locate the saddle from $f'(t)=\cosh t-\nu/x=0$ , near $\cosh t=0$ , i.e. $t_0=i\pi/2$ . Of the two saddles $\pm i\pi/2$ , the contour $C$ (ending at $+\infty+i\pi$ ) passes through $t_0=+i\pi/2$ . Why: the asymptotics are controlled by the saddle the contour can actually reach.

Step 3. Expand locally: $\sinh t_0=i$ and $f''(t_0)=\sinh t_0=i$ , so $x f(t)\approx ix+\tfrac12 x\,i\,(t-t_0)^2$ . The saddle value $e^{ix}$ — the outgoing phase — appears at once. Why: the constant term is the leading phase, the quadratic term sets the amplitude.

Step 4. The Gaussian integral along the descent direction is $\int e^{\frac12 x f''(t_0)(t-t_0)^2}\mathrm{d}t=\sqrt{2\pi/(-ix)}$ ; the decaying branch picks $\sqrt{-i}=e^{-i\pi/4}$ , giving both the $x^{-1/2}$ amplitude and the universal $-\pi/4$ Maslov phase. Why: the saddle width is the amplitude, its tilt in the complex plane is the constant phase.

Step 5. Restore the order term, $e^{-\nu t_0}=e^{-i\nu\pi/2}$ , and assemble $H^{(1)}_\nu\sim\tfrac{1}{i\pi}\,e^{ix}e^{-i\nu\pi/2}\sqrt{2\pi/x}\,e^{-i\pi/4}$ ; with $1/(i\pi)=e^{-i\pi/2}/\pi$ the phases collect to $x-\nu\pi/2-\pi/4$ and the prefactors to $\sqrt{2/\pi x}$ . Why: the slow factor contributes only its saddle value to leading order.

Go deeper · Derivation B · 1 minThe Liouville-Green route — amplitude and phase from WKB

Derivation BThe Liouville-Green route — amplitude and phase from WKB

The aim is to obtain the same leading form (3) without any integral, directly from the differential equation, so the two hallmark features — the $x^{-1/2}$ amplitude and the linear phase — appear as generic consequences of a second-order ODE with a slowly varying coefficient. This route shows why every Bessel-type function shares one envelope and one $\pi$ zero-spacing, and it exposes the exact validity condition $x\gg|\nu^2-\tfrac14|$ .

Step 1. Remove the first-derivative term with $y=u(x)/\sqrt{x}$ : the Bessel equation becomes the Schrödinger-like normal form $u''+\big(1-(\nu^2-\tfrac14)/x^2\big)u=0$ . Why: eliminating $y'$ turns the equation into a "particle in a potential", where WKB reasoning applies directly.

Step 2. The effective wavenumber is $q(x)=\sqrt{1-(\nu^2-\tfrac14)/x^2}\to1$ . Far out $u''+u\approx0$ , so $u\sim e^{\pm ix}$ and $y\sim x^{-1/2}e^{\pm ix}$ . Why: a nearly constant coefficient gives plane waves in $u$ ; the $1/\sqrt{x}$ from the substitution is the entire amplitude law.

Step 3. Read the amplitude as energy conservation: $|y|^2\propto1/x$ keeps the flux $x|y|^2$ through a ring of circumference $2\pi x$ constant — the 2D radiation condition. In 3D the substitution $y=u/x$ gives $|y|\propto1/x$ instead. Why: the power law is dimension-dictated, not order-dictated, hence universal across all $\nu$ .

Step 4. The next WKB order, $u\sim q^{-1/2}\exp(\pm i\int^x q\,\mathrm{d}x')$ with $\int^x q\,\mathrm{d}x'=x-\tfrac{\nu^2-1/4}{2x}+\cdots$ , plus the turning-point connection near $x\approx\nu$ , supplies the $-\nu\pi/2-\pi/4$ offset. The validity condition $|q'/q^2|\ll1$ is exactly $x\gg|\nu^2-\tfrac14|$ . Why: the turning-point analysis that gives Airy functions in the transition region gives the Maslov phase in the far region.

Go deeper · Derivation C · 1 minRecovering the oscillatory forms of the first and second kind

Derivation CRecovering the oscillatory forms of the first and second kind

The purpose is to descend from the compact Hankel asymptotic (3) back to the everyday statements about $J_\nu$ and $Y_\nu$ — the damped cosine and sine of (4) — and to extract the payoff most calculations use: McMahon's formula for the large zeros. Because the inversion (2) is exact, the only approximation is the single asymptotic (3), so any accuracy statement about $J_\nu$ inherits directly from the Hankel expansion.

Step 1. Start from the exact inversion $J_\nu=\tfrac12(H^{(1)}_\nu+H^{(2)}_\nu)$ , $Y_\nu=\tfrac1{2i}(H^{(1)}_\nu-H^{(2)}_\nu)$ , valid for all $x$ . Why: keeping the inversion exact isolates the approximation to one place, the Hankel asymptotic.

Step 2. Substitute $H^{(1,2)}_\nu\sim\sqrt{2/\pi x}\,e^{\pm i\chi}$ with $\chi=x-\nu\pi/2-\pi/4$ . The real combination collapses by $\tfrac12(e^{i\chi}+e^{-i\chi})=\cos\chi$ , giving $J_\nu\sim\sqrt{2/\pi x}\,\cos\chi$ . Why: adding the outgoing and incoming exponentials reconstructs the standing cosine — the cylindrical Euler identity.

Step 3. The imaginary combination $\tfrac1{2i}(e^{i\chi}-e^{-i\chi})=\sin\chi$ gives $Y_\nu\sim\sqrt{2/\pi x}\,\sin\chi$ , a quarter-cycle ahead of $J_\nu$ with identical amplitude. Why: the $1/(2i)$ weighting selects the sine, so $J_\nu$ and $Y_\nu$ are in phase quadrature, their zeros interlacing.

Step 4. Read off the zeros. $J_\nu(x)=0$ requires $\cos\chi=0$ , i.e. $\chi=(n-\tfrac12)\pi$ , giving McMahon's estimate $j_{\nu,n}\approx\big(n+\tfrac{\nu}{2}-\tfrac14\big)\pi$ ; consecutive zeros are spaced by $\pi$ as $n\to\infty$ . Why: the phase advances linearly, so zeros are equally spaced far out — the fact behind cylindrical mode-counting and Weyl's law. *McMahon's formula gives $j_{0,5}\approx14.923$ against the exact $14.931$ , a 0.06 percent error, improving with $n$ .*

Limiting cases

The behaviour of $H^{(1,2)}_\nu(x)$ is governed by the size of $x$ relative to the order $\nu$ ; the large-argument formula is only one of four regimes.

Case	Result	Meaning
Large argument, fixed order ( $x\gg\nu$ )	$H^{(1,2)}_\nu(x)\sim\sqrt{\tfrac{2}{\pi x}}\;e^{\pm i(x-\nu\pi/2-\pi/4)}$	pure travelling waves; the regime governing scattering, radiation, and mode-counting
Small argument ( $x\to0^+$ )	$H^{(1)}_\nu(x)\sim-\tfrac{i}{\pi}\Gamma(\nu)\left(\tfrac{2}{x}\right)^{\nu}\ (\nu>0)$	dominated by the $Y_\nu$ singularity; $H^{(2)}_\nu$ is its conjugate for real $\nu$
Imaginary argument (link to $K_\nu$ )	$K_\nu(x)=\tfrac{\pi}{2}\,i^{\nu+1}H^{(1)}_\nu(ix)$	rotating the contour turns oscillation into decay, $K_\nu(x)\sim\sqrt{\tfrac{\pi}{2x}}\,e^{-x}$
Transition region ( $x\approx\nu\gg1$ )	$J_\nu(\nu)\sim\dfrac{2^{1/3}}{3^{2/3}\Gamma(2/3)}\,\nu^{-1/3}\approx0.4473\,\nu^{-1/3}$	neither series applies; uniform Airy asymptotics are required at the turning point

Common trapThe large-x formula is not valid near x ≈ ν, and the outgoing label depends on your time convention

Two traps. First, the asymptotic (3) holds only for $x\gg|\nu^2-\tfrac14|$ ; using it when $x\lesssim\nu$ is simply wrong, because that is the turning-point region where the reduced equation's $1/x^2$ term is no longer negligible and one must switch to uniform Airy asymptotics — the same crossover that governs the rainbow and semiclassical scattering. A high-order function like $J_{50}$ does not become a damped cosine until $x$ is well past $50$ ; at $x=50$ it is still in the Airy regime, where $J_{50}(50)\approx0.121$ decays as $\nu^{-1/3}$ , not $\nu^{-1/2}$ . Second, which Hankel function is "outgoing" depends on the time convention: under $e^{-i\omega t}$ (physics) $H^{(1)}$ radiates outward, but under $e^{+i\omega t}$ (much engineering) it is $H^{(2)}$ . The mathematics is identical; only the label flips — always state your convention before quoting a radiation condition.

Worked examples

1 · Evaluating J₀(10) from the asymptotic series (confidence builder)

Use the large-argument expansion to estimate $J_0(10)$ at leading order and with the first correction, then compare with the exact value $J_0(10)=-0.245936$ .

Step 1. Assemble the leading term. The amplitude is $\sqrt{2/(\pi\cdot10)}=\sqrt{0.063662}=0.252313$ , and the phase is $\chi=x-\nu\pi/2-\pi/4=10-\pi/4=9.214602$ rad. Subtracting $2\pi$ , $\chi\equiv2.931417$ rad, so $\cos\chi=-0.977994$ and $\sin\chi=0.208632$ . The leading estimate is $J_0(10)\approx0.252313\times(-0.977994)=\mathbf{-0.24676}$ — already within $0.3\%$ of the exact value from a one-term formula.

Step 2. Add the first correction. For $\nu=0$ the coefficients are $a_1=(4\nu^2-1)/8=-\tfrac18$ and $a_2=(4\nu^2-1)(4\nu^2-9)/(2\cdot64)=\tfrac{9}{128}$ . Collecting the series (5) into its real form gives $J_0(x)\approx\sqrt{2/\pi x}\big[\cos\chi\,(1-\tfrac{9}{128x^2})+\tfrac{1}{8x}\sin\chi\big]$ . With $x=10$ : the cosine term is $-0.977994\times(1-0.000703)=-0.977306$ and the sine term is $0.208632/80=0.002608$ .

Step 3. Combine: $J_0(10)\approx0.252313\times(-0.974698)=\mathbf{-0.245932}$ , against the exact $-0.245936$ . One term gives 3 significant figures; two terms give 5. For $x\gtrsim5$ a handful of asymptotic terms beats the ascending power series, which needs dozens of terms and heavy cancellation near $x=10$ .

2 · Large zeros of J₀ and the density of cylindrical modes

Predict the fifth zero $j_{0,5}$ of $J_0$ from McMahon's formula and use the zero spacing to estimate how many radial modes fit in a drum of radius $a$ up to wavenumber $k$ .

Step 1. McMahon's estimate is $j_{\nu,n}\approx(n+\tfrac{\nu}{2}-\tfrac14)\pi$ . For $\nu=0$ , $n=5$ : $j_{0,5}\approx(5-\tfrac14)\pi=4.75\pi=\mathbf{14.923}$ , against the exact $14.931$ — a $0.06\%$ error. (For comparison the first zero gives $0.75\pi=2.356$ versus the exact $2.405$ , a $2\%$ error: the formula tightens rapidly as $n$ grows, since it is the leading term of an expansion in $1/n$ .)

Step 2. The spacing follows immediately: $j_{0,n+1}-j_{0,n}\to\pi$ . Physically, the radial nodes of a vibrating circular membrane clamped at $r=a$ occur where $J_0(ka)=0$ , i.e. at $ka=j_{0,n}$ , so the allowed radial wavenumbers are $k_n=j_{0,n}/a\approx(n-\tfrac14)\pi/a$ , equally spaced by $\pi/a$ .

Step 3. The number of radial modes with wavenumber below $k$ is $N(k)\approx ka/\pi$ , so the radial mode density is $\mathrm{d}N/\mathrm{d}k=a/\pi$ , independent of $n$ — the 1D Weyl law along the radius. Combined with the angular index $m$ it reproduces the leading $\text{Area}\times k^2/4\pi$ term of the 2D Weyl law for the drum, the starting point for Casimir-energy and heat-kernel sums.

3 · Where the large-x formula fails: the central value J₅₀(50)

Show that at $x=\nu$ the large-argument formula is inapplicable, and evaluate the true central value for $\nu=50$ .

Step 1. The validity condition $x\gg|\nu^2-\tfrac14|$ fails badly at $x=\nu=50$ : here $\nu^2-\tfrac14\approx2500$ , comparable to $x^2$ , so the potential term is not small. The point $x=\nu$ is the classical turning point, where the solution is neither oscillatory nor exponential but transitional.

Step 2. The correct description is the Airy-type uniform asymptotic, whose central value is $J_\nu(\nu)\sim\dfrac{2^{1/3}}{3^{2/3}\Gamma(2/3)}\,\nu^{-1/3}=0.44731\,\nu^{-1/3}$ . For $\nu=50$ : $J_{50}(50)\approx0.44731\times50^{-1/3}=0.44731\times0.27144=\mathbf{0.12142}$ , against the exact $0.121409$ — a match to four figures.

Step 3. Note the different power law: at the turning point the amplitude scales as $\nu^{-1/3}$ , not the far-field $\nu^{-1/2}$ . The turning-point amplitude is anomalously large, which is why grazing rays and glory scattering — the physical face of $x\approx\nu$ — are so prominent. Only once $x$ climbs well beyond $\nu$ does $J_{50}$ settle into the damped cosine of (4).

Physics of the portrait

The Hankel functions are named for a mathematician who died at 34 but whose short career fixed the standard language of cylinder functions; the asymptotics that carry the deeper physics were assembled by a lineage running from Poisson through Stokes to Debye.

Origins · 1 minHermann Hankel and how this was discovered

Hermann Hankel

1839–1873 · cylinder functions and their integral representations

Hermann Hankel was a German mathematician, a student of Riemann and Möbius, whose 1869 memoir Die Cylinderfunctionen erster und zweiter Art ("Cylinder functions of the first and second kind"), in the very first volume of Mathematische Annalen, gave the theory of Bessel functions its modern unified footing. It was Hankel who systematized the contour-integral representations — the ancestors of Sommerfeld's formula (7) — from which the large-argument behaviour drops out by saddle-point analysis, and the complex "third-kind" solutions were later named after him. He also gave physics the Hankel transform, the radial Fourier transform pervading cylindrically symmetric problems, before dying at 34 in 1873. The asymptotics themselves have a longer pedigree: Poisson found the leading oscillatory form around 1823, Stokes uncovered the discontinuities of the asymptotic constants (the Stokes phenomenon) in the 1850s, and Debye perfected the method of steepest descent for these functions in 1909 — the calculation of the first derivation above. Watson's 1922 Treatise on the Theory of Bessel Functions gathered the whole edifice, and Sommerfeld made the outgoing Hankel function the carrier of his radiation condition, which still defines "outgoing" in scattering theory.

History

1824

— Bessel systematizes the functions Jν in his theory of planetary perturbations

1857

— Stokes describes the discontinuous switching of asymptotic constants, the Stokes phenomenon

1869

— Hankel introduces the cylinder functions of the third kind and their integral representations

1909

— Debye derives the large-argument asymptotics by the method of steepest descent

1922

— Watson publishes A Treatise on the Theory of Bessel Functions

Going further

Texts. Stone & Goldbart, Mathematics for Physics, chapter on special functions and asymptotics · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 14 · Hassani, Mathematical Physics, chapters on second-order ODEs and complex analysis · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 6 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, chapter on special functions · Watson, A Treatise on the Theory of Bessel Functions, for the definitive account. Key numerical anchors: the steepest-descent phase assembles to $x-\nu\pi/2-\pi/4$ ; the Wronskian is $W\{H^{(1)}_\nu,H^{(2)}_\nu\}=-4i/(\pi x)$ ; the first series coefficient is $a_1=(4\nu^2-1)/8$ ; the worked value is $J_0(10)=-0.245936$ with two-term asymptotic $-0.245932$ ; McMahon's formula gives $j_{0,5}\approx14.923$ against the exact $14.931$ ; and the central value is $J_\nu(\nu)\sim0.4473\,\nu^{-1/3}$ , giving $J_{50}(50)=0.121409$ .

Prerequisites. Bessel Functions of the First and Second Kind · The Gamma Function · Recurrence Relations for Special Functions. Leads to. Spherical Bessel Functions · Modified Bessel Functions · Spherical Harmonics.

Ask the tutor: Ask me to walk through the steepest-descent saddle for $H^{(1)}_\nu$ line by line, or to show how choosing $H^{(1)}_0(kr)$ enforces the 2D radiation condition in a scattering problem.

Why introduce Hankel functions when Jν and Yν already solve the Bessel equation?

Because $J_\nu$ and $Y_\nu$ are standing waves, and physics on open domains needs travelling waves. Just as $\cos$ and $\sin$ suit a bounded string but $e^{\pm i\theta}$ suit propagation, the Hankel functions carry a definite direction of energy flow. A scattered field must be purely outgoing; expanding it in $J_\nu$ and $Y_\nu$ forces you to fine-tune their ratio at every order to cancel the incoming part, whereas $H^{(1)}_\nu$ builds the radiation condition in automatically. The complex basis also diagonalizes the asymptotics — one exponential each — which is what makes scattering amplitudes and Green's functions clean.

Is the asymptotic series in equation (5) convergent?

No — and that is a feature. Like most asymptotic expansions it diverges for every fixed $x$ : the coefficients $a_k(\nu)$ eventually grow factorially. What it offers is optimal truncation: the terms first shrink to a smallest term of size roughly $e^{-2x}$ , then grow, and stopping just before the smallest term gives an error of that order. For $x=10$ that is accuracy near $e^{-20}\approx10^{-9}$ from only a few terms — far better than the convergent ascending series, which needs dozens of terms and heavy cancellation there. The divergence is tied to the Stokes phenomenon: a subdominant exponential the tail cannot represent.

What is the Stokes phenomenon and does it affect these formulas?

The Stokes phenomenon is the discontinuous change in the constant multiplying a subdominant exponential as you cross certain rays (Stokes lines) in the complex plane. For the Hankel functions the single clean exponential of equation (3) is valid only in a sector of the complex $x$ -plane; rotate $x$ far enough and a second, exponentially smaller term switches on. On the positive real axis, where nearly all physics lives, $H^{(1)}_\nu$ is dominantly outgoing and $H^{(2)}_\nu$ incoming, so the simple formulas are safe. It matters when you analytically continue — connection formulas, or the modified functions $I_\nu$ and $K_\nu$ — where naive continuation gives wrong answers unless the switching is tracked.

How do the cylindrical Hankel functions relate to the spherical ones?

Through half-integer order: $h^{(1)}_\ell(x)=\sqrt{\pi/2x}\;H^{(1)}_{\ell+1/2}(x)$ . Because the series (5) terminates for half-integer $\nu$ , the spherical versions are elementary — finite combinations of $e^{ix}/x$ times polynomials in $1/x$ , for example $h^{(1)}_0(x)=-ie^{ix}/x$ exactly. The key difference is the amplitude law: cylindrical waves decay as $x^{-1/2}$ (energy on a ring of circumference $2\pi x$ ), spherical waves as $x^{-1}$ (energy on a sphere of area $4\pi x^2$ ). Both impose the radiation condition, but the 3D case gives the familiar $e^{ikr}/r$ of scattering theory and the 2D case gives $e^{ikr}/\sqrt{r}$ .

Where does the phase offset νπ/2 + π/4 come from physically?

Both pieces are accumulated WKB phase. The $-\pi/4$ is universal — the Maslov phase picked up on passing through the classical turning point at $x\approx\nu$ , the same shift that appears in the WKB quantization of any smooth well (and gives the oscillator its $\tfrac12\hbar\omega$ zero-point energy). In the steepest-descent picture it is the tilt $\sqrt{-i}=e^{-i\pi/4}$ of the descent direction. The $-\nu\pi/2$ piece encodes the centrifugal barrier $\nu^2/x^2$ : a higher-order wave is held off from the origin more strongly and lags in phase. Together they set the zeros of $J_\nu$ at $\big(n+\tfrac{\nu}{2}-\tfrac14\big)\pi$ .

Is there a Bessel analogue of an optical index below one, where phase speed exceeds c?

Yes — the phase velocity of a cylindrical wave can exceed the naive $c$ near the axis, and the turning-point structure explains it. Inside the classical turning point $x<\nu$ the solution is evanescent (the centrifugal barrier forbids propagation); only for $x>\nu$ does it oscillate and carry energy outward. The "faster than expected" phase behaviour near $x\approx\nu$ is the Airy transition region, not a violation: as with a superluminal phase velocity in optics, no signal outruns the group motion, set by $\mathrm{d}\omega/\mathrm{d}k$ , which stays subluminal. Both are turning-point (WKB) phenomena governed by the same Airy connection formula.