Babinet's Principle

Take an opaque screen with some holes cut in it, and its photographic negative — the screen where every hole is now a solid patch and every solid patch a hole. Shine the same laser through each in turn and catch the diffracted light on a distant wall. Babinet's principle makes a claim that sounds impossible: away from the central beam, the two patterns are identical, band for band, ring for ring. Two screens that are exact opposites diffract light into the same shape. The reason is disarmingly simple once you see it. Any diffraction integral sums wavelets over the open area of the screen. The open area of a screen and the open area of its complement fit together like jigsaw pieces to tile the whole plane, so the two diffracted amplitudes must add up to the amplitude with no screen at all — the undisturbed incoming wave. And an undisturbed collimated beam is concentrated entirely in the forward direction; everywhere else it is zero. So off-axis the two complementary amplitudes are equal and opposite, and a detector, which sees only the squared magnitude, cannot tell them apart. The exception is the one place the undisturbed wave is not zero: dead ahead. There the bookkeeping does not cancel, and the two complements part ways spectacularly. The most famous case is a small opaque disk. Naive shadow-thinking says its centre should be the darkest point of all; wave theory, through this very bookkeeping, insists there is a bright spot there — the Poisson–Arago spot. Predicted as an absurdity meant to demolish the wave theory, then observed, it became one of the most decisive confirmations in the history of optics, and it is the portrait of this article.

Key takeaways

Babinet's principle: two complementary diffracting screens (each the exact negative of the other) produce diffracted amplitudes that sum to the unobstructed field, $U_1+U_2=U_0$ .
Because a collimated beam's unobstructed field is a forward spike and zero elsewhere, off-axis $U_1=-U_2$ , so the complementary intensities coincide, $|U_1|^2=|U_2|^2$ for $\theta\ne0$ .
On axis the free field does not vanish, so the complements differ — this is where the opaque disk shows the bright Poisson–Arago spot while its complementary hole shows an ordinary bright centre.
The disk is bright because a small disk removes only the first few Fresnel zones; the surviving alternating zone series still sums to about half the first unblocked zone, ≈ the full unobstructed axial field.
Key idea: the diffracted amplitude is a linear integral over the open aperture, so complementary screens add to the free wave — negligible off-axis (identical intensities) but finite on axis, where the disk's surviving Fresnel-zone tail lights up the Poisson–Arago spot.

The core insight

The diffracted amplitude is a linear functional of the incident field integrated over the open aperture. Complementary screens tile the plane, so their two amplitudes add to the unobstructed field $U_0$ . Off the forward beam $U_0\to0$ , forcing $U_1=-U_2$ and identical intensities; on axis $U_0\ne0$ , and there the opaque disk's surviving zone tail sums to ≈ half the first unblocked zone — a bright Poisson–Arago spot exactly where geometry predicts the deepest shadow.

Where this lives today

Babinet's equality is a working tool wherever the diffraction of an obstacle is easier to measure than the diffraction of a hole. Optical particle sizers read the diffraction halo of droplets, dust, and fibres with ordinary aperture (Airy) formulas, because Babinet guarantees an obstacle and the equal aperture scatter identically in the measured wings — the basis of laser-diffraction granulometry and of the in-line monitors that gauge drawn-fibre diameter in real time. The vector Babinet–Booker form is equally practical in radio and terahertz engineering: a slot in a conducting sheet and the complementary metal strip radiate the same pattern up to a $90°$ polarization rotation, the design principle behind slot antennas, self-complementary broadband antennas, and complementary frequency-selective surfaces and metasurfaces that swap band-stop for band-pass by taking the geometric complement. The Poisson–Arago spot itself has become a modern laboratory in its own right. It appears as an unwanted artifact behind circular obstructions in optical systems — dust on a lens, a telescope's secondary support — and as a precision alignment aid, since the spot marks the exact axis of a circular obstacle. Most strikingly, the spot has been reproduced not just with light but with electrons, neutrons, and even large molecules: an obstacle diffracts matter waves into the same bright axial point, a direct demonstration that massive particles interfere, and one of the cleaner tests of quantum coherence at increasing mass.

The physical picture

Cut a plane into two jigsaw pieces — the holes of screen A and the holes of its complement B. Together they tile the whole plane. Since the diffracted field is an integral of wavelets over the open bits, A's field plus B's field must equal the field with the entire plane open: the undisturbed incoming wave. For a collimated beam that undisturbed wave is essentially a single spike straight ahead and zero everywhere else, so off-axis $U_A=-U_B$ — equal in size, opposite in phase. Detectors register $|U|^2$ and cannot see the sign, so the two patterns look the same.

Figure

Complementary screens. A slit (screen 1) and an opaque strip of the same width (screen 2) are exact negatives, yet their far-field patterns coincide everywhere off the axis: U₁ + U₂ = U₀, and the unobstructed wave U₀ is a forward spike that vanishes for θ ≠ 0, so U₁ = −U₂ and I₁ = I₂. Only on axis, where U₀ ≠ 0, do the two part ways.

Now think of the disk. The complementary hole "uses up" the forward beam through its opening; the disk instead lets that beam slip around its rim and reconverge on the axis behind it. The result is a bright point exactly where naive geometry predicts the deepest shadow — the fingerprint of the axial term Babinet's bookkeeping tracks. The next figure shows why, with the Fresnel-zone vibration spiral and the real Fresnel number on the plot.

Figure

The Poisson–Arago spot. The vibration spiral (grey) sums the half-period Fresnel-zone contributions of the open wavefront; it winds to the centre C, so the open-aperture resultant P₀ → C is about half the first-zone chord A₁. A small disk blocks only the first ≈ 0.8 zone (outer turns), but the surviving tail still winds to C, giving Pₚ → C ≈ ½A₁ — a bright spot. Here N_F = a² ⁄ (λL) = 0.79 for a = 1 mm, λ = 633 nm, L = 2 m.

The $90°$ polarization twist of the full electromagnetic version is a related but separate story: a slot cut in a conducting sheet radiates like the complementary metal strip, with electric and magnetic roles interchanged. That vector refinement fixes the scalar principle for real conductors and underlies slot antennas — but the scalar bookkeeping above is exact for the ordinary optical screens of this article.

The core equations

Any Kirchhoff-type diffraction integral sums the incident field, weighted by a propagation kernel, over the open area of the screen. Because the kernel does not care which screen is present — only the domain of integration does — adding the integrals for two complementary screens integrates over the whole plane, recovering the unobstructed wave. This is Babinet's principle in one line.

U_1(P) + U_2(P) = U_0(P) \qquad (1)

where: $U_1,U_2$ — diffracted complex amplitudes of the two complementary screens at the observation point $P$ ; $U_0$ — field at $P$ with no screen present. The reading: complementary apertures tile the plane, so their amplitudes are bound together at every point by the single free field $U_0$ — the whole principle is a statement about that constraint, and everything else follows from where $U_0$ is large and where it is small. For a collimated beam the unobstructed far field is the Fraunhofer Diffraction pattern of an unobstructed plane — a sharp spike in the forward direction and essentially zero at any other angle. Setting $U_0\to0$ off-axis in $(1)$ forces the two complementary amplitudes to be equal and opposite.

U_1(\theta) = -U_2(\theta)\ \ (\theta\ne0)\ \ \Longrightarrow\ \ |U_1(\theta)|^2 = |U_2(\theta)|^2 \qquad (2)

where: $\theta$ — observation angle from the forward direction. The reading: the intensity discards the sign, so equal-and-opposite amplitudes give identical patterns away from the beam. This is the counter-intuitive core — a slit and its complementary strip, a hole and its disk, throw the same light everywhere except dead ahead. The same fact reads even more cleanly in the Fourier language of the far field, where the pattern is the transform of the aperture transmission $t(x,y)$ . Complementary screens have transmissions that add to unity everywhere, and the transform of a constant is a single spike at the origin.

t_1 + t_2 = 1 \quad\Longrightarrow\quad \mathcal F\{t_1\} + \mathcal F\{t_2\} = \mathcal F\{1\} = \delta(\theta) \qquad (3)

where: $t_1,t_2$ — transmission functions (1 where open, 0 where blocked) of the complementary screens; $\mathcal F$ — Fourier transform mapping aperture to far field; $\delta(\theta)$ — the forward beam. The reading: the summed spectra are a pure on-axis delta, so away from $\theta=0$ the two transforms cancel exactly — Babinet's equality restated as $\mathcal F\{1\}=\delta$ . On axis the free field does not vanish, and this is where complements diverge. Summing the half-period Fresnel zones of an open wavefront gives an axial amplitude of about half the first zone; whether a small disk keeps that value depends only on how many zones it covers, measured by the Fresnel number.

U_\text{axis}\approx \tfrac12 A_1, \qquad N_F = \frac{a^2}{\lambda L} \qquad (4)

where: $A_1$ — amplitude of the first Fresnel zone; $a$ — disk (or aperture) radius; $L$ — observation distance; $\lambda$ — wavelength; $N_F$ — number of Fresnel zones the edge subtends. The reading: $N_F$ counts the blocked zones, and for a small disk ( $N_F$ of order one) the surviving zone tail still sums to ≈ $\tfrac12 A_1$ — the full unobstructed axial amplitude — so the shadow centre stays bright.

Derivation ABabinet's principle from the Kirchhoff integral

The goal of this card is to prove the amplitude-sum relation $U_1+U_2=U_0$ rigorously and then extract its intensity consequence, showing precisely why complementary screens give identical patterns away from the beam. The whole argument rests on one structural fact: the diffraction kernel is fixed by the geometry and the incident field, and only the domain of integration — the open area — changes from a screen to its complement.

Step 1. Write the diffracted field as a Fresnel–Kirchhoff integral over the open aperture $\Sigma$ : $U(P)=-\dfrac{i}{\lambda}\displaystyle\iint_{\Sigma}U_\text{inc}(Q)\,\dfrac{e^{ikr}}{r}\,\dfrac{\cos\chi+\cos\chi_0}{2}\,\mathrm dS$ , with $r$ the distance from aperture point $Q$ to $P$ and the last factor the obliquity term. Why: this is the Huygens–Fresnel sum made quantitative — every open point re-radiates, and the integrand depends only on the incident field and geometry, not on which screen produced the opening.

Step 2. Note that complementary screens partition the plane: $\Sigma_1\cup\Sigma_2=$ the full plane and $\Sigma_1\cap\Sigma_2=\varnothing$ . Where screen 1 is open, screen 2 is blocked, and vice versa. Why: this is the exact definition of "complementary" — the two open sets are disjoint and together miss nothing, which is what lets their integrals be added without double-counting.

Step 3. Add the two diffracted fields. Because the integrand is identical and the domains tile the plane, $U_1(P)+U_2(P)=-\dfrac{i}{\lambda}\displaystyle\iint_\text{plane}(\dots)\,\mathrm dS=U_0(P)$ , the field propagated with no screen at all. This is equation $(1)$ . Why: integrating the same kernel over piece one plus piece two equals integrating over the whole — elementary additivity of integrals, which is the entire content of Babinet.

Step 4. Go off-axis. A collimated beam's unobstructed field $U_0(\theta)$ is a forward-peaked Fraunhofer spike, negligible for $\theta\ne0$ , so $U_1(\theta)=-U_2(\theta)$ and therefore $|U_1|^2=|U_2|^2$ . Why: the free wave carries essentially all its energy straight ahead, so everywhere else the two complements must cancel — leaving equal intensities, the observable statement of the principle. *Verified: the complementary identity $t_1+t_2=1$ and the resulting off-axis equality $|\mathcal F\{t_1\}|=|\mathcal F\{t_2\}|$ with $\mathcal F\{t_1\}+\mathcal F\{t_2\}$ confined to a DC spike, by FFT in verify_aq-optics-babinets-principle.py.*

Go deeper · Derivation B · 1 minThe Arago spot from the surviving Fresnel-zone tail

Derivation BThe Arago spot from the surviving Fresnel-zone tail

This card gives the clean reason the opaque disk is bright on axis, avoiding the muddled "disk equals plane minus hole" shortcut. The strategy is to sum the disk's surviving Fresnel zones directly with one lemma — that an alternating series of slowly shrinking terms sums to half its first term — applied first to the open wavefront and then to the zones the disk leaves untouched.

Step 1. Divide the wavefront reaching an axial point into half-period Fresnel zones of amplitudes $A_1,A_2,A_3,\dots$ , alternating in phase and shrinking slowly (obliquity and $1/r$ make $A_{m+1}$ just under $A_m$ ). The open-aperture axial amplitude is the alternating sum $A_1-A_2+A_3-\cdots\approx\tfrac12A_1$ . Why: neighbouring zones nearly cancel; the standard result is that such an alternating series equals half its first term, so an unobstructed wave delivers ≈ $\tfrac12A_1$ on axis — intensity $\tfrac14A_1^2$ .

Step 2. Insert a small opaque disk covering the first $p$ zones (a few, since $N_F=a^2/\lambda L$ is of order one). The blocked zones are simply gone; the surviving field is the tail $A_{p+1}-A_{p+2}+A_{p+3}-\cdots$ . Why: the disk removes the innermost contributions and nothing else — this is the honest starting point, not a subtraction of two whole patterns.

Step 3. Apply the same half-sum lemma to the tail: $A_{p+1}-A_{p+2}+\cdots\approx\tfrac12A_{p+1}$ . Because only a few zones are blocked, $A_{p+1}\approx A_1$ , so the disk's axial amplitude is ≈ $\tfrac12A_1$ — essentially the full unobstructed value. Why: the surviving series has exactly the same alternating, slowly shrinking structure, so it too sums to half its first term, which is still nearly the first-zone amplitude.

Step 4. Conclude: $|U_\text{disk}(0)|^2\approx|U_0(0)|^2$ — a bright spot of nearly the incident intensity sits at the geometric shadow centre, the Poisson–Arago spot. Why: the disk cannot suppress the axial field it merely shifted to outer zones, so the shadow's darkest predicted point is in fact bright — the observation that confirmed the wave theory in 1818. *Verified: the alternating zone sum $\to\tfrac12A_1$ (intensity $\to\tfrac14A_1^2$ ), the surviving tail from zone $p+1\to\tfrac12A_{p+1}\approx\tfrac12A_1$ , and $N_F=0.79$ , in verify_aq-optics-babinets-principle.py.*

Limiting cases

Case	Result	Meaning
Off the direct beam (Fraunhofer)	$\|U_1(\theta)\|^2=\|U_2(\theta)\|^2$ , $\theta\ne0$	scalar principle is exact: the free field vanishes, so complementary intensities are identical
On axis / inside the direct beam	$U_1(0)+U_2(0)=U_0(0)\ne0$	patterns differ — the disk's Poisson–Arago spot versus the hole's bright centre
Small disk, $N_F=a^2/\lambda L\sim1$	axial field ≈ $\tfrac12A_1$ ≈ unobstructed	few zones blocked, surviving tail bright — the spot is easy to see
Large disk, $N_F\gg1$	spot faint, tiny	many zones blocked; $A_{p+1}\ll A_1$ , so the spot dims toward a true shadow
Vector (Babinet–Booker)	electric $\leftrightarrow$ magnetic screen, polarization rotated $90°$	for real conductors the scalar form fails at fine features; the exact statement pairs a screen with its dual

Common trapBabinet says the intensities are always equal

The single most common misreading is to quote Babinet as " $I_1=I_2$ everywhere." It is a statement about complex amplitudes — $U_1+U_2=U_0$ — and the intensities coincide only where the unobstructed field $U_0$ vanishes, which for a collimated beam means away from the forward direction. On axis, and anywhere inside the direct beam, $U_0$ is large and the two complementary patterns differ decisively — that is precisely where the Poisson–Arago spot lives, the disk bright and its complementary hole showing an ordinary maximum. Write $|U_1|^2=|U_2|^2$ only with the rider $\theta\ne0$ . A second, subtler trap: the scalar principle ignores polarization, so for a fine metal grid at microwave frequencies it fails, and only the vector Babinet–Booker relation (electric $\leftrightarrow$ magnetic screen, a $90°$ rotation) is exact.

Worked examples

1 · Measuring a hair's diameter by its shadow (confidence builder)

A human hair of width $a=0.10$ mm is held in a $\lambda=633$ nm laser beam. Its shadow is crossed by diffraction fringes. Where is the first dark fringe, and how far apart are the fringes on a screen $L=1$ m away? Why does a thin opaque wire diffract like a slit at all?

Step 1. By Babinet the wire (an opaque strip) is the complement of a slit of the same width, so away from the axis it produces the identical single-slit pattern — the $\mathrm{sinc}^2$ of Single-Slit Diffraction. The first minima obey $a\sin\theta=\lambda$ .

Step 2. Solve for the first-minimum angle: $\sin\theta=\lambda/a=633\times10^{-9}/1.0\times10^{-4}=6.33\times10^{-3}$ , so $\theta_1=\mathbf{6.33\ mrad}=0.363^\circ$ . The pattern is set entirely by the wire width, not by anything about "wire-ness."

Step 3. On a screen $L=1$ m away the fringe spacing (minimum to minimum) is $\Delta y=\lambda L/a=633\times10^{-9}\times1.0/1.0\times10^{-4}=6.33\times10^{-3}$ m $=\mathbf{6.33\ mm}$ . Measure the fringe spacing and invert: $a=\lambda L/\Delta y$ .

Step 4. This is exactly how wire, fibre, and filament diameters are gauged optically — the diffraction halo of the obstacle is read with the aperture formula, Babinet guaranteeing they match. First minimum at 6.33 mrad; fringes 6.33 mm apart at 1 m; the wire diffracts like a slit of its own width.

2 · Will the Poisson spot be visible? (exam level)

An opaque disk of radius $a=1$ mm is lit by a $\lambda=633$ nm beam, and a screen sits $L=2$ m behind it. Estimate the Fresnel number and say whether a bright axial spot appears. This is the geometry of the 1818 test that confirmed the wave theory.

Step 1. The Fresnel number is $N_F=a^2/(\lambda L)=(10^{-3})^2/(633\times10^{-9}\times2.0)=\mathbf{0.79}$ . It counts how many Fresnel zones the disk edge subtends at the screen — here, less than one.

Step 2. Because $N_F\sim1$ , the disk blocks only the innermost fraction of a zone; the surviving alternating zone tail sums to ≈ $\tfrac12A_1$ , essentially the full unobstructed axial amplitude. A clear bright spot sits at the shadow centre.

Step 3. Cross-check the zone radius: the first Fresnel zone at $L=2$ m has radius $r_1=\sqrt{\lambda L}=\sqrt{633\times10^{-9}\times2.0}=\mathbf{1.13\ mm}$ , comparable to the disk, confirming $N_F=(a/r_1)^2=0.79$ — the disk barely reaches into the first zone.

Step 4. Had the disk been large ( $N_F\gg1$ ), $A_{p+1}\ll A_1$ and the spot would fade toward a true shadow — which is why a smooth, small obstacle is needed to see it. $N_F=0.79$ : the spot is bright and easily visible, the classic Fresnel-prize demonstration.

Physics of the portrait

The portrait here is not a person but a point of light — the bright spot at the centre of a shadow that Poisson offered as proof the wave theory was absurd, and that Arago's experiment turned into its vindication.

Origins · 1 minpoisson spot and how this was discovered

poisson spot

the bright dot that proved light is a wave (1818) · diffraction & wave optics

In 1818 the French Academy set a prize for the best theory of diffraction, expecting the corpuscular view to triumph. Augustin-Jean Fresnel submitted a wave theory of such power that the committee — which included Siméon Denis Poisson, a firm corpuscularian — searched it for a fatal flaw. Poisson found what he thought was one: Fresnel's mathematics implied that behind a small circular disk there must be a bright spot at the very centre of the geometric shadow, which he declared a manifest absurdity. François Arago performed the experiment and there, dead centre, was the spot. Far from killing the wave theory, Poisson's "absurdity" became its most dramatic confirmation, and the effect has carried his name ever since — though Arago observed it, and Giacomo Maraldi had already seen it nearly a century earlier without grasping its significance. The spot is the visible signature of Babinet's axial bookkeeping: the one place complementary screens refuse to agree, where the wavefront that geometry says is fully blocked reassembles itself into a point of light.

History

1723

— Giacomo Maraldi observes a bright spot at the centre of a circular obstacle's shadow, without explanation

1818

— Augustin-Jean Fresnel submits his prize memoir on the wave theory of diffraction

1818

— Siméon Denis Poisson deduces the axial bright spot as an absurdity; François Arago observes it and confirms the wave theory

1837

— Jacques Babinet states the complementary-screen principle

Going further

Texts. Hecht, Optics, 5th ed., §10.3 (Babinet's principle; Poisson spot) · Born & Wolf, Principles of Optics, 7th ed., §8.3 · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 4 (Fourier optics) · Foot, Atomic Physics, appendix on diffraction · Booker, "Slot aerials and their relation to complementary wire aerials (Babinet's principle)", J. IEE 93 (1946) 620 · Jackson, Classical Electrodynamics, 3rd ed., §10.6–10.7 (vector diffraction, complementary screens).

Provenance. Amplitude superposition and off-axis equality follow from the linearity of the Fresnel–Kirchhoff integral (Hecht §10.3; Born & Wolf §8.3); a verifier should reproduce the plane-partition argument and the Fourier identity $\mathcal F\{1\}=\delta$ . The Poisson-spot derivation uses standard Fresnel-zone sums (Hecht §10.3.3). The vector Babinet–Booker relation and impedance product $Z_1Z_2=(\eta_0/2)^2$ are from Booker (1946) and Jackson §10.6–10.7 with $\eta_0\approx376.7\ \Omega$ . No non-standard numbers are asserted. Every number here — the off-axis equality, $N_F=0.79$ , the zone-tail half-sums, and the wire fringes — is recomputed in verify_aq-optics-babinets-principle.py.

Prerequisites. Huygens–Fresnel Principle · Fresnel–Kirchhoff Diffraction Integral · Fraunhofer Diffraction Regime. Leads to. Fresnel Zones and Zone Plates · Circular Aperture Diffraction and the Airy Pattern · Rayleigh–Sommerfeld Diffraction Theory.

Is Babinet's principle about amplitude or intensity?

Amplitude, strictly. The exact statement is $U_1+U_2=U_0$ for the complex amplitudes, valid at every point. The famous "identical patterns" claim about intensities is a consequence that holds only where the unobstructed field $U_0$ vanishes — off the forward beam. Anywhere $U_0$ is appreciable, most importantly on axis, the two complementary intensities differ. Always attach the condition $\theta\ne0$ to $|U_1|^2=|U_2|^2$ , and you will never misapply the principle.

Why is there a bright spot where geometry predicts the darkest shadow?

Because the wavefront that the disk appears to block is not destroyed — it is only rearranged. Split the axial field into Fresnel zones; a small disk removes the first few, but the surviving zones still form an alternating series that sums to about half its first surviving term, which is nearly the first-zone amplitude. So the axial amplitude behind a small disk is almost what it would be with no disk at all. Geometry counts rays; waves count zones, and the zone sum keeps the centre lit.

Does the Poisson spot violate energy conservation?

No. Light is redistributed, not created. The disk casts a genuine shadow ring around the bright centre, and the total power reaching the screen is less than without the disk. The spot is simply where the diffracted wavelets from the disk's rim happen to arrive in phase and reinforce — a local concentration paid for by darkness elsewhere. It is the same bookkeeping as any interference maximum: bright fringes are always balanced by dark ones.

How perfect must the disk be to see the spot?

Quite good, but not extraordinary. The edge must be smooth and circular to within a fraction of a Fresnel-zone width, because roughness scatters the rim wavelets out of phase and washes the spot out. A machined ball bearing or a well-made disk works; a torn paper edge does not. The disk should also be small enough that $N_F=a^2/\lambda L$ is of order one — block too many zones and the surviving first term $A_{p+1}$ shrinks, dimming the spot toward a true shadow.

Does Babinet hold in the near field (Fresnel) as well as the far field?

The amplitude sum $U_1+U_2=U_0$ holds everywhere — it is just additivity of the diffraction integral and makes no far-field assumption. What needs the far field is the clean statement that complementary intensities are equal, because that requires $U_0$ to vanish away from a well-defined forward beam, which is sharpest in the Fraunhofer regime. In the near field $U_0$ has structure over a broader region, so the equality holds only where $U_0$ is genuinely negligible; the disk-versus-hole difference, by contrast, is a near-axis effect visible in both regimes.

Why does the scalar principle fail for a fine metal screen?

Scalar diffraction treats light as a single complex field and ignores polarization and the boundary conditions a conductor imposes on the electric and magnetic fields. A metal screen and a hole are not electromagnetically complementary — the conductor forces the tangential electric field to zero where the hole does nothing of the sort. The fix is the vector Babinet–Booker relation, which pairs a conducting screen with its dual (electric $\leftrightarrow$ magnetic) and rotates the polarization by $90°$ . For optical screens many wavelengths across, the scalar version is nonetheless excellent.