The residual was a loop

Boron trifluoride’s force field has a flat direction: a combination of its bending constants that no frequency can see. Projecting it out brought four fits from four different starting points into agreement on the bend constant to 0.0027, where before they had spread over nearly nine tenths. It did not bring them into agreement on everything. The B–F stretch constant still differed by 0.115 between the four, and that residual has been explained twice without being measured.

The first explanation called it a property of the eigenvalue problem rather than of the coordinate set. The attempt to test that — refitting with a second isotopologue — ran into a different problem and narrowed the candidates to two: the search, which might not have converged, or the arithmetic, which might be ill-conditioned near the edge of the determined subspace. Both have tests, and both tests would come back clean. The residual is neither, and the count that says so needs no computation at all.

Two frequencies for three constants #

A planar molecule with three-fold symmetry separates its vibrations into symmetry blocks, and a force constant can only affect the frequencies of the blocks its combination belongs to. Boron trifluoride’s six constants — the B–F stretch $r$, the stretch–stretch coupling $rr$, the bend $t$, the bend–bend coupling $tt$, the stretch–bend coupling $rt$ and the out-of-plane wag $w$ — sort as follows.

The out-of-plane wag is alone in its block, $A_2''$, with one frequency at 719.5 wavenumbers: one constant, one equation. The symmetric stretch, $A_1'$, depends only on $r + 2rr$ and has one frequency at 888. The symmetric bend would be the second $A_1'$ motion, but three angles in a plane sum to 360 degrees, so there is no symmetric bend to have — that is the redundancy, and $t + tt$ is the flat direction it produces.

What is left is the doubly degenerate E′ block, in which a stretch and a bend are coupled. It depends on three combinations — $r - rr$, $rt$ and $t - tt$ — and it supplies two frequencies, at 480.4 and 1453.9. Three unknowns and two equations: one combination is free, and it is free in a block the projection does not touch, because the projection removes $t + tt$ and nothing else.

That is the same arithmetic water’s force field had in its $A_1$ block, where four constants met three frequencies. There it was the reason a second isotopologue was needed at all. Here it was hidden behind a redundancy that looked like the whole problem.

The family is a closed curve #

A free combination in a two-by-two block does not produce a line of solutions. The two frequencies of the block are fixed by the trace and the determinant of a product of two matrices, and the trace is linear in the force constants while the determinant is quadratic. The solutions are where a plane meets a quadric surface — a conic section — and for these constants it is an ellipse.

Traced exactly, the curve is large. Along it the B–F stretch constant runs from 3.94 to 12.11 millidyne per ångström, and the stretch–bend coupling from −0.36 to 6.71, with every one of the six frequencies unchanged. Every tenth of the 1,402 traced points was fed back through the normal-mode calculation, and all of them reproduce 480.4, 719.5, 888 and 1453.9 to a thousandth of a wavenumber.

The four fits are the dots, and at this scale they are one dot. They sit together in a patch near a stretch constant of 7.4 and a coupling of 0.4, on the lower side of the curve. A spectroscopist quoting a stretch constant of 7.4 is quoting one point of a closed curve whose stretch constant covers a factor of three.

The other constants move with it. The stretch–stretch coupling changes sign, running from −1.64 to 2.44; the bend combination rises from a half to 4.58. Around the curve the four constants rise and fall in a fixed relation, and the band marks the short stretch where the four fits landed. Nothing in the spectrum distinguishes any point of this curve from any other, and nothing in the projection does either, because every point on it lies inside the subspace the projection keeps.

The fits are four points on an arc #

The question the residual raised was whether the four fits disagree because they are imperfect or because they are different. The curve answers it by position.

Each fit lies on the curve, within a few thousandths of a traced point, and they lie along it rather than around it. The fits started from zero and plus three tenths in the bend constants landed almost on top of each other; the one started from minus three tenths landed a little further along; the one started from plus one landed furthest. The 0.115 spread in the stretch constant is a distance measured along a curve of exact solutions, and a search that had converged perfectly from each start would have produced exactly this spread.

That disposes of the search directly. The fit from the most distant start was repeated with its convergence criterion tightened by a factor of a million and its simplex rebuilt eight times rather than three. It took three and a half times as long and moved $r - rr$ by seven ten-thousandths. A search that stops early stops off the curve, in a direction that raises the residual; these stopped on it, where the residual has nothing left to lower. Where on the curve each one stopped was decided by the path its simplex took from its start, and that path is not even monotonic in the start: along the arc the four fall in the order plus three tenths, zero, minus three tenths, plus one. Nothing about the curve prefers any of those positions, so the order is a record of four searches and not a property of the molecule.

It disposes of the arithmetic too, in a way the proposed test could not have. The suggestion was to look at the smallest non-zero eigenvalue of the determined subspace and see whether the problem was ill-conditioned near its edge. That subspace is defined by which directions in the space of force constants change the Hessian, and every direction along this curve changes the Hessian — the constants change and so does the matrix. What stays fixed is its eigenvalues. A freedom in the eigenvalue problem is invisible to a test on the Hessian map, so the conditioning test would have reported a well-conditioned subspace and been right, and the residual would still have been there.

What a second isotopologue does to a curve #

The standard remedy for an underdetermined block is more data with the same constants and different masses. Substituting boron-10 for boron-11 changes how the B–F stretch mixes into the E′ motions, because the boron atom moves in them, while leaving the force constants alone.

To ask what the substitution can settle — rather than what a particular table of measurements says — the boron-10 frequencies here are the ones the first fit predicts: 482.3 and 1507.3 in the E′ block. Those are exact data from a known field, so the question is purely one of how many fields they are consistent with.

They are consistent with two. Going around the curve, the boron-10 mismatch falls to zero twice and nowhere else: once at the field the data came from, with a stretch constant of 7.43, and once at a quite different field with a stretch constant of 3.97. The two differ in every constant the E′ block depends on. Between them the mismatch reaches 56 wavenumbers, so the substitution is not weak; it simply has a second solution.

Why two and not one #

The count says the substitution should over-determine the block: two more frequencies, four equations in all, for three unknowns. It does not, and the reason is an identity already met in a different role — as the one isotope rule that is exact — which here turns out to be the reason an isotope cannot finish the job. It costs nothing to check, and it explains the pair completely.

In a two-by-two block the product of the two squared frequencies is the determinant of the product of the kinematic matrix and the force-constant matrix, and the determinant of a product is the product of the determinants. The kinematic determinant depends only on masses and geometry. So the product equation from boron-10 says exactly what the product equation from boron-11 says — that the force-constant matrix has a stated determinant — scaled by a ratio of kinematic determinants that involves no force constant at all. That is the Teller–Redlich product rule, which is why the ratio of the products of isotopologue frequencies can be predicted from masses alone; read the other way, it means a substitution adds no information through the product.

What the substitution does add is its trace: one new linear equation. The family of exact boron-11 fits is a plane meeting the quadric surface of fixed determinant; the boron-10 trace is a second plane; two planes meet in a line, and a line meets a quadric in two points. The second field is not an accident of boron trifluoride’s numbers. It is what any two-by-two block with three constants and two isotopologues must leave, and a third isotopologue — whose trace is a third plane — is the first data that can pick one point out.

So the field a spectrum picks out is determined up to a discrete choice, and which member of the pair is physical has to be decided by something outside the frequencies: a sign expected on chemical grounds, a Coriolis constant, a centrifugal distortion constant, or a third mass. The identity is the same one that fixed forty-five of methane’s constants as combinations rather than one by one: a spectrum is a set of invariants, and invariants determine what they are invariants of only up to what leaves them unchanged.

The second field is not an absurd one. Its stretch constant is about half the fitted one, its stretch–stretch coupling three times larger, and its bend combination four times larger; every number in it is of an ordinary size. Nothing about it would look wrong in a table.

And there is a harder point about precision. Measured boron-10 frequencies are not exact, and a real fit would use them to within an uncertainty. Allowed half a wavenumber, the substitution leaves a stretch constant anywhere from 7.32 to 7.52 near the fitted field — a range of 0.20, nearly twice the 0.115 over which the four fits happened to agree. The agreement between the four fits was a property of where their searches started, and it looked like a precision the data do not have.

How the curve was traced #

The frequencies come from the library’s own normal-mode calculation: a Wilson GF analysis of the valence force field in internal coordinates, with the redundancy handled as it is everywhere else. The four fits are the stored drift fits, which differ only in where the bend and bend–bend constants started.

Holding $w$ and $r + 2rr$ at the fitted values, each field in the family is written in its three E′ combinations. The sum of the two squared E′ frequencies is fitted as a linear function of those three, and their product as a quadratic, from thirty evaluations of the normal-mode calculation spread around the fit; the two polynomials reproduce the calculation to a part in 10¹⁰, which is what they must do if the structure of a two-by-two block has been read correctly. At each value of $rt$ the linear equation gives $t - tt$ as a function of $r - rr$, and the quadratic then gives two values of $r - rr$ or none; the edges of the curve are found by bisecting on whether a solution exists, and the points between are spaced by a cosine so that the turns are sampled densely.

The boron-10 comparison substitutes the central atom’s mass and recomputes the E′ frequencies at every traced point; its zeros are the local minima of the mismatch below five hundredths of a wavenumber, which is the resolution of the tracing.

The checks, run wherever these figures are drawn: the polynomials reproduce the calculation they were built from; every tenth traced point reproduces all six frequencies to five thousandths of a wavenumber; the stretch constant’s range along the curve exceeds five millidyne per ångström and forty times the fits’ spread; each fit lies on the curve and shares the symmetric-stretch combination; and the boron-10 frequencies select exactly two points, one being the fitted field. The refusal is a field moved a tenth off the curve in $r - rr$: it must stop reproducing the E′ frequencies by more than a wavenumber, because a family every nearby field belonged to would be a flat direction, and the flat direction is the one already removed. The tightened search is recorded rather than re-run: it takes fifty seconds.

What this curve does not include #

The valence force field is a model with six constants, and counting what a curve can determine before fitting it is the same discipline applied to a smaller model. A general harmonic force field for boron trifluoride has more, and its E′ block more couplings; the curve here is the family within this parameterisation, and a larger one would be a family of higher dimension, not a smaller one.

The frequencies are observed fundamentals, treated as harmonic. Anharmonicity shifts every one of them by a few per cent, and differently for different isotopologues, so a real two-isotopologue fit compares quantities that are not quite what the harmonic calculation predicts. That is why the half-wavenumber range matters: it is a generous allowance for measurement and a small one for anharmonicity.

And the boron-10 data are synthetic. Using the fit’s own predictions isolates identifiability from measurement; it cannot say which of the two fields a real boron-10 spectrum would favour.

A fit’s agreement with itself #

The pattern here is an old one in new clothes. Several fits agreeing is evidence about the fits, not about the answer. Four starts that differ only in the bend constants will land near each other on any curve of exact solutions, because a local search moves as little as it can; their agreement measures how far apart they started, and it can look like precision. It is the same trap as a correlation that is not an account, one level down: agreement among outputs that share an origin.

The count that exposed it — constants against frequencies, block by block — is the cheapest diagnostic available and it was not done, because a flat direction had already explained a disagreement and the explanation looked complete. A projection can remove a freedom it knows about. It says nothing about a freedom in a different place, and the determined subspace is determined only in the sense it was defined in.

Still open: which of the two, and a third mass #

The obvious open question is which of the two boron-10-consistent fields is physical, and the data that could decide it are of a different kind. A Coriolis coupling constant between the two E′ components depends on the eigenvectors of the block rather than its eigenvalues, and so on exactly the angle that parameterises the curve; for a symmetric top like boron trifluoride it is accessible from rotationally resolved infrared bands. Computing ζ along the curve would show whether it separates the two fields, and whether it does so by more than its measured uncertainty.

The nearer question is what fluorine substitution would add. Boron trifluoride has only one stable fluorine isotope, so the second isotopologue in practice is the one used here, but a hypothetical fluorine mass change moves the E′ block through the other atom — the one that moves in the bend rather than the stretch. Whether its curve meets the boron-10 pair at one of the two points or at neither is a statement about how differently two substitutions sample one block, and it can be answered with masses alone.