The third mass that cannot help

Boron trifluoride’s E′ force constants are not determined by its frequencies. The block couples the B–F stretches with the in-plane bends through three combinations of constants — the stretch minus its coupling, the stretch–bend coupling, and the bend minus its coupling — and supplies two frequencies. Every field that reproduces both lies on a closed curve, along which the stretch constant runs from 3.9 to 12.1 millidyne per ångström with no frequency changing.

A boron-10 substitution cuts the curve to two points. It adds two frequencies, and it would add two equations if both were new. They are not. The product of a two-by-two block’s squared frequencies is the determinant of its kinematic matrix times the determinant of its force-constant matrix, and the kinematic determinant contains only masses — that is the Teller–Redlich product rule, which is why isotope ratios of frequency products can be predicted from masses alone. Read the other way, the product equation from boron-10 repeats the product equation from boron-11. Only the sum — the trace — is new, a plane meeting the curve twice: at the field the data came from, and at a second field with a stretch constant of 3.97.

The essay that found the pair closed with the natural next step. A third isotopologue has a third trace; a third plane should pick one point of two. The suggestion was specific: change the fluorines rather than the boron, because the fluorines move in the bend where the boron moves in the stretch, so the block would be sampled through a different atom.

A third mass, tried every way #

Nothing about the question needs an isotope that exists. The frequencies depend on the force constants and the masses, and the masses can be anything. So the two fields — the fitted one and its partner — were each put through ten isotopologues, real and hypothetical, and their predicted spectra compared.

Six keep the molecule’s symmetry: boron-10; boron of mass 9 and of mass 13; all three fluorines of mass 17, and of mass 21; and boron-10 with all three fluorines of mass 21. Under every one of the six, the two fields give the same E′ frequencies to 1.5 × 10⁻⁶ cm⁻¹ — a millionth of a wavenumber, which is where the calculation’s own precision sits, since that is also how well the two members of each degenerate pair agree with each other.

No substitution that keeps the symmetry can tell the two fields apart. Not another boron, not the fluorines, not both together. The fluorine suggestion fails exactly as the boron one does, and to the same number of figures.

The trace is one plane in two masses #

The reason is in how the kinematic matrix depends on the masses. For a planar molecule with one central atom and three equivalent outer ones, every element of the E′ block’s kinematic matrix is a reciprocal boron mass times a number plus a reciprocal fluorine mass times another number. The trace equation — the sum of the two squared E′ frequencies — is therefore two fixed linear functionals of the force constants, weighted by $1/m_\mathrm{B}$ and $1/m_\mathrm{F}$:

$$\omega_1^2 + \omega_2^2 \;=\; \frac{\alpha(F)}{m_B} + \frac{\beta(F)}{m_F}.$$

That is checked rather than assumed. Fitted on three isotopologues as a function of the two reciprocal masses, the sum predicts two more to a few parts in a million million, and the constant term comes out at zero.

So every symmetric isotopologue’s trace equation says one thing about α and one thing about β, weighted by its own masses. Two isotopologues with different mass ratios fix both α and β, and every later one is a weighted sum of what is already known. The third plane is a member of the pencil the first two define — it passes through the same line — and a line meets the curve at the same two points it did before. The product equation adds nothing, by Teller and Redlich. A symmetric isotopologue contributes no new equation after the second, whatever its masses, and the two-fold ambiguity is permanent for every set of them.

The same identity that made the product rule exact makes the ambiguity permanent. Forty-five of methane’s constants are fixed as combinations of its spectrum’s invariants and not one by one; here the invariants run out one short, and every symmetric substitution produces the same invariants in different proportions.

Three equations for three constants, and one of them quadratic #

It is worth writing down what the count looks like once the structure is respected, because the naive count is what made a third isotopologue look sufficient. Each isotopologue supplies two E′ frequencies. Two isotopologues supply four, a third six, against three unknown combinations of constants: over-determined twice over, on the face of it.

Respecting the structure, the equations fall into two kinds. The trace equations are linear in the constants, and across every symmetric isotopologue there are exactly two independent ones, one per reciprocal mass — because the block contains two kinds of atom, the central one and the equivalent outer ones. The product equations are quadratic, and across every isotopologue there is exactly one, because Teller and Redlich make every product a mass factor times the same determinant. Three independent equations for three constants, and one of them quadratic. That is a determined system with two solutions, and no number of further symmetric isotopologues changes either count.

The general form of this is a rule about symmetry blocks rather than about boron trifluoride. A block of size k has k power sums of its squared frequencies, the first linear in the constants and the last the determinant. Symmetric isotopologues contribute to the j-th power sum only through the distinct products of j reciprocal masses of the symmetry-distinct kinds of atom, and the determinant only once. So the number of equations a whole family of symmetric isotopologues can ever supply to a block is fixed before any spectrum is taken, by the block’s size and the number of kinds of atom it contains. Counting what a curve can determine before fitting it is the same discipline at a different scale.

That is why adding data made it worse in an earlier fit, and why a second molecule had a blind spot of its own: in both, the data added were of a kind whose structure made them repeat equations already present, and the fit had no way to know. The flat direction of ten directions no frequency can see is the extreme case — a combination of constants no equation contains at all — and the two-fold choice here is the mildest case — a combination every equation contains in the same quadratic way.

What the two fields have in common #

The two fields differ in almost everything the block depends on. The stretch constant is 7.42 in one and 3.97 in the other; the stretch–stretch coupling 0.70 against 2.43; the bend 0.51 against 2.24. What they share is the wag, which the E′ block does not contain, and the stretch–bend coupling, 0.378649 in both to seven figures.

That shared constant is the plane’s doing too. The line along which every symmetric isotopologue’s trace is satisfied runs at constant stretch–bend coupling: the two trace functionals fix that coupling between them and leave a single combination of the stretch and the bend, which the product equation then meets twice. Every symmetric isotopologue determines the stretch–bend coupling exactly and cannot determine the stretch. The constant a chemist would most want from the block, the B–F stretch, is precisely the one left two-valued.

Breaking the symmetry adds an equation #

The argument has a hole in it, and the hole is the way out. It assumed all three fluorines had the same mass. Replace one fluorine, or two, and the molecule is no longer symmetric: the E′ pairs split, the E′ block and the totally symmetric block mix, and the kinematic matrix is no longer two functionals in two masses.

With one fluorine of mass 21, the two fields’ spectra differ. The lower member of the split bending pair moves by 0.42 cm⁻¹ between them and the upper by 0.25; the symmetric stretch, now mixed with the E′ block, by 0.45; one member of the split stretching pair by 0.22 and the other not at all; the out-of-plane wag not at all. One fluorine of mass 17 gives differences up to 0.66, boron-10 with one fluorine of mass 21 up to 0.42, two fluorines of mass 21 up to 0.45. An isotopologue that breaks the symmetry is the first data that can pick one of the two fields, and it does so by four to seven tenths of a wavenumber for a mass change of two.

How much it separates them depends on how badly the symmetry is broken. At the natural mass the difference is zero, because the molecule is symmetric again. A change of one mass unit gives 0.15 cm⁻¹ on the light side and 0.12 on the heavy side; a change of two, 0.66 and 0.45; a change of four, 3.3 and 1.5. The difference grows roughly as the square of the mass change, which is what a second-order effect of breaking a symmetry does: the first-order change in each frequency is the same under both fields, and only the second order tells them apart.

What that separation is worth #

Four to seven tenths of a wavenumber is small against what a real spectrum carries. Observed fundamentals are anharmonic, shifted from their harmonic values by a few per cent — tens of wavenumbers in the stretch — and the shift differs between isotopologues. A harmonic calculation compared with anharmonic data therefore has an uncertainty far larger than the half-wavenumber the broken isotopologue buys. The discrimination is real in the model and is available in practice only to a fit that carries anharmonic corrections accurate to a tenth of a wavenumber, or to an analysis that works with harmonic frequencies already extracted from overtone data.

The measurement itself is not the obstacle. A band origin from a rotationally resolved infrared spectrum is routinely known to a hundredth of a wavenumber or better, so half a wavenumber between two predictions is fifty times the measurement’s own uncertainty. The obstacle is the model the measurement is compared with. A valence force field predicts harmonic frequencies, and turning observed fundamentals into harmonic ones needs the anharmonic constants of every mode, each of them an input with its own uncertainty and its own isotope dependence. The data can resolve the ambiguity; the harmonic model of the data cannot resolve it unless its anharmonic corrections are known to a tenth of a wavenumber — a demand much stronger than the one the ambiguity itself makes, and the honest way to state what a broken isotopologue buys.

Fluorine has one stable isotope, so for boron trifluoride itself the broken isotopologue does not exist. The argument transfers to molecules where it does: boron trichloride, whose natural chlorine is a mixture of mass 35 and 37, is observed as a mixture of isotopologues in which one or two chlorines differ from the rest — the symmetry-broken kind, in natural abundance. For such a molecule the data that could resolve the E′ block are already in the spectrum, as the weak lines of the less symmetric species.

What was computed and how #

The two fields are the fitted boron-11 field, from the library’s valence force field fitted to the six observed frequencies, and its partner: the second exact intersection of the boron-10 trace with the family of exact fits, located on the traced curve and then polished by Gauss–Newton iteration on the four E′ frequencies of boron-11 and boron-10 together until they agree to 2 × 10⁻⁸ cm⁻¹. The polish matters: the curve is traced to five hundredths of a wavenumber, which is enough to find the partner and not enough to ask a question at the sixth decimal.

Every isotopologue is a Wilson GF normal-mode calculation with the same constants and changed atom masses, hypothetical masses included. For the symmetric ones the comparison is over the E′ modes; the other blocks agree between the two fields only to the family’s own tolerance of a few hundred-thousandths, because the partner was built by moving the E′ combinations with the others held. The linearity check fits the sum of squared E′ frequencies as $a/m_\mathrm{B} + b/m_\mathrm{F} + c$ on boron-11, boron-10 and a fluorine-21 molecule and predicts two more.

The claims are stated where they can fail: that each of the six symmetric substitutions gives the two fields E′ frequencies within 10⁻⁵ cm⁻¹ and every frequency within the family’s tolerance; that each of the four symmetry-breaking substitutions separates them by more than three tenths of a wavenumber and by more than ten thousand times the largest symmetric difference; and, as the refusal, that the trace is linear in the two reciprocal masses to a part in a million, because if it were not, the reason given for the first claim would be the wrong reason.

Where the argument stops #

A valence force field with six constants. A general harmonic field has more couplings in E′ and the family of exact fits is then of higher dimension; the argument about symmetric isotopologues goes through unchanged, since it concerns only the kinematic matrix, but the number of fields they leave is no longer two.

Harmonic frequencies throughout. The two fields are exact harmonic solutions; real data are anharmonic, and the half-wavenumber separation the broken isotopologue buys is smaller than the harmonic–anharmonic gap. What is established is identifiability in principle, not a recipe for a real fit.

And the other discriminators are untouched. Coriolis coupling constants depend on the eigenvectors of the block rather than its eigenvalues, so they see the mixing angle that distinguishes the two fields directly, and a symmetric top’s rotationally resolved bands carry them. Nothing here says how far apart the two fields put them.

When more data of the same kind is not more data #

The habit this argument calls for is to count equations by their structure before collecting them. More isotopologues felt like more data, and each one adds two frequencies. But a frequency is an equation only if it is not a function of equations already written down, and for a symmetric molecule’s isotopologues both the product and the trace are: one by an exact identity, the other because the masses enter through only two coefficients. Data of the same symmetry sample the same invariants, and the invariants run out one short.

It is a correlation that is not an account seen from the side of the data: agreement between many measurements that share an origin is one measurement, and a set of symmetric isotopologues shares its origin in two reciprocal masses. The force field is not in the spectrum was the first statement of this in general; the symmetric isotopologues are its most tempting counterexample, and they are not one.

What breaks the deadlock is data of a different kind: a lower symmetry, which mixes blocks the higher symmetry kept apart, or a different observable, which sees eigenvectors where frequencies see eigenvalues. The first is available in natural abundance for molecules with mixed isotopes. The second is what the ambiguity has been pointing to since the curve was found.

Still open: the Coriolis constant along the curve, and a molecule with two chlorines #

The obvious open question is the Coriolis constant. The E′ block’s Coriolis coupling depends on the angle that parameterises the curve of exact fits, so computing it along the curve would say how far apart the two fields put it, and whether that difference exceeds the uncertainty of a measured value. A planar symmetric top has sum rules that constrain its E′ Coriolis constants, and whether the sum rule leaves the one that distinguishes the fields free is the first thing to establish.

The nearer question is a molecule that has the broken isotopologues naturally. Boron trichloride’s mixed-chlorine species are observed alongside the symmetric ones, and the same construction — a fitted field, its partner under a boron substitution, and the spectra of the mixed species under both — would say whether the natural isotopic mixture of a real molecule already contains the equation that decides its E′ block.