The partner field was a swap

Boron trifluoride’s E′ block has three force constants and two frequencies, so its exact fits form a closed loop. A ¹⁰B spectrum cuts the loop to two fields, no isotopologue that keeps the symmetry separates them, and the Coriolis constant that was expected to do it turned out to be the boron isotope shift again. Each of those was a calculation on a loop traced numerically at some fourteen hundred points, with the partner field polished to the ninth decimal. Each ended with a fact that looked like an accident: two fields, three mdyn per ångström apart, sharing their interaction constant to a part in ten million.

The last essay in that line asked what boron trichloride would do. Chlorine has two stable isotopes, so the species fluorine could only imagine sit in every bottle. Setting up that calculation meant writing down the two fields for a second molecule, and the arithmetic of writing them down is where the accident stopped looking like one.

Two fields, as printed #

Take boron trichloride with a bond length of 1.742 Å, and the bands of ¹¹B³⁵Cl₃: $\nu_3$ at 958, $\nu_4$ at 243 and $\nu_1$ at 471 cm⁻¹. Write the E′ field in the three numbers of its symmetry block, $x$ for the stretching constant less its interaction, $y$ for stretch against bend and $z$ for the angle’s own. Cut with ¹⁰B’s $\nu_3$ at 995 and two fields come out:

$$x=4.329,\ \ z=0.7144\qquad\text{and}\qquad x=0.7062,\ \ z=4.379,$$

with $y$ the same in both. Those numbers look unrelated until the units are put right. An angle’s constant is in different units from a bond’s, and the natural exchange rate between them is the factor by which an angle turns into a displacement: rescale with $z'=3z/R^2$, and the first field has $(x,z')=(4.329,\,0.7062)$ and the second has $(0.7062,\,4.329)$. They are the same pair of numbers, the other way round, to as many digits as are carried.

Boron trifluoride does exactly the same: from its fitted field the exchange gives $x=1.53561$ where the partner polished off a traced loop has $1.53561$, a difference of two parts in a hundred million that belongs to the polishing. That partner was found by searching. It could have been found by exchanging two numbers.

One element always equals another #

The reason takes one line. In the rescaled coordinate the kinetic-energy matrix for the E′ block has equal diagonal elements:

$$G=\begin{pmatrix} g & h\\ h & g\end{pmatrix},\qquad g=\frac{1}{m_{\mathrm X}}+\frac{3}{2m_{\mathrm B}},\qquad h\propto\frac{1}{m_{\mathrm B}}.$$

Now let $\sigma$ be the matrix that exchanges the two coordinates. It commutes with this $G$, so if a field $F$ is replaced by $\sigma F\sigma$, which is exactly the exchange of $x$ and $z'$ with $y$ kept, then the product $G\,\sigma F\sigma$ equals $\sigma (GF)\sigma$. The two have the same eigenvalues at every pair of masses. That is the whole of why no third symmetric mass could help, the Teller–Redlich argument included: it is a theorem about one symmetry of a two-by-two matrix, with nothing particular to boron, fluorine or the numbers involved. Every planar XY₃ molecule has this pair at every point of its loop.

It also says something the frequencies alone never could. The eigenvectors of $\sigma(GF)\sigma$ are those of $GF$ with their two coordinates exchanged. Where one field’s $\nu_3$ is mostly stretch and a little bend, the other’s is mostly bend and a little stretch, by exactly the same amounts. In the fields above the first gives the bond a stiffness of 4.41 mdyn per ångström, and the second gives it 1.99 and makes the 958 cm⁻¹ band an angle deformation. No frequency at any symmetric mass can object to that.

The loop, without tracing it #

If the exchange is the structure, the loop should be elementary, and it is. The sum of the squared frequencies fixes $g(x+z')+2hy'$, with $y'$ the rescaled interaction constant. The product fixes $xz'-y'^2$. So at each $y'$ the numbers $x$ and $z'$ have a known sum and a known product: they are the two roots of a quadratic, and the two sides of the loop are that quadratic’s two roots, read either way. Eliminating them leaves

$$(x-z')^2=\left(\frac{S-2hy'}{g}\right)^2-4y'^2-\frac{4P}{g^2-h^2},$$

a conic, which here is an ellipse.

A second boron mass changes $g$ and $h$ differently and leaves two linear equations in $x+z'$ and $y'$, which is a statement that the cut fixes $y'$ and nothing else. That is the fact recorded as a measurement in the essay on the Coriolis constant: the fields share their interaction constant to ten million parts. Of course they do. It was the only thing the cut could reach, and the two roots at that $y'$ are the two fields. The straight line of the interaction constant in $\zeta_3$, which the same essay measured and said it could not explain, is one step further. At a fixed field the boron’s share of a mode is linear in the mass-derivative of the sum, and that derivative is linear in $y'$ once the sum itself is fixed, so the share is a straight line in $y'$. Differencing the frequencies around the loop gives a straight line to 7×10⁻¹¹. Both of those statements can now be proved rather than sampled.

The figure carries a warning for anyone reading a force field off the data. One wavenumber of error in that ¹⁰B band moves $y'$ by 0.23, because the ellipse is long and thin in that direction. At 985 cm⁻¹ the stiff field’s bond constant is well above five; at 1000 the two roots have nearly met, 2.81 against 2.30. A real band carries an anharmonic shift the harmonic model does not, so the distance between the two fields is itself uncertain by much more than the difference this essay goes on to look for.

What one heavy chlorine breaks #

Replace one chlorine with chlorine-37 and the threefold axis has gone. One mirror remains, through the heavy atom and boron, and each E′ pair splits into a component even in that mirror, species $A_1$, and one odd in it, species $B_2$. The first component has the heavy atom moving along its own bond. The second has it moving across.

A change of mass at that atom adds a term to the kinetic energy that sits in exactly two places. Motion along the bond belongs to the even stretching coordinate, with weight two thirds; motion across the bond belongs to the odd bending coordinate, rescaled, also with weight two thirds. Here is the point of the rescaling again. The even stretch and the odd bend receive the same term, so exchanging the fields carries the even block under one field onto the odd block under the other, with the species of the two blocks exchanged and everything else left alone.

So a natural chlorine does something a boron cannot do, and it is not what the question proposed. The frequencies stay almost equal, and the assignment reverses. In the most abundant mixed species the lower component of $\nu_3$ is at 954.45 under the stiff field and 954.42 under its partner, with species $A_1$ under the first and $B_2$ under the second. The upper component is at 957.96 under both, with the same reversal. $\nu_4$ splits by less than a wavenumber and reverses too.

The picture is the claim in its plainest form. Two fields, one band and nearly one frequency, and in one of them the heavy atom moves along its bond and in the other across it. The physical reason for the band’s lower frequency is quite different in the two: the bond carrying a heavy atom gets softer, or the angle does.

This is true of the imaginary molecules as well. One fluorine of mass 21, the case the earlier essay left at about half a wavenumber, has the same exchange in both doublets. So the half wavenumber was the small part of the difference between those fields. What that essay did not have was a label to look at.

A natural sample, row by row #

Boron is about four-fifths ¹¹B and chlorine three-quarters chlorine-35, so the species in a bottle can be listed.

More than half of any natural sample, 55.1%, is a species with only a mirror left. In all of them the two fields reverse the pair in each doublet, and the frequency agreement never gets worse than 0.062 cm⁻¹. The band that decides is available in bulk in every sample of the compound, with a splitting of 3.5 cm⁻¹ in $\nu_3$ and about one in $\nu_4$. What it needs from a spectrum is not the fifth significant figure. It needs the band type: a component even in the mirror has its dipole change along the axis through the heavy chlorine, one odd in the mirror has it across that axis, and rotational structure tells those apart. So this is an experiment about the shape of a band envelope rather than the position of a band centre. No such assignment has been read from the literature here, and whether the components can be disentangled in a sample with four species and two borons is a matter for the spectrum.

Why the frequencies stay so close, and the small part that is left #

The exchange above is exact on the E′ coordinates. The frequencies still differ, and a difference should have a source.

It starts at the second power: the fitted slope up to one mass unit is 1.99. At a step of two mass units that gives 0.06 cm⁻¹. First-order differences cannot exist, because at first order the frequency change is entirely the exchanged term.

The second-order source is a third coordinate the argument has so far ignored. Motion along the heavy atom’s bond also belongs to the symmetric stretch, the $A_1'$ mode at 471 cm⁻¹, which the mass change now mixes into the even block but cannot mix into the odd one. That breaks the exchange, because one block has an extra mode to talk to and the other does not. It can be tested directly by changing the symmetric constant without touching anything else.

As that constant goes to zero, the difference goes to zero in proportion: 1.66×10⁻⁵ cm⁻¹ at a ten-thousandth of it, 1.67×10⁻⁴ at a thousandth, which is exactly the shape a product of two matrix elements with the constant in one of them gives. The peak where $\nu_1$ passes close to the $\nu_4$ doublet is a near-resonance. And it does not go away at the other end. Sending the constant to infinity freezes the symmetric stretch, and a frozen coordinate is a constraint on the other two, so the difference levels at 0.066 rather than falling. That is the distinction between decoupling a mode and clamping it, and the limit that would have seemed obvious was the wrong one to try first.

The Coriolis constants, again #

The essay on the Coriolis constant closed by asking whether, in a species with only a mirror, the first-order couplings between components might separate the fields by more than the frequencies. They are two new numbers, one per doublet, no longer tied to anything by a sum rule. Between the two components of $\nu_3$ the fields give 0.75138 and 0.75142, and between those of $\nu_4$ 0.75127 and 0.75117. Those agree to about a part in ten thousand, which is about what the second-order frequency difference would allow, and the sign carries no information. A Coriolis constant is built from a pair’s motion, and the exchange exchanges the pair; the sum it computes barely notices. So those constants are the same weak witness as the frequencies. The two constants in ammonia that were one number are still a useful comparison: more kinds of measurement are not more independent ones, and the question to ask of each is what symmetry the instrument can see.

The calculation and its checks #

The field is assembled here for the plane directly: boron at the origin, three chlorines at 120°, three stretches, three angles with no force on their sum, and a matrix of second derivatives built from those coordinates and diagonalised in mass-weighted Cartesians. That assembly shares no code with the library that found the partner, and it is checked against it: on boron trifluoride it reproduces the published frequencies, the published partner and the published single-substitution numbers, 471.474, 474.948, 873.554, 1442.656 and 1453.824 cm⁻¹ for one fluorine of mass 21. Species are read from each mode’s shape by applying the remaining mirror.

The band positions used are gas-phase values rounded to the wavenumber, from the standard compilations; they are observed and not harmonic, the fit does not correct them, and they should be read as a realistic setting for the argument and not as a field for BCl₃.

The claims are tested where they can fail: equal diagonals at six masses and lengths, agreement of the two fields to 10⁻⁷ cm⁻¹ at four symmetric species, an exchange of species in every doublet of every mixed species, and the slope of two. The refusal is a field off the swap: raising $x$ by 0.05 mdyn per ångström with $z$ kept must move a symmetric species’ band by more than a tenth of a wavenumber, or the agreement would be the eigenvalue problem being insensitive to everything.

Where the argument stops #

This is the harmonic answer about symmetry, and a measurement is anharmonic. The exchange of species is a statement about the quadratic field. Anharmonic coupling can mix a component with a nearby combination band, and BCl₃ has an obvious candidate: $\nu_1+2\nu_4$ falls at 957 cm⁻¹ by simple addition, one wavenumber from $\nu_3$. If such mixing is large, a real band’s character is shared and a component’s type is not simply one or the other.

The model’s field has no terms the two-by-two argument did not. A general quadratic field for a planar XY₃ has only these constants in the E′ block, so that is not an approximation. But the two fields are compared only in-plane; the out-of-plane wag has one constant and one frequency and is not involved, and nothing is claimed about its anharmonic cross terms.

A prior could always have chosen. The partner of a sensible field usually has an absurd stretching constant, and chemists have rejected it on sight without a calculation. What is done here is to show that the rejection could rest on a measurement. It should not be read as if the choice had been open in practice.

Nothing has been measured. A band type at 954 cm⁻¹ in one isotopic species of a sample of eight ought to be within reach of a jet-cooled, high-resolution spectrum, and it has not been read from any assignment for this essay.

A symmetry in a two-by-two #

The force field is not in the spectrum has been the refrain of this line of argument, and adding data has sometimes made it worse. Those were true, and they read as though the difficulty were one of data. It was also one of a matrix. Equal diagonals are a very small thing to find after so many calculations, and they are why the ambiguity was there at every mass from the start. A quantity that agrees between two fields at every mass, to every digit, is a symmetry looking for its name, and the time to ask for that name is before tracing fourteen hundred points.

Who wrote it down #

The G-matrix method is Wilson’s, set out with Decius and Cross in 1955, and the planar XY₃ elements have been in the tables since; the equal diagonal is visible there to anyone who rescales them. Fits of a two-frequency block have long been known to admit more than one root, with the choice usually made on the size of the stretching constant, or by a Coriolis constant or isotope shift that this essay and the one before it show cannot help. The correlation of E′ with $A_1$ and $B_2$ when one substitution lowers the symmetry is ordinary group theory, in Herzberg among others.

What is computed here is the exact exchange, the loop as a conic, the reversal of species in every mixed species at nearly equal frequencies, and the second-order size of what is left. Whether the exchange has been recorded in this form was not checked exhaustively.

Still open: the band type itself, and a pyramid #

The obvious open question is the spectrum: the $A_1$ and $B_2$ components have different dipole directions, and their relative intensities under the two fields are a further computable difference, needing a dipole-derivative model rather than a force field. Each field’s reversal could be turned into a predicted envelope for ³⁵Cl₂³⁷Cl at a stated rotational temperature, which is the calculation that would make the proposal into something a measurement could reject.

The nearer question is why this required planarity. In a pyramid such as ammonia the rotation pair shares the E species and the off-diagonal geometry differs, so the rescaled diagonals need not be equal, and its fit carried a wide band of fields rather than a closed pair. Whether some other exchange exists there, between other coordinates, or whether its family has no such symmetry at all is the next two-by-two to write out.