The suspect that did not fit

Summing the bands established something clean and then guessed about a leftover. The clean part: the depolarisation reading, which turns out to be a direction rather than a number, stops being a direction as soon as one sums over the bands instead of taking the largest. The maximum over bands swings by ninety-eight per cent around the circle of distortions; the sum swings by four parts in ten thousand. The circle itself is the one the distortion the ratio cannot see established has a blind direction in it, and the reading being watched is the depolarisation ratio a spectrum uses to tell a symmetric band from the rest.

That is a factor of two and a half thousand, and it is the argument. A sum cannot notice the ordering of the things summed, so if each band’s departure were an isotropic quadratic form the sum would be exactly constant — and it is nearly exactly constant, which is as strong a confirmation as a numerical experiment gives.

The guess was about the “nearly”. Four parts in ten thousand is far above the numerical precision, so something is producing it, and the fourth-order term was named as the suspect with a plain statement that this was a guess and the residual was a number. It is the same care an infrared spectrum’s dipole reading needed, applied to a leftover rather than to a headline.

The suspect can be tested and it does not survive.

Why a scaling law decides it #

Write the sum at amplitude $a$ and direction $\theta$ as a power series in the distortion:

$$S(a, \theta) = a^2 C + a^3 g(\theta) + a^4 h(\theta) + \cdots$$

The leading term is isotropic — that is the quadratic-form argument, and it is why the sum is flat at all. Everything that depends on direction is in the higher terms.

What was measured is the relative spread of $S$ around the circle. Dividing through by the leading $a^2 C$:

$$\text{relative spread} \approx a\,\frac{\Delta g}{C} + a^2\,\frac{\Delta h}{C} + \cdots$$

So the candidates make different predictions about a quantity that is easy to measure. A cubic origin makes the relative residual scale as the first power of the amplitude. A quartic origin makes it scale as the second. And a residual that is really the solver’s floor scales as the zeroth power — it does not move at all.

Three hypotheses, three exponents, one sweep. That is a better position than most guesses about a leftover get to be in.

What the sweep says #

Nine amplitudes spanning a factor of sixteen, from 0.01 to 0.16 in the units used throughout. The residual falls from 3.31 × 10⁻³ to 1.05 × 10⁻⁴ — a factor of thirty-two.

That disposes of the third candidate immediately and is worth saying before the interesting part. The residual moves with the amplitude, so it is not the normal-mode solver’s precision leaking into the answer. There is a term in the expansion producing it.

The fitted exponent is 1.248.

Neither candidate is anywhere near it. The quartic predicts 2 and is off by 0.75. The cubic predicts 1 and is off by 0.25 — nearer, and still not close. To see whether “not close” is a real statement rather than an artefact of a bad fit, the thing to look at is how well any single power describes the data at all.

A single power fits to within seven per cent across the entire sixteen-fold range. That is close enough that the exponent is a real quantity — a fit through points that were not on a line would return a slope too, and it would mean nothing. Seven per cent over sixteen-fold is a genuine power law with something on top of it.

And the something on top is visible. The departures are not scattered around the fitted line; they run high at the small end, low through the middle, and high again at the large end. That shape is curvature the single power has not got, and curvature in a log–log plot is exactly what a mixture of two powers looks like.

The test that does not need a fit #

A fitted exponent between one and two is consistent with both terms being present, and consistency is not evidence. There is a direct test for the odd term, it needs no fitting at all, and it is cheap.

If the expansion contained only even powers of the amplitude — if reversing a distortion left the reading unchanged — then the cubic term would be identically zero and the leading anisotropic term would be the quartic, exactly as supposed. Reversing a direction in this plane means adding 180° to the angle, and the sweep already visits both.

Along a basis direction — one bond stretched against another — reversal changes the sum by two parts in ten million. That is nothing, and it is the control: an asymmetry that showed up everywhere equally would be a defect in how the distortion is applied rather than a term in the expansion.

Halfway between the basis directions it changes by 7.36 × 10⁻⁴. The isotropy residual at the same amplitude is 5.24 × 10⁻⁴.

Those are the same size, and only an odd power of the amplitude can produce either. So the cubic term is not merely permitted, it is present and it is the larger part of the residual. The suspect was wrong by one order.

That also explains the exponent. A residual made of a cubic term with a quartic on top scales as neither one nor two but as something between, weighted towards the cubic while the amplitude is small — which is 1.248 over this range, and which is what the two-term fit is trying to express.

Fitting both, which nearly works #

A mixture is now the model the evidence points at rather than a guess, since the reversal test has established the cubic term independently. The honest form for the relative residual is $A a + B a^2$, with both coefficients unknown.

That fit is not even nonlinear: dividing through by $a$ makes it a straight line in $a$, so ordinary least squares gives both coefficients at once. It gives $A = 1.014 \times 10^{-2}$ and $B = 6.93 \times 10^{-2}$, and the two terms would be equal at an amplitude of 0.146 — near the top of the range swept, so the cubic dominates almost everywhere that was measured.

The worst misfit falls from 7.0 per cent to 3.5. Halving the error by adding a parameter is not evidence of anything: a second free coefficient improves almost any fit, and the improvement here is about what one would expect from the extra degree of freedom alone.

What decides it is the shape of what is left, and the shape has not changed. The two-term model is still high at the small end, low through the middle and high at the large end — two sign changes across nine amplitudes, in the same places. A model with the right terms in it leaves residuals that wander; a model missing a term leaves residuals that march. These march.

So the two-term model is closer and is not complete either. The honest report has three parts: the leading anisotropic term is cubic, established directly; a quartic term is there too, which is what makes the fitted exponent exceed one; and something beyond both is still producing a systematic miss of three and a half per cent. The original suspect exists and is not the answer.

What is left standing #

It is worth being clear about how little this damages.

The finding that matters is that the sum is flat where the maximum is not, by a factor of two and a half thousand. Nothing here touches that. The residual not fully identified here is three orders of magnitude below the anisotropy the flatness is explaining away, and it is smaller still at smaller distortions. The quadratic-form argument holds exactly as well as it did.

What has changed is the status of one sentence. The residual was described as “presumably the fourth-order term”, flagged as a guess, with the number kept separate from the guess. The guess was wrong by an order — the leading term is the third — and had it been stated as a finding it would be a false explanation of a real effect, in prose, with nothing to catch it, because nothing else depends on the answer and so no later calculation would have come out wrong.

That is the case for testing guesses that nothing rests on. They are exactly the guesses that never get corrected.

What was computed, and how #

The molecule is boron trifluoride, whose spectrum counts environments rather than atoms, and the distortion is a direction in the plane of bond stretches that sum to zero — the same plane as the isotropy sweep, so that this measurement extends that one rather than a different one. At each amplitude the sweep takes seven directions covering the sixty-degree period, recomputes the normal modes at the distorted geometry with the force field held fixed, evaluates the Raman derivatives, and reads the departure of each depolarised band’s ratio from three-quarters. The sum over those bands is the quantity, and its relative spread over the seven directions is the residual.

The exponent is fitted by least squares in logarithms, over the amplitudes whose residual clears a floor of 10⁻¹². The fit refuses rather than reporting a number when fewer than four amplitudes clear it, which is the guard that would fire if the sweep were ever run at amplitudes small enough for everything to be precision.

The checks are six and two of them are refusals of named candidates rather than confirmations, which is the useful shape here. One checks that the residual moves with amplitude at all. One checks that a single power describes it well enough for its exponent to mean something — without that, the two refusals are refusals of nothing. Two state the refusals separately, because “not quartic” and “not cubic” are different claims and a single check covering both would pass if either failed. One checks that the two-term model does fit better, so the sign-pattern argument is being made about a model that was given its fair chance. And the last reads the sign pattern.

Why the reversal test is worth more than the fit #

The two measurements point the same way and they are not equally good, which is worth saying because the weaker one came first and was the one set out to make.

The scaling fit is an inference. It measures a quantity — an exponent — that no term in the expansion has, and then argues from its value that certain terms must be present in certain proportions. Everything about that argument depends on the model being a sum of two powers over the range fitted, which is an assumption the fit cannot test on itself. Its verdict is a refutation of the quartic-alone hypothesis, and refutations are what fits are good for.

The reversal test measures something a term in the expansion actually has. Odd powers change sign under reversal and even powers do not, so an asymmetry is a direct observation of odd content, with no model between the measurement and the conclusion. It also carries its own control — the basis directions, where the asymmetry vanishes to seven figures — so a systematic defect in how the distortion is applied would show up as an asymmetry that did not vanish anywhere, and it does vanish.

So the positive claim in this essay rests on the second test and the negative one rests on the first, which is the right way round. Identifying what a leftover is takes a measurement of the thing itself; ruling out what it is not can be done with a curve. And the general lesson is that a scaling exponent between two predicted values should prompt a search for a test that separates them directly, rather than an argument about which prediction it is nearer to.

Where the model stops #

The force field is fitted and held fixed at each distorted geometry, which is the harmonic approximation the ratio has been read under since it was first squared. A real distorted molecule has different force constants, and every amplitude here is the response of a fixed field to a changed geometry.

Seven directions is enough to find the spread over a sixty-degree period and is not a fine sampling of it — the period itself being a fact established separately rather than one assumed here. The residual is a difference of a maximum and a minimum over those seven, so it is a lower bound on the true spread, and a systematic dependence of that bound on amplitude is conceivable — though it would have to be a strange one, since the same seven directions are used at every amplitude.

And the amplitude range is bounded below by the solver’s precision and above by the harmonic approximation losing its meaning. Sixteen-fold is what fits between them. A wider range would separate the candidate exponents further, and there is nowhere for it to come from without changing the model.

The generalisation #

The transferable point is about what makes a scaling test decisive, and it is not the exponent.

Fitting a power to some numbers always returns an exponent. What makes the exponent evidence is that the candidates predicted different ones before the fit, and that a single power describes the data well enough for the fitted value to be a property of the data rather than of the fit. Both halves are needed. A sweep over too narrow a range gives a well-fitted exponent that cannot distinguish anything; a sweep over a wide range of badly-described data gives a number that means nothing.

The second point is about adding parameters. When a one-parameter model misses and a two-parameter model misses less, the improvement is not evidence, because it happens anyway. The evidence is in whether the structure of the miss goes away, and that is read off the signs rather than the magnitudes. Here the magnitudes halved and the signs did not move, which is the clearest possible statement that the second parameter absorbed some of the error without being the thing that was missing.

Who found it, and when #

The depolarisation ratio, its value of three-quarters for a non-totally-symmetric band, and the polarisability theory behind it are all standard and old. The sum over bands as a test of isotropy is a simple construction, and what is added here is the scaling test and the two refusals.

The habit worth naming is the one that made this test possible: write down a number that can be measured and a guess that cannot yet be tested, keep them separate in the prose, and say which is which. A guess kept visibly apart from its number can be checked; a guess folded into the finding cannot.

Still open: separating the halves, and the weights #

The obvious open question is the term beyond the second, and the reversal test says how to isolate it. Reversal separates the expansion cleanly: the average of the forward and reversed readings keeps only the even powers, and half their difference keeps only the odd ones. Running the amplitude sweep on each half separately would give two exponents instead of one mixed one, and each should be an integer — the odd half scaling as $a$ if the cubic leads it, the even half as $a^2$ if the quartic leads that. Any half that still refuses an integer is where the remaining structure lives, and that is a much narrower place to look than the whole residual.

The nearer algebraic question is why the reading is not a polynomial at all. It is $|0.75 - \rho|$, where $\rho$ is a ratio of two quadratic invariants of the Raman tensor — a ratio, so its expansion carries the denominator’s series, and an absolute value, which is not analytic anywhere its argument changes sign. Either could produce a residual no finite polynomial describes, and the sign pattern that would not go away here is the kind of thing they produce.

The other near question is the weights, which have not yet been tried. What an unresolved measurement reports is an intensity-weighted combination of the bands rather than their unweighted sum, and the intensities are computed at every direction already. Whether the weighted combination is flat too decides whether the flatness of the sum is about an observable or about a convenient stand-in for one — and unlike the residual, that question has something resting on it.