The six motions that are not modes

Every account of vibrational spectroscopy begins by subtracting six. A molecule of $N$ atoms has $3N$ Cartesian coordinates; three of them describe where the molecule is and three describe how it is turned; so $3N - 6$ describe its shape and there are that many vibrations.

The argument is correct and it is a counting argument. It says nothing about which six, and it offers no way to check that a particular calculation has separated them properly — which matters, because the separation is where a vibrational calculation most often goes wrong without saying so.

The six are computable. They are eigenvectors of the mass-weighted Hessian with eigenvalue zero, and the space they span can be written down independently, from the geometry and the masses and nothing else.

Writing the six down #

In mass-weighted coordinates, where each Cartesian displacement is multiplied by the square root of its atom’s mass, the six rigid motions are elementary.

The three translations are every atom moving together: the vector with $\sqrt{m_i}$ in each atom’s $x$ slot and nothing elsewhere, and the same for $y$ and $z$.

The three rotations are infinitesimal turns about the centre of mass: the vector whose entry for atom $i$ is $\sqrt{m_i}\,(\hat{\omega}\times\mathbf{r}_i)$, with $\mathbf{r}_i$ measured from the centre of mass and $\hat{\omega}$ the axis.

Not one of those is an eigenvector of anything, and none of them was obtained from a Hessian. They are six vectors written from the atom positions and the masses, and the claim to be tested is that the six zero eigenvalues of the Hessian span exactly the space they span.

Getting the centre of mass right is the part that has to be done and is easy to skip. A rotation is about a point, and a rotation taken about the wrong point is a rotation with a translation mixed into it. The mixture still lies inside the six-dimensional space, so a wrong centre would not break the projection test — which is a reason to note it rather than a reason to relax.

The test, and why it has two halves #

Two projections, and both are required.

Every zero eigenvector must lie entirely inside the rigid space. Orthonormalise the six written-down vectors, project each of the Hessian’s zero eigenvectors onto them, and the length of the projection must be one.

Every vibration must lie entirely outside it. Project each vibrational eigenvector onto the same space and the length must be zero.

Checking only the first would pass on a Hessian whose vibrations were also rigid, which is impossible — and is exactly the kind of impossibility a numerical error produces. Across six molecules with fitted force fields, the largest leak of any vibration into the rigid space is $2.4\times10^{-8}$, and the largest zero eigenvalue is $1.0\times10^{-7}$ of the largest eigenvalue in the same matrix.

The rank is checked as well, and it is where the linear case appears as a number.

Why a linear molecule has five #

The usual statement is that a linear molecule has only two rotational degrees of freedom because rotation about the molecular axis “does nothing”, and the word doing the work there is vague.

In the construction above it is not vague at all. The rotation vector about an axis $\hat{\omega}$ has entry $\hat{\omega}\times\mathbf{r}_i$ at each atom. For a linear molecule with $\hat{\omega}$ along the molecular axis, every atom lies on that axis, so $\mathbf{r}_i$ is parallel to $\hat{\omega}$ and the cross product is exactly the zero vector — at every atom, not approximately.

So the sixth vector is not a small vector or a nearly-dependent one. It is identically zero, and the six written-down vectors span five dimensions rather than six. Carbon dioxide’s rigid space has rank five, its Hessian has five zero eigenvalues, and it has four vibrations rather than three.

The check covers that too: for a linear molecule, exactly one of the three rotation vectors must have zero norm. One, not two, and not none.

This is the cleanest place on the site to see why a linear molecule is a special case in every vibrational count. It is not that the rotation is unobservable or that its energy is small. The motion does not exist.

What the check is really checking #

Honesty about this is important because the test is stronger than it first looks and weaker than it might be taken for.

This site builds its Hessians from internal coordinates — bond stretches and angle bends — transformed into Cartesian coordinates by the usual $\mathbf{B}$ matrix. An internal coordinate does not change when the molecule is translated or rotated, so the resulting Cartesian Hessian annihilates rigid motions by construction. The six zeroes are not a discovery about nature; they are a property of how the matrix was built.

That does not make the test empty, and here is what it does catch.

A wrong transformation. If the $\mathbf{B}$ matrix has an error — a sign, an index, a missing term — the resulting Hessian generally does not annihilate rigid motions, and the zero eigenvalues come out non-zero. It would take an unusual error to leave them at zero while getting the vibrations wrong.

A zero eigenvector that is not rigid. A Hessian can have an accidental zero eigenvalue that is a genuine soft mode rather than a rigid motion, and the projection separates the two. That is precisely the diagnostic used in real calculations to distinguish a stationary point from a saddle.

A wrong mass assignment. The mass weighting enters both the Hessian and the written-down vectors, and it enters differently — as $1/\sqrt{m_i m_j}$ in one and $\sqrt{m_i}$ in the other. A mass attached to the wrong atom breaks the agreement.

So the test is a check on the arithmetic rather than on the physics, and most checks of a Hessian are of that kind. The physics enters where the frequencies are compared with measurement, which the force field is not in the spectrum treats at length.

What each mode is made of in internal coordinates is a separate reading of the same eigenvectors, and the six rigid motions have no internal-coordinate content at all: they change no bond length and no angle, which is the same statement as their having zero frequency.

The numbers, molecule by molecule #

The check runs on six molecules with fitted force fields, and the numbers are worth setting out because their uniformity is the result.

Water, sulfur dioxide, ammonia, methane and boron trifluoride are non-linear and their rigid spaces have rank six. Carbon dioxide is linear and its rank is five. Every rank was computed by orthogonalising the written-down vectors and counting how many survived, not by asking whether the molecule is linear and looking the answer up.

The largest zero eigenvalue, relative to the largest eigenvalue in the same matrix, runs from $1.6\times10^{-9}$ for methane to $1.0\times10^{-7}$ for water. Those are ordinary rounding for a matrix of this size, and the spread between molecules reflects how large the force constants are rather than anything about the separation.

The largest leak of a vibration into the rigid space runs from $1.3\times10^{-10}$ for carbon dioxide to $2.4\times10^{-8}$ for ammonia. Every one is at least a factor of forty below the tolerance applied, and the tolerance was chosen from the arithmetic rather than fitted to the results.

Methane is the largest case, with fifteen coordinates, six rigid vectors and nine vibrations, and it is the one where the counting would be hardest to do by inspection. Its nine vibrations reduce to $\mathrm{A}_1 + \mathrm{E} + 2\mathrm{T}_2$ — four distinct frequencies from nine modes — and the six that were subtracted are accounted for individually.

Where the rotations go in a real spectrum #

The rotations come out at zero here because a Hessian describes how the energy changes when the shape changes, and turning a molecule does not change its shape. That is the harmonic approximation’s statement about them and it is exact within it.

A real molecule’s rotations are not at zero energy: they form a ladder of levels with spacing set by the moments of inertia, which is a separate calculation entirely. The rotational spectrum is a moment of inertia builds those levels from the same atom positions and the same masses used here — the moment of inertia tensor is, up to a factor, the same object as the metric of the three rotation vectors written above.

That is a connection worth drawing explicitly. The three rotation vectors have norms proportional to the square roots of the three principal moments of inertia, so the vector that vanishes for a linear molecule vanishes because one principal moment is zero. The same fact appears as a zero norm here and as a missing rotational constant there, and it is one fact.

The vibration–rotation coupling that a real spectrum shows — the fine structure on every band, the inertial defect, the centrifugal distortion — is where the separation between the two breaks down, and it is a correction not computed here. What it does compute is the two halves separately, from one geometry, with the arithmetic that links them stated.

The three rotations are rotations about the principal axes, and the moments about those axes are computed from the coordinates and the masses. That is where the rotational half of the six comes from, and it is the half that becomes five when the molecule is linear.

What a projection is doing, and why it is the right test #

Projecting one vector onto a space is elementary and it is worth saying why it is the right question rather than some other one.

The natural alternative is to compare eigenvectors directly: take the Hessian’s zero eigenvectors and the six written-down ones and check that they match up. That does not work, and the reason is instructive. The zero eigenvalues are degenerate, six-fold or five-fold, and a degenerate eigenvalue has no preferred eigenvectors at all — any orthonormal set spanning the same space is as good as any other, and a diagonaliser returns whichever set its arithmetic happened to produce.

So the individual zero eigenvectors are not comparable with anything. They are arbitrary combinations of translations and rotations, differing between runs and between machines.

What is not arbitrary is the space they span, and that is what the projection measures. A vector’s projection onto a space is independent of which basis the space is written in, so the test asks the one question that has a determinate answer.

This is the same distinction degeneracy is a group theorem draws for electronic states: within a degenerate set the individual functions are conventions and the set is the object. Here the set is six-dimensional and its members are whatever came out of a Jacobi sweep, and every meaningful statement about them is a statement about the space.

The imaginary case, which is the useful one #

The reason chemists look at the low eigenvalues of a Hessian at all is a case not computed here, and it should be named.

At a geometry that is a genuine minimum, every non-rigid eigenvalue is positive and every frequency is real. At a transition state exactly one is negative, and its square root is imaginary — the mode along which the structure falls apart. At a geometry that is neither, several are negative and the structure is not a stationary point at all.

Counting the negative eigenvalues is therefore how a computed structure is classified, and it only works if the six rigid ones have been correctly identified and set aside. A calculation that mistook a rigid motion for a low vibration would report a spurious near-zero frequency and a chemist would spend an afternoon on it.

That entire practice rests on the separation this essay computes. Transition states are not computed here — they belong to reactions rather than to stable structures — but the diagnostic is the same object, and it is worth knowing that the six zeroes are what makes it possible.

Drawn as a spectrum, water’s three bands sit where the fitted field puts them and the six rigid motions appear nowhere — not as weak features, not at zero, not at all. A mode with no restoring force has no frequency to be drawn at, which is the sense in which they are not modes.

Six is a maximum rather than a rule #

One edge case makes the counting argument look less inevitable than it usually does, and it is worth a paragraph.

A single atom has three coordinates and three translations and no rotations at all. A rotation vector needs $\mathbf{r}_i$ measured from the centre of mass to be non-zero somewhere, and for one atom it is zero everywhere. So the rigid space has rank three and there are no vibrations, which is right and which the $3N-6$ formula would report as $-3$.

A diatomic has six coordinates, five rigid motions and one vibration. The formula for a linear molecule gives $3(2) - 5 = 1$, correctly, and the reason is again that the rotation about the internuclear axis is the zero vector.

So the rank of the rigid space is at most six and is smaller when the geometry degenerates, and the vibration count is whatever is left. Writing it as a formula with two cases hides that there is one construction with a rank, and the rank is what a computation returns.

That reading also says where the next case would come from. A structure with all its atoms at one point would have rank three; nothing else in molecular geometry degenerates further, which is why two cases have always been enough.

The six are a convergence check as well as a count #

Six eigenvalues at arithmetic noise is a result about a correctly constructed calculation, and the same six are used in practice as an instrument — because when they are not at noise, their size measures something specific.

The six vanish only at a stationary point of the energy. Translating or rotating a molecule costs nothing wherever it sits, so those six directions have zero curvature everywhere; but the Hessian’s block structure only separates them cleanly when the gradient is zero, and at a geometry that has not been fully optimised the residual gradient leaks into them.

So a frequency calculation run on an incompletely optimised structure returns translations and rotations at frequencies that are small but not negligible — tens of wavenumbers rather than fractions of one — and the size of those spurious frequencies is a direct measure of how far the geometry is from the minimum.

That makes them the cheapest diagnostic in the whole calculation, and every program prints them for that reason. Six frequencies near zero says the geometry converged; six frequencies at forty wavenumbers says it did not, and the second is a common enough mistake that reading the six before reading anything else is standard practice.

It is also a check that fails informatively rather than silently. A badly converged geometry gives real vibrational frequencies that look entirely plausible — the calculation produces a spectrum, the bands are in sensible places, and nothing announces the problem except the six numbers a reader was going to skip.

Two more molecules make the count’s independence from the structure plain, and the second is the one where a substitution changes everything except the six.

What the six zeroes add #

Four familiar results about normal modes are that a mode is not a bond stretch, that an isotope shift is arithmetic, that a force field is not recoverable from a spectrum, and that a frequency is not a bond strength. Every one of those is about the $3N-6$ that are there.

This essay is about the six that are not, and its content is that they are computable rather than assumed. Six vectors written from the geometry, six eigenvalues at arithmetic noise, and a projection that accounts for each in terms of the other to eight decimal places — with the linear case falling out as a vector that is identically zero rather than as an exception to be remembered.

The value of that is mostly diagnostic and the diagnostic is the point. A count that is quoted cannot fail; a count that is computed can, and the projection above is what turns $3N-6$ from a piece of arithmetic into a statement about a particular matrix that a particular error would break.

The next question would need a geometry not available here. Every Hessian here is built at a structure that is stationary by construction, so the interesting cases — a saddle with one negative eigenvalue, a structure that has not been optimised, a molecule with a genuinely soft coordinate whose frequency is comparable with the numerical noise — never arise. Those are where the separation stops being clean and where the projection stops returning ones and zeroes, and they are also where the diagnostic earns its keep in practice. Reaching them means computing an energy surface rather than fitting a force field to a spectrum, which is a different subject and a much larger calculation.