Subject: RE:CCL potential surface

From: Marvin Waldman <[email-masked]>
Date: Fri, 19 Mar 1999 08:10:50 -0800

On 6-Mar-99, Konrad Hinsen wrote:

> > I would like to understand why when working in cartesian coordinates
> > and diagonalizing the hessian matrix on a potential surface at a point
> > which is not a stationnary point I find only 3 zero eigenvalues (what 
> > about the rotations).
>
> Not being at a stationary point means that there is a non-zero energy
> gradient. So when you rotate the whole system, the energy gradient
> follows the rotation as well. The difference between the rotated and
> the original gradient is thus not zero, meaning that this motion does
> not correspond to a zero eigenvalue.
> -------------------------------------------------------------------------
> Konrad Hinsen                            | E-Mail: [email-masked]
> Centre de Biophysique Moleculaire (CNRS) | Tel.: +33-2.38.25.55.69
> Rue Charles Sadron                       | Fax:  +33-2.38.63.15.17
> 45071 Orleans Cedex 2                    | Deutsch/Esperanto/English/
> France                                   | Nederlands/Francais
> -------------------------------------------------------------------------

Sorry for this delayed response to the original question.  I tried sending
this message last week, but it didn't get delivered due to some e-mail
problems.  Here is my analysis of what is happening.  The hessian matrix
represents the mass-normalized energy second derivatives of the molecule
with respect to atomic Cartesian coordinates.  The eigenvectors formed by
the diagonalization of this matrix represent atomic motions (the normal
modes) with respect to mass-normalized atomic Cartesian displacements.
Therefore, the modes corresponding to rotation are represented as
displacements in Cartesian coordinates.  Now, for an infinitesimal motion
of the atoms, this is OK.  However, if one were to take a rotational mode
expressed this way and move the atoms a finite amount along these Cartesian
displacements, one would find that more than a rotation has taken place and
the motions in the Cartesian directions are starting to cause bond
stretching, angle bending, etc.  This is because pure rotations need to be
expressed in curvilinear coordinates and expressing them in Cartesian
coordinates is only correct for infinitesimal displacements.  Now, at a
stationary point on the potential energy surface, it works out that the
eigenvalues for rotational modes are zero, but at non-stationary points
they are not zero because of this effect.

Here is a simple example to illustrate why this happens.  Consider a
diatomic molecule.  Place one atom at the origin and the other atom along
the X-axis.  Make the atom at the origin have extremely large (virtually
infinite) mass while the atom on the X-axis is very light.  Then,
rotational motion for this system corresponds to the light atom moving
(infinitesimally) in the Y-direction.  How do the energy derivatives of the
system behave for this type of motion.  Let r represent the bond length of
the molecule.  For any placement of the light atom in the X-Y plane, we have:

                    r = sqrt( x*x + y*y)

Let E be the energy of the system which is strictly a function of r, the
bond length.  Then, the derivative of the energy with respect to motion
along y (starting out with the light atom on the X-axis) is:

            dE/dy  = (dr/dy) * (dE(r)/dr)  = (y/r) * (dE/dr)

          d^2E/dy^2 = (d^2r/dy^2) * (dE/dr) + (dr/dy)^2 * (d^2E/dr^2)

                   =  (d^2r/dy^2) * (dE/dr) (since dr/dy = y/r and y = 0)

Now, at equilibrium or a stationary point, we have dE/dr = 0, so this last
term also drops out and we get a second derivative d^2E/dy^2 = 0.  However,
away from equilibrium this term is not zero, and d^2r/dy^2 is not zero, so
we get a non-zero value for d^2E/dy^2, which is the second derivative of
the energy with respect to rotational motion expressed in Cartesian
coordinates.  A similar, but more general, form of the argument/analysis
applies to the case of polyatomic molecules.  The jist of the argument is
that rotational motion expressed in Cartesian coordinates causes internal
coordinate distortions in second order (d^2r/dy^2) which multiplies
the first order energy derivative in internal coordinates (dE/dr).  At a
stationary point, the energy derivative vanishes, and we get a zero second
derivative of the energy with respect to rotation expressed in Cartesian
coords (and a zero eigenvalue in the corresponding eigenvector of the
hessian matrix).  Away from equilibrium, the energy first derivative with
respect to internal coords is non-zero, leading to a non-zero value for the
energy second derivative with respect to rotational motion, when the
rotation is represented in Cartesian coords.  Hope this helps clarify the
situation.

Marv Waldman