Ian Tickle wrote:
Ed, the screen z axis is not the same axis in the molecule for the first and
last
rotations, except in the special case beta = 0 or 180. The fallacy in your
argument is
that you're implicitly assuming that rotations commute, whereas of course they
don't i.e.
Rz.Ry.Rz is not the same as Rz.Rz.Ry unless Ry = unit matrix or 2-fold. The
first and
last rotations are both indeed around the screen z axis but the orientation of
the
molecule has changed because of the intervening y rotation, so the two z
rotations are not
additive unless beta = 0. Indeed if beta = 180 the net effect is the
difference of the
two z rotations. For other values of beta the net z rotation is a more
complicated
function of the Eulerian angles.
OK, I need to think about this more when I have time, but at this point
I think it is a semantic difference- For me the first and last rotation are
about the same Z axis because as you say they are both around the screen Z axis
and both operators look like cos,sin, 0, -sin, cos, 0, 0, 0, 1; i.e. rotation
about THE z axis; and it is not helpful to consider it a different z axis just
because the atoms moved.
We come up with the same conclusions with our different ways of thinking about
it:
for one, deriving the concatenated simple operators to represent a general
rotation,
and the commutativity: I would say the operators do not commute as long as the
axes
they rotate about are kept fixed, but if the axes rotate the same as the
molecule
then the z axis will always be passing through the atoms the same way.
Then rotations would commute, because the z axis would always represent the same molecular
axis. Which I am sure is NOT what you meant by saying new z axis.
Thanks,
eab
HTH!
Cheers
-- Ian
On 29 March 2014 21:22, Edward Berry ber...@upstate.edu
mailto:ber...@upstate.edu wrote:
Thanks, Ian!
I agree it may have to do with being used to computer graphics, where x,y,z
are fixed
and the coordinates rotate. But it still doesn't make sense:
If the axes rotate along with the molecule, in the catenated operators of
the polar
angles, after the first two operators the z axis would still be passing
through the
molecule in the same way it did originally, so rotation about z in the
third step
would have the same effect as rotating about z in the original orientation.
Or in eulerian angles, if the axes rotate along with the molecule at each
step, the z
axis in the third step passes through the molecule in the same way it did
in the first
step, so alpha and gamma would have the same effect and be additive. In
other words
if the axes we are rotating about rotate themselves in lock step with the
molecule, we
can never rotate about any molecular axes except those that were originally
along x,
y, and z (because they will always be alng x,y,z) (I mean using simple
rotations about
principle axes: cos sin -sin cos).
Maybe I need to think about the concept of molecular axes as opposed to lab
axes. The
lab axes are defined relative to the world and never change. The molecular
axis is
defined by how the lab axis passes through the molecule, and changes as the
molecule
rotates relative to the lab axis. But then the molecular axis seems
redundant, since
I can understand the operator fine just in terms of the rotating
coordinates and the
fixed lab axes. Except the desired rotation axis of the polar angles
would be a
molecular axis, since it is defined by a line through the atoms that we
want to rotate
about. So it rotates along with the coordinates during the first two
operations, which
align it with the old lab Z axis (which is the new molecular z axis?) . . .
You see
my confusion.
Or think about the math one step at a time, and suppose we look at the
coordinates
after each step with a graphics program keeping the x axis horizontal, y
axis
vertical, and z axis coming out of the plane. For Eulerian angles, the
first rotation
will be about Z. This will leave the z coordinate of each atom unchanged
and change
the x,y coordinates. If we give the new coordnates to the graphics
program, it will
display the atoms rotated in the plane of the screen (about the z axis
perpendicular
to the screen). The next rotation will be about y, will leave the y
coordinates
unchanged, and we see rotation about the vertical axis. Final rotation
about z is in
the plane of the screen again, although this represents rotation about a
different
axis of the molecule. My view would be to say the first and final rotation
are
rotating about the perpendicular to the screen which we have kept equal to
the z axis,
and it is the same z axis.
Ed
Ian Tickle __ 03/29/14 1:39 PM
Hi Edward
As far as Eulerian rotations go, in the 'Crowther' description the 2nd
rotation can
occur either about the new (rotated) Y
