[Pw_forum] monoclinic cell definition question

[Michael Mehl]

jasius_1 at yahoo.com wrote:
> thanks. the problem is that the new gamma between new a and b is still 90o!

Sigh.  This is what comes of trying to answer a question near my
bedtime.  I'm afraid that I've totally fraked the explanation.  Let's
try again:

The page http://cst-www.nrl.navy.mil/lattice/struk/g0_6.html shows a
lattice with unique axis "b", not "c", as is obvious from the "beta"
angle.  Last night I was thinking it was a "c" lattice, but like I
said, it was near my bedtime.

So the problem is: how to convert a monoclinic lattice from a unique
axis b representation to a unique axis c representation, and visa
versa, while preserving the shape of the unit cell.

Apologies to the list, this is going to take a bit of time:

By convention, when we say "a monoclinic lattice with unique axis b"
we mean the second vector of the set of three primitive vectors is
perpendicular to the other two, and those other two are not
perpendicular, but are pointed at some non-90 degree angle relative to
one another.

So let's look at a set of lattice vectors for this system.  One choice
(and there are an infinite number) would be:

a1 = [    s     , 0 ,    0     ]
a2 = [    0     , t ,    0     ]
a3 = [ u cos(v) , 0 , u sin(v) ]

where v <> 90 degrees and I deliberately didn't use
a,b,c,alpha,beta,gamma.  Note that this is the choice made on the web
page in question.

Now in the traditional expression of lattice vectors:

a = Sqrt[a1.a1] = s
b = Sqrt[a2.a2] = t
c = Sqrt[a3.a3] = u

and the angles are defined by

cos(alpha) = a2.a3/(|a2| |a3|) = 0
cos(beta)  = a3.a1/(|a3| }a1|) = cos(v) => beta = v
cos(gamma) = a1.a2/(|a1| |a2|) = 0

That's the unique axis b monoclinic lattice: vector b (a2) is
perpendicular to vectors a (a1) and c (a3).

Now let's take the same system and express it as a monoclinic system
with unique axis c.  That is, we want vector a3 to be perpendicular to
a1 and a2, and a1 and a2 to have the same non-right angle between them
as vectors a1 and a3 did in the old system, while preserving the shape
of the primitive cell.  One more or less obvious
choice, which preserves the right-hand nature of the system, is to just 
permute (not rotate, duh) the vectors, and list them as:

a1' = [ u cos(v) , 0 , u sin(v) ]
a2' = [    s     , 0 ,    0     ]
a3' = [    0     , t ,    0     ]

And now:

a = Sqrt[a1'.a1'] = u
b = Sqrt[a2'.a2'] = s
c = Sqrt[a3'.a3'] = t

and

cos(alpha) = a2'.a3'/(|a2'| |a3'|) = 0
cos(beta)  = a3'.a1'/(|a3'| }a1'|) = 0
cos(gamma) = a1'.a2'/(|a1'| |a2'|) = cos(v) => gamma = v

However (and this is where the "rotate" comment came in), my personal
quirk is that (a1,a2,a3) should be as close as possible to (X,Y,Z).
So NOW lets rotate that system about the (111) axis, taking
Z->X->Y->Z.  We get

a1" = [ u sin(v) , u cos(v) , 0 ]
a2" = [    0     ,    s     , 0 ]
a3" = [    0     ,    0     , t ]

where we leave it as an exercise for the reader to show that this
is the same combination of (a,b,c,alpha,beta,gamma) as the primed
vectors above.

What about the basis of the lattice?  Suppose we have a basis vector 
that was written in original crystal (aka lattice) coordinates as

B1 = p a1 + q a2 + r a3

In Cartesian space, this becomes

B1 = [p s + r u cos(v), q t , r sin(v) ]

and a little bit of thought shows that this can be written in terms of
the primed lattice as:

B1 = r a1' + p a2' + q a3'

which is the permutation that someone mentioned.

If you then want to go to my rotated space, you get

B1 = r a1" + p a2" + q a3"
    = [r u sin(v) , r u cos(v) + p s, q t]

I hope this clears things up, but once again it's getting near my
bedtime, so it's possible I've made another mistake.

This is a lot easier to see if you draw some pictures of the
individual sets of primitive vectors.  You'll see that all the sets
(a1,a2,a3), (a1',a2',a3') and (a1",a2",a3") define the same lattice,
and the the corresponding B1 vector is at the same place in the
primitive cell.

-- 
Mike Mehl
Naval Research Laboratory, Washington DC
(At least until someone reads this post)
Disclaimer:  If you've been following this discussion, you should know
to take everything I've said with a grain of salt.