[Pw_forum] monoclinic cell definition question

[Jonas Baltrusaitis]

Wow, talk about comprehensive

Let me summarize it (mainly so I am sure I understand correctly)

say my cif file says"

 a=2; b=3; c=4; alpha=90; beta=123; gamma=90

So If I move it into  pwscf, they would become:

a=4; b=2; c=3, alpha=90; beta=90; gamma=123

ADDIONALLY
(cif) to (pwscf)
x=y
y=z
z=a

the latter is the part I didn't quite get: do we move coords as well?

thanks for your help

Jonas
--- On Wed, 5/13/09, Michael Mehl <rcjhawk at gmail.com> wrote:

> From: Michael Mehl <rcjhawk at gmail.com>
> Subject: Re: [Pw_forum] monoclinic cell definition question
> To: jasius_1 at yahoo.com, "PWSCF Forum" <pw_forum at pwscf.org>
> Cc: "Lorenzo Paulatto" <paulatto at sissa.it>
> Date: Wednesday, May 13, 2009, 6:32 PM
> 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.
> 
>