[Pw_forum] help on how to compute celldm(4), celldm(5) and celldm(6)

Ridwan Agbaoye agbaoyeridwan at yahoo.com
Tue Feb 23 13:07:52 CET 2016


Thank you very much sir, now I have gotten a cl ear picture of how to calculate celldm(4)-celldm(6)
Now my question is that  in the case where celldm(4)=0.00 for monoclinic structure can we say the crystal structure is similar or same as that of orthorhombic sinc
 
 
  On Mon, Feb 22, 2016 at 9:54, Giovanni Cantele<giovanni.cantele at spin.cnr.it> wrote:   
> On 20 Feb 2016, at 20:34, Ridwan Agbaoye <agbaoyeridwan at yahoo.com> wrote:
> 
> Please I need clarification on how to compute celldm(4), celldm(5) and celldm(6) for ibrav 12, 13 and 14 
> 
> I have conflicting opinion on how it should be calculated though it is stated as for ibrav =12 and 13 celldm(4)=cos(ab),for ibrav 14 celldm(4)=cos(bc), celldm(5)=cos(ac)and celldm(6)=(ab).  Although cos(ab) is the cosine of angle between lattice parameters a and b as stated in INPUT_PW.HTML of quantum espresso
> 
> Please I need clarifications
> 
> Agbaoye Ridwan Olamide
> Federal University of Agriculture, Abeokuta
> M.Sc. (in view)
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Example for ibrav=12

cosAB = cosine of the angle between axis a and b (gamma)

So if you know the lattice vector coordinates, that is, a=(ax,ay,az) and b=(bx,by,bz), just calculate 
cosAB = a . b / ( |a| |b| )
where a . b = dot product of a and b = ax bx + ay by + az bz
|a| = length of vector a = sqrt( ax^2 + ay^2 + az^2 ) and same for |b|.

Instead, if you know, directly gamma, just compute cos( gamma ).

Similar consideration apply to cos(bc) and cos(ac) for the other lattices.

Giovanni

-- 

Giovanni Cantele, PhD
CNR-SPIN
c/o Dipartimento di Fisica
Universita' di Napoli "Federico II"
Complesso Universitario M. S. Angelo - Ed. 6
Via Cintia, I-80126, Napoli, Italy
e-mail: giovanni.cantele at spin.cnr.it
Phone: +39 081 676910
Skype contact: giocan74

ResearcherID: http://www.researcherid.com/rid/A-1951-2009
Web page: http://people.na.infn.it/~cantele



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