<html><body><div style="color:#000; background-color:#fff; font-family:times new roman, new york, times, serif;font-size:14pt"><div>Dear Prof. <span class="">Stefaan Cottenier<br>I want to make a struct file for a rhombohedral compound but with hexagonal symmetry. How can i do that?<br>the positions in rhomboherdral </span>are (0 0 0) (2/9 2/9 2/9) (7/9 7/9 7/9) with 166-R3m<br>the <span id="misspell-5" class=""><span>equivalent</span></span> positions by <span id="misspell-5" class="">using rhom</span>2hex are (0 0 0) (0 0 0.22) (0 0 0.77) with H space group, with this positions WIEN2K suggest 183 (P 6 m m) space group which seems it is not correct because i think in regard to <span id="misspell-7" class=""><span>crystallography</span></span>, the number of atom during the conversion from rhombohedral to hexagonal may change.<br></div> <div><span><br></span></div><div style="display: block;" class="yahoo_quoted"> <br> <br> <div style="font-family:
times new roman, new york, times, serif; font-size: 14pt;"> <div style="font-family: times new roman, new york, times, serif; font-size: 12pt;"> <div dir="ltr"> <font face="Arial" size="2"> On Thursday, 6 February 2014, 9:50, negin kamali <negin_kamali6184@yahoo.com> wrote:<br> </font> </div> <div class="y_msg_container"><div id="yiv2204994267"><div><div style="color:#000;background-color:#fff;font-family:times new roman, new york, times, serif;font-size:14pt;"><div><span id="yiv2204994267misspell-0" class="yiv2204994267">Dear Wien</span>2K Users,</div><div>I want to run the <span id="yiv2204994267misspell-1" class="yiv2204994267mark">sm</span>-<span id="yiv2204994267misspell-2" class="yiv2204994267">element</span> in hexagonal symmetry. Sm-element is a <span id="yiv2204994267misspell-2" class="yiv2204994267mark">rhombohedral</span> structure with 166(R-3m space group) and a=8.966A and <span id="yiv2204994267misspell-3"
class="yiv2204994267">alpha</span>=23.13 and the atomic position are (0 0 0) (2/9 2/9 2/9) (7/9 7/9 7/9).</div><div>I know, In WIEN2K, when we want to run <span id="yiv2204994267misspell-5" class="yiv2204994267mark">rhombohedral</span> structure, we must to change the <span id="yiv2204994267misspell-6" class="yiv2204994267">lattice</span> constant to hexagonal which is in <span id="yiv2204994267misspell-7" class="yiv2204994267mark">sm</span> case a=b=3.621 and c=26.25 A (<span id="yiv2204994267misspell-8" class="yiv2204994267">aphlfa</span>=beta=90 and gamma=120), but i could not determine the
positions in hexagonal axes.As you know in crystallography the R-3m space group has two <span id="yiv2204994267misspell-9" class="yiv2204994267">sub-state</span> first r-axes an second h-axes.</div><div>So, i want to <span id="yiv2204994267misspell-10" class="yiv2204994267">know</span>;<br></div><div> First: how the number of atoms are changed in this conversion (from <span id="yiv2204994267misspell-11" class="yiv2204994267mark">rhombohedral</span> to hexagonal symmetry)? (it means, is it <span id="yiv2204994267misspell-12" class="yiv2204994267"><span>necessary</span></span> to change number of atom when i want to convert a <span id="yiv2204994267misspell-12" class="yiv2204994267mark">rhombohedral</span> symmetry to hexagonal) Second: What are the atomic positions in h-axes?<br></div><div><br></div><div><br></div></div></div></div><br><br></div> </div> </div> </div> </div></body></html>