I want to get 11-24 reciprocal space mapping, the sample surface is 11-22, so it is an asymmetric scan. After I get 2theta/omega vs rel_omega, how do I transfer it to the qz vs qx map? Normally, the XRD software can do it automatically, but for ours it seems problematic. I want to use the formula: qz= [dwcosw + (dv-dw)cos(phi-w)]. In this formula, dw-dv is the rel_omega, right? phi is known, but how do we get w and dw?
Qx=K[cos("Theta"-w)-cos("Theta"+w)]
Qz=K[sin("Theta"-w)+sin("Theta"+w)]
where K=2pi/"lambda"
Ref: M. Birkholtz, Thin FIlm Analysis by X-Ray Scattering
Hello Zhi,
Which XRD system do you use for RSM measurements (Bruker or Panalytica or Rigaku or )??
You can find below the text book snapshot of the conversion, of course, Mikhail has mentioned above the same.
Thanks to both of you.
I use Bruker. and I found someone to use the following formulas to converse ( see book snapshot formulas, first for symmetric scan, second for asymmetric scan).
it seems qx and qz here are deviation vectors with respect to the reciprocal lattice spot. so, it is different with Qx and Qz you showed above.
do you understand the differences? both of the conversion are right?
Both transformations should be right. For the last one it looks to be different scan axis and different reference coordinate system. For the Conversion that I have gave You is for conversion to the coordinate system of the lab that exclude inclination of the surface normal out from the X-ray incidence plane and where scan axis are 2theta and w (set of rocking curves at different 2theta angles).
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