I believe:
delta_n=n_zz-n_xx
However, you may check it for yourself:
Birefringence := delta_n=n_e-n_o [1,2]
Orthorhombic ɛ_xx and ɛ_yy perpendicular and ɛ_zz parallel [3] a bit
similar to tetragonal that has ɛ_xx = ɛ_yy and ɛ_zz tensor (Table 1.1) [4]
n_x = sqrt(ɛ_xx/ɛ_o) and n_z = sqrt(ɛ_zz/ɛ_o) [5,6]
Section 1.4 of [4]:
n_e = n_z (or n_zz) and n_o = n_x (or n_xx)
[1] https://en.wikipedia.org/wiki/Birefringence#Uniaxial_materials
[2]
http://shodhganga.inflibnet.ac.in/bitstream/10603/4722/13/13_chapter%203.pdf
(Equation 3.24)
[3] https://pubs.acs.org/doi/abs/10.1021/jp410786w
[4]
http://www.physics.ucc.ie/fpetersweb/FrankWeb/courses/PY4118/Notes/Symmetry%20of%20Tensor%202.pdf
[5] https://www.as-photonics.com/book_page/book-preview.pdf (Equations
2.13 and 2.15)
[6]
https://nanoed.tul.cz/pluginfile.php/1598/mod_resource/content/1/Optick%C3%A9%20vlastnosti%20krystal%C5%AF.pdf
(slide 5)
On 12/4/2018 5:54 PM, Wien2k User wrote:
Dear Wien2k users;
for a tetragonal material, the birefringence is given by :
delta_n=n_xx-n_zz
or
delta_n=n_zz-n_xx
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