已知晶格常数

$a=|vec a|, b=|vec b|, c=|vec c|$
$alpha=vec b wedge vec c=arccos{vec b cdotvec c over bc}, beta=vec a wedge vec c=arccos{vec a cdot vec c over ac}, gamma=vec a wedge vec b=arccos{vec a cdot vec b over ab}$

取x轴沿$vec a$方向, y轴处于$vec a vec b$平面内, z轴处于$vec a times vec b$方向, 则晶格矢量

$vec a= (a_x,a_y, a_z)=a(1, 0, 0) $
$vec b= (b_x, b_y, b_z)=b (cosgamma, singamma, 0) $
$vec c= (c_x, c_y, c_z)=c (cosbeta, {cosalpha-cosbetacosgamma oversingamma},{sqrt{1+2cosalphacosbetacosgamma-cos^2alpha-cos^2beta-cos^2gamma}over singamma}) $

设空间某点分数坐标为$(u,v, w)$, 直角坐标为$(x, y, z)$, 则

$begin{Bmatrix} x \ y \z end{Bmatrix} = mathbf A begin{Bmatrix} u \ v \wend{Bmatrix}=begin{pmatrix} a_x & b_x & c_x \ a_y & b_y &c_y \ a_z & b_z & c_z end{pmatrix} begin{Bmatrix} u \ v \wend{Bmatrix} $
$ begin{Bmatrix} u \ v \w end{Bmatrix} = mathbf A^{-1} begin{Bmatrix} x\ y \x end{Bmatrix} $

证明

由C点向面OAB做垂线, 交点为D, 连接OD,OD将$gamma$为$gamma_1,gamma_2$. 作DE垂直于OA,DF垂直于OB, 连接CE,CF, 则CE$perp$OA, CF$perp$OB.

设OD长度为$rho$,则

$rho cos gamma_1=c cos beta =c_x$
$rho cos gamma_2=c cos alpha$

令$k={c overrho}$, 有

$cosgamma=cos(gamma_1+gamma_2)=cos gamma_1 cosgamma_2-singamma_1singamma_2=k^2 cosalpha cosbeta -sqrt{1-k^2 cos^2alpha} sqrt{1-k^2cos^2beta} $

可解得

$k^2=c^2/rho^2={cos^2alpha + cos^2beta-2cosalpha cosbetacosgamma over sin^2gamma} $

由$ sin^2gamma_1=1-cos^2gamma_1=1-k^2 cos^2beta$

可得$ c_y^2 =rho^2 sin^2gamma_1 =c^2(k^2-cos^2beta)={(cosalpha-cosbetacosgamma)^2 over sin^2gamma }$

若取$alpha lt beta$, 则开方取正号, 故

$c_y=c{ cosalpha-cosbetacosgamma over singamma }$

而

$begin{split} c_z^2 &=c^2-c_x^2-c_y^2=c^2[cos^2beta-(k^2-cos^2beta)] \&=c^2(1-k^2)={sin^2gamma-cos^2alpha-cos^2beta+2cosalpha cosbetacosgamma over singamma} \&=c^2{1-cos^2gamma-cos^2alpha-cos^2beta+2cosalpha cosbetacosgamma over sin^2gamma} end{split}$

由上面所得公式, 我们亦可得到平行六面体的体积

$begin{split}V&=abc_zsingamma \&=abcsqrt{1-cos^2gamma-cos^2alpha-cos^2beta+2cosalpha cosbetacosgamma} \ &=begin{vmatrix} vec a cdot vec a & vec a cdotvec b & vec a cdot vec c \ vec b cdot vec a & vec b cdot vec b & vec b cdot vec c \ vec c cdot vec a & vec c cdot vec b& vec c cdot vec c end{vmatrix}^{1/2} \ &=begin{vmatrix} a^2& abcosgamma & accosbeta \ abcosgamma & b^2 &bccosalpha \ accosbeta & bccosalpha & c^2 end{vmatrix}^{1/2}end{split} $

此式多有论述, 不赘述.