参考书

Cai, W., & Nix, W. (2016). Imperfections in Crystalline Solids (MRS-Cambridge Materials Fundamentals). Cambridge: Cambridge University Press. doi:10.1017/CBO9781316389508

macroscopic degree of freedom=5

宏观上晶界的自由度总共有5个.其中3个来自interface operation,可以用orentational matrix或者rotation axis+angle表示.剩下两个来自于grain boundary plane normal

interface operation-orientational matrix

例:

立方晶体1是立方结构的Li,假设其晶格常数a=b=c=1

令参考系的xyz轴与晶体1的abc轴分别对齐

则\(a_1=(1,0,0),b_1=(0,1,0),c_1=(0,0,1)\)

让晶体1图中红点为旋转点,相对自己的001轴顺时针旋转53.13度,产生了晶体2.

图从001角度观察晶体.红点左侧区域(包括红点)为晶体1,红点右侧为晶体2,需要自行添加周期性条件脑补

$Image

通过图中可以看出,晶体2相当于a轴和b轴顺时针旋转了53.13度,而c轴不变.旋转之后晶体2的abc轴分别是

\[a_2=(cos(-53.13),sin(-53.13),0)=(0.6,-0.8,0)\] \[b_2=(sin(53.13),cos(53.13),0)=(0.8,0.6,0)\] \[c_2=(0,0,1)\]

可以通过orientational matrix来描述他们之间的方向关系.

\[R_{12}=\left\{\begin{matrix} a_1\cdot a_2&a_1\cdot b_2&a_1\cdot c_2\\ b_1\cdot a_2&b_1\cdot b_2&b_1\cdot c_2\\ c_1\cdot a_2&c_1\cdot b_2&c_1\cdot c_2\\ \end{matrix}\right\}\]

代入可得

\[R_{12}=\left\{\begin{matrix} (1,0,0)\cdot (0.6,-0.8,0)&(1,0,0)\cdot (0.8,0.6,0)&(1,0,0)\cdot (0,0,1)\\ (0,1,0)\cdot (0.6,-0.8,0)&(0,1,0)\cdot (0.8,0.6,0)&(0,1,0)\cdot (0,0,1)\\ (0,0,1)\cdot (0.6,-0.8,0)&(0,0,1)\cdot (0.8,0.6,0)&(0,0,1)\cdot (0,0,1)\\ \end{matrix}\right\}\] \[R_{12}=\left\{\begin{matrix} 0.6&0.8&0\\ -0.8&0.6&0\\ 0&0&1\\ \end{matrix}\right\}\]

这个矩阵意义在于,晶体1中任意矢量可以通过此矩阵转化为晶体2中的矢量

即\(vec_2=R_{12}vec_1\)

比如晶体1中的(210)方向在晶体2中是

\[R_{12}\left\{\begin{matrix} 2\\ 1\\ 0\\ \end{matrix}\right\}=\left\{\begin{matrix} 0.6&0.8&0\\ -0.8&0.6&0\\ 0&0&1\\ \end{matrix}\right\}\left\{\begin{matrix} 2\\ 1\\ 0\\ \end{matrix}\right\}=\left\{\begin{matrix} 0.6*2+0.8*1+0*0\\ -0.8*2+0.6*1+0*0\\ 2*0+1*0+0*1\\ \end{matrix}\right\}=\left\{\begin{matrix} 2\\ -1\\ 0\\ \end{matrix}\right\}\]

interface operation:axis-angle representation

两个方向任意的晶体A和B,B一定可以通过将A绕某个轴旋转一定角度并且进行一定的平移得到(旋转轴-角度的组合不唯一,由于周期性,平移也不唯一)

例如\(\Sigma 5(210)\)晶界,可以通过绕001轴旋转53.13度,也可以绕210轴旋转180度

虽然旋转轴有三个坐标,但是只有两个自由度.因为旋转轴是单位矢量,其第三个坐标没有自由度.或者换一种角度,放在球坐标中,只有其角度部分是自由度,长度部分不算自由度.即,在表示自由度的时候需要这么理解旋转轴

旋转轴与角度的关系通过如下方式定义:

令旋转轴\(t_R=\left\{\begin{matrix} t_{R_x}\\ t_{R_y}\\ t_{R_z}\\ \end{matrix}\right\}=\left\{\begin{matrix} cos\alpha_rsin\phi_r\\ sin\alpha_rsin\phi_r\\ cos\phi_r\\ \end{matrix}\right\}\)

旋转角度为\(\theta\)

首先定义

\[Z_{12}=\left\{\begin{matrix} cos\alpha_rcos\phi_r&sin\alpha_rcos\phi_r&-sin\phi_r\\ -sin\alpha_r&cos\alpha_r&0\\ cos\alpha_rsin\phi_r&sin\alpha_rsin\phi_r&cos\phi_r\\ \end{matrix}\right\}\]

其具有性质

\[Z_{12}\cdot t_R=\left\{\begin{matrix} 0\\ 0\\ 1\\ \end{matrix}\right\}\]

以及绕001旋转theta度的旋转矩阵是

\[g(\theta)=\left\{\begin{matrix} cos\theta&-sin\theta&0\\ sin\theta&cos\theta&0\\ 0&0&1\\ \end{matrix}\right\}\]

则有

\[R_{12}=Z_{12}^T\cdot g(\theta)\cdot Z_{12}\] \[R_{12}=\left[\begin{matrix}\sin^{2}{\left(\alpha \right)} \cos{\left(\theta \right)} + \sin^{2}{\left(\phi \right)} \cos^{2}{\left(\alpha \right)} + \cos^{2}{\left(\alpha \right)} \cos^{2}{\left(\phi \right)} \cos{\left(\theta \right)} & - \sin{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} \cos{\left(\alpha \right)} \cos{\left(\theta \right)} + \sin{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} \cos{\left(\alpha \right)} - \sin{\left(\theta \right)} \cos{\left(\phi \right)} & \left(\sin{\left(\alpha \right)} \sin{\left(\theta \right)} - \cos{\left(\alpha \right)} \cos{\left(\phi \right)} \cos{\left(\theta \right)} + \cos{\left(\alpha \right)} \cos{\left(\phi \right)}\right) \sin{\left(\phi \right)}\\- \sin{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} \cos{\left(\alpha \right)} \cos{\left(\theta \right)} + \sin{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} \cos{\left(\alpha \right)} + \sin{\left(\theta \right)} \cos{\left(\phi \right)} & - \sin^{2}{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} \cos{\left(\theta \right)} + \sin^{2}{\left(\alpha \right)} \sin^{2}{\left(\phi \right)} + \cos{\left(\theta \right)} & \left(- \sin{\left(\alpha \right)} \cos{\left(\phi \right)} \cos{\left(\theta \right)} + \sin{\left(\alpha \right)} \cos{\left(\phi \right)} - \sin{\left(\theta \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\phi \right)}\\\left(- \sin{\left(\alpha \right)} \sin{\left(\theta \right)} - \cos{\left(\alpha \right)} \cos{\left(\phi \right)} \cos{\left(\theta \right)} + \cos{\left(\alpha \right)} \cos{\left(\phi \right)}\right) \sin{\left(\phi \right)} & \left(- \sin{\left(\alpha \right)} \cos{\left(\phi \right)} \cos{\left(\theta \right)} + \sin{\left(\alpha \right)} \cos{\left(\phi \right)} + \sin{\left(\theta \right)} \cos{\left(\alpha \right)}\right) \sin{\left(\phi \right)} & \sin^{2}{\left(\phi \right)} \cos{\left(\theta \right)} + \cos^{2}{\left(\phi \right)}\end{matrix}\right]\]

其关系为

\[t_{R_x}=\frac{b_1\cdot c_2-c_1\cdot b_2}{2sin\theta}\] \[t_{R_y}=\frac{a_1\cdot c_2-c_1\cdot a_2}{2sin\theta}\] \[t_{R_z}=\frac{a_1\cdot b_2-b_1\cdot a_2}{2sin\theta}\]

可以验证\(t_R=\left\{\begin{matrix} t_{R_x}\\ t_{R_y}\\ t_{R_z}\\ \end{matrix}\right\}=\left\{\begin{matrix} cos\alpha_rsin\phi_r\\ sin\alpha_rsin\phi_r\\ cos\phi_r\\ \end{matrix}\right\}\)

验证:

对应于旋转53.5度

其旋转轴是\(t_R=\left\{\begin{matrix} \frac{(0,1,0)\cdot (0,0,1)-(0,0,1)\cdot (0.8,0.6,0)}{2sin53.5}\\ \frac{(1,0,0))\cdot (0,0,1)-(0,0,1)\cdot (0,6,-0.8,0)}{2sin53.5}\\ \frac{(1,0,0)\cdot (0.8,0.6,0)-(0,1,0)\cdot (0,6,-0.8,0)}{2sin53.5}\\ \end{matrix}\right\}=\left\{\begin{matrix} \frac{0-0}{2sin53.5}\\ \frac{0-0}{2sin53.5}\\ \frac{0.8+0.8}{2*0.8}\\ \end{matrix}\right\}=\left\{\begin{matrix} 0\\0\\1\\ \end{matrix}\right\}\)

也就是说,比如通过一个已知旋转轴和角度得到了一个orientational matrix,想要得到在另一个旋转轴的等效角度,那么就带进去Z求g矩阵,反则反之

\[Z_{12}^{-1}R_{12}Z_{12}^{T-1}=g(\theta)\]

grain boundary plane normal

首先:晶界平面法线:grain boundary plane normal

这个在参考书《Cai, W., & Nix, W. (2016). Imperfections in Crystalline Solids (MRS-Cambridge Materials Fundamentals). Cambridge: Cambridge University Press. doi:10.1017/CBO9781316389508》中第437页图13.1有极好的表述.展示了这个grain boundary plane normal和旋转轴的区别

这部分带来了额外的两个自由度

grain boundary plane的表示法:

\[n_B=\left\{\begin{matrix} cos\alpha_B sin\phi_B\\ sin\alpha_B sin\phi_B\\ cos\phi_B\\ \end{matrix}\right\}\]

\(n_B=t_R\),twist \(n_B\cdot t_R=0\),tilt

其他情况下:混合

microscopic degree of freedom

晶界还有几个微观自由度.

除此之外,两个晶体还可以在旋转之后做相对的平移.有两个自由度是在grain boundary平面内的平移,比如对于NaCl,平移之前可能是Cl对Cl,Na对Na,平移之后是交错着.而垂直于grain boundary plane方向的平移,不知道算不算自由度.因为垂直方向只有一个特定距离是能量最低的,很难将这个是自由度.

Energy

grain boundary formation energy

adhesion energy

interface formation energy:

energy gain to create one unit of interface from bulk material

surface energy