同一矢量和张量在不同坐标系下的转换

坐标系定义球坐标系(R,θ,ϕ)(R,\theta,\phi)(R,θ,ϕ)，直角坐标系(x,y,z)(x,y,z)(x,y,z)x=Rsin⁡θcos⁡ϕ, y=Rsin⁡θcos⁡ϕ, z=Rcos⁡θx=R\sin\theta\cos\phi,\;y=R\sin\theta\cos\phi,\;z=R\cos\thetax=Rsinθcosϕ,y=Rsinθcosϕ,z=Rcosθ球坐标

洋芋片炒肉

5613人浏览 · 2020-07-25 21:12:09

洋芋片炒肉 · 2020-07-25 21:12:09 发布

坐标系定义

球坐标系 $(R,θ,ϕ)(R,\theta,\phi)$ ，直角坐标系 $(x, y, z)$

$z=Rcos⁡θx=R\sin\theta\cos\phi,\;y=R\sin\theta\cos\phi,\;z=R\cos\theta$

球坐标系和直角坐标系单位矢量转换

( $R^,θ^,ϕ^\hat{R},\hat{\theta},\hat{\phi}$ )为球坐标系的局部直角坐标单位矢量， $(x^,y^,z^)(\hat{x},\hat{y},\hat{z})$ 为全局坐标单位矢量

$(R^θ^ϕ^)=(sin⁡θcos⁡ϕsin⁡θsin⁡ϕcos⁡θcos⁡θcos⁡ϕcos⁡θsin⁡ϕ−sin⁡θ−sin⁡θcos⁡φ0)(x^y^z^)\left(\begin{array}{c}\hat{R}\\\hat{\theta}\\\hat{\phi}\end{array}\right)=\left(\begin{array}{ccc}\sin\theta\cos\phi&\sin\theta\sin\phi&\cos\theta\\\cos\theta\cos\phi&\cos\theta\sin\phi&-\sin\theta\\-\sin\theta&\cos\varphi&0\end{array}\right)\left(\begin{array}{c}\hat{x}\\\hat{y}\\\hat{z}\end{array}\right)$

记 $A=(sin⁡θcos⁡ϕsin⁡θsin⁡ϕcos⁡θcos⁡θcos⁡ϕcos⁡θsin⁡ϕ−sin⁡θ−sin⁡θcos⁡φ0)\mathbf{A}=\left(\begin{array}{ccc}\sin\theta\cos\phi&\sin\theta\sin\phi&\cos\theta\\\cos\theta\cos\phi&\cos\theta\sin\phi&-\sin\theta\\-\sin\theta&\cos\varphi&0\end{array}\right)$ ， $A\mathbf{A}$ 是正交矩阵， $ATA=I\mathbf{A}^T\mathbf{A}=\mathbf{I}$

那么 $(R^θ^ϕ^)=A(x^y^z^)\left(\begin{array}{c}\hat{R}\\\hat{\theta}\\\hat{\phi}\end{array}\right)=\mathbf{A}\left(\begin{array}{c}\hat{x}\\\hat{y}\\\hat{z}\end{array}\right)$ ， $(x^y^z^)=AT(R^θ^ϕ^)\left(\begin{array}{c}\hat{x}\\\hat{y}\\\hat{z}\end{array}\right)=\mathbf{A}^T\left(\begin{array}{c}\hat{R}\\\hat{\theta}\\\hat{\phi}\end{array}\right)$

不同坐标系的矢量转换

矢量 $g=gxx^+gyy^+gzz^=(x^y^z^)(gxgygz)=(R^θ^ϕ^)A(gxgygz)\mathbf{g}=g_x\hat{x}+g_y\hat{y}+g_z\hat{z}=\left(\begin{array}{ccc}\hat{x}&\hat{y}&\hat{z}\end{array}\right)\left(\begin{array}{c}g_x\\g_y\\g_z\end{array}\right)=\left(\begin{array}{ccc}\hat{R}&\hat{\theta}&\hat{\phi}\end{array}\right)\mathbf{A}\left(\begin{array}{c}g_x\\g_y\\g_z\end{array}\right)$

所以 $(gRgθgϕ)=A(gxgygz)\left(\begin{array}{c}g_R\\g_{\theta}\\g_{\phi}\end{array}\right)=\mathbf{A}\left(\begin{array}{c}g_{x}\\g_{y}\\g_{z}\end{array}\right)$

不同坐标系的张量转换

直角坐标系下的张量 $Tc=(TxxTxyTxzTyxTyyTyzTzxTzyTzz)\mathbf{T}_{c}=\left(\begin{array}{ccc}T_{xx}&T_{xy}&T_{xz}\\T_{yx}&T_{yy}&T_{yz}\\T_{zx}&T_{zy}&T_{zz}\end{array}\right)$

写成分量形式

$T=Txxx^x^+Txyx^y^+Txzx^z^+Tyxy^x^+Tyyy^y^+Tyzy^z^+Tzxz^x^+Tzyz^y^+Tzzz^z^=(x^y^z^)(TxxTxyTxzTyxTyyTyzTzxTzyTzz)(x^y^z^)\mathbf{T}=\begin{array}{c}T_{xx}\hat{x}\hat{x}+T_{xy}\hat{x}\hat{y}+T_{xz}\hat{x}\hat{z}\\+T_{yx}\hat{y}\hat{x}+T_{yy}\hat{y}\hat{y}+T_{yz}\hat{y}\hat{z}\\+T_{zx}\hat{z}\hat{x}+T_{zy}\hat{z}\hat{y}+T_{zz}\hat{z}\hat{z}\end{array}=\left(\begin{array}{ccc}\hat{x}&\hat{y}&\hat{z}\end{array}\right)\left(\begin{array}{ccc}T_{xx}&T_{xy}&T_{xz}\\T_{yx}&T_{yy}&T_{yz}\\T_{zx}&T_{zy}&T_{zz}\end{array}\right)\left(\begin{array}{c}\hat{x}\\\hat{y}\\\hat{z}\end{array}\right)$

代入 $(x^y^z^)=AT(R^θ^ϕ^)\left(\begin{array}{c}\hat{x}\\\hat{y}\\\hat{z}\end{array}\right)=\mathbf{A}^T\left(\begin{array}{c}\hat{R}\\\hat{\theta}\\\hat{\phi}\end{array}\right)$ 得

$T=(R^θ^ϕ^)A(TxxTxyTxzTyxTyyTyzTzxTzyTzz)AT(R^θ^ϕ^)\mathbf{T}=\left(\begin{array}{ccc}\hat{R}&\hat{\theta}&\hat{\phi}\end{array}\right)\mathbf{A}\left(\begin{array}{ccc}T_{xx}&T_{xy}&T_{xz}\\T_{yx}&T_{yy}&T_{yz}\\T_{zx}&T_{zy}&T_{zz}\end{array}\right)\mathbf{A}^T\left(\begin{array}{c}\hat{R}\\\hat{\theta}\\\hat{\phi}\end{array}\right)$

所以，球坐标系下张量为

$Ts=ATcAT\mathbf{T}_s=\mathbf{A}\mathbf{T}_c\mathbf{A}^T$

通过矢量的方向导数推导梯度张量的坐标系转换关系

$u^\hat{u}$ 是一个空间中的任意方向的单位矢量

矢量 $g\mathbf{g}$ 沿着 $u^\hat{\mathbf{u}}$ 的方向导数为
$gu=(u^⋅∇)g\mathbf{g}_u=(\hat{\mathbf{u}}\cdot \nabla)\mathbf{g}$

$(u^⋅∇)g=(ux∂∂x+uy∂∂y+uz∂∂z)g=(ux∂gx∂x+uy∂gx∂y+uz∂gx∂z)x^+(ux∂gy∂x+uy∂gy∂y+uz∂gy∂z)y^+(ux∂gz∂x+uy∂gz∂y+uz∂gz∂z)z^=(TxxTxyTxzTyxTyyTyzTzxTzyTzz)(uxuyuz)=Tu\begin{aligned}(\hat{\mathbf{u}}\cdot \nabla)\mathbf{g}=&(u_x\frac{\partial}{\partial x}+u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z})\mathbf{g}\\ =&(u_x\frac{\partial g_x}{\partial x}+u_y\frac{\partial g_x}{\partial y}+u_z\frac{\partial g_x}{\partial z})\hat{x}\\ &+(u_x\frac{\partial g_y}{\partial x}+u_y\frac{\partial g_y}{\partial y}+u_z\frac{\partial g_y}{\partial z})\hat{y}\\ &+(u_x\frac{\partial g_z}{\partial x}+u_y\frac{\partial g_z}{\partial y}+u_z\frac{\partial g_z}{\partial z})\hat{z}\\ =&\left(\begin{array}{ccc}T_{xx}&T_{xy}&T_{xz}\\T_{yx}&T_{yy}&T_{yz}\\T_{zx}&T_{zy}&T_{zz}\end{array}\right)\left(\begin{array}{c}u_{x}\\u_{y}\\u_{z}\end{array}\right)=\mathbf{T}\mathbf{u}\end{aligned}$