dyadic product
There are several equivalent terms and notations for this product:
- the dyadic product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted by \(\mathbf{a b}\) (juxtaposed; no symbols, multiplication signs, crosses, dots, etc.)
- the outer product of two column vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted and defined as \(\mathbf{a} \otimes \mathbf{b}\) or \(\mathbf{a} \mathbf{b}^{\top}\), where \(\mathbf{T}\) means transpose,
- the tensor product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is denoted \(\mathbf{a} \otimes \mathbf{b}\),
Three-dimensional Euclidean space
To illustrate the equivalent usage, consider three-dimensional Euclidean space, letting: \[ \begin{aligned} & \mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k} \\ & \mathbf{b}=b_1 \mathbf{i}+b_2 \mathbf{j}+b_3 \mathbf{k} \end{aligned} \] be two vectors where \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) (also denoted \(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\) ) are the standard basis vectors in this vector space (see also Cartesian coordinates). Then the dyadic product of \(\mathbf{a}\) and \(\mathbf{b}\) can be represented as a sum: \[ \begin{aligned} & \mathbf{a b}=a_1 b_1 \mathbf{i} \mathbf{i}+a_1 b_2 \mathbf{i} \mathbf{j}+a_1 b_3 \mathbf{i} \mathbf{k} \\ & +a_2 b_1 \mathbf{j i}+a_2 b_2 \mathbf{j} \mathbf{j}+a_2 b_3 \mathbf{j k} \\ & +a_3 b_1 \mathbf{k i}+a_3 b_2 \mathbf{k j}+a_3 b_3 \mathbf{k k} \\ & \end{aligned} \] or by extension from row and column vectors, a \(3 \times 3\) matrix (also the result of the outer product or tensor product of \(\mathbf{a}\) and \(\mathbf{b}\) ): \[ \mathbf{a b} \equiv \mathbf{a} \otimes \mathbf{b} \equiv \mathbf{a b}^{\top}=\left(\begin{array}{c} a_1 \\ a_2 \\ a_3 \end{array}\right)\left(\begin{array}{lll} b_1 & b_2 & b_3 \end{array}\right)=\left(\begin{array}{ccc} a_1 b_1 & a_1 b_2 & a_1 b_3 \\ a_2 b_1 & a_2 b_2 & a_2 b_3 \\ a_3 b_1 & a_3 b_2 & a_3 b_3 \end{array}\right) \]
\(\mathrm{N}\)-dimensional Euclidean space
If the Euclidean space is \(\mathrm{N}\)-dimensional, and \[ \begin{array}{r} \mathbf{a}=\sum_{i=1}^N a_i \mathbf{e}_i=a_1 \mathbf{e}_1+a_2 \mathbf{e}_2+\ldots+a_N \mathbf{e}_N \\ \mathbf{b}=\sum_{j=1}^N b_j \mathbf{e}_j=b_1 \mathbf{e}_1+b_2 \mathbf{e}_2+\ldots+b_N \mathbf{e}_N \end{array} \] where \(\mathbf{e}_i\) and \(\mathbf{e}_j\) are the standard basis vectors in \(N\)-dimensions (the index \(i\) on \(\mathbf{e}_i\) selects a specific vector, not a component of the vector as in \(a_i\) ), then in algebraic form their dyadic product is: \[ \mathbf{a b}=\sum_{j=1}^N \sum_{i=1}^N a_i b_j \mathbf{e}_i \mathbf{e}_j . \]
This is known as the nonion form of the dyadic. Their outer/tensor product in matrix form is: \[ \mathbf{a b}=\mathbf{a b}^{\top}=\left(\begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_N \end{array}\right)\left(\begin{array}{llll} b_1 & b_2 & \cdots & b_N \end{array}\right)=\left(\begin{array}{cccc} a_1 b_1 & a_1 b_2 & \cdots & a_1 b_N \\ a_2 b_1 & a_2 b_2 & \cdots & a_2 b_N \\ \vdots & \vdots & \ddots & \vdots \\ a_N b_1 & a_N b_2 & \cdots & a_N b_N \end{array}\right) \]
double-dot product
Letting \[ \mathbf{A}=\sum_i \mathbf{a}_i \mathbf{b}_i, \quad \mathbf{B}=\sum_j \mathbf{c}_j \mathbf{d}_j \]
Double dot product
\[ \mathbf{A}: \mathbf{B}=\sum_{i, j}\left(\mathbf{a}_i \cdot \mathbf{c}_j\right)\left(\mathbf{b}_i \cdot \mathbf{d}_j\right) \]
For example, introduce a cartesian basis, so \(\bar{a} \times \bar{T}\) is \[ \begin{gathered} \bar{a} \times \bar{T}=\left(a_i \hat{e}^i\right) \times\left(T_{p q} \hat{e}^p \otimes \hat{e}^q\right) \\ =a_i T_{p q}\left(\hat{e}^i \times \hat{e}^p\right) \otimes \hat{e}^q \end{gathered} \] where \(\otimes\) is the tensor product. If the second step seems unfamiliar, think about what you would do with a dot product \(\bar{a} \cdot(\bar{b} \otimes \bar{c})\).
Now we rewrite the cross product itself. \[ \begin{gathered} \bar{a} \times \bar{T} \\ =a_i T_{p q}\left(\epsilon_{l m n} e_m^i e_n^p \hat{e}^l\right) \otimes \hat{e}^q \\ =a_i T_{p q} \epsilon_{l m n} \delta_{i m} \delta_{p n}\left(\hat{e}^l \otimes \hat{e}^q\right) \\ =\epsilon_{l i p} a_i T_{p q}\left(\hat{e}^l \otimes \hat{e}^q\right) \end{gathered} \] which returns an object with 2 free indices.
star pieces
- 并乘(并矢): \(P\cdot Q\)
\[ \boldsymbol{e}_i \times \boldsymbol{e}_j=\sum_{k=1}^3 \varepsilon_{i j k} \boldsymbol{e}_k \]
\[ \varepsilon_{i j k} \varepsilon_{k l m}=\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l} \]
surface element
To apply this to the present case, one needs to calculate how \(\mathbf{r}\) changes with each of the coordinates. In the conventions used, \[ \mathbf{r}=\left[\begin{array}{c} r \sin \theta \cos \varphi \\ r \sin \theta \sin \varphi \\ r \cos \theta \end{array}\right] \] Thus, \[ \frac{\partial \mathbf{r}}{\partial r}=\left[\begin{array}{c} \sin \theta \cos \varphi \\ \sin \theta \sin \varphi \\ \cos \theta \end{array}\right]=\hat{\mathbf{r}} \]
\[ \frac{\partial \mathbf{r}}{\partial \theta}=\left[\begin{array}{c} r \cos \theta \cos \varphi \\ r \cos \theta \sin \varphi \\ -r \sin \theta \end{array}\right]=r \hat{\boldsymbol{\theta}} \]
\[ \frac{\partial \mathbf{r}}{\partial \varphi}=\left[\begin{array}{c} -r \sin \theta \sin \varphi \\ r \sin \theta \cos \varphi \\ 0 \end{array}\right]=r \sin \theta \hat{\boldsymbol{\varphi}} . \]
The desired coefficients are the magnitudes of these vectors: \[ \left|\frac{\partial \mathbf{r}}{\partial r}\right|=1, \quad\left|\frac{\partial \mathbf{r}}{\partial \theta}\right|=r, \quad\left|\frac{\partial \mathbf{r}}{\partial \varphi}\right|=r \sin \theta \] The surface element spanning from \(\theta\) to \(\theta+\mathrm{d} \theta\) and \(\varphi\) to \(\varphi+\mathrm{d} \varphi\) on a spherical surface at (constant) radius \(r\) is then \[ \mathrm{d} S_r=\left\|\frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \varphi}\right\| \mathrm{d} \theta \mathrm{d} \varphi=|r \hat{\boldsymbol{\theta}} \times r \sin \theta \hat{\boldsymbol{\varphi}}|=r^2 \sin \theta \mathrm{d} \theta \mathrm{d} \varphi \] 注意理解叉乘也用来求解平行四边形面积
一定要注意，曲线坐标系中坐标基矢对坐标的导数不为0！！！
GDT Gauss Divergence Theorem: 在math review里面
球坐标 $0 0 $
注意使用课本上已用过的符号，比如\(\boldsymbol{\Omega}\) 就是 vorticity tensor,计算他没说也是！
张量的转置
\(\nabla \underset{\sim}{u} =\frac{\partial u_j}{\partial x_i} e_i e_j\)
\(\nabla \underset{\sim} u^{\top}=\frac{\partial u_i}{\partial x_j} e_i e_j\)
others
\[ \boldsymbol{e}_i \times \boldsymbol{e}_j=\sum_{k=1}^3 \varepsilon_{i j k} \boldsymbol{e}_k \]
\[ \varepsilon_{i j k} \varepsilon_{k l m}=\delta_{i l} \delta_{j m}-\delta_{i m} \delta_{j l} \]
\[ \varepsilon_{i j k}=\left\{\begin{aligned} +1 & \text { if }(i, j, k) \text { is }(1,2,3),(2,3,1), \text { or }(3,1,2) \\ -1 & \text { if }(i, j, k) \text { is }(3,2,1),(1,3,2), \text { or }(2,1,3) \\ 0 & \text { if } i=j, \text { or } j=k, \text { or } k=i \end{aligned}\right. \]
- 非常有用的解答！！！！如何通俗地理解格林公式? - 数学达人上官正申的回答 - 知乎 https://www.zhihu.com/question/46442061/answer/3462627428
The-Helmholtz-decomposition
\[ \frac{\mathrm{d}}{\mathrm{d} t} \delta \boldsymbol{r}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\boldsymbol{r}-\boldsymbol{r}_0\right)=\boldsymbol{v}-\boldsymbol{v}_0=\delta \boldsymbol{v}, \]
the differential symbol \(\frac{\mathrm{d}}{\mathrm{d} t}\) and variation symbols \(\delta\) can be used interchangeably. that is \(\frac{d}{d t} \delta \vec{r}=\delta \frac{d}{d t} \vec{r}\) , thus \[ \frac{\mathrm{d}}{\mathrm{d} t} \delta \boldsymbol{r}=\frac{\partial \boldsymbol{v}}{\partial x} \delta x+\frac{\partial \boldsymbol{v}}{\partial y} \delta y+\frac{\partial \boldsymbol{v}}{\partial z} \delta z . \] in variation principle \[ \delta f=\frac{\partial f}{\partial x} \delta x+\frac{\partial f}{\partial y} \delta y+\frac{\partial f}{\partial z} \delta z \] It exhibits similarities to the total differential operator \(d\)