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Chapter 1 - Mathematics
1.5 - Dyads and Tensors
Definition. A dyadic is a representation of two-ish vectors.
\stackrel{\leftrightarrow}{\vb{D}} = \begin{matrix}
D_{xx} \vu{x}\vu{x} &+ D_{xy} \vu{x}\vu{y} &+ D{xz} \vu{x}\vu{z} \\
+ D_{yx} \vu{y}\vu{x} &+ D_{yy} \vu{y}\vu{y} &+ D{yz} \vu{y}\vu{z} \\
+ D_{zx} \vu{z}\vu{x} &+ D_{zy} \vu{z}\vu{y} &+ D{zz} \vu{z}\vu{z}
\end{matrix}
Definition. If a dyadic can be written as a composition of two vectors \vb{A} and \vb{B}, it is called a dyad.
\vb{AB} = \begin{matrix}
A_x B_x \vu{x}\vu{x} &+ A_x B_y \vu{x}\vu{y} &+ A_x B_z \vu{x}\vu{z} \\
+ A_y B_x \vu{y}\vu{x} &+ A_y B_y \vu{y}\vu{y} &+ A_y B_z \vu{y}\vu{z} \\
+ A_z B_x \vu{z}\vu{x} &+ A_z B_y \vu{z}\vu{y} &+ A_z B_z \vu{z}\vu{z}
\end{matrix}
The dot product of a dyad \stackrel{\leftrightarrow}{\vb{D}} = \vb{AB} and vector \va{v} can be written as follows:
(\vb{AB}) \vdot \va{v} = \vb{A} (\vb{B} \vdot \va{v})
Definition. A symmetric/antisymmetric dyadic is defined the same way that a matrix is.
Definition. The identity dyadic is \stackrel{\leftrightarrow}{\vb{I}} = \vu{x}\vu{x} + \vu{y}\vu{y} + \vu{z}\vu{z}.
Definition. FOr a tensor, with coordinates u^i, we have two sets of basis vectors:
\vb{e}_i = \pdv{\vb{r}}{u^i}
\vb{e}^i = \grad{u^i}
1.9 - Helmholtz Theorem
Given an arbitrary vector field \vb{F}(\vb(r)), we can write said field as a composition of a curl-free component \vb{\Phi}(\vb{r}) and a divergence-free component \vb{A}(\vb{r}) as follows:
\vb{F}(\vb{r}) = - \grad{\vb{\Phi}(\vb{r})} + \curl{\vb{A}(\vb{r})}
Definition. Here, the gradient of the scalar potential is \grad{\vb{\Phi}(\vb{r})} and the curl of the vector potential is \curl{\vb{A}(\vb{r})}. Thus, the scalar potential is \vb{\Phi}(\vb{r}) and the vector potential is \vb{A}(\vb{r}).
Letting said field be over bounded volume V with closed surface \partial V, and the functions \vb{C}(\vb{r}) = \curl{\vb{F}(\vb{r})} and \vb{D}(\vb{r}) = \div{\vb{F}(\vb{r})} are known, we can say that
\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \frac{\vb{F}(\vb{r}') \vdot \va{n}'}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'} - \frac{1}{4 \pi} \int_{\partial V} \va{n}' \cross \frac{\vb{F}(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{S'}
Now, assume that \lim(\frac{\vb{F}(\vb{r})}{\vb{r}}) = 0 as \vb{r} \rightarrow \infty, with a large enough volume, we see that the second terms vanish.
\vb{\Phi}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{D(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}
\vb{A}(\vb{r}) = \frac{1}{4 \pi} \int_V \frac{C(\vb{r}')}{\abs{\vb{r}-\vb{r}'}} \dd{V'}