## Tuesday, December 15, 2009

### Twist And Spin

A matrix $M$ is skew-symmetric if and only if $M^{tr} = -M$. Let $S$ be the set of all 4x4 skew-symmetric matrices.

## Twist

Suppose $F:\mathbb{R}^4 \to \mathbb{R}^4$ is a differentiable vector field in $\mathbb{R}^4$, we define the operation "twist" that maps each point in $\mathbb{R}^4$ to $S$ as follows:

If $F(x,y,z,w) = A\vec i + B \vec j + C \vec k + D \vec l$ then

$(\bold{twist} F)(x,y,z,w) = \left( \begin{array}{cccc} 0 & B_x - A_y & C_x - A_z & D_x - A_w \\ A_y-B_x & 0 & C_y - B_z & D_y - B_w \\ A_z-C_x & B_z - C_y & 0 & D_z - C_w \\ A_w - D_x & B_w - D_y & C_w - D_z & 0 \end{array} \right)$

Where $B_x$ denotes $\partial B / \partial x$ and so on. One way that makes it easier to memorize the formula is to associate the first column and row with $x$, second column and row with $y$, third with $z$, and last with $w$.

## Spin

If $G : \mathbb{R}^4 \to S$ is a differentiable function that takes a point in $\mathbb{R}^4$ to a skew-symmetric matrix, we define the operation "spin" that produces a vector field in $\mathbb{R}^4$ as follows:

If $G(x,y,z,w) = \{ \sum_{i,j} G_{ij}(x,y,z,w) \}$ (where $G_{ij}$ is the entry at the $i$-th row and $j$-th column, and since $G$ is skew-symmetric, then $G_{i,i} = 0$ and $G_{ji} = -G_{ij}$)

Then

$(\bold{spin} G)(x,y,z,w) = (G_{23_w} + G_{34_y} + G_{42_z}) \vec i$

$+(G_{13_w} + G_{34_x} + G_{41_z}) \vec j$

$+(G_{12_w} + G_{24_x} + G_{41_y}) \vec k$

$+(G_{12_z} + G_{23_x} + G_{31_y}) \vec l$

where again $G_{ij_x}$ denotes $\partial G_{ij} / \partial x$ and so on.

## Problems:

Prove the following:

1. $(\bold{spin}(\bold{twist} F)) = \vec 0$

2. $\bigtriangledown \cdot ( \bold{spin} G) = 0$

3. If $F$ is a conservative vector field, then $\bold{twist} F = \vec 0$