Saturday, August 9, 2014

Index Notation Partials and Rotations, a Cool Gradient Trick: EMII Notes 2014_08_09 Part I

Summary of what's gone on before.  Finally got the Levi-Civita to Kronecker delta identity down yesterday, 2014/08/08.  Today we're making more use of the index notation.  There are however, some notational stumbling points

$r^2 = x^2 + y^2 + z^2$

$\vec{r} = \left(x, y, z\right)$

Now if we want to take the derivative of both sides of the magnitude equation above, first remember we can write $r^2$ as

$r^2 = x_jx_j$

Now, finally taking the derivative of both sides o the above we get

$2r \partial_i r = 2x_j\partial_i x_j$

remembering the partial differentiation rules and the kronecker delta we can write the above down as

$2r \partial_i r = 2x_j \delta_{ij} = 2x_i$

which finally gives:

$\partial_i r = \dfrac{x_i}{r}$

The Kronecker trick above is crucial. Also remember, not one of the r's is a vector, they're all the magnitude of the vector.



\section{Rotations}

Any rigid rotation of a vector can be defined as:

$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix} = M\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix} $

Here are a few of the important parts.  The Matrix $M$ is orthogonal and, $M^TM = 1$

The transpose actually defines orthogonality.  If such a matrix is of dimension n, then it is n $)\left(n\right)$ matrix.

The footnote has all the cool kid stuff about rotation matrices and how to name them:

First of all, if the determinant of a matrix is +1, then it's a special orthogonal matrix, $SO\left(n\right)$.  The other sort, the sort with a determinant of -1 are actually rotations with a reflection of the coordinates.

There are also some identities in the footnote that will come in handy

$det\left(AB\right) = \left(det A\right)\left(B\right)$

and

$det\left(A^T\right) = det\left(A\right)$

Using the above two identities, we can see that $\left(det M\right)^2 = 1$

Memorize these!!!

In index notation, the rotation above can be written as
$x_i^{\prime} = M_{ij}x_j$

The orthogonality condition becomes

$M_{ki}M_{kj} = \delta_{ij}$







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