## Monday, August 18, 2014

### More Tensor Index Identity Proofs: EM II Notes 2014_08_18

Summary:  Having worked through the examples that looked the most difficult, today's notes contain examples that are pick-up work from the easy problems.  These are simple-ish tensor index identities, including the divergence of the position vector, the cross product of the position vector, the Laplacian of one over the displacement squared, and the curl of a gradient.

$\nabla \cdot \vec{r} = 3$
$= \dfrac{\partial}{\partial x_i} r_i$
Keep in mind that $r_1 = x$, $r_2 = y$, and $r_3 = z$.  Using the rules of partial differentiation, when the partial operates on the variable it is with respect to it will return 1, and when it operates on any other variable, it will return 0.  The results sum to 3.

$\vec{\nabla} \times \vec{r} = 0$
$=\epsilon_{ijk} \partial_j r_k$
$= 0$

For the $\epsilon{ijk}$ to evaluate to a non-zero result, $j$ and $k$ have to not be equal.  However, as discussed above, if $J \ne k$, then the partial derivative evaluates to zero.  Consequently, the entire expression evaluates to zero.

$\nabla^2 \dfrac{1}{r} = 0$
The trick here is to do the derivatives one at a time, keeping things in index notation and look for things to cancel out.  There's also one other identity we'll need $r^2 = x_i x_i$, where the $x_i$ are the Cartesian components of the coordinate system.

So,

$\nabla^2 \dfrac{1}{r} = \partial_i \left(- \dfrac{x_i}{r^3}\right) = -\dfrac{3}{r^3} + \dfrac{3x_i x_i}{r^5}$,

but, $x_i x_i = r^2$, so the r.h.s. above is 0.

For multipole work where you're taking partial derivatives in multiple dimensions, this comes in handy for expressions like:

$\delta_{ij}\partial_i \partial_j \partial_k \dfrac{1}{r}$,

because the terms can be rearranged to show that any such expression is 0.  For way more detail, check out the material near equation 6.26 in https://drive.google.com/file/d/0B30APQ2sxrAYcHl2R3pCSG1HQXM/edit?usp=sharing

$\vec{\nabla} \times \vec{\nabla}f = 0$

$= \epsilon_{ijk} \partial_j \partial_k f$

The trick here is to think about what terms will survive and what the Levi-Civita symbol will do to them negative sign-wise.  Only pairs of derivatives where $j \ne k$ will survive the Levi-Civita.  There will be two of each of these terms, but they will be of opposite signs and will cancel, for example,

$\epsilon_{i23}\partial_2 \partial_3 = -\epsilon_{i32}\partial_3 \partial_2$.

Hence, all terms will cancel and we have a zero result, and a handy identity moving forward:

$\epsilon_{ijk}\partial_j\partial_k = 0$

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