Using the method of index notation

In this post I present examples of application of the method of index notation and Einstein's summation convention. I find the technique extremely useful especially when I don't have access to a book of formulae. As a student, I was taught that it's always best to be mathematically self-sufficient - meaning should the need arise, one should be able to derive formulae or construct an analytical framework from scratch using a few established axioms in mathematics. Such proficiency in mathematics enables one to focus on the subject that one needs analyzed rather than on mathematics which only provides tools to facilitate analysis. I find that this, unfortunately, is not a universal view. Below, I show how some of the most widely applied equations in physics can be processed or derived readily using the index notation method.

I derive these equations every time I need them. I must have done so a million times. So, I find the process easy to use. However, I have a very vivid memory of my first encounter with this technique: It was a struggle! But once I had it down, it was super easy and fun to use. I've tried to avoid shortcuts in the following examples for the sake of clarity. Once you get used to working with indices you'll be able to work these problems out quickly in minimal number of steps.

DEFINITIONS:

$\nabla = \hat{x}_{1} \frac{\partial}{\partial x_{1}} + \hat{x}_{2} \frac{\partial}{\partial x_{2}} + \hat{x}_{3} \frac{\partial}{\partial x_{3}}$ in 3-dimensional space.

$\delta_{ij}$ is Kronecker delta function: $\delta_{ij} = 0$ when $i = j$
$\hspace{5.9cm} = 1$ when $i \ne j$

Remember $\partial_{i} \partial_{j} = \partial_{j} \partial_{i}$ where $\partial_{i} = \frac{\partial}{\partial x_{i}}$. For instance, $\frac{\partial}{\partial x_{1}} \frac{\partial f}{\partial x_{2}} = \frac{\partial}{\partial x_{2}} \frac{\partial f}{\partial x_{1}}$

EXAMPLES:

$\nabla \cdot (f \textbf{A}) = \partial_{i} \, (fA)_{i}$
$\hspace{1.7cm} = f \, \partial_{i} A_{i} + A_{i} \, \partial_{i}f$
$\hspace{1.7cm} = f \, (\nabla \cdot \textbf{A}) + \textbf{A} \cdot \nabla f$

$\Rightarrow \nabla \cdot (f \textbf{A}) = f \, (\nabla \cdot \textbf{A}) + \textbf{A} \cdot \nabla f$

$\nabla \times (f \textbf{A}) = \epsilon_{ijk} \, \partial_{j} \, (fA)_{k}$
$\hspace{1.9cm}= \epsilon_{ijk} \, ( f \, \partial_{j} A_{k} + A_{k} \, \partial_{j}f )$
$\hspace{1.9cm}= \epsilon_{ijk} \, f \, \partial_{j} A_{k} + \epsilon_{ijk} \, A_{k} \, (\partial_{j}f )$
$\hspace{1.9cm}= f \, \epsilon_{ijk} \, \partial_{j} A_{k} - \epsilon_{ikj} \, A_{k} \, (\partial_{j}f )$
$\hspace{1.9cm}= f \, (\nabla \times \textbf{A}) - \textbf{A} \times (\nabla f)$

$\Rightarrow \nabla \cdot (f \textbf{A}) = f \, (\nabla \times \textbf{A}) - \textbf{A} \times (\nabla f)$

$\nabla \cdot (\textbf{A} \times \textbf{B}) = \partial_{i} \, ( \epsilon_{ijk} \, A_{j} B_{k} )$
$\hspace{2.3cm} = \epsilon_{ijk} \, \partial_{i} \, ( A_{j} B_{k} )$
$\hspace{2.3cm} = \epsilon_{ijk} \, ( A_{j} \, \partial_{i} B_{k} + B_{k} \partial_{i} A_{j} )$
$\hspace{2.3cm} = \epsilon_{ijk} \, A_{j} \, \partial_{i} B_{k} + \epsilon_{ijk} \, B_{k} \, \partial_{i} A_{j}$
$\hspace{2.3cm} = - \epsilon_{jik} \, A_{j} \, \partial_{i} B_{k} + \epsilon_{kij} \, B_{k} \, \partial_{i} A_{j}$
$\hspace{2.3cm} = -\textbf{A} \cdot (\nabla \times \textbf{B}) + \textbf{B} \cdot (\nabla \times \textbf{A})$

$\Rightarrow \nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B} \cdot (\nabla \times \textbf{A}) -\textbf{A} \cdot (\nabla \times \textbf{B})$

In cases where more than one cross-products are involved, we encounter more than one Levi-Civita symbol. Let's look at the following combination of Levi-Civita symbol which we come across when dealing with two vector products.
$$\begin{array}{l} \epsilon_{ijk} \epsilon_{klm} &=& \epsilon_{ijk} \epsilon_{lmk} \\ &=& \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} \end{array}$$ To derive this result, consider the facts that indices $i,j,k,l,m$ can take values in the range $\{1,2,3\}$; index $k$ is a dummy index, hence must be summed over, whereas indices $i,j,k,l$ and $m$ are free and take specific values. Therefore, $\epsilon_{ijk} \epsilon_{lmk} = \epsilon_{ij1} \epsilon_{lm1}+\epsilon_{ij2} \epsilon_{lm2}+\epsilon_{ij3} \epsilon_{lm3}$; but only one of these terms will survive because if $i,j$ take values, say $2,3$, then terms with $k=2$ and $3$ will vanish (since $\epsilon_{ijk}=0$ when two or more indices take same values). Let's say the term $\epsilon_{ij3} \epsilon_{lm3}$ survives because $i,k,l$ and $m$ are either $1$ or $2$. Specifically, if $i,j=1,2$ and $l,m=1,2$, then $\epsilon_{ij3} \epsilon_{lm3}=\epsilon_{123} \epsilon_{123}=+1$. Alternatively, if $i,j=1,2$ and $l,m=2,1$, then $\epsilon_{ij3} \epsilon_{lm3}=\epsilon_{123} \epsilon_{213}=-1$. We can express this result as $\epsilon_{ijk} \epsilon_{lmk}=\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$ (think about it). Now let's apply this result.


$\nabla \times (\nabla \times \textbf{A}) = \epsilon_{ijk} \, \partial_{j} (\nabla \times \textbf{A})_{k}$
$\hspace{2.6cm} = \epsilon_{ijk} \, \partial_{j} (\epsilon_{klm} \partial_{l} A_{m})$
$\hspace{2.6cm} = \epsilon_{ijk} \epsilon_{klm} \, \partial_{j} \partial_{l} A_{m}$
$\hspace{2.6cm} = \epsilon_{ijk} \epsilon_{lmk} \, \partial_{j} \partial_{l} A_{m}$
$\hspace{2.6cm} = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \, \partial_{j} \partial_{l} A_{m}$
$\hspace{2.6cm} = \delta_{il} \delta_{jm} \, \partial_{j} \partial_{l} A_{m} - \delta_{im} \delta_{jl} \, \partial_{j} \partial_{l} A_{m}$
$\hspace{2.6cm} = \partial_{j} \partial_{i} A_{j} - \partial_{j} \partial_{j} A_{i}$
$\hspace{2.6cm} = \partial_{i} (\partial_{j} A_{j}) - (\partial_{j} \partial_{j}) A_{i}$
$\hspace{2.6cm} = \nabla (\nabla \cdot \textbf{A}) - \nabla^{2} \textbf{A}$

$\Rightarrow \, \nabla \times (\nabla \times \textbf{A}) = \nabla (\nabla \cdot \textbf{A}) - \nabla^{2} \textbf{A}$

$\nabla \times (\textbf{A} \times \textbf{B}) = \epsilon_{ijk} \,\partial_{i} \, (\epsilon_{klm} A_{l} B_{m})$
$\hspace{2.6cm} = \epsilon_{ijk} \,\epsilon_{klm} \partial_{i} \, ( A_{l} B_{m} )$
$\hspace{2.6cm} = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) \,(A_{l} \partial_{i} \, B_{m} + B_{m} \partial_{i} \, A_{l} )$
$\hspace{2.6cm} = \delta_{il} \delta_{jm} \, A_{l} \partial_{i} \, B_{m} - \delta_{im} \delta_{jl} \, A_{l} \partial_{i} \, B_{m} + \delta_{il} \delta_{jm} \, B_{m} \partial_{i} \, A_{l} - \delta_{im} \delta_{jl} \, B_{m} \partial_{i} \, A_{l}$
$\hspace{2.6cm} = A_{i} \partial_{i} \, B_{j} - A_{j} \partial_{i} \, B_{i} + B_{j} \partial_{i} \, A_{i} - B_{i} \partial_{i} \, A_{j}$
$\hspace{2.6cm} = (\textbf{A} \cdot \nabla) \textbf{B} - \textbf{A}(\nabla \cdot \textbf{B}) + \textbf{B} (\nabla \cdot \textbf{A}) - (\textbf{B} \cdot \nabla) \textbf{A}$

$\Rightarrow \, \nabla \times (\textbf{A} \times \textbf{B}) = (\textbf{A} \cdot \nabla) \textbf{B} - \textbf{A}(\nabla \cdot \textbf{B}) + \textbf{B} (\nabla \cdot \textbf{A}) - (\textbf{B} \cdot \nabla) \textbf{A}$

$\nabla \times (\nabla f ) = \epsilon_{ijk} \, \partial_{j} \, \partial_{k} f$
$\hspace{1.9cm} = 0 \hspace{1.5cm}$ since $\partial_{j} \, \partial_{k} f = \partial_{k} \, \partial_{j} f$ and $\epsilon_{ijk}$ is antisymmetric under interchange of $j$ and $k$

$\nabla \cdot (\nabla \times \textbf{A} ) = \partial_{i} \, (\epsilon_{ijk} \, \partial_{j} A_{k} )$
$\hspace{2.3cm} = \epsilon_{ijk} \, \partial_{i} \, (\partial_{j} A_{k} )$
$\hspace{2.3cm} = \epsilon_{ijk} \, \partial_{i} \partial_{j} A_{k}$
$\hspace{2.3cm} = 0$