# PlanetPhysics/Examples of Einstein Summation Notation

Some examples of applying the Einstein summation notation.
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Example 1 . Let us consider the quantity
$S=a_{\alpha \beta }x^{\alpha }x^{\beta }$
for a three dimensional space. Since the index $\alpha$ occurs as both a subscript and a superscript, we sum on $\alpha$ from 1 to 3. This yields
$S=a_{1\beta }x^{1}x^{\beta }+a_{2\beta }x^{2}x^{\beta }+a_{3\beta }x^{3}x^{\beta }$
Now each term of $S$ is such that $\beta$ is both a subscript and superscript. Summing on $\beta$ from 1 to 3 as prescribed by our summation convention yields the quadratic form
   ${\begin{matrix}S=a_{11}x^{1}x^{1}+a_{12}x^{1}x^{2}+a_{13}x^{1}x^{3}\\+a_{21}x^{2}x^{1}+a_{22}x^{2}x^{2}+a_{23}x^{2}x^{3}\\+a_{31}x^{3}x^{1}+a_{32}x^{3}x^{2}+a_{33}x^{3}x^{3}\end{matrix}}$
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Example 2 . If $x^{1},x^{2},x^{3},\dots ,x^{n}$ is a set of independent variables, then
${\frac {\partial x^{1}}{\partial x^{1}}}={\frac {\partial x^{2}}{\partial x^{2}}}={\frac {\partial x^{3}}{\partial x^{3}}}=\dots ={\frac {\partial x^{n}}{\partial x^{n}}}=1$
and if $i\neq j$
${\frac {\partial x^{1}}{\partial x^{2}}}=0,\qquad {\frac {\partial x^{i}}{\partial x^{j}}}=0$
We may write
   ${\frac {\partial x^{i}}{\partial x^{j}}}\equiv \delta _{j}^{i}{\begin{cases}=&1\qquad =if=\quad i=j\\=&0\qquad =if=\quad i\neq j\end{cases}}$
The symbol $\delta _{j}^{i}$ is called the Kronecker delta. We have
$\delta _{\alpha }^{\alpha }=\delta _{1}^{1}+\delta _{2}^{2}+\cdots +\delta _{n}^{n}=n$
Let us now assume that the quadratic form at the end of example 1 vanishes identically for all values of the independent variables $x^{1}$,$x^{2}$, $x^{3}$, and $a_{ij}$ to be constant. Differentiating $S=a_{\alpha \beta }x^{\alpha }x^{\beta }=0$ with respect to a given variable, say $x^{i}$, yields
   ${\begin{matrix}{\frac {\partial S}{\partial x^{i}}}=a_{\alpha \beta }x^{\alpha }{\frac {\partial x^{\beta }}{\partial x^{i}}}+a_{\alpha \beta }x^{\beta }{\frac {\partial x^{\alpha }}{\partial x^{i}}}=0\\{\frac {\partial S}{\partial x^{i}}}=a_{\alpha \beta }x^{\alpha }\delta _{i}^{\beta }+a_{\alpha \beta }x^{\beta }\delta _{i}^{\alpha }=0\\{\frac {\partial S}{\partial x^{i}}}=a_{\alpha i}x^{\alpha }+a_{i\beta }x^{\beta }=0\end{matrix}}$
Now differentiating with respect to $x^{i}$ yields
${\frac {\partial ^{2}S}{\partial x^{j}\partial x^{i}}}=a_{\alpha i}\delta _{j}^{\alpha }+a_{i\beta }\delta _{i}^{\beta }=0$
so that $a_{ji}+a_{ij}=0$ or $a_{ij}=-a_{ji}$ for $i,j=1,2,3$.
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Example 3 . We define $\epsilon ^{ij},i,j=1,2$, to have the following numerical values: Let $\epsilon ^{11}=\epsilon ^{22}=0,\epsilon ^{12}=1,\epsilon ^{21}=-1$. We now consider the expression
   $D=\epsilon ^{ij}a_{i}^{1}a_{j}^{2}$
Expanding (2) by use of our summation convention yields
$D=\epsilon ^{11}a_{1}^{1}a_{1}^{2}+\epsilon ^{12}a_{1}^{1}a_{2}^{2}+\epsilon ^{21}a_{2}^{1}a_{1}^{2}+\epsilon ^{22}a_{2}^{1}a_{2}^{2}=a_{1}^{1}a_{2}^{2}-a_{2}^{1}a_{1}^{2}$
The reader who is familiar with second-order determinants quickly recognizes that
   $\epsilon ^{ij}a_{i}^{1}a_{j}^{2}=\left|{\begin{matrix}a_{1}^{1}&a_{2}^{1}\\a_{1}^{2}&a_{2}^{2}\end{matrix}}\right|$
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Example 4 . The system of equations
   $\left({\begin{matrix}y^{1}&=&y^{1}(x^{1},x^{2},\cdots ,x^{n})\\y^{2}&=&y^{2}(x^{1},x^{2},\cdots ,x^{n})\\\vdots &\vdots &\vdots \\y^{n}&=&y^{n}(x^{1},x^{2},\cdots ,x^{n})\end{matrix}}\right)$
represents a coordinate transformation from an $(x^{1},x^{2},\cdots ,x^{n})$ coordinate system to a $(y^{1},y^{2},\cdots ,y^{n})$ coordinate system. From the calculus we have
$dy^{i}={\frac {\partial y^{i}}{\partial x^{1}}}dx^{1}+{\frac {\partial y^{i}}{\partial x^{2}}}dx^{2}+\cdots +{\frac {\partial y^{i}}{\partial x^{n}}}dx^{n}\qquad i=1,2,\cdots ,n$
$dy^{i}={\frac {\partial y^{i}}{\partial y^{\alpha }}}dx^{\alpha }$
The $\alpha$ in the term ${\frac {\partial y^{i}}{\partial x^{\alpha }}}$ is to be considered as a subscript. If, furthermore, the $x^{i}$, $i=1,2,\cdots ,n$, can be solved for the $y^{1},y^{2},\cdots ,y^{n}$, and assuming differentiability of the $x^{i}$ with respect to each $y^{i}$, one obtains
${\frac {\partial y^{i}}{\partial y^{j}}}\equiv \delta _{j}^{i}={\frac {\partial y^{i}}{\partial x^{\alpha }}}{\frac {\partial x^{\alpha }}{\partial y^{j}}}$
Differentiating this expression with respect to $y^{k}$ yields
$0={\frac {\partial y^{i}}{\partial x^{\alpha }}}{\frac {\partial ^{2}x^{\alpha }}{\partial y^{k}\partial y^{i}}}+{\frac {\partial ^{2}y^{i}}{\partial x^{\beta }\partial x^{\alpha }}}{\frac {\partial x^{\beta }}{\partial y^{k}}}{\frac {\partial x^{\alpha }}{\partial y^{j}}}$
Multiplying both sides of this equation by ${\frac {\partial x^{\sigma }}{\partial y^{i}}}$ amd summing on the inex $i$ yields
$0={\frac {\partial x^{\sigma }}{\partial y^{i}}}{\frac {\partial y^{i}}{\partial x^{\alpha }}}{\frac {\partial ^{2}x^{\alpha }}{\partial y^{k}\partial y^{j}}}+{\frac {\partial ^{2}y^{i}}{\partial x^{\beta }\partial x^{\alpha }}}{\frac {\partial x^{\beta }}{\partial y^{k}}}{\frac {\partial x^{\alpha }}{\partial y^{j}}}{\frac {\partial x^{\sigma }}{\partial y^{i}}}$
or
$0=\delta _{\alpha }^{\sigma }{\frac {\partial ^{2}x^{\alpha }}{\partial y^{k}\partial y^{j}}}+{\frac {\partial ^{2}y^{i}}{\partial x^{\beta }\partial x^{\alpha }}}{\frac {\partial x^{\beta }}{\partial y^{k}}}{\frac {\partial x^{\alpha }}{\partial y^{j}}}{\frac {\partial x^{\sigma }}{\partial y^{i}}}$
which yields
${\frac {\partial ^{2}x^{\sigma }}{\partial y^{k}\partial y^{j}}}=-{\frac {\partial ^{2}y^{i}}{\partial x^{\beta }\partial x^{\alpha }}}{\frac {\partial x^{\beta }}{\partial y^{k}}}{\frac {\partial x^{\alpha }}{\partial y^{j}}}{\frac {\partial x^{\sigma }}{\partial y^{i}}}$
In particular, if $y=f(x)$, then
${\frac {d^{2}x}{dy^{2}}}=-{\frac {d^{2}y}{dx^{2}}}\left({\frac {dx}{dy}}\right)^{3}$