PlanetPhysics/Examples of Einstein Summation Notation

Some examples of applying the [[../EinsteinSummationNotation/|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

$\begin{array}{c}S=a_{11}{x}^1{x}^1+a_{12}{x}^1{x}^2+a_{13}{x}^1{x}^3\\ +a_{21}{x}^2{x}^1+a_{22}{x}^2{x}^2+a_{23}{x}^2{x}^3\\ +a_{31}{x}^3{x}^1+a_{32}{x}^3{x}^2+a_{33}{x}^3{x}^3\end{array}$

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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 $i\ne j$

$\frac{\partial x^1}{\partial x^2}=0,\frac{\partial x^i}{\partial x^j}=0$

We may write

$\frac{\partial x^i}{\partial x^j}\equiv {\delta }_j^i\{\begin{array}{cc}=& 1=if=i=j\\ =& 0=if=i\ne j\end{array}$

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

$\begin{array}{c}\frac{\partial S}{\partial x^i}=a_{\alpha \beta }{x}^{\alpha }\frac{\partial x^{\beta }}{\partial x^i}+a_{\alpha \beta }{x}^{\beta }\frac{\partial x^{\alpha }}{\partial x^i}=0\\ \frac{\partial S}{\partial x^i}=a_{\alpha \beta }{x}^{\alpha }{\delta }_i^{\beta }+a_{\alpha \beta }{x}^{\beta }{\delta }_i^{\alpha }=0\\ \frac{\partial S}{\partial x^i}=a_{\alpha i}{x}^{\alpha }+a_{i\beta }{x}^{\beta }=0\end{array}$

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 ${ϵ}^{ij},i,j=1,2$, to have the following numerical values: Let ${ϵ}^{11}={ϵ}^{22}=0,{ϵ}^{12}=1,{ϵ}^{21}=-1$. We now consider the expression

$D={ϵ}^{ij}{a}_i^1{a}_j^2$

Expanding (2) by use of our summation convention yields

$D={ϵ}^{11}{a}_1^1{a}_1^2+{ϵ}^{12}{a}_1^1{a}_2^2+{ϵ}^{21}{a}_2^1{a}_1^2+{ϵ}^{22}{a}_2^1{a}_2^2=a_1^1{a}_2^2-a_2^1{a}_1^2$

The reader who is familiar with second-order [[../Determinant/|determinants]] quickly recognizes that

${ϵ}^{ij}{a}_i^1{a}_j^2=\left|\begin{array}{cc}{a}_1^1& a_2^1\\ a_1^2& a_2^2\end{array}\right|$

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Example 4 . The [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]] of equations

$\left(\begin{array}{ccc}{y}^1& =& y^1(x^1,x^2,\cdots ,x^n)\\ y^2& =& y^2(x^1,x^2,\cdots ,x^n)\\ ⋮& ⋮& ⋮\\ y^n& =& y^n(x^1,x^2,\cdots ,x^n)\end{array}\right)$

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

$d{y}^i=\frac{\partial y^i}{\partial x^1}d{x}^1+\frac{\partial y^i}{\partial x^2}d{x}^2+\cdots +\frac{\partial y^i}{\partial x^n}d{x}^{n}i=1,2,\cdots ,n$

$d{y}^i=\frac{\partial y^i}{\partial y^{\alpha }}d{x}^{\alpha }$

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

$\frac{d^{2}x}{d{y}^2}=-\frac{d^{2}y}{d{x}^2}{\left(\frac{dx}{dy}\right)}^3$

References

[1] Lass, Harry. "Elements of pure and applied mathematics" New York: McGraw-Hill Companies, 1957.

This entry is a derivative of the Public [[../Bijective/|domain]] [[../Work/|work]] [1].