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

$$S = a_{1 \beta}x^1x^{\beta} + a_{2 \beta}x^2x^{\beta} + a_{3 \beta}x^3 x^{\beta}$$

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

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

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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 $$i \ne j$$

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

We may write

$$ \frac{\partial x^i}{\partial x^j} \equiv \delta_j^i \begin{cases} = & 1 \qquad =if= \quad i = j \\ = & 0 \qquad =if= \quad i \ne j \end{cases} $$

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

$$D = \epsilon^{11}a_1^1a_1^2 + \epsilon^{12}a_1^1 a_2^2 + \epsilon^{21}a_2^1 a_1^2 + \epsilon^{22}a_2^1a_2^2 = a_1^1a_2^2 - a_2^1a_1^2 $$

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

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

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

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

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