View source for 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

<math>S = a_{\alpha \beta} x^{\alpha}x^{\beta}</math>

for a three dimensional space. Since the index <math>\alpha</math> occurs as both a subscript and a superscript, we sum on <math>\alpha</math> from 1 to 3. This yields

<math>S = a_{1 \beta}x^1x^{\beta} + a_{2 \beta}x^2x^{\beta} + a_{3 \beta}x^3 x^{\beta}</math>

Now each term of <math>S</math> is such that <math>\beta</math> is both a subscript and superscript. Summing on <math>\beta</math> from 1 to 3 as prescribed by our summation convention yields the quadratic form

<blockquote><math>\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}</math></blockquote>

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'''Example 2''' . If <math>x^1, x^2, x^3, \dots, x^n</math> is a set of independent variables, then

<math>\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</math>

and if <math>i \ne j</math>

<math>\frac{\partial x^1}{\partial x^2} = 0, \qquad \frac{\partial x^i}{\partial x^j} = 0</math>

We may write

<blockquote><math>
\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}
</math></blockquote>

The symbol <math>\delta_j^i</math> is called the Kronecker delta. We have

<math>\delta_\alpha^\alpha = \delta_1^1+\delta_2^2 + \cdots + \delta_n^n = n</math>

Let us now assume that the quadratic form at the end of example 1 vanishes identically for all values of the independent variables <math>x^1</math>,<math> x^2</math>, <math>x^3</math>, and <math>a_{ij}</math> to be constant. Differentiating <math>S=a_{\alpha \beta}x^\alpha x^\beta = 0</math> with respect to a given variable, say <math>x^i</math>, yields

<blockquote><math>\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}</math></blockquote>

Now differentiating with respect to <math>x^i</math> yields

<math>\frac{\partial^2 S}{\partial x^j \partial x^i} = a_{\alpha i} \delta_j^\alpha + a_{i \beta}\delta_i^\beta = 0</math>

so that <math>a_{ji} + a_{ij} = 0</math> or <math>a_{ij} = -a_{ji}</math> for <math>i,j = 1,2,3</math>.

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'''Example 3''' . We define <math>\epsilon^{ij}, i, j = 1,2</math>, to have the following numerical values: Let <math>\epsilon^{11} = \epsilon^{22} = 0, \epsilon^{12} = 1, \epsilon^{21} = -1</math>. We now consider the expression

<blockquote><math>
D = \epsilon^{ij} a_i^1 a_j^2
</math></blockquote>

Expanding (2) by use of our summation convention yields

<math>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 </math>

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

<blockquote><math>
\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 |
</math></blockquote>

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

<blockquote><math>
\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 )
</math></blockquote>

represents a coordinate transformation from an <math>(x^1, x^2, \cdots, x^n)</math> coordinate system to a <math>(y^1, y^2, \cdots, y^n)</math> coordinate system. From the calculus we have

<math>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</math>

<math> dy^i = \frac{\partial y^i}{\partial y^\alpha} d x^\alpha</math>

The <math>\alpha</math> in the term <math> \frac{\partial y^i}{\partial x^\alpha} </math> is to be considered as a subscript. If, furthermore, the <math>x^i</math>, <math>i = 1, 2, \cdots, n</math>, can be solved for the <math>y^1, y^2, \cdots, y^n</math>, and assuming differentiability of the <math>x^i</math> with respect to each <math>y^i</math>, one obtains

<math> \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}</math>

Differentiating this expression with respect to <math>y^k</math> yields

<math>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}</math>

Multiplying both sides of this equation by <math> \frac{\partial x^\sigma}{\partial y^i} </math> amd summing on the inex <math>i</math> yields

<math>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}</math>

or

<math>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}</math>

which yields

<math>\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}</math>

In particular, if <math>y = f(x)</math>, then

<math> \frac{d^2x}{dy^2} = - \frac{d^2y}{dx^2}\left(\frac{dx}{dy} \right)^3</math>

===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].

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