# Thornton & Marion, Classical Dynamics, Fifth Edition

## Chapter 1. Matrices, Vectors, and Vector Calculus

### Problem 06. Another proof of the orthogonality condition

##### The problem asks you to
prove that Equation (1.15) can be obtained by using the fact that the transformation matrix preserves the length of the line segment.
##### This problem assumes
that the coordinate systems are both orthogornal.
1) Transformation (rotation) matrix
2) Orthogonality condition
##### Solution
We assume a point $P$ $P$ is represented in the $x_i$ $x_i$ coordinate system by $P(x_1, x_2, x_3)$ $P(x_1, x_2, x_3)$, and it can be also represented in the $x'_i$ $x'_i$ coordinate system by $P(x'_1, x'_2, x'_3)$ $P(x'_1, x'_2, x'_3)$. These coordinate system have the same origin.
The length of the line segment from the origin and the point $P$ $P$ is $l^2 = x^2_1 + x^2_2 + x^2_3 = \sum_{i=1}^{3} x^2_i$ $l^2 = x^2_1 + x^2_2 + x^2_3 = \sum_{i=1}^{3} x^2_i$ $l'^2 = x'^2_1 + x'^2_2 + x'^2_3 = \sum^3_{i=1} x'^2_i$ $l'^2 = x'^2_1 + x'^2_2 + x'^2_3 = \sum^3_{i=1} x'^2_i$
Since the transformation matrix preserves the length of the line segment, $l^2 = l'^2$ $l^2 = l'^2$
thus, $\sum^3_{i=1} x^2_i = \sum^3_{i=1} x'^2_i$ $\sum^3_{i=1} x^2_i = \sum^3_{i=1} x'^2_i$
In the equation of transformation, $x'_i = \sum_j \lambda_{ij} x_j$ $x'_i = \sum_j \lambda_{ij} x_j$
using the fact that the index $j$ $j$ is a dummy variable, so we can write the above equation like this: $\sum^3_{i=1} x'^2_i = \sum^3_{i=1} ( \sum^3_{j=1} \lambda_{ij} x_j) ( \sum^3_{k=1} \lambda_{ik} x_k)$ $\sum^3_{i=1} x'^2_i = \sum^3_{i=1} ( \sum^3_{j=1} \lambda_{ij} x_j) ( \sum^3_{k=1} \lambda_{ik} x_k)$
Therefore, $\sum^3_{i} x^2_i = \sum^3_{i=1} ( \sum^3_{j=1} \lambda_{ij} x_j) ( \sum^3_{k=1} \lambda_{ik} x_k) = \sum_{j,k}^3 x_j x_k (\sum_i^3 \lambda_{ij} \lambda_{ik})$ $\sum^3_{i} x^2_i = \sum^3_{i=1} ( \sum^3_{j=1} \lambda_{ij} x_j) ( \sum^3_{k=1} \lambda_{ik} x_k) = \sum_{j,k}^3 x_j x_k (\sum_i^3 \lambda_{ij} \lambda_{ik})$
This equation is satisfied only if $\sum_i \lambda_{ij}\lambda_{ik} = \delta_{jk}$ $\sum_i \lambda_{ij}\lambda_{ik} = \delta_{jk}$