(a) Defining the scalar product r s by Equation (1.7), r s | StudySoup

Textbook Solutions for Classical Mechanics

Chapter 1 Problem 1.16

Question

(a) Defining the scalar product \(\mathbf{r}\cdot\mathbf{s}\) by Equation (1.7), \(\mathbf{r}\cdot\mathbf{s}=\sum_{ }^{ }r_is_i\), show that Pythagoras's theorem implies that the magnitude of any vector \(\mathbf{r}\) is \(r=\sqrt{\mathbf{r}\cdot\mathbf{r}}\). (b) It is clear that the length of a vector does not depend on our choice of coordinate axes. Thus the result of part (a) guarantees that the scalar product \(\mathbf{r}\cdot\mathbf{r}\), as defined by (1.7), is the same for any choice of orthogonal axes. Use this to prove that \(\mathbf{r}\cdot\mathbf{s}\), as defined by (1.7), is the same for any choice of orthogonal axes. [Hint: Consider the length of the vector \(\mathbf{r}+\mathbf{s}\)]

Solution

Step  1 of 2

Applying Pythagorean theorem

a) Consider the two dimensional case

From the definition

 Where  is the magnitude of

Now, since we are dealing with an orthogonal coordinate frame, adding another coordinate is just using Pythagoras theorem between the new coordinate and all of the rest combined, giving that the sum of squares of components is equal to the length squared.

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full solution

Title Classical Mechanics 0 
Author John R Taylor
ISBN 9781891389221

(a) Defining the scalar product r s by Equation (1.7), r s

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