Chapter 2, Selected problems and solutions to NFCM, 2nd Ed.

 

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Section 4:

Note: This is a new websection on my website. I expect to be updating this section from time to time.

Click on this link to find Problem (4.5) on page 71 of NFCM. Problem (4.5) on page 71 of NFCM.

The Problem

On page 71 of NFCM (See References), we find problem (4.5): Given $\ba$ and $\bb$ as unit vectors and that \begin{equation} \ba\bb = e^{\bi\theta}\,,\label{eq:First} \end{equation} show that

  (a) $(\ba\bb)^\dagger = \bb\ba = e^{-\bi\theta}$,

  (b) $(\ba-\bb)^2 = 4\sin^2\half \theta$,

   (c) $(\ba+\bb)^2 = 4\cos^2\half \theta$.

Solution to Part (a)

We begin with the observing that $\bi^\dagger = -\bi$. Then, taking the reverse operation on the LHS of (\ref{eq:First}): \begin{equation} (\ba\bb)^\dagger = \bb^\dagger\ba^\dagger = \bb\ba \,.\label{eq:Part(a)1} \end{equation} Next, we take the reverse operation on the RHS of (\ref{eq:First}): \begin{equation} (e^{\bi\theta})^\dagger = (\cos\theta + \bi\sin\theta)^\dagger = \cos\theta - \bi\sin\theta = e^{-\bi\theta} \,,\label{eq:Part(a)2} \end{equation} and we have established Part (a).

For Parts (b) and (c) we will make use of the following figure.

         

Figure 1. We've constructed a parallelogram with all sides equal, hence, a
rhombus. By symmetry, we can prove that the diagonals, which intersect at
point $P$, bisect each other, and have right angles all around. This latter result
can also be proven algebraically by showing that $(\ba-\bb)\cdot(\ba+\bb)=0$.

First, let's prove that the two angles at point $O$ are actually equal by using vector methods. Let $\bu$ be a unit vector along the vector $\ba+\bb$. Then we need to show that \begin{equation} \ba\cdot\bu = \bb\cdot \bu\,.\label{eq:c.u} \end{equation} But \begin{equation} \bu = \frac{\ba+\bb}{\abs{\ba+\bb}}\,. \end{equation} So, (\ref{eq:c.u}) becomes \begin{equation} \ba\cdot \frac{\ba+\bb}{\abs{\ba+\bb}}\ \overset{?}{=}\ \bb\cdot \frac{\ba+\bb}{\abs{\ba+\bb}}\,. \end{equation} On multiplying through by $\abs{\ba+\bb}$, we have \begin{equation} \ba\cdot (\ba+\bb)\ \overset{?}{=}\ \bb\cdot (\ba+\bb)\,. \end{equation} On expanding this and simplifying, we see that the equality holds. What this means is that the angle between $\ba$ and the long diagonal is the same measure as that between $\bb$ and the long diagonal; hence, the two angles are equal, and equal to $\theta/2$.

Solution to Part (b)

Next, we are to show that $(\ba-\bb)^2 = 4\sin^2\half \theta$. We'll need only trigonometry at this point. Referring to Fig.\ 1, from the right triangle $\Delta OPQ$, we have that \begin{equation} \sin\half \theta =\frac{\half\abs{\ba-\bb}}{b} = \half\abs{\ba-\bb}\,. \end{equation} On squaring this and multiplying through by 4, we get \begin{equation} (\ba-\bb)^2 = 4\sin^2\half \theta\,, \end{equation} which is what we were to show.

Solution to Part (c)

This next proof is similar to the last one. This time we take the cosine of the angle, to get \begin{equation} \cos\half \theta =\frac{\half\abs{\ba+\bb}}{b} = \half\abs{\ba+\bb}\,. \end{equation} On squaring this and multiplying through by 4, we get \begin{equation} (\ba+\bb)^2 = 4\cos^2\half \theta\,, \end{equation} which is what we were to show.

Click on this link to find Problem (4.9) on page 71 of NFCM. Problem (4.9) on page 71 of NFCM.

References

D. Hestenes, New Foundations in Classical Mechanics, 2nd Ed., Kluwer Academic Publishers, 1999.

NFCM Problems and Solutions