Skip to main content

Section 1.12 Orthonormality of Basis Vectors

In this book we will only work with orthonormal coordinates, such as rectangular, cylindrical, or spherical coordinates. Each such coordinate system is called orthogonal because the basis vectors adapted to the three coordinates point in mutually orthogonal directions, i.e. the basis vectors adapted to a particular coordinate system are perpendicular to each other at every point. In particular,

\begin{align*} \shat\cdot\phat \amp = \phat\cdot\zhat = \zhat\cdot\shat = 0 \qquad\hbox{(cylindrical)} ,\\ \rhat\cdot\that \amp = \that\cdot\phat = \phat\cdot\rhat = 0 \qquad\hbox{(spherical)} . \end{align*}

Figure 1.10.2 shows this orthogonality in the case of polar basis vectors.

We can also choose the basis vectors to be normalized, i.e. they are unit vectors:

\begin{align*} \shat\cdot\shat \amp = \phat\cdot\phat = \zhat\cdot\zhat = 1 \qquad\hbox{(cylindrical)} ,\\ \rhat\cdot\rhat \amp = \that\cdot\that = \phat\cdot\phat = 1 \qquad\hbox{(spherical)} . \end{align*}

A basis with both of the orthogonal property and the normalization property is called orthonormal.

Arbitrary vectors can be expanded in terms of a basis; this is why they are called basis vectors to begin with. The expansion of an arbitrary vector \(\vv\) in terms of its components in the three most common orthonormal coordinate systems is given by:

\begin{align*} \vv \amp = v_x \,\xhat + v_y \,\yhat + v_z \,\zhat \qquad\hbox{(rectangular)}\\ \amp = v_s \,\shat + v_\phi \,\phat + v_z \,\zhat \qquad\hbox{(cylindrical)}\\ \amp = v_r \,\rhat + v_\theta \,\that + v_\phi \,\phat \qquad\hbox{(spherical)} \end{align*}

It is straightforward to find the components of an arbitrary vector field \(\vv\) in terms of an orthonormal basis. Since the vectors in an orthonormal basis are mutually orthogonal, it is just a matter of using the dot product to figure out how much of \(\vv\) points in each of those directions.

Activity 1.12.1. Finding Coefficients.
Find the \(\phat\)-component of the vector \(\vv\text{.}\)
Solution

Since the basis is orthonormal, it is straightforward to compute the components algebraically. For example,

\begin{align} \vv\amp = v_s \,\shat + v_\phi \,\phat + v_z \,\zhat ,\label{eq-component}\tag{1.12.1}\\ \phat\cdot\vv \amp = \phat\cdot \left(v_s \,\shat + v_\phi \,\phat + v_z \,\zhat\right)\notag\\ \amp = \left(\phat\cdot v_s \,\shat + \phat\cdot v_\phi \,\phat + \phat\cdot v_z \,\zhat\right)\notag\\ \amp = \left(0 + v_\phi\, 1 + 0\right) ,\tag{1.12.2}\\ \Rightarrow v_\phi \amp = \phat\cdot \vv ,\tag{1.12.3} \end{align}

with similar expressions holding for the other components.

Of course, this calculation is overkill if you already have the vector \(\vv\) visibly decomposed as in Equation (1.12.1). In that case, you should just read off the coefficient of \(\phat\text{.}\) This method becomes useful if the expansion in Equation (1.12.1) is algebraically hidden in some way.

You can also find coefficients geometrically:

\begin{align*} v_\phi \amp = \phat\cdot\vv \\ \amp = \vert \phat\vert\, \vert \vv \vert\, \cos\alpha \end{align*}

where \(\alpha\) is the angle between \(\vv\) and \(\phat\) and \(\vert \phat\vert=1\text{.}\)

In situations with high symmetry, many physical quantities may be simpler to understand or interpret if they are written in terms of curvilinear basis vectors. However, you must be very careful anytime you are adding or subtracting (or integrating, which is just a continuous form of adding) vectors expressed in curvilinear basis vectors.

Warning 1.12.1. Never add or integrate expressions involving curvilinear basis vectors at more than one point.

When using basis vectors adapted to curvilinear coordinates in sums and differences (including integrals), it is essential to remember that these basis vectors are not constant. For example, if you try to add two vectors that are expanded in terms of the basis vectors appropriate to two different points, e.g. \(\{\rhat_1,\that_1,\phat_1\}\) and \(\{\rhat_2,\that_2,\phat_2\}\text{,}\) you cannot simply add the components:

\begin{align*} \vv_1+\ww_2 \amp = (v_r\rhat_1+v_{\theta}\that_1+v_{\phi}\phat_1) +(w_r\rhat_2+w_{\theta}\that_2+w_{\phi}\phat_2)\\ \amp \ne (v_r+w_r)\rhat +(v_{\theta}+w_{\theta})\that+(v_{\phi}+w_{\phi})\phat \nonumber \end{align*}

The last line is not correct because \(\{\rhat_1,\that_1,\phat_1\}\ne\{\rhat_2,\that_2,\phat_2\}\text{,}\) so that the coefficients of the basis vectors cannot be factored out. The only way out of this dilemma is to rewrite the curvilinear basis vectors in terms of rectangular basis vectors. Since rectangular basis vectors do not change from point to point, they can be factored out of the relevant expressions.

Similarly, later in this text, when we discuss taking various derivatives of vector fields, we will need to show how to take the derivatives of these curvilinear basis vectors.