The scalar product

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The scalar product

A scalar quantity is invariant under all possible rotational transformations. The individual components of a vector are not scalars because they change under transformation. Can we form a scalar out of some combination of the components of one, or more, vectors? Suppose that we were to define the ``ampersand'' product,

$\begin{displaymath} {\bf a} \& {\bf b} = a_x b_y + a_y b_z + a_z b_x = {\rm scalar number}, \end{displaymath}$

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for general vectors ${\bf a}$ and ${\bf b}$ . Is ${\bf a} \& {\bf b}$ invariant under transformation, as must be the case if it is a scalar number? Let us consider an example. Suppose that ${\bf a} = (1, 0, 0)$ and ${\bf b} = (0, 1, 0)$ . It is easily seen that ${\bf a} \& {\bf b}= 1$ . Let us now rotate the basis through $45^\circ$ about the

-axis. In the new basis, ${\bf a} = (1/\sqrt{2}, -1/\sqrt{2}, 0)$ and ${\bf b} = (1/\sqrt{2}, 1/\sqrt{2}, 0)$ , giving ${\bf a} \& {\bf b} = 1/2$ . Clearly, ${\bf a} \& {\bf b}$ is not invariant under rotational transformation, so the above definition is a bad one.

Consider, now, the dot product or scalar product:

$\begin{displaymath} {\bf a} \cdot {\bf b} = a_x b_x + a_y b_y + a_z b_z = {\rm scalar number}. \end{displaymath}$

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Let us rotate the basis though $\theta$ degrees about the

-axis. According to Eqs. (10)-(12), in the new basis ${\bf a} \cdot {\bf b}$ takes the form

$\displaystyle {\bf a} \cdot {\bf b}$	$\textstyle =$	$\displaystyle (a_x \cos\theta+a_y \sin\theta) (b_x \cos\theta + b_y \sin\theta)$
		$\displaystyle +(-a_x \sin\theta + a_y \cos\theta) (-b_x \sin \theta + b_y \cos\theta) +a_z b_z$	(18)
	$\textstyle =$	$\displaystyle a_x b_x + a_y b_y + a_z b_z.$

Thus, ${\bf a} \cdot {\bf b}$ is invariant under rotation about the

-axis. It can easily be shown that it is also invariant under rotation about the

- and

-axes. Clearly, ${\bf a} \cdot {\bf b}$ is a true scalar, so the above definition is a good one. Incidentally, ${\bf a} \cdot {\bf b}$ is the only simple combination of the components of two vectors which transforms like a scalar. It is easily shown that the dot product is commutative and distributive:

$\displaystyle {\bf a} \cdot {\bf b}$	$\textstyle =$	$\displaystyle {\bf b} \cdot {\bf a},$
$\displaystyle {\bf a}\cdot({\bf b}+{\bf c})$	$\textstyle =$	$\displaystyle {\bf a} \cdot {\bf b} + {\bf a}\cdot {\bf c}.$	(19)

The associative property is meaningless for the dot product, because we cannot have $({\bf a}\cdot{\bf b}) \cdot{\bf c}$ , since ${\bf a} \cdot {\bf b}$ is scalar.

We have shown that the dot product ${\bf a} \cdot {\bf b}$ is coordinate independent. But what is the physical significance of this? Consider the special case where ${\bf a} = {\bf b}$ . Clearly,

$\begin{displaymath} {\bf a} \cdot {\bf b} = a_x^{ 2}+a_y^{ 2} + a_z^{ 2} = {\rm Length} (OP)^2, \end{displaymath}$

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if ${\bf a}$ is the position vector of

relative to the origin

. So, the invariance of ${\bf a} \cdot {\bf a}$ is equivalent to the invariance of the length, or magnitude, of vector ${\bf a}$ under transformation. The length of vector ${\bf a}$ is usually denoted $\vert a\vert$ (``the modulus of

'') or sometimes just

, so

$\begin{displaymath} {\bf a} \cdot {\bf a} = \vert a\vert^2 = a^2. \end{displaymath}$

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**Figure 5:**
$\begin{figure} \epsfysize =1.75in \centerline{\epsffile{fig5.eps}} \end{figure}$

Let us now investigate the general case. The length squared of

(see Fig. 5) is

$\begin{displaymath} ({\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a} ) = \vert a\vert^2 + \vert b\vert^2 - 2 {\bf a} \cdot {\bf b}. \end{displaymath}$

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However, according to the ``cosine rule'' of trigonometry,

$\begin{displaymath} (AB)^2 = (OA)^2 + (OB)^2 - 2 (OA) (OB) \cos\theta, \end{displaymath}$

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where

denotes the length of side

. It follows that

$\begin{displaymath} {\bf a} \cdot {\bf b} = \vert a\vert \vert b\vert \cos\theta. \end{displaymath}$

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Clearly, the invariance of ${\bf a} \cdot {\bf b}$ under transformation is equivalent to the invariance of the angle subtended between the two vectors. Note that if ${\bf a} \cdot {\bf b} =0$ then either $\vert a\vert=0$ , $\vert b\vert=0$ , or the vectors $\bf a$ and $\bf b$ are perpendicular. The angle subtended between two vectors can easily be obtained from the dot product:

$\begin{displaymath} \cos\theta = \frac{{\bf a} \cdot {\bf b}}{\vert a\vert \vert b\vert }. \end{displaymath}$

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The work performed by a constant force $\bf F$ moving an object through a displacement $\bf r$ is the product of the magnitude of $\bf F$ times the displacement in the direction of $\bf F$ . If the angle subtended between $\bf F$ and $\bf r$ is $\theta$ then

$\begin{displaymath} W = \vert F\vert (\vert r\vert \cos\theta) = {\bf F}\cdot {\bf r}. \end{displaymath}$

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The rate of flow of liquid of constant velocity $\bf v$ through a loop of vector area $\bf S$ is the product of the magnitude of the area times the component of the velocity perpendicular to the loop. Thus,

$\begin{displaymath} {\rm Rate of flow} = {\bf v}\cdot{\bf S}. \end{displaymath}$

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Next: The vector product Up: Vectors Previous: Vector areas

Richard Fitzpatrick 2006-02-02