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 “percent” product,

$${\bf a}\,\%\,{\bf b} \equiv a_x \,b_z + a_y \,b_x + a_z \,b_y = {\rm scalar~number},$$ (1.23)

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 ${\bf a} \equiv (0,\,1,\,0)$ and ${\bf b} \equiv (1,\,0,\,0)$. It is easily seen that ${\bf a}\,\%\,{\bf b}= 1$. Let us now rotate the coordinate axes through $45^\circ$ about $Oz$. In the new coordinate system, ${\bf a} \equiv (1/\sqrt{2},\, 1/\sqrt{2},\, 0)$ and ${\bf b} \equiv (1/\sqrt{2},\,- 1/\sqrt{2},\, 0)$, giving ${\bf a}\,\%\,{\bf b} = 1/2$. Clearly, ${\bf a}\,\%\,{\bf b}$ is not invariant under rotational transformation, so the previous definition is a bad one.

Consider, now, the dot product or scalar product:

$${\bf a} \cdot {\bf b} \equiv a_x \,b_x + a_y \,b_y + a_z \,b_z = {\rm scalar~number}.$$ (1.24)

Let us rotate the coordinate axes though $\theta$ degrees about $Oz$. According to Equations (A.20)–(A.22), ${\bf a} \cdot {\bf b}$ takes the form

$${\bf a} \cdot {\bf b} = (a_x\, \cos\theta+a_y\,\sin\theta)\,(b_x\,\cos\theta + b_y\,\sin\theta)$$

$$\phantom{=}+(-a_x\,\sin\theta + a_y\,\cos\theta)\,(-b_x\,\sin \theta + b_y\,\cos\theta) +a_z\, b_z$$

$$= a_x\, b_x + a_y\, b_y + a_z\, b_z$$ (1.25)

in the new coordinate system. Thus, ${\bf a} \cdot {\bf b}$ is invariant under rotation about $Oz$. It can easily be shown that it is also invariant under rotation about $Ox$ and $Oy$. We conclude that ${\bf a} \cdot {\bf b}$ is a true scalar, and that the previous definition is a good one. Incidentally, ${\bf a} \cdot {\bf b}$ is the only simple combination of the components of two vectors that transforms like a scalar. It is easily shown that the dot product is commutative and distributive; that is,

$${\bf a} \cdot {\bf b} = {\bf b} \cdot {\bf a},$$

$${\bf a}\cdot({\bf b}+{\bf c}) = {\bf a} \cdot {\bf b} + {\bf a}\cdot {\bf c}.$$ (1.26)

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

We have shown that the dot product ${\bf a} \cdot {\bf b}$ is coordinate independent. But what is the geometric significance of this? Well, in the special case where ${\bf a} = {\bf b}$, we get

$${\bf a} \cdot {\bf b} = a_x^{\,2}+a_y^{\,2} + a_z^{\,2} = \vert{\bf a}\vert^2=a^{\,2}.$$ (1.27)

So, the invariance of ${\bf a} \cdot {\bf a}$ is equivalent to the invariance of the magnitude of vector ${\bf a}$ under transformation.

Let us now investigate the general case. The length squared of $AB$ in the vector triangle shown in Figure A.6 is

$$({\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a} ) = \vert{\bf a}\vert^{\,2} + \vert{\bf b}\vert^{\,2} - 2\,{\bf a} \cdot {\bf b}.$$ (1.28)

However, according to the “cosine rule” of trigonometry,

$$(AB)^2 = (OA)^2 + (OB)^2 - 2 \,(OA)\,(OB)\,\cos\theta,$$ (1.29)

where $(AB)$ denotes the length of side $AB$. It follows that

$${\bf a} \cdot {\bf b} = \vert{\bf a}\vert\, \vert{\bf b}\vert\, \cos\theta.$$ (1.30)

In this case, 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 $\vert{\bf a}\vert=0$, $\vert{\bf b}\vert=0$, or the vectors $\bf a$ and $\bf b$ are mutually perpendicular. The angle subtended between two vectors can easily be obtained from the dot product; in fact,

$$\cos\theta = \frac{{\bf a} \cdot {\bf b}}{\vert{\bf a}\vert \,\vert{\bf b}\vert }.$$ (1.31)

Figure A.6: A vector triangle.

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

$$W = \vert{\bf F}\vert \,(\vert{\bf r}\vert\,\cos\theta) = {\bf F}\cdot {\bf r}.$$ (1.32)

The work $dW$ performed by a non-constant force ${\bf f}$ that moves an object through an infinitesimal displacement $d{\bf r}$ in a time interval $dt$ is $dW={\bf f}\cdot d{\bf r}$. Thus, the rate at which the force does work on the object, which is usually referred to as the power, is $P = dW/dt = {\bf f}\cdot d{\bf r}/dt$, or $P={\bf f}\cdot {\bf v}$, where ${\bf v}=d{\bf r}/dt$ is the object's instantaneous velocity.