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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,

$\displaystyle {\bf a}\,\%\,{\bf b} \equiv a_x \,b_z + a_y \,b_x + a_z \,b_y = {\rm scalar~number},$ (A.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:

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

$\displaystyle {\bf a} \cdot {\bf b}$ $\displaystyle = (a_x\, \cos\theta+a_y\,\sin\theta)\,(b_x\,\cos\theta + b_y\,\sin\theta)$    
  $\displaystyle \phantom{=}+(-a_x\,\sin\theta + a_y\,\cos\theta)\,(-b_x\,\sin \theta + b_y\,\cos\theta) +a_z\, b_z$    
  $\displaystyle = a_x\, b_x + a_y\, b_y + a_z\, b_z$ (A.25)

in the new coordinate system. Thus, $ {\bf a} \cdot {\bf b}$ is invariant under rotation about $ Oz$ . It is easily demonstrated 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 definition (A.24) 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 readily shown that the dot product is commutative and distributive: that is,

$\displaystyle {\bf a} \cdot {\bf b}$ $\displaystyle = {\bf b} \cdot {\bf a},$    
$\displaystyle {\bf a}\cdot({\bf b}+{\bf c})$ $\displaystyle = {\bf a} \cdot {\bf b} + {\bf a}\cdot {\bf c}.$ (A.26)

The associative property is meaningless for the dot product, because we cannot have $ ({\bf a}\cdot{\bf b}) \cdot{\bf c}$ , as $ {\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 property? In the special case where $ {\bf a} = {\bf b}$ , we get

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

$\displaystyle ({\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a} ) = \vert{\bf a}\vert^{\,2} + \vert{\bf b}\vert^{\,2} - 2\,{\bf a} \cdot {\bf b}.$ (A.28)

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

$\displaystyle (AB)^2 = (OA)^2 + (OB)^2 - 2 \,(OA)\,(OB)\,\cos\theta,$ (A.29)

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

$\displaystyle {\bf a} \cdot {\bf b} = \vert{\bf a}\vert\, \vert{\bf b}\vert\, \cos\theta.$ (A.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:

$\displaystyle \cos\theta = \frac{{\bf a} \cdot {\bf b}}{\vert{\bf a}\vert \,\vert{\bf b}\vert }.$ (A.31)

Figure A.6: A vector triangle.
\epsfysize =1.5in

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$ times the displacement in the direction of $ \bf F$ . If the angle subtended between $ \bf F$ and $ \bf r$ is $ \theta $ then

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

next up previous
Next: Vector Area Up: Vectors and Vector Fields Previous: Coordinate Transformations
Richard Fitzpatrick 2016-03-31