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Vector Algebra

Figure 1: A directed line element.
\epsfysize =1.5in
In applied mathematics, physical quantities are (predominately) represented by two distinct classes of objects. Some quantities, denoted scalars, are represented by real numbers. Others, denoted vectors, are represented by directed line elements in space: e.g., $\stackrel{\displaystyle \rightarrow}{PQ}$ in see Fig. 1. Note that line elements (and, therefore, vectors) are movable, and do not carry intrinsic position information: i.e., in Fig. 2, $\stackrel{\displaystyle \rightarrow}{PS}$ and $\stackrel{\displaystyle \rightarrow}{QR}$ are considered to be the same vector. In fact, vectors just possess a magnitude and a direction, whereas scalars possess a magnitude but no direction. By convention, vector quantities are denoted by bold-faced characters (e.g., ${\bf a}$) in typeset documents. Vector addition can be represented using a parallelogram: e.g., $\stackrel{\displaystyle \rightarrow}{PR}
\stackrel{\displaystyle \rightarrow}{PQ}+ \stackrel{\displaystyle \rightarrow}{QR}$ in Fig. 2. $\stackrel{\displaystyle \rightarrow}{PR}$ is said to be the resultant of $\stackrel{\displaystyle \rightarrow}{PQ}$ and $\stackrel{\displaystyle \rightarrow}{QR}$. Suppose that ${\bf a}\equiv \,\stackrel{\displaystyle \rightarrow}{PQ}\,\equiv\,\stackrel{\displaystyle \rightarrow}{SR}$, ${\bf b} \,\equiv \,\stackrel{\displaystyle \rightarrow}{QR}\,\equiv
\,\stackrel{\displaystyle \rightarrow}{PS}$, and ${\bf c} \equiv
\,\stackrel{\displaystyle \rightarrow}{PR}$. It follows, from Fig. 2, that vector addition is commutative: i.e., ${\bf a} + {\bf b} = {\bf b}+{\bf a}$ (since $\stackrel{\displaystyle \rightarrow}{PR}$ is also the resultant of $\stackrel{\displaystyle \rightarrow}{PS}$ and $\stackrel{\displaystyle \rightarrow}{SR}$). It can also be shown that the associative law holds: i.e., ${\bf a} +
({\bf b} + {\bf c}) = ({\bf a} + {\bf b}) + {\bf c}$.
Figure 2: Vector addition.
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There are two general approaches to vector analysis. The geometric approach is based on drawing line elements in space, and then making use of the theorems of Euclidian geometry. The coordinate approach assumes that space is defined by Cartesian coordinates, and uses these to characterize vectors. In Physics, we generally adopt the second approach, because it is far more convenient.

In the coordinate approach, a vector is denoted as the row matrix of its components along each of the Cartesian axes (the $x$-, $y$-, and $z$-axes, say):

{\bf a} \equiv (a_x,\, a_y,\, a_z).
\end{displaymath} (1)

Here, $a_x$ is the $x$-coordinate of the ``head'' of the vector minus the $x$-coordinate of its ``tail,'' etc. If ${\bf a} \equiv (a_x, a_y, a_z)$ and ${\bf b} \equiv (b_x, b_y, b_z)$ then vector addition is defined
{\bf a} + {\bf b} \equiv (a_x+b_x,\, a_y+b_y,\, a_z+b_z).
\end{displaymath} (2)

If ${\bf a}$ is a vector and $n$ is a scalar then the product of a scalar and a vector is defined
n\,{\bf a} \equiv (n\,a_x,\, n\,a_y,\, n\,a_z).
\end{displaymath} (3)

The vector $n\,{\bf a}$ is interpreted as a vector which points in the same direction as ${\bf a}$ (or in the opposite direction, if $n<0$), and is $\vert n\vert$ times as long as ${\bf a}$. It is clear that vector algebra is distributive with respect to scalar multiplication: i.e., $n\,({\bf a} + {\bf b}) = n\,{\bf a} + n\,{\bf b}$.

Unit vectors can be defined in the $x$-, $y$-, and $z$-directions as ${\bf e}_x \equiv (1,0,0)$, ${\bf e}_y \equiv (0,1,0)$, and ${\bf e}_z \equiv
(0,0,1)$. Any vector can be written in terms of these unit vectors: i.e.,

{\bf a} = a_x\, {\bf e}_x + a_y \,{\bf e}_y + a_z\,{\bf e}_z.
\end{displaymath} (4)

In mathematical terminology, three vectors used in this manner form a basis of the vector space. If the three vectors are mutually perpendicular then they are termed orthogonal basis vectors. However, any set of three non-coplanar vectors can be used as basis vectors.

Examples of vectors in Physics are displacements from an origin,

{\bf r} = (x,\, y,\, z),
\end{displaymath} (5)

and velocities,
{\bf v} = \frac{d {\bf r}}{dt} = \lim_{\delta t\rightarrow 0}
\frac{ {\bf r}(t+\delta t) - {\bf r}(t) }{\delta t}.
\end{displaymath} (6)

Figure 3: Rotation of the basis about the $z$-axis.
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Suppose that we transform to a new orthogonal basis, the $x'$-, $y'$-, and $z'$-axes, which are related to the $x$-, $y$-, and $z$-axes via a rotation through an angle $\theta$ around the $z$-axis--see Fig. 3. In the new basis, the coordinates of the general displacement ${\bf r}$ from the origin are $(x', y', z')$. These coordinates are related to the previous coordinates via the transformation
$\displaystyle x'$ $\textstyle =$ $\displaystyle x\cos\theta + y\sin \theta,$ (7)
$\displaystyle y'$ $\textstyle =$ $\displaystyle -x\sin\theta + y\cos\theta,$ (8)
$\displaystyle z'$ $\textstyle =$ $\displaystyle z.$ (9)

Now, we do not need to change our notation for the displacement in the new basis. It is still denoted ${\bf r}$. The reason for this is that the magnitude and direction of ${\bf r}$ are independent of the choice of basis vectors. The coordinates of ${\bf r}$ do depend on the choice of basis vectors. However, they must depend in a very specific manner [i.e., Eqs. (7)-(9)] which preserves the magnitude and direction of ${\bf r}$.

Since any vector can be represented as a displacement from an origin (this is just a special case of a directed line element), it follows that the components of a general vector ${\bf a}$ must transform in an similar manner to Eqs. (7)-(9). Thus,

$\displaystyle a_{x'}$ $\textstyle =$ $\displaystyle a_x\cos\theta + a_y\sin \theta,$ (10)
$\displaystyle a_{y'}$ $\textstyle =$ $\displaystyle -a_x\sin\theta + a_y\cos\theta,$ (11)
$\displaystyle a_{z'}$ $\textstyle =$ $\displaystyle a_z,$ (12)

with analogous transformation rules for rotation about the $y$- and $z$-axes. In the coordinate approach, Eqs. (10)-(12) constitute the definition of a vector. The three quantities ($a_x$, $a_y$, $a_z$) are the components of a vector provided that they transform under rotation like Eqs. (10)-(12). Conversely, ($a_x$, $a_y$, $a_z$) cannot be the components of a vector if they do not transform like Eqs. (10)-(12). Scalar quantities are invariant under transformation. Thus, the individual components of a vector ($a_x$, say) are real numbers, but they are not scalars. Displacement vectors, and all vectors derived from displacements, automatically satisfy Eqs. (10)-(12). There are, however, other physical quantities which have both magnitude and direction, but which are not obviously related to displacements. We need to check carefully to see whether these quantities are vectors.

next up previous
Next: Vector Area Up: Vectors Previous: Vectors
Richard Fitzpatrick 2007-07-14