Vector Area

Figure 4: A vector area.

Suppose that we have a plane surface of scalar area $S$. We can define a vector area ${\bf S}$ whose magnitude is $S$, and whose direction is perpendicular to the plane, in the sense determined by a right-hand grip rule on the rim--see Fig. 4. This quantity clearly possesses both magnitude and direction. But is it a true vector? Well, we know that if the normal to the surface makes an angle $\alpha_x$ with the $x$-axis then the area seen looking along the $x$-direction is $S\,\cos\alpha_x$. Let this be the $x$-component of ${\bf S}$. Similarly, if the normal makes an angle $\alpha_y$ with the $y$-axis then the area seen looking along the $y$-direction is $S\,\cos\alpha_y$. Let this be the $y$-component of ${\bf S}$. If we limit ourselves to a surface whose normal is perpendicular to the $z$-direction then $\alpha_x = \pi/2-\alpha_y=\alpha$. It follows that ${\bf S} = S\,(\cos\alpha,\, \sin\alpha,\, 0)$. If we rotate the basis about the $z$-axis by $\theta$ degrees, which is equivalent to rotating the normal to the surface about the $z$-axis by $-\theta$ degrees, then

$$S_{x'} = S\,\cos\,(\alpha-\theta) = S\,\cos\alpha\,\cos\thet... ...S\,\sin\alpha\,\sin\theta = S_x\,\cos\theta + S_y\,\sin\theta,$$ (13)

which is the correct transformation rule for the $x$-component of a vector. The other components transform correctly as well. This proves that a vector area is a true vector.

According to the vector addition theorem, the projected area of two plane surfaces, joined together at a line, looking along the $x$-direction (say) is the $x$-component of the resultant of the vector areas of the two surfaces. Likewise, for many joined-up plane areas, the projected area in the $x$-direction, which is the same as the projected area of the rim in the $x$-direction, is the $x$-component of the resultant of all the vector areas: i.e.,

$${\bf S} = \sum_i {\bf S}_i.$$ (14)

If we approach a limit, by letting the number of plane facets increase, and their areas reduce, then we obtain a continuous surface denoted by the resultant vector area

$${\bf S} = \sum_i \delta {\bf S}_i.$$ (15)

It is clear that the projected area of the rim in the $x$-direction is just $S_x$. Note that the vector area of a given surface is completely determined by its rim. So, two different surfaces sharing the same rim both possess the same vector area.

In conclusion, a loop (not all in one plane) has a vector area ${\bf S}$ which is the resultant of the vector areas of any surface ending on the loop. The components of ${\bf S}$ are the projected areas of the loop in the directions of the basis vectors. As a corollary, a closed surface has ${\bf S} = {\bf0}$, since it does not possess a rim.