I'm confused as to how does one find out the basis vectors of a curvilinear co-ordinate system. In the context of a general, arbitrary curvilinear co-ordinate system, the textbook I'm reading states that:

If Cartesian co-ordinates $x, y, z$ are expressible in terms of the three curvilinear coordinates $a, b, c$, then:

$$ \frac{1}{h_i} \frac{\partial \mathbf{r}}{\partial i}, $$

defines the $i^{th}$ basis vector, where $i = a, b, c$ and $h_i$ is the modulus of the relevant partial derivative..

I'm not sure what's the justification behind this way of defining the basis vectors. The textbook tries to justify it in the following manner:

The surfaces $a = m_1$, $b = m_2$ and $c = m_3$, where $m_1, m_2, m_3$ are constants, are called the coordinate surfaces and each pair of these surfaces has its intersection in a curve called a coordinate curve or line. If at each point in space the three coordinate surfaces passing through the point meet at right angles then the curvilinear coordinate system is called orthogonal. For example, in spherical polars $a = r, b = \theta, c = \phi$ and the three coordinate surfaces passing through the point $(R, \gamma, \beta)$ are the sphere $r = R$, the circular cone $\theta = \gamma$ and the plane $\phi = \beta$, which intersect at right angles at that point. Therefore spherical polars form an orthogonal coordinate system.

Also, the next paragraph makes the following statement and I'm not so sure about it as well:

If $r(a, b, c)$ is the position vector of the point $P$ then $\vec{a_1} = \frac{\partial \mathbf{r}}{\partial a}$ is a vector tangent to the $a$-curve at $P$ (for which $b$ and $c$ are constants) in the direction of increasing $a$.


$\newcommand{\dd}{\partial}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Sph}{\mathbf{r}}$A mathematician might denote spherical coordinates by $$ \left[\begin{array}{@{}c@{}} x \\ y \\ z \\ \end{array}\right] = \Sph(\rho, \theta, \phi) = \left[\begin{array}{@{}c@{}} \rho\cos\theta\sin\phi \\ \rho\sin\theta\sin\phi \\ \rho\cos\phi \\ \end{array}\right]. $$ The derivative $D\Sph$ is represented by the matrix of partial derivatives $$ \left[\begin{array}{@{}ccc@{}} \frac{\dd x}{\dd \rho} & \frac{\dd x}{\dd \theta} & \frac{\dd x}{\dd \phi} \\ \frac{\dd y}{\dd \rho} & \frac{\dd y}{\dd \theta} & \frac{\dd y}{\dd \phi} \\ \frac{\dd z}{\dd \rho} & \frac{\dd z}{\dd \theta} & \frac{\dd z}{\dd \phi} \\ \end{array}\right] = \left[\begin{array}{@{}c@{}} \cos\theta\sin\phi & -\rho\sin\theta\sin\phi & \rho\cos\theta\cos\phi \\ \sin\theta\sin\phi & \rho\cos\theta\sin\phi & \rho\sin\theta\cos\phi \\ \cos\phi & 0 & -\rho\sin\phi \\ \end{array}\right]. $$ The (normalized) columns $$ \frac{\dd\Sph}{\dd \rho} = D\Sph\, \Basis_{\rho},\qquad \frac{1}{\rho}\,\frac{\dd\Sph}{\dd \theta} = \frac{1}{\rho}\,D\Sph\, \Basis_{\theta},\qquad \frac{1}{\rho}\,\frac{\dd\Sph}{\dd \phi} = \frac{1}{\rho}\,D\Sph\, \Basis_{\phi} $$ are the "coordinate vector fields" for spherical coordinates.

Coordinate curves for spherical coordinates

In the diagram, the red and blue curves lie in a coordinate surface $\rho = \rho_{0}$ (i.e., a sphere); the blue and green curves lie in a coordinate surface $\theta = \theta_{0}$ (a longitudinal plane); the red and green curves lie in a coordinate surface $\phi = \phi_{0}$ (a cone about the $z$-axis). Small portions of the coordinates curves for $\rho$, $\theta$, and $\phi$ are green, red, and blue respectively. The coordinate fields (not shown) are unit tangent fields along the respective curves.

Generally, if $F$ is an arbitrary change of coordinates (in space, say), then the domain of $F$ is some open subset $U$ of $\Reals^{3}$, and $F$ maps $U$ "diffeomorphically" (smoothly and with smooth inverse) into $\Reals^{3}$. If we write $$ (x, y, z) = F(u, v, w), $$ then for each $u_{0}$ we may "restrict" $F$, obtaining a parametric surface $(v, w) \mapsto F(u_{0}, v, w)$. The image of this mapping is the "coordinate surface" $u = u_{0}$.

Similarly, we might consider $(u, w) \mapsto F(u, v_{0}, w)$ for some $v_{0}$, or $(u, v) \mapsto F(u, v, w_{0})$ for some $w_{0}$. The intersection of two coordinate surfaces, say $v = v_{0}$ and $w = w_{0}$, is the "coordinate curve" $u \mapsto F(u, v_{0}, w_{0})$. This parametric curve has a normalized velocity vector at each point, the $u$-coordinate vector field along the curve.

Incidentally, the notation suggests you're reading an engineering text. One cultural difference between engineers and mathematicians is that:

  • Engineers tend to denote mappings (functions) by assigning letters to input and output values, which can lead to profusions of letters (as in $a, b, c$, $m_{1}, m_{2}, m_{3}$, $r, \theta, \phi$, $R, \gamma, \beta$).

  • Mathematicians tend to focus on the functional relationship between inputs and outputs, which sometimes goes so far as to actively suppress the names of input and output variables (as in $(x, y, z) = \Sph(\rho, \theta, \phi)$, and speaking of $D\Sph$ instead of the partial derivatives $\dd x/\dd \rho$, etc.)

Among other things, engineering notation tends to be global: Each quantity has a single symbol, sometimes across an entire discipline.

By contrast, mathematical notation tends to be local and context-dependent: The meanings of symbols can change even from paragraph to paragraph, though symbols are chosen to obey loose cultural assumptions. (The beginning of the alphabet ($a$ through $c$) is generally reserved for constants, the middle ($i$ through $n$) for discrete (integer) indices, the late middle ($r$ through $t$ or $w$) for continuous parameters, the end ($x$ through $z$) for coordinate functions.)

Each notational convention has advantages and disadvantages. The better you're able to understand both (even if you spend most of your career in one camp or the other), the easier a time you'll have with the literature (textbooks and papers).

  • 1
    $\begingroup$ The normalized middle column is missing a factor of $1/\sin\phi$. $\endgroup$ – amd Apr 7 '18 at 7:55

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