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Coordinate Systems

The following are straight up summaries of important relations for the three most important coordinate systems: Cartesian, Spherical Polar, and Cylindrical. I don't derive the various expressions, but in a few cases I indicate how one could do so.

Some of the following you should work to just ``learn'', so you know it forever. Other parts you can probably look up when you need them. I've tried to concentrate on the former here, and will likely provide a formula sheet with the latter for use on exams. However, you still have to learn to work with stuff off the formula sheets, and that takes practice.

The key to understanding (and most simply deriving) differential operators in all coordinate systems besides flat Euclidean Cartesian coordinates is the definition of the directed length element $ d\vell$ . In the most general terms, for a coordinate frame with orthonormal components of a point $ P = (u,v,w)$ , $ d\vell$ at the point $ P$ is given by:

$\displaystyle d\vell = f\ du \hu + g\ dv \hv + h\ dw \hw $

where $ f$ , $ g$ , and $ h$ are functions of $ (u,v,w)$ evaluated at the point $ P$ . A coordinate system is characterized by the three functions. For example:

$\displaystyle f = g = h = 1$ $\displaystyle \quad\quad$ $\displaystyle \textrm{Cartesian}(x,y,z)$  
$\displaystyle f = 1, g = r, h = r\sin\theta$ $\displaystyle \quad\quad$ $\displaystyle \textrm{Spherical
$\displaystyle f = h = 1, g = s$ $\displaystyle \quad\quad$ $\displaystyle \textrm{Cylindrical}(s,\phi,z)$  

In order to do a gradient operator, we have to use the general expression for a total derivative of a scalar function $ T$ :

$\displaystyle dT = \grad T\cdot d\vell = (\nabla T)_u f\ du + (\nabla T)_v g\ dv +
(\nabla T)_w h\ dw $

where $ \displaystyle (\nabla T)_u = \frac{1}{f}\partialdiv{T}{u}$ , etc. The gradient is then given generally by:

$\displaystyle \grad = \hu \frac{1}{f}\partialdiv{}{u} +
\hv \frac{1}{g}\partialdiv{}{v} +
\hw \frac{1}{h}\partialdiv{}{w} $

where we have placed the unit vectors to the left to make it very clear that the partials do not act on them.

We need the Divergence Theorem to hold for our definition of divergence. To do the divergence of a vector valued function $ \vB = B_u
\hu + B_v \hv + B_w \hw$ expressed in curvilinear coordinates, we thus have to start with a volume element for the space:

$\displaystyle dV = d\tau = d\ell_u\ d\ell_v\ d\ell_w = (fgh) du\ dv\ dw $

where we have to use the components of $ d\vell$ because in general, the curvilinear coordinates may well not have dimensions of length (which is one of several things the functions $ f,g,h$ do for us in $ d\vell$ ). The boundary of this infinitesimal volume is a sort of deformed rectangular solid. Its directed surfaces are things like (in the $ \pm u$ direction):

$\displaystyle \hn\ dA = d\va = - gh\ dv\ dw\ \hu \quad\textrm{or}\quad + gh\ dv\ dw\
\hu $

for the back or front surfaces respectively. If we examine the flux through such a surface:

$\displaystyle \vB \cdot d\va = \pm (gh B_u)\ dv\ dw $

where the function $ F = gh B_u$ must be evaluated at either $ (u,v,w)$ (for (-)) or $ (u + du,v,w)$ (for (+)) respectively! Note well that these are both outward directed normal contributions! If we subtract these two in the direction of $ u$ and take the usual limits:

$\displaystyle F(u + du) - F(u) = \ddu{F}\ du $

or the total contribution to $ \vB \cdot d\va$ in this direction is:

$\displaystyle \left(\partialdiv{gh B_u}{u}\ du\right)\ dv\ dw =
...)}{u}\right) d\tau \frac{1}{(fgh)}
\left(\partialdiv{(gh B_u)}{u}\right) d\tau $

Summing over all six faces, we get for this infinitesimal curvilinear volume:

$\displaystyle \oint \vB \cdot d\va = \frac{1}{(fgh)}
\left[\left(\partialdiv{(g... +
\left(\partialdiv{(fg B_w)}{w}\right) \right]
d\tau = \deldot{\vB} d\tau $

We thus arrive at the general (local) differential form for the divergence of a vector valued function:

$\displaystyle \deldot \vB = \frac{1}{(fgh)}
\left[\left(\partialdiv{(gh\ B_u)}{...
...rtialdiv{(fh\ B_v)}{v}\right) +
\left(\partialdiv{(fg\ B_w)}{w}\right) \right]

This is, incidentally, a general derivation of the divergence theorem, since it is easy to show that one can build integrals over a general (non-local) volume with its bounding surface by summing over internal curvilinear chunks with cancellation of flux contributions across all infinitesimal surfaces inside the volume. For a finite volume $ \mathbb{V}$ , then:

$\displaystyle \oint_{\partial \mathbb{V}} \vB \cdot d\va = \int_\mathbb{V}
\deldot{\vB} d\tau $

where the non-cancelling part of the surface integral is only over the external surface $ \partial \mathbb{V}$ with an outward directed normal.

To do the curl, one has to repeat this argument for the formula:

$\displaystyle \oint \vB \cdot d\vell $

evaluated around an infinitesimal directed rectangular area $ d\va$ along each of the $ (u,v,w)$ directions. The algebra is tedious (but is reviewed in similar detail in Griffith's, Appendix A if ou want to see it) and leads to:

$\displaystyle \curl{\vB} = \frac{1}{gh} \left[ \partialdiv{(h B_w)}{v} -
\parti...{1}{fg} \left[ \partialdiv{(g B_v)}{u} -
\partialdiv{(f B_u)}{v} \right] \hw $

which applies to a general infinitesimal surface as $ \oint \vB \cdot
d\vell = \curl{\vB}\cdot d\va$ .

As before, if we chop a finite surface $ S$ bounded by a closed curve $ C$ up into many differential pieces, all of the internal loop contributions between adjacent infinitesimal pieces cancal and one gets Stoke's Theorem:

$\displaystyle \oint_C \vB \cdot d\vell = \int_{S/C} \curl{\vB}\cdot d\va $

This outlines how to evaluate the gradient, divergence and curl for the three primary coordinate systems we will use in this course. Below we summarize the result

next up previous contents
Next: Cartesian Up: Mathematical Physics Previous: Integration By Parts in   Contents
Robert G. Brown 2017-07-11