I'm working through Introduction to Electrodynamics by Griffiths, and there's a derivation of gradient, divergence, and curl in (orthogonal) curvilinear coordinates (Appendix A) which I don't entirely follow.

The derivation for divergence starts as follows. Consider a vector function $\mathbf{A}=A_u\mathbf{\hat{u}}+A_v\mathbf{\hat{v}}+A_w\mathbf{\hat{w}}$ and the solid found by starting at $(u,v,w)$ and changing each of the coordinates by $\text{d}u$, $\text{d}v,$ and $\text{d}w$ respectively.enter image description here

The volume of this solid is $\text{d}V = fgh \ \text{d}u \ \text{d}v \ \text{d}w$ where $f$, $g$, and $h$ are the functions that determine the infinitesimal displacement vector, i.e., $\text{d}\mathbf{l}=f\mathbf{\hat{u}} + g \mathbf{\hat{v}} + h \mathbf{\hat{w}}$. The flux out of the front and back surfaces respectively are $-A_u g h|_{u} \text{d}v\text{d}w$ and $-(A_ugh)|_{u+\text{d}u}\text{d}v\text{d}w$ so the total flux contributed by these two surfaces is $\frac{\partial A_ugh}{\partial u}\text{d}u\text{d}v\text{d}w$

There's a couple of questions I have with this part of the derivation.

  1. Firstly, how is it known that the normal vector from the back surface will be $\mathbf{\hat{u}}$ if the solid can 'curve' in an arbitrary manner? I understand that the unit vectors may be functions of position, but I don't see how that can ensure that the faces will always be orthogonal to $\mathbf{\hat{u}}$ (it would be true if all surfaces were planes, but there's no reason for that to be true in general coordinates).
  2. Furthermore, how can $A_u$ be considered constant on each of the surfaces, this seems to imply it can be considered constant over distances $\text{d}v$ and $\text{d}w$ (along the face) but can vary significantly over the distance $\text{d}u$ (between the faces).
  3. Also, how can the difference in length for each of the sides be neglected when calculating the volume, but matter when finding the flux out of the surfaces? These two calculations seem inconsistent with each other. In general, is there a criterion for how accurate these approximations have to be when deriving formulas like this?

Any insight would be appreciated.


1 Answer 1


Griffiths' calculations are a formal manipulation that can be done more rigorously through other approaches.

For the first issue, he seems to be defining the solid in this way. The vectors are orthogonal at every point, but, as you mentioned, they changed from point to point. The surface he has constructed has sides of constant values of $u$, $v$, and $w$, and as a consequence one of the unit vectors is always constant to each surface. Think in spherical coordinates, for an example: the vector $\mathbf{\hat{r}}$ is always orthogonal to a sphere.

For the second issue, he is considering the difference because in here the difference is the leading order term. In other words, if he did not consider it, the expression would vanish. For an example, suppose you are considering the expression $f(x + \epsilon) + f(x)$ for $\epsilon$ quite small. Taylor expanding, we have \begin{align} f(x + \epsilon) + f(x) &= f(x) + f(x) + \epsilon f'(x) + \mathcal{O}(\epsilon^2), \\ \approx 2f(x), \end{align} for the corrections $\epsilon f'(x) + \mathcal{O}(\epsilon^2)$ are being assumed to be much smaller than $f(x)$. However, consider now $f(x + \epsilon) - f(x)$. This time, we have \begin{align} f(x + \epsilon) - f(x) &= f(x) - f(x) + \epsilon f'(x) + \mathcal{O}(\epsilon^2), \\ \approx 0 + \epsilon f'(x), \end{align} and we keep the $\epsilon$ term because it is now the leading term. In both computations we are keeping the most important term, while dropping along the way those terms which are just small corrections (rigorously, they would vanish when taking the appropriate limits).

The third issue is, deep down, just the same as the second one, but written in a slightly different manner.

In all cases, the weirdness in these computations comes from the formal manipulation. See also this answer I've written a while ago to a similar issue in a similar topic. To obtain the expressions with less mathematical awkwardness, one could use the differential geometric expressions for the differential operators and work their way around the exterior derivatives, Christoffel symbols, and so on. This is actually a cool exercise in Differential Geometry, but it does require a bunch of more complicated Math. Griffiths' approach is a bit rough and depends more on the awkward formal manipulations, but gets to the same conclusions through an easier route.

  • $\begingroup$ Thanks for the answer, considering the quantities in terms of the leading terms makes the manipulations make much more sense. When considering the unit vector as the normal vector to the face, is the idea that any variations in the face that would make this not true are higher order, meaning that they vanish in the limit? $\endgroup$
    – QED
    Dec 25, 2021 at 22:39
  • 1
    $\begingroup$ @QED this bit is slightly different. Griffiths uses a particular choice of solid that is exactly normal to the unit vectors (I suggest drawing how this solid would look like in cylindrical and/or spherical coordinates to get a grasp). He then goes on to construct an arbitrary solid by gluing together pieces of that same original shape, similarly to how we split arbitrary solids into rectangles when integrating in Cartesian coordinates. The limit is taken in this second part of the process $\endgroup$ Dec 25, 2021 at 22:46
  • 1
    $\begingroup$ I guess we can think of the changes as being of higher order, but there is this subtle difference in which he starts with a solid which is identically normal to the vectors to after use these as pieces to building an arbitrary solid. However, this is quite similar to what we already do in Cartesian $\endgroup$ Dec 25, 2021 at 22:47
  • $\begingroup$ How is it known that the surface will be orthogonal to the unit vectors? I understand how it works in the case of cylindrical and spherical coordinates, but I don't see what part of this set up ensures it works in a general set of coordinates. $\endgroup$
    – QED
    Dec 25, 2021 at 22:50
  • 1
    $\begingroup$ @QED Griffiths defines $\mathbf{\hat{u}}$, for example, to the the unique unit vector pointing in the direction of greatest increase of the value of $u$, i.e., if we consider $u$ as a function on space, we have $\mathbf{\hat{u}} \parallel \mathbf{\nabla} u$. It holds that the gradient is always normal to the surface levels of a function. The faces of the volume element are surface levels of the functions (one of the faces has a constant value of $u = u_0$, the opposite one has $u = u_0 + \text{d}u$, and so on) and hence $\mathbf{\hat{u}}$ is built to be orthogonal to two of the faces $\endgroup$ Dec 25, 2021 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.