In appendix A of Griffiths' Electrodynamics, he sketches proofs of certain theorems of vector calculus.

My question is related to this question, indeed the latter concerns the exact passage of the book my question refers to. Yet I would like to revisit it in more detail here.

First let me show his notation before I ask my question.

"For the sake of generality", he uses arbitrary (orthogonal) curvilinear coordinates $(u,v,w)$.

$\hat{u}$,$\hat{v}$, and $\hat{w}$ are the three unit vectors pointing in the direction of the increase of the corresponding coordinates.

Any vector can be expressed in terms of $\hat{u}$,$\hat{v}$, and $\hat{w}$. In particular, the infinitesimal displacement vector from $(u,v,w)$ to $(u+du,v+dv,w+dw)$ can be written


where $f$, $g$, and $h$ are functions of position, specific to whatever the particular coordinate system is.

Now suppose we have a vector function $A(u,v,w)=A_u\hat{u}+A_v\hat{v}+A_w\hat{w}$ and we wish to evaluate the surface integral $\oint_S \vec{A} \cdot d\vec{a}$, where $S$ is the infinitesimal volume generated by starting at $(u,v,w)$ and increasing each of the coordinates in succession by an infinitesimal amount, as in the following picture:

Griffiths' Electrodynamics, Appendix A

He calls the infinitesimal volume

$$d\tau = dl_udl_vdl_w=(fgh)dudvdw$$

Then he investigates what $\vec{A} \cdot d\vec{a}$ is for two faces of the volume in the figure above: the one facing the reader, and the one on the opposite side, facing into the screen.

For the front surface,


Note that more accurate would be to write



$$\vec{A}\cdot \vec{a_{front}} = -g(u,v,w)h(u,v,w)A_u(u,v,w)dvdw$$

The calculation for the opposite face is similar:


$$\vec{A}\cdot \vec{a_{front}} = g(u+du,v,w)h(u+du,v,w)A_u(u+du,v,w)dvdw$$

Now to the part that concerns my question:

Griffith now mentions that for any differentiable function $F(u)$, in the limit when $du \to 0$,

$$F(u+du)-F(u)=\frac{\partial F}{\partial u} du$$

Therefore, the front and back faces of the infinitesimal volume "together amount to a contribution":

$$\left [ \frac{\partial}{\partial u} (ghA_u) \right ] dudvdw= \frac{1}{fgh}\frac{\partial}{\partial v} (fhA_v)d\tau\tag{1}$$

How was $(1)$ obtained?


1 Answer 1


The equation in my edition of Griffiths (4th ed., printed in USA, Section A.4, page 577), which gives the contribution from the "front and back" of the infinitesimal parallelepiped, is not what you've written; it is $$ \left [ \frac{\partial}{\partial u} (ghA_u) \right ] dudvdw= \frac{1}{fgh}\frac{\partial}{\partial u} (ghA_u)d\tau $$ which follows because $d\tau = f g h \, du \, dv \, dw$. The quantity $$ \frac{1}{fgh}\frac{\partial}{\partial v} (fhA_v)d\tau, $$ given in the next line of the text, corresponds to the contributions from the "right and left" sides; it can be derived by applying similar logic to those sides.

As far as obtaining the left-hand side of your equation (1), it can be done by defining $F(u,v,w) \equiv g(u,v,w) h(u,v,w) A_u(u,v,w)$. Under this definition, we have $$ \frac{\partial F}{\partial u} du = g(u + du,v,w) h(u +du,v,w) A_u(u + du,v,w) \\- g(u,v,w) h(u,v,w) A_u(u,v,w) $$ and so the sum of $\vec{A}\cdot \vec{a}_{front}$ and $\vec{A}\cdot \vec{a}_{back}$ is equal to $$ \frac{\partial F}{\partial u} du \,dv\, dw = \frac{\partial (ghA_u)}{\partial u} du \,dv\, dw. $$

  • $\begingroup$ The difference you spotted in my question relative to the book is simply a typo on my part. The question still remains: how does Griffiths obtain the correct expression? $\endgroup$
    – xoux
    Mar 24, 2022 at 16:05
  • $\begingroup$ In particular, he seems to be adding $\vec{A} \cdot \vec{a_{front}}$ with $\vec{A} \cdot \vec{a_{back}}$. These look similar, but in fact have terms ($g$, $h$, $A_u$) that are evaluated at different points (in particular, $u$ vs $u+\Delta u$ difference). $\endgroup$
    – xoux
    Mar 24, 2022 at 16:09
  • $\begingroup$ @evianpring: See my edited answer. $\endgroup$ Mar 24, 2022 at 16:34

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