Mathematical Tools for Physics

9—Vector Calculus 1 259

In the limit that the all the∆x,∆y, and∆zshrink to zero the terms with a second derivative vanish, as do all
the other higher order terms. You are left then with a rather simple expression for the divergence.

divergence of~v= div~v=

∂vx

∂x

+

∂vy

∂y

+

∂vx

∂x

(11)

This is abbreviated by using the differential operator∇, “del.”

∇=xˆ

∂

∂x

+yˆ

∂

∂y

+ˆz

∂

∂z

(12)

Then you can write the preceding equation as

divergence of~v= div~v=∇.~v (13)

The symbol∇will take other forms in other coordinate systems.
Now that you’ve waded through this rather technical set of manipulations, is there an easier way? Yesbut,
without having gone through the preceding algebra you won’t be able to see and to understand which terms are
important and which terms are going to cancel or otherwise disappear. When you need to apply these ideas to
something besides rectangular coordinates you have to know what to keep and what to ignore. Once you know
this, you can go straight to the end of the calculation and write down only those terms that you know are going
to survive.This takes practice.

Simplifying the derivation
In the long derivation of the divergence, the essence is that you find~v.ˆnon one side of the box (maybe take it
in the center of the face), and multiply it by the area of that side. Do this on the other side, remembering that
ˆnisn’t in the same direction there, and combine the results. Do this for each side and divide by the volume of
the box.

[

vx(x 0 + ∆x,y 0 + ∆y/ 2 ,z 0 + ∆z/2)∆y∆z−vx(x 0 ,y 0 + ∆y/ 2 ,z 0 + ∆z/2)∆y∆z

]

÷

(

∆x∆y∆z

)

(14)

the∆yand∆zfactors cancel, and what’s left is, in the limit∆x→ 0 , the derivative∂vx/∂x.
I was careful to evaluate the values ofvxin the center of the sides, but you see that it didn’t matter. In
the limit as all the sides go to zero I could just as easily taken the coordinates at one corner and simplified the