### An Intuitive Way to the Spherical Gradient and Laplacian

It's that time of year again when physics students everywhere are deriving the spherical and cylindrical del, nabla, gradient, or Laplacian operators.  Every derivation I saw prior to this week involved lots of algebra and the chain rule... even mine.  Fortunately for me, a comment on my derivation, and a homework assignment from Rutgers [pdf] led me to a far simpler and more intuitive way of doing things.  You just start from the differential displacement in a given coordinate system and go from there.

The differential displacement in spherical coordinates is:

The element in the r direction is easy to understand.  A small displacement along the r direction is represented as dr.  The theta and phi displacements might not be as obvious.  The graphic to the left illustrates what's going on.  With small displacement along the theta direction you're moving along a circle with radius r.  The distance you've moved is equal to the length of the arc which is equal to the radius of the circle, r times the small angle, hence, .  The element in the in the phi direction is similar to the element in the theta direction in that you're moving along a circle.  However, this time you have to keep in mind that the radius of the circle is not r.  The r vector is propped up by angle of theta and so, the projection of r on the r/phi plane is , and the distance traveled along the phi displacement is .
Once we have our differential displacement element, the rest is kind of easy.  First, we get the gradient operator.  The gradient is just the derivative of a function with respect to the differential displacement, so we get:

The next step is to derive the divergence in spherical coordinates.  First, we need the differential surface area element in all three dimensions.  I don't have a great picture for this yet, so you'll just have to visualize a small square on the surface of a sphere.  The surface area element in the r direction is a rectangle on the surface of a sphere with a length of  on one side and a length of  on the other.  So, the element of area is the product of the sides or  .  In the theta direction, the sides are a small displacement in the r direction and a small displacement in the phi direction giving us .  Finally, in the phi direction the sides of the rectangle are a small displacement along r and a small displacement along theta giving us , (.

You might be wondering what all the surface elements had to do with divergence.  This is where the tip I mentioned at the start of the post comes in.  Divergence is defined to be the flux of a vector field, (the vector field dot producted with the surface area), per unit volume.  Now that we have a differential surface area vector, we can dot product our vector of interest, (say A), with it to get the flux through a differential surface area.  We wind up with:

Now, we need the differential element of volume for spherical coordinates.  Once again, we can use purely geometric considerations to get where we want to go.  Visualize making a small cube out of the small rectangles we defined above for the surface area differentials.  If you took the volume of that cube, you'd just multiply the three sides:

Now to get the divergence, we just divide each term of the flux by the differential volume element.  For the r term we get:

the theta and phi differentials cancel out.  You can bring the sin theta term outside of the derivative and cancel it since it doesn't depend on r.

For the theta term we get:

The r and phi differentials cancel.  The r can be taken outside the derivative since it is constant with respect to the remaining theta differential.

Finally, for the phi term we get:

The r and theta differentials cancel and we can bring the r outside the remaining phi derivative.

All together, we now have the three terms of the divergence in spherical coordinates:

We didn't use the chain rule once, just simple geometry!

The final task is to derive the Laplacian.  The Laplacian is the divergence of the gradient.  We simply substitute our gradient result in place of the vector A in our divergence experession to get the following three terms:
r:

theta:

phi:

WillemHekman said…
Dear Blogger, first of thanks for this post, it got me started in the right direction.

But, unfortunately your way of deriving had two connected oversimplifications which I think to have solved:

First of, you need to account for the flux going through the box. To do so you dot the surface of ONE side with the vector. But here you do not account for the other side. A constant vector in the dot product with both sides would yield no net flux...

Solution: You actually skip a step where you should say that the difference in the vector A in the direction of the surface on opposite sides of the box is the infinitesimal difference dA between A(r) and A(r + dr) for instance.

The other problem was to me a manifestation of the previous simplification:

In calculating the end result you do the following (1/dr) x r^2 x Ar = (d/dr) x r^2 x Ar. This is a big oversimplication/ error to make.

Solution: Use the dA instead from before this way you`ll get 1/dr x r^2 x dA = d(r^2 x A)/dr.

I would really appreciate if you responded.
Hamilton Carter said…
Willem,
Thanks very much for all the info. I think you're correct and it clears up a bit of confusion I had regarding some of the steps. Thanks very much! I'm going to write more when I have time to review what you've said more completely. Thanks again!
Lau Tzehaygary said…
Sorry >__< when i try to use the equation for Flux/volume

i didnt get what you get in the equation

for example: (Ar r^2 sin(theta) d(theta) dphi)/(r^2 sin(pheta) dr d(theta) d(phi))

=Ar/(dr) ~~ what happen????(not the above one of d(Ar r^2)/dr * (1/r^2))

thank you so much

### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…

### Lab Book 2014_07_10 More NaI Characterization

Summary: Much more plunking around with the NaI detector and sources today.  A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building.  The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment?  Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts.  Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM

Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.

The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…

### Unschooling Math Jams: Squaring Numbers in their own Base

Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself.  Last week, we discovered that square of any number in its own base is 100!  Pretty cool!  As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all.  Here’s how it went.

I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1.  She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own.  This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.

"I can sing that song... about the letters? all by myself now!"  2 meant the alphabet song.  His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)

After 2 had worked his way through the so…