Monday, September 10, 2012

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:





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...

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