These are notes on how to express the Laplacian in spherical coordinates.
Here is the radius from the origin, is the angle from the -axis and is the azimuthal angle from the -axis. The map is a one-to-one invertible map from onto minus the -axis. The mapping is singular on the -axis since the value of is undetermined there.
If we use (1) to express in terms of , then the chain rule leads to the identities:
In principle we could use these identities to express the Laplacian in terms of by first inverting the formulae to obtain , , , in terms of , , etc. and then squaring. However, this computation is boring and not very enlightening. There is a more geometric approach.
To obtain this approach it is useful to think first about the gradient operator We are used to thinking of the gradient of a function as the vector field
where , , are unit vectors in the directions of the positive , and axes respectively. However, it is also useful to think of somewhat differently. Specifically given and in , gives the directional derivative of at in direction , i.e.,
That is , , is an orthonormal frame. It is a moving frame because the choice of frame depends on our position in configuration space.
Since , , is an orthonormal frame, it follows that
In other words, we can express the gradient operator as follows
However in expanding this expression we need to be mindful of the fact that , , is a moving frame and thus certain derivatives of these vectors are non-zero:
Consider, for example, the term When this acts on a function we have, by the product rule,
since and . Thus
When expanded fully, the right hand side to (4) has nine terms:
Adding these up leads to the expression
Observing that and this can be re-expressed as