Not too surprisingly, I suppose, a homework problem in EM recently asked us to find the Laplacian in a different coordinate system. The system of choice this time was parabolic cylindrical coordinates. At first, I was unconcerned, I had my handy example on how to do this using the line element for the coordinate system[1]. As it turned out, that trick was disallowed by the assignment itself. We were required to use the chain rule. The parabolic coordinate system was defined as the set of equations to the left, (picture 1).

The chain rule of calculus for some odd reason has always been my arch-nemesis. This is particularly problematic for studying physics because we use it to exhaustion. I'd hoped to provide a more coherent presentation, but more midterms are coming up. For now I'll just capture the solution for finding the Laplacian with a few notes as to what can be learned, and the motivations for various steps.

**Step one**

Write down the u and v differentials in terms of x and y abstractly (picture 2):

The above is essentially the multi-variable chain rule. If you're me, and you can get to this step, the rest is pretty easy. One thing to notice here. As physicists, we're often fond of 'cancelling' differentials like fractions, (as a memory aid if for not other reason). Our mathematician buddies frequently caution us that this is just wrong. If you 'cancel' the partial x or partial y differentials above and add, you'll see an example of why you have to be careful with this.

**Step 2**

Calculate the x and y differentials with respect to u and v, (the missing pieces above). (picture 3)

**Step 3**

**Step 4**

I've done a horrible, horrible thing here, one of the things I hate the most. I've skipped steps somewhat wholesale. Do be fair, I did them in my head, so I know they can be done, but there you have it. I did it. (picture 5)

**Here's what we're up to. We want a Laplacian that we can apply to scalar fields in the parabolic cylindrical coordinate system. If you look at the last set of results above, you'll notice that all the u and v differentials are bound up behind other differential operators. That won't do. We need to be able in the end to be able to apply them directly to functions written in the u, v system. So, we collect our blind faith again and apply 'partial partial u' and 'partial partial v' calculated in step two to the x and y differentials. After that, all that remains is to collect like terms and make the whole mess look like a Laplacian. The z term gets tacked on a the end since it was orthogonal to the other two coordinates in each system and had not u or v dependence.**

Hopefully, there will be more on this later.

**References:**

1. http://copaseticflow.blogspot.com/2012/09/an-intuitive-way-to-spherical-gradient.html

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