Transform the equation

to polar coordinates , andspecialize the resulting equation to the case when u does not not depend on the angular variable theta.

Changing to polar.

How do I make u not depend on theta?

Printable View

- February 3rd 2011, 12:16 PMdwsmithTransform Polar
Transform the equation

to polar coordinates , and**specialize the resulting equation to the case when u does not not depend on the angular variable theta.**

Changing to polar.

How do I make u not depend on theta? - February 3rd 2011, 12:51 PMAckbeet
How about setting

- February 3rd 2011, 12:52 PMdwsmith
- February 3rd 2011, 12:53 PMAckbeet
Sure. If does not depend on it follows that and then of course,

- February 4th 2011, 02:15 PMJester
- February 4th 2011, 02:22 PMAckbeet
I think you would have

because you essentially have the Laplacian in two dimensions. If you ignore the z dimension, you can just copy down the version for cylindrical coordinates, which is what I did. - February 4th 2011, 03:28 PMdwsmith
- February 4th 2011, 03:31 PMAckbeet
Let me challenge you right back: how do you get

The units don't work for that expression. I don't buy it. - February 4th 2011, 03:34 PMAckbeet
See here for a derivation. You'll notice that they've multiplied out the product rule that I have not multiplied out.

- February 4th 2011, 03:55 PMdwsmith
- February 4th 2011, 04:10 PMAckbeet
Please provide a reference for the equation

Like I said before, I think this equation is incorrect. The units don't work out. For the first term, the and units cancel out, because they're both length. That leaves you with units of divided by units of length. On the second term, the and units cancel out, as before, but the units of are NOT units of divided by units of length. Therefore, you can't add the terms because they don't have the same units.

Quote:

On the site, they have

...

Operators are very slippery creatures. There's a HUGE difference between

and

Think of the first as an operator: it's going to look to its right and try to find something to operate on (the operation in question is partial differentiation with respect to phi). On the other hand, the second is not an operator, because the differentiation has already taken place. It's just a function.

Does that make sense? - February 4th 2011, 04:15 PMdwsmith
- February 4th 2011, 05:17 PMAckbeet
Hmm. If you look carefully at the link I gave in post # 9, you will see that

Note that

and

implying that

You can see that the units work out correctly in this expression. - February 4th 2011, 10:08 PMdwsmith
Starting from scratch we have:

I found my error.

I forgot to type r^2 for

Geez, guys where was the help in finding that mistake :p

Just kidding. Thanks. - February 4th 2011, 10:09 PMdwsmith