How can the curl be calculated in polar or spherical coordinates?

[Amok]

Can anyone show me how you get the curl in polar or spherical coordinates starting from the definitions in cartesian coordianates? I haven't been able to do this.

 

[I like Serena]

Well, curl in spherical coordinates is for instance available here:
http://en.wikipedia.org/wiki/Nabla_in_cylindrical_and_spherical_coordinates

Or do you really need to derive them from cartesian?

 

[Amok]

Or do you really need to derive them from cartesian?

Yes, I've never seen that done rigorously anywhere, unfortunately.

 

[I like Serena]

Yes, I've never seen that done rigorously anywhere, unfortunately.

Well, someone worked it out here:
http://nl.wikipedia.org/wiki/Nabla_in_verschillende_assenstelsels



Alternately, you can apply the definition of curl

to derive it.

To do it, you need to define rectangular curves matching with the coordinates of infinitesimal size, and work it out. This starts with a good drawing of what cylindrical and spherical coordinates look like.

This is the easiest, shortest, and most elegant way.
However, I haven't seen someone do it that way on the internet yet.

 

[Amok]

Well, someone worked it out here:
http://nl.wikipedia.org/wiki/Nabla_in_verschillende_assenstelsels



Alternately, you can apply the definition of curl

to derive it.

To do it, you need to define rectangular curves matching with the coordinates of infinitesimal size, and work it out. This starts with a good drawing of what cylindrical and spherical coordinates look like.

This is the easiest, shortest, and most elegant way.
However, I haven't seen someone do it that way on the internet yet.

But is there a general way to go about this? I mean, I could ask the same question about the gradient. Do I have to start drawing stuff every time?

 

[I like Serena]

But is there a general way to go about this? I mean, I could ask the same question about the gradient. Do I have to start drawing stuff every time?

Ah, but this is a general way. That is the beauty of it.
You have to draw it only once, and then you can derive the gradient, the divergence, and the curl.
This works for any coordinate system. Note that curl is the most work.

You could try cartesian coordinates first to get the hang of it.
Then cylindrical, and if you get that, spherical.[EDIT]Note that, as I said before, someone did the derivations on this page:
http://nl.wikipedia.org/wiki/Nabla_i..._assenstelsels

That is, without drawing any pictures, but just doing the math.
Perhaps you didn't see, because the derivation is done below the table.[/EDIT]

 

[Amok]

Ah, but this is a general way. That is the beauty of it.
You have to draw it only once, and then you can derive the gradient, the divergence, and the curl.
This works for any coordinate system. Note that curl is the most work.

You could try cartesian coordinates first to get the hang of it.
Then cylindrical, and if you get that, spherical.

Yes, I get it, thank you. Anyway, divergence and laplacian are very easy to derive. They are not vector quantities so it's not harder than computing a jacobian.

 

[I like Serena]

Yes, I get it, thank you. Anyway, divergence and laplacian are very easy to derive. They are not vector quantities so it's not harder than computing a jacobian.

Well, if you look at curl one component at a time, it's not a vector quantity either.
And that is the way to do it.

 

[Amok]

Well, if you look at curl one component at a time, it's not a vector quantity either.
And that is the way to do it.

I don't think you get the right results because you don't account for change in direction. But I have to try this again (last time I tried, was a when I took calculus).