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Could anyone please define the physical meaning of [Laplacian(f) = 0; f is a potential function of a vector field] ..

I don't know whether it's easy or not, but I'm a noob in vector analysis, so I thought I'd better ask :)

Regards,

MTarek

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- Thread starter mtarek16
- Start date

- #1

- 10

- 0

Could anyone please define the physical meaning of [Laplacian(f) = 0; f is a potential function of a vector field] ..

I don't know whether it's easy or not, but I'm a noob in vector analysis, so I thought I'd better ask :)

Regards,

MTarek

- #2

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- 14

- #3

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When the laplacian is zero you get the n-dimensional laplace equation.

http://en.wikipedia.org/wiki/Laplace's_equation

- #4

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Oops, my bad. Ignore everything I said.

- #5

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If the laplacian is zero, that means that f(p) is equal to the local average of f. Imagine f as a 2D surface, and (px,py,f(p)) is a point on the surface - If the surface is curving up in one direction around p, it must be curving down in another direction. Thus functions where laplacian f is zero everywhere are ones where every point looks like a saddle point.

As a result, when the laplacian is zero, f can have no local maxima or minima - if f had a local maxima at q, then f(q) > average of f around q, which would make the laplacian nonzero.

- #6

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maze, I don't quite get it, but I understand a little bit of what you're getting at. That's satisfactory for now .. finals start in two days so I will investigate in this later.

Would appreciate it though if you, or anyone else could provide a graphical example or something to clarify it more.

- #7

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http://www.math.hmc.edu/~jacobsen/sirev-flat-as-possible.pdf

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