The Physical Meaning of Laplacian(f) = 0 in Vector Analysis Explained - MTarek

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The discussion centers on the physical meaning of the equation Laplacian(f) = 0, where f represents a potential function in vector analysis. It is clarified that when the Laplacian is zero, the function f at a point equals the local average of f in its vicinity, indicating that the function exhibits no local maxima or minima. This situation suggests that the function behaves like a saddle point, curving up in one direction and down in another. The conversation also touches on the distinction between the Laplacian and the Laplace transform, emphasizing their different applications. Overall, understanding this concept is essential for grasping the behavior of functions in multi-dimensional spaces.
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Hello all,

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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What function has 0 as its Laplace transform? Well, f=0 will work, and the inverse transforms are unique, so f=0 is the only solution. Or am I totally wrong?
 
Oops, my bad. Ignore everything I said.
 
The laplacian of a function at a point, \Delta f(p) measures how much f(p), deviates from the average of f on a small circle surrounding p. This is similar to how the second derivative measures whether a function of a single variable is concave up or concave down, except extended to functions of many variables. In a sense, it measures how much the function is "curving up" or "curving down" around a point.

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.
 
Thank you all ..

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.
 
Here is a pdf I've found on the subject. Seems pretty good, and has some pictures.
http://www.math.hmc.edu/~jacobsen/sirev-flat-as-possible.pdf
 
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