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

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Discussion Overview

The discussion centers on the physical meaning of the equation [Laplacian(f) = 0], where f is described as a potential function of a vector field. The scope includes conceptual understanding and technical explanation related to vector analysis.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • MTarek seeks a definition of the physical meaning of [Laplacian(f) = 0] in the context of vector analysis.
  • One participant mistakenly conflates the Laplacian with the Laplace transform, suggesting that f=0 is the only solution.
  • Another participant clarifies that the Laplacian and Laplace transform are distinct concepts and mentions that [Laplacian(f) = 0] corresponds to the n-dimensional Laplace equation.
  • A later post explains that the Laplacian at a point measures how much the function deviates from the average value in a surrounding area, likening it to the second derivative in single-variable calculus.
  • This participant further elaborates that if the Laplacian is zero, f must equal the local average, indicating that functions with zero Laplacian everywhere resemble saddle points and cannot have local maxima or minima.
  • MTarek expresses some confusion but indicates a willingness to explore the topic further, especially in light of upcoming finals.
  • Another participant shares a PDF resource that includes graphical examples related to the topic.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the topic, with some clarifications made but no consensus reached on the physical meaning of [Laplacian(f) = 0].

Contextual Notes

Some participants demonstrate uncertainty about the distinction between the Laplacian and Laplace transform, and there are unresolved aspects regarding the implications of the Laplacian being zero.

mtarek16
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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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