# Help on physical meaning

1. May 15, 2008

### mtarek16

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

2. May 15, 2008

### qspeechc

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?

3. May 15, 2008

### Vid

4. May 15, 2008

### qspeechc

Oops, my bad. Ignore everything I said.

5. May 15, 2008

### maze

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.

6. May 22, 2008

### mtarek16

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.

7. May 24, 2008