# Physical significance of the laplacian operator?

1. Dec 15, 2008

### Tac-Tics

What is the physical significance of the laplacian operator?

The laplacian operator is the divergence of the gradient. I understand the intuitive meanings of both. The gradient when dotted against a unit vector gives the rate of change in that direction. The divergence is the flow in or out of an infinitesimal sphere surrounding a point. But I can't quite figure out what they mean when used together.

I was wondering if there was a simple way of describing it visually.

2. Dec 15, 2008

### lurflurf

The Laplacian is an averaging operator (actually an average difference). Think of the divergence theorem. The divergence of the gradient is the average over a surface of the gradient.
say
Laplacian(f)~(1/h^2)(f(x+h)-2f(x)+f(x-h))=[(f(x+h)+f(x-h))/2-f(x)]/h^2
=(average(f)-f)/h^2

So the famous Laplace equation
Laplacian(f)=0
says f at each point is equal to its average at nearby points.

3. Dec 16, 2008

### HallsofIvy

Mathematical concepts do NOT have "physical meaning". They may have have physical meaning in a specific physical application. lurflurf's answer is the best general meaning you will get.

4. Dec 16, 2008

### Tac-Tics

You don't need to be so pedantic. A firm physical intuition of a concept is more useful than knowing its rigorous definition when it comes to reasoning about it.

lurflurf's answer was pretty much the kind of answer I was looking for.

5. Dec 16, 2008

### NoMoreExams

This probably deserves a different thread but in line with this thinking I was wondering what the Laplacian operator (probably different from what's mentioned here?) really does as far as phase portraits and equil. analysis. I'm refering to the fact that under certain conditions you can apply "L" and convert PDE into a DE and DE into an algebraic equation. Now I'm not sure about phase portraits and PDEs but I was pretty sure that every DE induces a diff. vector field(?) so what does the Laplacian operator do or rather how does it relate to that concept?

6. Dec 17, 2008

### Tac-Tics

I don't know either subject very well, so I could be wrong, but I think you are referring to the Laplace Transform, which is distinct from the Laplacian operator.

7. Dec 17, 2008

### NoMoreExams

You are correct, it's a different thing but I'm still curious :). Thank you.

8. Dec 17, 2008

### Hootenanny

Staff Emeritus
I have to disagree with you there. A physical intuition is useful (but not essential) in understanding a concept, but it is imperative that one fully understands the rigorous definition when reasoning about it.

9. Dec 17, 2008

### arildno

Physics (and, reality in general) is merely a special case of maths, Tac-Tics, that's why we should understand, and master, the maths.

10. Dec 17, 2008

### Tac-Tics

There is a reason counting is taught with number lines, algebra is taught with graphs, and geometry with shapes. The mind works best in pictures. I don't think there's a single school that teaches the formal rigorous definition of a limit until after the student understands limits in terms of wholes and asymptotes in a graph. The physical intuition dictates the rigorous definition, not the other way around.

I understand what you are trying to say, that mathematics, once the definitions are chosen, are independent of the physical world. But it was a lesson in mathematical philosophy I was looking for.

11. Dec 17, 2008

### Hootenanny

Staff Emeritus
If you re-read my post, you'll find that I make the distinction between learning about a concept and reasoning about it. I fully appreciate the advantages of physical intuition with respect to learning new concepts, what I object to is your assertion that physical intuition is more useful than a mathematically rigorous definition when reasoning about a topic or discussing it's consequences.

12. Jan 16, 2009

### Nolen Ryba

One way to think about the Laplacian of a function is that it is a multi-dimensional version of concavity. In the 1-d case it is just the second derivative. Think about the wave equation for a string- this states the vertical acceleration of a point on the string is proportional to the concavity of the string at that point. The easy to visualize 2-d case is a membrane (say drum head). Here again we have the wave equation, which states that the vertical acceleration of point on the membrane is proportional to the Laplacian of the membrane shape at that point- the sum of the concavity in both directions. In 3-d it's a little harder to think about this but it still makes sense. Think about a sound field- the wave equation states the acceleration of the pressure at a point (second derivative in time) is proportional to the laplacian of the pressure field at that point- changes in the change of the pressure in time are caused by second order spatial changes in the pressure.

13. Jan 16, 2009

### Tac-Tics

After going back to it a few weeks ago, something finally clicked. It's hard to put it into words exactly why it suddenly starts to make sense. Maybe it's just simple familiarity. The laplacian is simply the force instead of the flow out of each point. Or something like that. I think that in looking at it in general terms (divergence of the gradient), The idea that it's just the sum of the second derivatives in each orthogonal direction just didn't sink in until I came back to it.

14. Jan 20, 2009

### barton

The distinction wasn't very clear in your previous post. And on the contrary, having a good mental model of the the math topic you are trying to reason is an absolute must. You just may not be sharing it with another person say, when you write out a formal proof.

15. Jun 5, 2009

### endopol

I think of the Laplacian as the rate at which a function at a point differs from the average at the points surrounding it. In fact, in n dimensions, it can be written
$$\Delta f (x) = 2n \lim_{h\to 0} \frac{\operatorname{Avg}\{f(z) :\; |z-x|=h\}-f(x)}{h^2}$$
(I may be off by a constant factor). That's why it's used in equations like the heat equation - Newton's law of cooling states that an object cools at a speed proportional to the difference between itself and its surroundings.

Last edited: Jun 5, 2009
16. Jun 6, 2009

### HallsofIvy

The Laplacian is the simplest second order differential operator that is "invariant" under rigid motions. That is, it has the same form under translations or rotations.

17. Aug 9, 2009

### Gaco

On divergence

I think it's a bit hard to come from a cartesian understanding of divergence (and Laplacian operator for that matter) and then understand the same in terms of spherical polar coordinates intuitively?

I mean, say that a vector field is v = x i where i is the x-direction unit normal vector. This vector field is very easy to visualize. It's divergence is 1 because it's x value is increasing in the x direction.

Thus if v is a velocity field, the divergence value of 1 can mean one of two things:

Case 1: The fluid density is constant. The velocity vectors of the fluid entering the left-half of any closed surface are smaller in magnitude than the velocity vectors in the right-half, i.e. more fluid is entering any closed surface than is leaving it, which means that there must be a source of fluid everywhere.

Case 2: There are no fluid sources or sinks. Again the velocity vectors of the fluid entering in the left-half of any closed surface are smaller in magnitude less than the velocity vectors in the right-half, i.e. the fluid is diverging in the x-direction, it's speeding up making the fluid density less and less in the x-direction.

Now lets look at the vector field v = k*1/r^2 where k is just a constant and r is the distance from the origin. The formula for divergence in spherical polar coordinates gives a ZERO divergence everywhere except the origin, at which it is undefined. I'm sure that if you calculate the divergence of the field v = k*1/sqrt(x^2+y^2+z^2) in cartesian coordinates, you'll get zero as well, but if you look at any closed surface with the fluid entering as before, the velocity vectors entering the surface closest to the origin are larger in magnitude than the ones leaving the surface, therefore you should think that the density should be increasing / there should be a source everywhere?

I think this divergence intuition inherited from the cartesian coordinate system is confusing me when trying to appy it to spherically symmetrical situations and spherical polar coordinates.

On the Laplacian

Tac-Tics recently wrote: "The idea that it's just the sum of the second derivatives in each orthogonal direction just didn't sink in until I came back to it."

I kind of got the hang of thinking about it like this as well, but again when I apply it to the eletrostatic potential, phi = k*1/r, I just don't find it that intuitive anymore that the Laplacian of phi should give zero on the basis on my intuition of Laplacian from the cartesian world.

Take an arbitrary point along the x-axis (x=x0, y=0, z=0) in the electrostatic potental scalarfield of a positive point charge in the origin, then d^2(phi)/dx^2 is negative, but both d^2(phi)/dy^2 and d^2(phi)/dz^2 are also negative since all three increases distance to origin, all three which makes the value of the scalar potential lower. How can this give zero?

Again, you can probably show it's zero by during the proper calculation, but it shouldn't be too hard to point out the the fault in my logic and make it to make sense intuitively?

Thanks in advance to anyone who can help me sort this out :)