# Use of curl of gradient of scalar

1. Nov 28, 2003

### Mike2

I'm wondering if physics ever uses a differential equation of the form of a curl of a gradient of a scalar function. Or is this too trivial?

Thanks.

2. Nov 28, 2003

### chroot

Staff Emeritus
The curl of the gradient of a scalar function is always zero. So, yeah, it's pretty trivial.

- Warren

3. Nov 28, 2003

### Mike2

well yes, but then don't they use the fact that the Laplacian is zero?

4. Nov 28, 2003

### chroot

Staff Emeritus
I think I see what you're asking. How about electrostatics? The electric field can be represented by a scalar potential field. Therefore, the curl of the gradient of this field is zero -- the electric field is curl-free.

- Warren

5. Nov 28, 2003

### Ambitwistor

Yes, but the Laplacian of an arbitrary function isn't automatically zero, so only certain functions (the harmonic ones) satisfy the condition that their Laplacian is zero. Every function satisfies the condition that the curl of its gradient equals zero, so that equation is not too useful on its own.

6. Nov 28, 2003

### lethe

as ambitwistor correctly points out: the laplacian is not zero.

the the curl of a gradient is zero.

physicists use both facts.

7. Nov 28, 2003

### Mike2

In other words, you could never find a "unique" function that satisfies the conditions. For ALL functions satisfy the conditions, right? Thanks.

8. Nov 28, 2003

### Mike2

Yes, but the question is HOW do they use the fact that the curl of the gradient of a scalar. Yes they use it in Stoke's theorem, but do they use it in its differential form?

9. Nov 28, 2003

### Ambitwistor

Yes, all functions satisfy the condition that the curl of their gradient is zero. In the case that the Laplacian is zero, there isn't a unique function that satisfies that condition (unless you also supply boundary conditions), but nor is it true that every function satisfies it, either. That's what makes it more interesting than an equation to which every function is a solution.

10. Nov 28, 2003

### lethe

you bet they do. this fact is the basis of de Rham cohomology. it provides a powerful tool for topological classification of manifolds, which rests on Poincaré duality, which is really just an application of Stokes theorem.

11. Mar 20, 2009

### Triatticus

Yeah I'm glad as physicists we never really run into those weird functions that don't obey Clairaut's Theorem. Because if we did then technically the curl of a gradient of a scalar function wouldn't necessarily be zero. Since the curl of a gradient gives us second order mixed partial derivatives of the function in question like Fxy - Fyx as the z component, this may or may not be zero for Every function out there, though is certainly true for smooth twice continuously differentiable functions at least. SO the statement should be not that every function has a zero curl gradient but that only continuous smooth functions do, because a piecewise continuous function is certainly a function whether its smooth or not.