# Curl of Gradient of a Scalar Field

1. Aug 12, 2015

### Nishant Garg

Hello, new to this website, but one question that's been killing me is how can curl of a gradient of a scalar field be null vector when mixed partial derivatives are not always equal??

consider Φ(x,y,z) a scalar function
consider the determinant [(i,j,k),(∂/∂x,∂/∂y,∂/∂z),(∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z)] (this is from ∇×(∇Φ))
When you expand this you will get
[(∂^2Φ/∂y∂z)-(∂^2Φ/∂z∂y)]i-[(∂^2Φ/∂x∂z)-(∂^2Φ/∂z∂x)]j+[(∂^2Φ/∂x∂y)-(∂^2Φ/∂y∂x)]k
Now this can only be null vector when individual components are 0, and that's only when mixed partials are equal, but they are not always equal now are they?

2. Aug 12, 2015

### BvU

Hello Nishant, welcome to PF !

If I (like you probably also did) simply google
proof that curl of a gradient is zero
then I find (e.g. here , but in many other places) that cross derivatives are equal.
You place a question mark there, though. Perhaps the definition of derivative -- writing out the lot in limit terms -- can help you see they really are equal. Or do you know a counter-example ?

Wiki on this subject.

3. Aug 12, 2015

### Nishant Garg

The last line of that post, like I said

If f is twice continuously differentiable, then its second derivatives are independent of the order in which the derivatives are applied. All the terms cancel in the expression for curl∇f, and we conclude that curl∇f=0.

So it must satisfy that condition above for curl of gradient to be null vector? It's not always null vector right, it must satisfy that condition?

Counter example, I try hard but can't think of a function whose second partial derivatives are different depending on respect to which you took derivative first

4. Aug 12, 2015

### BvU

Last edited by a moderator: May 7, 2017
5. Aug 12, 2015

### Nishant Garg

So is it safe to assume that our scalar function has continuous second order mixed partials?

6. Aug 12, 2015

### Nishant Garg

Ok, thank you. How do I close this thread?

7. Aug 12, 2015

### BvU

If we stop posting, that's closing the thread. It stays on the forum for the benefit of all !