# Calculus: I can't understand why curl of gradient of a scalar is zero

1. Sep 5, 2008

### chingcx

(Sorry, the title should read "...why curl of gradient of a scalar "function" is zero)

Of course I know how to compute curl, graident, divergence. Algebrically I know curl of gradient of a scalar function is zero.

But I want to know the reason behind this...and also the reason why gradient of divergence of a vector function is always zero.

This really makes me feeling bad for a long time. Thanks in advance.

2. Sep 5, 2008

### Defennder

The gradient of a scalar function would always give a conservative vector field. Now think carefully about what curl is. If you've done an E&M course with vector calculus, think back to the time when the textbook (or your course notes) derived $$\nabla \times \mathbf{H} = \mathbf{J}$$ using Ampere's circuital law. What is the closed path integral of a conservative field?

3. Sep 5, 2008

### atyy

Is this true? In Gauss's law, the divergence of the electric field is to equal an arbitrary charge density. I would be surprised if the gradient of an arbitrary charge density is zero.

4. Sep 5, 2008

### Ben Niehoff

$$\nabla(\nabla \cdot \vec F) = 0$$

is certainly false. I think you mean

$$\nabla \cdot (\nabla \times \vec F) = 0$$

which is true.

5. Sep 5, 2008

### tiny-tim

Hi chingcx!

(curl grad f)x = ∂/∂y(∂f/∂z) - ∂/∂z(∂A/∂y) = 0

Similarly, div curl A = 0
But (grad div A)x = ∂/∂x(∂Ax/∂x) + ∂/∂x(∂Ay/∂y) + ∂/∂x(∂Ax/∂z) ≠ 0

6. Sep 5, 2008

### Defennder

Hey you're right. Wow I can't believe I didn't even bother thinking about whether it might be correct, as opposed to why it might be correct.

7. Sep 5, 2008

### chingcx

ya, sorry, I mean divergence of curl of vector function is always zero. Why is that true?

8. Sep 5, 2008