# Divergence of the Curl

1. Jul 15, 2011

### MotoPayton

For there to be curl is some vector field fxy cannot equal fyx.
Where fx= P, and fy=Q. Since the (partial of Q with respect to x)-(Partial of P with respect to y) is a non zero quantity giving curl.

I understand that the terms will cancel due to the right-handedness of the definition but we are assuming the these second order partials of P,Q,R follow Clairaut's theroem?

How can clairauts theorem be unsatisfied in order for the vector field to have curl and the clairauts theorem to be satisfied in order for their to be zero divergence?

I hope that makes sense. Thanks

Last edited: Jul 15, 2011
2. Jul 15, 2011

### HallsofIvy

I have no clue what you mean by this. What "scaler field" are you talking about?
If F is a any vector valued function with differentiable components, then "curl F" is defined.

Clairaut's theorem simply says that if F(x,y) is any function of two variables, with continuous second derivatives, then
$$\frac{\partial^2F}{\partial x\partial y}= \frac{\partial^2F}{\partial y\partial x}$$

3. Jul 15, 2011

### MotoPayton

Vector field... my mistake