# Every conservative vector field is irrotational

I have done an example where I needed to show that every conservative $C^2$ vector field is irrotational. However, there is something unclear in the solutions: Namely, I am uncertain what does the following sentence at the end of the solution mean:

"since second partial derivatives are independent of the order (for smooth functions)", and I was wondering how does that imply that the equality before that is 0?

• Note that $\partial_y \phi_z = \frac{\partial^2 \phi}{\partial y\partial z}$. – user99914 May 13 '18 at 23:33

This is normally called symmetry of second derivatives, or Clairaut's theorem. It means that for functions that have continuous second partial derivatives, $$\frac{\partial^2f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x} = 0.$$ But each term in the line above is of precisely this form (Presumably the document's notation has $\phi_x = \partial_x \phi$ and so on).
• Thanks! Also, why does it have to be a $C^2$ field, would not the same thing apply for $C^1$ field, since we are calculating derivatives of the potential, and in the end only calculate the 1st derivatives of the field? – Relax295 May 14 '18 at 0:07
• Yes, that should be sufficient. Presumably it's just to make it consistent with the second part, which does need $C^2$ since 2 derivatives really are taken. – Chappers May 14 '18 at 0:15
• But, regarding the 2nd part, first a derivative of potential of F is taken, and then, in fact, a derivative of F, which means that we only require $C^1$, since only the 1st derivative of F is calculated(in other words: 2nd derivative of potential of F) or? – Relax295 May 14 '18 at 0:23