1
$\begingroup$

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? enter image description here

$\endgroup$
  • 1
    $\begingroup$ Note that $\partial_y \phi_z = \frac{\partial^2 \phi}{\partial y\partial z}$. $\endgroup$ – user99914 May 13 '18 at 23:33
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Relax295 May 14 '18 at 0:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Chappers May 14 '18 at 0:15
  • $\begingroup$ 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? $\endgroup$ – Relax295 May 14 '18 at 0:23
  • $\begingroup$ I mean part (b) of the question. $\endgroup$ – Chappers May 14 '18 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.