1
$\begingroup$

If a vector field $\vec{E}(\vec{x})$ in $\mathbb{R}^3$ has no closed integral curves, does this imply that the field is conservative, i.e. there is some scalar function $V(\vec{x})$ such that $\vec{E}(\vec{x}) = -\nabla V(\vec{x})$?

$\endgroup$
3
  • $\begingroup$ Are you mixing up the concepts of "integral curve which is closed" with "integration along a closed curve"? They are different things. $\endgroup$ Feb 8, 2020 at 15:11
  • $\begingroup$ Do you consider stationary orbits to be closed? $\endgroup$
    – GReyes
    Feb 8, 2020 at 19:17
  • $\begingroup$ I mean "an integral curve which is closed". $\endgroup$ Feb 8, 2020 at 21:12

1 Answer 1

1
$\begingroup$

No, consider the vector field $\mathbf{E}(x,y,z)=\left< 0, 1 + x^2, 0 \right>$. You can check that this has non-zero curl and cannot be a gradient, and has no closed integral curves.

$\endgroup$

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.