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I know that a vector field is conservative if for any closed path $C$, the integral with respect to $dr =0$. Are these assumptions below correct?

To prove a vector field $F$ is conservative, all I need to do is find a gradient field for the vector field $F$? That will be enough?

To prove a vector field $F$ isn't conservative, all I need to do is find one example of a closed curve $C$ where the integral is not equal to $0$? Alternatively, can I just show that the curl is nonzero?

But how do I know that a gradient field exists without finding it explicitly? As in, how do I prove that a vector field has a potential gradient field without calculating it? Is it something to do with the curl?

Dam
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    Finding/calculating of the curl is one of options. Let $\vec{F}$ is a three-dimensional vector field, $\vec{F}:\mathbb{R}^3\to\mathbb{R}^3$. It will be conservative if its curl is zero (vector). – Anton Vrdoljak Dec 06 '23 at 17:55
  • @AntonVrdoljak And for a vector field $F = (P,Q)$, if $P_x - Q_y=0$, then $F$ is a gradient field? – Dam Dec 06 '23 at 17:59
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    To @Dam: that theorem is correct. – Anton Vrdoljak Dec 06 '23 at 18:12
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    The condition is $P_{\color{red}y}-Q_{\color{red}x}=0,.$ The famous vector field $(-y,x)/(x^2+y^2)$ satisfies that condition on $\mathbb R^2\setminus{0}$ but on that domain it is not a gradient field. See this post for details. – Kurt G. Dec 06 '23 at 19:53
  • @KurtG. So Anton is wrong? How am I supposed to prove that a gradient field exists then without explicitly calculating it? – Dam Dec 06 '23 at 20:22
  • Not wrong when Anton silently assumed the field is irrotational on a simply connected domain. The Poincare lemma tells you when a potential function exists. The Helmholtz decomposition may allow you to calculate it. Please don't expect simple answers. – Kurt G. Dec 06 '23 at 20:29
  • @KurtG. What does irrotational mean? Does it mean it’s defined and differentiabke everywhere in the domain? – Dam Dec 06 '23 at 20:39
  • curl zero. you must have heard of google. – Kurt G. Dec 06 '23 at 20:57
  • @KurtG. But isn't curl zero the same as $P_y - Q_x=0$ – Dam Dec 07 '23 at 08:50
  • Did I say anything that contradicts that? – Kurt G. Dec 07 '23 at 09:28
  • @KurtG. Well what is your definition of curl on a $2$ dimensional vector field ? Isn't it the same as $P_y-Q_x=0$. You said before that there's a famous vector field that satisfies that relationship but isn't a gradient field, which contradicts your statement. – Dam Dec 07 '23 at 14:47
  • $d\omega=0$ (closed 1-form). Apparently you are not reading the links I strongly recommend. There was one 18 hours ago. – Kurt G. Dec 07 '23 at 14:51

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