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 $\vec{E}(\vec{x}) = -\nabla V(\vec{x})$?
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$\begingroup$ Are you mixing up the concepts of "integral curve which is closed" with "integration along a closed curve"? They are different things. $\endgroup$– Paul SinclairFeb 8, 2020 at 15:11
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$\begingroup$ Do you consider stationary orbits to be closed? $\endgroup$– GReyesFeb 8, 2020 at 19:17
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$\begingroup$ I mean "an integral curve which is closed". $\endgroup$– creillyuclaFeb 8, 2020 at 21:12
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1 Answer
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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.
