# closed integral curves and conservative fields

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 $$\vec{E}(\vec{x}) = -\nabla V(\vec{x})$$?

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

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.