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There are several conditions that must be met in order for a force to be conservative.
One of them is that the curl of that force must be equal to zero?
What is the physical intuition behind this?
If you can, please explain it to me via the magnetic force fields because i have read that time-varying magnetic fields are not conservative because they do not meet that condition.I do not understand this.

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    $\begingroup$ What about the Wikipedia article, where the equivalence of three usual definitions of a conservative force is proven, doesn't satisfy you? A conservative force is the gradient of a potential, and gradients don't have curls. $\endgroup$ – ACuriousMind Mar 31 '15 at 21:09
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    $\begingroup$ no,because it does not fully cover the whole intuition behind it. $\endgroup$ – TheQuantumMan Mar 31 '15 at 21:57
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"Curl" is a pretty well named mathematical term--it denotes the degree of "rotation" in the vector field. For this reason, if you go all the way around in a vector field, you'll find that the total integral along that path will depend on the curl of the field in question. If a force had a curl, you could go all the way around and have some net work done, and so it would be nonconservative. A conservative force, on the other hand, cancels itself back out as you go on a closed loop.

Think of a curl-ful field as a whirlpool--you could imagine going around and around and building up speed in it. But a curl-free field might be more like a river. You can flow down the river, but if you go back and forth down the river you spend as much time going up as you do going down, so you can't get anything out of it. (This is a highly non-mathematical analogy, but it's how I think of it.)

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A force field is called conservative if its work between any points $A$ and $B$ does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the work from $A$ to $B$ will be greater than from $B$ to $A$ leading a non vanishing circulation. This is true even if you are able to write this force as $\vec F(\vec r,t)=-\vec \nabla U(\vec r,t)$ or $\vec \nabla\times\vec F(\vec r,t)=0$. Moreover one can show that this would imply $\frac{dE}{dt}=\frac{\partial U}{dt}\neq 0$.

Regarding the curl, I like to visualize it as an infinitesimal circulation. Let me try to elaborate the answer given by zeldredge. Imagine a two dimensional fluid flow. You can think of the curl as an infinitesimal paddle wheel put with its axis perpendicular to the fluid. Whenever the the fluid makes it rotate, the curl is different from zero. The angular velocity of the paddle wheel gives the magnitude of the curl. It is not hard to visualize that this paddle will rotate when put in a whirlpool and remain static when put in a laminar flow of a river. When you go to more abstract vector fields, like the electric or magnetic ones, you just to think about an abstract paddle wheel.

To finish let me mention that a vanishing curl (even for a position only dependent force) does not implies, in general, that the force is conservative. Those things are equivalent only when the space is simply connected. When this is not the case, the Stokes and Green theorems fail and zero curl does not imply zero circulation. The classic example is the two dimensional force $\vec F(x,y)=\frac{-y\hat i+x\hat j}{x^2+y^2}$, which has vanishing curl and circulation $2\pi$ around a unit circle centerd at origin. If this vector field is meant to be a flow velocity field it clearly means the fluid is rotating around the origin. However it gets slower as we go away from the origin. Imagine an infinitesimal closed path, $l_1+l_2+l_3+l_4$, in polar coordinates do not containing the origin, as shown in the figure. We have zero "work" over $l_3$ and $l_4$ and mutually canceling "work" over $l_1$ and $l_2$. The longer path over $l_2$ compensate the slower flow. enter image description here

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  • $\begingroup$ Great answer, but i have one question: what is the intuition behind having zero curl but having non-zero circulation? $\endgroup$ – TheQuantumMan Apr 16 '16 at 21:06
  • $\begingroup$ Please have another look at the answer. I edited and tried to elaborate the example given. I hope it might help. $\endgroup$ – Diracology Apr 16 '16 at 22:02
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the heart of a force being conservative is that it is integrable, that, if we have a force ${\vec F}$, then it is possible to find a potential $\phi({\vec x})$ such that ${\vec F} = - {\vec \nabla}\phi$. The reason for this is that if we pick out two points $p$ and $q$, we want the difference in energy between the two points to be $\phi(p) - \phi(q)$, and this has to be path-independent.

However, it turns out that for all functions $f({\vec x})$, we have ${\vec \nabla} \times \left({\vec \nabla}f\right) = 0$. Therefore, if ${\vec F}$ is to be integrable, it's necessary (but not sufficient) that ${\vec \nabla} \times {\vec F} = 0$

so, in essence, path independence -> curl-free.

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Just to add to Jerry Schirmer's answer, I find it helpful to think of a potential as being a more compact form of expressing $\vec F$ because $\vec F$ is 3-dimensional and $\phi$ is one-dimensional. If $\vec F$ is conservative, it would seem to only contain "one dimension" worth of information; it has some redundancy in its form (i.e. is not its simplest expression).

On the flip-side, you could have some curled functions $\vec F$ (in other words, $\nabla\times\vec F\ne\vec0$) which do not have a corresponding "compact version" $\phi$.

Indeed there might be some potentials $\phi$ that are not simpler than their corresponding gradients $\vec F$, but then again this is more about intuition.

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protected by Qmechanic Apr 16 '16 at 14:36

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