Conservative field vs conservative force

Question

For a conservative field (e.g. electrostatic field) the circulation of the field (along a closed line) is zero.

For a conservative force (e.g. macroscopic elastic force) the work performed on a particle along a closed path is zero.

But it seems to me that the two conceps are pretty different:

1) In case of the field, the circulation is calculated at a specific instant of time (the value of the field at each point in space is "frozen" during the calculation). The operation is performed on a test particle, whose influence on the field is negligible.

2) in case of the force, the work integral is performed accounting for the time evolution of the whole system during the motion of the particle that is feeling the force. So in this case when we talk about a "closed loop" we really mean that the particle returns to the original RELATIVE position with respect to another element of the system (the "source" of the force).

Is this correct?

Answer 1

The two conditions you have listed are equivalent. In other words, for a vector field $\mathbf V(\mathbf r)$ the condition that $\nabla\times\mathbf V=0$ everywhere is the same as saying $\oint_C\mathbf V\cdot\text d\mathbf r$ for any closed path $C$. Either of these also means that the field can be written as the gradient of a scalar field $\mathbf V=-\nabla\phi$.

The proof of the equivalence of these statements should be found in any vector calculus or possibly classical mechanics text, or by looking online.

As for the second condition relying on the time evolution of the system, it actually does not matter here. Note how the above line integral doesn't involve time. In other words, if you have an actual physical object traveling along a closed path where the vector field in question is the force acting on the object, it does not matter how it traverses this path. If it follows a closed path in any way, the work done by the conservative force is $0$.