Why does an exact differential mean a force is conservative?

Question

If you can express an integrand as an expression of just one variable i.e. $xdy + ydx = d(xy) = df$ then why does that mean that a loop integral on that will equal 0? Is it because if it is just a function then there is no reason we expect force to depend on velocity etc. (i.e. we immediately know friction is not present)?

Answer 1

From a purely mathematical viewpoint, if you have a function f(x,y) then: $$df=f(x+dx, y+dy) -f(x,y)$$ and so $$\Delta f = f_f-f_i=\int_{f}^{i}df$$Since f is a well-defined, singular function, the integral can only depend upon the initial and final points. This would be true for any physical force defined by a singular function, such as gravity. Every point in space has a well-defined, single, value of the force of gravity.

Edit (I've shortened my answer here.)

Any process, therefore, which does depend upon the path taken between two points (e.g. the work done in sliding a block between two points with friction present) cannot be written as a well-defined singular function. To do so would lead to a mathematical inconsistency.

The nature of a frictional force that leads to its non-conservative nature is the fact that it always points in a direction that opposes the direction of motion. Since I can move though a point in different directions, that means that the friction force, a vector, could take multiple values at the same point.

Answer 2

By definition, a Conservative Force is one such that, $$ W\equiv \oint_C F\cdot dr = 0 $$ In other words, the work performed over a closed path equals zero. For the following we assume all surfaces are smooth (infinitely differentiable).

Using Stoke's Theorem (Fundamental theorem of Calculus in higher dimensions), we can rewrite the equation as $$ W= \int\int_S (\nabla \times F) \cdot dS = 0 $$ where the integral is now over the area inside the original contour and the integrand is the Curl (rotation) of the Force. Since the total Curl of the Force is zero, it can be easily shown that, $$ \nabla \times F = 0 $$ Either equation implies that (sign is arbitrary), $$ F=-\nabla \phi $$

Notice that $\nabla\times\nabla\phi = 0$ (always); this is what is meant by exact differential (being able to write F as a derivative). In differential forms language, $d^2\phi=0$ where d is an exterior derivative.

$\phi$ is usually called the potential and in one dimension we have $F=-\partial_x\phi(x)$. In the case of gravity $F=-\partial_y (mgy)= -mg$. Conservative Forces lead, of course, to the concept of conservation of energy, and, the idea of being able to rewrite a physical quantity as the derivative of another, to Noether's theorem and the 'first integral' technique.