Why is the force being the differential of a potential equivalent to it being a conservative force?

Question

I was reading Goldstein's book on mechanics and came across this theorem:

$F(r) = - \nabla V(r)$ is a necessary and sufficient condition of the force field being conservative.

So far, I have understood the condition of a force being conservative as path independence: $\int_{\text{closed loop}} F \cdot ds = 0$.

The new condition was justified by a brief argument which I don't follow:

The existence of V can be inferred intuitively by a simple argument. If $W_{12}$ is independent of the path of integration between the end points 1 and 2, it should be possible to express $W_{12}$ as the change in quantity that depends only upon the positions of the end points.

I follow that for the work done must depend only upon the end points $W = W(start, end)$ and to ensure the path integral is always zero the "return trip" must cancel out the "outgoing trip", i.e., $W = W(start, end) = -W(end, start)$. But how do I go from this to the form given above?

Answer 1

If a force can be represented as $$\mathbf{F}=-\nabla U(\mathbf{r}), $$ we can calculate the work over a closed circuit as $$ \oint\mathbf{F}\cdot d\mathbf{r}=-\oint\nabla U(\mathbf{r}) \cdot d\mathbf{r}= -\oint dU(\mathbf{r})=0, $$ since $U(\mathbf{r})$ is a uniquely valued function. Note that I used here the full differential (see here for more background): $$ dU(\mathbf{r})=\nabla U(\mathbf{r}) \cdot d\mathbf{r}= \frac{\partial_xU(x,y,z)}{\partial x}dx + \frac{\partial_xU(x,y,z)}{\partial y}dy + \frac{\partial_xU(x,y,z)}{\partial z}dz $$

Another amusing (although non-canonical way) to use this is by looking at the first integral of the Newton's second law:
$$ m\ddot{\mathbf{r}}=-\nabla U(\mathbf{r}), $$ multiply it by $\dot{\mathbf{r}}$ and perform some algebraic transformations, we obtain: $$ 0=\left[m\ddot{\mathbf{r}} +\nabla U(\mathbf{r})\right]\dot{\mathbf{r}}= m\ddot{\mathbf{r}}\dot{\mathbf{r}} +\nabla U(\mathbf{r})\dot{\mathbf{r}}= \frac{d}{dt}\left[\frac{m\dot{\mathbf{r}}^2}{2}+U(\mathbf{r})\right]=\frac{d}{dt}E(t) $$ That is, the quantity $E(t)=\frac{m\dot{\mathbf{r}}^2}{2}+U(\mathbf{r})$ does not change with time (i.e., conserved) along the trajectories described by the equation of motion. This is just a mathematical fact - there is no much flexibility in adjusting it.