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?

  • $\begingroup$ Since any scalar $V$ gives $\nabla \times \nabla V = 0$, you can use stoke's theorem to show $\int_{C}{\nabla V \cdot d\vec{r}}=\int_{A}{(\nabla \times \nabla V) \cdot d\vec{s}}=0$ where $A$ is the area enclosed by the loop $C$. $\endgroup$
    – Andy Chen
    Jul 6, 2022 at 14:09
  • $\begingroup$ @AndyChen Thanks, I can now see why the $F = - \nabla V$ implies $\int_c F \cdot ds = 0$ but I can't see why the reverse is true... Is the reverse true? $\endgroup$
    – physBa
    Jul 6, 2022 at 14:24
  • $\begingroup$ Yes, the reverse is true. We can define some scalar $V$ and show $F=-\nabla V$: Set some point $x_0$ and without loss of generality , we make $V(x_0)=0$. For any point $x$, we define $V(x)=-\int_{L}{F \cdot ds}$ where $L$ is the straight line starting from $x_0$ and stopping at $x$. You can show $F=-\nabla V$ by definition with the condition that $\int_{c}{F \cdot ds}=0$ where $c$ is a closed and non-intersecting curve. Also, sorry for the notation difference in my previous comment. $\endgroup$
    – Andy Chen
    Jul 6, 2022 at 14:40

1 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.


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