In mechanics course, when the idea of equilibrium was introduced they included the idea of a force that derives from potential energy which is the force $F$ which is related to the potential energy $E_p$ by the relation:

$$F=-\nabla E_p$$

I didn't understand at all the physical meaning of such a definition. Any help in such an explanation (physical meaning) is appreciated.

  • 2
    $\begingroup$ Welcome to PhysicsSE! Note that this site supports MathJax. You can click the link to learn the basics. I've taken the liberty of typesetting this post for you. $\endgroup$ – Chris May 22 '19 at 22:27
  • $\begingroup$ Have you seen the analogy of a ball rolling on a bumpy surface under the action of gravity? It's not exact but it can help form some intuition. $\endgroup$ – Triatticus May 23 '19 at 1:57
  • $\begingroup$ This is the magnitude of force along the direction of displacement. $\endgroup$ – Shreyansh Pathak May 23 '19 at 6:39

Any force $\vec{F}$ that can be represented by a gradient $\nabla$ of a scalar field $V$ is called a "conservative force". But why is this definition important?

Let's try to derive the work done by this force $\vec{F}$. For any force, we have the following from the definition of "work": $$W = \int_\mathcal{C} \vec{F}\cdot\vec{d}l$$ for a given path $\mathcal{C}$.

Now, if our force can be written as $\vec{F}=-\nabla V$, we have $$W = \int_\mathcal{C} (-\nabla V)\cdot d\vec{l}$$

There is a theorem called "gradient theorem", or "the fundamental theorem of calculus for line integrals", that claims the following for path integrals on a curve $\gamma$ from $x_1$ to $x_2$: $$\int_{\gamma[x_1,x_2]} \nabla\phi(\vec{r})\cdot{}d\vec{r} = -(\phi(x_2) - \phi(x_1))$$

Using this theorem, we can find the following interesting result: $$W = \int_\mathcal{C} (-\nabla V)\cdot d\vec{l} = V(x_2) - V(x_1)$$ for $(x_1,x_2)$ the starting and ending points of path $\mathcal{C}$.

What does this mean? This means that, for conservative forces, the work done by this force is "path independent", i.e. it only depends on the initial and final points of the path rather than how to path moves.

Notice that this final result is the definition of a potential field, that is, $V(\vec{r})$, for this force. If you have a not conservative force, you simply cannot define a potential to it.

For example: Existence of electric potential means that the electrostatic force is a conservative force. But, for example, since the work done by friction is always path dependent, we cannot define a potential energy to it.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.