From classical mechanics, the force on a spring is given by the negative gradient of the potential energy with respect to position or displacement.

Can we also say $F=-\partial U / \partial x$, where $U$ is the "internal" energy of the spring?

  • $\begingroup$ Yes; why wouldn't you be able to say this? $\endgroup$ – Chemomechanics Sep 18 '18 at 22:34
  • $\begingroup$ Of course it is. $\endgroup$ – Bob D Sep 18 '18 at 22:37
  • 1
    $\begingroup$ @Chemomechanics because the "internal energy" is defined for the material of the spring as well, and includes thermal energy of the atoms. It therefore depends on temperature as well. For an imcompressible substance we usually see $U = mcT$ for internal energy and we typically see $U = \frac{1}{2}mkx^2$ for a non-dissipative spring (so I guess this becomes potential energy). But in reality, the internal energy of a spring depends on displacement and temperature. When calculating $F=\partial U / \partial x$, do we simply keep temperature constant in this case? $\endgroup$ – Drew Sep 18 '18 at 22:57

The relation you have stated, $F = -\partial U/\partial x$ is valid for all potentials in 1D to determine the conservative force associated with the potential. Note that in the more general case we have the following (for any dimension):

$$\mathbf{F} = -\mathbf{\nabla} U $$

Where $\nabla$ is the gradient operator. Note that the force arises due to a gradient or difference in the potential in space, not due to a changing potential in time. Therefore, temperature would be implicitly included since either $T = const$ everywhere or $T = T(\mathbf{r})$ changes in space in which case the gradient captures the change.

  • $\begingroup$ So generally $ U = U(\mathbf{r}, T)$, and it's possible a function of other properties, and the gradient captures the spatial change. I was unsure because we usually see $U=U(\mathbf{r})$ for a spring and call it "potential energy", but that's an approximation assuming no temperature dependence right? $\endgroup$ – Drew Sep 19 '18 at 2:19

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