According to the strong law of action and reaction for internal forces (Goldstein):

$\mathbf F_\mathrm {ij}=-\mathbf F_\mathrm{ji}$ and the forces lie along the direction joining the particles.

Now consider the statement

If those internal forces are conservative we can associate the internal forces with a potential of the form $V_\textrm{ij}(|\mathbf{r_i}-\mathbf{r_j}|)$.

How can one justify this statement mathematically though it seems intuitively obvious?

  • 1
    $\begingroup$ Are you asking about the third law or conservative forces? $\endgroup$
    – jinawee
    Oct 21, 2013 at 8:02
  • 1
    $\begingroup$ conservative forces. $\endgroup$ Oct 21, 2013 at 8:03
  • $\begingroup$ You've got everything here: en.wikipedia.org/wiki/Conservative_force $\endgroup$
    – jinawee
    Oct 21, 2013 at 8:06
  • $\begingroup$ No,not really,my question was why the potential of conservative internal forces which follow the strong law has the form $V(|\vec{r_i}-\vec{r_j}|)$ and not anything else? $\endgroup$ Oct 21, 2013 at 8:10

2 Answers 2


The full mathematical statement is as follows:


If two particles exert a mutual conservative force $\mathbf{F}_{12}$ and $\mathbf{F}_{21}$ which is independent of any other degree of freedom of any bigger system they're part of, and obeys Newton's third law as $\mathbf{F}_{12}+\mathbf{F}_{21}=\mathbf{0}$, with the forces collinear to the particles' relative orientation, then this mutual force can be written in the form $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|)$$ for some appropriate potential $V$.

It is important to assume that the force is independent of any degrees of freedom other than the particles' DOFs. This is quite natural, as otherwise it would be very difficult to interpret the force as one caused by one particle on another.

Under this assumption, the requirement that the force be conservative requires the existence of some potential $V$, which depends only on $\mathbf{r}_1$ and $\mathbf{r}_2$, such that $$\mathbf{F}_{ij}=-\nabla_j V(\mathbf{r}_1,\mathbf{r}_2).$$

This function can be transformed, without losing any information, into a function of the relative position $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$ and some sort of "mean" position, say the average $\mathbf{R}=\tfrac12\mathbf{r}_1+\tfrac12\mathbf{r}_2$, though any linear combination (e.g. COM position) independent to $\mathbf{r}$ is acceptable. Under this change of variables, the gradients transform as $$ \left\{\begin{array}{} \nabla_{\mathbf{r}_1}=\phantom- \nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}},\\ \nabla_{\mathbf{r}_2}=-\nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}}, \end{array}\right. $$ so that the statement of Newton's third law in this setting becomes $$ \mathbf{0} =\nabla_{\mathbf{r}_1}V(\mathbf{r},\mathbf{R})+\nabla_{\mathbf{r}_2}V(\mathbf{r},\mathbf{R}) =\nabla_{\mathbf{R}}V(\mathbf{r},\mathbf{R}). $$

There is thus no dependence on the absolute position $\mathbf{R}$, and the force can be written as $$\mathbf{F}_{ij}=-\nabla_j V(\mathbf{r}).$$

For the final step, as QMechanic pointed out, you use the fact that the force is collinear to the relative orientations. If $V$ depends on the angular variables of $\mathbf{r}$, then its gradient will not be purely radial. If you require a formal proof, note that forcing the non-radial components of the gradient in spherical coordinates, $$ \nabla f =\frac{\partial f}{\partial r}\hat{{r}} +\frac1r\frac{\partial f}{\partial \theta}\hat{{\theta}} +\frac{1}{r\sin\theta}\frac{\partial f}{\partial r}\hat{{\phi}}, $$ to vanish, requires the partial derivatives with respect to $\theta$ and $\phi$ to vanish; there is thus only a dependence in $r$.

With this, then, you have your final result: the force can be written as $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|).$$


OP is essentially asking

If a conservative force $$\vec{F}(\vec{r})~=~-\vec{\nabla} V(\vec{r})~\parallel~\vec{r}$$ is collinear, how do we know that the potential $V(\vec{r})$ only depends on the length $|\vec{r}|$, and not the angular variables $\Omega$?

Answer: This is because dependence on $\Omega$ would yield a non-collinear gradient.


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