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?
Best Answer
The full mathematical statement is as follows:
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 $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|).$$