Internal potential energy and relative distance of the particle Today, I read a line in Goldstein Classical mechanics and got confused about one line. 

To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance between the particles: $V_{ij} = V_{ij}(|{\bf r}_i-{\bf r}_j|)$.

What confuses me is that I can't see the logic between these two statements. Obviously, I understand strong law of action and reaction and Internal energy. But why the strong law of action and reaction leads to internal energy only depending on relative distances?
I prefer to receive mathematical proof (not thorough, but provide a direction so that I can know where I'm going); yet, intuitive illustration is also welcome.
 A: The quote is taken from just above eq. (1.32) in Ref. 1: 

[...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance between the particles: $$V_{ij} ~=~ V_{ij}(|{\bf r}_i-{\bf r}_j|). \tag{1.32}$$

The structure of internal forces among $N$ point particles can be quite rich in general, see e.g. this Phys.SE post. However the first sentence in the quote makes it clear that Ref. 1 is additionally assuming:


*

*that the internal forces on one particle is a sum of forces from the other particles. Thus it is enough to study the internal force ${\bf F}_{ij}$ from the $i$th particle on the $j$th particle. 

*that ${\bf F}_{ij}({\bf r}_i,{\bf r}_j)$ only depends on the two positions ${\bf r}_i$ and ${\bf r}_j$ of the $i$th and $j$th particles, respectively. 

*that ${\bf F}_{ij}({\bf r}_i,{\bf r}_j)$ is a conservative force, meaning that there exists a potential $V_{ij} = V_{ij}({\bf r}_i,{\bf r}_j)$ such that 
$$ {\bf F}_{ij}~=~-{\bf \nabla}_{j} V_{ij}. $$ 

*that the potential $$V_{ij}({\bf r}_i,{\bf r}_j)~=~V_{ji}({\bf r}_j,{\bf r}_i)$$ is symmetric.
The weak form of Newton's 3rd law then implies that 
$$ {\bf 0}~=-{\bf F}_{ij}-{\bf F}_{ji}~=~ {\bf \nabla}_{j} V_{ij}+{\bf \nabla}_{i} V_{ji}~=~ ({\bf \nabla}_{i}+ {\bf \nabla}_{j}) V_{ij},$$
which in turn implies that the potential $V_{ij} = V_{ij}({\bf r}_{ij})$ only depends on the difference ${\bf r}_{ij}:={\bf r}_j-{\bf r}_i$ in positions.
Finally the strong form of Newton's 3rd law implies that 
$${\bf r}_{ij}~\parallel~ {\bf F}_{ij}~=~-{\bf \nabla}_{j} V_{ij}({\bf r}_{ij}),$$
which in turn implies that the potential $V_{ij} = V_{ij}(|{\bf r}_{ij}|)$
only depends on the distance $|{\bf r}_{ij}|$. [The last point follows from the fact that a scalar function $V:\mathbb{R}^3\to \mathbb{R}$, with the property that the gradient $${\bf \nabla} V({\bf r})~\parallel~ {\bf r}$$ is parallel to the position vector ${\bf r}$, can only depend on the length $|{\bf r}|$. Can you see why? Hint: Decompose the gradient in spherical coordinates.]
References:


*

*Herbert Goldstein, Classical Mechanics, Chapter 1.

A: The strong law of action and reaction says that the forces that two bodies exert on each other have the same magnitude, opposite direction and act along the line joining the particles. When you want that last bit to be true and you want to write the force on particle $i$ as $-\nabla_i V_{ij}$, then the potential has to be a function of the relative distance.
You can see this the following way: 
Take a simple potential depending on the relative distance like
$$V_{12}=|\vec r_1 - \vec r_2|$$
and compute $\vec F_1= -\nabla_1 V_{12}$. You should get $$\vec F_{12} = -|\vec r_1 - \vec r_2|^2 (\vec r_1 - \vec r_2) $$ and $\vec F_{21}=-\vec F_{12}$. 
Now do the same with a potential depending on the relative velocities: 
$$\hat V_{12}=|\vec v_1 - \vec v_2|$$
Somewhere you should get expressions like $\frac{\partial}{\partial x_1}v_{x1}$ but we will set them all to 1, so you get as an answer
$$\vec F_{12} = -|\vec v_1 - \vec v_2|^2 (\vec v_1 - \vec v_2) $$ and $\vec F_{21}=-\vec F_{12}$ again. 
Now you might ask "the forces have same magnitude and opposite direction, shouldn't the strong law of action and reaction hold as well?" 
The answer is no because of the "last bit" I mentioned above: since $\vec v_1 - \vec v_2$ is not $\vec r_1 - \vec r_2$, the force from the potential $\hat V$ doesn't point along the line joining the particles. Just think of two particles, one at x=-1 and the second at x=1, both flying in the y-direction: while $\vec r_1 - \vec r_2$ points from particle one to the other, $\vec v_1 - \vec v_2$ and the force calculated from it is zero.
This is why the strong law of action and reaction leads to the internal energy only depending on relative distances: only forces which arise from a potential depending on the relative distances point along the line joining the particles.
