newtonian mechanics - 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.

Answer

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:

1. 
   

   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.

   
   


2. 
   

   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.

   
   
   


3. 
   

   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 $$ {\bf F}_{ij}~=~-{\bf \nabla}_{j} V_{ij}. $$

   
   


4. 
   

   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:

1. Herbert Goldstein, Classical Mechanics, Chapter 1.