Goldstein equation 1.33

Question

I am trying to read from Goldstein for self-study but I am stuck on equation 1.33. Let me restate some of the lines from Goldstein (with some modification):

If $\textbf{F}_{ij}$ (internal force, force exerted on particle $i$ by particle $j$) depends only on the relative positions $\textbf{r}_{ij}$ and can be derived from a scalar potential energy function $V_{ij}(\textbf{r}_{ij})$ with $V_{ij}=V_{ji}$ then

$$\textbf{F}_{ij}=-\nabla_{i}V_{ij}$$ and

$$\textbf{F}_{ij}=-\nabla_{i}V_{ij}=+\nabla_{j}V_{ij}=+\nabla_{j}V_{ji}=-\textbf{F}_{ji}.\tag{1.33}$$

Now, I am not able to understand how I can prove $-\nabla_{i}V_{ij}=+\nabla_{j}V_{ij}$. After trying a lot I think the problem is that probably I didn't understand $\textbf{F}_{ij}=-\nabla_{i}V_{ij}$ properly. Though I do understand that for conservative forces we can express the force as negative gradient of potential. But I guess the indices here is what I don't understand. My understanding of equation $\textbf{F}_{ij}=-\nabla_{i}V_{ij}$ is that we are taking gradient of $V_{ij}$ with respect to the coordinates of the $i$th particle. But then I don't know how to prove $-\nabla_{i}V_{ij}=+\nabla_{j}V_{ij}$. Please help.

Answer 1

Newton's third axiom states $F_{ij}=-F_{ji}$. The potential energy between the particles is the pair potential satisfying $V(r_{ij})=V(r_{ji})$

The force acting on particle i due to particle j is given by $F_{ij}=-\nabla_{i}V_{ij}$

and the force acting on particle j due to particle i is $F_{ji}=-\nabla_{j}V(r_{ij})$

And now use Newton's third axiom to obtain

$F_{ij}=-\nabla_{i}V_{ij}=-F_{ji}=\nabla{j}V(r_{ij})$

I hope this helps