Prove forces derived from a velocity-dependent potential are not central

[Happiness]

I don't see why the last sentence in the attachment is true. It claims that if $V_{ij}$ or simply $V$ is also a function of the difference of velocities of particles $i$ and $j$, then the force derived from $V$ is not central. In other words, if $V=V(|r_i-r_j|, |v_i-v_j|)$, then (1.34) is not satisfied.

Let $p=|r_i-r_j|$ and $q=|v_i-v_j|$ and $\partial_x=\frac{\partial}{\partial x}$.

$\nabla_iV(p, q)=\partial_{x_i}V(p, q)\vec{e_x}+\partial_{y_i}V(p, q)\vec{e_y}+\partial_{z_i}V(p, q)\vec{e_z}$. Definition of $\nabla_i$ is given by the first sentence of the attachment.

Since $\partial_{x_i}V(p, q)=\partial_{x_i}V(p)$, the RHS above should be the same as the RHS of (1.34). Then, (1.34) will be satisfied.

$\partial_{x_i}V(p, q)=\partial_{x_i}V(p)$ for the same reason as $\partial_x(x+\dot{x}^2)=1=\partial_xx$.

EDIT: I found the mistake. $\partial_{x_i}V(p, q)\neq\partial_{x_i}V(p)$ for the same reason as $\partial_x(x+x\dot{x})\neq\partial_xx$.

 

[Happiness]

Nonetheless, I am still able to prove that the force derived from $V=V(p, q)$ satisfies (1.34), contrary to what is claimed by the book.

Recall that $p=|r_i-r_j|=\sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$ and $q=|v_i-v_j|$.

Thus, $\frac{\partial p}{\partial x_i}=\frac{1}{2p}2(x_i-x_j)=\frac{1}{p}(x_i-x_j)$.

Let $V'(p, q)=\partial_{p}V(p, q)$.

$\nabla_iV(p, q)=\partial_{x_i}V(p, q)\vec{e_x}+\partial_{y_i}V(p, q)\vec{e_y}+\partial_{z_i}V(p, q)\vec{e_z}$

$=V'(p, q)\frac{\partial p}{\partial x_i}\vec{e_x}+V'(p, q)\frac{\partial p}{\partial y_i}\vec{e_y}+V'(p, q)\frac{\partial p}{\partial z_i}\vec{e_z}$

$=\frac{V'(p, q)}{p}(x_i-x_j)\vec{e_x}+\frac{V'(p, q)}{p}(y_i-y_j)\vec{e_y}+\frac{V'(p, q)}{p}(z_i-z_j)\vec{e_z}$

$=\frac{V'(p, q)}{p}(\vec{r_i}-\vec{r_j})$,

which is equivalent to the RHS of (1.34).

What's wrong?