# Vector calculus simplification in calculation of generalized force

Consider a system of $$N$$ particles subject to forces $$\vec F_i\ (i=1\dots N)$$ that derive from a potential $$V$$. My lecture notes propose a simple proof that

$$Q_j = -\frac{\partial V}{\partial q_j}$$

where the generalized forces are defined as $$Q_j = \sum_i \vec F_i\cdot\frac{\partial\vec r_i}{\partial q_j}$$. It goes like this:

$$Q_j = \sum_i \vec F_i\cdot\frac{\partial\vec r_i}{\partial q_j} = -\sum_i\vec\nabla_i V\cdot\frac{\partial\vec r_i}{\partial q_j} = -\frac{\partial V}{\partial q_j}$$

I'm trying to understand the last step in detail, but I get a wrong answer by a factor $$N$$. For example with two particles, and writing $$\vec r_i = (x_i,y_i,z_i)$$, I have \begin{aligned} \sum_i\vec\nabla_i V\cdot\frac{\partial\vec r_i}{\partial q_j} &= \vec\nabla_1V\cdot\frac{\partial\vec r_1}{\partial q_j} + \vec\nabla_2V\cdot\frac{\partial\vec r_2}{\partial q_j} \\ &= (\tfrac{\partial V}{\partial x_1}, \tfrac{\partial V}{\partial y_1}, \tfrac{\partial V}{\partial z_1}) \cdot(\tfrac{\partial x_1}{\partial q_j}, \tfrac{\partial y_1}{\partial q_j}, \tfrac{\partial z_1}{\partial q_j}) + (\tfrac{\partial V}{\partial x_2}, \tfrac{\partial V}{\partial y_2}, \tfrac{\partial V}{\partial z_2}) \cdot(\tfrac{\partial x_2}{\partial q_j}, \tfrac{\partial y_2}{\partial q_j}, \tfrac{\partial z_2}{\partial q_j}) \\[1ex] &= \frac{\partial V}{\partial q_j} + \frac{\partial V}{\partial q_j} \\[1ex] &= 2\frac{\partial V}{\partial q_j} \end{aligned} What did I do wrong to get this factor 2?

• There's already a question on the same result (or very close) but a with a different proof: physics.stackexchange.com/q/271213 . But I'm really trying to understand this proof. Commented Apr 16, 2019 at 14:17
• This is just the chain rule of calculus applied to a function of more than one variable. There is no factor 6.
– user197851
Commented Apr 16, 2019 at 14:40
• Thanks, that corrects a factor 3 (I edited the question accordingly). But I'm still wrong by a factor $N$ (factor 2 in the example). Commented Apr 16, 2019 at 15:03
• (As Pedro Fernando made clear, I was confused about the $V$ function, it's "just the chain rule" as you say :-). ) Commented Apr 16, 2019 at 15:21

$$\frac{\partial V}{\partial q_{j}}=\sum_{i}\frac{\partial V}{\partial x_{i}}\frac{\partial x_{i}}{\partial q_{j}}=\frac{\partial V}{\partial x_{1}}\frac{\partial x_{1}}{\partial q_{j}}+\frac{\partial V}{\partial y_{1}}\frac{\partial y_{1}}{\partial q_{j}}+\frac{\partial V}{\partial z_{1}}\frac{\partial z_{1}}{\partial q_{j}}+\frac{\partial V}{\partial x_{2}}\frac{\partial x_{2}}{\partial q_{j}}+\frac{\partial V}{\partial y_{1}}\frac{\partial y_{2}}{\partial q_{j}}+\frac{\partial V}{\partial z_{2}}\frac{\partial z_{2}}{\partial q_{j}}$$
• Thanks, I didn't think clearly about $V$ being a 6-variable function. Commented Apr 16, 2019 at 15:19