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?

  • $\begingroup$ 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. $\endgroup$
    – alfba
    Commented Apr 16, 2019 at 14:17
  • 1
    $\begingroup$ This is just the chain rule of calculus applied to a function of more than one variable. There is no factor 6. $\endgroup$
    – user197851
    Commented Apr 16, 2019 at 14:40
  • $\begingroup$ 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). $\endgroup$
    – alfba
    Commented Apr 16, 2019 at 15:03
  • $\begingroup$ (As Pedro Fernando made clear, I was confused about the $V$ function, it's "just the chain rule" as you say :-). ) $\endgroup$
    – alfba
    Commented Apr 16, 2019 at 15:21

1 Answer 1


The problem is that you are making incorrect use of the chain rule for derivatives. In the last step

$$\frac{\partial V}{\partial q_{j}}=\sum_{i}\frac{\partial V}{\partial x_{i}}\frac{\partial x_{i}}{\partial q_{j}}=\frac{\partial V}{\partial x_{1}}\frac{\partial x_{1}}{\partial q_{j}}+\frac{\partial V}{\partial y_{1}}\frac{\partial y_{1}}{\partial q_{j}}+\frac{\partial V}{\partial z_{1}}\frac{\partial z_{1}}{\partial q_{j}}+\frac{\partial V}{\partial x_{2}}\frac{\partial x_{2}}{\partial q_{j}}+\frac{\partial V}{\partial y_{1}}\frac{\partial y_{2}}{\partial q_{j}}+\frac{\partial V}{\partial z_{2}}\frac{\partial z_{2}}{\partial q_{j}}$$

  • $\begingroup$ Thanks, I didn't think clearly about $V$ being a 6-variable function. $\endgroup$
    – alfba
    Commented Apr 16, 2019 at 15:19

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