# Constraints and time derivative

Consider a system of $$N$$ particles. There are $$C$$ holonomic time independent constraints, \begin{aligned} f_1(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0 \\ f_2(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0 \\ \vdots \\ f_C(\mathbf{r}_1,\dots,\mathbf{r}_N) & =0,\end{aligned} which can be written in a shorter form: $$f_n(\mathbf{r}_i)=0$$.

For an allowed trajectory $$\{\mathbf{r}_i(t)\}$$ of the particles we have $$f_n(\mathbf{r}_i(t)) = 0, \forall t$$.

When we take the derivative with respect to time, we find the conditions for the allowed velocities: $$\sum_i \nabla_i f_n(\mathbf{r}_i(t))\cdot \mathbf{\dot{r}}_i(t) = 0.$$

I don't see where the result from the blockquote comes from. It's clear that all $$\mathbf{r}_i$$ are functions of time, so we have $$f_n(\mathbf{r}_1(t), \dots, \mathbf{r}_N(t))$$ for all the contraints $$(n=1,\dots, C)$$. In fact, $$\mathbf{r}_i(t) = (x_i(t),y_i(t),z_i(t)), i=1,\dots,N.$$ It seams like the $$i$$ of the summation is not the same as the index $$i$$ of $$\mathbf{r}$$ in $$f_n(\mathbf{r}_i)$$.

• Block quote from which reference? Feb 23 '19 at 12:37
• It comes from a syllabus my professor wrote himself, but I believe he got his inspiration from 'Classical Mechanics' (3rd edition, by Goldstein, Poole & Safko). Feb 23 '19 at 12:44
• There everything is correct, if we put $\nabla _i=(\frac {\partial }{\partial x_i},\frac {\partial }{\partial y_i }, \frac {\partial }{\partial z_i}), i=1,...,N$, then $\dot {f_n}=\sum _i \nabla _if_n.\vec {r_i}=0, n=1,...,C$. Feb 23 '19 at 13:04
• @AlexTrounev That should be an answer! as long as you replace your $\vec{r}_i$ by its time derivative.
– user197851
Feb 24 '19 at 14:26
• @LonelyProf you're right. must be $\dot {f_n}=\sum _i\nabla _if_n.\frac {d\vec {r}}{dt}$ Feb 24 '19 at 15:11

$$\dot {f_n}=\sum _i\nabla _if_n(\vec {r_1},...,\vec {r_N} ).\frac{d\vec r_i}{dt}=0, n=1,...,C$$
$$\nabla _i=(\frac {\partial}{\partial x_i},\frac {\partial}{\partial y_i},\frac {\partial}{\partial z_i}), i=1,...,N$$