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A particle's position is given by $r_i=r_i(q_1,q_2,...,q_n,t)$. So velocity: $$v_i=\frac{dr_i}{dt} = \sum_k \frac{\partial r_i}{\partial q_k}.\dot q_k + \frac{\partial r_i}{\partial t} $$

In my book it's given $$\frac{\partial v_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}$$ without any proof. So I tried to take derivative of $v_i$ w.r.t $\dot q_k$. I can only arrive at the proof if $$\frac{\partial r_i}{\partial \dot q_k} = 0?$$ Why is that? Is it because of explicit dependence of $r_i$ on $\dot q_k$? Sorry if this question is too basic but I'm confused because I believe it could be written as $$\frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}$$

Weezy
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  • I think your book makes an abuse of notation by writing this. It could perhaps make sense in some context (like in Euler-Lagrange equations). Perhaps you should specify what the book does with these identities afterwards? – H. H. Rugh Oct 01 '16 at 12:03
  • The books is classical mech. by Goldstein and the equation is needed to write D'alembert's principle in terms of generalized coordinates instead of cartesian coordinates. I found this link as possibly related to this question – Weezy Oct 01 '16 at 12:06
  • Yes. That's what causing my confusion. What I want to know why is $\frac{\partial r_i}{\partial \dot q_k} = 0$ despite being possible to write it as the last eqn. using chain rule. I guess it has something to do with explicit/implicit dependence which I don't understand and would be good to have someone clarify it for me. – Weezy Oct 01 '16 at 12:15

1 Answers1

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A very partial explanation:

In order to eliminate indices, consider the free motion of one particle on a curve in 3D. The equation is $\ddot{r} \cdot \delta r = 0$ (say scalar product in 3D, and $\delta r$ in the tangent direction of the curve). When transforming to a generalized coordinate $r=r(q)$ this becomes first $\dot{r}=\frac{\partial r}{\partial q} \dot{q}$ and then $$ \left[ F(q,\dot{q}) \right]\ \delta q = \left[\left( \frac{\partial^2 r}{\partial q^2} \dot{q}^2 + \frac{\partial r}{\partial q} \ddot{q} \right) \cdot \frac{\partial r}{\partial q}\right] \delta q=0$$ So the equation of motion becomes $F(q,\dot{q})=0$ in the $q$ coordinate. Now suppose you would like to derive the equivalent Euler-Lagrange equations for that motion starting with the kinetic energy $T(\dot{r}) = \frac12 (\dot{r})^2$. In the generalized coordinate this becomes: $$ T(q,\dot{q}) = \left( \frac{\partial r(q)}{\partial q} \dot{q} \right)^2$$ If you now define this to be an expression in the two variables $q,\dot{q}$ then you may calculate $\partial T / \partial \dot{q} = (\partial r/\partial q)^2 \dot{q}$ and then after a small calculation find: $$ F = \frac{d}{dt} \frac{\partial T}{\partial \dot{q}} - \frac{\partial T}{\partial q} $$ And $F=0$ is then the Euler-Lagrange equation for $T$. Upshot: We think of $T$ as a function of two indepdent variables, say $q$ and $y$ but when calculating $d/dt$ we insert $y=\dot{q}$ before taking the derivative.

H. H. Rugh
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