Proof of independence of position and velocity

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Homework Help Overview

The discussion revolves around the independence of position and velocity in the context of classical mechanics, specifically examining the relationship between a particle's position, given as a function of generalized coordinates and time, and its velocity derived from that position.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the derivative of velocity with respect to generalized velocities and question the implications of the independence of position from time derivatives. There is confusion regarding the assumption that position does not depend on the time derivatives of coordinates.

Discussion Status

Some participants have provided insights into the assumptions underlying the problem, particularly the nature of the dependence of position on coordinates versus their time derivatives. There is ongoing exploration of the implications of these assumptions without reaching a consensus.

Contextual Notes

Participants note the importance of adhering to the homework template and express concerns about the clarity of the problem setup and the definitions involved.

weezy
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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 $$\frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}$$ and that $$\frac{\partial q_k}{\partial \dot q_k}$$ is not zero.
 
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You said it yourself, ##r_i## is assumed to be a function of the coordinates ##q_k##, not of their time derivatives.
 
In future posts, please do not delete the homework template with its three parts.
weezy said:
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 $$\frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}$$ and that $$\frac{\partial q_k}{\partial \dot q_k}$$ is not zero.
 
Orodruin said:
You said it yourself, ##r_i## is assumed to be a function of the coordinates ##q_k##, not of their time derivatives.
yes but can't that be written like ##
\frac{\partial r_i}{\partial \dot q_k} = \frac{\partial r_i}{\partial q_k}.\frac{\partial q_k}{\partial \dot q_k}##?
##
\frac{\partial r_i}{\partial \dot q_k} ## is not zero that's obvious so does it imply that ##
\frac{\partial q_k}{\partial \dot q_k}
## is zero?
 
Mark44 said:
In future posts, please do not delete the homework template with its three parts.
Sorry for any inconvenience. I'll be careful.
 

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