Lagrangian where time is a dependent coordinate

[dipole]

Homework Statement

I don't know why I'm having trouble here, but I want to show that, if we let $t = t(\theta)$ and $q(t(\theta)) = q(\theta)$ so that both are now dependent coordinates on the parameter $\theta$, then

$$L_{\theta}(q,q',t,t',\theta) = t'L(q,q'/t',t)$$

where $t' = \frac{dt}{d\theta}, q' = \frac{dq}{d\theta}$

The Attempt at a Solution

Writing $L = \frac{m}{2} \dot{q}^2 - V(q)$, we let $\frac{d}{dt} \to \frac{d\theta}{dt}\frac{d}{d\theta}$ and then,

$$L = \frac{m}{2} \frac{q'^2}{t'^2} - V(q)$$

Which clearly doesn't agree with what I need to show... where am I going wrong here?

 

[Nate810]

A:We have $L_{\theta}(q,q',t,t',\theta)=\frac{\partial L}{\partial q}\frac{\partial q}{\partial \theta}+\frac{\partial L}{\partial q'}\frac{\partial q'}{\partial \theta}+\frac{\partial L}{\partial t}\frac{\partial t}{\partial \theta}+\frac{\partial L}{\partial t'}\frac{\partial t'}{\partial \theta}+\frac{\partial L}{\partial \theta}$ By the chain rule, $\frac{\partial q}{\partial \theta}=\frac{\partial q}{\partial t}\frac{\partial t}{\partial \theta}=\frac{q'}{t'}$ and $\frac{\partial t}{\partial \theta}=t'$ Also, $\frac{\partial q'}{\partial \theta}=\frac{\partial q'}{\partial t}\frac{\partial t}{\partial \theta}+\frac{\partial q'}{\partial q}\frac{\partial q}{\partial \theta}=\frac{q''}{t'}+\frac{q'}{t'}\frac{q'}{t'}=\frac{q''t'-q'^2}{t'^2}$And, $\frac{\partial t'}{\partial \theta}=\frac{\partial t'}{\partial t}\frac{\partial t}{\partial \theta}+\frac{\partial t'}{\partial q}\frac{\partial q}{\partial \theta}=\frac{t''}{t'}+\frac{0}{t'}\frac{q'}{t'}=\frac{t''}{t'}$Finally, we compute the last partial derivative.$\frac{\partial L}{\partial \theta}=-\frac{\partial V}{\partial q}\frac{\partial q}{\partial \theta}=-\frac{\partial V}{\partial q}\frac{q'}{t'}$So, $L_{\theta}(q,q',t,t',\