Where Did I Go Wrong with Conserved Quantities in Double Pendulum Lagrangian?

[Physgeek64]

Homework Statement

Hi, I'm doing the double pendulum problem in free space and I've noticed that I get two different conserved values depending on how I define my angles. Obviously, this should not be the case, so I'm wondering where I've gone wrong.

Homework Equations

The Attempt at a Solution

For simplicity m=1 and r=1 in both cases for both bobs
If I choose to define both angles from the vertical then the Lagrangian is

$L=\dot{\theta}^2 +\dot{\phi}^2 +\dot{\theta}\dot{\phi}\cos{\theta - \phi}$
for $\theta \rightarrow \theta + \delta$ and $\phi \rightarrow \phi + \delta$ $f_{\theta}=1$, $f_{\phi}=1$

Then the conserved angular momentum is

$Q= \frac{\partial L}{\partial \dot{\theta}}+\frac{\partial L}{\partial \dot{\phi}}$
$Q= 2\dot{\theta}+\dot{\phi}+(\dot{\theta} +\dot{\phi})\cos{\theta - \phi}$

If we define the upper angle from the vertical and the second angle as measured from the first then the Lagrangian is

$L=\dot{\theta}^2+\frac{1}{2}(\dot{\theta} +\dot{\alpha})^2+\dot{\theta}(\dot{\theta}+\dot{\alpha})\cos{\alpha}$

The conserved quantity is then

$Q= 4\dot{\theta}+2\dot{\alpha}+3\dot{\theta}\cos{\alpha} +\dot{\alpha}\cos{\alpha}$

However if I sub in $\phi= \theta + \alpha$ into Q it turns out these are not the same. Wondering where I have gone wrong.. Many thanks

 

[Physgeek64]

I have worked out my mistake. $\alpha$ should not be varied in the second case. Thanks :)