Conserved quantity for lagrangian $L=\frac{1}{2}(m_ar_1^2\dot{\theta_1^2}+m_br_2^2\dot{\theta_2^2})+kr_1r_2\cos(\theta_1-\theta_2)$ I am considering the lagrangian
$$L(\theta_1,\theta_2,\dot{\theta_1^2},\dot{\theta_2^2})=\frac{1}{2}(m_ar_1^2\dot{\theta_1^2}+m_br_2^2\dot{\theta_2^2})+kr_1r_2\cos(\theta_1-\theta_2)$$
It can be noted that the lagrangian is invariant for rotations of both points of the same angle
$$\varphi_{\epsilon}(\theta_1,\theta_2) \longrightarrow(\theta_1+\epsilon,\theta_2+\epsilon)$$
According to Noether's theorem:
$$J=\dfrac{\partial L}{\partial\dot{\theta_1}}+\dfrac{\partial L}{\partial\dot{\theta_2}}=m_ar_1^2\dot{\theta_1}+m_b r_2^2\dot{\theta_2}$$
Consider a linear transformation of coordinates
\begin{align*}
\theta_1&=b\phi_1+a_{12}\phi_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, b,a_{12},a_{22} \, \in \mathbb{R}\\
\theta_2&=b\phi_1+a_{22}\phi_2
\end{align*}
So new Lagrangian
$$L(\phi_2,\dot{\phi_1^2},\dot{\phi_2^2})=\frac{m_ar_1^2}{2} \left( b^2\dot{\phi_1}+2ba_{12}\dot{\phi_1}\dot{\phi_2}+a^2_{12}\dot{\phi_2^2}\right)  +\frac{m_br_2^2}{2} \left( b^2\dot{\phi_1^2}+2ba_{22}\dot{\phi_1}\dot{\phi_2}+a^2_{22}\dot{\phi_2^2}\right)+kr_1r_2\cos(\phi_2(a_{12}-a_{22}))$$
$\phi_1$ is a cyclic coordinate so the respective conjugated momentum is conserved
\begin{align*}p_{\phi_1}=\dfrac{\partial L}{\partial \dot{\phi_1^2} } &= m_ar_1^2b^2\dot{\phi_{1}}+m_aba_{12}r_1^2\dot{\phi_{2}}+m_br_2^2b\dot{\phi_{1}}+m_bba_{22}r^2_2\dot{\phi_{2}}= \\ &= b^2(m_ar^2_1+m_br_2^2)\dot{\phi_1}+b(m_aa_{12}r_1^2+m_ba_{22}r^2_2)\end{align*}
Considering the result found above with Noether's theorem and replacing the transformation
$$J=\dfrac{\partial L}{\partial\dot{\theta_1}}+\dfrac{\partial L}{\partial\dot{\theta_2}}=m_ar_1^2\dot{\theta_1+m_br_2^2\dot{\theta_2}}$$
$$J_1=b(m_ar_1^2+m_br^2_2)\dot{\phi_1}+(m_ar_1^2a_{12}+m_br^2_2a_{22})\dot{\phi_2}$$
I find that the two results for the conserved quantity do not coincide. 
Why?
$$J_1 \neq p_{\phi_1}$$
Are they two different motion constants? Did I make any mistakes doing the math? Should the two paths, theoretically speaking, lead to the same result?
 A: You state that
\begin{align*}
\theta_1&=b\phi_1+a_{12}\phi_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, b,a_{12},a_{22} \, \in \mathbb{R}\\
\theta_2&=b\phi_1+a_{22}\phi_2
\end{align*}
now, 
\begin{align*}
\dot{\theta}_1&=b\dot{\phi}_1+a_{12}\dot{\phi}_2\\
\dot{\theta}_2&=b\dot{\phi}_1+a_{22}\dot{\phi}_2
\end{align*}
Now, from the Lagranigian
\begin{align}
\frac{1}{2}m_a r^2_{1}\dot{\theta}_{1}^2 + \ldots&= \frac{1}{2}m_a r^2_{1}(b\dot{\phi}_1+a_{12}\dot{\phi}_2)^2 +\ldots \\
&=\frac{1}{2}m_a r^2_{1}(b^2 \dot{\phi}_{1}^2+2a_{12}b\dot{\phi}_{1}\dot{\phi}_{2}+a_{12}^2\dot{\phi}_{2}^2)+\ldots
\end{align}
See how I find that the cross terms $\dot{\phi}_{1}\dot{\phi}_{2}$ is not quadratic? I can't see how you get quadratic cross terms in your transformed Lagrangian? When you perform the partial derivatives with respect to $\dot{\phi}_{1}^2$ now all should be conserved...
A: Another than the mistake in the expansion of $\dot\theta_1^2$ and $\dot\theta_2^2$, be careful that Lagrangian uses $\theta_1,\theta_2,\dot\theta_1,\dot\theta_2$ as input variables, not $\theta_1,\theta_2,\dot\theta_1^2,\dot\theta_2^2$. Similarly, the partial derivatives for defining conjugate momentum should be $\frac{\partial L}{\partial \dot\theta_1}$, not $\frac{\partial L}{\partial \dot\theta_1^2}$.
Fix all these mistakes, do the calculations again and see if there is still discrepancy in the conserved quantity.
As for whether the two conserved quantities should be the same, I would say they should be, because the symmetry $(\theta_1,\theta_2)\mapsto(\theta_1+\epsilon,\theta_2+\epsilon)$ is essentially the same as $(\phi_1,\phi_2)\mapsto(\phi_1+\frac{\epsilon}{b},\phi_2)$.
