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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?

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    $\begingroup$ Your linear transformation of coordinates doesn't seem invertible. Specifically, $\phi_2$ is missing in the transformation. $\endgroup$
    – edm
    Dec 6, 2017 at 12:26
  • $\begingroup$ Sorry there was a mistake, I hope I have not committed others in transcribing $\endgroup$ Dec 6, 2017 at 12:30
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    $\begingroup$ You calculated $\dot\theta_1^2= b^2\dot{\phi_1^2}+2ba_{12}\dot{\phi_1^2}\dot{\phi_2^2}+a^2_{12}\dot{\phi_2^2}$, and $\dot\theta_2^2= b^2\dot{\phi_2^2}+2ba_{22}\dot{\phi_1^2}\dot{\phi_2^2}+a^2_{22}\dot{\phi_2^2}$, which are wrong. $\endgroup$
    – edm
    Dec 6, 2017 at 12:37
  • $\begingroup$ Sorry I'm a very distracted person and I made several mistakes in transcribing now I hope to have them corrected, the two results are not the same but almost, maybe there is still some error in the calculations. $\endgroup$ Dec 6, 2017 at 13:39
  • $\begingroup$ I would also like to know whether the two results should be the same from a theoretical point of view or not $\endgroup$ Dec 6, 2017 at 13:41

2 Answers 2

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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...

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  • $\begingroup$ Are you missing a 2 factor in the last formula? $\endgroup$
    – Jon
    Dec 6, 2017 at 12:51
  • $\begingroup$ @Jon, Yes Jon, yes I am. I will edit, thank you for noticing,.. $\endgroup$
    – user284001
    Dec 6, 2017 at 12:52
  • $\begingroup$ Sorry I'm a very distracted person and I made several mistakes in transcribing now I hope to have them corrected, the two results are not the same but almost, maybe there is still some error in the calculations. $\endgroup$ Dec 6, 2017 at 13:38
  • $\begingroup$ I would also like to know whether the two results should be the same from a theoretical point of view or not $\endgroup$ Dec 6, 2017 at 13:42
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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 $(\phi_1,\phi_2)\mapsto(\phi_1+\frac{\epsilon}{b},\phi_2)$.

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  • $\begingroup$ thank you very much, I'm trying to remove all the transcription mistakes, I would also like to know whether the two results should be the same theoretically speaking or not $\endgroup$ Dec 6, 2017 at 13:45
  • $\begingroup$ @StefanoBarone see the new paragraph. $\endgroup$
    – edm
    Dec 6, 2017 at 14:05

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