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My original Lagrangian is this, but I want to obtain nonlinear terms considering small oscillations : $$ L = ma^2[\dot \theta^2(1+ 2\sin^2\theta) + \Omega^2\sin^2\theta + 2\Omega_0\cos\theta] . $$ Now, equilibrium point of potential energy $U$ is $\cos\theta_0 = \frac{\Omega_0^2}{\Omega^2}$. Now if $\Omega_0 = \Omega$ and, $x = \theta - \theta_0 $ where $x$ is angular displacement. Then Taylor expansion around equilibrium point $\theta_0$ for potential energy is: $$ U = ma^2(-2 +\frac{x^4}{4})$$ and Kinetic enrgy is: $$ T=ma^2\dot x^2(1+ sin^2x)=ma^2\dot x^2(1 + 2x^2)$$ and finally Lagrangian is:$$ L= T - U =ma^2\dot x^2(1 + 2x^2) - ma^2(-2 +\frac{x^4}{4}) $$ is it procces correct, and if so, can I solve with successive approximation?

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The potential energy divided by $m\,a^2$ is:

$$U={\Omega}^{2} \left( \sin \left( \vartheta \right) \right) ^{2}+2\, \Omega_{{0}}\cos \left( \vartheta \right) $$

the Taylor series for a small $\vartheta$ at $\vartheta_0$ and $\vartheta^n=0~,n=3,4,...~$is:

$$U_T=U(\vartheta_0)+\frac{\partial U}{\partial \vartheta} \bigg|_{\vartheta_0}\left(\vartheta-\vartheta_0\right)+\frac{1}{2}\,\frac{\partial^2 U}{\partial \vartheta^2}\bigg|_{\vartheta_0}\left(\vartheta-\vartheta_0\right)^2$$

edit

notice that for $~\vartheta=\vartheta_0~~$ you don't obtain static equilibrium

Why:

the equation of motion :

$$\frac{d}{dt}\left(\frac{\partial T_T}{\partial \dot \vartheta}\right)+\underbrace{\left(\frac{\partial U_T}{\partial \vartheta}\right)}_{-F}=0$$

thus F is:

$$F=f_1(\vartheta_0)+f_2(\vartheta_0)\,(\vartheta-\vartheta_0)$$

with $f_1=U'\bigg|_{\vartheta_0}~,f_2=U''\bigg|_{\vartheta_0}$

now for $~\vartheta=\vartheta_0~,F=f_1(\vartheta_0)$

thus to obtain static equilibrium at$~\vartheta=\vartheta_0~$ you have to add to the equation of motion the static force $f_1(\vartheta_0)$

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  • $\begingroup$ Presumably the fixed point is at $\vartheta=\vartheta_0$ so you'd need to keep the quadratic term in the series for $U$, so the EOM is linear and non-trivial. $\endgroup$ Sep 25 '20 at 15:48
  • $\begingroup$ yes i will correct it!! $\endgroup$
    – Eli
    Sep 25 '20 at 15:49

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