Double pendulum Lagrangian using small angle approximation formula

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this part (b) of this problem,

From (a), we know that
$\mathcal{L}\left(\phi_{1}, \phi_{2}, \dot{\phi}_{1}, \dot{\phi}_{2}\right)=\frac{1}{2} m \ell^{2}\left[2 \dot{\phi}_{1}^{2}+\dot{\phi}_{2}^{2}+2 \cos \left(\phi_{1}-\phi_{2}\right) \dot{\phi}_{1} \dot{\phi}_{2}\right]+m g \ell\left(2 \cos \phi_{1}+\cos \phi_{2}\right)$

And we want $\mathcal{L}\left(\phi_{1}, \phi_{2}, \dot{\phi}_{1}, \dot{\phi}_{2}\right)=\frac{1}{2} m \ell^{2}\left[2 \dot{\phi}_{1}^{2}+\dot{\phi}_{2}^{2}+2 \dot{\phi}_{1} \dot{\phi}_{2}\right]-\frac{1}{2} m g \ell\left(2 \phi_{1}^{2}+\phi_{2}^{2}\right)$

We use the following formula for the small angle approximation of cosine

$\cos \phi_1 = 1 - \frac{\phi^2_1}{2}$

$\cos \phi_2 = 1 - \frac{\phi^2_2}{2}$

There imply that,

$\cos(\phi_1 - \phi_2) = 1 - \frac{(\phi_1 - \phi_2)^2}{2}$

$\cos(\phi_1 - \phi_2) = 1 - \frac{\phi_1^2 - 2\phi_1\phi_2 + \phi^2_2}{2}$

Thus, this proves that $\cos(\phi_1 - \phi_2)$ is a second order term to we must remove it from the expression.

However, why don't we remove $\dot \phi_2 \dot \phi_1$ instead?

This would mean $\mathcal{L} = \frac{1}{2}ml^2(2\dot \phi_1^2 + \dot \phi_2^2 + 1 - \frac{\phi^2_1}{2} + \phi_1\phi_2 - \frac{\phi^2_2}{2}) - \frac{1}{2}mgl(2\phi^2_1 + \phi^2_2)$

I also have ao confusion about transfomring one of the other terms namely the $mgl(2\cos\phi_1 + \cos \phi_2)$

I get $mgl(3 - \phi_1^2 + \frac{\phi^2_2}{2})$ instead of $\frac{1}{2}mgl(2\phi_1^2 + \phi_2^2)$

This is from

$2\cos\phi_1 + \cos\phi_2 = 2[1 - \frac{\phi_1^2}{2}] + 1 - \frac{\phi_2^2}{2} = 3 - \phi_1^2 + \frac{\phi_2^2}{2}$

Does anybody please know what I have done wrong?

Thanks!

 

[haruspex]

Thus, this proves that $\cos(\phi_1 - \phi_2)$ is a second order term to we must remove it from the expression.

However, why don't we remove $\dot \phi_2 \dot \phi_1$ instead?

The complete term is $\cos(\phi_1 - \phi_2)\dot \phi_2 \dot \phi_1$.
Expanding $\cos(\phi_1 - \phi_2)$ produces a second order term and a fourth order term.

 

[haruspex]

I also have ao confusion about transfomring one of the other terms namely the $mgl(2\cos\phi_1 + \cos \phi_2)$

I get $mgl(3 - \phi_1^2 + \frac{\phi^2_2}{2})$ instead of $\frac{1}{2}mgl(2\phi_1^2 + \phi_2^2)$

No, you should get $mgl(3 - \phi_1^2 - \frac{\phi^2_2}{2})$, instead of $-\frac{1}{2}mgl(2\phi_1^2 + \phi_2^2)$. The difference is a constant.

 

[Orodruin]

The difference is a constant.

… and at the risk of stating the obvious, a constant addition to the Lagrangian does not affect the equations of motion and can therefore be removed.

Also, to state #2 slightly different: You have a term on the form $\dot \phi^2 g(\phi)$. The only second order term from such an expression comes from the zero order contribution of $g(\phi)$ as
$$
\dot \phi^2 g(\phi) = \dot \phi^2 [g(0) + \mathcal O(\phi)]
$$
(In this particular case $\mathcal O(\phi^2)$)

 

[member 731016]

No, you should get $mgl(3 - \phi_1^2 - \frac{\phi^2_2}{2})$, instead of $-\frac{1}{2}mgl(2\phi_1^2 + \phi_2^2)$. The difference is a constant.

… and at the risk of stating the obvious, a constant addition to the Lagrangian does not affect the equations of motion and can therefore be removed.

Also, to state #2 slightly different: You have a term on the form $\dot \phi^2 g(\phi)$. The only second order term from such an expression comes from the zero order contribution of $g(\phi)$ as
$$
\dot \phi^2 g(\phi) = \dot \phi^2 [g(0) + \mathcal O(\phi)]
$$
(In this particular case $\mathcal O(\phi^2)$)

Thank you for your replies @haruspex and @Orodruin!

We want $$\mathcal{L}\left(\phi_{1}, \phi_{2}, \dot{\phi}_{1}, \dot{\phi}_{2}\right)=\frac{1}{2} m \ell^{2}\left[2 \dot{\phi}_{1}^{2}+\dot{\phi}_{2}^{2}+2 \dot{\phi}_{1} \dot{\phi}_{2}\right]-\frac{1}{2} m g \ell\left(2 \phi_{1}^{2}+\phi_{2}^{2}\right)$$

Taking the first part of the Lagrangian and given that $$\cos(\phi_1 - \phi_2) = 1 - \frac{\phi_1^2 - 2\phi_1\phi_2 + \phi_2^2}{2}$$

We write that $$\frac{1}{2}ml^2[2\dot \phi_1^2 + \dot \phi_2^2 + 2\cos(\phi_1 - \phi_2)\dot \phi_1 \dot \phi_2]$$ term as $$\frac{1}{2}ml^2[2\dot \phi_1^2 + \dot \phi_2^2 + 2(1 - \frac{\phi_1^2 - 2\phi_1\phi_2 + \phi_2^2}{2})\dot \phi_1 \dot \phi_2]$$

Which is same as,

$$\frac{1}{2}ml^2[2\dot \phi_1^2 + \dot \phi_2^2 + (2 - \phi_1^2 + 2\phi_1\phi_2 - \phi_2^2)\dot \phi_1 \dot \phi_2]$$

Thus $$\dot \phi_1 \dot \phi_2$$ is a second order so we omit it by setting it equal to 1, thus, $$\dot \phi_1 \dot \phi_2 = 1$$. One can see that this relates the two time derivatives. Then we can integrate with respect to time to find $$\phi_1$$ in terms of $$\phi_2$$ or $$\phi_2$$ in terms of $$\phi_1$$. Is this please correct?

Thanks!

 

[Orodruin]

Thus $$\dot \phi_1 \dot \phi_2$$ is a second order so we omit it by setting it equal to 1,

Absolutely not! You want to keep up to second order terms and you definitely cannot put them equal to one! The point was that it is already a second order term, meaning that any terms higher than the constant term in the expansion of the cosine will result in quartic terms or higher - which shoukd be ignored. Only the constant term in the cosine expansion contributes.