# Why is the kinetic energy an elliptic quadratic form? (double pendulum)

Here is one question concerning the kinetic energy of the double pendulum I've used the coordinates $$x_1=L_1\sin\theta_1,~~x_2=L_1\cos\theta_1$$ for the first bob with mass $m_1$ and $$x_3=L_1\sin\theta_1+L_2\sin\theta_2,~~x_4=L_1\cos\theta_1+L_2\cos\theta_2$$ for the second bob with mass $m_2$.

Now the kinetic energy $T$ is given by $$T=\frac{1}{2}((m_1+m_2)L_1^2w_1^2+m_2L_2^2w_2^2)+m_2L_1 L_2\cos(\theta_1-\theta_2)w_1 w_2.$$

Now in a book I read the following:

T is (for any values $\theta_1,\theta_2$) an elliptic quadratic form in $w_1, w_2$. Because $T$ is bounded, therefore $w_1$ and $w_2$ are bounded.

This is not clear to me!

1.) Why is $T$ an elliptic quadratic form in $w_1,w_2$?

2.) Why are $w_1$ amd $w_2$ bounded if $T$ is an elliptic quadratic form and bounded?

It would be nice to get an explanation! Thank you, with kind regards.

• What is $w$? Do you mean the angular speed $\mathrm d\theta/\mathrm dt$, which is typically denoted omega ($\omega$, \omega)? – Rahul Jun 9 '14 at 16:27
• Yes, that is what I mean. – math12 Jun 9 '14 at 16:29