0
$\begingroup$

In Arnolds book Mathematical Methods of Classical Mechanics he defines the system

$$\ddot{x}_1=-x_1,\,\,\,\ddot{x}_2=-\omega^2x_2^2.$$

The potential energy is $U(x_1,x_2)=\frac{1}{2}(x_1^2+\omega^2x_2^2)$. Because both energies $E_1=\frac{1}{2}\dot{x}_1^2+\frac{1}{2}x_1^2$, $E_2=\frac{1}{2}\dot{x}_2^2+\frac{1}{2}\omega^2x_2^2$ preserved the variable $x_i$ is bounded by $|x_i|<\sqrt{2E_i}$ for $i=1,2$. This defines a rectangle in $\mathbb{R}^2$.

Now the problem given in the book is to show that this rectangle lies completely in the ellipse given by $U(x_1,x_2)=E$.

But if I check at the corners of the rectangle $(\sqrt{2E_1},\sqrt{2E_2})$ I get that $U(\sqrt{2E_1},\sqrt{2E_2})=E_1+\omega^2 E_2 > E$, so they dont lie in the ellipse.

Can anyone help finding where Ive done something wrong?

$\endgroup$
1
$\begingroup$

I think instead of $x_i \le E_i$, we have $x_1 \le E_1$ and $\omega x_2 \le E_2$, from which the conclusion readily follows.

$\endgroup$
  • $\begingroup$ Which also makes more sense in the context of the problem than $x_i\leq E_i$. Thank you very much! $\endgroup$ – TwoStones Oct 27 '19 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.