# Problem 2.1 in Birger Bergersen's and Michael Plischke's Equilibrium Statistical Mechanics

I have the solution to problem 2.1 from Bergersen's and Plischke's textbook which I don't quite understand.

I'll post the question itself and its solution.

a) Consider a harmonic oscillator with Hamiltonian $H = 1/2(p^2+q^2)$ show that any phase space trajectory $x(t)$ with energy $E$, on the average, spend equal time in all regions of the constant energy surface $\Gamma(E)$.

b) Consider two linearly coupled harmonic oscillator: $$H=1/2(p_1^2+p_2^2+q_1^2+q_2^2+q_1q_2)$$Express the phase space trajectory $x=(p_1,p_2,q_1,q_2)$ in terms of the initial values of the amplitude and phase of the normal coordinates. Show that there are regions of the constant energy surface which aren't visitedfor any particular trajectory $x(t)$.

Now, for the solution:

a) Let $$p=r\cos \phi ; q=r\sin \phi$$ The constant energy surface $\Gamma(E)$ in the $q-p$ plane is the circle $r=const$. The equations of motion can be written $\dot{r}=0 ; \dot{\phi}=const$, i.e., the phase space point moves with constant velocity and covers the whole "surface". It thus spends equal time in all regions of $\Gamma(E)$.

b) The solutions to the equations of motion can be written $$q_1=r_1\sin \phi_1 + r_2 \sin \phi_2$$ $$q_2 = r_1\sin\phi_1 - r_2\sin\phi_2$$ where $r_1,r_2$ are constants of the motion and $$\phi_1 = \sqrt{3/2}t + \phi_1(0)$$ $$\phi_2 = \sqrt{1/2}t + \phi_2(0)$$ The projection of the constant energy surface on the $r_1,r_2$ surface is the ellipse $$3/4 r_1^2+1/4 r_2^2=const$$ However, with given initial conditions only a single point on this ellipse is visited.

I don't understand the solution to part b), I assume that $p_i = r_i \cos \phi_i$, and then to derive $3/4 r_1^2+1/4 r_2^2=const$ I need to use the Hamiltonian given in b) and then equate what I get to some constant, but I get if I am not mistaken the following:

$$1/2(r_1^2+2r_1^2\sin^2 \phi_1 + r_2^2) = constant$$

Which is off by far, what do I need in order to derive this ellipse equation, I don't see how did they derive it out of thin air.

Thanks.

In general, imagine you have a Hamiltonian of the form

$$H = \sum_i \frac{p_i^2}{2} + \sum_{i,j}q_iO_{ij}q_j \tag{1}$$

where ${\bf O}$ is a positive definite symmetric matrix, you can diagonalize ${\bf O} = {\bf U}\Omega{\bf U}^{t}$, where $\Omega = {\rm diag} (\omega_1^2, \omega_2^2\cdots)$ and define

$$q_i' = \sum_j U_{ij}q_j , ~~~~p_i' = \sum_jU_{ij}p_j \tag{2}$$

you can check that in this coordinates, the Hamiltonian (1) becomes

$$H = \sum_{i}\frac{p_i'^2}{2} + \sum_{i}\omega_i^2q_i'^2 \tag{3}$$

That is, a set of uncoupled harmonic oscillators, each with solution of the form

$$q_i'(t) = r_i\sin(\omega_it + \phi_i) \tag{4}$$

In you case, the matrix ${\bf O}$ is simply

$${\bf O} = \left(\begin{matrix} 1 & 1/2 \\ 1/2 & 1\end{matrix}\right) \tag{5}$$

which can be diagonalized with

$${\bf U} = \frac{1}{\sqrt{2}}\left(\begin{matrix} 1 & -1 \\ 1 & 1\end{matrix}\right) ~~~ \Omega = \left(\begin{matrix} 3/2 & 0 \\ 0 & 1/2\end{matrix}\right) \tag{6}$$

Can you take it from here?

• I guess what was missing or not mentioned here is that $p=\dot{q}$, in which case we get: $p_i' = \omega_i r_i\cos(\omega_i t+\phi_i(0))$. – MathematicalPhysicist Oct 11 '17 at 22:20