1
$\begingroup$

I have the following phase space $$ M = \{ (p, q) \in \mathbb{R}^2 \mid q \geq 0 \} $$ and the Hamiltonian $H = q^2p^2t$. How does one solve for $q(t)$, with $q(0) = q_0 > 0, p(0) = p_0$ using the HJ equation?

I tried writing the generating function $S = S(q, t) = a(t)b(q)$, as I don't think additive separation of variables works in this case. The problem here is that plugging this guess into HJ, I get two differential equations (one for $a$ and one for $b$) and hence two arbitrary integration constants. But I only need one as the system is one-dimensional. I feel like I'm going nowhere with this. Any suggestions? Am I on the right track?

EDIT: here is my attempt. The HJ equation reads $\partial_tS + H(q, \partial_qS, t) = 0$. With $S(q, t) = a(t)b(q)$, we have $$ \dot{a}b + q^2a^2b'^2t = 0 \implies -\frac{\dot{a}}{a^2t} = \frac{q^2b'^2}{b} $$ Introducing the sepration constant $k$, we have \begin{cases} -\frac{\dot{a}}{a^2t} = k \\ \frac{q^2b'^2}{b} = k \end{cases} Here is where I'm stuck, because I get two integration constants, say $\alpha_1, \alpha_2$. Then, there are two more constants of motion, namely $\beta_i = \partial_{\alpha_i}S, i =1, 2$. Do I need them to determine $k$ maybe? I think there are two many constants for me to relate them to the initial conditions $(q_0, p_0)$.

EDIT 2: following Qmechanic's hint, I set $$ -\frac{\partial_tS}{t} = \alpha^2 = (q \partial_qS)^2 $$ From the first equality, I get $S = -\frac{\alpha^2t^2}{2} + f(q)$. Substituing into the second one, I get $f(q) = \alpha \log(q)$. From here, the solution is straightforward.

$\endgroup$
0
1
+50
$\begingroup$

Hint: Use the separation of variable trick on the HJ equation: $$-\frac{\partial_t S}{t} ~=~\alpha^2~=~ (q\partial_qS)^2.$$

$\endgroup$
1
  • $\begingroup$ Ok, I think I got it. Thanks! I'll post my solution when I'm done working out all the details. $\endgroup$ – fresh Feb 20 at 9:11

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.