1
$\begingroup$

So I have a very complicated Hamiltonian given by:

$$ H_R(\psi, P_\psi) = \frac{-B_0 R^3}{2 \pi} (2 \psi - sin(2 \psi) + \sqrt{B_0^6 R^6 V(\psi)^2 - P_\psi^2} $$

where

$$V(\psi) = \frac{1}{\pi} \sqrt{b_0^4 sin^4(\psi) + (\pi P/M - \psi + \frac{1}{2}sin(2\psi))^2}.$$

The potential makes this Hamiltonian VERY hard (according to me) to evaluate. My goal is to get the equation of motion on form $\psi(R)=$ something, for arbitrary $P/M$. $B_0$ and $b_0$ are just some constants.

In principle I could start from

$$ \frac{\partial H_R(\psi, P_\psi)}{\partial P_\psi} = \partial_R \psi \equiv \dot{\psi}.$$

Then I would get

$$ \dot{\psi} = \frac{-P_\psi}{\sqrt{B_0^2 R^6 V(\psi)^2 - P_\psi}}. $$

Then I could integrate $\psi(R)$ w.r.t $R$. But I don't know the limits of this integral. And even if I did, it would be a next to impossible task to integrate? Not even Mathematica can handle that integral.

So I will have to make some approximations I guess.

I could in principle solve it numerically but I want a solution for arbitrary $P/M$.

Any input on how I can ultimately get the equation of motion on form $\psi(R) = $ something?

I am grateful for any help.

$P_{\psi}$ is canonical momenta in $\psi$ which is the polar angle spanning a 3-sphere. P and M is the momentum and mass of a D-brane or graviton. The range of $\psi$ is $[0, 2 \pi]$

$\endgroup$
5
  • $\begingroup$ What are $P_{\psi}, P, M$ ? Are they related ? Presumably $\psi$ is an angle, but covering what range of values ? $\endgroup$ Commented Dec 21, 2020 at 10:48
  • $\begingroup$ Thanks! Will edit to make this clearer. $P_psi$ is canonical momenta in $\psi$ which is the polar angle spanning a 3-sphere. P and M is the momentum and mass of the field on the question. The range of $\psi$ is $[0, 2 \pi]$ $\endgroup$ Commented Dec 21, 2020 at 11:02
  • $\begingroup$ You said "Then I could integrate $\psi(R)$...w.r.t R". Where is the equation for $d\psi/dR$? R seems to be just a parameter. Perhaps you mean to integrate the $d_t\psi=...$ equation to get $\psi(R,t)$? $\endgroup$ Commented Dec 21, 2020 at 11:36
  • $\begingroup$ Thanks for your answer! So I believe that equation is simply Hamilton's equation of motion: $ (d/dR) \psi = \partial H / \partial P_{\psi} $. R is a radial coordinate in $\mathbf{R}^4$ $\endgroup$ Commented Dec 21, 2020 at 12:06
  • $\begingroup$ Use Hamilton-Jacobi equation, substitute: $H \to \frac{\partial S}{\partial t}$ and $P_\psi \to \frac{\partial S}{\partial \psi}$ and look for a solution $S=-Et+W(\psi)$. This way you'll get $W$ as an integral. Differentiate so obtained result with respect to $E$ (the only integral of motion) and equate the derivative to a constant. This way you'll have an equation relating time and coordinate, schematically: $t-t_0=f(\psi)$ $\endgroup$
    – nwolijin
    Commented Dec 21, 2020 at 12:34

1 Answer 1

0
$\begingroup$

from the Hamiltonian :

$$H=H(\psi,P_\psi)$$

you get two first order differential equations :

$$\dot{P}_\psi=-\frac{\partial H}{\partial \psi}=f_1(P_\psi\,,\psi)\tag 1$$ $$\dot{\psi}=\frac{\partial H}{\partial P_\psi}=f_2(P_\psi\,,\psi)\tag 2$$

if you linearized Eq. (1) and (2), you obtain analytical solution.

$$\begin{bmatrix} \Delta{\dot P}_\psi \\ \Delta{\dot{\psi}} \\ \end{bmatrix}=\left[ \begin {array}{cc} {\frac {\partial }{\partial P_{{\psi}}}}f_{ {1}} \left( P_{{\psi}},\psi \right) &{\frac {\partial }{\partial \psi} }f_{{1}} \left( P_{{\psi}},\psi \right) \\ {\frac { \partial }{\partial P_{{\psi}}}}f_{{2}} \left( P_{{\psi}},\psi \right) &{\frac {\partial }{\partial \psi}}f_{{2}} \left( P_{{\psi}}, \psi \right) \end {array} \right]_{[P_\psi=P_{\psi_0},\psi=\psi_0]} \,\begin{bmatrix} \Delta{ P}_\psi \\ \Delta{{\psi}} \\ \end{bmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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