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Given the phase space evolution of a system, $x(t)$ and $p(t)$, is there any way of getting the hamiltonian to make a later study of the system under the hamiltonian formalism?

My first thought was to take the time derivatives of the phase space coordinates, $\dot{x}(t)$ and $\dot{p}(t)$, and then try to integrate the Hamilton's equations $$ \dot{x} = \dfrac{\partial H}{\partial p} \quad\text{and}\quad \dot{p} =- \dfrac{\partial H}{\partial x}\ , $$ which gives me a hamiltonian with the generic form $$ H(x,p,t) = f(x,p,t) + g(t)\ , $$ where $f$ is the result of direct integration of the phase space space coordinates and $g$ is an arbitrary function of time.

However I think that this is not a correct approach, since I think that I should get rid of the time dependence of $x$ and $p$ prior to integrating the Hamilton's equations.

How would you get a hamiltonian when the trajectories $x(t)$ and $p(t)$ are given without any extra information?

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  • $\begingroup$ Do you know the trajectories for every possible choice of initial condition, or do you only know a single trajectory? $\endgroup$
    – J. Murray
    Oct 26, 2020 at 16:47
  • $\begingroup$ @J.Murray I have single expressions for x(t) and p(t) only, so I suppose that they represent a single trajectory where the initial conditions were already applied. $\endgroup$
    – Jaime_mc2
    Oct 26, 2020 at 16:54

1 Answer 1

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I want to answer the question "if you know the equations $\dot{x}=\ldots~,\dot{p}=\ldots~$ how you get the Hamiltonian?"

To obtain the Hamiltonian, you have to solve these partial differential equations :

$$-{\frac {\partial }{\partial x}}H \left( x,p \right) =f \left( x \right) $$ and

$${\frac {\partial }{\partial p}}H \left( x,p \right) =g \left( p \right) $$

which give you the solution:

$$H \left( x,p \right) ={ F_2} \left( x \right) +{ F_1} \left( p \right) $$

with :

$$f \left( x \right) =-{\frac {d}{dx}}{ F_2} \left( x \right)\tag 1$$ $$g \left( p \right) ={\frac {d}{dp}}{ F_1} \left( p \right)\tag 2$$

Example:

$$\dot{p}=-k\,x~,\dot{x}=\frac pm$$

$\Rightarrow$

Eq. (1)

$$-kx=-{\frac {d}{dx}}{F_2} \left( x \right) ~\Rightarrow~,F_2=\frac 12 k\,x^2+c_2$$

Eq. (2)

$$\frac pm=\frac{d}{dp}\,F_1~\Rightarrow~,F_1=\frac 12 \frac{p^2}{m}+c_1$$

and the Hamiltonian is:

$$H=\frac 12\left(k\,x^2+\frac {p^2}{m}\right)+c$$

Edit

in case that you know $x=x(t)~,p=p(t)$ you have to solve these equations

$$-{\frac {\partial }{\partial x}}H \left( x,p \right) =f \left( t \right) $$ and

$${\frac {\partial }{\partial p}}H \left( x,p \right) =g \left( t \right) $$ ($f(t)=\dot{x}~,g(t)=\dot{p}$)

which give you the solution:

$$H=F_3(t)\,x+F_4(t)\,p+F_5(t)$$

where :

$$f(t)=-F_3~,g(t)=F_4$$

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  • $\begingroup$ I know what you say but, in my case what I have is the time evolution of x and p. I mean, the functions x(t) and p(t) that one usually get after solving the differential equations of movement. I don't have the relations x = x(p) and p = p(x). $\endgroup$
    – Jaime_mc2
    Oct 26, 2020 at 20:07
  • $\begingroup$ I will write you the solution soon $\endgroup$
    – Eli
    Oct 26, 2020 at 20:13
  • $\begingroup$ @Jaime_mc2 see new document $\endgroup$
    – Eli
    Oct 26, 2020 at 20:22

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