I'm trying to prove that the initial values of the coordinates, $q(t_0=0)=q_0$, and the momentums, $p(t_0=0)=p_0$, can be expressed in terms of the general form of the Hamilton's equations as $$q_0=F(q,-p,-t)\\p_0=-G(q,-p,-t) \tag{1}$$

With $F$ and $G$ being the general form of the solution of Hamilton's equations $\dot {q} = \frac{\partial H}{\partial p}$ and $\dot {p} = -\frac{\partial H}{\partial q}$:


My attempt

Since the Hamiltonian is a function that deppends on the time and the variables $q$ and $p$, which also deppends on the time, I see that the general form of the solutions for $q(t)$ and $p(t)$ can be deffined as some functions $F$ and $G$ deppending on the initial values $q_0$ and $p_0$ and the time:

$$q(t)=\int_{t_0 =0}^t\frac{\partial H}{\partial p}dt =f\left(q(t),p(t),t\right)\bigg|_{t_0=0}^t \equiv F(q_0,p_0,t)$$ $$p(t)=\int_{t_0 =0}^t-\frac{\partial H}{\partial q}dt =g\left(q(t),p(t),t\right)\bigg|_{t_0=0}^t \equiv G(q_0,p_0,t)$$

However, how could equations $(1)$ be derived from equations $(2)$? Are there time-inversion properties that allows it?


1 Answer 1


I think the answer is easier that that as we see later. Meanwhile, I just wish to tell something about time reversal.

Once you solved the Hamilton equations of motion, you clearly get two functions as in (2). They depend on $p_{0}$ and $q_{0}$ and obviously on time. Now, the time reversal means that if you reverse the direction of time, you do not distinguish future from the past, in other words, the mathematical form of your trajectory is identical in both the directions (towards the future or back to the past). This means that you just need to invert the sign of time and not the sign of momenta (it actually changes as soon as you change the former).

In order to better explain that, let us consider a very simple system, namely, harmonic oscillator. The Hamiltonian is $$ H(p,q):= \frac{p^{2}}{2}+\frac{\omega^{2}}{2}q^{2} $$ where the mass is put to one and $\omega$ is the frequency. The solutions are trivially: $$ q(t)=A\cos(\omega t+\phi),\qquad (1)\\ p(t)=\frac{dq}{dt}(t)=-A\omega\sin(\omega t+\phi) $$ where $A$ and $\phi$ can be fixed once that the initial conditions $(p_{0},q_{0})$ have been imposed. Given an explicit values for $(p_{0},q_{0})$, we get a system $$ q_{0}=q(t=0)=A\cos(\phi)\\ p_{0}=p(t=0)=-A\omega\sin(\phi) $$ which can be solved for $(A,\phi)$ so that you have two functions: $$ A(p_{0},q_{0}):=\frac{\sqrt{p_{0}^{2} + q_{0}^{2} \omega^{2}} }{\omega} $$ and $$ \phi(p_{0},q_{0}):=arctan\left(-\frac{p_{0}}{q_{0}\omega}\right)+2\pi n,\qquad n\in\mathbb{Z}. $$ which can be plugged into the solution (1). In other words, your $F$ function is: $$ F(p_{0},q_{0},t):=\frac{\sqrt{p_{0}^{2} + q_{0}^{2} \omega^{2}} }{\omega} \;\cos\left(\omega\,t+arctan\left(-\frac{p_{0}}{q_{0}\omega}\right)\right) $$ where I have chosen $n=0$ for the sake of simplicity and similar for $G$.

The answer to your question is actually trivial, in fact, you can find the initial values, let us say, $q_{0}$ from (2) just evaluating the $F$-function at $t=0$ as follows: $$ F(q_{0},p_{0},t=0)\equiv q_{0} $$ What you probably mean in (1), in a formal way, is to show that given a solution and inverting the sign of the time direction, you can follow back your trajectory. However, if you define $F$ (as well as $G$) to be a function of the initial conditions and time, it does not make sense to write $F(q,p,t)$ as in (1).

If you wished to understand how the time reversal works, you can use the example above and compute $F(q_{0},p_{0},-t)$. You will see that it exactly coincides with $q_{0}$. In fact, still by inspection, $p(-t)=-p(t)$.

I hope to have clarified your doubt.

  • $\begingroup$ Ok, thanks. But why wouldn't make sense to write $F(q,p,t)$ as in $(1)$? (By the way, there was a mistake in my $(1)$, the statement of the problem says $p_0=-G(q,-p,-t)$) $\endgroup$
    – Invenietis
    Aug 9, 2021 at 18:35
  • $\begingroup$ What I meant is that the functions $G$ and $F$ in (1) cannot be the same as in (2) for the following reason. You defined in the function $q(t):=F(p_{0},q_{0},t)$ in (2) (this is the correct definition) thus in the first and second variables you just have constants, namely, the initial values for momentum and position. Now, the relation $q(t)=F(p_{0},q_{0},t)$ holds for any $t$, hence, also at $t_{0}=0$ which gives $q(t_{0}=0) F(p_{0},q_{0},t_{0}=0)$ which is different than (1). Thus, the initial value $q_{0}$ is just the value of the function $F$ at the point $t_{0}=0$. $\endgroup$
    – RTS
    Aug 9, 2021 at 18:53

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