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The hamiltonian for a free falling body is $$H = \dfrac{p^2}{2m} + mgy$$ and since we are using cartesian coordinates that do not depend on time and the potential only depends on the position, we know that $H=E$. For this hamiltonian, using the Hamilton's equations and initial conditions $y(0)=0$ and $p(0)=0$, we get the evolution in the phase space: $$y(t) = -\dfrac{1}{2}gt^2\quad p(t)=-mgt$$

Now, imagine the opposite problem: we don't know anything about the system and the potentials involved, but someone gives us the phase space evolution, $x(t)$ and $p(t)$, for the same initial conditions. Can we get the energy using the hamiltonian formalism?.

From the phase space evolution, we know that $\dot{y}=-gt = p/m$ and $\dot{p} = -mg$. Then $$ \dot{y}=\dfrac{\partial H}{\partial p} \ \Rightarrow\ H = \dfrac{p^2}{2m} + f(y,t) $$ $$ \dot{p} = -\dfrac{\partial H}{\partial y} = -\dfrac{\partial f}{\partial y} \ \Rightarrow\ f(y,t) = mgy + g(t) $$ Concluding that $$H = \dfrac{p^2}{2m} + mgy + g(t) $$

Apparently, we don't have enough information to determine the form of $g(t)$. Two questions came to my mind:

  1. Were the Hamilton's equations integrated correctly? This seems to work when I put $\dot{y}$ as a function of $p$, but woud it work expressing $\dot{y}$ in terms of other combinations of $y$, $p$ or $t$?. When is it mathematically correct to get rid of the time variable to integrate the equations?
  2. How can we know the expression for $g(t)$, and how can we know the relation of the found hamiltonian with the energy if we don't have any explicit information about the potentials?.

Extra example

If I don't get rid of the time at the beginning, then $$ \dot{y} = \dfrac{\partial H}{\partial p} = -gt \ \Rightarrow\ H = -gtp + f(y,t) \ \Rightarrow\ H = \dfrac{p^2}{m} + f(y,t) $$ We have lost the $1/2$ factor, and the equations of motions derivated from this hamiltonian won't be the same.

What is mathematically incorrect there?

SrJaimito
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1 Answers1

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So we are given the following information about the system : $$y(t)=-\frac{1}{2}gt^2$$ and $$p(t)=-mgt$$ with initial condition $y(0)=0$ and $p(0)=0$. You want to know the energy of the system. That's too is the information if you are asking for energy that means the system must conservative so that energy is conserved, and thus Hamiltonian should not depend on time.

But let's suppose we want to know the Hamiltonian ($H=H(q,p,t)$) of the system given such information, The Hamilton's equation given by

$$\frac{\partial H}{\partial q}=-\dot{p}, \ \ \ \ \ \frac{\partial H}{\partial p}=\dot{q}, \ \ \ \ \ \frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}$$

Now if it's not given that whether or not the system is conservative, we don't know about the last term. So as long as you don't know the third term you can not construct Hamiltonian uniquely.

Now write a way to do construct Hamiltonian is through its total derivative of Hamiltonian

$$dH=\frac{\partial H}{\partial p}dp+\frac{\partial H}{\partial q}dq$$

Now put values of these partial derivatives $$dH=mgdq+\frac{p}{m}dp$$

$$H=\frac{p^2}{2m}+mgq$$

Note: Again I'm writing this phase trajectory is useful for time-independent but if the game of rules is changing then it's not as great.

Extra Example

In your extra example

What is mathematically incorrect there?

$$\frac{\partial H}{\partial p}=-gt$$

you can not integrate here with respect to $p$ as $p$ is changing with time or vice verse.

Himanshu
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