0
$\begingroup$

I am stuck with the basic question on the classical Hamiltonian for free fall particle (let's say from the infinity). The Hamiltonian can be represented as the total energy of the system, which is the sum of kinetic and potential energy, $$\mathcal{H}=T+V.\tag{1}$$ In the case of particle's free fall in the gravitational field $$V(r)=-T(r),\tag{2}$$ therefore $\mathcal{H}$ is to be assumed as zero.

The same question applies to a free charge $q$ initially at rest then accelerated by uniform electric field $E$, so its total energy remains zero, so $\mathcal{H}=0$?

$\endgroup$
  • 2
    $\begingroup$ Could you elaborate how you got eq. (2)? $\endgroup$ – Qmechanic Oct 3 at 11:17
  • $\begingroup$ I assume that $V + T=const$ is correct. Then if at the initial moment $T=0$ and the particle was far away from the center of gravity then $V \approx 0$. I see your point then.. $\endgroup$ – Eddward Oct 3 at 11:37
  • $\begingroup$ You can't plug the solution of the EOM back into the Hamiltonian $\endgroup$ – Aaron Stevens Oct 3 at 11:42
  • $\begingroup$ So what is your issue here, exactly? Why is it a problem that $H=0$ for what you are wanting to do here? $\endgroup$ – Aaron Stevens Oct 3 at 11:53
  • 1
    $\begingroup$ Also note that if you want to solve the equations of motion, you do not really care about the value of the Hamiltonian rather than the derivatives with respect to the coordinates and the momenta. The value of the Hamiltonian is often constant (energy is often conserved after all), but the derivatives still can be quite complex expressions. $\endgroup$ – Ezze Oct 3 at 11:56
1
$\begingroup$

The problem here is that Eq. 2 is not true. It is true that the change in potential energy is equal to the negative change of kinetic energy and viceversa, since if energy has to be conserved those two "energy containers" have to obey this. When you lose some potential energy it gets entirely transfered to the kinetic energy container.

So yeah, $\Delta V = -\Delta T$. But this does not mean that $V = -T$ at all. You could have one million joules stored as potential energy and the particle not moving at all ($100000\neq 0$ so $V\neq -T$) and a moment after you could still have $99000$ joules in potential energy and $1000$ in kinetic energy. You would indeed have $\Delta V = 99000-100000 = -1000 = -(1000) = - (1000-0) = - \Delta T$ but you wouldn't have in general that $V=-T$.


Edit:

I assume that V+T=const is correct. Then if at the initial moment T=0 and the particle was far away from the center of gravity then V≈0. I see your point then..

That is indeed right. $V+T= constant$ is correct. And that's another reason why Eq. 2 is, in general, wrong. $V+T= constant$ means that $V = constant-T\neq -T$. The Hamiltonian would be $H = constant$ which makes sense since energy is indeed conserved.

It is also true that you can make that $contant=0$ since you are free to chose whatever zero value for the potential you want, and yes in that case you would indeed get Eq. 2 and have $H = 0$ which has nothing wrong about it and has perfect physical meaning if you have a well defined zero for the potential energy. But in general the answer should be $H = constant$ (assuming conservation of energy).

$\endgroup$

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.