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Stream: deprecated: thermodynamics

Topic: Hamiltonian puzzle

Owen Lynch (Jan 28 2022 at 17:00):

In our last meeting, @John Baez proposed a puzzle: what physical system has the Hamiltonian

$H(q,p) = \frac{1}{2} k q^2$

This is the limit as $m \to \infty$ of the Hamiltonian of the mass-spring system

$H(q,p) = \frac{1}{2m} p^2 + \frac{1}{2} k q^2$

So one way of thinking about this is a particle of infinite mass on a spring, which is rather strange.

A more realistic example of a system like this is an inductor connected to a voltage source.

circuit.png

The current just keeps going up forever in proportion to the current source.

Owen Lynch (Jan 28 2022 at 17:04):

This is in accordance with Hamilton's equations

$\dot{q} = 0$
$\dot{p} = -kq$

which describe $q$ (the magnetic momentum of the inductor) going up forever.

Owen Lynch (Jan 28 2022 at 17:06):

Interestingly, the description of this as a circuit is more intuitively achieved by letting $k \to \infty$ as $m$ stays fixed. That is, we think of the voltage source as a capacitor with infinite capacitance.

Owen Lynch (Jan 28 2022 at 17:07):

Thus, by analogy, we could also think of this as a particle with finite mass, being pushed by a constant force.

Owen Lynch (Jan 28 2022 at 17:09):

This is strange because these two examples, from a physical standpoint, don't have constant energy!

Owen Lynch (Jan 28 2022 at 17:13):

Ah, one answer is that if we scale capacitance to infinity, we must also scale charge to infinity to retain a constant voltage.

Owen Lynch (Jan 28 2022 at 17:14):

So the system has infinite energy at all times, the seeming addition of energy in the magnetic momentum of the inductor does not change anything.

John Baez (Jan 28 2022 at 17:55):

Great!

Just to lay everything out: what is the time evolution of $p(t)$ and $q(t)$ with the Hamiltonian

$H = \frac{1}{2} k q^2$

John Baez (Jan 28 2022 at 17:56):

Well, you basically said, but remind me, and more important explain why a particle of infinite mass with a force $-k q$ on it should have momentum and position evolving this way.

John Baez (Jan 28 2022 at 17:58):

I don't think infinite energy plays a role here. The energy is just $\frac{1}{2} kq^2$.

Owen Lynch (Jan 28 2022 at 17:59):

Well, the equations of motion are

$\dot{q} = 0$
$\dot{p} = -kq$

So, the time evolution is simply

$q(t) = \mathrm{const}$
$p(t) = -kqt + \mathrm{const}$

A force is acting on the particle of infinite mass to give it momentum, but at any point in time, it has only accumulated finite momentum, and thus has a speed of 0.

John Baez (Jan 28 2022 at 17:59):

Right!

There should be a reasonably good way to understand this problem as the $m \to \infty$ limit of the finite-mass problem

$H = \frac{1}{2m} p^2 + \frac{1}{2} k q^2$

with the energy always taking some particular constant value.

John Baez (Jan 28 2022 at 18:00):

Basically, as $m \to \infty$ the kinetic energy goes to zero for a fixed $p$ and $q$, and only potential energy is left.

Owen Lynch (Jan 28 2022 at 18:01):

What I'm saying is that this is the same behavior as if we stretch the spring to infinity while making the spring constant go to 0. Then the particle is infinitely far away from the origin always, but is accumulating more and more momentum.

John Baez (Jan 28 2022 at 18:01):

Hmm.

Owen Lynch (Jan 28 2022 at 18:02):

Which is analogous to the inductor attached to a voltage source.