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Hamiltonian for a free particle in polar coordinates

Physics Asked on March 13, 2021

If we consider a particle that starts at $(0,d)$ and moves along $y=d$ at constant speed $v=vhat i$ then in polar co-ordinates, $r=((vt)^2 + d^2)^{1/2}$ and $tan theta = frac {d}{vt}$

Its $a_r=0$ and $a_theta=0$ which is easily calculated by using the formulas $$a_r =ddot r – r dottheta^2$$
and
$$a_theta= r ddot theta+2dot rdot theta$$

The Hamiltonian for this case in polar coordinates can be written as:
$$H=frac{p_r^2}{2m} + frac{p_theta^2}{2mr^2}$$
where ${p_r}=m dot r$ and ${p_theta}=mr^2 dot theta $

From this, $$frac{partial H}{partial r}=-dot p_rneq 0 $$ since there is $r^2$ in denominator of the 2nd term , this implies that $a_r$ shouldn’t be zero. But in fact, it is.

What is the discrepancy when I apply the Hamiltonian approach?

I got this doubt due to a discussion in Liboff’s "Introductory QM" where he writes:

We may also consider the dynamics of a free particle in spherical coordinates.
The Hamiltonian is
$$H=frac{p_r^2}{2m} + frac{p_theta^2}{2mr^2}+frac{p_phi^2}{2mr^2sin^2 theta}$$
Only $phi$ is cyclic, and we immediately conclude that $p_phi$ (or equivalently, $L_z$) is
constant. However, $p_r$ and $p_theta$ are not constant. From Hamilton’s equations, we
obtain
$$dot p_r=frac{p_theta^2}{mr^3}+frac{p_phi^2}{mr^3sin^2 theta}$$
$$dot p_theta=frac{p_phi^2 cos theta}{mr^2sin^3 theta}$$
These centripetal terms were interpreted above. In this manner we find that the
rectilinear, constant-velocity motion of a free particle, when cast in a spherical
coordinate frame, involves accelerations in the r and $theta$ components of motion.
These accelerations arise from an inappropriate choice of coordinates. In simple
language: Fitting a straight line to spherical coordinates gives peculiar results.

Doesn’t this reasoning also apply for a 2D case as I have mentioned above?

The considerations are same : a rectilinearly moving point at constant velocity. In the 3D case mentioned in the book, the non-zero radial and azimuthal components of acceleration come out non-zero but they come out zero in my case. What’s the difference? (apart from moving from 3D to 2D which I think shouldn’t make a difference)

One Answer

I cannot imagine what discrepancy you are talking about. The EOM yields $$ partial_ r left (frac{p_r^2}{2m} + frac{p_theta^2}{2mr^2}right )=-dot p_r, $$ which amounts to $$ frac{p_theta^2}{mr^3} = dot p_r. $$ It is quite misleading to think of it as "the" radial force, as it willfully skips the centripetal acceleration.

The radial acceleration you found first, written in terms of canonical momenta, is $$ ma_r= dot p_r -frac{p_theta^2}{mr^3}, $$ so, given the EOM above, it vanishes. Where is the discrepancy?

Correct answer by Cosmas Zachos on March 13, 2021

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