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If we consider a particle that starts at $(0,d)$ and moves along $y=d$ at constant speed $v=v\hat i$ 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 $$a_r =\ddot r - r \dot\theta^2$$ and $$a_\theta= r \ddot \theta+2\dot r\dot \theta$$

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, $$\frac{\partial H}{\partial r}=-\dot p_r\neq 0 $$ 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 $$H=\frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2}+\frac{p_\phi^2}{2mr^2\sin^2 \theta}$$ 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 $$\dot p_r=\frac{p_\theta^2}{mr^3}+\frac{p_\phi^2}{mr^3\sin^2 \theta}$$ $$\dot p_\theta=\frac{p_\phi^2 \cos \theta}{mr^2\sin^3 \theta}$$ 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)

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    $\begingroup$ Can you explain why that implies $a_r \neq 0$? It might help to see things a bit clearer if you integrate your second equation $a_\theta$ wrt time and substitute $\dot{\theta}$ into your radial equation. $\endgroup$ Jan 24 at 22:25
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    $\begingroup$ Did you blow a sign? $ma_r= \dot{p}_r-\dot{p}_\theta^2/mr^3=0$ by your EOM. $\endgroup$ Jan 24 at 23:53
  • $\begingroup$ @Rumplestillskin Since, $\dot p$ is non-zero, then radial force and acceleration is non-zero. This is what I reasoned. What is wrong with it? $\endgroup$
    – Lost
    Jan 25 at 0:19
  • $\begingroup$ @Rumplestillskin Please also check the edit I made. Also, do you mean that having a non-zero $\dot p_r$ doesn't imply a $a_r=0$ since $\dot p_r=m\ddot r$ while $a_r$ has a term of $r \dot \theta^2$ as well? I think this was the mistake I was making. But then how does Liboff conclude from non -zero expressions of $p_\theta$ that there is an azimuthal acceleration? $\endgroup$
    – Lost
    Jan 25 at 0:43
  • $\begingroup$ @CosmasZachos Sorry. I do not understand. Where did I blow up the sign? $a_r$ comes zero. That I have calculated and it is correct. $\endgroup$
    – Lost
    Jan 25 at 0:47
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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?

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    $\begingroup$ Understood. Now its obvious. It was a stupid mistake. Also now I understand that the extra 2nd term in the 1st eqn given in Liboff is the reason $a_r$ is not equal to zero there. This is because thats the 3-D Hamiltonian. $\endgroup$
    – Lost
    Jan 25 at 1:29
  • $\begingroup$ I had thought that since writing EOM of linear trajectory in 3-D soherical coordinate system yields radial and azimuthal accelerations, same would be the case with 2-D polar coordinates. Though it seems it cannot be extended like that. Any other input as to why this lazy extension of mine was not valid apart from the obvious mathematics? $\endgroup$
    – Lost
    Jan 25 at 1:32
  • $\begingroup$ Will radial and azimuthal remain zero for any linear trajectory? Non-constant velocity and non-parallel to X-Y axes? If this is so, then is there a way of thinking about it without actually solving for the accelerations? $\endgroup$
    – Lost
    Jan 25 at 1:36

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