Two methods to obtain the Hamiltonian equations of motion but they ain't the same

Question

Given the Hamiltonian $$H=\frac{1}{2}m (\dot{r}^2 +(r\dot{\theta})^{2}) - \frac{k}{r},\tag{1}$$

And the Lagrangian $$L=\frac{1}{2}m (\dot{r}^2 +(r\dot{\theta})^{2}) + \frac{k}{r},\tag{2}$$

its generalized momentum of $\bf r$:

$${p_{r}} = \frac{\partial L}{\partial \dot{r}} =m\dot{r}$$

To obtain the Hamiltonian equation of motion:

For $\bf{r}$: $$\frac{\partial H}{\partial r} = -\dot{p_{r}} = mr\dot{\theta^2} + \frac{k}{r^2} = \frac{p_{\theta}^2}{mr^3}+\frac{k}{r^2} $$

However, if we substitute the generalized momentum to Hamiltonian at the beginning

$$\implies H =\frac{1}{2}\left(\frac{p_{r}^2}{m}+ \frac{p_{\theta}^2}{mr^2}\right) -\frac{k}{r}$$

Then we differentiate Hamiltonian with respect to $\bf{r}$:

$$\frac{\partial H}{\partial r}= -\dot{p_{r}}= -\frac{p_{\theta}^2}{mr^3} + \frac{k}{r^2}$$

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Why do the two different methods have two answers? Shouldn't they be the same equation? Because we are using exactly the same way to obtain the equation of motion? The only different is we substitute the generalized momentum at the the answer and another at the beginning.

Answer 1

Your first Hamiltonian is wrong. If you follow the correct procedure $$ H=\sum_i p_i \dot{q}_i -L $$ where you replace $\dot{q}_i$ by the corresponding expression for $p_i$ you will get $$ H =\frac{1}{2}\left(\frac{p_{r}^2}{m}+ \frac{p_{\theta}^2}{mr^2}\right) -\frac{k}{r} $$ which is correct. This is because the kinetic term in the Lagrangian is not of the form $$ T=\sum_{ij} m_{ij} \dot{q}_i\dot{q}_j. \tag{1} $$ for constant $m_{ij}$ because of the (non-constant) $r^2$ factor in front of the $\dot{\theta}^2$ term in the kinetic energy. Only when $T$ is of the form given by (1) can you unambiguously identify a pure kinetic term and use $H=T+V$. There's a related discussion in this thread.