- #1
Michael Korobov
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- TL;DR
- Why Lagrangian derivative over angle isn't included in the Hamiltonian calculation?
Hi,
In David Morin's "Introduction to classical mechanics", Problem 6.8, when deriving Hamiltonian of the bead rotating on a horizontal stick with constant angular speed, the Lagrangian derivative over angular speed isn't included.
Why is that?
Specifically, the Lagrangian takes form $$L=T=\frac{m {\dot r}^2}{2}+\frac{mr^2\omega^2}{2}=\frac{m {\dot r}^2}{2}+\frac{mr^2{\dot \theta}^2}{2}$$
The Hamiltonian is then calculated as
$$H=\left(\sum_{i=1}^N \frac{\partial L}{\partial {\dot q_i}}{\dot q_i} \right)-L=\frac{m {\dot r}^2}{2}-\frac{mr^2\omega^2}{2}$$
This implies either $$\frac{\partial L}{\partial {\dot \theta}}{\dot \theta}=0$$
or that the only generalized coordinate is ##r## and therefore ##N=1##
Why is that? I understand this relates to ##{\dot \theta}=const## but didn't manage to understand why the partial derivate should be 0.
Later on, in the "Analytical Mechanics" by Hand and Finch, in Question 14 on p.22, there is identical question and then they ask how Hamiltonian looks like if ##\omega=\omega (t) \neq const## which I presume should include Lagrangian dependency on ##\dot \theta##
Thanks a lot,
Michael
In David Morin's "Introduction to classical mechanics", Problem 6.8, when deriving Hamiltonian of the bead rotating on a horizontal stick with constant angular speed, the Lagrangian derivative over angular speed isn't included.
Why is that?
Specifically, the Lagrangian takes form $$L=T=\frac{m {\dot r}^2}{2}+\frac{mr^2\omega^2}{2}=\frac{m {\dot r}^2}{2}+\frac{mr^2{\dot \theta}^2}{2}$$
The Hamiltonian is then calculated as
$$H=\left(\sum_{i=1}^N \frac{\partial L}{\partial {\dot q_i}}{\dot q_i} \right)-L=\frac{m {\dot r}^2}{2}-\frac{mr^2\omega^2}{2}$$
This implies either $$\frac{\partial L}{\partial {\dot \theta}}{\dot \theta}=0$$
or that the only generalized coordinate is ##r## and therefore ##N=1##
Why is that? I understand this relates to ##{\dot \theta}=const## but didn't manage to understand why the partial derivate should be 0.
Later on, in the "Analytical Mechanics" by Hand and Finch, in Question 14 on p.22, there is identical question and then they ask how Hamiltonian looks like if ##\omega=\omega (t) \neq const## which I presume should include Lagrangian dependency on ##\dot \theta##
Thanks a lot,
Michael