- #1
AmirSafavi
- 10
- 0
Hi all,
I'm trying to understand something about the lagrangian.
My resources for learning a currently limited to Landau's mechanics and anything which is on the internet.
[tex] L = L(q,\dot{q},t) [/tex]
Now, here is a simple question: what are these generalized co-ordinated exactly?
For example, in Landau, does the "component-wise" lagrangian as such:
[tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i } -
\frac{\partial L}{\partial{q_i}} = 0[/tex]
and several pages later he says:
[tex] \frac{d}{dt} \frac{\partial L}{\partial \vec{v} } [/tex]
but in general, the [tex] i^{\mbox{th}} [/tex] component of [tex]\vec{v}[/tex] isn't equal to the time derivative of [tex] q_i [/tex].
So what is it? [tex] \dot{q_i} \neq \frac{dq_i}{dt} [/tex] ??
Could someone recommend some books which would better explain this stuff?
Thanks,
Amir
I'm trying to understand something about the lagrangian.
My resources for learning a currently limited to Landau's mechanics and anything which is on the internet.
[tex] L = L(q,\dot{q},t) [/tex]
Now, here is a simple question: what are these generalized co-ordinated exactly?
For example, in Landau, does the "component-wise" lagrangian as such:
[tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i } -
\frac{\partial L}{\partial{q_i}} = 0[/tex]
and several pages later he says:
[tex] \frac{d}{dt} \frac{\partial L}{\partial \vec{v} } [/tex]
but in general, the [tex] i^{\mbox{th}} [/tex] component of [tex]\vec{v}[/tex] isn't equal to the time derivative of [tex] q_i [/tex].
So what is it? [tex] \dot{q_i} \neq \frac{dq_i}{dt} [/tex] ??
Could someone recommend some books which would better explain this stuff?
Thanks,
Amir