Talk:Action (physics)
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Two things:
- The Euler-Lagrange equations' box looks really odd. I think it would look fine without the box, and with the text "Euler-Lagrange equations" converted to a wiki link (Euler-Lagrange equations).
- The second E-L equation doesn't look properly derived. I get:
- <math>\frac{d}{dt}\left( \frac{\partial L}{\partial\dot{\phi}} \right)
- \frac{\partial L}{\partial\phi} = 0
\qquad \Rightarrow \qquad
\ddot{\phi}r^2 + 2\dot{r}\dot{\phi} = 0<math>
-- Taral
Currently Euler-Lagrange equations is an indirect self link, but Euler-Lagrange equation redirects to Lagrangian b4hand 20:08, 20 Jul 2004 (UTC)
At some point, Euler-Langrange equations might end up its own page, but until then, both plural and singular can point to action (physics), which is the simpler of the pair. -- Taral 02:52, 21 Jul 2004 (UTC)
division by <math>r^2<math> yields the result as stated in the article :-) -- Unknown editor
No, it doesn't. That gives <math>\frac{2}{r^2}\dot{r}\dot{\phi}<math>, and only if r is nonzero. -- Taral 08:23, 18 May 2005 (UTC)
Here's how you get it:
- <math> \frac{\partial L}{\partial \phi} = 0<math>
- <math> \frac{d}{d t} \frac{\partial L}{\partial \dot{\phi}} = \frac{d}{dt} (m r^2 \dot{\phi} )<math>
- <math> \frac{d}{dt} (r^2 \dot{\phi}) = \left(\ddot{\phi} \frac{\partial}{\partial \dot{\phi}} + \dot{r} \frac{\partial}{\partial r} \right) (r^2 \dot{\phi}) = \ddot{\phi}r^2 + \dot{r}2r\dot{\phi}<math>
so
- <math> r^2 \ddot{\phi} + 2 r \dot{r} \dot{\phi} = 0 <math>
or for nonzero r
- <math>\ddot{\phi} + \frac{2}{r}\dot{r}\dot{\phi} = 0<math>
as stated in the article. --Laura Scudder | Talk 22:03, 18 May 2005 (UTC)
- There are infinitely many actions which give rise to Newton's laws.
- There are quantum field theories which are NOT "derived" from an action.
- This article only seems to be about the use of the action in classical Newtonian mechanics.
Phys 01:22, 31 Jul 2004 (UTC)
Feel free to fix it.
Taral 21:21, 16 Aug 2004 (UTC)