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Q&A Prove differential form of Lagrangian

1 answer  ·  posted 1y ago by deleted user  ·  last activity 1y ago by celtschk‭

#3: Post edited by user avatar deleted user · 2021-08-24T10:16:31Z (about 1 year ago)
  • How to derive the Lagrangian differential force?
  • $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
  • I was trying to do something.
  • $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
  • $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
  • $$\frac{\partial L}{\partial x}=-kx$$
  • $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
  • So, I can write that
  • $$m\ddot{x}(t)+kx$$
  • But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?
  • How to derive the Lagrangian differential force?
  • $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
  • I was trying to do something.
  • $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
  • $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
  • $$\frac{\partial L}{\partial x}=-kx$$
  • $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
  • So, I can write that
  • $$m\ddot{x}(t)+kx$$
  • But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?
  • >![Lagrangian](https://physics.codidact.com/uploads/z26XyqKjrj2xqwbS9PnAqnhB) ~ https://people.umass.edu/~bvs/601_Lagrange.pdf <br/> Which Euler's equation?
#2: Post edited by user avatar deleted user · 2021-08-24T10:03:59Z (about 1 year ago)
typo
  • How to derive the Lagrangian differential force?
  • $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
  • I was trying to do something.
  • $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
  • $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
  • $$\frac{\partial L}{\partial x}=-kx$$
  • $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
  • So, I can write that
  • $$m\ddot{x}(t)-kx$$
  • But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?
  • How to derive the Lagrangian differential force?
  • $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$
  • I was trying to do something.
  • $$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
  • $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
  • $$\frac{\partial L}{\partial x}=-kx$$
  • $$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
  • So, I can write that
  • $$m\ddot{x}(t)+kx$$
  • But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?
#1: Initial revision by user avatar deleted user · 2021-08-24T10:00:27Z (about 1 year ago)
Prove differential form of Lagrangian
How to derive the Lagrangian differential force?

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})+\frac{\partial L}{\partial x}=0$$

I was trying to do something.

$$L=T-U=\frac{1}{2} m\dot{x}^2-\frac{1}{2}kx^2$$
$$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$
$$\frac{\partial L}{\partial x}=-kx$$

$$\frac{d}{dt}\frac{\partial L}{\partial \ddot{x}}=m\ddot{x} (t)$$
So, I can write that 
$$m\ddot{x}(t)-kx$$

But, how can I prove that it's equal to $0$. I should find the equation from somewhere else. What's that?