- #1
acegikmoqsuwy
- 41
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Hi, I have a very basic question about the Lagrangian that I can't seem to understand: why is it dependent on both the position function and the time derivative? I know that it is the difference between the kinetic and potential energy, but why? Is there a derivation of this, is it a definition, or somehow based on experimental evidence?
In fact, when I try to derive the Euler-Lagrangian equations (based on the assumption that the Lagrangian is a function of both position and velocity), I get [itex]\displaystyle A=\int_{t_1}^{t_2} \sum\limits_{i}\left(\frac{\partial \mathcal L} {\partial x_i}-\frac d{dt} \frac{\partial \mathcal L} {\partial \dot x_i}\right) dt[/itex] and using that kinetic energy is dependent on velocity, and potential energy is dependent on position, I seem to arrive at the conclusion that [itex]\mathcal L (t,x,\dot x)=U-T[/itex].
Where have I gone wrong? And why doesn't the Lagrangian depend on [itex] \ddot x[/itex] or only [itex] x[/itex] ?
In fact, when I try to derive the Euler-Lagrangian equations (based on the assumption that the Lagrangian is a function of both position and velocity), I get [itex]\displaystyle A=\int_{t_1}^{t_2} \sum\limits_{i}\left(\frac{\partial \mathcal L} {\partial x_i}-\frac d{dt} \frac{\partial \mathcal L} {\partial \dot x_i}\right) dt[/itex] and using that kinetic energy is dependent on velocity, and potential energy is dependent on position, I seem to arrive at the conclusion that [itex]\mathcal L (t,x,\dot x)=U-T[/itex].
Where have I gone wrong? And why doesn't the Lagrangian depend on [itex] \ddot x[/itex] or only [itex] x[/itex] ?