derive Euler Lagrange

40 Views Asked by At

I was trying to derive Euler Lagrange Form which looks like

$$\frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$$

I was using a function which I got from here,

$$S(q)=\int_{t_0}^{t_1} L(q,\dot{q},t)dt$$

I was using Feynman technique since I know that rather than simple integration.

$$\frac{dS}{dt}=\int_{t_0}^{t_1} (\frac{\partial L}{\partial q}\dot{q}+\frac{\partial L}{\partial \dot{q}}\frac{\partial }{\partial t}\dot{q} )dt$$

Total time derivative of that function is $0$ so, writing $\frac{dS}{dt}=0$ $$0=\dot{q}(\frac{\partial L}{\partial q}+\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{q}})$$

$$ \frac{\partial L}{\partial q}+\frac{\partial }{\partial t}\frac{\partial L}{\partial \dot{q}}=0$$

I believe there's difference between total time derivative and short ($\frac{\partial}{\partial t}$) time derivative. Without caring of that I found that my equation is closely related to Euler Lagrange. We have only difference between sign.