classical mechanics - Adding a total time derivative term to the Lagrangian

This is proof that $L'$ represents same equation of motion with $L$ through Lagrange eq. I understand $L'$ satisfies Lagrange eq, but how does this proof mean $L'$ and $L$ describe same motion of particle? In other words, why does total time derivative term which is added to $L$ make no difference in equation of motion?

Answer

You have seen that the substitution

$$L\longrightarrow L':= L+\frac{\mathrm{d}F}{\mathrm{d}t}$$ does not change the Euler-Lagrange equations. Now, this happens because the time derivative satisfies the Euler-Lagrange equations identically.

Let us consider a concrete example. Take the Lagrangian of a simple harmonic oscillator:

$$L_\text{HO}=\frac{1}{2}m\dot q^2-\frac{1}{2}m\omega^2q^2$$ which gives the Euler-Lagrange equations $$\ddot q=-\omega^2 q$$ Now consider the modified Lagrangian $$L'=\frac{1}{2}m\dot q^2-\frac{1}{2}m\omega^2q^2+\dot q=L_\text{HO}+\dot q$$ The Euler-Lagrange equations are obviously linear. Thus $$\text{EL}[L']=\text{EL}[L_\text{HO}]+\text{EL}[\dot q]$$ As was shown above, $\dot q$'s Euler-Lagrange equation will vanish, but we can verify this: $$\text{EL}[\dot q]:=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial \dot q}{\partial\dot q}-\frac{\partial\dot q}{\partial q}=\frac{\mathrm{d}1}{\mathrm{d}t}-0=0$$ Thus, $$\text{EL}[L']=\text{EL}[L_\text{HO}]$$ i.e. the modified Lagrangian still implies $\ddot q=-\omega^2q$.