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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?

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  • $\begingroup$ The geometric/intuitive reason why this is so, is e.g. explained in this Phys.SE post (in the more general framework of field theory). $\endgroup$ – Qmechanic Apr 5 '15 at 20:27
  • $\begingroup$ Your algebraic steps are confusing in places. Additionally you did not state that you are assuming F is not an explicit function of the time derivative of the coordinates - which is required for the step after "This is shown to be true because" $\endgroup$ – user3728501 Jul 27 '16 at 18:32
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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 $\ddot q=-\omega^2q$.

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Well you just showed ${d \over dt } { \partial L' \over \partial \dot q}- { \partial L' \over \partial q}= {d \over dt } { \partial L \over \partial \dot q}- { \partial L \over \partial q}=0$ right? ${d \over dt } { \partial L \over \partial \dot q}- { \partial L \over \partial q}=0$ is the equation of motion for $q$, in other words this equation means exactly that: $L$ and $L'$ give the same equation of motion for q.

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Here is another way to think about it, using the variational principle version of the Euler-Lagrange equations.

The action of $L$ and $L'$ differ by $\dot F$.

$$\mathcal S = \int^{t_f}_{t_i} dt\ L$$

$$\mathcal S' = \int^{t_f}_{t_i} dt\ L' = \int^{t_f}_{t_i} dt\ (L+\dot F) = \mathcal S + F(t_f)-F(t_i) $$

Since $F(t_f)-F(t_i)$ is a constant, the paths that extremize $\mathcal S$ and $\mathcal S'$ are the same.

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