enter image description here

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?

  • $\begingroup$ 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
    Commented Apr 5, 2015 at 20:27
  • 1
    $\begingroup$ 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" $\endgroup$ Commented Jul 27, 2016 at 18:32

5 Answers 5


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 $\ddot q=-\omega^2q$.


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.


If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

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

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked. So, adding to the Lagrangian a total time derivative of a function of $q$ and $t$ does not change the equations of motion


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.


This can be easily calculated as seen in the attached image.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.