Lagrangian total time derivative - continues second-order differential

Question

In the lagrangian, adding total time derivative doesn't change equation of motion. $$L' = L + \frac{d}{dt}f(q,t).$$

After playing with it, I realize that this is only true if the $f(q,t)$ function has "continuous second partial derivatives". If it doesn't have, then the proof that equations are motion are the same for $L'$ and $L$ fail. The reason why proof would fail will take me a long time to explain, but if necessary, let me know and will try to update this.

Though, I have not seen anywhere such a restriction about $f(q,t)$. Shouldn't we be careful not to take $f(q,t)$ which doesn't have "continuous second partial derivatives"? In that case, it doesn't seem general to assume for any $f(q,t)$, it would yield the same EOM.

Update

Let me show why proof would fail.

$F$ is a function of $q(t), t$.

Since $L' = L + \frac{d}{dt}f(q,t)$, since we hear that it yields the same EOM, then let's see if it does.

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

$\frac{d}{dt}\frac{\partial}{\partial \dot q} (L + \frac{df}{dt}) - \frac{\partial}{\partial q}(L + \frac{df}{dt}) = 0$

$\frac{d}{dt}\frac{\partial L}{\partial \dot q} + \frac{d}{dt}\frac{\partial }{\partial \dot q} \frac{df}{dt} - \frac{\partial L}{\partial q} - \frac{\partial }{\partial q}\frac{df}{dt} = 0$

$\frac{d}{dt}\frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} + \frac{d}{dt}\frac{\partial }{\partial \dot q} \frac{df}{dt} - \frac{\partial }{\partial q}\frac{df}{dt} = 0$

If people say that $L'$ and $L$ yield the same EOM, then first 2 members are equal to 0 which means that we have to show that:

$\frac{d}{dt}\frac{\partial }{\partial \dot q} \frac{df}{dt} - \frac{\partial }{\partial q}\frac{df}{dt} = 0$

$\frac{d}{dt}\frac{\partial }{\partial \dot q} \frac{df}{dt} = \frac{\partial }{\partial q}\frac{df}{dt}\tag{1}$

We can figure out what $\frac{df}{dt}$ is. since it's a function of $q,t$, we get: $\frac{df}{dt} = \frac{\partial f}{\partial q} \dot q + \frac{\partial f}{\partial t}$. Plugging this in (1), we get:

$\frac{d}{dt}\frac{\partial }{\partial \dot q} (\frac{\partial f}{\partial q} \dot q + \frac{\partial f}{\partial t}) = \frac{\partial }{\partial q}(\frac{\partial f}{\partial q} \dot q + \frac{\partial f}{\partial t})$

$\frac{d}{dt}(\frac{\partial }{\partial \dot q} (\frac{\partial f}{\partial q}\dot q)) = \frac{\partial }{\partial q}(\frac{\partial f}{\partial q} \dot q + \frac{\partial f}{\partial t})$

$\frac{d}{dt}(\frac{\partial f}{\partial q}) = \frac{\partial }{\partial q}(\frac{\partial f}{\partial q}) \dot q + \frac{\partial }{\partial q}\frac{\partial f}{\partial t}$

Left side is total time derivative, so we can do the following:

$\frac{\partial }{\partial q} \frac{\partial f}{\partial q} \dot q + \frac{\partial }{\partial t}\frac{\partial f}{\partial q} = \frac{\partial }{\partial q}(\frac{\partial f}{\partial q}) \dot q + \frac{\partial }{\partial q}\frac{\partial f}{\partial t}$

One can see the first member of left side and first member of right side cancel, so we're left with:

$\frac{\partial }{\partial t}\frac{\partial f}{\partial q} = \frac{\partial }{\partial q}\frac{\partial f}{\partial t}$

So this must be equal otherwise, it fails. Then we know that the final equation is a mixed partial differential - Check here So you see $f$'s all the second partial derivatives must exist and must be continuous)

Answer 1

After playing with it, I realize that this is only true if the $f(q,t)$ function has "continuous second partial derivatives".

$\frac{\partial }{\partial t}\frac{\partial f}{\partial q} = \frac{\partial }{\partial q}\frac{\partial f}{\partial t}$

So this must be equal otherwise, it fails. Then we know that the final equation is a mixed partial differential - Check here So you see $f$'s all the second partial derivatives must exist and must be continuous)

This looks correct to me.

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In the lagrangian, adding total time derivative doesn't change equation of motion. $$L' = L + \frac{d}{dt}f(q,t).$$

I think you have just shown that this is only true for a certain (very wide and very typical) class of transformations. So, this is not true in all generality.

What is true, however, is that this transformation ($L \to L' = L + \frac{df}{dt}$) does not change the variation in the action: $\delta S \to \delta S' = \delta S$. Therefore, the form of the Hamiltonian principle equation remains the same regardless of which $L$ or $L'$ is used.

This is true because the respective actions are: $$ S = \int_{t_1}^{t_2}dt L\;, $$ and $$ S' = \int_{t_1}^{t_2}dt L' = \int_{t_1}^{t_2}dt \left(L + \frac{df}{dt}\right) = S + f(t_2) - f(t_1) $$

Clearly $$ \delta S' = \delta S\;, $$ since $$ \delta (f(t_2) - f(t_1)) = 0\;, $$ since $f(t_2) - f(t_1)$ does not depend on the path. (The variation $\delta$ is with respect to the path.)

Answer 2

I think that the textbook 'Classical Dynamics of Particles and Systems' by Marion and Thornton has mentioned this restriction (Section 7.4 in the fifth edition):

However, certain transformations that change the Lagrangian but leave the equations of motion unchanged are allowed. For example, equations of motion are unchanged if $L$ is replaced by $L + \frac{d}{dt}[f(q_i, t ) ]$ for a function $f(q_i, t)$ with continuous second partial derivatives.