Understanding gauge in Lagrangian mechanics [duplicate]

Question

I know given a Lagrangian $\mathcal{L}$ satisfying the Euler-Lagrange equations, then the Lagrangian $\mathcal{L}'=\mathcal{L}+\frac{d}{dt}f(q,t)$ is also a solution of said equations. Nonetheless, I keep getting that not every function of this form can be added to the original Lagrangian. Introducing $\mathcal{L}'$ in the Euler-Lagrange equations:

$$\frac{\partial \mathcal{L}'}{\partial q}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}'}{\partial \dot q}\right)=\frac{\partial \mathcal{L}}{\partial q}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot q}\right)+\frac{\partial}{\partial q}\frac{d}{dt}f(q,t)-\frac{d}{dt}\frac{\partial }{\partial \dot q}\frac{d}{dt}f(q,t)$$

The first two terms on the right hand side add up to zero since we established $\mathcal{L}$ is a solution, and since we know $\mathcal L'$ is also a solution, then this equality must also be zero. Hence:

$$\frac{\partial}{\partial q}\frac{d}{dt}f(q,t)-\frac{d}{dt}\frac{\partial }{\partial \dot q}\frac{d}{dt}f(q,t)=0$$

Rearranging the operators, I end up with:

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

Which is zero in the case of the Lagrangian, but I'm not sure how this is also guaranteed to be zero for any arbitrary $f(q,t)$.

Answer 1

You are almost there. You just need to expand the equation.

$$\left(\frac{\partial}{\partial q} - \frac{d}{dt}\frac{\partial}{\partial\dot{q}} \right)\frac{d f(q,t)}{dt} = 0 $$

Continuing, we get that

\begin{split} \frac{\partial}{\partial q}\frac{d f(q,t)}{dt} &= \frac{\partial}{\partial q} \left( \frac{\partial f}{\partial q} \dot{q}+\frac{\partial f}{\partial t} \right) \\ \end{split}

and similarly, \begin{split} \frac{d}{dt}\left( \frac{\partial}{\partial \dot{q}}\frac{d f(q,t)}{dt} \right) &= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{q}} \left( \frac{\partial f}{\partial q} \dot{q}+\frac{\partial f}{\partial t} \right)\right) \\ &= \frac{d}{dt}\left(\frac{\partial}{\partial \dot{q}} \left( \frac{\partial f}{\partial q} \dot{q} \right)\right)\\ & = \frac{d}{dt}\left( \frac{\partial f}{\partial q} \right) \\ &= \dot{q}\frac{\partial}{\partial q}\frac{\partial f}{\partial q}+ \frac{\partial}{\partial t}\frac{\partial f}{\partial q} \end{split} Putting everything together, we have that \begin{split} \frac{\partial}{\partial q} \left( \frac{\partial f}{\partial q} \dot{q}+\frac{\partial f}{\partial t} \right) - \dot{q}\frac{\partial}{\partial q}\frac{\partial f}{\partial q}+ \frac{\partial}{\partial t}\frac{\partial f}{\partial q} &= 0 \end{split} which clearly shows that \begin{equation} 0 = 0. \end{equation}