Follow-up question on "Adding a total time derivative term to the Lagrangian"

Question

I have a question to this proof here: Adding a total time derivative term to the Lagrangian

I was asking myself why $$\frac{\partial \dot{F}}{\partial \dot{q}} = \frac{\partial F}{\partial q}.\tag{1}$$ So to add a bit of context: We consider the Lagrangian $\mathcal L^{*} = \mathcal L(q, \dot{q}, t) + \frac{dF}{dt},$ where $F(q, t)$ is only a function of $q$ and $t$.

Now, $$\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q} \Leftrightarrow \frac{d}{dt}\frac{\partial F}{\partial \dot{q}} - \frac{\partial F}{\partial q} = 0, \tag{2}$$ implying that $F(q,t)$ has to satisfy the E.L. equation.

But why should it, if it is chosen arbitrarily? I hope you know what I mean, thanks!

Answer 1

You can prove it in the following way $$dF=dF(q,t)=\frac{\partial F}{\partial q}dq+\frac{\partial F}{\partial t}dt$$ $$\frac{dF}{dt}=\dot{F}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ $$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$$

Cheers!

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On OP's comment from the second last to last equation.

$$\dot{F}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ The only way for $\dot{q}$ to enter in this equation is through the first term on the right, the factor $\partial F/\partial q$ and $\partial F/\partial t $ are a function of $(q,t)$. On differentiating with respect to $\dot{q}$, We get

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