I'm presented with the following question:
From the Lagrangian for a relativistic particle in an electromagnetic field: $$L=-\frac{m}{\gamma} - eV + e\vec A \cdot \vec v$$ show that the gauge transformation $\vec A \rightarrow \vec A + \vec \nabla \phi$ and $V \rightarrow V - \partial_{t}\phi$ does not change the EL equations. Explain this invariance from the point of view of the action.
As far as I know, $V, A, \phi$ depend on $(r(t), t)$- it really isn't explicit.
So, I began by transforming $L$, and got the following: $$L' = L + e(\partial_{t}\phi + \vec \nabla\phi \cdot \vec v)$$ I realised that the second term could be rewritten as an exact derivative (I think this is right?):
$$L'=L+e\left(\frac{d\phi}{dt}\right)$$ From here, I then substituted $L'$ into the EL equations:
$$\frac{\partial L'}{\partial r_i}=\frac{\partial L}{\partial r_i} + \frac{\partial}{\partial r_i} \left(\frac{d\phi}{dt}\right)$$
$$\frac{d}{dt} \frac{\partial L'}{\partial \dot r_i}=\frac{d}{dt}\frac{\partial L}{\partial \dot r_i} + \frac{d}{dt}\frac{\partial}{\partial \dot r_i} \left(\frac{d\phi}{dt}\right)$$
Now, I know these things are supposed to be equal $$\frac{\partial L'}{\partial r_i} = \frac{d}{dt} \frac{\partial L'}{\partial \dot r_i}$$ Implying that: $$\frac{\partial L}{\partial r_i} + \frac{\partial}{\partial r_i} \left(\frac{d\phi}{dt}\right)=\frac{d}{dt}\frac{\partial L}{\partial \dot r_i} + \frac{d}{dt}\frac{\partial}{\partial \dot r_i} \left(\frac{d\phi}{dt}\right)$$
In there, I can see the original EL equations (which I can then cancel)
What I'm stuck with, is how to prove that: $$\frac{\partial}{\partial r_i} \left(\frac{d\phi}{dt}\right) = \frac{d}{dt}\frac{\partial}{\partial \dot r_i} \left(\frac{d\phi}{dt}\right)$$
Is it as simple as just treating it as a set of EL equations in themselves, thereby being equal to $0$ ?
Further, to finish the question:
$$S = \int L' dt = \int L dt + \int e\left(\frac{d\phi}{dt}\right)dt = S + e\int \frac{d\phi}{dt} dt$$ Which is just S plus some extra term.
Just as a sanity check, is the (physical) reason this becomes invariant (and the extra term plays no role in the variational principle) just because it (the extra term) is fixed? Or is there more to it?
- I can see that this term should just be $0$ due to $S$ cancelling from both sides, but I'm thinking more contextually here.
Thanks in advance, and any help / clues / further resources will be much appreciated!