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!


The new action is $$ S' = \int_{t_1}^{t_2}L' \text{d}t = \int_{t_1}^{t_2}\left(L + e\frac{\text{d}\phi}{\text{d}t}\right)\text{d}t = S + e\left[\vphantom{1_1^1} \phi\right]_{t_1}^{t_2}. $$ Since the extra term is constant, the variation of the action hasn't changed: $\delta S' = \delta S$, and therefore the EL equations remain the same.


How about this. We wish to show $$\frac{\partial}{\partial r_{i}}\left(\frac{d \phi}{d t}\right)=\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d t}\right).$$

Consider the right hand side. $$\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d t}\right)=\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d r_i}\frac{d r_i}{d t}\right)$$

by the chain rule

$$=\frac{d}{d t} \frac{\partial}{\partial \dot{r}_{i}}\left(\frac{d \phi}{d r_i}\dot{r_i}\right)$$

$$=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{r}_i}\left(\frac{d\phi}{dr_i}\right)\dot{r}_i+\frac{d\phi}{dr_i}\frac{\partial \dot{r}_i}{\partial \dot{r}_i}\right)$$

$$=\frac{d}{dt}\left(\frac{\partial}{\partial \dot{r}_i}\left(\frac{d\phi}{dr_i}\right)\dot{r}_i+\frac{d\phi}{dr_i}\right).$$

If $\phi=\phi(r(t),t),$ then$^1$ $$\frac{\partial}{\partial \dot{r}_i}\left(\frac{d\phi}{dr_i}\right)=\frac{d}{dr_i}\left(\frac{\partial\phi}{\partial\dot{r}_i}\right)=0,$$

so the right hand side reduces to


if we allow ourselves to interchange the order of differentiation. The right hand side is then the same as the left hand side (up to a $\partial/\partial t \leftrightarrow d/dt$) and we are done. I'm not sure about the subtleties of $\partial/\partial t$ vs. $d/dt$, but maybe you can come up with an argument. I hope this helps.

$1.$ In general if $\phi=\phi(r(t),\dot{r}(t),t))$ this step is not obvious and I'm not sure what the argument would be for this term to vanish.

  • $\begingroup$ For your footnote- I believe it's implied by the Gauge transformations to be some additional scalar potential added arbitrarily- I'm assuming (though I don't know) that it would only ever depend on r(t), t; or, in worst case scenario, this is true for my course, at least. $\endgroup$ – Frankie S. Palmer Oct 28 '19 at 21:25
  • $\begingroup$ @FrankieS.Palmer Great! Were you able to solve the problem? $\endgroup$ – aRockStr Oct 28 '19 at 23:23
  • $\begingroup$ As far as I can tell, these answer my problem. The only thing that stands in my was are the small subtleties between normal and partial derivatives here... I'll have to try and work that one out. (I wonder if, because it's with respect to just some singular r_i it forms only a partial derivative, as a component of gradient?) $\endgroup$ – Frankie S. Palmer Oct 30 '19 at 8:11

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