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I am reading Strominger's lecture notes "Lectures on the infrared structure of gravity and gauge theory" (https://arxiv.org/abs/1703.05448). At some point, following (I guess) the authors of the paper "New symmetries in massless QED" (https://arxiv.org/abs/1407.3789), the author of the lecture notes defines future and past conserved charges at near spatial infinity. The one for the future is

$$Q_{\varepsilon}=\frac{1}{e^2} \int_{\mathcal{I}^+_-} d^2z\gamma_{z\bar{z}}\varepsilon(z,\bar{z})F_{ru}^{(2)}$$

(the same definition was made in the paper the author follows, but with a different notation). I am trying to understand how or why this is a sensibly defined charge, in the sense of it being the zeroth (i.e. time)-component of a conserved current associated with the large gauge transformations: $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\varepsilon(x)$. So, my guess (and please correct me if I am wrong) is that I have to start with this current, which at Minkowski space is $J^{\mu}=\partial_{\nu}[\varepsilon(x)F^{\nu\mu}]$ and transform to retarded coordinates with the next step being to integrate over space (that is over the coordinates $\{r,z,\bar{z}\}$). The result I get is $$Q_{\varepsilon}=\frac{1}{e^2}\int d^2zdr\ r^2\gamma_{z\bar{z}}\nabla^i[\varepsilon(x)F_{i0}]$$ How do I proceed from here, such that I obtain an expression identical to the one above?? What considerations/assumptions must I make?

P.S.: The space of integration in my attempt of deriving the expression for the charge is still all the space. How do I go from there to being the past of future lightlike infinity ($\mathcal{I}^+_-$)?

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The current under consideration is $$ J \sim \star d ( \varepsilon \star F ) $$ It is clear that this current is conserved $d \star J = 0$. The Noether charge is then given by $$ Q_\Sigma = \int_\Sigma \star J \sim \int_\Sigma d ( \varepsilon \star F ) \sim \oint_{\partial\Sigma} \varepsilon \star F . $$ where $\Sigma$ is any Cauchy slice of the theory. We see that the Noether charge corresponding to this current is a boundary term evaluated on the boundary of the Cauchy slice.

In the paper, the authors take $\Sigma = {\cal I}^+ \cup i^+$. The boundary of this Cauchy slice is $\partial \Sigma = {\cal I}^+_-$. It follows that $$ Q_\varepsilon^+ = \oint_{{\cal I}^+_-} \varepsilon \star F \sim \int_{\mathcal{I}^+_-} d^2z\gamma_{z\bar{z}}\varepsilon(z,\bar{z})F_{ru}^{(2)}(-\infty,z,{\bar z}). $$

In your question, you are constructing the charge on a constant $t$ slice instead. The integrand $\nabla^i[\varepsilon(x)F_{i0}]$ is a total derivative so it reduces to a simple boundary integral. The only boundary in your case is the one located at $r \to +\infty$ (namely, spatial infinity) so the charge in your question reduces to $$ Q_{\varepsilon}(t) \sim \frac{1}{e^2}\int d^2z \gamma_{z\bar{z}} \varepsilon(t,z,{\bar z})F^{(2)}_{tr}(t,z,{\bar z}) $$ This is the charge at a constant $t$ slice. The charges constructed in the paper are then obtained by taking $t \to \pm\infty$, $$ Q_\varepsilon^\pm = \lim_{t \to \pm\infty} Q_{\varepsilon}(t) . $$

PS - I have not kept track of any overall factors in this answer. That's for you to work out!

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    $\begingroup$ Ok Prahar, thanks. I think it is fairly clear now how the two charges are connected! Appreciate it. One question though (it may be silly). You say that the boundary in my case is located at $r\rightarrow\infty$ and that makes sense. However, by doing some calculations, one can show that $\sqrt{-g}\nabla^{i}[\varepsilon(x)F_{i0}]=\partial^i [\sqrt{-g}\varepsilon(x)F_{i0}]$ integrated over r and the angular coordinates. $\endgroup$
    – schris38
    Commented May 12, 2022 at 13:41
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    $\begingroup$ $S^2$ is compact. There is no boundary in the angular directions. $\endgroup$
    – Prahar
    Commented May 12, 2022 at 13:45
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    $\begingroup$ Okay so is the process of writting the expression in terms of two total derivatives wrong?? Or is the fact that we assume that there exist boundaries while performing the integrations of of these two terms wrong?? $\endgroup$
    – schris38
    Commented May 12, 2022 at 13:49
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    $\begingroup$ Writing out the expression as you did is not wrong. You just then have to use the fact that $\int_{S^2} \partial_A W^A = \oint_{\partial S^2} n_A W^A = 0$ where the last equality is true since $S^2$ is closed so $\partial S^2 = \emptyset$. $\endgroup$
    – Prahar
    Commented May 12, 2022 at 13:50
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    $\begingroup$ Sorry for asking again, but why do the authors take that to be the Cauchy slice of the theory?? If $\Sigma$ were simply $\mathcal{I}^+$, I could understand that, because this is where photons and other charged particles go. But why is it the union of $\mathcal{I}^+$ with $i^+$?? And why in your answer of $Q_{\varepsilon}(t)$ is your "time-like" component the $F_{tr}$ instead of being the $F_{ur}$ since we supposedly pefrormed general transformations on the charge?? $\endgroup$
    – schris38
    Commented May 12, 2022 at 15:43

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