Discussion between schris38 and Prahar

9:00 AM

1

A: Conserved charge at null infinity associated with Large gauge transformation

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 \s...

schris38

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.

This yields two boundary terms, one over the radial coordinate and one over the angular ones, correct?? For example, $\partial^r[\sqrt{-g}\varepsilon(x)F_{r0}]+\partial^A[\sqrt{-g}\varepsilon(x)F_{A0}]$, where $A$ is an index labelling the angular coordinates. Doesnt' this contradict the fact that the volume has only one boundary, namely at $r\rightarrow\infty$??

Prahar

$S^2$ is compact. There is no boundary in the angular directions.

schris38

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??

Prahar

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 $\partial S^2 = \emptyset$.

schris38

Okay thank you so much!!!

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??

Prahar

9:00 AM

The authors are interested in describing a Ward identity for the $S$-matrix. This is an overlap between states living on the infinite past boundary (${\cal I}^- \cup i^-$) and the infinite future boundary (${\cal I}^+ \cup i^+$) so it makes sense to consider these Cauchy slices. ${\cal I}^+$ is NOT A CAUCHY SLICE whereas ${\cal I}^+ \cup i^+$ is! On the other hand, a constant $t$ slice on the other hand, IS a Cauchy slice. To answer your second question, we have $F_{tr}=F_{ur}$ since $t=u+r$.

In the paper, the authors do not discuss $i^+$ since that paper is focussing on massless QED. In this case, since everything is massless, $i^+$ is trivial (i.e. all fields vanish there) so only ${\cal I}^+$ needs to be discussed. In other words, in this special case $\int_{{\cal I}^+\cup i^+} = \int_{{\cal I}^+}$ since $\int_{i^+} (\text{anything}) = 0$.

schris38

Okay!! Once more, thank you so much!!

Prahar

hi

schris38

9:18 AM

Hello again Prahar. I am sorry for sending you a personal message, but as time goes by (and I read your answer), more questions come in mind, despite the fact that the questions I asked previously are answered in a very satisfactory way. So, if you please answer a couple more questions, I would be grateful. If you can not, that is okay. I will not trouble you much, I promise. Okay, so there goes: my questions are listed below
1) you say that I can recover the charges the authors define if from the constant $t$ charge we usually define in Minkowski space, I send $t$ to $\infty$. I understand…

Prahar

1) The pictorial way you presented is I think the easiest way to understand this. If you would like to be more precise, you should draw the Penrose diagram using the explicit coordinate transformations. Then draw precisely the curves corresponding to constant $t$ slice (use Mathematica). Take $t \to \pm \infty$ and verify that in such a limit you indeed end up on ${\cal I}^\pm \cup i^\pm$ as expected.

2) There are many different kinds of Cauchy slices. A Cauchy slice is ANY hypersurface such that by providing some data on that surface, we can determine the solution in the entire spacetime UNIQUELY.

A constant time slice ("all of space at a given time") is a Cauchy slice

A PAIR of orthogonal null surfaces is also a Cauchy slice

${\cal I}^+ \cup i^+$ is also a Cauchy slice as is ${\cal I}^- \cup i^-$

By providing appropriate data on ANY of these slices, one can determine the full solution to any wave equation uniquely in the entire spacetime.

You can work on ANY hypersurface you wish as long as its a Cauchy slice. Some surfaces are more convenient for some purposes. For the author's goal, ${\cal I}^+$ is a convenient Cauchy slice to work on.

${\cal I}^+$ is three dimensional and $i^+$ is also three dimensional

3) I'm "ADDING" the two spaces, not taking a tensor product

so the dimension of ${\cal I}^+ \cup i^+$ is also 3-dimensional.

4) ${\cal I}^+$ has two boundaries, ${\cal I}^+_+$ and ${\cal I}^+_-$. However, the Cauchy slice ${\cal I}^+ \cup i^+$ has only one boundary ${\cal I}^+_-$.

Most of your questions can be answered by simply drawing the Penrose diagram of Minkowski space and looking how each of the slices under consideration behave

hope this helps! Ask more questions if you have.

schris38

10:49 AM

I Can't thank you enough Prahar. Wow. Your answers are very clear and precise.
1) May I ask what is meant by the $\cup$ though? Is it the union between all the sets belonging in two (or more) surfaces/volumes?? I understand from reading your replies that you do not sum the dimensionalities of the sets you take the union of, so you simply put together the points in those sets??
2) Also, is the reason for taking the union $\mathcal{I}^+\cup i^+$ simply because, one can backwards "time" evolve of all the points inside this union, given that he or she has some date, and obtain the whole Minkows…

Prahar

$\cup$ is just union of sets. $R \cup R$ still has dimension 1 ($R$ is the set of real numbers) whereas $R X R$ has dimension 2.

2) yes!

3) Draw the slice on a Penrose diagram and you should be able to see this

another way to see this is to note that ${\cal I}^+ \cup i^+$ is obtained by taking a $t \to \infty$ of a constant $t$ slice

The boundary of a constant $t$ slice is the sphere at spatial infinity.

In the $t \to \infty$ limit, this boundary limits to ${\cal I}^+_-$.

schris38

11:26 AM

Okay now I think everything is clear about the Cauchy slices and their boundaries (I believe). Thank you so much!! I am very grateful for all the help you have provided! If you ever come to Cyprus, drinks on me ;)
Oh by the way, can you recommend some book on material covering Cauchy slices and other relevant material?? (It may be useful in the future)

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Discussion between schris38 and Prahar