The h Bar

Manas Dogra

12:15 AM

Is there a modern equivalent book of "Gravitation" by MTW?
or something like it?

Charlie

12:54 AM

A new version or a book that is a similar style?

Oh they're gone

Manas Dogra

1:08 AM

A similar style

@Charlie Weinberg's QFT books combined with the special QED,QCD-all collected are apparently like what MTW is to gravitation.

0xDBFB7

2:53 AM

Hey, does anyone here happen to know a thing or two about the FDTD method?

John Rennie

4:43 AM

@ManasDogra Gravitation has recently been republished as a new edition.

Manas Dogra

@JohnRennie Does it have any new content?

or just reprint?

John Rennie

It is still basically the same book but has been updated with new stuff since the original printing. I don't have a copy of the new edition so I cam't comment on how much has been updated. It's worth a Google for reviews of the new edition.

0xDBFB7

5:04 AM

Zerror=6.5%, Ferror=4.8%

pardon me...

hell yes

okay carry on

apt45

7:46 AM

Hi all! Consider a wave in curved space with four-momentum $k^\mu$. If $g^{\mu\nu}k_\mu k_\nu=0$, we say that this wave moves along null trajectories, and then it moves at the speed of light. While $g^{\mu\nu}k_\mu k_\nu=0$ is the definition of "null trajectories," it is not clear to me what is the meaning of velocity of a wave in curved space.

Very naively, if you consider a wave propagating along the $x-$direction, it should be $v=k^0/|k^x$, which, using the null-trajectory condition, is $v=g_{xx}/g_{00}$, very different from v=1 for non-trivial background solutions.

In which sense then waves propagating along null-trajectories move at the speed of light?

Slereah

8:04 AM

In the optical approximation of waves

As for "how they move at the speed of light", it depends on what you want, but a thing to keep in mind is the front velocity

apt45

@Slereah this is the approximation where you give the meaning of velocity of light?

Slereah

ie a wavepacket in a compact neighbourhood will spread along the light cone

apt45

can you be more precise on your last statement?

Slereah

Take some compactly supported wave

ie, in a compact region $U$, the field is non-zero, and outside of $U$, it is zero

(I'm talking about space here, not spacetime)

Then the time evolution of the wave will be such that it will be contained in $J^+(U)$, the causal future of $U$

And will be zero outside of $J^+(U)$

apt45

that's ok

what is your definition of front velocity?

Slereah

8:12 AM

Hm

I'm not sure about a coordinate independent definition

With coordinates it's simple enough, it's the change of the boundary $U$

apt45

this definition is too abstract...

Slereah

Well consider for instance a sphere

At time $t_1$, the field is zero outside of a sphere of radius $R_1$

And at time $t_2$, the field is zero outside of a sphere of radius $R_2$

The front velocity is $(R_2 - R_1) / (t_2 - t_1)$

apt45

yes I understood, but I would like a definition using dispersion relations of waves

Slereah

Well yes, but as you may recall, phase and group velocity aren't necessarily $c$!

(the proper theory to deal with wave equation velocities is the theory of characteristic surfaces btw, if you're interested)

apt45

I was excepting the group velocity to be $c$ for gravitational waves

Slereah

8:21 AM

The vacuum solution is certainly so, yes

I wouldn't venture a guess in the general case, though

Waves travelling at the speed of light in GR is theorem 5.1.2 of Friedlander's "The wave equation on a curved space-time"

But I'm afraid it's nothing as nice as dispersion relations

apt45

by vacuum solution you mean Minkowski?

Slereah

@apt45 Solutions without matter

apt45

of $R_{\mu\nu}=0$?

Slereah

yes

apt45

so, this is also true on Schwartzchild?

Slereah

8:29 AM

Well Schwazschild has no gravitational waves :p

apt45

I mean, you can surely consider the propagation of gravitational waves around a black hole

not generated by a single Schwarzschild blackhole, but GWs can travel on Schwarzschild space-time

Slereah

Sure, but that's a different issue then!

apt45

that's the case I am Interested in

Slereah

Perturbation of a Ricci flat spacetime may not be Ricci flat

apt45

you have a background spacetime and GWs travelling on top of it

Slereah

8:31 AM

Well, it is true for perturbative Minkowski, yes

apt45

I want to cover cases which are not perturbative Minkowski

for sure, I can give a definition of velocity locally

but I want to avoid to use local coordinates

Slereah

I think Hawking-Ellis shows that gravitational waves propagate in the light cone, too

Assuming the dominant energy condition

Captain Bohemian

9:24 AM

so hot

what's dominant energy condition?

Slereah

$R_{\mu\nu} t^\mu t^\nu > 0$ I think?

Captain Bohemian

I think that's weak energy condition.

Slereah

WEC is $T_{\mu\nu} t^\mu t^\nu$

Captain Bohemian

@Slereah ok. Then I have not ever seen the dominant energy condition you refer to, or maybe I have seen but fortgotten.

Slereah

There are many energy conditions

Captain Bohemian

9:33 AM

what are the purposes of those energy conditions?

I forget.

it's for coupling matter to gravity or what?

Slereah

Energy conditions are a way to encode salient features of matter fields without using the actual matter fields

it's simpler to use an energy condition to prove a theorem than showing that it's true for the standard model

or whatever else

Captain Bohemian

@Slereah then an energy condition needs to be general enough to apply to sufficiently many kinds of matters to be useful.

Slereah

Yes, pretty much

Or just some

After all, sometimes you're only interested in a specific case

The WEC is generally enough for all classical forms of matter

Captain Bohemian

what kind of matter does the dominant energy condition apply to?

Slereah

10:01 AM

oh wait, DEC is actually $T_{\mu\nu} V^\mu V^\nu \geq 0$, $T^\mu_\nu V^\nu T^\sigma_\mu V_\sigma \leq 0$

The DEC basically means $\rho \geq 0$ and $\rho \geq p$

So most reasonable matter will obey it

Captain Bohemian

@Slereah assuming the signature (-1,1,1,1)?

Slereah

yes

user434058

10:35 AM

Hey, someone's out there, adding a method-of-images tag on every question method of images question that was untagged until now. This involves old questions which were inactive for years. Is such an editing spree encouraged, and is there any reason I shouldn't approve their edits?

Captain Bohemian

10:47 AM

can PDF Architect comment on PDF?

PM 2Ring

@FakeMod On most sites, mass retagging is coordinated via the meta site. A low rep user probably shouldn't be doing this. But the mods definitely need to know about this. In the mean time, I advise you to not approve those edits, but I guess telling you that won't stop other edit reviewers from approving them.

That tag doesn't even have a tag wiki. Creating one should be the first step...

user434058

@PM2Ring Alright, let me flag one of the edited questions. And I am sorry, I approved about 10 such suggestions till now :( Though many of them also did other worthwhile improvements.

PM 2Ring

11:03 AM

Never mind. It's not a major catastrophe, just a minor annoyance. It's great that this person is trying to do stuff to improve the site, but it'd be better if it were discussed on meta first.

Urb

11:14 AM

Sorry, I was the one making the retags. I just realized that every time an edit is made it gets bumped into the main page, so it is indeed annoying to other users. I'll stop. Thanks for pointing this out @FakeMod and @PM2Ring.

user434058

@Urb no worries :-)

PM 2Ring

@Urb No worries. This tagging does soynd like a good idea, though, once it has been discussed on meta. So you can post a question about it on meta. And compose a suitable tag wiki for it.

You can get an idea of what a tag wiki looks like by clicking on the tag, eg electrostatics

@Urb For more info, please see meta.stackexchange.com/q/214337/334566

Urb

11:45 AM

Thanks for the links @PM2Ring. I'll take a look at them.

Captain Bohemian

12:00 PM

what on earth is a central charge? Is it a number or a group?

bolbteppa

It's an operator that commutes with everything so it can be simultaneously diagonalized with everything to give eigenvalues we call 'charges'

Captain Bohemian

@bolbteppa so a central charge is an eigenvalue of the operator commuting with everything?

I read a central charge of the Virasoro algebra is $3L_{AdS}/2G_N$. Is this central charge the only one central charge of the Virasoro algebra or just one of many central charges of the Virasoro algebra?

bolbteppa

12:18 PM

I'm not sure where that comes from

Virasoro algebra

In mathematics, the Virasoro algebra (named after the physicist Miguel Angel Virasoro) is a complex Lie algebra, the unique central extension of the Witt algebra. It is widely used in two-dimensional conformal field theory and in string theory. == Definition == The Virasoro algebra is spanned by generators Ln for n ∈ ℤ and the central charge c. These generators satisfy [ c , L n ] = 0 {\displaystyle [c,L_{n}]=0} and The factor of 1⁄12 is merely a...

Beliod

1:39 PM

I'm not a physicist or mathematician and I can't interpret this information. I'm trying to find some information about light-by-light (gamma-gamma) scattering. Apparently, at sufficiently high energies, photons can indirectly interact with each other. This has apparently been observed at CERN (atlas.cern/updates/physics-briefing/…)

So, I'm trying to find out at what energies these things happen (meaning, are we talking about the energy of a sun, Hawking Radiation, etc.). And, what would this look like? Meaning, if they indirectly interact with each other, and I had to sufficiently-high-energy light sources ramming into each other, would I see them interact?

PM 2Ring

2:38 PM

@Beliod See en.wikipedia.org/wiki/Two-photon_physics The 2 photons need to have sufficient energy between them to produce an electron + positron pair, i.e., 1.022 MeV.

Beliod

2:53 PM

@PM2Ring I know, but what's the energy required?

John Rennie

@Beliod it's not the energy of the photons that matters, as long as the combined energy of the two photons is greater than 1022 keV. It's the energy density i.e. the intensity.

That's because the probability of photon-photon scattering is extremely low, so you need extremely large numbers of photons to have a chance of it happening.

Charlie

Sorry to interject, does this sentence make sense? "I don't think I have anything specific to ask further about."

I'm sending an email to a lecturer and I want to use this phrase but I'm not sure if it actually makes sense

John Rennie

"I have no more specific questions ..."

Charlie

ahh, nice

ty

John Rennie

Presumably you have some more general questions, otherwise you'd just say "I have no more questions".

Charlie

3:06 PM

Well I've just got feedback on my project, and don't really have anything I want to dispute or points to raise about the feedback

Beliod

@JohnRennie I'll try to keep up since I'm not scientifically literate. I really appreciate the response.

So, the energy of an individual proton in this circumstance doesn't actually matter. But if two photons sufficiently close together have a combined energy greater than 1022 keV, they have a probability of interacting?

Charlie

Didn't mean to interrupt the conversation, ty John

Beliod

No worries @Charlie, I'm new here anyway! I'm the one intruding with silly questions lol.

John Rennie

@Beliod Yes. The energy of an electron and a positron is 511 keV, i.e. 1022 keV for the two together, so if you want the reaction photon + photon -> electron + positron to go then each photon has to have an energy of at least 511 keV. Any more than that is just a bonus.

But the probability of the reaction happening in any one photon + photon interaction is vanishingly small, so you need gazillions of interactions a second to observe it happening.

And to get gazillions of photons per second interacting you need an enormous energy density i.e. an enormous number of photons per unit volume.

Beliod

@Johnrennie Thank you so much. That's one part of this overall puzzle.

So, is there anything in the universe which might be able to supply enough protons with that combined energy density to create quite a lot of interactions?

Again, scientifically illiterate here, but quasars must have an enormous amount of photons in the area, and I assume that the energy density is insane. Would there likely be many interactions here?

John Rennie

3:12 PM

@Beliod good question, and I don't know the answer to that.

Charlie

The astophysics SE might be able to answer that @Beliod

Beliod

@Johnrennie Glad to have asked a good question haha. I'm also emailing ATLAS at Cern in hopes that I can get some answers.

You've been very helpful, John. I sincerely appreciate it. Scratching an itch that I've had for a weekend now lol.

@Charlie I'm heading there right now. I need answers haha. Thanks a bunch!

PM 2Ring

Photons, not protons! You need lots of photons, and those photons need high energy. You could get a visible light photon scattering with a gamma photon of >1022 MeV, though.

Beliod

I keep accidentally saying protons, or my thing is autocorrecting. But I do mean photons. :D

PM 2Ring

@Beliod One place in nature where photon-photon interactions happen is in the core of large stars that are on the verge of going supernova.

@Beliod Yeah, I figured that was the case. :)

BTW, if a photon with sufficient energy (>1022 keV) does pass close to a proton then the photon energy can produce an electron + positron pair.

@Beliod Yes. Active black holes are the other place in nature where you can find large numbers of high energy photons where photon-photon interactions can happen.

FWIW, I've been reading a bit about high energy photons in the last day or so, while polishing up this answer: physics.stackexchange.com/a/561217/123208

Annoyingly, it's hard to find good info about very low energy gamma photons.

A normal star core (like that of our Sun) does produce 511 keV photons, but the density of those photons is pretty small, so the photon-photon scattering is negligible.

Charlie

3:31 PM

If I could ask an unrelated question, when we make a quantum measurement and get an eigenvalue that is degenerate, the state collapses to the projection of the original state onto the eigenspace of that eigenvalue. Is this analogous to taking a vector in $\Bbb{R}^3$ and just projecting it straight down onto say the $xy-$plane?

So where exactly the state collapses to depends on which part of the plane it happens to be hovering over at that moment in the $\Bbb{R}^3$ analogy

Beliod

3:45 PM

@PM2Ring That's immensely helpful. And that last bit was even more of what I was looking for.

And to ask a follow-up, when you say that a small star's core produced 511 keV photons, does that mean that the overall photons in the sun's core have a combined keV of 511 or that each photon's individual keV averages 511?

Sorry if I'm being dense here, this is several orders magnitude above my current education.

PM 2Ring

@Beliod No worries! I love talking about this stuff. Give me a minute while I make a coffee...

Beliod

@PM2Ring I appreciate the kindness. I decided to eat while I read your replies, so that'll work nicely haha.

PM 2Ring

@Charlie Kind of. Remember, states correspond to rays in the Hilbert space, not points. But I'd prefer ACM, or another QM expert to answer this.

Charlie

Oh yeah

PM 2Ring

4:01 PM

@Beliod There's a large, continuous range of photon energies in a star core. A hot body is approximately a black body, which emits a spectrum of energies that depends on its temperature.

ACuriousMind

@Charlie Yes (modulo philosophical issues with the word "collapse" ;P)

PM 2Ring

In a star core, you have a bunch of hot nuclei and electrons, buzzing around with various amounts of kinetic energy. The temperature is a measure of that kinetic energy. A star forms through the gravitational collapse of a bunch of gas. That creates a lot of heat & pressure, and that kicks off fusion reactions. Those reactions produce energy in a couple of ways.

Firstly, the reaction products receive some kinetic energy. Secondly, in the process of turning 4 hydrogen atoms into 1 helium atom, 2 protons are converted to neutrons. That conversion process emits a positron from each of the 2 protons. Each of those positrons quickly annihilate with an electron, and that annihilation reaction creates a pair of photons, and each of those photons has 511 keV of energy.

Yuvraj

@PM2Ring hi long time

PM 2Ring

Hi, @Yuvraj

Yuvraj

4:17 PM

Hi sir

PM 2Ring

@Beliod In case you aren't familiar with that notation, 511 keV is 511,000 electron-volts. A diagnostic x-ray machine produces x-rays by slamming high speed electrons into a metal target, and the energy of the highest x-rays produced is limited by the voltage used to accelerate those electrons. So an x-ray tube running on 20,000 volts can produce x-rays with a maximum energy of 20 keV.

Johan Liebert

-2

A: How do we determine orbital names?

The second letter to the p orbital series denotes an axial position in a three dimensional space. The y axis is vertical, the x horizontal and the z longitudinal. If you press f3 in minecraft you'll notice a simile system to plot position in a three dimensional space.

Minecraft reference for real world things!

Charlie

4:38 PM

Nice ty! :) @ACuriousMind

Beliod

@PM2Ring This has been absolutely invaluable. Without the prerequisites, it becomes impossible to penetrate the literature.

PM 2Ring

4:55 PM

@Beliod In a star like our Sun, those 511 keV photons are much higher energy than the typical "thermal" photons. But in a really big star, in the last phases before it blows up in a supernova, it can get so hot that the thermal photons have enough energy to induce pair production.

That can have very drastic effects: you get a pair-instability supernova that totally blows the star apart, without leaving a neutron star or black hole remnant. See en.wikipedia.org/wiki/Pair-instability_supernova

user434058

5:12 PM

Someone's got a lot of time :-) Jokes apart, glad to see that some users go to these lengths to make this site a beter place. Great work, Thormund :-)

Thormund

xD

Thanks for helping me fix the latex

its holiday for me now too, i dont really have much to do either

i feel like not many people answer the atomic questions on stack ex, so i decided, why not!

i had such a poor experience being lectured in this topic too, so it didnt hurt to help

user434058

@Thormund no worries, though I do kind of feel that the drudgy stuff should be left up to the OP. As long as it's just a few lines of editing, that's alright, but devoting extended periods in editing doesn't require the OP improve their question (they need to learn how to ask a good question, because you aren't going to be there for them always), so that the next time, they just don't depend on anyone else's kindness.

user434058

@Thormund good thing :-)

Thormund

good point oops

user434058

In the above vein, I also left a related comment under the question 👇

user434058

5:25 PM

Your question has been drastically edited for good and there are chances that some errors might have crept in while editing, please ensure that the question in its current form does not miss out or misinterpret the original image which you used. As a side note, using images is discouraged because of the loss of searchability in images, thus from now on, I strongly recommend you to not use images, rather type it out. Also, you might also not get the services of kind editors in future, so don't expect your image to be typesetted by someone else. — FakeMod 6 mins ago

PM 2Ring

Here's a recent question on the mother Meta about transcribing text from images:

11

Q: Should I replace terminal screenshots by plain text?

In some questions or answers, users post screenshots of their terminal to highlight a terminal application or simply to share the output of a command. In some cases, I find using a screenshot useless. If the terminal output can be copied and pasted in answer/question without losing any informatio...

Transcribing hand-written text & equations is a little different, though. It's a lot of work, and the OP should be doing that work. OTOH, because it's a lot more work than merely copying and pasting something from the terminal it can be harder to motivate OPs to do it.

user434058

By the way, I was thinking of having a post on Meta regarding the disadvantages of using images. Kind of a community wiki post covering all the aspects of images in posts. Though I am not free enough currently, however if anyone is, then please make a meta post about it (or if anyone knows that such a question already exists, please just link it here).

user434058

I am tired of writing that "Images hinder the searchability of text within. They also do not work well for people who use text-to-speech engines. And moreover they might also mess up the display in mobile devices thus reducing readability."

PM 2Ring

@FakeMod Does Physics meta really need its own page for that? Here's an older MSE post that is more specific:

17

Q: Why are images of text, code and mathematical expressions discouraged?

Often I see new users posting images, containing the text of the question they were intending to ask, or a part of it. This is especially true for the science and math based sites where new users don't know how to use MathJax. I tried searching through the previous questions on Meta SE, but could...

discussion images accessibility

user434058

5:41 PM

@PM2Ring Ah, the link you provided will work completely fine, thanks!

user434058

23 hours ago, by ACuriousMind

May 3 at 9:46, by Loong

Jan 5 '16 at 22:43, by ACuriousMind

Jun 17 '15 at 6:43, by DanielSank

I regard online chat as an asynchronous communication protocol :-)

JingleBells

@ACuriousMind Is there somewhere private where I can message you?

user434058

@ACuriousMind Don't worry, I am going to continue your legacy when you're not present :P

JingleBells

@FakeMod Somewhere private where I can message u?

ACuriousMind

@JingleBells If it is related to physics.SE moderation I can open a private chat room, otherwise no.

user434058

5:49 PM

@JingleBells I am nowhere on the internet, and AFAIK, SE doesn't allow such private chats (except for mods).

JingleBells

Skype, Discord, Facebook...?

I want to talk to you as friends, not SE slaves

user434058

Though, we can chat in SoundCloud comments, I am on SoundCloud :P

JingleBells

@FakeMod cool, can we chat there?

I want to tell you the idea of the product I'm releasing :P

Get a feedback or two

user434058

@JingleBells I dunno, but I definitely intended that to be a joke.

Qwid

@FakeMod, just found some memes on Patanjali

Really hiarious

user434058

5:52 PM

@Qwid can't see now, will surely chck dem out later :-)

Qwid

Ok

Charlie

If I make a quantum measurement, get a degenerate eigenvalue and so my state is plopped somewhere in the eigenspace of that operator, if the time evolution operator causes the state to stay in the eigenspace indefinitely (i.e the system evolves in such a way that it just "rotates" around in the eigenspace) is there anything special about this? Is it even possible?

ACuriousMind

@Charlie It means that your eigenspace is a sum of eigenspaces of the Hamiltonian

JingleBells

@ACuriousMind Do you mind joining a third party private chatroom? I need some feedback on the idea and an advice.

Charlie

@ACuriousMind Oh ok nice, thanks

ACuriousMind

6:00 PM

@JingleBells If I was interested in communicating privately with you I'd have offered a way to do so, sorry. I do not provide consulting services.

JingleBells

@ACuriousMind Alright, no worries

ACuriousMind

@Charlie Do you see how to see that? (Think about how the Hamiltonian acts on its eigenstates)

JingleBells

Btw I'm currently 3d printing a Dogma Jesus hopefully it turns out nicely

Charlie

@ACuriousMind I'm getting a little ahead of myself, I haven't actually reached the time evolution section in Shankar yet, it is the next section though I'll have a think once I'm there

JingleBells

6:14 PM

why is iron man in the chatroom

@FakeMod That's called e-mailing

user434058

@JingleBells what are you talking about, I can only see a Sherlock Holmes...

user434058

@JingleBells Say that to ACM :P

JingleBells

Building the MVP of the idea, unfortunately, takes 4 months :P

Chances are, it will be a failure. It's a weird idea with a weird execution, but still, I've come so far, I'll not stop until it's released and I've done everything to market it properly.

It should come up with ideas which MVP takes a month or less :P

@ACuriousMind If the notion of "I'm the first to come up with this idea" is unrealistic, then where do ideas come from :D?

ACuriousMind

@JingleBells "unlikely" $\neq$ "impossible"

JingleBells

then I can be the first one to come up with my idea yay

my idea combines two incredibly simple and obvious things and yet I can't find anything when I google. The idea is either complete crap, or it's complete crap :D

Therefore, I can conclude, my idea is not crap. Because 1 + 1 = -2

where's my nobel prize

user434058

6:25 PM

Get some rest boi.

JingleBells

If Steve had 32 melons, and Alex had 25 apples, then how is combinatorics?

32 + 25 = 30 + 2 + 20 + 5 = 50 + 7 = 56

@Charlie I feel comfortable sharing my idea with u. Wanna hear?

i need feedback ;(

Charlie

Sure let me just get my notepad

user434058

@JingleBells No, the correct final question to be asked is : "What is the mass of the sun?"

Charlie

jk, if you want feedback I'm happy to help

user434058

@JingleBells QUICK MATHZ

JingleBells

6:28 PM

@Charlie yay, do u have skype, facebook or any third party private chat thing?

@FakeMod 2 + 2 = 4 - 2 = 3

Charlie

I'm not super comfortable giving any of that out, if we're able to private chat on here that would be best

JingleBells

@Charlie Do you mind if I send you a private chat room of a third party (chatzy)?

Charlie

sure

Hammad

7:33 PM

Could you see why this question got a down-vote? physics.stackexchange.com/questions/561522/…

Charlie

Seems like a reasonable question to me

Hammad

me too

Charlie

It's only a single downvote though, there's no saying why that person decided to do so

ACuriousMind

It's not really clear to me what the question is asking for. The asker is a) already aware that the error incurred from taking $g$ to be constant is miniscule and b) already knows how $g(r)$ should depend on $r$ in Newtonian gravity (evident from the formula at the end).

So what exactly is the question here?

Hammad

Maybe he wants additional supporting (critics and confidence)?!

ACuriousMind

7:43 PM

I don't know what that's supposed to mean

user434058

I'm too old to learn a new language. My word processor does superscripts and subscripts. It would help if your system would accept them. — R.W. Bird 56 mins ago

user434058

I don't know how to respond, I feel pity but at the same time I also acknowledge the importance of MathJax...

ACuriousMind

@FakeMod 1. You're under no obligation to respond at all. 2. See physics.meta.stackexchange.com/q/10284/50583.

user434058

8:03 PM

@ACuriousMind I am probably going to use 1. as a reason to not respond.

user434058

Well, I had a change of heart and I did post a comment because of sympathy.

PM 2Ring

FWIW, Anna V has been using MathJax recently. And she turns 80 this year...

PM 2Ring

9:06 PM

@ACuriousMind I ran into that OP earlier, when they posted this answer: physics.stackexchange.com/a/561415/123208 Maybe he just has a strange style of communicating. I can't tell whether that answer is just plain wrong, or if I'm misunderstanding him.

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