The h Bar

Secret

08:07

sciencealert.com/…

user116211

@Secret What does that mean?

user116211

@FenderLesPaul Congrats man! You deserve that :D

Secret

Basically, they said when they tried to simulate a black hole in 5D spacetime, they got naked singularities

I have not read the actual article yet to check the details, however

user116211

then I gonna star it.

FenderLesPaul

08:42

@user36790 Thanks!

Hari Prasad

@Secret what happens in 6D spacetime

Secret

No idea, they have not reported it

Hari Prasad

@Secret i mean with respect to what they did, if we do that with 6D space-time what would happen? maybe hyper relativity?

Secret

I am not sure, if naked singularities already formed in 5D, then adding one more spatial dimensions is not going to pack them back into an event horizon again

since 6D you have a lot more room to wiggle around, regardless of how subatomic that dimension is

That's my guess

Hari Prasad

@Secret hmmm nice guess!

Slereah

09:09

arxiv.org/abs/hep-ph/9810439

neat

Valentin Tihomirov

11:09

Don't you know when you have equation exp(some energy over kT)/T0? The T0 puzzles me.

Slereah

11:47

Help I am stuck in a nightmare of being sleepy at work

user116211

@Slereah what happened?

user116211

Boogeyman??

Slereah

12:27

The monster of having to wake up every day and go to work

user116211

O.O

DanielSank

13:14

@ChrisWhite That's the whole point.

@ChrisWhite Well yeah dude, but the basis functions tell you what temperature profiles are time independent and we were trying to come up with a physical reason for the exponential in $x$.

John Duffield

@0celo7 : re "it's too close to the particle electron". There is no magic. Electromagnetic-wave bosons do not magically pop out of existence whilst charged-particle fermions magically pop into existence.

DanielSank

@JohnDuffield Oh! I was sure it was magic!

0celo7

@DanielSank I'm sorry for being rude last night.

Qmechanic

0

Q: Which law of physics is applicable for communication?

What is the physics behind communication like internet, mobile phones etc. What are the equations?

Too broad?

user116211

@Qmechanic Very broad.

Slereah

13:30

Let's use string theory to do a radio

user116211

lol

0celo7

@Slereah it could be like one of those tin can + string contraptions

But with string and branes

John Duffield

@Secret : see this bit? "...singularity is thought to be the point of a black hole where gravity is at its most intense - the centre..." The force of gravity at some location is related to the local gradient in the "coordinate" speed of light. You can plot this using optical clocks at different elevations. However at the event horizon the coordinate speed of light is zero. And it can't go lower than this.

Slereah

Maybe that is how we can communicate with other universes

We must oscillate our brane so that the cosmic string will communicate it to the next brane

Secret

@0celo7 I don't know much, except that some second order ODE can be rearranged into stum lioville form which then allow the solution to be expressed as a series of eigenfunctions where its coefficients can be computed by the weighted inner product of the eigenfunctions

0celo7

13:38

@Secret Oh, I figured out my problem.

I had to do a strange manipulation to get the wife functions I needed.

Slereah

The wife function?

Is that how u get a wife

Secret

that's good

0celo7

I had to invoke that regular SL problems have a minimal eigenvalue.

*eigenfunctions

Secret

yup that's one of the results of stum liuville theory

I am not sure how that works in PDEs though, because the heat equation is a system of PDE

0celo7

When you separate variables one of the ODEs you get is SL.

Secret

13:41

ok that makes sense

John Duffield

@DanielSank : sadly you aren't alone. A lot of people believe in garbage like this. As if a 511keV photon magically morphs into a 511keV electron and a 511kev positron spontaneously, like worms from mud.

0celo7

I don't think I've ever eaten a worm.

Slereah

Well you know what you have to do now

Secret

0

Q: Where does the $i$ come from in the Schrödinger equation?

I am currently trying to follow Leonard Susskind's "Theoretical Minimum" lecture series on quantum mechanics. (I know a bit of linear algebra and calculus, so far it seems definitely enough to follow this course, though I have no university physics education.) In general, I find these lectures f...

quantum-mechanics schroedinger-equation hamiltonian

someone is going to put together that 3 weeks ago stuff in this chat room in order to answer this question, I think

0celo7

13:57

Maybe PSE should have a "GDP" tag.

Slereah

None-of-that-useless-bullshit tag

0celo7

Maybe "useless bullshit" should be a tag.

Slereah

I can ask all my category theory questions with it

0celo7

And GR, of course.

Holy crap I need that book.

Slereah

which book

0celo7

14:02

Beem et al

BBE from now on.

I think those are the initials.

yuggib

15:07

@0celo7 you can believe a part of those pages if you have not time to check everything ;-þ

0celo7

@yuggib Time/life energy

FenderLesPaul

15:19

sup

0celo7

Oh my god there's an old ass dude in my algebra class

He keeps trying to call out the prof and just embarrasses himself and wastes class time

FenderLesPaul

o.0

0celo7

@FenderLesPaul what

FenderLesPaul

15:41

<3 you

ACuriousMind

Sigh...I'm not keeping this up for a week, the chatroom is mainly people talking to themselves when one can't see 0celo7's posts.

0celo7

@ACuriousMind If I'm that terrible, you can leave...

ACuriousMind

@0celo7 The point is that you're not that terrible that I'd make this chat essentially non-sensical to me just to teach you a lesson.

0celo7

Oh, so I am terrible :/

Danu

@ACuriousMind Just talk to me

Ignore 0celo7 :D

skull petrol

15:52

He already did.

FenderLesPaul

@0celo7 you still have mee

0celo7

@Danu I can piss him off again if you want

@FenderLesPaul it's not the same

FenderLesPaul

:(

Secret

LAst night, I dreamt about this. I probably have spent too much time on surreals

Danu

^draw it in TeX

Secret

15:56

The man in the dream said he found a novel and efficient way to find the root of any polynomial using this kind of thing. But as I look at his workings, I was thinking "wait, that's factorisation, how could that possibly have anything to do with the roots of an (irreducible) polynomial?"

@Danu I wonder if tex has \fractals?

skull petrol

TeX has everything.

Danu

48

Q: How to create a Sierpinski triangle in LaTeX?

I have been trying to recreate the following image in LaTeX and for me personally it has to be done with pdflatex. Although answers using pst-fractals From Jake I got a good start which led to the following code \documentclass{standalone} \usepackage{tikz} \def\level{5} \usetikzlibrary{linden...

tikz-pgf pstricks

Better link^

skull petrol

see^

:)

There are other ways to teach him a lesson @ACuriousMind :P

0celo7

Now he hates me :(

That's cruel and unreasonable punishment, not a lesson

skull petrol

He has his reasons.

I've witnessed this whole drama unfold and remained silent for my own reasons.

0celo7

16:07

What drama?

skull petrol

You.

0celo7

@skullpetrol Huh?

Read the paper, maybe? — ACuriousMind 4 mins ago

@ACuriousMind A little harsh...

ACuriousMind

@0celo7 Someone who links the original paper and then asks about mathematical details deserves a little harshness :P

0celo7

@ACuriousMind How do you feel about prime factorization

:)

FenderLesPaul

@0celo7 no one cares

ACuriousMind

16:19

@0celo7 I don't care about it.

0celo7

@FenderLesPaul you're on the couch tonight

FenderLesPaul

@0celo7 no

I refuse

not this time

I shall not be a victim of abuse any longer!

skull petrol

Cyber bullying?

FenderLesPaul

domestic abuse

0celo7

@FenderLesPaul Getting relegated to the couch is not abuse

FenderLesPaul

16:21

@0celo7 it is a product of the patriarchy

0celo7

I love the patriarchy

skull petrol

Typical patriot.

0celo7

What?

FenderLesPaul

Holy shit Harper Lee died

0celo7

I didn't like To Kill a Mockingbird, tbh.

FenderLesPaul

16:24

Of course you didn't.

0celo7

@FenderLesPaul Sigh...what does that mean?

FenderLesPaul

@0celo7 Nothing!

nothing

Just a flippant remark

0celo7

@FenderLesPaul this is a safe space

you can tell me

FenderLesPaul

@0celo7 I can't it's too traumatic

0celo7

@FenderLesPaul then text it to me

FenderLesPaul

16:30

Okay

0celo7

D:

@FenderLesPaul I...

It's too painful

FenderLesPaul

I can only imagine how awkward this conversation must be for the others watching

0celo7

Imagine if they saw the other half, too

FenderLesPaul

hahaha

0celo7

@FenderLesPaul Ok, so what questions are there about Wald (1984)

Prop 8.1.1

That superradiance thing

The exercise about falling into a black hole has the wrong answer

FenderLesPaul

16:37

@0celo7 which black hole exercise?

but yes prop 8.1.1. and superradiance

0celo7

@FenderLesPaul It's in chap 6

FenderLesPaul

The one about the rope?

0celo7

There's one about a rope?

FenderLesPaul

And the redshifted tension due to the black hole?

yeah in ch6

0celo7

Oh, well that's bullshit too

FenderLesPaul

16:38

Nah that one is right

0celo7

But I was thinking about something else

@FenderLesPaul Chris White couldn't solve it

How do you do it?

FenderLesPaul

I can show you how to solve it

after my class is over

0celo7

Ok

No, the maximal time you can survive inside of a black hole

FenderLesPaul

oh

0celo7

It's not $\pi M$

FenderLesPaul

16:39

that one is correct too

0celo7

Nope.

FenderLesPaul

under the right approximation

I can show you the solution for that as well

0celo7

Negative.

FenderLesPaul

Fite me

0celo7

Ok, but it will be wrong.

FenderLesPaul

16:39

Challenge assepted

0celo7

You can use a jetpack to survive longer.

FenderLesPaul

Let's talk about it after my class is over

@0celo7 you can't afford a jetpack

0celo7

Is that your solution?

FenderLesPaul

No that one is wrong

There are hidden assumptions that need to be made explicit

0celo7

@FenderLesPaul I can't afford your bills either

FenderLesPaul

16:41

I'll show you later just trust me

0celo7

@FenderLesPaul There exist at least one paper that explicitly constructs a trajectory longer than $\pi M$.

FenderLesPaul

I'll explain to you why that doesn't apply within the context of the problem

just wait

0celo7

Then Wald's hint is wrong.

FenderLesPaul

Possibly

0celo7

Secret

16:47

progress is slow, will get back to you later

0celo7

@Secret That's more than I will ever learn

I don't think I can do my senior thesis on mathy GR

Too many spacetime diagrams

It can't be anything that requires diagrams of any kind.

So no diff geo

0celo7

17:23

Chat is dead.

@ACuriousMind Play any good games lately

Did you finish TW3?

DanielSank

@JohnDuffield How do you personally think about nonlinear corrections to Maxwell's equations if you don't imagine spontaneous particle pair production?

ACuriousMind

@0celo7 Long ago

It is taken as axiom how a quantum field transforms under a Lorentz transformation. Without showing how this fact is reflected from the transformation of the underlying Hilbert space. I want to see how tranformation of coordinates implies tranformation of fields in a mathematically consistent manner. While doing so and relating to the above "axioms" that we assumed, I would like to know how they fit in each other. — Sai krishna Deep 4 mins ago

Anyone understanding what OP is asking?

DanielSank

@ACuriousMind ::hides::

@ACuriousMind do you have any advice on how to deal with Hamiltonians that have periodic potentials and a momentum offset?

0celo7

@DanielSank my physics prof didn't know what the exponential means

ACuriousMind

@DanielSank What do you mean by "momentum offset"?

DanielSank

17:36

$H = (p - p_0)^2 + V(x)$

ACuriousMind

@DanielSank Yeah, most people seemed to think that. There's a single leave open vote by @JohnRennie, otherwise everyone in the review queue seems to have skipped that one.

DanielSank

@ACuriousMind I haven't looked at the review queue in months.

ACuriousMind

@DanielSank Do you have a reason for not transforming to $p' = p - p_0$?

DanielSank

@ACuriousMind Nope. So I should use some kind of $\exp(i p_0 x)$ thingy?

ACuriousMind

@DanielSank I'd really just redefine $p' := p-p_0$ and treat that as the momentum from then on, and then remember at the very end to resubstitute

DanielSank

17:40

But $p'$ and $x$ don't commute in the usual way. Surely this will lead to confusion.

Suppose this were a harmonic oscillator.

ACuriousMind

@DanielSank Of course they commute in the usual way - or is your $p_0$ not just a number?

DanielSank

@ACuriousMind Perhaps you're right.

ACuriousMind

Since eveerything commutes with the identity, $[p_0,x] = 0$, so $[p,x] = [p',x]$, unless $p_0$ is not meant to be just a number implicitly multiplied with the identity

DanielSank

@ACuriousMind I think you're right.

Ok, so if $V(x)$ is periodic, then $p_0$ essentially shifts all the Bloch wave vectors within each band.

Interesting.

Can any of you guys explain Bloch states in a way that's not confusing?

Danu

Aren't those one of these things that can be explained in a precise way using topology? :P

DanielSank

17:50

Dunno.

This is one of those things I never really understood.

Let's see if we can work it out.

Danu

I never thought about it---it was mentioned in passing in my QM3 course

DanielSank

Suppose $T_n$ translates a state by $n$ lattice units.

Danu

Are you talking about the qubit thing as well, or not?

DanielSank

Obviously $[T_n, H]=0$, which means that we can pick our energy eigenstates to be eigenstates of $T_n$.

@Danu Qubit?

Danu

No, lol wait

It's this other thing

The theorem of decomposing everything

into some form right

For a periodic potential

So what's your problem with it

DanielSank

17:53

@Danu Oh you were thinking Bloch sphere?

Danu

Bloch wave

A Bloch wave (also called Bloch state or Bloch function or Bloch wave function), named after Swiss physicist Felix Bloch, is a type of wavefunction for a particle in a periodically-repeating environment, most commonly an electron in a crystal. A wavefunction ψ is a Bloch wave if it has the form: where r is position, ψ is the Bloch wave, u is a periodic function with the same periodicity as the crystal, k is a vector of real numbers called the crystal wave vector, e is Euler's number, and i is the imaginary unit. In other words, if you multiply a plane wave by a periodic function, you get a Bloch...

@DanielSank Yeah, but no, I corrected myself :)

DanielSank

@Danu Yes I'm aware of Wikipedia.

Danu

I also know the Bloch states

@DanielSank That was just to make sure we are talking about the same thing :)

DanielSank

@Danu Yes.

Danu

(you passive-aggressive ass haha)

0celo7

17:54

I'm also aware of it, and Google.

Doesn't mean I can use it.

Danu

ORLY?

Oh, right

0celo7

:P

Danu

So you agree that this "decomposition" is possible right, @DanielSank

ACuriousMind

@DanielSank: To make us explain it in a not confusing way, please first state what confuses you ;)

DanielSank

@Danu What decomposition?

Danu

17:55

The Bloch form

DanielSank

@Danu Well, I know it's possible, but I'd like to go through everything from the beginning.

@ACuriousMind, I'm confused about what it means for $k$ to live in not the first Brillouin zone.

ACuriousMind

@Danu Don't start socratic questioning until you know what he actually wants to know :P

Danu

LMAO

DanielSank

I'm also confused about what an umklapp process really is physically.

Danu

I love this

umklapp? haha

Going full German

DanielSank

17:57

I don't think I have a chance at understanding this unless I go through everything from the start.

@Danu What's funny?

Danu

Also god-fucking-damnit chatroom, let me type quicker

@DanielSank I like talking to you both :)

DanielSank

@Danu Don't vent your frustrations at having a biological thought-electrical transducer. If you're displeased go upgrade to the latest cybernetics.

Danu

I am slightly inebriated

DanielSank

"slightly"

Danu

@DanielSank Lololol

@DanielSank I will slightly slap you if you don't shaddap with your smart remarks :)

DanielSank

17:59

Anyway, we surely can choose energy eigenstates which are also eigenstates of $T_n$.

Danu

Yes

DanielSank

Suppose we have such a thing $\Psi$, and suppose $T_n \Psi = C \Psi$.

Danu

Also I have to go for dinner

I WILL RETURNNNNNNN (soon-ish)

ACuriousMind

@DanielSank Oh god I have never found anyone who could explain those things. Most of the people who care about Brillouin zones and crystals seem to have never really heard of a Fourier transform :P

Danu

Sorry :\

I think I know this stuff---I had a thorough condensed matter 101 course

ACuriousMind

18:01

And the "reciprocal lattice vector" lingo is just plain obfuscatory

DanielSank

@ACuriousMind I know. I aim to be the hero this area of physics needs.

0celo7

@Danu wtf a 101 course that does QM?

DanielSank

I remember asking my solid state physics professor if all this business were similar to aliasing and he looked like I were speaking Greek.

@ACuriousMind That part doesn't bug me. Anyway, let's see if we can sort this all out.

ACuriousMind

@DanielSank Okay, let's try it ;)

DanielSank

@ACuriousMind That part doesn't bug me. Anyway, let's see if we can sort this all out.

Well, it seems that it should be the case that $T_n^{-1}\Psi = C^* \Psi$, but I'm not sure why I think that.

I'm fishing for a reason to write $C = \exp(i \theta)$.

ACuriousMind

18:05

@DanielSank The translation operator is unitary, so $T^{-1} = T^{\dagger}$, so $T\psi = C\psi \implies T^{-1}\psi = T^\dagger \psi = C^\ast\phi$.

DanielSank

Is it obvious that it's unitary?

Ah, yes it is obvious.

Ok!

ACuriousMind

You can see it either by saying it's a symmetry or by saying it's generated by the momentum operator.

DanielSank

Right, so $T_n \Psi = e^{i \theta} \Psi$.

Secret

@DanielSank How do you experimentally prepare coherent quantum states from non coherent ones for your quantum experiments?

DanielSank

@Secret We... don't.

The system is put in a cryogenic environment such that $kT \ll \hbar \omega$ where $\omega$ is the lowest energy excitation in the system.

Therefore, the system sits in its quantum ground state.

Ok @ACuriousMind so I think we should define a "non-phase" part of $\Psi$ as $u(x) \equiv e^{-i \theta} \Psi(x)$.

Actually, screw coordinates, $|u\rangle \equiv e^{-i \theta} |\Psi\rangle$.

0celo7

18:10

He's going full abstract now!!

DanielSank

That seems wrong.

Then $T_n |u\rangle = e^{-i \theta}T_n |\Psi\rangle = |\Psi\rangle$.

ACuriousMind

@DanielSank Why should we define that? The $\theta$ is the value for the specific translation you chose. You should define it as $\mathrm{e}^{-\mathrm{i}\theta/L x}\psi(x)$ to remove the phase "uniformly", for $L$ the length by which your $T$ shifted.

DanielSank

wow, chat messages out of order.

Right, let's define $u(x) \equiv e^{-ikx}\Psi(x)$.

There's a coordinate independent way to write that, isn't there?

$|u\rangle = e^{i k \hat{x}}|\Psi\rangle$?

ACuriousMind

@DanielSank And what's $x$?

DanielSank

@ACuriousMind Position.

ACuriousMind

18:16

Ah, as an operator

Didn't see the little hat

DanielSank

@ACuriousMind Yes.

@ACuriousMind That's why I usually inline this stuff: $|u\rangle = \exp(i k \hat{x}) |\Psi\rangle$.

Is that right? Did I get the sign right?

ACuriousMind

Well, let's evaluate the wavefunction: $u(x) = \langle x \vert u\rangle = \langle x\vert \mathrm{e}^{\mathrm{i}kx}\vert\psi\rangle = \mathrm{e}^{-\mathrm{i}kx}\langle x\vert \psi\rangle = \mathrm{e}^{-\mathrm{i}kx}\psi(x)$

Looks good to me.

DanielSank

Right.

Groovy.

Ok, now let us find out if $|u\rangle$ has any interesting translation properties.

$T_n |u\rangle = T_n(\exp(i k \hat{x}) |\Psi\rangle)$...

ACuriousMind

Looks like a case for writing $T_n = \exp(\mathrm{i}L_n \hat{p})$ and using Zassenhaus

DanielSank

What's $L_n$?

ACuriousMind

18:22

The length by which $T_n$ translates

DanielSank

Ok, then we do:

$T_n \exp(ik \hat{x}) T_n^{-1}T_n |\Psi\rangle$.

That does something simple to the $\exp$ part, does it not?

ACuriousMind

@DanielSank Ah, yes (it's directly below the Zassenhaus formula in the article I linked)

It gives $T_n\exp(\mathrm{i}kx) T_n^{-1} = \exp(\mathrm{i}kx) + [\mathrm{i}L_n p,\exp(\mathrm{i}kx)]$, I think.

DanielSank

Reading...

ACuriousMind

Hm, wiat, I have to write this out somewhere...

DanielSank

The article says:

$\exp(X) Y \exp(X)^{-1} = \exp([X,Y])$, I think.

But my notes say:

If $[a, b] = \eta$ where $\eta$ is a number, then $e^{ia}be^{-ia} = b + i \eta$.

ACuriousMind

18:36

@DanielSank What about: $T_n \exp(\mathrm{i}kx) T_n^\dagger$ is the transformation of the operator in the middle under the shift $x\mapsto x+L_n$, so the result is $\exp(\mathrm{i}k(x+L))$.

DanielSank

Yes, that matches what my notes say.

actually, I think the sign is backwards.

@ACuriousMind This expression may have the wrong sign.

ACuriousMind

Yes, I am currently checking that

0celo7

@ACuriousMind Are there baked beans in Germany?

DanielSank

Hah, you've also ignored $\hbar$, but I'm ok with that.

0celo7

@DanielSank $\hbar=1$

DanielSank

18:41

Is there a way to vote to close chat messages?

I vote to close @0celo7's last message.

0celo7

:(

DanielSank

@0celo7 You cannot set dimensionful things to dimensionless numbers. it makes no sense.

I don't want to get into that here though.

0celo7

@DanielSank Imagine the horror if everyone thought like you

Can you imagine writing $\frac{8\pi G}{c^4}$ all the time in GR!!

Secret

particle physicists tend to use natural units

PS one of my physics classmate always said "fuck hbar" whenever I told him to add hbars

DanielSank

@Secret I really really don't want to get into this right now, but calling what particle physicists do "natural units" is a horrible abuse of language that confuses the shit out of countless students.

0celo7

18:44

Does it?

Secret

ok then, focus on the problem first, we will chat later about that

DanielSank

Setting dimensionful things to dimesionless numbers should be illegal.

0celo7

@DanielSank Personally, I only get confused when you try to explain to me why I'm confused.

And I get confused by what I'm supposed to be confused by.

DanielSank

@ACuriousMind The operator $T_n$ as you defined it converts $x$ to $x - L_n$.

So we should just reverse the sign and be happy.

Doing this, we get:

$T_n |u\rangle = \exp(ik(x+L_n)) T_n |\Psi\rangle$.

yuggib

@ACuriousMind you can't use BCH or zassenhaus for unbounded operators... T__T

DanielSank

18:50

@yuggib Sure you can.

Danu

@yuggib Lol

0celo7

@yuggib A physicist can

ACuriousMind

@DanielSank Arrrrgh, I found my mistake. Yes, I agree (and this should have been this hard to get :D )

DanielSank

Well ok, $T_n |\Psi\rangle = \exp(i \theta) |\Psi)$, so in the end we have:

ACuriousMind

@yuggib How does the mathematician compute $[x,\exp(\mathrm{i}Lp)]$, then?

DanielSank

18:52

$T_n |u\rangle = \exp(ik(x + L_n)) \exp(i \theta) |\Psi\rangle$

$=\exp(ik L_n + i \theta) \exp(ikx) |\Psi\rangle = \exp(i(kL_n + \theta)) |u\rangle$.

ACuriousMind

@DanielSank Yep, and since $\theta = kL_n$...the sign is wrong :D

DanielSank

Oh, yeah we'd like those to cancel, wouldn't we?

ACuriousMind

Yeah. $\lvert u\rangle$ is supposed to have eigenvalue 1 for $T_n$

DanielSank

Right but now I'm confused because I did my math assuming that $T_n \hat{x} T_n^{-1} = x + L_n$, didn't I, and isn't that what it should do?

In other words, where did I mess up the sign?

Oh wait, this isn't a real problem. I can just say that $T_n |\Psi\rangle = \exp(-i \theta)|\Psi\rangle$.

Now, @ACuriousMind why do you say that $\theta = k L_n$?

ACuriousMind

@DanielSank Because we defined it that way

DanielSank

18:59

@ACuriousMind We did?

ACuriousMind

46 mins ago, by ACuriousMind

@DanielSank Why should we define that? The $\theta$ is the value for the specific translation you chose. You should define it as $\mathrm{e}^{-\mathrm{i}\theta/L x}\psi(x)$ to remove the phase "uniformly", for $L$ the length by which your $T$ shifted.

I wrote that, and then you decided to write $k$ for my $\theta/L$.

DanielSank

Interesting, I didn't notice that. I need to re-read.

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