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12:01 PM
@ACuriousMind thanks, to be honest I was expecting to be told that. The post was at least partly an attempt to clarify my own thinking at at least it has some value in showing my thinking was wrong :-)
@SirCumference: did you see my reply to the question you asked last night?
 
12:17 PM
Yep :D
@JohnRennie But by what you said, time dilation must increase as you approach the apparent horizon, not event/absolute horizon, right?
I think I'll ask the question on the site and see what other responses I get
 
Time dilation increases as you descend into the gravitational potential well. The true horizon is the point where the time dilation becomes infinite. An apparent horizon is a point where it only looks as if the time dilation might be infinite because you haven't been watching for long enough.
True horizons are stranger than you think. A true horizon is a property of the entire spacetime.
 
@JohnRennie Hm, I thought the apparent horizon was simply the boundary between whether light directed outwards would move outwards or inwards
And as it grows it would approach the size of the absolute horizon?
So the absolute horizon is where time dilation becomes infinite?
 
Precisely defining a true horizon is a complicated business in all but the simplest geometries, and I freely admit I don't understand the details.
 
@JohnRennie All right, as far as I know, and this might be an incredible oversimplification: the apparent horizon is what an outside observer would call the "black" part of a black hole at a given time. The absolute horizon is what the future apparent horizon will look like, right?
 
I think the definition is basically that light emitted from a true horizon will never real null infinity. Light emitted from an apparent horizon will reach null infinity eventually though it may take an arbitrarily long time to do so.
The horizon is a property of the whole spacetime because you cannot tell whether an outgoing light ray will reach null infinity unless you know the spacetime out to null infinity.
 
12:31 PM
On the contrary, I thought it was like this: "the apparent horizon is the boundary between light rays that are directed outwards and moving outwards, and those directed outward but moving inward. If the light is within the absolute horizon, it will not escape to infinity."
It seems like the apparent horizon must be within the absolute horizon
Better yet, Wikipedia sums this up
"Within an apparent horizon, light is not moving away from the black hole, whereas in an event horizon, light cannot escape from the black hole."
"It is possible for light to be currently moving away from the black hole (and so outside the apparent horizon), but in the future will not be able to escape (e.g. because the mass of the black hole is growing) and therefore inside the event horizon"
So, er, not sure what to make of that
 
Ok that makes sense.
 
Does it to you?
Not really to me
If light is moving away from the absolute horizon, then why shouldn't it escape to infinity?
Would the black hole's apparent horizon increase to meet the size of the absolute horizon, before the photon can escape?
Sigh, I have no idea what Wikipedia was saying
 
Suppose you are inside the event horizon falling towards the centre of a black hole. You hold a torch at arms length nearer the centre than you are.
The light from the torch can reach you i.e. it looks to you as if the light is moving outwards.
 
Yeah, ok
 
But the light isn't moving outwards. What happens is that you are falling inwards faster than the light is, so you catch up with the light even though the light is moving inwards. So far so good?
 
12:45 PM
Er, I'm falling faster than light?
 
You're falling inwards faster than light that has been shone outwards.
Light that has been shone inwards is moving faster still.
 
All right
 
Now suppose you put the torch on the end of a long selfie stick so it's a lot nearer the singularity than you are.
 
There comes a point where even light directed outwards falls inwards so quickly that it will never reach you i.e. to you it looks as if that light is falling inwards.
 
12:48 PM
And that point is the start of the apparent horizon?
 
The boundary between where the light that is shone outwards appears to you to be moving outwards and the point where it appears to you to be moving inwards is the apparent horizon.
 
All right, that makes sense
But how about this
I take a huuuuge selfie stick holding a torch, and I put the torch inside the event horizon of a black hole, while I comfortably remained outside the event horizon.
Would the light from the torch reach my eyes?
 
Incidentally, I wrote an answer on this here:
4
A: Taking selfies while falling, would you be able to notice a horizon before hitting a singularity?

John RennieThis is largely the same answer as Rob's, though rather than use Eddington-Finkelstein coordinates I'm going to use Kruskal-Szekeres coordinates because I think this makes the argument easier to understand. This is what the situation looks like in Kruskal-Szekeres coordinates: For the non-nerd...

@SirCumference Are you falling inwards or stationary relative to the black hole?
 
Stationary
 
In that case no the light from inside the horizon will not reach you.
 
12:53 PM
So, essentially, the entire area within the absolute horizon seems black when I'm outside of it
But once I'm inside and holding a torch, it no longer looks black?
All right, better question:
What would someone outside of the black hole call the "black part"?
 
It's more complicated than that because apparent horizons are not invariants. They depend on the observer. Feel falling observers will report a different apparent horizon to accelerated observers.
Take you example of you outside the true horizon, the torch inside the true horizon and you falling freely.
 
Crap, I gotta go
But I must leave this
23
A: Does any particle ever reach any singularity inside the black hole?

G. 't HooftIndeed you made one mistake: the infalling observer does not see the outside universe "speed up". Look at what happens in a space-time diagram. At the spacetime point where your astronaut passes the horizon, he can only see what's in his past light cone, and that's the universe at early times onl...

 
The light from the torch cannot move outwards and come out through the horizon. However you can fall through the horizon and catch up with the light.
 
Jim
One of these days, scientists will have to just throw someone into a black hole and have them constantly report what they see as they see it. That way we can put an end to all this speculating on what an in-falling observer sees.
 
This answer seems contradictory
 
12:58 PM
I don't think any of the answers to that question are good ones. Note that I commented contradicting that answer and my comment got 8 upvotes.
This is a fearsomely complicated area to understand.
 
Only one thing to do then
Reload the page
Now there's a huge bounty
And I'm out of points
And...crap, now I can't comment...
So I can't even write why some answers aren't good
 
@SirCumference That's what the bounty text is for
Which you can't edit now :P
 
@ACuriousMind I know. I'm stupid. It's too late.
I see "(optional)" and I think it's a waste of my time
Now I realize I can't comment
And now I have this on my profile
Two locks
 
@SirCumference If you write the comment you want to post here I can post it for you or I imagine one of the moderators could do it.
 
1:14 PM
@SirCumference that's why you should be careful about how you set bounties ;-)
I'll make you a one-time offer: I can refund the bounty and you can re-set it for a smaller amount. If you want.
 
Jim
@DavidZ It's good to be the king
 
"king" is a stretch, but I suppose
 
@SirCumference I strongly recommend you accept David's offer, and actually you might want to think twice about placing the bounty. I'm not sure any of the current site members can comment authoritatively on the subject and the bounty will be wasted.
I don't think 't Hooft has answered the question that was asked. Or at least if he has his answer is so far above my head that I've failed to understand his point.
 
Jim
@DavidZ Well, as far as I know, the phrase isn't "It's good to be not really the king, but someone of superior authoritative power"
 
haha :-P
 
1:20 PM
I've always fancied being a dark lord myself. One day ...
 
Jim
:Jim bows in reverence to the future dark lord:
 
@Jim I'll put you on the list to be slaughtered last :-)
 
What is it with all the slaughter? Everyone with dark lord ambition is so morbid.
 
I think that's kinda part of the job description
 
I always thought it would be more evil to hide in the shadows (figuratively as well as possibly literally) and convince other people to implement your evil plan.
...killing only as necessary to move the plan along.
 
1:28 PM
@DavidZ Ah, but then you're a gray eminence, not a dark lord
 
@ACuriousMind: re that attempted answer of mine about virtual particles: is there any hope of writing an answer at that level that provides a coherent explanation of how virtual particles arise? Or is the whole idea a non-starter?
 
@JohnRennie Well, they say "never say never", but I can't see any explanation being at that level, correct and conveying non-trivial information.
Doesn't mean you should stop trying, though ;)
 
Hmm, that's what I thought you'd say. That's pretty much what David said when I asked him.
Well you said It's in that evolution of the asymptotic past state to the asymptotic future state that the virtual particles hide so I guess the target would be to explain in broad terms how this calculation is done and what role the VPs play in it.
 
...which is what all introductions to QFT basically are about
 
So I've got to learn QFT :-)
 
1:35 PM
I guess you could explain it simply as "the internal lines in Feynman diagrams", but that would be profoundly useless to anyone who doesn't already know QFT.
 
@SirCumference : I know what happens when you fall into a black hole.
 
@Qmechanic: you closed this as non-mainstream:
0
Q: Would a sudden mass decrease would cause antigravitation?

DimsSuppose some mass $M$ is located in rest in some point of space. It creates gravitational potential well, which attracts test bodies towards $M$: Now suppose that the mass of a body drastically decreases several times to $m << M$, for example if the mass is radiated away as gravitational waves...

But I had edited it to make it a mainstream question. In its current form I can't see anything non-mainsteam about it.
It's just asking about the evolution of the gravitational potential energy as mass/energy is ejected from the system.
 
@DavidZ All right, I think I'll take the offer
Thanks. Proves my idiocy though...
 
OK, done
 
@SirCumference I would be willing to go halves with you. I think two people can place a bounty on the same question (probably best to check first though). My only concern is that I don't think we'd get a good answer even with a humungous bounty. I'm not even sure if anyone knows the answer.
 
1:48 PM
You could hold off on placing a bounty until you track down someone who would give a satisfactory answer
 
@JohnRennie Only one bounty at a time
 
With Hawking radiation we now have a time dependent metric, and they are a lot harder to deal with than stationary ones. I certainly don't feel confident to comment, though it seems blatantly obvious that the black hole must evaporate before the observer falls in, which is what I said in my comment.
 
Morning
 
@ACuriousMind ah, OK, thanks.
 
Day 6 of not understanding flows has begun
It's basically a meme at this point.
 
1:58 PM
@JohnRennie : I know the answer.
 
Can someone explain censoring of words
Does't "f*ck" just make you read "fuck"
 
@0celo7 yeah, probably. I guess it has something to do with overly literal application of rules against swearing, or maybe beating automatic filters :-/
 
2:31 PM
0
Q: How do we judge the flow of time from our frame of reference?

NationWidePantsIs there a means to determine our frame of reference on past events in space-time? How do I know my current reference (today) is the same as it was two days ago for the rate at which time is perceived? A satellite sits in space and has an atomic clock while another is on earth there will be a per...

Not sure what to said about this one...
 
O_O"~ I'm not even sure I understand him or he understands the concepts he is using - could be either or both
 
@ACuriousMind Should I ask my flow question on MO?
 
2:46 PM
@0celo7 How am I supposed to know that?
 
3:01 PM
@ACuriousMind Because you know such things...
 
3:18 PM
Spin is mental
 
@bolbteppa Ricci flow is flat out insane
 
Where is your flow question
 
You mean where is his flow of questions?\
:D
 
Even a drip of one would be fine ;)
 
@bolbteppa yeah you're right. :D .. that's why you have won various gold, silver and bronze medals in olympics.. striving for questions and knowledge ..
:D
 
user54412
3:37 PM
@JohnRennie I would put in the time to write a full answer, but as I have a thesis to defend in 5 days...
 
@ChrisWhite congrats on your thesis defending
 
3:55 PM
@ChrisWhite as and when you come back to us, basking in the glow of a successful defence, let us know and we'll put a bounty on the question.
 
4:07 PM
hello, again
 
4:32 PM
does anyone know how atomic force microscopy works?
 
@0celo7 : You're reading a book on homology that doesn't mention mapping cones?
 
@WillO I'm not reading a book on homology
 
Okay...you referred to a book you were reading that had an exercise on homology, so I assumed....
 
It does not appear in the index either
 
May I ask what book?
 
4:40 PM
It's in the OP.
I still don't know what you mean by "horizontal homology is zero"
Do you mean the cohomology when $q$ is fixed
I don't know what that diagram you drew means
 
I apologize for not having read the OP more carefully.
Yes, by horizontal homology I mean the (co)homology of the rows.
 
But those are not zero
In the case where they are zero, I can prove it
 
Okay, here is what you are missing:
You have a map f:K\rightarrow L, yes?
 
Yes
 
Now let's suppose for a moment that you had a third complex C, and maps K->L->C, inducing a long exact homology sequence.
(Are we on the same page with the phrase "long exact (co)homology sequence"?
 
4:47 PM
Yes
 
Okay. Suppose this map gives you a long exact cohomology sequence for the total complexes, and for each individual row.
Now three steps:
1) Because the map K->L induces isomorphism on the horizontal cohomology for each q, the LES shows that the horizontal cohomology of C is zero for each q.
 
@WillO What map?
 
2) I thought you were given a map?
 
Oh, f?
 
Yes, f.
So again: Step 1: f is iso on the homology of the rows, therefore the LES shows that the homology of the rows of C is zero.
Step 2: Because the horizontal homology of C is zero, you can use the result you say you already know to conclude that the total homology of C is zero.
Step 3: Because the total homology of C is zero, the LES for total homology shows that f induces an isomorphism on total homology.
So you're done, provided we can find a map L->C that induces long exact sequences as promised.
 
4:52 PM
Shit, I made an error in my Ricci flow thing
 
The C you are looking for has C_{ij}=L_{ij}+K_{i,j+1}. The differential is the differentia from L, plus the differential from K, plus a cross term given by f. (Up to signs, anyway.)
That C is called the mapping cone and you can read about in any of a kajillion places
 
@WillO cross term?
 
The map from the K at one stage to the L at the next stage
 
@DavidZ In a real sense that's what got Plagueis and Sidious in the end.
They almost had it.
 
@WillO never heard of it, and it's not in any of the three algebraic topology books I own
 
4:57 PM
This is a problem in pure algebra. You need an algebra book.
 
@JohnRennie Noooooooooooooooooooooo
@DavidZ What's wrong with the intermediate states in a time dependent perturbation series?
 
Or....you can just work through the details of what I've just said. It's all there in the scroll, except for some routine verifications.
 
@WillO uhhh
I'm very confused
 
@JohnRennie I think a problem with the QFT that it is new, and changes often. Thus, different sources often say contradictory statements.
 
I think our horizontal and verticals are different?
 
5:00 PM
Could be.
 
The $d$ cohomology, where I have the isomorphism already, is the vertical cohomology
 
You don't need QFT to understand virtual particles.
 
Okay, then change all of my horizontals to verticals and the same argument works.
 
Well, you need QFT (but not relativistic QFT) to understand "particle", but once you get that you just need vanilla quantum mechanics to understand the "virtual" part.
 
You are in the first quadrant, so nothing is changed by switching horizontal with vertical
 
5:03 PM
So I have a SES $0\to K\to L\to C\to 0$
 
No.
You have maps K->L->C, no zeros on the ends
 
well then I can't apply the snake lemma to get an LES
 
Yes, you can, for this particular choice of C
You can't, of course, in general.
Or more precisely: You get the LES; you dont need the snake lemma.
Try it for a single complex before you try it for a double complex:
C_i= L_i + K_{i+1}. (+ means direct sum)
 
what is the connecting homomorphism
 
Map C_i to C_{i+1} by taking taking (x,y) to (dx+f(y),dy)
(You might need to adjust the signs; maybe dx-f(y).....)
 
5:10 PM
jesus
 
It is a good exercise to check that K->L->C now gives you a long exact sequence
 
0
Q: Serial up and down voting by a moderator

JenHow many times can a moderator or super user vote a user over time? I know if I go to another user's profile and up and down voted multiple Q&A then the votes could get retracted as serial votes. So is there a time between votes that would not red flag? I believe I may be serial down voted slowly...

 
What is the map $L\to C$?
 
No snake lemma involved.
The map L to C just takes x to (x,0)
 
ok
 
5:13 PM
It's pretty straightforward to check that this gives a LES. Try working that out first. Then it's not hard to see you can do the same sort of thing with a double complex that we just did with an ordinary complex. Then use my Steps 1,2,3 above.
 
where does $\delta$ come in
oh I guess it doesn't for the single complex
 
Gotta step out. I promise that all you need is in the above (though I understand that it can take some time to digest).
 
Wait
By connecting homomorphism I mean the map H^p(C)-> H^{p+1}(K)
in the LES
 
5:45 PM
what is a length vector?
 
5:56 PM
@DanielSank I'm not sure I understand that
 
Jim
6:07 PM
@ItachíUchiha it is a value that contains a magnitude of length along with an associated direction for the length
 
6:30 PM
@JohnRennie : I removed some non-causal non-mainstream elements, and reopened it.
 
@JohnRennie By the way, I'm up for that
 
@JohnRennie doesn't work, thanks
 
user54412
I should have asked Hitler. — Jen 16 mins ago
 
user54412
6:46 PM
The list of people whose every post will be obliterated for sheer stupidity if we change the VLQ policy continues to grow.
 
user54412
You know, I'm kind of warming up to the idea.
3
 
@ACuriousMind Posted on MO
I should probably email the author
 
@ChrisWhite What? Are we trapped in the Q&A version of The Producers?
 
user54412
Masses on Springs for Hitler
 
This is how you sing "Least Action Anzatz"!
 
6:57 PM
aaaaand I got a downvote
@ACuriousMind Was that you
 
@0celo7 what?
downvote on what?
 
my MO question
 
@0celo7 I don't even have an MO account, silly
 
@0celo7 you mean the parallel port card? What do you mean by doesn't work? Does it show up in device manager?
 
@JohnRennie drivers won't install on Windows 10
@ACuriousMind Huh?
 
7:05 PM
@0celo7 It's a Startech card, if you look on the Startech site I bet there are w10 drivers for it.
 
@JohnRennie Too late, we installed it on another computer
Less convenient
But $30 down the drain thanks to you
 
@0celo7 Sorry :-(
 
@JohnRennie they have W10 instructions on the website
@ACuriousMind Adventuring into tech...
help
 
7:22 PM
As well as a manual.
The card is supported on Windows 10
 
Already did all of that
Still doesn't work
 
All right, can someone explain why the light cone isn't completely contained in the Hubble sphere?
 
@SirCumference this looks relevant:
9
Q: Is the edge of our Hubble Sphere within our Cosmic Event Horizon and why?

DilithiumMatrixI was recently shown a pretty cool video about common cosmological misconceptions. It got me reviewing the different between event horizon (current distance within which we will see/interact), particle horizon (current distance within which we have seen/interacted, and the Hubble Sphere (distanc...

 
Tried that, too much jargon
 
One last chance @JohnRennie
Nope
You broke the PC
 
7:30 PM
Well you bought the card from Dell, for use in a Dell PC. If it doesn't work send it back to Dell and get your money back.
 
How do I tell if it's the card or the dongle
The PC recognizes the card and lets me install drivers
But the software does not recognize the dongle plugged in
It seems to work on another PC
 
@0celo7 Aha! So the card does work. You might need to set the parallel port mode for the dongle to work. Does the software come with any instructions e.g. whether the port mode should be SPP, ECP or whatever.
And check what port number the card appears as - the software probably expects LPT1.
@0celo7 on that other PC is it using the built in parallel port?
 
user54412
7:44 PM
> The teardrop shape of our past light cone in the top panel of Figure 1 shows why we can observe objects that are receding superluminally. Light that superluminally receding objects emit propagates towards us with a local peculiar velocity of c, but since the recession velocity at that distance is greater than c, the total velocity of the light is away from us (Equation (20)).
 
user54412
> However, since the radius of the Hubble sphere increases with time, some photons that were initially in a superluminally receding region later find themselves in a subluminally receding region. They can therefore approach us and eventually reach us. The objects that emitted the photons however, have moved to larger distances and so are still receding superluminally. Thus we can observe objects that are receding faster than the speed of light (see Section 3.3 for more detail).
 
user54412
 
Ah! So a photon that is emited by a superluminally receding galaxy and directed towards us will find that their region, which was once expanding faster than light, now expands slower than light?
After enough time?
 
user54412
yes
 
Oh my god yes!
 
user54412
7:48 PM
or at least it can happen
 
You put me out of my misery
So, how come the Hubble sphere will eventually meet up with the event horizon?
Also, is the Hubble sphere growing faster than the objects outside of it can retreat?
 
8:05 PM
@ChrisWhite Er...you there?
 
@SirCumference The man has a thesis defense in less than a week, you'll have to forgive him intermittent presence :P
4
 
@JohnRennie what
@JohnRennie yes.
 
@ChrisWhite This seems contradictory
13
Q: As the universe ages, will we see more stars or less?

YairAfter a very long time will we see more stars (due to the fact that more light is get to us) or less stars (as the universe expends and light have to pass larger distance)? In general, can stellar objects go outside of the scope of the observable universe or is it only growing with time? Calculat...

You said the Hubble sphere is growing, while the answerer here said
> If a galaxy is receding from us faster than light, a photon emitted from said galaxy towards us will initially be receding from us, but it will crawl to regions of space - at local light speed - which are receding from us a bit slower than the one from which it was emitted, and so it will recede a bit slower from us than before, etc., until it finally enters a region of space that is receding from us slower than light, from which point it is trivial that it will eventually reach us.
 
@DanielSank Halp pliz?
 
lol, why did you buy an unknown item at the asia store?
 
8:15 PM
Because that's fun
 
@Danu That path leads to bean curd, bean curd leads to scrunchy-face, and scrunchy-face leads to school girls laughing at you.
 
Lololol
 
...what is a scrunchy face and why does tofu lead to it?
 
@ACuriousMind Not just tofu, but also the nasty brown bean paste that is put on pastries as a trap for unwary gaijin who facy chocolate.
Not that I have any experience with these things, of course. None. At all.
 
8:27 PM
Tofu is not so weird
the bean paste is funny
 
I'm actually OK with firm tofu, if it has been given some flavor and cut into pieces small enough that I don't have to experience the texture. But the soft stuff is another matter entirely.
 
@Danu Yes?
Oh.
Uh, I have no idea what that is.
 
It's super weird
When eating it, the texture is really like a soft plastic for a little bit
 
"Thank you for reviewing 20 first posts today; come back in 3 hours to continue reviewing." O_o there is a cap?
 
Yup.
 
8:41 PM
wow :O I'm surprised
 
@DavidZ If you write the Dyson series (time dependent perturbation theory) you see that any quantum process where the system goes from one state to another involves a sum over intermediate states.
Those intermediate states can do all kinds of crap, like not conserve energy/momentum.
Those are virtual states, and you need nothing beyond the Schrodinger equation to see them.
@Danu gross
 
@DanielSank Yeah, but really intriguing. I kinda liked it, after a while.
 
Have any of the regulars in here taken, TA'ed or taught a gen ed level astronomy class? If so what labs (observations or practical) did you do?
3
 
Sigh...I can't stand people who read too much sci-fi
@SirCumference No one can say when it will render the Earth uninhabitable. Even your answer, which gives 1 billion years as a guess, whilst a good answer, is wild speculation, at best. 1 billion years is a hell of a long time in terms of human technological development. I doubt something as trivial as a 47 degrees and water at the poles would be an issue for a human race that has survived and advanced another billion years. We've only been around for about 0.02% of a billion years, and already we can live in extremely harsh conditions such as Antarctica, or the ISS. — JBentley 41 mins ago
"Something as trivial as 47 degrees"
Jesus christ do you know what that will do to the planet?
 
Do you?
 
8:46 PM
Of course not
Just six degrees wiped out 95% of species about 250m years ago
God knows what kind of inferno the Earth will be
 
They didn't have engineers 250m years ago
 
@0celo7 Just increasing four degrees, ice sheets will vanish from both poles, rainforests will burn up and turn to desert, the dry and lifeless Alps will resemble the High Atlas, and rising seas will scour deep into continental interiors.
Volcanic eruptions would be much more prominent and powerful, to the point where they would put nuclear weapons to shame
 
user54412
^ uhhhhhhh
 
@ChrisWhite Er, yeah?
 
user54412
I'm really hoping that was sarcasm
 
8:56 PM
@ChrisWhite Ya ever heard of the Yellowstone supervolcano?
 
user54412
apparently I had too much hope
 
Just look up what will happen if it erupts
 
user54412
you need to stop visiting conspiracy pseudoscience sites
 
user54412
(1) that volcanic region is dead
 
user54412
(2) there's no reason to expect all the volcanic energy to be released at one time -- in fact this would require miraculous uniformity in the structure of the crust
 
8:59 PM
Dude
 
user54412
(3) volcanoes are in no way affected by climate. no. way.
 
The Yellowstone Caldera is the volcanic caldera and supervolcano located in Yellowstone National Park in the United States, sometimes referred to as the Yellowstone Supervolcano. The caldera and most of the park are located in the northwest corner of Wyoming. The major features of the caldera measure about 34 by 45 miles (55 by 72 km). The caldera formed during the last of three supereruptions over the past 2.1 million years: the Huckleberry Ridge eruption 2.1 million years ago (which created the Island Park Caldera and the Huckleberry Ridge Tuff), the Mesa Falls eruption 1.3 million years ago...
This thing is in the category "Active volcanoes"
 
user54412
(4) all volcanoes of any note are already more powerful than nuclear explosions
 

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