The h Bar

fqq

01:22

@imbAF it can't be correct, what's $p_x$?

Slereah

10:13

Apparently there is a nice halfway point between having a purely geometric connection and having a very general Lagrangian sort of connection

Where you take a physical theory that has the structure of a Randers manifold

Since that's essentially the action of a charged point particle

Qmechanic

11:27

Should the bounty here be removed and post closed for being homework-like?

Slereah

15:17

I think the Polyakov action is written wrong

It should be $$S = \int e^{-1} \langle e\dot{x}, e\dot{x} \rangle dt$$

Doesn't matter too much since $e$ and $e^{-1}$ can be related, but still

Or maybe not idk

Wikipedia says that the volume measure on a manifold should be a density of weight 1 and those are given by $e \to \det(e)^{-1}$

Or is the $-1$ for the transformation of the frame and not the frame itself

Manas Dogra

17:15

Where can I find details about the chordal distance of AdS? It appears in equation (2) of "Local bulk operators in AdS CFT_A boundary view of horizons and locality by HKLL"

I learned that the chordal distance is invariant under AdS isometries...is this a special feature of AdS space or is it valid for some other special spaces also?

ACuriousMind

17:28

@ManasDogra "chordal distance" is a general concept for embedded manifolds, not exactly special to AdS - it's just the length of a straight line in the ambient space that joins two points

I don't think it comes up all that often

Feynman_00

ACM, what's your take on eagles?

ACuriousMind

overhyped (stop copying Tolkien and make some other giant birds, lazy writers!), but otherwise a decent bird

user4539917

and the band?

Feynman_00

You don't like my favorite birds

Eagles and owls

Manas Dogra

17:54

@ACuriousMind Is this distance generally invariant under the isometries of the embedded manifold?

Slereah

ACuriousMind

@Feynman_00 I just said eagles are decent birds!

that's not not liking them :P

@ManasDogra I don't know

Slereah

Is there an equation of deviation in GR that's not for geodesics

If I have two curves, with the equations $\nabla_{u_{i}} u_i = A_i$, that have the same initial conditions

What is their deviation like?

if necessary I can make them a family of geodesics parametrized by some parameter

Such that the acceleration depends on it

Presumably it would be something like $$\nabla_{u_s} \partial_s u_s = A_s$$

Wait no, $$\nabla_{u_s} \partial_s u_s = \partial_s A_s $$

cbpfindex.cbpf.br/publication_pdfs/…

I think this is related

Feynman_00

18:41

@ACuriousMind Decent is not majestic :(

imbAF

@fqq You are correct, it's not right, because you cannot remove p_x outside of the integral, since we are also integrating in relation to p_x, but I still don't know how to solve it, keeping the same expression, that one uses for any other operator

Feynman_00

I think I'll go nuts these day: one of my courses uses Lorentz-Heaviside units, another uses Gaussian units and then natural units sometimes pop up

I mean, I can barely convert between Gaussian and SI, that's too much for my poor mind

Sir Cumference

19:18

@Slereah here i am listening to the band as i read this

Avantgarde

19:28

Hey physicists

imbAF

19:49

Can anyone explain to me, how equation 04, proves hermicity of the momentum operator? physics.stackexchange.com/questions/280178/…

ACuriousMind

@imbAF the answer does not claim that eq. (04) "proves Hermiticity"

it's just part of the (lengthy) argument there at the end of which you know it's Hermitian

imbAF

"This is achieved" is meant for what?

ACuriousMind

The passage you're looking at is the introduction to the argument that follows

what do you think the rest of that answer is for if Hermiticity followed directly from eq. (04)?

imbAF

If 04 is an initial step to prove hermicity

ACuriousMind

a less ambiguous formulation would perhaps be "This will be achieved" - the author is just saying they're gonna use that equation

imbAF

19:55

then where does 04 comes from? Cuz I am pretty sure, you find out the result of the comutator between momentum and position, once you know how the momentum ie expressed in position space, something that we want to find, or idk, it's known in this case?

ACuriousMind

usually canonical quantization postulates that QM commutators are the same as the classical Poisson brackets

imbAF

the classical Poisson brackets between momentum and position, contains $\hbar$ and the complex i ?

ACuriousMind

no, of course not

see en.wikipedia.org/wiki/…

you have to get used to people using technically incorrect language :P

Feynman_00

Oh, quantization :)

What terminology should I use to avoid "first" and "second" quantization?

ACuriousMind

just say "quantization"

Feynman_00

20:01

Oh, it is that simple

ACuriousMind

in field theory terms "second quantization" is just "first quantization" applied to a field theory, really

Feynman_00

Yes, in fact that word has always been confusing to me

ACuriousMind

unless you're talking about specific steps when dealing with many-body systems, I see little use in distinguishing the two notions

it's just historical baggage

Feynman_00

I'm talking about introducing the field operators in non relativistic theory

imbAF

@ACuriousMind I made a threat about how to find the momentum operator component, in position space, using $\langle \vec r|$ and $|\vec r'\rangle$ . But someone said, and rightfully so, you cant find it that way, because you can't move p_x out of the integral. So I guess consistency is thrown out of the windon in this case. Considering how the basis switch happens for a component of an arbitrary operator

Feynman_00

20:03

Oct 8 at 9:27, by Feynman_00

ACuriousMind

I don't know what you're talking about but please don't make threats ;P

imbAF

We talked about it yesterday

ACuriousMind

yes, and I already linked you to at least one question where the correct derivation of the momentum matrix elements is discussed

Feynman_00

Only threats by fight birds owners are accepted here

imbAF

Yes

but the solution of the link you gave me, still used to typical

$\hat P \psi(x)$

how is that similar to using $\langle \vec r|$ and $|\vec r'\rangle$ ?

Feynman_00

20:08

@imbAF $\hat{P}\psi(x):=\langle x\lvert\hat{P}\lvert\psi\rangle$

ACuriousMind

I have no idea what you're asking

Feynman_00

For some reasons my message looks broken

Oh the escape

Oh, regarding the *classical Schrödinger field we've recently talked about it seems I can't find any place where it is discussed, only KG or Dirac field

ACuriousMind

probably because it doesn't describe any useful system

Feynman_00

So I wondered since the Heisenberg equations look exactly like the Schrödinger equation, does it mean that the field operators above are the quantized version of the classical field?

I mean, I could consider it that way, right?

imbAF

Eh, I already tried to explain the obvious difference between This $\hat{P}\psi(x):=\langle x\lvert\hat{P}\lvert\psi\rangle$, which gives me an expression for the momentum operator in position space and this: $\langle \vec r|P_x|\vec r'\rangle$ which, should,as you said that it has as a result, a component. And I want this last one, which is not what the link you gave me has,

since it utilizes this :$\hat{P}\psi(x):=\langle x\lvert\hat{P}\lvert\psi\rangle$ and not this: $\langle \vec r|P_x|\vec r'\rangle$. But it's ok.

ACuriousMind

20:14

@imbAF The link I gave very obviously derives an expression for $\langle x\vert p\vert x\rangle$ which except for using different symbols is exactly what you want. I don't know what your problem with that is.

Feynman_00

Maybe their problem is that a component of momentum is acting on $\lvert \vec{r}\rangle$ and not on $\lvert x\rangle$

In that case, nothing changes

I mean that $\langle\vec{r}\lvert\hat{P}_x\lvert\vec{r}'\rangle=-i\hbar\frac{\partial}{\partial x}\delta^3(\vec{r}-\vec{r}')=-i\hbar\left[\frac{\partial}{\partial x}\delta(x-x')\right]\delta(y-y')\delta(z-z')$

imbAF

I mean, how do you make the first jump

and how does it make sense to use the position representation of the momentum operator, in order to find the expression for a matrix element of it, in the position space.

Feynman_00

I used the formula I wrote above for $\lvert \psi\rangle=\lvert \vec{r}'\rangle$

$P_x$ only acts on the x part

Asklepian

The universe is flat at the largest scales, correct?

ACuriousMind

20:31

@Asklepian no, due to the non-zero cosmological constant it's a deSitter space

Avantgarde

de Sitter :)

Asklepian

because I'm (trying) to learn about spacetime curvature here: youtube.com/watch?v=0e3h9w9ldaA

and I'm confused about the relationship between spacetime curvature d/t gravity and the overall curvature of the universe

one is local, the other is global

Avantgarde

How are they different?

ACuriousMind

maybe instead of "gravity" you should say "mass-energy that isn't dark energy" :P

Asklepian

is the local/global difference not the case?

or are we saying that the overall curvature arises from the distribution of "mass-energy that isn't dark energy"?

ACuriousMind

20:42

I'm not sure what exactly you mean by "overall curvature"

Asklepian

sorry

curvature of the universe

ACuriousMind

curvature (whether scalar or tensor) is always a function of position

now, if there was only the cosmological constant (=dark energy) in the universe, then it would be a de Sitter space with constant positive scalar curvature

Asklepian

but what's the position of the entire universe?

ACuriousMind

@Asklepian that phrase is meaningless

Asklepian

if curvature is always a function of position, how can you refer to the curvature of the universe without defining a position?

ACuriousMind

20:45

as I just said - pure cosmological constants lead to constant scalar curvatures

Asklepian

oh it's scalar

Avantgarde

The curvature of the universe is sourced by every matter/energy present in it. This curvature is a local quantity, that is, it is time and position-dependent. de Sitter is the solution our universe asymptotes to at late times.

ACuriousMind

...the crucial word there is constant, not scalar

scalar just means it's a single number at each position, not a vector or higher tensor

Asklepian

this is definitely breaking my brain... I understand the example of holonomy via parallel transport and how that can be used to measure curvature. I also understand that curvature is a local quantity (i.e. more curvature near a star). However, you then refer to the "curvature of the universe" which I am failing to understand as being "local" in any way...

ACuriousMind

it's really just the constant part of the curvature that comes from the cosmological constant/dark energy

the curvature function for that is constant, i.e. assigns to every point the same value

so you don't need to specify a position because it's the same everywhere

Asklepian

20:59

ok

so when we refer to curvature there are really two components: 1.) local, which arises from mass-energy densities and 2.) global, which arises from a "background" cosmological constant

?

ACuriousMind

I wouldn't use local/global because we use those words already for too many different things, but yes

but the point about "dark energy" is that we do think it's a mass-energy density, too!

so this isn't two different mechanisms

Asklepian

how could a cosmological constant be a mass-energy density?

ACuriousMind

it's just a constant energy density throughout space(time)

like "dark matter", we call it dark energy because we really don't know anything more about it

Asklepian

and how do we know that it's there?

which measurement?

ACuriousMind

well, the universe expands

Asklepian

21:05

this is hubble's law?

ACuriousMind

and since we know Einstein's equations work in most other circumstances, we have to add a constant energy density to the stress-energy tensor to produce the cosmological constant through them

Avantgarde

Accelerating expansion of the universe

Observations show that the expansion of the universe is accelerating, such that the velocity at which a distant galaxy recedes from the observer is continuously increasing with time. The accelerated expansion of the universe was discovered during 1998 by two independent projects, the Supernova Cosmology Project and the High-Z Supernova Search Team, which both used distant type Ia supernovae to measure the acceleration. The idea was that as type Ia supernovae have almost the same intrinsic brightness (a standard candle), and since objects that are further away appear dimmer, we can use the observed...

Asklepian

so, really, when someone is referring to the "curvature of the universe", they are simply referring to the cosmological constant

not including local mass-energy densities as well?

ACuriousMind

very likely, yes

Asklepian

even though, technically, the "curvature of the universe" would include both, correct?

ACuriousMind

21:09

I mean...physical models almost always simplify, i.e. ignore effects that are negligible for whatever specific thing you're doing

Asklepian

so ALL the mass-energy densities in the universe are negligible?!

ACuriousMind

in the context where you're interested in the really large-scale behavior, sure

Avantgarde

Certainly not. You're on earth, right?

Asklepian

seriously?

Avantgarde

It depends on what problem you're talking about.

ACuriousMind

21:10

68% of the total mass-energy content of the universe is dark energy

Asklepian

I mean how negligible?

ACuriousMind

only 5% are observable mass-energy

so yes, when it comes to the really large scale, that doesn't matter all that much

Asklepian

wow

ok, I get it

it's a scaling problem

curvature at large scales is dominated by dark energy

curvature at small scales is dominated by gravity

or not?

ACuriousMind

(the scale here is one where a single galaxy is just a miniscule dot)

Asklepian

right but that's really what's changing is the scale of what's considered "local" in order to define the curvature

ACuriousMind

21:13

@Asklepian if by "gravity" you mean "ordinary matter" then...sometimes yes, sometimes no (there's still dark matter, after all, depends how small your scale is and what you're looking at)

Asklepian

so, at smaller scales still, electromagnetism dominates, correct?

ACuriousMind

what scale do you have in mind?

and electromagnetism isn't like gravity - it doesn't act on all mass-energies, there's no scale at which a bunch of uncharged objects will start interacting electromagnetically

Asklepian

ok nevermind that idea then

so, how does a cosmological constant lead to an accelerated expansion?

is it the constant addition of energy everywhere?

ACuriousMind

I'm not sure there's a good explanation for that other than "that's what you get when you plug a constant energy density into the Einstein equations"

I mean, I can wave my hands that an energy density should be associated with a "pressure" and that pressure is what expands space but that doesn't even convince myself :P

Asklepian

that's unusual as well, because my assumption of the effect of an energy density would be as a negative pressure that contracts space, causing curvature (like a star)

but perhaps these are two sides of the same coin?

ACuriousMind

21:25

you need to be careful - just because in the Newtonian limit masses attract objects, that doesn't mean they "contract space"

Asklepian

well they cause 4D geodesics that look contracted compared to nearby empty space

ACuriousMind

"space expansion/contraction" refers to a factor in front of the spatial part of the metric that depends on time

i.e. with time distances become larger or smaller

that's not how gravitational attraction works - there is no time-dependence in the Schwarzschild metric

Asklepian

oh now you're getting weird

ACuriousMind

nature is weird, I'm afraid :P

Asklepian

EVERY animation I've ever seen of gravity curving spacetime looks like 'spacetime contracts' near massive objects

ACuriousMind

21:28

@Asklepian but not temporally

Asklepian

please explain this

and be gentle

totally confused

ACuriousMind

you're playing a word game here where you use "expand/contract" in two different meanings: 1. In the context of universe expansion/contraction, it's that distances change over time. 2. In the context of gravity, you're talking about distances changing when you get closer to the mass, but this difference stays the same forever (unless the mass moves, gets bigger, whatever)

i.e. in the first case, there is a difference between distances at different times, in the second case, there is a difference between distances at different positions

these two phenomena are not at all the same, and so we shouldn't use the same words to describe them

(I don't mean the "word game" as an accusation: I just mean that describing physical facts in natural language is a game where we have to try to pick the least misleading option)

Asklepian

I understand you

you described that very well

so, the overall curvature is a result of how distances change between different positions and times and that's why both effects are included?

ACuriousMind

that's almost the whole truth ;)

there's also a "rotational" aspect to curvature that doesn't mean the distances as such get longer or shorter, cf. frame dragging

Asklepian

that's what frame dragging is

ok, so overall curvature really takes 3 inputs: 1.) how distances change between different positions 2.) how distances change between different times and 3.) something something rotational curvature

ACuriousMind

21:39

yes

Asklepian

you're an excellent teacher

so, now, where does deSitter space fit into this?

ACuriousMind

(mathematically, this is most directly connected to the vielbein formalism, where you express the metric not via a tensor but via a field that assigns to every point an orthonormal basis, the basis vectors of a coordinate system: the distances are represented by the vectors' length, the rotation by the coordinate system rotating)

@Asklepian "de Sitter space" is what we call a universe has that has just a positive cosmological constant and nothing else

Asklepian

so, an anti-deSitter space has a negative cosmological constant?

ACuriousMind

yes

and flat space has zero c.c.

Asklepian

if a positive cosmological constant corresponds to an expanding universe, would an anti-deSitter space correspond to a contracting one?

ACuriousMind

21:48

uh

Asklepian

so, how is an anti-deSitter space not just a time-reversed deSitter space?

ACuriousMind

actually, I think anti-de Sitter doesn't "contract"

you're right that a contracting universe would just be de Sitter, but time running the other direction

Asklepian

so then the positive c.c. isn't related to expansion?

ACuriousMind

anti de Sitter is much weirder because it's closed, i.e. the spatial extent is finite

Asklepian

it's hyperbolic, if I remember

ACuriousMind

21:53

@Asklepian well...it is, but there's also the "mystery" of why time runs in the direction where we perceive the space as expanding and not contracting

it's the "cosmological arrow of time"

Asklepian

I think that strominger and cotler paper gave a good reason for that one

ACuriousMind

what decides the direction of time is a whole other can of worms

Asklepian

you can encode a small history within a larger future but you cannot encode a larger history within a smaller future

ACuriousMind

yeah, there's all sorts of arguments about that (note that arguments about "encoding information" require quantum arguments - in classical GR you can evolve the universe both "forwards" and "backwards" from any spatial slice)

it's one of these questions where you can find plenty of people who consider them settled but they won't agree about what settles them ;)

Asklepian

So, here's another paper that I have a question about: arxiv.org/abs/1310.4691

In 4D GR, the universe is static

Is GR from the perspective of an unentangled observer?

ACuriousMind

22:01

@Asklepian what do you mean by "static"?

static has a technical meaning in GR that applies to de Sitter but not to "the universe" in general in GR

Asklepian

not that kind of static

ACuriousMind

I suspect you rather mean the block universe view

Asklepian

yes, that's what I mean

or, more precisely, does gravity approximate GR in the unentangled limit?

ACuriousMind

that's really philosophy, not physics - you could also interpret classical mechanics (pre-GR) that way because, after all, the initial conditions at one point in time determine the state at all other times

i.e. it's definitely not the case that GR forces you to take the block universe interpretation - GR has, like classical mechanics, a formulation as an initial value problem (cf. Cauchy surface) with unique solutions (for most/nice spacetimes)

Asklepian

well, in the sense that time does not arise from the dynamics in GR, GR is a classical theory in the same deterministic sense

since we know that in QG time must arise from the dynamics, my question is really about whether GR is a good approximation of gravity because this is what an unentangled observer would see

In that paper, they show that time arises from entanglement

ACuriousMind

22:12

I would not say that we know that "time must arise from the dynamics"

Asklepian

ok, fair enough

ACuriousMind

this idea that the arrow of time is quantum informational is certainly one view people take, but it is far from universally accepted

Asklepian

but still, what's the difference between an observer that is entangled with the universe they are observing and an observer that isn't? How do their perspectives of that universe differ?

ACuriousMind

I feel like this would require us discussing the measurement problem first

because you seem to assume a lot about observers and entanglement that presuppose a particular quantum interpretation

Asklepian

this is my understanding, edit where you like:

entanglement arises from interaction. as interactions accumulate, entanglement spreads until the entire system is described by a single wave function. these entanglement events result in the energy equipartitioning that explains the statistical mechanical arrow of time.

ACuriousMind

22:25

that's decoherence, except that " described by a single wavefunction" is not the result of decoherence

Asklepian

what is it the result of?

ACuriousMind

what you get from decoherence is that after interaction with the environment, it is plausible that there are several possible states and their probabilities acts "classically"

the selection of one of these states as the "realized one" still requires a measurement of the system, and the question of a) what constitutes a "measurement" and b) how measurement induces that selection is the measurement problem

Asklepian

I see

ACuriousMind

the original inventors of decoherence theory claim that it resolves the measurement problem, but not everyone agrees

Asklepian

I think measurement is the same as interaction

I don't think it makes sense those could be separate processes

as for the induction part, I don't know enough about entanglement dynamics

ACuriousMind

22:33

I mean, there are several interpretations in which essentially every interaction is a measurement if "you"/the observer is one of the entities interacting

that's not the hard part :P

Asklepian

no, I think everything is an observer

all interaction is observation

dividing the world into observers and observed doesn't really make sense

there are only different entanglement networks bounded by some Markov blanket, which then interact and become entangled themselves

so defining an observer is merely defining an entanglement network

defining an environment is the same thing

ACuriousMind

I don't know what an "entanglement network is", but sure, in most interpretations everything can function as an observer

Asklepian

it's just my own language to describe what I imagine decoherence to be like

ACuriousMind

all of this doesn't really nail down the resolution of the measurement problem: What happens during a measurement? Why do we on one hand (the external viewpoint) describe the outcome of the measurement interaction as an entangled state (in which multiple different outcomes of the measurement are superposed) but to the observers involved it seems that their measurement yielded a definite value?

Asklepian

because they belong to the same entanglement network

superpositions are only visible because they are potentials

ACuriousMind

22:39

(this is a version of Wigner's friend)

examples of possible consistent answers include: "Every time we get such an entanglement, the "observer" too splits into as many versions of itself as the possible result of measurements and each of these versions is simultaneously realized but they cannot interact with each other" (many worlds), "there is no such thing as a "quantum state" or objective state of the universe, all that a "quantum state" is is a subjective quantification of our knowledge of a quantum system" ($\psi$-epistemic)

"I don't care, this correctly predicts what my lab instruments will measure" (operationalist)

etc.

precisely because there are so many different consistent interpretations I would be very careful to pick a particular one to try to argue about quantum gravity or the nature of time :P

Asklepian

then which perspective is mine?

ACuriousMind

you haven't yet said anything that really resolves the measurement problem (or I didn't understand it)

Asklepian

superpositions are potential futures

ACuriousMind

and what is it about interaction/measurement that eliminates all but one of these potential futures?

Asklepian

that I can't answer

ACuriousMind

22:47

then you have no answer to the measurement problem!

Asklepian

and to the best of our knowledge, this selection is random?

ACuriousMind

it's either random or implies faster-than-light communication or implies something even weirder (according to Bell's theorem)

Asklepian

ok, wait a minute

if superpositions (potential futures) are only visible while unentangled, then entanglement is the process whereby a single history becomes encoded from these potential futures

and so "observers involved" would already be entangled and only able to access the already encoded history

ACuriousMind

that's not how entanglement/decoherence works

Asklepian

how does it work?

ACuriousMind

22:57

if you measure the spin of a particle and that spin is either up or down, then when you describe the interaction of the measurement apparatus and the particle from the outside the result is "apparatus says spin up and particle is spin up" superposed with "apparatus says spin down and particle is spin down". This is an entangled state because you cannot assign a unique state of "up" or "down" to either the apparatus or the particle.

both of these are consistent histories, since the particle and the apparatus agree

but the process of interaction doesn't decide which one of the states of the apparatus is the one the poor grad student that we're forcing to repeat this experiment for the thousandth time observes when they look at the apparatus

entanglement alone doesn't solve this (again, Wigner's friend) - if you describe the grad student as a quantum system, too, you get the entangled state of "student sees spin up and apparatus says spin up and particle is spin up" superposed with "student sees spin down and apparatus says spin down and particle is spin down"

Asklepian

what it sounds like you're describing is the growth of an entanglement network (particle > apparatus > grad student)

as each new interaction occurs the network grows

if you are not entangled with this network, you must consider all potential futures

if you are entangled with this network, you will only be able to access a single already-encoded history

ACuriousMind

yes, but why

what is it that makes me perceive only a single history

Asklepian

because the other futures aren't possible anymore... at least not from within your network

you've interacted

which perspective on the measurement problem does my viewpoint correspond with?

ACuriousMind

you're still sidestepping the problem ;)

the problem is what makes the other futures "impossible"

Asklepian

23:12

they aren't

they just didn't happen

ACuriousMind

nothing about our description of measurement as interaction that leads to these entangled states explains how I find myself in the history where the particle was spin-up and not in the one where it was spin-down

Asklepian

but they could've and you have to account for their potential

ACuriousMind

it sounds to me like you want to argue for a classical collapse interpretation similar to the folklore often called Copenhagen - when I interact with the entangled system, one history is chosen at random (according to the weights of the superposition) and the others are discarded

Asklepian

actually, I'm curious about how energy plays a role in selection

a system with more free energy could be said to have more available future states

I don't know how this corresponds to the quantum world

ACuriousMind

that's not generally the case

Asklepian

23:19

a system with more free energy doesn't have more available future states?

I can drive my car to many more places with 10 liters of gas than with 5

I'm almost curious whether you could define free energy in terms of available future states?

ACuriousMind

this would require another detour into thermodynamics that I'm not really up for after 1 am :P what you say is sort of true for most thermodynamical systems but it is not a general feature of quantum mechanics

Asklepian

sorry it's only 7:30 here

where are you from?

ACuriousMind

also this notion of "future states" doesn't directly connect to the possible measurement results of measurement

@Asklepian Germany; I live in Heidelberg

Asklepian

somehow, I was already guessing Germany lol

if superpositions are different possible futures, then "future states" would connect to the possible measurement results... whether or not this feature is a part of quantum mechanics though is something you can explain another time

thanks always for your discussion

I learn a lot when we talk

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