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16:00
I'm not even sure any spacetime could
It would at least need to be non-compact, so the entire universe would be the computer
the spacetime seams reasonable in a standard kerr metric, but it doesnt have finite dimensions it has one infinite dimension time so Bekenstein bound isnt violated
they make reference to the kerr metric, but they also note that it ignores black hole evaporation
sooo I just have to shrug my shoulders
the actual paper decides to use a black hole large enough that the evaporation is negliglible
but it still ignores it and maybe for infinitly long which may be problematic
Well yes, since evaporation time is always finite
Sorry there's no magical computer
the maybe is i'm not sure about the perspective of the black hole evaporation. for example we expierience a finite time while the computer expieriences an infinite time. im not sure what time the evaporation expieriences
16:07
The infalling observer wouldn't have access to infinite memory though
Hello
Is the schrodinger equation and the making of a measurement in QM inconsistent?
To access arbitrary amount of memory you need to get out of whatever region you're in to access it
Which takes time
stretching the boundaries a bit. What exactly does a nonlocal signalling interaction look like in a spacetime diagram?
16:09
at least penrose's speculations are in the direction of 'perhaps the brain is influenced by quantum mechanics' rather than 'perhaps quantum mechanics is influenced by the brain'
Depends on what kind of signaling it is
Either spacelike curves or just discontinuous events
@Semiclassical yeah the latter is quantum mysticism a well known pseudoscience that does not even make sense when you actually analyse in detail
positing that quantum mechanics is influenced by consciousness is something i hate hate hate
i take the principal as "we are not special clumps of matter"
that quantum mechanics allows nonlocal correlations does not imply it allows nonlocal signaling, much less telekinesis
16:11
@Slereah I am thinking about two spacelike events and some nonlocal signalling occurs
@Semiclassical indeed
I have telekinesis
But it is only magic
Not QM
you are influencing my mind from afar...
you have technotelekenisis. most commonly referred to as telephone but lately means internet as well
For spacelike curves, it probably reduces to (classical) tachyons and the situation is well explored in the literature. But for discontinuous events, I am not sure if the two spacelike events will share the same inertial frame due to the nonlocal signal
your all fine and dandy with imaginary mass but i bring up negative mass and everyone freaks out
depends what you mean by negative mass
16:15
imaginary mass are tachyonic, which is more well explored despite still does not known to exist
in the sense of effective mass, I'm perfectly happy to talk about negative mass
we do so all the time when we talk about hole states in solid state
and in QFT tachyons are non signalling, so you cannot do causality violation with it, which removes any media wow factor
on the other hand, negative mass as a property of fundamental particles? nope, not interested
hence no freakout
are you happy to talk about this $m=cos(t)/c^2$ and $v=csin(t)$ in special relativity? ive been trying to figure what the domain of possble t is
16:16
no
i don't see why that'd be useful.
I don't think rest mass of individual types of particles can change
@Secret ehhh, i think there's at least one subtlety there coming from renormalization in particle physics
Otherwise, e.g. muons tauons and electron will be one and the same
but that really comes to is the origin of mass in QFT
boo...
16:19
the use is i'm trying to take a photon and alter it to give it rest mass in some way. i have good but long worded reasons to belive this possible. but only for a group of photons
@CooperCape it's a ghost!
group of entangled photons
Oh no
The ghost of h bar present.
dunno what that entails
Why will entangled photons have rest mass if individual photons don't have rest mass to start with?
16:20
we all gain a few pounds be fair...
A kinetic ghost
interestingly, there is such a concept as 'thermal photon mass': journals.aps.org/prd/abstract/10.1103/PhysRevD.28.908
but that's again an effective mass, not a fundamental one
to put a huge paper shortly, information can reverse entropy and do work. therefore information represents an energy. this energy is inherit in a entangled system
The ghost action?
if the photons are on timelike curves (expieriences no time, may have used the wrong word)then we could not extract or add that information which is not observed
16:24
$KE = \frac{1}{2}mv\ \mathfrak{scared}$
puns for life!
Shai: there is entanglement entropy, and mutual information can indeed do work as a paper from last year about cooled electrons showed, but I think something is missing in the above statement of yours
can you link me that paper by the way?
the reason i'm doing all this is i'm doing active research into an information engine and its consequences for the minimal volume of a computer with nqubits
at a certain temperature that is
lemme find a more representative paper
i literally just googled 'thermal photon mass'
Okay on a more serious note does anyone know the role CY-manifolds play in string theory? These pop sci books are leaving me confuzzled. Are they just where the strings are housed so to speak? Iirc from my book it said something about strings wrapping round them... all very confusing...
the reason for temperature comes from szilards engine and landaurs principle by the way
or more originally maxwells demon
16:30
what even is a calabi yau manifold
the more temperature the more energy each bit is worth
hrm, not finding a good example right now
for reference:
In algebraic geometry, a Calabi–Yau manifold, also known as a Calabi–Yau space, is a particular type of manifold which has properties, such as Ricci flatness, yielding applications in theoretical physics. Particularly in superstring theory, the extra dimensions of spacetime are sometimes conjectured to take the form of a 6-dimensional Calabi–Yau manifold, which led to the idea of mirror symmetry. Calabi–Yau manifolds are complex manifolds that are generalizations of K3 surfaces in any number of complex dimensions (i.e. any even number of real dimensions). They were originally defined as compact...
@BalarkaSen a Ricci flat Kahler manifold
why should i care
16:33
so the bottom line seems to be "a really nice manifold"
@BalarkaSen Unless you're into string theory, I don't think you should care about complex manifolds at all.
riemann surfaces are neat, though
@Semiclassical Yeah, but only because complex analysis is useful in differential equations
yes thanks @secret. from the abstract it seems it will make a good read and refrence
16:34
heh
it's part and parcel, really
it's silly to say complex manifolds are interesting only because of string theory
it's probably the stupidest statement i have heard in weeks
sometimes it's useful to approach from the differential equations side of things, sometimes it's useful to approach from the contour integral side of tings
It's a common sentiment among analysts @BalarkaSen
(and sometimes it's best to approach them from the Riemann-Hilbert perspective, but that requires more knowledge than I've got)
@BalarkaSen you must not talk to many people
16:36
@0ßelö7 trash
@Semiclassical Riemann-Hilbert?
I'll take that as a nope then... Have to ask the big JR then
I think it is fair to say that the main reason CY manifolds have gotten so much press is the connection to string theory
@BalarkaSen Wow you're getting butthurt
that doesn't mean people don't care about CY manifolds outside of string theory, but it's not 'fashionable' in the same way
16:37
no lol ur being silly
@Semiclassical 0celo7's statement has nothing to do with calabi yau manifolds
@0ßelö7 basically, you describe a complex function in terms of how it jumps across certain boundaries in the complex plane
e.g. you could describe the principal branch of the log function as being analytic everywere except for the negative real axis, and it has an additive jump of 2pi*i across that cut
@BalarkaSen let me clarify
if you're doing scalar functions of a complex variable, that's not too strange
No one varies about CY folds outside of some larger ST context. I don't care about complex manifolds because it's all ST or algebra.
but you can also consider a matrix function of M(z)
16:39
@0ßelö7 That is much better
with it being analytic everywere except on a set of boundaries, and jump information $M_1(z)=A(z)M_2(z)$ along each boundary
there's a nice survey article in the Mathematics Intelligencer by Its, if memory serves
Do you mean Tits
Hey, if anyone could answer this simple question I'd appreciate it
in Logic, 17 hours ago, by Phase
If we have some hermitian matrix $A$, and call $\mathcal A$ the set of eigenvectors of $A$ found through the standard $\Delta (Av - \lambda I v) = 0$ characteristic equation, you just make a matrix out of the eigenvectors and sandwich A between it, and it's inverse to diagonalise it and make a new representation $A'$ in it's eigenbasis.

Now if $A$ and $B$ commute, it's pretty easy to diagonalise them both, it's just as trivial, but what if $B$ (and / or A) is degenerate? How does one go about formally approaching this? Does one just compute the eigenvectors of both to see which is less deg
oh yeah rip it doesn't paste the full question
oh, it was in "Notices of the AMS"
not sure why I was thinking it was MI
16:43
"...enerate / non degenerate? And once one has found that, if it's say, threefold degeneracy, do you just pick three eigenbasis vectors for that? The same amount of choice as a single degenerate operator?"
What do you mean by degenerate?
same eigenvalue @balarka
Multiple eigenvalues?
Eigenvalue
@Semiclassical Okie-dokie.
16:44
O and today there are maxwell zombies
Maxwell Zombies (MZ) are a class of thermodynamic paradoxes closely related to the famous 19th-century thought experiment, Maxwell's Demon (MD), which tests the foundations of the second law of thermodynamics. MZs differ from MDs insofar as they are not prone to the same physical shortcomings that ultimately foil the demons. The original Maxwell's demon is an imaginary creature that sorts gas molecules on an individual basis, creating a thermodynamic disequilibrium (either in pressure or temperature), whereby work can be performed on a perpetual basis, thereby subverting the second law. MDs fail...
You can't diagonalize if you don't have distinct eigenvalues, in general.
What?
@BalarkaSen he's doing Qm
I thought all Hermitians were diagonalisable by the spectral theorem
Presumably everything is hermitian
16:45
the key phrase there is 'hermitian matrix'
I wrote hermitian though
Oh, you didn't say that
Balarka doesn't know QM
he did, actually -_-
at the beginning though I wrote hermitian
16:45
Ah, you did, I just can't read.
OK, I agree.
I guess I forgot to specify that B was hermitian but I may have wrongly assumed that was implicit
ah ok fair enough
non-hermitian matrices are worse
16:46
Yeah I was mostly reading the last paragraph.
do you know QM, @0ßelö7
you can get some interesting stuff if you assume $A^\dagger = J A$ where $J^2=I$, though
I have a shiny transcript that says I do
fun fact: i'm typing this while sitting in on the QM lecture I'm TAing for
Ah yes, the teacher's ass
16:47
frown
Tool Assiting?
(What's TAing?)
like as a non-joke
or is it $A^\dagger = J A J$? I forget
teaching assistant
@CooperCape 0.o
Oh right
16:49
I guess I also want $J^\dagger=J$
$v^\dagger A v=\lambda v^\dagger v\implies v A^\dagger v = v^\dagger J A J v=(J v)^\dagger A (Jv)=\lambda^* v^\dagger v$
Any.. takers on my question? D:
I know I'm remembering something but I don't remember what it is :(
I don't remember how simultaneous diagonalization works so I'd have to pass, @Phase.
Glad to be of no help
@0ßelö7 pls
Why me?
I don't see a doubt
16:51
Because despite how mean you are you're also the one who's helped answer most of my other stupid questions
Mean?
@0ßelö7 the Tyrant
Lack of Rigour carries a death penalty
[reports]
I swear @0ßelö7 is Mr. Notation though
wtf does that mean
@Phase you need to present evidence for your claim
For the second time in two days...
Sep 3 at 13:20, by 0ßelö7
Jesus, please use correct notation
Like a month ago though lel
I'm still scarred
Hope you know that.
16:54
you need to see a doctor if that scarred you
@0ßelö7 I'm pretty sure I can envision a SAW like movie, where you capture victims who have lacked mathematical rigour and put them in torturous lose/lose challenges for their lives
@ACuriousMind Is comparing me to a vicious serial killer "Nice"?
Jigsaw wasn't a vicious serial killer... He was a teacher
Just wanted people to learn how much they value their life
And you want people to value notation
it fits.
You clearly have not seen the movie @0ßelö7
@BalarkaSen I don't like horror movies.
16:56
Saw is a great series
excpet V
and VII
VI was good
I will get nightmares if I watch it.
sorta decline past III though
Right now I only have deadline nightmares
can someone pls help me :(
I have no idea what your doubt is
16:57
What do you mean
the only worthwhile horror franchise is the hellraiser series
It's not a doubt, I have a question about how to do it
I'm asking, if $A$ and $B$ are both hermitian, and one [or both] are degenerate, what does one do? Compute eigenvecotrs of both by hand to see which is less degenerate, then choose the eigenbasis for the degenerate one / parts?
And how much freedom for choice is there with choosing the degenerate eigenbasis, the same as if just diagonalising one degenerate hermitian?
what does one do for what?
presumably run away because no one likes linear algebra

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