The h Bar

Avantgarde

00:14

hellooo

user76284

01:22

What was the final consensus with regard to physics.stackexchange.com/questions/2865/…?

David Z

01:44

@user76284 oh dear that question is a horrible mess

I don't know if I've looked at it before

well, at least not since shortly after it was first posted, when it was a lot more manageable

uhoh

04:11

3

Q: Which astronaut has experienced the largest relativistic shift in time?

Following this answer and then this question (where I've linked to Cosmonaut Sergei Krikalev; the World's Most Prolific Time Traveller) I've noticed that currently Krikalev does not hold the precisely longest record in space. Relativistic effects include both velocity, and gravitation. Moving fa...

would anybody like to check the math I've just added to the question?

Dvij Mankad

Is there some nice conditions (in terms of the prepared state and the relevant operators) as to when the expectation value of the commutator of operators will vanish and when not? Since it is the expectation value of the commutator of two operators that ultimately decides the limit on the minimum value of the product of their variance, it seems to be of particular fun to look for some qualitative understanding of when that would vanish and not.

John Rennie

@DvijMankad The commutator vanishes only when the state is an eigenstate of both operators

Dvij Mankad

Yes but even if the commutator is non-zero in itself, the expectation value can certainly be zero for a lot of states of interest. For example, a spin 1/2 state prepared in an eigenstate of Sx will have zero expectation value for Sz which is essentially the commutator of Sx and Sy.

spin 1/2 system*

John Rennie

Ah, oops, sorry I missed the fact you were interested in the expectation value.

Dvij Mankad

Correct me if I am missing something.

Yeah, I was looking for some qualitative argument to help me look through the expectation value and to understand what we would ultimately measure, the variances--I mean their product

Semiclassical

04:31

So $\langle \psi | AB-BA|\psi\rangle = \langle \psi | AB|\psi\rangle - \langle \psi |AB|\psi\rangle ^*=2i \text{ Im}(\langle \psi|AB|\psi\rangle)$

so the expectation value of the commutator being zero is the same as the expected value of the product being real

Anonymous

11:18

Eh, to first order they're all identical spherical astronauts of uniform density, then. ¯\_(ツ)_/¯ — Russell Borogove 7 hours ago

Anonymous

Gotta love this

ACuriousMind

12:31

@Semiclassical If the commutator is non-zero, the product is not Hermitian, and hence not an observable. We don't usually speak of an expectation value for non-observables, do we?

user1732

> When asked what it was like to set about proving something, the mathematician likened proving a theorem to seeing the peak of a mountain and trying to climb to the top. One establishes a base camp and begins scaling the mountain's sheer face, encountering obstacles at every turn, often retracing one's steps and struggling every foot of the journey.

ACuriousMind

@DvijMankad I don't think there is anything inherently special about states where the expectation value of the commutator vanishes. Since the UP merely provides a lower bound on the variance, it must vanish on states where both observables have well-determined values (i.e. no variance), but it can also vanish on states where the variances are arbitrarily large. There doesn't seem to be anything special to me about the latter case.

user1732

> Finally when the top is reached, one stands examining the peak, taking in the view of the surrounding countryside and then noting the automobile road up the other side!

Robert J. Kleinhenz

danielunderwood

12:46

Do other universities use a course numbering system similar to MIT? It seems strange to me every time I look at the OCW site

Dvij Mankad

12:59

@ACuriousMind Okay, yes, seems alright but I was thinking that there might be something special about them because it appears to me that Landau uses UP between momentum and position to conclude that there is no consistent notion of the trajectory of an electron. Or can we conclude that simply by saying that they don't commute and can't have shared eigenstates? Or even simply by saying that no physical state would be the eigenstate of either of the operators?

bolbteppa

13:26

@DvijMankad Landau states that the UP is the claim there is no notion of a trajectory for an electron

ACuriousMind

@DvijMankad Yes, the argument is supposed to be that since the expectation value of their commutator is never zero, no state can be a simultaneous eigenstate of both position and momentum. Note that this is strictly stronger than the claim that the two operators do not commute: Non-commuting operators can't share an eigenbasis, but they can share eigenstates.

That there is something meaningful to say about the case where it is never zero does not imply that there is something meaningful to say about the generic case where it is zero.

Semiclassical

hmm. "parametrizing" or "parameterizing"

doesn't seem to be consistent online

ACuriousMind

13:42

@Semiclassical parametrize. We don't have "parameteric equations" or the "meteric system" either, after all.

2

Semiclassical

eh. parameter-izing as "express in terms of parameters"

I like the former spelling better, but there doesn't seem to be a consensus online

ACuriousMind

@Semiclassical ...and "parameter-ic" as in "pertaining to the use of parameters"?

Semiclassical

hmm

ACuriousMind

I advocate to be consistent in the method of deriving words - if you elide the 'e' in some derivatives of parameter, elide it in all!

Semiclassical

at the very least, it's not consistent in the wild. see for instance this book published by Springer: books.google.com/…

also: merriam-webster.com/dictionary/parameterize

ACuriousMind

13:47

Oh sure, you'll find both spellings used. I just don't see why anyone would actively decide to use the second one :P

Semiclassical

some discussion on the English SE:

20

Q: "Parametrise" or "parameterise" a curve?

In British English, which one is correct? Does one parameterise a curve or parametrise it?

verbs british-english orthography mathematics

ACuriousMind

'I see my argument is the top voted one :)

Semiclassical

yeah

though when it comes to linguistics I always come back to this: smbc-comics.com/comic/know-your-linguistic-philosophies

(on the other hand I couldn't care less about -ize vs. -ise)

Emilio Pisanty

yo @ACuriousMind, how's your Epic progress going?

::grin::

ACuriousMind

@EmilioPisanty Well, I already earned three-digit rep today, so there's hope :P

Emilio Pisanty

14:00

@ACuriousMind keep at it, then =D

I'll race you to the rep cap

danielunderwood

@Semiclassical isn't ize vs ise just a regional difference?

Semiclassical

yeah

hence why I don't care at all :)

Emilio Pisanty

wait, too late

danielunderwood

there's a rep cap?

Slereah

@ACuriousMind dropped Planescape avatars?

Emilio Pisanty

14:09

@danielunderwood 200 rep daily

accepts and bounties exempt

90

Q: What is the daily reputation cap and how can I hit it?

I've found some questions and those two badges "Epic" and "Legendary" regarding the "daily reputation cap". What is it and how can I hit it?

support reputation daily-reputation-limit daily-reputation-badges

danielunderwood

ahhh I thought you meant like total rep

Emilio Pisanty

oh, no, total rep is unbounded

I guess it's probably some finite-bitlength integer in the actual implementation, so if you earn too much rep too fast so that SE can't re-code, then you'd cycle back to negative rep

though in practice the maximal reputation is on the order of a million

Jon Skeet, Reading, United Kingdom

1047k 646 7724 8295

c# java .net linq generics asp.net xml

"1047k"

danielunderwood

wow

how do people manage to find time to do that much?

Emilio Pisanty

well

this is SO you're talking about

it's a different story

danielunderwood

yeah that's true. Not a ton of in-depth stuff over there

Emilio Pisanty

14:17

what?

no, not at all

Jon Skeet has a ton of amazing in-depth software-engineering answers

but I imagine the bulk of his rep comes from simpler answers

also, keep in mind that he's been around for 9 years, and that SO is rather older than PSE

danielunderwood

Yeah I guess that's true. Most of my time on SO is "I have X error"

Emilio Pisanty

yep, that's probably the bulk of that rep

with the added qualifier that those answers are highly visible to google and they continue to accrue rep over the years

danielunderwood

Yeah I continue to gain rep on SO though I'm not really active with questions or answers now

Nearly 35k answers though. That's insane

Emilio Pisanty

over 9 years, though

puts it down to about 3k answers a year, or about 10 answers a day

danielunderwood

Still close to 10 answers a day

Emilio Pisanty

14:22

still pretty insane

but hey

programmers are pretty intense at what they do

Semiclassical

I'm not at all active now on the main site (which for me is MSE), so I'm not surprised that my rep is at a trickle now

danielunderwood

He also has close to 10 contributions per day on github

Though github contributions aren't really a great metric

Although I'll admit that I find it pretty crazy that someone can take a good bit of their free time over a year or two to write a textbook

Emilio Pisanty

@danielunderwood you don't get fake internet points if you write a textbook

well, I guess that if you count Amazon reviews then you do, but who looks at those?

danielunderwood

Someone had a comment about amazon reviews in their textbook...Zee maybe?

And looking at amazon reviews on physics textbooks, it seems like they're either people who already know the subject or bought the book because they want to think they know it

I once saw a review of "Having it on the coffee table makes friends think I'm smart"

Semiclassical

logic check: Suppose I've got a spin-s particle and I want to compute the matrix element $\langle s s|e^{-i\beta S_y}|s-s\rangle$

Emilio Pisanty

14:34

why are my $\Omega$s growing extra legs?

curiosity

Hi everyone

Emilio Pisanty

@Semiclassical $\langle s s|e^{-i\beta S_y}|0\rangle$?

curiosity

and hi @EmilioPisanty

danielunderwood

uhhh

Semiclassical

$|0\rangle$?

oh, $|sm_s\rangle=|s-s\rangle$

curiosity

14:36

i have a really burning question and i don't wanna sound demanding but can someone please take a look at (physics.stackexchange.com/questions/425697/…) ?

@EmilioPisanty it is a follow up question to an answer that you gave yesterday

Emilio Pisanty

@Semiclassical oh, you meant $|s{-s}\rangle$

Semiclassical

so that's $m_s=\pm s$

Emilio Pisanty

hmmm

still not great

$|s\, {-s}\rangle$

Semiclassical

stoopid braket notation

Emilio Pisanty

@Semiclassical nah, it's just LaTeX thinking everything is a binary operator

Semiclassical

14:37

yeah, that's probably better

fair

Emilio Pisanty

I mean, compare $|-\alpha\rangle$ to $|{-\alpha}\rangle$

but anyways

Semiclassical

right

Emilio Pisanty

go on

danielunderwood

$\lvert s, -s \rangle$?

Do people not use that?

Semiclassical

no, you typically omit the comma

danielunderwood

14:39

Ahh that's what I always did when there were numbers or signs involved

Emilio Pisanty

so, you have a sin-$s$ particle and you want to calculate $⟨s|e^{i\beta S_y}|{-s}⟩$?

Semiclassical

ya

or, well: I know the answer, I'm just trying to find the simplest derivation of it

Emilio Pisanty

what's the answer?

some Wigner matrix?

Semiclassical

literally, yes

Emilio Pisanty

ugh

Semiclassical

14:41

however, I think you can get to the final answer immediately by writing $|s\rangle = |\uparrow\rangle^{\otimes 2s}$, if that notation makes sense

Emilio Pisanty

I think the simplest derivation in things like that is normally "go look in Edmonds"

Semiclassical

And then writing $S_y=\sum_{k=1}^{2s} S_y^{(k)}$, with all these spin-1/2 operators commuting

Emilio Pisanty

@Semiclassical that sounds legit

Semiclassical

if I'm following the logic right, this immediately collapses the matrix element to $\prod_{k=1}^{2s}\langle \uparrow | e^{-i \beta S_y^{(k)}}|\downarrow\rangle$

Emilio Pisanty

though it will give you an absolute mess, I should think

Semiclassical

14:44

all the spin-1/2 operators commute, so $e^{-i\beta S_y}$ should factorize

and since none of these matrix elements actually depend on $k$, we just get $\langle \uparrow| e^{-i \beta S_y^{(1)}}|\downarrow\rangle^{2s}$

Emilio Pisanty

or maybe not? each factor is a single number so you don't have a combinatoric explosion of terms

Semiclassical

right.

Emilio Pisanty

@Semiclassical you mean \right.?

Semiclassical

of course

Emilio Pisanty

@Semiclassical Also: compare $\prod_{k=1}^{2s}\langle \uparrow | e^{-i \beta S_y^{(k)}}|\downarrow\rangle$ and $\prod_{k=1}^{2s}\langle {\uparrow} | e^{-i \beta S_y^{(k)}}|{\downarrow}\rangle$

Semiclassical

14:47

nice

I wasn't liking that extra whitespace

Emilio Pisanty

it's like the minus sign above

Semiclassical

yeah

Emilio Pisanty

though I'm not sure if it thinks \downarrow is a unary or a binary operator

Semiclassical

I'm used to using \, for integrals, e.g. \int f(x)\,dx

Emilio Pisanty

it doesn't treat it as an atom, that's for sure

Semiclassical

14:48

but I hadn't realized the utility of {} in this context

Emilio Pisanty

@Semiclassical it's just to atomize things

Semiclassical

gotcha

Emilio Pisanty

like e.g. $a-b$ vs $a{-}b$

Semiclassical

yep. anyways, all that remains is to note that, since it's a spin-1/2 spin operator, we have $e^{-i\beta S_y^{(1)}}=\cos\beta-i S_y^{(1)}\sin\beta$

Emilio Pisanty

(full disclaimer: I'm not sure "atom" is the way the TeX docs treat the concept. But it's a good way to think about it.)

Semiclassical

14:52

and therefore we get $$\langle {\uparrow}|e^{-i\beta S_y^{(1)}}|{\downarrow}\rangle =\langle{\uparrow}|{\downarrow}\rangle \cos\frac{\beta}{2} - i \langle {\uparrow}|S_y^{(1)}|{\downarrow}\rangle \sin \frac{\beta}{2}=-i(i\hbar/2)\sin \frac{\beta}{2} = \frac12 \hbar \sin \frac{\beta}{2}$$

too few curly brackets, ugh

so one should have $\langle s\,{s}|e^{-i\beta S_y}|s\,{-s}\rangle = \left(\frac{\hbar}{2}\sin\frac{\beta}{2}\right)^{2s}$

oh, but I'm being a bit silly. should be $e^{-i\beta S_y/\hbar}$ if I'm not taking $\hbar=1$

ACuriousMind

@Slereah They always were from more than just Planescape :P

Semiclassical

so $\hbar$ won't actually appear in that last expression

also, should've been $\beta/2$ in the earlier formula as well

So that's my derivation. Seems legit? (stoopid algebra errors aside)

curiosity

we all know the procedure by which a set of coupled quantum oscillators can be decoupled, and the hamiltonan brought in the form $H
=\sum_i^{3M}\hbar\omega_i\left(a_i^\dagger a_i+\frac{1}{2}\right)$

the eigenstates are now interpreted as phonons in several modes with different frequencies

Emilio Pisanty

@Semiclassical looks legit to me

curiosity

but are all occupations of these phonos automatically an allowed state of the system?

Semiclassical

15:06

cool

Emilio Pisanty

if the answer is right, that is

Semiclassical

lol

Emilio Pisanty

i.e. if it coincides with the Wigner-matrix answer from the literature

@curiosity yes. why wouldn't they be?

curiosity

for example if we only have two coupled oscillators i think it can be shown that the $|00\rangle$ is symmetric with respect to particle exchange

Semiclassical

If they're eigenstates of the Hamiltonian and there's no further constraints, then they're all allowed states.

curiosity

15:07

and if the two particles were ferminon, it would not be an allowed state

Semiclassical

If you do have further constraints, of course, then that need not be true.

curiosity

@Semiclassical are you saying if the further contraint is that that particles are fermions, we cant use the phonon picture?

i mean, not all states would be allowed?

Emilio Pisanty

@curiosity we're saying that you haven't specified enough information to get a full answer.

Semiclassical

i'm saying that, if you impose that further constraint, you'll only be able to use those photon states which satisfy it

Emilio Pisanty

Generically they are all allowed. There are situations where additional restrictions can rule some sectors out. You can't say what sectors get ruled out without saying what the restrictions are.

curiosity

15:10

@EmilioPisanty say i have two fermions with the hamiltonian $$H=\frac{p_1^2+p_2^2}{2}+\frac{1}2{}kq_1^2+\frac{1}{2}kq_2^2+\frac{1}{2}\kappa(‌q_2-q_1)^2 \tag{1}$$

Emilio Pisanty

@curiosity ... and these fermions have spin?

no spin?

curiosity

1/2

and are in a spin triplett state

so the position wf has to be antisymmetric

Emilio Pisanty

@curiosity this is the kind of information that needs to be supplied up front

Semiclassical

It's like any multi-particle state. If you say nothing about them being distinguishable/bosons/fermions, then there's no reason to assume any further constraints.

Emilio Pisanty

@curiosity so there you have it.

Semiclassical

15:11

The moment you do impose such a constraint, of course, you'll be restricted to some subspace.

Emilio Pisanty

you've successfully applied an external constraint.

curiosity

so in this case, not all phonon states would be allowed?

Emilio Pisanty

that is correct.

Semiclassical

Note, though, that your earlier statement is not correct:

5 mins ago, by curiosity

for example if we only have two coupled oscillators i think it can be shown that the $|00\rangle$ is symmetric with respect to particle exchange

5 mins ago, by curiosity

and if the two particles were ferminon, it would not be an allowed state

what's required is that the total wavefunction be symmetric under exchange

curiosity

but when we are dealing with phonons in condensed matter theory, the individual particles are the atoms of the lattice, which may well be identical fermions, right?

Semiclassical

15:14

sure, but you already stated a case in which the position wf would be antisymmetric under exchange

curiosity

@Semiclassical i was in this case supposing that the wf of the spin dof is in a symmetric state

Semiclassical

in which case, the rest of the wf had better be symmetric under exchange

...wait. hang on, I think I may have mixed myself up

spin triplet = symmetric under exchange, so the position wf has to be antisymmetric under exchange. hence |00> is not allowed

bleh, that was silly of me

curiosity

that is interesting

when we are dealiong with condensed matter theory

Emilio Pisanty

Keep in mind, though, that when you write $|00⟩$ for a phonon state, that does not mean that the first factor is for the first fermion and the second factor is for the second one.

curiosity

we may have a lattice of fermions, which we want to describe by phonon states

@EmilioPisanty yes im aware of that, but $|00\rangle$ is an element of the two particle hilbertspace right?

Emilio Pisanty

15:18

the fermion exchange operator does not coincide with the phonon exchange operator

curiosity

that might be an entangled state

but we require the wavefunction to be antisymmetric under the fermion exchange op

doesnt that mean that our phonon soolution is almost useless unless we find out which occupations are allowed by that constraint?

Akash. B

15:54

@JohnRennie sir are you there ?

John Rennie

16:05

@Akash.B hi

Avantgarde

16:21

@JohnRennie Sir John Rennie :P

Semiclassical

Ser John Rennie (not not really)

Sir Cumference

Surgeon Rennie

Semiclassical

Sturgeon Rennie

ACuriousMind

Sirloin Rennie.

4

(yes, I'm hungry)

Avantgarde

His holiness John Rennie I

Semiclassical

16:24

ACuriousCarnivore

Avantgarde

Fullyclassical

Semiclassical

Sortaclassical

Avantgarde

Almostclassical

Barelyclassical

Semiclassical

Microclassical

Sir Cumference

Avacadogarden

God we're bored

Secret

16:27

Holyholyholyholyholyholyholy...

Semiclassical

@SirCumference Ser Radians

Sir Cumference

@ACuriousMind You don't say parameterize?

@Semiclassical Steradians?

Semiclassical

yes, that was the joke

Secret

semiradians

Sir Cumference

K good

ACuriousMind

16:27

@SirCumference Nope

Secret

semiclassical mind

Semiclassical

Sir Facearea

John Rennie

@SirCumference That's what you take to bulk up your muscles isn't it?

Semiclassical

@JohnRennie your trig muscle, yes

Sir Cumference

@JohnRennie Steroidians are my new favorite unit

John Rennie

16:29

Lots of pie

ACuriousMind

I think my strength is about 0.01 steroidians :P

Semiclassical

steroids help you do tri sets, steradians help you do trig sets

(that joke almost works)

Sir Cumference

@Semiclassical Brilliant

I'll be stealing that if you don't mind ;)

Semiclassical

works for me

user54826

16:44

I need points. There are so many questions I wanted to give a bounty on. But I ask too many 0 points questions, I answer nothing and when I do, nobody upvote. I need to answer more questions damn it

Sir Cumference

@user54826 You need imaginary internet points to award other people imaginary internet points? :P

user54826

no

I need imaginary internet points to get interesting questions answered

Sir Cumference

If they're interesting they'll be answered on their own. I don't think bounties do much tbh

user54826

wrong

look at that question for instance physics.stackexchange.com/questions/284444/…

Sir Cumference

That was probably on the hot network. I don't see any points awarded

user54826

16:48

the current accepted and top answer is moot, doesn't even adress the question "does that also mean that there is a minimum-possible frequency (and thus, a maximum-possible wavelength)?"

yet it has 28 points, it's over. nobody is going to try a new good answer that deals with the question asked

it's just over without a bounty IMO. I'll gather point and offer 500 of them

John Rennie

I have to say I don't see why you think that is a good enough question to be worth a bounty.

user54826

well , according to ana v, there cannot exist a photon whose wavelength is greater than the size of the universe. to me, this is wrong, because it's not like the photon has to satisfy a kind of a boundary condition as say, a photon in a microwave oven

the fact that a giant in physics (ana v) claims this whilst others on IRC have different opinions, makes me think the question is good . and i find it interesting too

John Rennie

11

Q: Making a physicist

Materials: CV step-up transformer with coil ratio 1:5 Method: Apply the CV to the primary coil and spin it. Result: A physicist is output. Who?

riddle physics

ACuriousMind

@user54826 I don't see what you want to achieve with a bounty: Josephus' answer is already correct, you aren't going to get a much better response than that to this (not very interesting :P) question. If you think Josephus' answer is so excellent it deserves to get a bounty, sure, go ahead.

John Rennie

@user54826 the question asks:

> since energy is quantized, presumably there exists some quantum of energy that is the smallest possible. Is there truly such a universally-minimum quantum of E

The answer is no. End of question.

Anonymous

17:13

@user54826 "I answer nothing and when I do, nobody upvote" --- that's sort of obvious isn't it? Not many people here care about current electricity and solid state questions. Try answering other popular topics to gain quick points :P

Anonymous

(I'm not saying that you shouldn't answer those type of question either ;))

Anonymous

We do need more solid state people (albeit it's a very very boring subject)!

Anonymous

One way to gain cheap points is to answer basic Newtonian mech questions btw

Sir Cumference

@Blue Yowch. "Your subject of interest is very very boring" :P

Anonymous

17:20

It's my subject too ;_; (well, at least one of my subjects)

Anonymous

We had this whole solid state electronics book in the first year. Never attended any of the classes ( except the initial few) and somehow managed to pass the exams by studying in the last 2 weeks :P

Secret

17:38

[Random]

Theorem X: Any magma M with division by zero is only power associative

Proof: See math chat

user54826

18:05

thanks John Rennie, Blue, Sir cumference and ACuriousone

Emilio Pisanty

3

Q: Who or what is George Rajna?

I don't know maybe this question seems off topic but anybody get the chance to look at vixra.org archive? It seems it is blown up by some manuscripts from a guy, which is called "George Rajna", and more interestingly all of them are about some ridiculously (excuse me I didn't find another appro...

online-publication

finally, someone in Stack Exchange with the courage to tackle some of the truly great questions facing humanity today

Anonymous

2-3 50+ page papers per day...doesn't sound very human to me...

Anonymous

"I don't think this is a bot. It looks more like a (somewhat misguided) real human, who may or may not be operating under a pseudonym."

Anonymous

I mean, even typing that much takes time

Anonymous

Hmm, might be simply a copy-paste compilation

Emilio Pisanty

18:15

@Blue see? it is a mystery

Anonymous

If it's just a copy-paste thing, that shouldn't be a mystery. We could check them with some academic plagiarism software :P

Emilio Pisanty

or what if it's a copy-paste thing that's also supplemented with some fraction of pure crank?

where's the crank, and what is it about?

Anonymous

A part of me agrees with knzhou that it is not worth our time looking for their real identity, but now you made me curious :P

rob

> anybody get a chance to look at the vixra.org archive?

NO

Anonymous

I do have a mental list of *****'s whom I'd be glad to meet face to face and just observe how their real persona and online persona differ. They are intriguing after all (mostly because I can't justify their actions in cost-profit terms)

ACuriousMind

18:28

@Blue If you find all human actions intriguing that can't be justified in cost-profit terms, you must live in a constant state of amazement!

Anonymous

@ACuriousMind I'm including "mental satisfaction" and "guilty pleasures" in the "profit" category, so that cuts it down a bit

Anonymous

But it is indeed a mystery why someone would post crackpotish papers at that rate, without making much attempts at spreading them (those papers seem to have 0 downloads mostly)

Anonymous

The 0 downloads might be a bit misleading tho (as most browsers allow you to view temporarily without downloading)

Anonymous

The older ones have a bit more downloads

Anonymous

vixra.org/abs/1201.0063

Anonymous

18:35

Goddamn, 1608 downloads

Anonymous

Must be the attractive title

enumaris

thicc title

Emilio Pisanty

18:51

@rob ah, c'mon, you gotta check in at least once to see what it's like

if only to be able to dismiss it correctly

enumaris

too many articles to dismiss

just summarily dismiss everything in there lol

rob

@EmilioPisanty Is every couple of years enough?

Anonymous

"Artificial Intelligence Bring Sun Power to Earth"

Emilio Pisanty

@rob I'd say that's too much

when I said once I meant once

Semiclassical

vixra is proof for me of the concept of 'negative credibility'

a random blog post may not carry any positive credibility as such, but it doesn't actively associate itself with a sea of wrong

if I look at a random one of Rajna's recent articles, it reads more like someone copy-pasted a bunch of sentences from other abstracts: vixra.org/abs/1808.0606

without much attention to whether the sentences so combined actually fit together

user54826

19:29

what on Earth

looking at the papers on vixra about condensed matter. they're like almost all from that dude

Semiclassical

well, his papers seem to be bits and pieces of actual original work from across the web, amalgamated in a haphazard way. so in that sense I guess he is an expert on condensed matter

user54826

i have a question related to onsager's famous paper

Semiclassical

which?

user54826

if i remember well, it is stated that a system reaches thermalization in a time proportional to l^2, in other words, its length squared

i guess he means the length in the direction of heat propagation

my question is, how this apply only for particular boundary conditions like dirichlet ones (maintain constant T on both ends)?

I think so. but I am not 100% sure

i imagine a big cube of metal and a small rod of metal, both having the same height

if i maintain constant the temperature at the base of both object, it's obvious (neglecting radiation) that they will thermalize at the same time

on the other hand, if I use the same heat flux through each object, this won't be true anymore, i.e. if I pass 10 W to heat up the rod and 10 W to heat up the immense cube, it's obvious the rod will thermalize faster

just want to make sure my intuition is correct

page 419 of journals.aps.org/pr/pdf/10.1103/PhysRev.37.405

Semiclassical

19:46

that sounds reasonable. with an intuitive argument like the one he's making you have to be careful to think through what assumptions are being made

user54826

yeah

Semiclassical

whereas with a more formal argument it's usually more obvious what's being assumed

user54826

wait, it should be easy to check i think

the problem is easily solved analytically by hand. the time dependent part is a decaying exponential

or increasing one, like charging a capacitor

i just need to check the argument of the exponential, I think

the factor in front of t, the time

if there's an L^2 involved then bingo we're done

Semiclassical

I wouldn't be surprised if his argument works a bit weirdly on a system with geometric constraints, e.g. thermalization on a loop would presumably proceed differently once the heat has spread out enough to know about the geometry

user54826

hmm

i tihnk his argument holds

in the case of the loop, the direction of propagation is not fix

it changes, so it's longer than what a straight rod of the same height would have

but the argument is still valid I think

Semiclassical

19:56

hmm

that seems reasonable enough

I think you could probably cook up geometries which are weird enough to make that fail

but that's more of a headache than I care for

user54826

I might try to derive it tomorrow. should come off pretty easily if separation of variables works for a finite length rod

on this, good night all!

Semiclassical

night

enumaris

whomp

just finished another all marketing meeting

mmm

user76284

20:26

So for physics.stackexchange.com/questions/2865/…, is the MTY mechanism for converting a wormhole into a time machine valid? Their mechanism is described here on page 1447.

ZeroTheHero

20:51

@Semiclassical There are algorithms that will produce reasonable synthax for an original manuscript based on providing the algorithm with “real” papers, i.e. the algorithm searches for some recurring keywords in the input papers and can then stitch something readable from those.

... however glancing at papers by your friend Rajna some of the postings aren’t even that sophisticated. More like copy-paste from some random blogs somewhere.

Semiclassical

21:19

there's at least one paragraph I found which seems to have been lifted from a Wikipedia comment, which is...weird

anyways, i was attempting a joke: he creates so much 'condensed matter' in the form of his many many stitched-together papers that he must be considered an expert of it

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