The h Bar

ACuriousMind

00:01

@enumaris should work :)

enumaris

You are on the corporate slack right? Is your name somewhere...or can you send it to me via pm?

ACuriousMind

09:25

@enumaris You can find me in the company address book as Bjoern Jueliger (they're playing it safe and aren't using the correct umlauts :P). Not on the Slack, but most other channels should work. When in doubt, send an email.

skullpatrol

nice new avatar pal

ACuriousMind

@Knight I'm not old! ::sobs old man tears:: To me, 'video games' is just the superset of both 'PC games' and 'console games'.

skullpatrol

the moustache joined to the side burns has a 70s look to it

ACuriousMind

It's very disco.

skullpatrol

John Travolta was the disco look

re: Saturday night fever

but yes, any look could be disco

@Loong thanks for sharing

Loong

09:41

sure

skullpatrol

I still haven't watched Chernobyl :(

ACuriousMind

I hear it's pretty rad.

skullpatrol

extremely

They're saying that human life on earth could have ended?

(worst case senerio)

ACuriousMind

I'm not sure who They are

skullpatrol

pessimists

Loong

09:50

@skullpatrol No. The accident already released 33 % of the Cs-137 inventory to the environment. How much worse could it get? So maximum three times worse.

skullpatrol

What would have three times worst have done to Europe?

Loong

I guess still less than 1 mSv in the first year.

I would have to check a few numbers.

skullpatrol

please, if you have some spare time :-)

I wonder if it was done in the book the series was based on...

Loong

10:08

The total dose in the 50 years following the accident is about 2 mSv.

Almost all of it is caused by Cs-137.

So the release of the total inventory would result in a dose of about 6 mSv.

The dose from iodine wouldn't change since the accident already released almost all the iodine that was available.

Doses from Cs-137 are highest in the first year because of inhalation, ground contamination and direct contamination of plants.

skullpatrol

thanks

Loong

So maybe 0.4 mSv after one year, maybe less than 0.05 mSv per year today.

So maybe about 10 000 additional cancer cases in Europe from the peak in the worst case.

Not really "human life on earth could have ended".

skullpatrol

10:52

@Loong thanks for the fact check

Why We Couldn't Stop Watching Chernobyl

►

we are left on our couch "thinking" (incorrectly) the world almost ended and I didn't know...

Knight

@ACuriousMind Are you a family man? I think it is not possible to play video games in house if junior Jueligers are present :)

ACuriousMind

@Knight I don't have a family, but I have plenty of friends who have kids and still play video games.

The first time I saw a game was because my father plays them, too!

Knight

11:09

Hahahha. Great! Did you ever completed Mario fully?

Mario was a console video game, everyone of my generation played it.

ACuriousMind

I didn't own many consoles, so most Nintendo games I've only played with friends who had one

Knight

You know there was an app through which you could video call to strangers who are using that app. It connects two peoples from anywhere in the world.

skullpatrol

video game playing is a competitive sport nowadays

could be nominated for the Olympics soon :P

Knight

Once I got connected with a German and after some little Hi and Hello, I asked him about Aryans, the race which Mr. Adolf Hitler tried to increase and he suddenly became furious and said to me that I can get two years of jail for talking about Nazism in Germany. Why he got furiated? And is it true that I Will be imprisoned if were to do Something like that

skullpatrol

two years?

come on, think about it

that's a major felony

Knight

11:23

Yeah

Just for making a sound of H-I-T-L-E-R

skullpatrol

then again, trump has convicted felons working for him

Knight

What? I'm unaware of that

Loong

no, not for "talking about", but for dissemination of propaganda material

ACuriousMind

@Knight The usage of symbols unambiguously associated with the Nazis or other unconstitutional organizations can be punished with up to three years of imprisonment. Incitement to hatred can earn you up to five years. Neither of these is specific to expressing views associated with Nazism.

Loong

and it's up to three years

skullpatrol

11:26

:O

ACuriousMind

@Knight And if you describe Hitler one more time as someone who "tried to increase the Aryan race" or any other favourable terms, you will be no longer welcome in this chat.

Knight

@Loong what does that mean?

@ACuriousMind Sir i'm not saying anything I just asked him what I was taught by my history teachers.

Loong

> (1) Whoever disseminates in Germany or produces, stocks, imports or exports or makes publicly available through data storage media for dissemination in Germany or abroad the propaganda material
1. of a political party which has been declared unconstitutional by the Federal Constitutional Court or a political party or organisation which has been held by final decision to be a surrogate organisation of such a party,
2. of an organisation which has been banned by final decision because it is directed against the constitutional order or against the concept of international understanding or …

Knight

@Loong Can you please tell me why ACM got hurt so much by those words of mine?

ACuriousMind

@Knight I'm not in the mood to educate you on the evils of Hitler and the Nazis. Calling him "Mr Adolf Hitiler" and choosing from all his goals just that he "wanted to increase the Aryan race" borders on expressing sympathy with Nazism. Don't do that here.

skullpatrol

11:34

ok, ok, calm down please

Knight

All right. I won't do anything like that again.

skullpatrol

enough said

24 hours later...

:-)

Knight

Pal is it you who has starred my message?

skullpatrol

yup

Knight

Thank you so much.

skullpatrol

11:38

np

Knight

Didn't you sleep whole night?

skullpatrol

12 mins ago, by Knight

What? I'm unaware of that

google search^

Knight

Okay

skullpatrol

16 mins ago, by skullpatrol

then again, trump has convicted felons working for him

he's a gangsta

Knight

Yeah

I will be back after few hours

skullpatrol

11:43

cya pal

skullpatrol

11:55

> Originally called, No Truce with the Furries...

interesting

ACuriousMind

Furies! "No Truce with the Furries" is an entirely different kind of game...

skullpatrol

sorry

source

what exactly is a "disco washout"?

ACuriousMind

Someone who's washed out and also very disco.

skullpatrol

hmm

tpg2114

@ACuriousMind If that isn't made yet, somebody could make a fortune with that one

ACuriousMind

12:22

@tpg2114 The sequel will be named Mad Cats: Furry Road

skullpatrol

discos haven't washed out, they've evolved into techno

skullpatrol

12:40

> Vince Aletti was one of the first to describe disco as a sound or a music genre. He wrote the feature article "Discotheque Rock Paaaaarty" that appeared in Rolling Stone magazine in September 1973

TIL

not to be confused with "discography"

Knight

14:17

@AaronStevens Please see my answer one again and tell if it has any demerits even now.

Novalium Company

15:21

-1

Q: What is a common and cheap PIC that supports hardware UART and software SPI?

I've been looking all day for the right PIC for me. I want to be using a hardware UART to control an HC-05 Bluetooth module and also have a few pins for software SPI that will communicate with two MCP3008-I/SL chips. What PICs could work, that are also commonly found in a local store?

ACuriousMind

Jan 11 at 15:04, by ACuriousMind

> Don't advertise your recent questions. If you just posted something on the main site, give it some time, and don't tell people to go there and look at it. Particularly by pinging people. That's rude. Those who can answer are already watching the queue on the main site!

Abhas Kumar Sinha

15:43

@ACuriousMind ya!

@ACuriousMind Still got no idea about that proof?

ACuriousMind

?

Abhas Kumar Sinha

@ACuriousMind $$\nabla R_{\mu u} = \nabla 1/2 g_{\mu u} R$$

ACuriousMind

If you've found that equation, I'm sure you can find a proof of it. What do you need me for?

Abhas Kumar Sinha

@ACuriousMind I asked that to u looong ago and you didn't respond

@ACuriousMind You seem a bit like god of war to me.

ACuriousMind

I...don't know what that means

Abhas Kumar Sinha

15:56

@ACuriousMind R is ricci tensor and g is metric tensor

ACuriousMind

Not that - the "god of war" remark

Abhas Kumar Sinha

@ACuriousMind i'm talking about eqn.

@Knight shit!! That's so deep...

ACuriousMind

I'm not really interested in talking about that equation or its proof.

Abhas Kumar Sinha

@ACuriousMind then what?

@ACuriousMind see the lead character, you are that...

God of War – Story Trailer | PS4

►

@ACuriousMind you don't know what your pic is...

ACuriousMind

I'm pretty sure I do since I chose it myself. What makes you say that?

Abhas Kumar Sinha

16:02

@ACuriousMind see the video, compare yourself...

ACuriousMind

22 hours ago, by ACuriousMind

@Knight It's the protagonist of Disco Elysium

Your proposal has more beard and less hair :P

Abhas Kumar Sinha

@ACuriousMind ok hahaha...

ya

@JohanLiebert ya hello

Johan Liebert

@AbhasKumarSinha hello! how are you ?

Abhas Kumar Sinha

@JohanLiebert JEE Main results are up

Johan Liebert

@AbhasKumarSinha so what is your result?

Abhas Kumar Sinha

16:13

@JohanLiebert 92 percentile...

Johan Liebert

@AbhasKumarSinha great! Though I don't know how to properly evaluate percentile!

Abhas Kumar Sinha

@JohanLiebert The top scorer is 100 percentile, rest are evaluated relative to him/her/

Johan Liebert

@AbhasKumarSinha ah OK!

Abhas Kumar Sinha

@JohanLiebert What's up?

Heard the new BTS Song?

skullpatrol

it means only 8% did better than him :P

Abhas Kumar Sinha

16:15

@skullpatrol yap.

Although I had boards and I'm not in a dummy school, I've to cover multiple exams, I'll try to get into 99.9 percentile next time.

Johan Liebert

@AbhasKumarSinha percentile system is new for me.

Abhas Kumar Sinha

@JohanLiebert me too.

skullpatrol

It's a statistical idea.

Abhas Kumar Sinha

yep.

Johan Liebert

@AbhasKumarSinha so these two exams would be separately evaluated?

Abhas Kumar Sinha

16:21

@JohanLiebert yap, highest in the two will be taken into consideration.

@JohanLiebert physics was bahoot hard!!

Johan Liebert

@AbhasKumarSinha like diamond hard?

Abhas Kumar Sinha

@JohanLiebert yep... :'(

Johan Liebert

@AbhasKumarSinha you visited physics meta?

Abhas Kumar Sinha

@JohanLiebert why?

skullpatrol

en.wikipedia.org/wiki/Percentile_rank?wprov=sfti1

Johan Liebert

16:25

@AbhasKumarSinha dmckee sir is stepping down as moderator :-(

Abhas Kumar Sinha

@JohanLiebert whi?

@skullpatrol yap pal yap...

@JohanLiebert He was one of the best quantum mechanist here.

Johan Liebert

@AbhasKumarSinha I don't know much but on the mother meta his resignation has been linked to the recent firing of CMs.

Abhas Kumar Sinha

@JohanLiebert CM?

Johan Liebert

@AbhasKumarSinha community managers

Abhas Kumar Sinha

@JohanLiebert link?

Johan Liebert

16:29

645

Q: Firing Community Managers: Stack Exchange is not interested in cooperating with the community, is it?

Less than a month ago, there was some settling of the incident with firing Monica. It's obvious that there was not a lot of approval from the community on the course of action chosen by SE, or the settlement. Today, we are seeing a new wave of SE actions targeted on work with the community: firi...

discussion support community company community-managers

@AbhasKumarSinha

Abhas Kumar Sinha

@JohanLiebert Sad...

ACuriousMind

Please Be Nice. We're better than this.

skullpatrol

Third cm within a week...

Abhas Kumar Sinha

@ACuriousMind Hope Dmcee sir hears your condolences.

Johan Liebert

Who knows what else is going to come.

Abhas Kumar Sinha

16:36

@JohanLiebert Those who are gone back are never going to come.

ACuriousMind

@AbhasKumarSinha I wish him all the best and will miss him, but I don't think condolences are necessary.

Johan Liebert

@AbhasKumarSinha and it's even worse than you think. The mod resignation post would not be featured for more than 24 hours. How else would community know that they loosed a diamond?

Abhas Kumar Sinha

@JohanLiebert means?

ACuriousMind

@JohanLiebert Note that dmckee's post has been featured for more than 2 days now and still is. Not all is lost.

Abhas Kumar Sinha

@ACuriousMind dmcee sir is one of the best quantum mechanist on the site.

Johan Liebert

16:40

163

Q: How long can Moderator Resignation notices be featured?

How long are moderator resignation posts allowed to be featured in the sidebars of our sites?

discussion per-site-meta stack-exchange-team featured-tag

Abhas Kumar Sinha

@JohanLiebert physics.meta.stackexchange.com/questions/12654/… It's more than 24 hours, I think, it's still featured.

@ACuriousMind reason behind this creep?

Johan Liebert

@ACuriousMind because he has good friends like you and other mods!

ACuriousMind

@AbhasKumarSinha What 'creep'?

Abhas Kumar Sinha

@ACuriousMind not behind featured for more than 24 hours

Johan Liebert

@AbhasKumarSinha the reason is there on the post I shared.

ACuriousMind

16:43

Sorry, can you try to speak in complete sentences? I find it very hard to tell what you're saying with these fragments.

skullpatrol

ok, ok, let's not lose focus of the jee...it requires all your attention now!

Abhas Kumar Sinha

@skullpatrol oh yes, pal.

skullpatrol

That is why they make it so hard.

Abhas Kumar Sinha

yes bro yes.

@ACuriousMind When all the moderator diamonds have cracked, someone will have to realize that unlike real ones, they can't be formed by pressure alone. well said!

@skullpatrol pal, I've some homework now, I'll be back tomorrow.

Bye everyone.

Johan Liebert

@ACuriousMind "real ones"= community chosen ones or staff?

ACuriousMind

16:51

@JohanLiebert The 'ones' refers to 'diamonds', i.e. 'real diamonds'.

skullpatrol

cya pal

ACuriousMind

(that comment makes no sense unless you know how diamonds are formed)

Constantine Black

Hi to all. Is there a standard way to normalize distributions. In particular the one-particle density profile of a system? I am running a brownian dynamics simulation, and I calculate the histogram of the distances of the particles from the center of the simulation box, but I am not sure what the normalization should be. If the thing is to make the histogram's integral equal to one, then how on implements the idea logarithmically? Or is there some other idea here?

Thanks

tpg2114

@ConstantineBlack What do you mean "implements the idea logarithmically"?

skullpatrol

have you seen the movie blood diamond?

tpg2114

16:53

If you want a probability distribution function, then it needs to be normalized so the integral is 1

Johan Liebert

@ACuriousMind I have read some old meta post on how diamonds are elected though it still isn't much clear.

Constantine Black

@tpg2114 sorry, algorithmically

sorry

I mean the algorithm, not the logarithm

tpg2114

Okay -- still not sure what you mean then. To normalize it, you would compute the integral of your histogram (which will be not == 1 unless you are really lucky), and then divide all the bins by the value of the integral you computed

Then if you re-integrate, you'll find the integral is 1

Constantine Black

Ok.

tpg2114

That will give you a discrete pdf

Constantine Black

16:55

But are there any specific cases where one needs to also do some other calculation to get a physically sensible result?

Otherwise, ok, I will see if that's the only thing needed

skullpatrol

@JohanLiebert Actual diamonds can be made in a chemistry lab using very high pressure.

tpg2114

Sometimes, particularly if I just want to visualize something, I will normalize it so the maximum bin is 1. That way, I can use a contour map or something in log scale

@ConstantineBlack I guess that depends on what kind of physical things you're trying to make sense of... Dividing by the total counts, for example, will give you a "relative amount of counts spent in bin X to X+dX"

Which could be useful

Constantine Black

Yes. But it seems I am not getting the correct behavior for the system, and by just dividing with a constant, will just move the values on the y axis.

The problem is that the distribution should be constant after some value, and almost linear before that one, but instead I get a gaussian-like result

Johan Liebert

@skullpatrol now I get it. Thanks!

Constantine Black

Ok, yes. I get that.

tpg2114

17:00

@ConstantineBlack Not sure why it would be constant -- if you're talking about a pdf of a particle's distance from the center of a system, a constant would mean A) stationary particles at a fixed distance or B) particles orbiting around the center like a planet, all at the same radius

Constantine Black

Yes

tpg2114

Unless I am mis-understanding your system

Constantine Black

I think for this system

skullpatrol

@JohanLiebert en.wikipedia.org/wiki/Synthetic_diamond?wprov=sfti1

for more info

Constantine Black

the distribution of the distances becomes like that of an ideal gas at long distances and after diffusion of the system

so after some distance, the probability to find a particle is constant.

Does that make any sense?

tpg2114

17:01

Sure, but that looks kinda like a Gaussian

Mean inter-particle distance

Mean inter-particle distance (or mean inter-particle separation) is the mean distance between microscopic particles (usually atoms or molecules) in a macroscopic body. == Ambiguity == From the very general considerations, the mean inter-particle distance is proportional to the size of the per-particle volume 1 / n {\displaystyle 1/n} , i.e., ⟨ r ⟩ ∼ 1 / n 1 /...

For example

Constantine Black

Yes,yes, you are correct, ofcourse

Can I show you an image from some notes I have?

skullpatrol

Also, when a person "cracks" it means a mental breakdown @JohanLiebert

tpg2114

Sure

Constantine Black

ok, one moment

tpg2114

Is that figure showing the mean distance from the center?

I'm not sure what I'm looking at there

Constantine Black

17:08

The system's density in the first image becomes constant after some distance.

The system evolves under brownian dynamics and there is a sink in the center of the system, that's why there is a forbidden area at the start. When a particle enters that region, the algorithm relocates it on the surface of a sphere inside the simulation box

We try to calculate the distribution of distances from the center

For an infinite system one should get $\rho (r ) = \rho _{\infty} (1- R/ r) ,$ where R is the radius of the sink.

tpg2114

Hrm... Maybe since there is a sink, that's why it becomes a constant. Cause anything too close will just fall in? I'm not sure -- I haven't thought about a system with a sink in the middle

Constantine Black

Yeah. Ok. Thanks anyway @tpg2114 I just kind find out what goes wrong. I get something like a gaussian and I have no idea why

Thank you.

*can' t no kind

Yuvraj Singh...

17:38

Am I right

String theory is a kind of QFT but it breaks with QFT in that it's elementary 'things' are strings. And we quanitze those strings. That is how it deals with no locality, it says a point does not exist, only strings and higher dimensional Planck sized objects.

Or It is a misconception.

Jake Rose

Trying to trivialise anything so complicated into just short sentences like that will always probably lead to misconceptions. @YuvrajSingh...

QFT is fundamentally a field theory. It doesn't (necessarily) deal with strings

And while there is some overlap in the mathematical formalism, they are different fields.

Saying all of this, I haven't studied QFT nor string theory. But I do next year xoxoxo

ACuriousMind

@YuvrajSingh... The only thing in there that is true is "We quantize those strings". In physics.stackexchange.com/a/293983/50583 I talk a bit about a non-technical view of string theory, but in physics.stackexchange.com/a/458354/50583 I also talk about how the conception of string theory being about actual strings moving in some space is not necessary at all from certain viewpoints.

Jake Rose

18:04

@Semiclassical You around by any chance?

With the transformation we were discussing yesterday, doesn't it result in the (1,0) vector being none time dependent?

Naveen Balaji

18:37

Hey all! I was working on a problem in (1+1)-D and wanted to know how we could demand that the metric, $g = d\rho^{2} + g_{tt}dt^{2}$, is regular at the origin by demanding absence of a conical deficit? Are there any papers you know that elucidates this specific topic?

ZeroTheHero

18:49

Yet a new avatar for @ACuriousMind

ACuriousMind

New year, new face! (a bit late, I admit but still... :P )

Sir Cumference

19:10

man why does string theory have to have a cool premise but so little evidence

it's almost like a guilty pleasure for anyone to study it

ZeroTheHero

Chinese New Year perhaps... ;)

Sir Cumference

@ACuriousMind isn't that the guy from the movie 300

ACuriousMind

3 hours ago, by ACuriousMind

22 hours ago, by ACuriousMind

@Knight It's the protagonist of Disco Elysium

Sir Cumference

eh it kind of looks like him

ACuriousMind

Why does no one seem to see the difference between a full beard and massive sideburns here?!

Semiclassical

20:22

@JakeRose in the rotated basis, yes. but the unitary transformation you use is itself time-dependent, so the eigenvectors in the original basis are time-dependent

looksquirrel101

21:04

New user here. I just had a question closed for being a HW question. However, it actually was meant to be a conceptual question. In fact, a user that has a high reputation here made a comment on it that was actually wrong. So I actually believe that it does have some conceptual value, even though it is a rather simple question. Is there anyone I can contact to see if this question can be opened back up?

ACuriousMind

@looksquirrel101 When you edit a question for the first time after it is closed, it is automatically placed in a review queue for reopening

So try and edit it so that your conceptual problem becomes clearer, and our reviewers might reopen it if they agree with you.

looksquirrel101

OK. I attempted to do so. I guess we will see.

Jake Rose

23:02

@Semiclassical but then you get
$$\begin{pmatrix}
1 &0 \\
0& e^{-i\omega t}
\end{pmatrix}
\begin{pmatrix}
1\\1

\end{pmatrix}

=\begin{pmatrix}
1\\ e^{-i\omega t}\end{pmatrix}$$

Mhmm

Why is this not working

ACuriousMind

You forgot an \end{pmatrix}, I fixed it :)

Jake Rose

Thanks @ACuriousMind !

@Semiclassical I used used the 1,1 vector as an example to show that the first component doesn't receive any time dependence. Meaning that overall it can't be normalisable right?

ACuriousMind

I don't know the context, but that time-dependence doesn't make the norm time-dependent since $\lvert \mathrm{e}^{-\mathrm{i}\omega t} \rvert = 1$.

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