12:01 AM
@enumaris should work :)

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

9 hours later…
9:25 AM
@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.

nice new avatar pal

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

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

It's very disco.

John Travolta was the disco look
re: Saturday night fever
but yes, any look could be disco
@Loong thanks for sharing

9:41 AM
sure

I still haven't watched Chernobyl :(

extremely
They're saying that human life on earth could have ended?
(worst case senerio)

I'm not sure who They are

pessimists

9:50 AM
@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.

What would have three times worst have done to Europe?

I guess still less than 1 mSv in the first year.
I would have to check a few numbers.

please, if you have some spare time :-)
I wonder if it was done in the book the series was based on...

10:08 AM
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.

thanks

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".

10:52 AM
@Loong thanks for the fact check
we are left on our couch "thinking" (incorrectly) the world almost ended and I didn't know...

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

@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!

11:09 AM
Hahahha. Great! Did you ever completed Mario fully?
Mario was a console video game, everyone of my generation played it.

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

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.

video game playing is a competitive sport nowadays
could be nominated for the Olympics soon :P

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

two years?
that's a major felony

11:23 AM
Yeah
Just for making a sound of H-I-T-L-E-R

then again, trump has convicted felons working for him

What? I'm unaware of that

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

@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.

and it's up to three years

11:26 AM
:O

@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.

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

> (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

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

@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.

11:34 AM

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

enough said
24 hours later...
:-)

Pal is it you who has starred my message?

yup

Thank you so much.

11:38 AM
np

Didn't you sleep whole night?

12 mins ago, by Knight
What? I'm unaware of that

Okay

16 mins ago, by skullpatrol
then again, trump has convicted felons working for him
he's a gangsta

Yeah
I will be back after few hours

11:43 AM
cya pal

11:55 AM
> Originally called, No Truce with the Furries...
interesting

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

sorry
what exactly is a "disco washout"?

Someone who's washed out and also very disco.

hmm

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

12:22 PM

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

12:40 PM
> 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"

2 hours later…
2:17 PM
@AaronStevens Please see my answer one again and tell if it has any demerits even now.

1 hour later…
3:21 PM
-1

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?

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!

3:43 PM
@ACuriousMind ya!
@ACuriousMind Still got no idea about that proof?

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

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

@ACuriousMind I asked that to u looong ago and you didn't respond
@ACuriousMind You seem a bit like god of war to me.

I...don't know what that means

3:56 PM
@ACuriousMind R is ricci tensor and g is metric tensor

Not that - the "god of war" remark

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

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

@ACuriousMind then what?
@ACuriousMind see the lead character, you are that...
@ACuriousMind you don't know what your pic is...

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

4:02 PM
@ACuriousMind see the video, compare yourself...

22 hours ago, by ACuriousMind
@Knight It's the protagonist of Disco Elysium
Your proposal has more beard and less hair :P

@ACuriousMind ok hahaha...
ya
@JohanLiebert ya hello

@AbhasKumarSinha hello! how are you ?

@JohanLiebert JEE Main results are up

4:13 PM
@JohanLiebert 92 percentile...

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

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

@JohanLiebert What's up?
Heard the new BTS Song?

it means only 8% did better than him :P

4:15 PM
@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.

@AbhasKumarSinha percentile system is new for me.

@JohanLiebert me too.

It's a statistical idea.

@AbhasKumarSinha so these two exams would be separately evaluated?

4:21 PM
@JohanLiebert yap, highest in the two will be taken into consideration.
@JohanLiebert physics was bahoot hard!!

@JohanLiebert yep... :'(

@JohanLiebert why?

4:25 PM
@AbhasKumarSinha dmckee sir is stepping down as moderator :-(

@JohanLiebert whi?
@skullpatrol yap pal yap...
@JohanLiebert He was one of the best quantum mechanist here.

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

@JohanLiebert CM?

4:29 PM
645

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...

Please Be Nice. We're better than this.

Third cm within a week...

@ACuriousMind Hope Dmcee sir hears your condolences.

Who knows what else is going to come.

4:36 PM
@JohanLiebert Those who are gone back are never going to come.

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

@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?

@JohanLiebert means?

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

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

4:40 PM
163

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

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

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

@ACuriousMind not behind featured for more than 24 hours

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

4:43 PM
Sorry, can you try to speak in complete sentences? I find it very hard to tell what you're saying with these fragments.

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

@skullpatrol oh yes, pal.

That is why they make it so hard.

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.

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

4:51 PM
@JohanLiebert The 'ones' refers to 'diamonds', i.e. 'real diamonds'.

cya pal

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

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

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

have you seen the movie blood diamond?

4:53 PM
If you want a probability distribution function, then it needs to be normalized so the integral is 1

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

@tpg2114 sorry, algorithmically
sorry
I mean the algorithm, not the logarithm

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

That will give you a discrete pdf

4:55 PM
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

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

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

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

@skullpatrol now I get it. Thanks!

Ok, yes. I get that.

5:00 PM
@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

Unless I am mis-understanding your system

I think for this system

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?

5:01 PM
Sure, but that looks kinda like a Gaussian
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

Yes,yes, you are correct, ofcourse
Can I show you an image from some notes I have?

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

Sure

ok, one moment

Is that figure showing the mean distance from the center?
I'm not sure what I'm looking at there

5:08 PM
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.

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

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

5:38 PM
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.

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

@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.

6:04 PM
@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?

6:37 PM
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?

6:49 PM
Yet a new avatar for @ACuriousMind

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

7:10 PM
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

Chinese New Year perhaps... ;)

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

3 hours ago, by ACuriousMind
22 hours ago, by ACuriousMind
@Knight It's the protagonist of Disco Elysium

eh it kind of looks like him

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

1 hour later…
8:22 PM
@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

9:04 PM
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?

@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.

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

2 hours later…
11:02 PM
@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

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

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?

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