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12:25 PM
Well I've just heard some interesting news.
share with us then @JohnRennie
Apparently we have a Physics SE moderator election coming real soon.
Time to polish those manifestos.
You know what? Screw the U.S.
@SirCumference what's up?
hmm...i guess only 3-4 candidates
12:27 PM
I've spent seventeen years confused on whether pounds are a measurement of mass or force
Today I found out it can mean both
Our country makes no sense whatsoever
you knew about forces when you were 1 year old?
You know what I mean...
@Xasel actually I have no idea who is going to stand. I won't be standing.
hmm.there aren't many contenders
@Xasel who do you think will stand?
@SirCumference Doesn't the US education system use SI for teaching physics?
12:31 PM
hmm..i'm new here but if i have to place bets then its Qmechanic
Qmechanic is already a moderator
It's an election for a new moderator. The existing moderators don't have to reapply.
@JohnRennie When? Why?
@JohnRennie @DavidZ ??
Or Manish?
@MAFIA36790 Soon - I have this from an SE staff member
12:34 PM
Or do we need more mods?
I don't know the reasons. Either we've got busy enough to need a new mod or one of the mods has chosen to stand down. I assume there'll be an official announcement soon.
@JohnRennie ohh....sure.
Well, that is surprising; this would be my first election here at PSE.
Vote for me. I'll ban all discussion of thermodynamics! Hear that @KaumudiHarikumar :-)
@JohnRennie Seriously, you are in my list of nominees.
12:37 PM
@JohnRennie please, optics too.
@MAFIA36790 Nope, I'm too divisive a figure. I tend to make it obvious who I think the twits amongst us are. Moderators have to be more measured than I could manage.
Maybe we might be missing Kyle since he was also in my list for any future elections. He was active in meta as well as reviews.
@JohnRennie:Well why you enemity with thermodynamics
@Xasel Who likes thermodynamics O.o
12:41 PM
@Xasel I'm joking about thermodynamics. It's an exceedingly important part of physics. It's just that I found the thermodynamics courses deadly dull at university.
Let's see when we hear the official announcement.
@JohnRennie How did you get that information from a staff member? >100k rep privileges? :P
And the thermodynamics problems seemed to be just excuses for getting us to solve ridiculously complicated situations involving ideal gases, heart engines and walruses.
@JohnRennie nods, nods
12:43 PM
@JohnRennie:so what shall we do to add poetic beauty to the subject so that it can allure/"bewitch" those on the quest of queching insatiable thirst of knowledge(9too much poetics)
What sort of thermodynamics exercise involves walruses?
@ACuriousMind I might have misremembered the walruses
@JohnRennie Global Warming homework problems ;P
@ACuriousMind I'd love to be able to boast about some exclusive access to the bowels of the SE, but it just came up in conversation.
12:45 PM
I don't think I want exclusive access to the bowels of anything
Couldn't you have chosen a...prettier metaphor? :)
things happen when phsicst gets poetic
Anyhow having dropped my bombshell I now have to go score myself some lunch. See you all in a couple of hours.
@Xasel Walruses and bowels - not amongst the most poetic subjects even for a physicist :-)
@JohnRennie Noooo :P Y you do this?
12:47 PM
@JohnRennie:it's all about perspective :)
@KaumudiHarikumar you're safe - there is zero chance of me being elected a modrator :-)
@JohnRennie At least you've managed to stay room owner longer than Chris, so maybe you're overestimating your divisiveness ;) (CW's account is gone entirely btw, for those that hadn't noticed)
@JohnRennie why
@ACuriousMind He got banned?
@0celo7 The deletion was his own wish
@0celo7 principally because I'm not going to stand. That's going to make it kind of hard to elect me :-)
12:48 PM
He got chat banned for a day, though
Holy [REDACTED], what happened?
I don't even understand all this about moderators. It was my understanding that all high rep. users became mods after a point. Somebody please explain?
@0celo7 He spoke his mind about a certain user, a flag was cast, he was banned. Appearently that upset him enough to stop participating in the site entirely, although of course there may be stuff I'm not privy to
@JohnRennie Yes, but many engineers still use pounds. In 1998 the U.S. spent $125 million on a Mars probe. Problem was, some of the engineers have their measurements in pounds, and the other engineers assumed it was in newtons. The discrepancy caused the probe to stray way off course, lost in space.
This is why I love this country...
@ACuriousMind Do you know which user
12:50 PM
I hate imperial unit :(
@0celo7 who do think?
Looks like CW's a bad loser
@JohnRennie No idea.
I don't pay attention to drama around here.
@Xasel Preaching to the choir, buddy
nor participate, etc.
@KaumudiHarikumar High rep users get some extra powers, but they are still only limited.
12:51 PM
@KaumudiHarikumar You are right that many functions of moderation - like closing and reopening questions, editing, protecting question - are available to all high rep users, but a "moderator" is something more: They've got those diamonds next to their names on the site, and they are the people who deal with all the flags you raise that don't go into a high rep review queue
There's an enormous gulf between what I can do and what a moderator can do.
Q: When do moderator elections take place?

ThursagenI'm really confused on this. I have no idea on several things: When do elections begin? Is there a set time each year? Do elections actually take place each year? Can I vote? Where do I vote? Etc. Can someone please help me?

maybe @ACuriousMind canstan for mod
Ohh, I seee. Thanks!
12:53 PM
@Xasel in my list too.
@JohnRennie Enormous? Okay...
High rep users can vote to close, delete, etc but it takes five votes to close a question. Modrators can singlehandedly close and delete questions.
Wow! Cool.
They can also suspend users for talking about thermodynamics and other antisocial behaviour.
think of them as titans to greek gods
12:54 PM
@KaumudiHarikumar Not to mention, when a mod deletes a post, it is absolute. Only another mod can undelete it, not a high rep user.
If high rep users were mods, Lumo would be a mod. And we'd all be dead, probably.
@Xasel You can edit within 2 min.
@JohnRennie Noo :P
edit what @MAFIA36790
12:55 PM
@SirCumference Oh, I see...
@KaumudiHarikumar :-)
@0celo7 Not in my list; he doesn't participate in meta and review.
@JohnRennie Quite a heated subject
or supergods to still mortal norse god
I'd say it takes too much energy to ban people though
12:56 PM
You can't just go aroujnd banning people at will as a mod, at some point someone will complain and you'll lose your diamond
@ACuriousMind What if you delete everyone's account first?
Losing a diamond was what happened to Tildal, though I dunno the actual reason
The music streaming service?
@0celo7 Stack Overflow staff bans you
Being a mod is a really hard job. It's about trying to keep this site running as smoothly as possible whilst ruffling as few feathers as possible. People underestimate just how much work our mods do.
12:58 PM
@0celo7 Mods cannot delete accounts, I think, that's a SE-employee level power.
They're the overlords in the end
@ACuriousMind I need to become an SE employee
where do I sign up
It's a paid job 0celo
Though that is kind of obvious...
@SirCumference I'm better qualified than @ACuriousMind
@0celo7 You're not a physicist
How can you be better qualified than ACM?
12:59 PM
@JohnRennie That. Being an invisible janitor most of the time strikes me as a thankless job.
@JohnRennie Se employee doesn't need to be a physicist ;)
@0celo7 You'd ban people for disrespecting the holiness of GR
@ACuriousMind damn right
Or misspelling y'all
1:00 PM
Anyway I must go.
@JohnRennie Have a nice lunch!
@JohnRennie He isn't either.
Q: A naive question over some General Relativity quantities

EntropyReally naively: what kind of information do those quantities give us? $$\text{Ricci Scalar} ~~~~~~~ R$$ $$\text{Ricci Tensor} ~~~~~~~ R_{\mu\nu}$$ $$\text{Riemann Tensor} ~~~~~~~ R_{\mu\nu\lambda\rho}$$ I mean: for the purposes of the description of some phenomenon or event in a gravitational...

OP should have googled
People who ask questions that can easily be found on google are the worst
@0celo7 Aha.
1:16 PM
@ACuriousMind Yes?
So, it took me a couple of days...
But I have now realized that I can finally run the Intel C Compiler on this machine
You guys are all doomed
Aug 15 at 13:11, by Bernard Meurer
The only reason the human race still exists is the fact that I have an AMD processor
Brace yourselfs
1:29 PM
What the hell
@BernardMeurer my birthday is coming up
You need to figure out what Reb is getting me
@0celo7 Lol, tell her to text me
@BernardMeurer I'm very afraid you'll find a way to contact her.
If you really wanted to, you could already.
@0celo7 If I had no respect for you or your privacy I would've contacted her a long time ago lol
1:42 PM
@ACuriousMind what is the use/point of the newton-wigner localization/operator in qft?
my prof talked about it but didn't say what's used for.
@IceLord Well, those things are the closest to a "position operator" that you can get
@ACuriousMind what is the use of a position operator in qft
And so they are what you might use if you want to know whether a particle is inside a certain extent of space
since he said we're doing away with it
1:45 PM
Well, we're doing away with it because it turns out even the N-W operators behave a bit strangely and don't really respect relativistic invariance
@JohnRennie : Due to increased traffic, the phys mods have asked the SE team to increase the number of phys mods from 4 to 6, i.e. to organize an election for 2 new phys mods. [edited by DZ since this is a pinned post]
@ACuriousMind btw when would you want to do that?
can you give an example please?
1:46 PM
@IceLord Because you want to know whether your particle will hit a detector located at a certain spot or not, for instance
@ACuriousMind Isn't it just $\hat{x}$?
@hwlau Well, what does that mean in QFT? Remember, we made the fields $\phi(x)$ into operators, but not the coordinates $x$ themselves.
@ACuriousMind Then it should be $\hat{\psi}(x)$
ok, I am not sure
@hwlau Well, that's a field operator, but what has that to do with a position operator?
1:52 PM
@ACuriousMind So they are usable as soon as we fix a specific Lorentz frame, but then we have lost Lorentz invariance, correct?
@Bass I think so, yes, but I never really worked with them either
It's best not to try and localize stuff ;)
OK I see.
Strange world. We started physics with "this thing is here, and later it will be there". Now the most fundamental theory we have cannot even localize stuff :)
ok, i now know the reason for it
2:14 PM
ACM will probably become a mod.
@ACuriousMind Don't do it
@0celo7 What are you on about? I haven't even said whether I might run or not.
@Qmechanic We're doomed
@JohnRennie This thing is a tank
@BernardMeurer I've never tried using the Intel compiler because I've always been developing for Windows so I used MSVC. What's the Intel compiler like to use?
@Secret Is that your method? Asking to Google instead of asking to people who may be really into this field, knowing a LOT about, and knowing how to say it properly without confusion? Interesting. — Entropy 58 mins ago
Ok it seems I am too rude and my comment got deleted. Ok then next time I am not goign to remidn others to google things
@JohnRennie It's a lot like GCC, or Clang, nothing special on the usage side of things IMO
The magic happens in the assembly it generates
It does some hardcore black magic and tends to create binaries that are significantly faster than GCC or Clang, even if you do GCC with -O2
2:31 PM
@BernardMeurer Cool :-) What debuggers will work with it's binaries? (Please don't say "printf statements" :-).
The single coolest thing about MSVC is that the debugger is ridiculously good.
@JohnRennie I think only the IDB works with the ICC
IDB = Intel Debugger
That's a command line debugger isn't it?
I don't know about windows dev at all, apart from the fact that supposedly VS is an amazing IDE
It's like the early 80s all over again :-)
Yep, commandline
They see me rolling writing commands, they hatin'
2:33 PM
@ACuriousMind ;_;
guyas it seems to me that i have forgootent the basic algebra :(
@BernardMeurer ???
2x - x= 0
=> x = 0
Is chamillionaire still alive
@JohnRennie The other day my graphics driver failed and I restarted it with no image b/c of the command line
2:34 PM
but if you add x to bth side
2 = 1
It's amazing
@Xasel Wat?
@Xasel you're dividing by zero.
$2=2$ I assure you
@BernardMeurer $2+2=5$. Everyone knows that.
@ACuriousMind Only Radiohead fans know that :p
2:36 PM
@BernardMeurer Please report to room 101.
huh..hmm...but what will you tell to a child who just started out algebra?w=how will you explain it to them
@ACuriousMind :)
@ACuriousMind Do you want to come to lisbon and bring me a Flensburger chair?
I am in desperate need
@Xasel Tell them you can't divide by zero without breaking everything
@BernardMeurer ^^^
2x - x = 0 => x=0
2x - x + x = 0 + x
2x = x
0 = 0
@BernardMeurer No time to do so
2:38 PM
@ACuriousMind Lies, everyone knows grad students don't do anything
@BernardMeurer Recall I have at least 5 years of semithesis that I can explain why you cannot divide by zero in nearly all conceivable algebraic structures
@Secret So I'm right, you can't divide by zero without blowing shit up
Well the short answer is, because zero is such a 'good' absorber, any attempt to divide by zero you collapse the structure into the trivial structure {0}
@JohnRennie You're a seasoned man
Tell me what is this guy on
Because David Byrne had something before this show, that's for sure
I never really got the Talking Heads. If you're male and in your fifties you're supposed to regard the Talking Heads and REM as gods. But somehow I just never quite got what they were trying to do.
2:42 PM
I don't like REM much
I do absolutely wholeheartedly love Talking Heads
So mathematically yes, division by zero, loosely speaking, does create a balck hole and collapse reality
They've done some cracking good tracks. A Best of Talking Heads album is a good album.
The Live one?
But I still struggle to raise much interest in them.
I'm just wondering what drug does that to people because there's something wrong with that man lol
2:44 PM
@ACuriousMind why in $\langle \mathbf{k} | e^{-iHt} | \mathbf{x} = 0 \rangle$ in the answer the plane wave has an eigenvalue of the hamiltonian instead of the hamiltonian?
@IceLord I don't understand the question.
so it's $ = \frac{1}{2\pi^3}e^{\omega_k t}$
what did I say wrong?
$k$ is the basis state for spinless mesons.
and is an eigenstate of the hamiltonian
2:46 PM
1. You've given no context - which system are you looking at and what is $H$? 2. What do you mean by "the plane wave has an eigenvalue of the Hamiltonian instead of the Hamiltonian" - I just don't understand what you're trying to say with that
H is the relativistic hamiltonian.
and $\omega$ is the eigenvalue of it in the k basis.
Are we doing QFT or some rQM hack job?
Because I'm not good with those ad hoc rQM computations
2:48 PM
This chat needs more balls
::Toss in a geodesic ball (whatever that means...)::
@0celo7 I'll show you the balls ( ͡° ͜ʖ ͡°)
@ACuriousMind second one
@ACuriousMind Interesting result: an isometry of a Riemannian manifold in the metric space sense is an isometry in the Riemannian sense, in particular it must be smooth.
Why are you telling me that?
2:53 PM
@ACuriousMind I'm just asking how do you compute $\langle \mathbf{k} | e^{-iHt} | \mathbf{x} = 0 \rangle$
how does the $\omega_k$ come in in the answer?
I have a question, thankyou in advance for your help
@IceLord Well, if $\lvert k\rangle$ is an eigenstate of $H$ with eigenvalue $\omega_k$, then $\langle k \rvert \mathrm{e}^{-\mathrm{i}Ht} = \langle k\rvert \mathrm{e}^{\mathrm{i}\omega_k t}$.
what does it mean at the beginning of a book: "TEXT FLY WITHIN THE BOOK ONLY "
2:55 PM
@ACuriousMind wait is there a proof as to why that is true
what's there to prove?
why is that true
why does it work
2:59 PM
don't laugh
you're in a QFT class asking about eigenvalues
Because for any function $f(H)$, $f(H) \lvert k\rangle = f(\omega_k)\lvert k\rangle$ by definition of what $f(H)$ means. If you define $\mathrm{e}^{\mathrm{i}Ht}$ by its power series instead, then expanding into the series and using $H^n\lvert k\rangle = \omega_k^n\lvert k\rangle$ yields the result.
@0celo7 this was not in shankar
@IceLord Such a lie lol
First chapter
3:00 PM
@0celo7 I took the question to be the same one as the one you asked recently: "What does $f(H)$ do?"
I was almost going to ask how does the eigenvalue equation $H^n\lvert k\rangle = \omega_k^n\lvert k\rangle$ somehow become an exponential euqation. Looks like when going back to quanutm , I need to study more about exponential maps
I am still not very good at series manipulations. They are simply not as chunky as algebra stuff
@ACuriousMind Yeah, that's basic material everyone knows.
@Secret No, you just need to think a bit harder :P You don't need any knowledge about "exponential maps" for this method, you just write $\mathrm{e}^{\mathrm{i}Ht} = \sum_n \frac{(\mathrm{i}Ht)^n}{n!}$, apply the eigenvalue equation for each summand and then write it back as an exponential.
I was asking you about the functional analysis.
3:03 PM
brb gonna print a book on the work printer, hope I don't get fired
ACM: I know, this why I said I am bad at series. Even though I have learned them before in my class, I tend to end up having a fixation of solving problems by trying to use the operators directly without expanding into series
@0celo7 where in shankar does it say what acm said?
I read ch1 doood
He discusses functions of operators.
I'll bet you a copy of Lee
Page 55
Equation 1.9.6
@IceLord Do you know why there's a minus on the left and a plus on the right?
3:17 PM
@0celo7 I personally prefer it to hear from a good teacher rather than reading a book to learn the stuff. and I think people here have a good knowledge and understanding of various physics issues which can teach us better than any mere text. that's why you better jump off the building now :D
@2physics obe doesn't want to take a proper class on qm
or maybe drink some HF..
a proper class is nothing more than what you see here
@ACuriousMind What the heck does $dp$ mean where $p\in M$
@Secret I wouldn't qualify your comment as rude. Though in the future, it couldn't hurt to be a bit more informative.
@0celo7 If you mean in an integral measure, that's just a notation for the volume form.
I.e. don't try to interpret it :P
3:21 PM
DavidZ: Noted, thanks
@ACuriousMind they say it's an "infinitesimal movement"
this is in a legitimate mathematics text
I'm very confused
And for the record, we (current mods) kind of figured it was going to be another few weeks or so before election proceedings got started. That's why we didn't bother to make an announcement. No secrecy, just lack of motivation ;-)
Maybe the SE team had a more accelerated timeline in mind.
I vote for @BalarkaSen
@0celo7 Well, I don't know any better what you're reading about than you do, and I can't tell from your scarce description
@ACuriousMind That scare description is all I have!
I blame physicists
@ACuriousMind What's up with French and Russians having German names?
Lang, Berger, etc.
Is Berger pronounced the German way or a butchered French way?
3:26 PM
@0celo7 I don't know in those specific cases, but most such German names either come from people that fled the Nazi regime or from German settlers (either during occupation or from earlier times)
Q: reflection in the micro world.

Shivam KaseraI have a hypothetical question in my mind. Suppose I have a single photon and an atom and I throw the photon towards the atom. So if the photon gets reflected, from where is it likely to do so?? From the nucleus or the orbiting electrons?

Scattering problem?
I think that's a surprisingly deep question. The answer is that it's primarily the interaction with the electrons that does the scattering, but I'm not sure I know how to prove it.
That is, if you imagine removing all the electrons to just leave a grid of nuclei they would scatter much less strongly.
I felt like the scattering amplitude of the photon as it interacts with the electron density need to be calculated. However I am not really sure how this will be done, as for this scale QM is not enough and we need to go QED
@JohnRennie I'm not sure I understand the issue here: Unless the photon matches one of the energy level differences (of either the electron or the nucleus states), nothing will happen at all.
Rayleigh scattering (pronounced /ˈreɪli/ RAY-lee), named after the British physicist Lord Rayleigh (John William Strutt), is the (dominantly) elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation. Rayleigh scattering does not change the state of material, hence it is a parametric process. The particles may be individual atoms or molecules. It can occur when light travels through transparent solids and liquids, but is most prominently seen in gases. Rayleigh scattering results from the electric polarizability of the particles...
3:42 PM
@JohnRennie That's molecular, not atomic scattering
@JohnRennie You just picked that because the guy is british
@ACuriousMind if the photon matches that/is resonant with any two levels of the electron and nuclear states, then I guess the photon will be absorbed instead and this result in the state to transit to an excited state
Molecules have additional d.o.f. (rotational/vibrational) that are much less selectively spaced and allow for greater occurence of scattering
Although the question does mention reflection so I wonder if they were thinking about metals ...
@ACuriousMind Monatomic gases, e.g. noble gases, would Rayleigh scatter as well.
It's meaningless to ask whether molecular scattering is "off the electrons" or "off the nuclei" because the molecule as a whole is involved
@JohnRennie Are you sure?
3:44 PM
@ACuriousMind Absolutely sure. A Rayleigh scatterer requires only a non-zero dimension.
The Wikipedia article that you didn't read says: The particles may be individual atoms or molecules
Honestly,you theoreticians :-)
Then I don't understand the mechanism behind this scattering at all.
Do I detect a HNQ effect here:
A: Why gases have weight?

John RennieImagine a gas molecule in a closed box bouncing vertically between the top and boottom of the box. Let's suppose the mass of the gas molecule is $m$ and its speed at the top of the box is $v_t$. When the gas molecule moving upwards hits the top of the box and bounces back the change in momentum ...

@JohnRennie What is the mechanism of rayleigh scattering, as all I can see are just equations that describe how the intensity will be changed. Is it classical?
Yes, it's just Maxwell's equations. Actually I don't really know the details, but don't tell the theoretician that! :-)
3:48 PM
Ohhhhh, it's not an absorb/reemit scattering at all, it's scattering off the induced dipole. Nothing quantum to see at all :/
This would be one case in which one of the more persistent anti-QM users on this site would be right with his "classical antenna theory" :P
:: John makes the sign of the cross in classic Hammer Horror style ::
::googles "Hammer Horror"::
Hammer Horror films!!
Hammer Films or Hammer Pictures is a British film production company based in London. Founded in 1934, the company is best known for a series of Gothic "Hammer Horror" films made from the mid-1950s until the 1970s. Hammer also produced science fiction, thrillers, film noir and comedies — and, in later years, television series. During its most successful years, Hammer dominated the horror film market, enjoying worldwide distribution and considerable financial success. This success was due, in part, to distribution partnerships with major United States studios, such as Warner Bros. During the late...
Awesomely bad films from the 70s.
Typically with Christopher Lee as Dracula and Peter Cushing as Van Helsing.
What's going on in here?
Oooohhhhhh, scattering!
@ACuriousMind As is so often the case.

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