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4:00 PM
@0celo7 maybe...but I have no time to waste, I am not a student ;-P
 
I guess PDE people care more about infinite dimensional manifolds. No idea.
 
@BalarkaSen is a manifold also a topoi
 
@BalarkaSen I personally care more about infinite dimensional spaces, but it is my personal taste
@0celo7 topos, not topoi
latin is important ;-P
 
@0celo7 I have no idea what a topos is.
 
@BalarkaSen extremely useless stuff a bridge between any field of science apparently
 
4:05 PM
@yuggib Greek?
 
@yuggib I took Latin
 
@JohnRennie ouch...indeed ;-P
 
But data is also a singular word
Even though it's a Latin plural
 
@0celo7 datum is a singular word
@BalarkaSen at least if you are to believe this b*****t very interesting theory
 
@yuggib no one who speaks English as a first languages agrees.
 
4:09 PM
Nah.
 
@0celo7 well, according to the dictionary I have on the computer, data is either the plural of datum, or a "mass noun"
like people?
anyways, it means many "datum"
 
Mass noun?
 
I have not written the dictionary
 
No one says "look at the datum" if there's one data point
 
I am simplying reporting
@0celo7 I do
"The initial datum is ..."
I almost surely used it in scientific papers
 
4:12 PM
Since when are scientists linguists
 
@0celo7 I am just proud of my latin lineage
 
6
Q: What's the adjective form of "data"/"datum"?

qazwsx"Informative" is the adjective form of "information". What's the adjective form of "data"/"datum"?

 
@yuggib too bad the Saxons destroyed your shitty little empire
 
@0celo7 they did not
they converted as soon as they arrived
 
Learn history, bub
 
4:14 PM
then the saxons were romans
 
@0celo7 I think that was the Visigoths.
 
What would a Brit know about that anyway
 
Public school education. Big on classics.
 
@0celo7 bub?
 
Yeah, old woman
 
4:15 PM
Be nice pal
 
I'll fite all of you IRL
Except @yuggib, don't wanna be crushed.
What
That's IRL not IRS
 
@0celo7 and technically, there was a roman empire, in one form or another, from 27 BC to 1806 AD
 
@yuggib Holy Roman Empire does not count
 
and a roman republic from 509 BC to 27 BC
@0celo7 it is called roman empire
 
4:18 PM
It's neither holy nor Roman nor an empire
 
@yuggib After the fall of Constantinople??
 
and as I told you, northeners came, and changed their habits and names to fit roman culture
 
@yuggib yeah well guess who won the war
America
So be quiet.
 
@JohnRennie the eastern continued until 1453
 
No, you be quiet @0celo7 :P
 
4:20 PM
@skillpatrol just ban me for a month
@yuggib I'll fight you
Give me a time and place
 
poor little victim?
 
I'm not a victim
 
calm down then pal
 
@0celo7 america copied romans already in naming its congress house
 
I'm a strong independent black womyn
 
4:21 PM
@yuggib That was my point. In what form did the Roman empire exist after that?
 
@JohnRennie the holy roman empire
 
@JohnRennie Holy Romab!!
I just said it
Cheerio
 
Google, Google, hmm that's a somewhat technical definition of Roman Empire.
 
0
Q: Is there simple way to color latex equations?

AchmedIs there simple way to color background, highlight some parts of background, equation or some parts of equation? something like these Latex codes: {\bgcolor{color name}{equation}} or {\bg{color name}{equation}} for background and {\color{color name}{equation}} or {\mathcolor{color name}{equ...

 
What's the definition of the Briish Empire :P
 
4:25 PM
Us and the Falkland Islands I think ...
 
@skillpatrol you're forgetting letters pal
 
At least we have a leader now.
 
@JohnRennie gibraltar?
isle of man?
@JohnRennie norsefire?
 
Well Gibraltar isn't at all impressed with the leave vote. I imagine they're having quiet talks with the Spanish as we speak.
 
Roman empire always reminds me of Asterix :P
 
4:28 PM
The Empire Strikes Back
 
This conversation is getting silly :-)
 
user116211
what would Sir Arthur Pendragon think of Brexit?
 
user116211
@JohnRennie You understood now ;P
 
@JohnRennie Yeah, and HM Government will be happy to just take the note and let the base go.
@Qmechanic Ok, so here's a full-on campaign
 
wow a 19 mb .djvu file turned into 200 mb when converted to pdf. I wonder how djvu is so compressed..
 
4:30 PM
0
A: Trackbacks from SE to the arXiv.org?

E.P.This really should be implemented: it is good for the sites, it is good netizenship, people want this, and an implementation already exists. As far as I can tell - from any publicly available information - all that's needed is (in essence) for SE to flip a switch. Just to sum up some responses, ...

0
Q: Please consider supporting arXiv trackbacks on the mother Meta

Emilio PisantyOne long-standing feature requests for the Stack Exchange engine is the implementation of trackbacks for mentions of the arXiv on Stack Exchange sites. In general, if you have a blog and you blog about an arXiv eprint, you can send the arXiv a specially crafted http message, and then your blog ge...

0
Q: Please consider supporting arXiv trackbacks on the mother Meta

E.P.One long-standing feature requests for the Stack Exchange engine is the implementation of trackbacks for mentions of the arXiv on Stack Exchange sites. In general, if you have a blog and you blog about an arXiv eprint, you can send the arXiv a specially crafted http message, and then your blog ge...

0
Q: Please consider supporting arXiv trackbacks on the mother Meta

E.P.One long-standing feature requests for the Stack Exchange engine is the implementation of trackbacks for mentions of the arXiv on Stack Exchange sites. In general, if you have a blog and you blog about an arXiv eprint, you can send the arXiv a specially crafted http message, and then your blog ge...

0
Q: Please consider supporting arXiv trackbacks on the mother Meta

Emilio PisantyOne long-standing feature requests for the Stack Exchange engine is the implementation of trackbacks for mentions of the arXiv on Stack Exchange sites. In general, if you have a blog and you blog about an arXiv eprint, you can send the arXiv a specially crafted http message, and then your blog ge...

 
@EmilioPisanty I bugged the CMs about it; they say a response is forthcoming
 
Similar posts coming on TCS and Cross Validated once that 'you can only post once every 40 minutes' banner goes away
@DavidZ Well, there's been public pressure for ages and no response
I'm going to keep bugging until there's a public response
 
Yeah, hopefully this time they actually post something
From what I heard in private, I think their stance is that this is somewhere in the middle of a long list of features they would like to implement
 
@DavidZ Yeah, but the 'would like to implement' is a bizarre thing to say if the implementation already exists and they don't say why it's harder than just flipping a switch.
 
1
Q: Please consider supporting arXiv trackbacks on the mother Meta

Emilio PisantyOne long-standing feature requests for the Stack Exchange engine is the implementation of trackbacks for mentions of the arXiv on Stack Exchange sites. In general, if you have a blog and you blog about an arXiv eprint, you can send the arXiv a specially crafted http message, and then your blog ge...

 
4:36 PM
Well, I'd say wait for the public response by the CMs (which I was told is coming) to see whether they address that.
I mean, don't go by my description.
 
@DavidZ Sure.
In the meantime, there's plenty of incentives now for them to respond.
I should think.
 
@JohnRennie I think it's great
 
Or maybe I'll be downvoted to oblivion, who knows.
 
@Slereah Lily The Pink or Hawkwind? Or both?
 
The former
 
4:43 PM
@DavidZ Yeah, there it is
 
I like me a good jolly silly song
 
1
A: Trackbacks from SE to the arXiv.org?

PopsThe bounty asks (taking things a bit out of order) Can the SE team comment on the status of the roll-out to other SE sites? Alright, here goes. If this is status-declined, can we have an official statement and some thoughts as to the reason? First off, the good (or less bad) news: thi...

 
@Slereah The thing about silly songs is that they're funny the first time you hear them. Sometimes they're funny the second time.
 
I listened to it about 20 times today
 
See I warned you about playing with those closed timelike curves
 
4:46 PM
Silly song? Tom Lehrer.
 
Tom Lehrer has some great one
Including math ones
 
Yep.
 
My favorite math one is probably "Any question"
 
@BalarkaSen They aren't silly, they are witty! It's an important distinction that not everyone gets.
 
It is quite close to home
 
4:48 PM
@JohnRennie That's fair.
 
It's a song that you need to sing as a professor in your last class
 
Whereas Lily the frakking Pink ...
 
I guess satirical and parodic songs are different than joke songs.
 
@Slereah Mine is "Lobachevsky".
 
4:48 PM
Now then, are there any questions,
Any problems, any questions?
If there are none, then I am done,
And I can bid you all good day.
For there's no reason I should stay here,
Since I've said all I have to say here.
If there are none then I am done,
I wish you luck on the examination.
And so my friends, I bid you all goodbye,
I hope you liked this course as much as I.
Goodbye, goodbye,
Goodbye to one and all I say goodbye.
Just one more thing -- and do not laugh,
I hope you'll take the second half:
Ha, he asks if there are any questions.
Holy smoke have I got questions!
I've got a ton, and every one,
Would take him half a day to do.
But I don't really want to stay here
Since he's said all he has to say here
But it's agreed that I shall need
Much more than luck on the examination.
And so I think I'll let him say goodbye
I guess that he is as relieved as I
Goodbye, goodbye,
Thank God the course is over now, goodbye
One thing he said that makes me laugh
He hopes I'll take the second half
Ha ha, ha ha, ha ha,
 
Plagiarise, plagiarise, let no-one's work evade your eyes
 
Remember why the good lord made your eyes!
 
New Math is also pretty good
 
Though I don't know why Lehrer picked on Lobachevsky. As far as I know he didn't have a reputation for plagiarism.
 
user116211
@Slereah WOW! ! !! !!!! !!!!!!
 
4:54 PM
Maybe he just thought it was a funny name
 
@JohnRennie Perhaps because of the purported dispute between Gauss, Lobachevsky and Bolayi on the discovery of hyperbolic geometry.
But he did mention it was just for prosodic purposes.
 
user116211
Hey @Bernard o/
 
user116211
when is your college starting?
 
@MAFIA36790 Howdy, september
 
user116211
ah.
 
5:01 PM
That is an odd meme
 
Emma?
 
I dunno
 
user116211
@Slereah talking about $\mathbb R^3?$
 
Any metric space, from context
 
user116211
Well, really odd ;/
 
5:06 PM
34
Q: When is the closure of an open ball equal to the closed ball?

Alex LapanowskiIt is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$ d(x,y)= \begin{cases} 0\qquad&\text{if and only if $x=y$}\\ 1&\text{otherwis...

 
user116211
It is not true in general and we are asked to find metric space for which this doesn't hold as an exercise in Kreyszig's book.
 
user116211
@Slereah: Got a comment mentioning Sommerfield's model.
 
user116211
But I take it as an expansion of Bohr's model.
 
Yeah my question was kinda on the classical models
Bohr is already somewhat quantum
 
for $(\mathbb{Z}/2^{n}\mathbb{Z})^{\times}$ to not be cyclic, does it mean that there cannot be a generator $x$ whose order is equal to the order of the group $|2^n|$?
 
5:42 PM
@JohnRennie wow
I would have not thought that
 
5:54 PM
@0celo7 There are much easier examples where this fails.
Easier as in geometric.
 
@EmilioPisanty : Wow. Thanks a bunch. Hopefully that should get the SE team talking and hopefully acting. Not implementing this proposal means lots of lost quality internet traffic for SE.
 
6:13 PM
Is there some philosophy behind why avogadro's number is constant of that particular value?
Does it reflect some property about the nature of our universe?
 
No
 
Not really. It just reflects the amount of mass someone decided to call a kilogram. (Or gram, if you prefer)
 
@deostroll No. It defines the quantity "one mole", which is an arbitrary human measurement unit.
 
We just picked the number of atoms in 12 grams of carbon
It's just a number that is roughly $1g / m_p$
 
His hypothesis...how did avogadro conclude it?
 
6:26 PM
It's not a hypothesis, it's a definition
The hard part is to estimate it
I think it was done with the Millikan experiment?
Wait no
In chemistry and physics, the Avogadro constant (named after the scientist Amedeo Avogadro) is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by one mole. Thus, it is the proportionality factor that relates the molar mass of a compound to the mass of a sample. Avogadro's constant, often designated with the symbol NA or L, has the value 7023602214085700000â™ 6.022140857(74)×1023 mol−1 in the International System of Units (SI). Previous definitions of chemical quantity involved Avogadro's number, a historical term closely related...
 
I saw an explanation of how 2 litres of water created from 2 litres of hydrogen gas and a litre of oxygen gas...
To me...that example makes me conclude the presence of molecules...
Not the statement avogadro made...
I forgot the clause - under standard conditions
 
6:44 PM
So anyway
What would be the expectation value of something of the form $\hat \varphi^3(x)$
And is it renormalized by the energy renormalization
I wonder
Let's find out perhaps!
 
@Slereah In what model?
 
Free scalar field
 
Isn't 0 the obvious answer?
That expectation value is just the three-point function, no?
And there are no diagrams contributing to that in the free theory
 
Well $\varphi^4$, then
my question is if the polynomial terms of the field are all renormalized by the rescaling of $\Lambda$
$\mathcal H$ is, but is it general
 
@Slereah Can you draw a $\varphi^4$ diagram that contributes to the three-point function?
 
6:48 PM
I mean $\langle \phi^4 \rangle$
In the free theory
 
That's...just two propagators.
The only possible diagram for that is two straight lines
 
The Hamiltonian isn't just a bunch of propagators, tho
The point is that this is considered at the same spacetime point
And from hence the singular structure arises
 
@Slereah Well, but the "vacuum energy" is not really $\langle H\rangle$.
Examine the derivation of the "vacuum energy" e.g. in LSZ as the 0-point functions
There's limits involved and stuff
 
Sure
But the renormalization constant for the vacuum energy is the vacuum energy
That is $\mathcal L = \mathcal L_\varphi + \rho$
 
Yes, it is.
 
6:51 PM
Is that constant enough to renormalize every such observable?
If I have an observable of the form $\sum a_n \varphi^n(x)$
Is it renormalizable
Just with $\rho$
 
What do you mean when you ask of an observable whether it is "renormalizable"?
 
Is the expectation value finite
After proper rescaling
 
Anyway, in the free theory all of your n-point functions are either zero (in the odd case) or factor as propagators (in the even case)
 
Hm
I think that makes it likely that it is?
 
That they might become singular when you move the fields' argument together is not something that needs "renormalization", I think
 
6:54 PM
But that's just a gut feeling
Why not
We do it for the hamiltonian
 
@Slereah Because I think that singularity is the $\delta$ (i.e. unit probability per spacetime volume) for "nothing happens"
 
Isn't the singularity just the $1/|x-y|$ bit of the propagator
 
I suspect that it is renormalizable thusly
 
But you have to think what e.g. $\langle \phi(x)\phi(x)\rangle$ signifies - the amplitude for a particle going from x to x.
 
6:57 PM
I should probably try to make a decent proof
 
There's nothing to renormalize here because that quantity is physically rather non-sensical to begin with
 
But it is used without problem for the stress energy tensor
Also you know me
No fan of particles in QFT me :p
 
QFT is not a fan of your "stress energy tensor", though :P
 
Well then maybe it should stop living in the universe
Stress energy tensor is perfectly well defined!
$\lim_{x\to y}\langle \hat T_{\mu\nu}(x,y) \rangle_\omega$
 
@Slereah Yes, and as a Noether current it will obey the Ward-Takahashi identity associated to it
If you look into the W-T identity, there are $\delta$s in it - the "contact terms". This is not something that needs to be renormalized.
 
7:00 PM
But you are assuming that it is a Noether current
Which isn't true in the general case
Don't try to nitpick fight with me
 
...are you doing QFT in curved spacetime or something?
What even is your notion of "vacuum", then?
 
It is frame dependant!
Tho that doesn't really matter here
 
Ordinary QFT has full Poincaré symmetry and therefore of course the stress-energy as a Noether current.
 
Since the renormalization constant is $\Lambda$
Which is all good
Wait, I forget
 
If you're doing QFT in cirved spacetime nothing of this dialogue made any sense because I was assuming you were talking about QFT. :P
 
7:03 PM
In QFT on curved spacetime, do you renormalize with $\Lambda$ or also use $G$
Well
AS A LIMIT IN MINKOWSKI SPACE
These should be the same arguments :p
 
I have no idea that one can even do renormalization on curved spacetime
 
You can
The usual method is point splitting
 
@Slereah I'll believe that, but I don't know how :D
 
Take the quantity $T(x, x + dx)$
Expand around $x$ in Riemann normal coordinates
That will give you the divergences
I think you can also do dimensional regularization
otoh apparently using normal ordering or regulators doesn't work
Don't remember why tho
 
@Slereah In my understanding, QFT in curved spacetime breaks down far before I could worry about renormalization :P
 
7:06 PM
Well you still need to renormalize free fields
IIRC you need up to four renormalization constants for semiclassical gravity
 
I mean, I can probably just say it's defined by the path integral, but I'm not sure how to do physics with that in the generic case
 
$\Lambda, G, A, B$
Where $A$ and $B$ are constants of the Lagrangian terms $R^2$ and $\Box R$ IIRC
Or something
 
@ACuriousMind is it correct to say that in a Yang-Mills theory the gauge bosons are the generators of the gauge group, i.e. a basis of the Lie algebra, and the gauge fixings are the possible bases of the Lie algebra? In other words, fixing a gauge is nothing more or less than choosing a base in the Lie algebra.
 
The gauge bosons are the connection of the principal bundle
 
@Bass That depends on what you mean by "the gauge bosons". The fields? The particle states? Something else? And fixing a gauge has nothing to do with fixing a basis of the Lie algebra.
 
7:10 PM
@ACuriousMind yes the fields. $A_\mu$ in pretty much all the literature I've seen.
 
It's a lie algebra valued form or some shit
lie group valued?
 
@Slereah Lie algebra. I've read about that, but I can't yet match it with the physics I've learned.
 
@Bass The gauge field $A_\mu$ is Lie-algebra valued and once you've chosen a basis of generators $T^a$, you may expand as $A_\mu = A^a_\mu T^a$. The $T^a$ are the basis, the $A^a_\mu$ are the fields.
 
It is the curvature of ur bundle~
Just bend the bundle a bit
Bam!
Fields
 
@ACuriousMind The $A_\mu^a$ are the fields, okay. What are the $A_\mu$?
 
7:14 PM
@Slereah The curvature is $F$, not $A$.
 
You know what I mean
 
@Bass It's the Lie-algebra valued gauge field.
 
You old fuddy duddy
 
You call both "the gauge field".
 
user116211
@Slereah: Here comes BKS theory.
 
user116211
7:16 PM
There was the BKS theory, which was a mess and died at the age of about three: en.wikipedia.org/wiki/BKS_theory The fundamental issue that was confusing people in this period was that Bohr kept insisting that the atom should be quantized but that radiation should be classical. This leads to nonconservation of energy and momentum, except on a statistical basis, and that prediction was how the theory was disproved by Bothe and Geiger ca 1925. — Ben Crowell 1 min ago
 
@ACuriousMind I see. For $SU(N)$ we have $N^2-1$ gauge bosons. Let's write those indices: $A_{\mu,a}$ where $a$ tells us the index of the gauge field, ok?
 
Sure
They are paired up with multiplet indexes for the matter fields
 
@Bass ok
 
Keep writing more indexes, btw
@ACuriousMind loves indexes
 
So we have $A_{\mu,a}=A_{\mu,a}^bT^b$ if they are expanded. What if we choose the basis $T_a$ such that $A_{\mu,a}^b=\delta_a^b$? Is that anything special? You see why I said the gauge bosons are the Lie algebra basis?
 
7:21 PM
There is only one gauge index?
 
After all, we have both $N^2-1$ gauge fields and $N^2-1$ generators...
 
@Bass ???
The $a$ on the field already labels the $A_{\mu,a} T^a$ expansion
 
It is just $A_\mu = A_\mu^a \tau_a$
For instance $SU(2)$ is $W^+_\mu \sigma_1 + W^-_\mu \sigma_2 + Z_\mu \sigma_3$
 
@Slereah Don't we have eight gluons, so eight times a $A_\mu=A_\mu^aT^a$?
 
You don't have $N^2-1$ gauge fields that are $\mathfrak{su}(N)$-valued. You have one $\mathfrak{su}(N)$-valued gauge field that has $N^2-1$ real components.
 
7:23 PM
Ohhh
And one specific gauge fixing is the same as one specific choice of generators?
 
@Bass No
 
@Qmechanic No worries. We both care about this.
 
A gauge fixing is choosing some condition that has exactly one solution in each equivalence class of fields that are related by a gauge transformation
 
@ACuriousMind is that the section in the principal bundle?
 
@Bass What section?
A non-trivial principal bundle doesn't have global sections
There are many different possible field configurations. Some of them are related by gauge transformations, and we declare those that are equivalent. A gauge fixing is a way to pick a unique representant from every such equivalence class.
 
7:35 PM
On the other hand
The gauge field for gravity is $\omega^{ab}_\mu$
Why is it
Why two indexes instead of one
 
@Slereah Because you use the conventional anti-symmetric indices to write elements of $\mathfrak{so}(n)$.
 
Oh right
 
That way, the $ab$ really correspond to the $ab$-th entry of the matrix
 
Does that mean I can write it as $\omega^a_\mu$
Also $\mathfrak{so}(n-1,1)$ please :p
 
@Slereah If you switch to just numbering the generators, then sure
 
7:39 PM
Hm
But then what would the geodesic equation look like, I wonder
This sounds like a job for @0celo7
 
@Slereah Yeah, there's another point - the action of diffeomorphisms is clear on the antisymmetric indices, but not obvious on your new variant.
 
@ACuriousMind I thought it sounded like a section (maybe local) of a principal bundle, because choosing exactly one point for each equivalence class under the gauge group operation would mean to choose a local section, doesn't it?
 
What are even the generators of $\mathfrak{so(3,1)}$
 
@ACuriousMind I see, makes sense.
 
Oh wait
A group has the same generator as its double cover, right?
 
7:41 PM
@Slereah Yes.
The explicit forms for the generators of boosts and rotations should be found in the standard texts
 
It's just the $\sigma^{ab}$ bullshit
 
@ACuriousMind so how is "gauge fixing" described in the principal bundle language?
 
A gauge is just a section of the principal bundle
 
@Slereah No
 
Isn't it?
Dang
then what is a section of the principal bundle
 
7:49 PM
There are no sections of non-trivial principal bundles
 
:O
 
So QFT must be trivial. I knew it.
 
The worst definition of QFT I ever saw used a Hilbert bundle to define it
 
@Bass You can form the space $\mathcal{A}$ of all allowed connection forms on the principal bundle. The group $\mathcal{G}$ of gauge transformations acts on this space, and $\mathcal{A}/\mathcal{G}$ is the space of physically different field configurations. "Gauge fixing" is reducing the path integral from an integral over $\mathcal{A}$ to one over the quotient.
 
Just looking at it I had a headache
 
7:54 PM
Or it might also be choosing a representant in $\mathcal{A}$ for every element in the quotient
It depends on what exactly you understand as "gauge fixing", really.
There's also a "gauge fixing fermion" in the BRST formulation
 
is it the spooky popov fermion
 
@Slereah no
The ghosts are different things
 

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