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18:00
@Slereah Is Hawking even alive
@JohnRennie I know many books. But there isn't what I want in any of them.
@0celo7 Somewhat
@lucas Then maybe you're making the wrong question
@BernardMeurer harsh
@JohnRennie Isn't he almost 80 now?
No. You think my question is nothing.
18:02
After 60 you're already less-than-entirely alive to me
@BernardMeurer so I only have five years of life left :-)
Of course when I'm 60 I'll say it's 70, and so on :p
Hawking is 74
@JohnRennie Your internet points give you extra life
@BernardMeurer Bob is 69.
18:03
@0celo7 He's sucking life off you
I don't want answer because of my job or education. I am unemployment and have free time. I am interested in.
@0celo7 He's been dead for years
The computer does all the talking now
btw @ACuriousMind
1
Q: Contour integral of the retarded Klein Gordon propagator

SlereahI've been trying to prove by hand the Peskin's formula for the retarded propagator of the Klein Gordon equation, that is, $$\int_{x^0 > y^0} \frac{d^4p}{(2\pi)^4} \frac{-e^{-ip(x-y)}}{i(p^2 - m^2)} = \int \frac{d^3p}{(2\pi)^3} \frac{e^{-ip(x-y)} - e^{ip(x-y)}}{2E_p}$$ With the condition that i...

if u know what's up
Have I mentioned I hate vector bundles
@JohnRennie May you take a look to my questions an answers and then say what do you think about me?
Is there anywhere for private conversation?
@Slereah We should learn category theory together
18:09
Should we
I mean
Yes.
I don't think I need ALL category theory to learn AQFT
@lucas not on this site (except in special cases for moderation purposes only)
The original AQFT paper doesn't even mention it
18:10
Categorical differential geometry @Slereah
nothx
@DavidZ Can I ask some questions from you?
You can always ask, but no guarantees I can answer
Can I ask are you a physics professor?
You can ask that :-P
and because I suspect you actually wanted to ask that: no, I'm not
18:12
@DavidZ How old are you? Please
@DavidZ What are you wearing
I'm not going to answer that
(those)
Why you answer people's questions that you never know them?
@DavidZ sexy
@lucas What?
That's not a proper sentence.
18:16
@ACuriousMind Lol.
Are you good people that want to help the others?
@lucas No.
@lucas No one in here is a good person
I know like one good person
@BernardMeurer :(
18:17
@lucas Sometimes. Depends on what "the others" need help with
Which doesn't mean I don't love you all <3
@BernardMeurer :/ I'll take it.
@ACuriousMind Yeah, if it's Riemannian geometry you never help
@NeuroFuzzy You better, it's more than my ex ever got
Jesus would want us to help people on Riemannian geometry
Obviously
18:18
the 11th commandment
@ACuriousMind Are you a physics professor?
lol
implying he's ever gonna have a "job"
@lucas He's a a my little pony stripper
@lucas No, I am a master's student.
@BernardMeurer uhhh
18:18
Does that mean he dresses up like a pony and takes off his clothes
@BernardMeurer Only on Saturdays.
11. Thou shalt help people with geometry
The bible doesn't contain a lot of math
@Slereah Yep
Unlike the Vedas
Even in the Vedas, it's very specific math
Geometry for altar building
18:19
@ACuriousMind Do you think classic mechanics is finished?
Why do you keep asking random questions to people
How do I put a Pepsi can inside a boot?
pick up can
put in boot
@lucas ...what? What has that to do with anything?
@Slereah Does that work for a Pepsi Max can?
18:21
Is it oddly shaped?
@ACuriousMind If you see a question connected with classic mechanics, you ignore it?
@Slereah Star shaped.
Well
With Pepsi inside
@lucas No. Why would you think so? Why are you asking me that?
18:21
As long as the diameter of its circumscribing circle is small enough
Should be fine
@ACuriousMind What's your favorite berry?
@BernardMeurer I don't think he wants to go to Memphis with us.
Not sure if you can put a Faxe can in a boot, though
@Slereah Would you like to go
@Slereah Faxe is my fav canned beer
18:22
Is it
It's one of the worst beer
@0celo7 Hoping I can join
I can only find it at truck stops
@Slereah I love it
You awful awful man
Not that one though, the blue canned one
18:22
Do you also like Ottinger
@Slereah Aren't there other names for "extra strong beer"? "liquor" iirc?
@ACuriousMind I think you ignore questions that you think they are old knowledge. Do you know all things about classic mechanics?
Liquor is a specific distillation method
Beer is just fermented
@lucas That's plain rude. You're seeking out any old user to insult because your question wasn't answered?
@lucas Uh, no, I don't ignore questions about "old knowledge". What are you talking about?
@ACuriousMind Aren't you going to answer me?
Peskin seems to do something like $\delta'(x) f(x) = \delta(x) f'(x)$
Is that the same principle
@BernardMeurer About the berry? Let me think
Pre-emptive application of deltas
18:25
@ACuriousMind It's clearly the most important question in this conversation so far
@BernardMeurer Elderberry, I guess.
@ACuriousMind That's surprisingly fitting
@Slereah that's essentially called definition of distributional derivative
Well yes, but usually, they do that in the integral
Hence why preemptive
If my question is not good, you can at least take a look to it and after solving it say to me that it is not a good question but not before solving.
18:27
yeah, but delta has meaning only inside integrals
I suppose yeah
I suspected but I prefered asking
Is anyone chatting with me?
you know, blah blah distribution is not a function blah blah
yeah yeah
I know
I guess technically it should be $\delta'[f] = \delta[f']$
Ok. I am sorry for myself. good luck everyone. bye.
18:30
or even more simply $\delta'(f)=f'(0)$
of course $f$ has to be rapidly decreasing or compactly supported and smooth
Wow, lots of stuff happened while I was gone
Errrr
It's the propagator
Well the Pauli-whatever function
I think it's rapidly decreasing?
$\approx e^{-ipx}$
Or something
@lucas What did you want to chat about?
@Slereah it's not rapidly decreasing
Oh
Hm
Then I dunno
Maybe it's because QFT is shit
And poorly defined
18:33
but probably is physics bullshit
probably is a multiplication of distributions
and you need a proper way to define it
(it's not completely impossible)
I'm sure Jaffe has a rigorous demonstration of that
@yuggib Umm, you can define a product of distributions. Shouldn't be a problem.
Well you can, but it is a problem
Like it is not trivial at all
Integrate[pdf1*cdf2] or something
@barrycarter only if the support is convenient
18:35
@yuggib The support is always -Infinity to Infinity :P
Doing something like $\delta(x) \delta(x)$ is a lot harder
(IIRC it is 0)
Oh, the dirac delta function. Just deal with its CDF, it doesn't have a PDF.
We mean functional distributions, not probability distributions :p
18:50
@Danu do you know about horizon supertranslations and near Killing horizon metrics by any chance?
@FenderLesPaul I know plenty, ask away
@BernardMeurer <3
@FenderLesPaul <3 :D
I wish I did though
19:06
@FenderLesPaul I'm only taking SUSY this semester, so no.
@lucas D'awww :\
@ACuriousMind are you sure about that comment you posted?
@Danu Pretty sure, yes
when measuring something with respect to some variable, does the variable have to be a scalar?
Or do you have a non-perturbative definition of the string scattering?
(for the usual 10D formulation)
I never see something like $\frac{da}{d\vec b}$ so I'm just wondering'
@ACuriousMind I don't know much string theory---but I can pull up the quote in Erdmenger's book on AdS/CFT if you like...
> A crucial point in view of correspondences between quantum field theory and string theory is that the large N expansion is formally the same as a perturbation expansion of closed oriented strings with string coupling 1/N
19:18
Perturbation expansion of what though ;)
If I ever dated a Susy would you guys bully her?
2
To my knowledge, there is no generally accepted way to define the string S-matrix non-perturbatively - one always defines it by the sum over worldsheet topologies.
> So far we have considered only non-interacting strings and we have discussed their different mass states in detail. Let us now allow for interactions among the strings. Assuming a small coupling constant between strings, the idea of the perturbative expansion in terms of Feynman diagrams, similar to those introduced within quantum field theory, may be carried over to string theory in a natural way.

> So far, we have considered worldsheets with the topology of a cylinder for closed strings and world sheets with the topology of a strip for open strings. To describe interactions of strings,
^First paragraphs of the section called "string perturbation theory"
If you look at the section (4.1.3, in Ammon & Erdmenger - Gauge-Gravity Duality), it very much looks like a perturbative expansion to me.
No, I don't see any non-perturbative object there.
The partition function (?)
19:24
4.49 is an empty statement - what was $\int_\Sigma$ supposed to denote if not $\sum_g \int_{\Sigma_g}$?
^that's the point (it defines $\int_\Sigma$)
Yeah, so you just defined the partition function as the sum over stringy Feynman diagrams
That's inherently perturbative - you rely on that sum being tractable to define Z
@ACuriousMind Okay, but then it's still fair enough to say that the lowest genus calculations are the lowest orders in this perturbative expansion (of something which is defined by its pert. exp.)
@Slereah It may bother you on lack of rigor grounds, but it still scores the single best theory/experiment agreement in all of physics measure in digits that agree. (g-2 for an electron)
@Danu Well, yes. But I don't like the word "expansion" here, it implies some other function is being expanded, it's not - that power series is the definition of what we want to compute.
19:27
@dmckee Well it wouldn't be used all that much otherwise!
@ACuriousMind Okay---I think we've reduced it to semantics ^^
I guess I agree with you; thanks for the comment (I'll be deleting my comments now)
Although a lot of QFT feels a bit
Stilted?
Some parts of QFT are just arbitrarily fitted to fit experiments it feels a bit
There's a big feel of a lack of a deeper theory
mostly when it comes to neutrino matrices or parity violation
It's just kind of there
Wait don't we produce enough parity violation in the Standard Model?
@Danu Actually, that definition has always bothered me. Apart from the analogy to QFT and our particles now being strings, there is no reason to write down that thing. It's totally ad hoc.
Well the CP violation one
Cabibo matrix?
Or is it Kobayashi
19:30
I think it's both :P
Cabibbo–Kobayashi–Maskawa
and more!
But I think there are at least already "theories waiting in line" for experimental testing as soon as we get there
the neutrino one is Pontecorvo–Maki–Nakagawa–Sakata
for CP violations and shit
@ACuriousMind I mean... It's useful in motivating AdS/CFT
This article is a good read: A Fight for the Soul of Science.
19:32
@JohnDuffield This was at my uni---some of the speakers are my teachers :D
I'm with Ellis, not with Carroll.
By the way @ACuriousMind your semi-philosophical view on "expansions" and when to use the word runs into trouble for simple things like $e^x=1+x+\dots$. Are you really going to say that one shouldn't use "expansion" for that? :P
@Danu Well, define $\mathrm{e}^{x}$ as the solution to $f' = f$ and not by its power series and that becomes a non-trivial fact :P
@ACuriousMind Lame game ;)
It's the exponential map from R to R
19:40
@Danu You're just mad because you didn't think of that!
2
@ACuriousMind In any case, whenever your terminology becomes dependent on switching out equivalent definitions, you know you're being a physicist ;)
Now, go wash your hands and keyboard, you dirty boy.
@Danu As a physicist, I have to be glad whenever I have an actual definition
heheh
the exponential is just ⊢ exp = (𝑥 ∈ ℂ ↦ Σ𝑘 ∈ ℕ0 ((𝑥↑𝑘) / (! ‘𝑘)))
That notation is the worst.
Yeah
They kinda had to pick a really bad notation to make it fit into the program
to avoid ambiguities
19:45
@Danu What are you talking about?
@Danu ikr
@0celo7 Read up.
19:58
I learned a new (to me) verb today. To brook. Great. Thanks!
Ooops
Not helpful @0celo7
Tfw your solid state laptop makes a horrible scratching noise
Fuuuuuuuu
@0celo7 I brook that.
20:01
@0celo7 Yiiikes :\
"tfw when" = "that face when when"?
That feeling when.
Oh.
20:02
I also brook that.
I'm obviously distraught.
Also hey @Keepthesemind I haven't seen you around here before---or are you the new alias of someone I know?
Also my shirt has these weird fuzzies on it
@0celo7 >laptop still made of solids
Step up your game man
@Danu Yes. I changed it when chat brought a faulty worded pop-up.
20:03
@Slereah ?
Now it's all gaseous laptops
@Keepthesemind So... who are you? :P
I'm saying I don't have a hard drive, so what the hell could have made a scratching noise??
@Danu FKA Glen The Udderboat, amongst others.
20:04
Fucking known as?
@Keepthesemind Ah, okay. I remember Glen ;)
@0celo7 I don't brook that. Take it back.
No.
@Keepthesemind lol
@Keepthesemind I will not brook your brooking much longer.
I don't like that word.
I know a person named "Brooks" who is probably the most annoying person I know.
20:06
You know what's amazing? Sphere fibrations: Fucking homotopy groups, what do they mean?!
@0celo7 I'm not brooking it myself much longer.
@Danu No, you know what's amazing? Your mom.
@0celo7 Her cooking skills are 12/10, it must be said.
Can't spell.
20:08
"ArtOfCode has invited you to join Trash." Wut? :)
lol
I'm trash :(
1 message moved to Trash
byebye, if you can't be nice then go away
I like this trash room :) No more (removed) messages! :D
@Danu Trash-talking...
@Keepthesemind It's what I do best.
20:11
@Danu Well maybe if the mods stopped removing every other message this wouldn't be an issue!
Is the trash a new feature?
Or did you mods only now discover the power of trash?
2
@0celo7 #DrunkOnPower
@ACuriousMind I must admit I was unaware of this...
@ACuriousMind Not only that... but also the power of Trashcan!
@Slereah Suppose $M$ is compact, even-dimensional and has positive sectional curvature. Then for each Killing field $X$ there is at least one $p\in M$ s.t. $X(p)=0$.
I'm not sure I believe that.
What about the angular vectors on the sphere?
20:17
Ahem, hairy ball theorem.
I assume the proof will end up mapping to the sphere and then using the hairy ball theorem.
@Danu No.
You define $f=|X|^2$ and use compactness to find a minimum.
You then show it's $0$.
@0celo7 I still suspect it can definitely be done like that.
@Danu Maybe, but it would be way harder.
Especially because this works on manifolds where the hairy ball theorem does not apply.
Yeah but it's obvious that the assumptions you're making are restricting it to be "almost a sphere".
@Danu But what about the compact manifolds with $\chi(M)=0$?
Hairy ball does not apply there.
20:20
Am I being silly or is the accepted answer on this totally wrong? physics.stackexchange.com/questions/248855/…
@0celo7 Do they have everywhere positive sectional curvature? I want to say no, by Gauss-Bonnet, but I forgot the def. of sec. curv.
Hmm, can a $\chi(M)=0$ manifold even have positive curvature?
^ lel
in terms of being "not much brighter than the brightest star in the sky"
@Danu What's Gauss-Bonnet in $2n$ dimensions?
Don't you need the Euler characteristic?
20:21
Google it
It's like $\chi(M)=\frac{1}{something}\int e$
Anyways, I'm sure you see what I mean now by my remark that you're always going to be on "almost a sphere"
@Danu Sure, in 2 dimensions.
@0celo7 No, in $2n$ dimensions.
But that's a harder proof than is necessary
@Danu I have no feeling for $3+$ manifolds.
20:22
@0celo7 Did you try it?
@Danu Not interested.
Ask your supervisor if the approach should work or not.
Just to prove me... wrong ;)
@Danu Currently avoiding him.
"sounds good"
What?
@Danu He would probably say "it's trivial if I had some time to think about it"
@Danu I think you would need higher-dim Gauss-Bonnet to justify excluding $\chi=0$ manifolds. So that's not fun
20:28
@0celo7 "not fun"? Gauss-Bonnet is da bomb
@Danu You need the Euler class dood
And you have to link the Euler class to the sectional curvature
You probably learned about the Euler class in your Chern-Weil class?
@Danu If I had Kobayashi and Nomizu I would accept your challenge and look for that proof.
We didn't do any (co)homology, so no.
We only did G-B for $d=2$, with generalizations to "manifolds with edges" and some other small expansions.
We had very, very little time for "big theorems" after all the preparatory work.
In mathematics, the generalized Gauss–Bonnet theorem (also called Chern–Gauss–Bonnet theorem) presents the Euler characteristic of a closed even-dimensional Riemannian manifold as an integral of a certain polynomial derived from its curvature. It is a direct generalization of the Gauss–Bonnet theorem (named after Carl Friedrich Gauss and Pierre Ossian Bonnet) to higher dimensions. Let M be a compact orientable 2n-dimensional Riemannian manifold without boundary, and let be the curvature form of the Levi-Civita connection. This means that is an -valued 2-form on M. So can be regarded as a skew...
The Pfaffian of the curvature form, oo-lala
21:13
Stupid question from a non-physicist: Is this an important breakthrough?
@SEJPM If it does what they claim then it is a step that was always going to be necessary.
@dmckee is such a paper usually a result of a very long time of work (like 1 year+)?
@JohnRennie I have the full proof of the closed geodesic thing on my blog
21:39
is there only one unique linear combination for a g.c.d?
@Obliv no
@SEJPM one year is not that much for a paper.
@0celo7 how would you get others? I know working the euclidean algorithm backwards gets one combination
wait
@0celo7 $(a,b) = ax + by$ are there other $x$'s and $y$'s to get $(a,b)$? that's more specific I think
@Obliv dunno, we proved the Euclidean algorithm in my algebra class but the prof said there's an infinite number of combs
@Obliv I know what you mean
ok
I feel that way too. Just not sure how to find the others :o
21:43
@Obliv dunno
@Obliv Now if they're relatively prime, I think there might only be one?
I don't even think there would be a linear combination between them
@Obliv wrong
think of 100 and 7. $(100,7) = 1 = 100x + 7y$
@Obliv $(a,b)=1$ iff $\exists r,s$ s.t. $ra+sb=1$.
yeah actually why wouldn't there be more than 1..
21:45
@Obliv use the Euclidean algorithm to find $x,y$
that ones guaranteed i'm going to try to find another one
lol
@Obliv take smaller numbers
yeah
2 and 3?
7 and 3 are good
or even better
$1 = 2x + 3y$
works when $x = 2$ and $y = -1$ and when $x = 5$ and $y = -3$
so i guess there are infinitely many maybe
21:51
@SEJPM A mere year isn't that long a time in experimental work. It's probably the result of years of steady build up in expertise and equipment sophistication.
22:05
@Slereah What the hell is a star-shaped neighborhood, anyway
could someone suggest a strategy for this please? ctrlv.in/737044
@JoeStavitsky Well, think about it - what has changed for the electron at 3cm compared to the electron at 9cm?
@ACuriousMind yea I saw it was a conservation problem ty
@ACuriousMind now I feel kinda dumb
22:21
@ACuriousMind just dropping in to say I'm still puzzled, but making progress.
Thanks again for the discussion yesterday.
23:14
Good evening peeps
@ACuriousMind What does one study diffeologies for?
hey.
@0celo7 I have no idea what "diffeologies" are
23:39
In mathematics, a diffeology on a set declares what the smooth parametrizations in the set are. In some sense a diffeology generalizes the concept of smooth charts in a differentiable manifold. The concept was first introduced by Jean-Marie Souriau in the 1980s and developed first by his students Paul Donato (homogeneous spaces and coverings) and Patrick Iglesias (diffeological fiber bundles, higher homotopy etc.), later by other people. A related idea was introduced by Kuo-Tsaï Chen (陳國才, Chen Guocai) in the 1970s, using convex sets instead of open sets for the domains of the plots. == D...
23:51
@0celo7 Oh, that. I think the main purpose is to get a notion of "smooth space" that forms a nice category of smooth spaces, in particular that has such things as quotients or internal hom.
1
Q: Symbolic dynamics of a multidimensional system

SKMLet $x_t = F(x_{t-1})$ be a discrete-time dynamical system in the chaotic regime. Starting from an initial condition $x_0$, we can generate a time series $(x_t)$ where $t =1,2,...,T$ indicates the time index. From this, we can generate symbols as follows: $s_t = 1$ if $F^t(x_0) > c$, and $s_t = ...

Temporarily locked question to avoid migration cf. meta discussion about on-topicness of math questions. Input to how we proceed in this question is welcomed. ...[June 15th, 2017: Unlocked.]

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