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obe
obe
00:00
I used kittel's solid state physics.
solid state is awesome though
do QM dammit
get your shit together
obe
obe
right, fail.
especially once you get into modern condensed matter
00:00
your mom is a modern condensed matter
fite me irl
I'm down
you know where I'm at
that's true I do
the building with the centipedes, 8AM
but can you come here instead
00:01
or 9AM
I don't have a car
me neither
right now
next year then
great Windows defender tells me I have malware but won't tell me where it's coming from
I haven't even done anything shady
should I be worried?
I'm new to this PC thing
"haven't done anything shady" he says
snickers
eats snickers
obe
obe
Download a trial of an antivirus.
why won't it tell me where I downloaded it
00:08
Does anybody know if Reimer's law is the current accepted mass loss relation in astronomy?
@obe just ask your questions here
there's no shame
obe
obe
I will.
ask away then
obe
obe
I just started ch3.
Why is the christoffel symbol defined the way it is?
pretty sure he explains that
in good detail
obe
obe
00:13
Right, I didn't read that far yet.
because babies
+10 for reference get
@obe read his explanation and see if it makes sense
if it doesn't then we can try to explain
obe
obe
What page is it explained in?
@0celo7 jeez, what book are you talking about now?
obe
obe
Whatever I'll find it myself.
00:15
@NeuroFuzzy Relativity on Curved Manifolds
it has some stuff that HE was unclear on
@obe hmm he doesn't seem to explain it the way I would like for someone learning GR for the first time
he basically does it the way Wald does it
which is by defining covariant derivatives abstractly
but that doesn't necessarily offer much physical insight
the best way to define the covariant derivative is Straumann
@0celo7 Oh, that looks really good.
@NeuroFuzzy It's hard.
obe
obe
@FenderLesPaul I'm on pg95 and so far it's ok.
00:18
@obe basically the point is that you need a way to compare vectors that live in different tangent spaces
in flat space this means we need to transport a vector $V = V^i e_i$ from one point to another in some coordinate system
obe
obe
in 3.6 that is for a (1,1) tensor
what would it be for a different type of tensor?
0
Q: How to suggest an edit to fix a 1-character typo

PaulI tried to make a suggested edit for a post1 that was otherwise great, but the OP had made a single-character typo. When I tried to correct the typo, I was told that the edit needed to be 6 characters. As it was, the OP had an image there with the default [image description goes here] text, so I...

but $e_i$ can vary with the coordinates so the $\Gamma^i_{jk}$ take this into account as you consider $DV/Dt$ along some curve $\gamma(t)$
@0celo7 Yeaaah and I'm not adding anything else GR to my todo list. But it does look fun.
obe
obe
what do you make "takes this into account"?
00:21
crap my internet is spazzing out
BRB, installing ooboontoo.
erm, ooh-bun-two
yay virus free
@obe it might be easier to just read it in a book
read section 20.4 of Hartle
obe
obe
ok.
I understand 3.8 but not 3.7
@FenderLesPaul what is wrong with sect. 15.2 in Straumann
obe
obe
00:23
why is there a second $V^{\nu}$ term?
@0celo7 :p
oh that's a fighting face
raises fists
obe
obe
fight later, help me first.
stay outta this child
go eat your broccoli
obe
obe
00:25
::turns head down and walks away::
@FenderLesPaul throws Abramowitz and Stegun
pats obe
@0celo7 throws Weinberg Vol 1
it was super effective
@obe one sec sorry
that was low
@obe which term is confusing?
obe
obe
3.8 makes sense to me.
3.7 does not.
you have the first one from the partial hitting the fraction
then the second one from the second term in the first equality
obe
obe
00:30
huh
remember how the partial derivative of a vector transforms
the first equality. the first term turns into the first two terms in the second line, the second term in the first line is the last term on the second line
obe
obe
i do.
that's what I don't understand
why does the first term turn into the first two terms?
specifically the second term.
it's either idk how the second term forms or idk how the first term forms
why is the first one transforming like a covariant derivative?
because the transformation of a partial derivative acting on a vector
remember that gives you two terms
(Zee makes a big deal out of this)
obe
obe
hmm
though i remember from 2.18
that it's only one term.
no you need the one acting on a vector
I thought we did that yesterday
2.38
obe
obe
00:34
right.
fail.
2.18 isn't a vector?
failure is the road to success
and success is the road to Insomnia cookies
mmmm Insomnia cookies
no that's how the operator transforms
well partials are vectors
ok Carroll is shit
I would like to say Straumann defines this all very nicely using parallel transport
@FenderLesPaul You have a problem. ;)
obe
obe
I think I sort of understand it.
@ACuriousMind maybe you can explain this to the lad
obe
obe
00:37
It's ok I'm re-reading after ch3 to fix my issues.
@0celo7 to me?
unless you understand it of course
@0celo7 My "explanation" of Christoffels is that they are a $\mathfrak{gl}(n)$ gauge field :P
obe
obe
idk if I do
@ACuriousMind this is true
@ACuriousMind well yes
obe
obe
00:39
@ACuriousMind I'm drowning now.
I need a good fiber bundle book!!!
dives in to save obe
Halp!!!
obe
obe
00:40
@skillpatrol Sorry pal if you're patrolling for skill here, now is not a good time as their is no evidence of skill in my questions.
Christoffel symbols are very nicely explained here books.google.ie/…
"viewing limit"
how is that calculated
@0celo7 Just reload
00:42
wtf
that's so dumb
GTM Fibre Bundles should be interesting
@0celo7 nice!
That probably took a while to draw in TeX
beautiful
00:51
> 7. Functorial Description of the Homotopy Classification of Vector Bundles
ok wrong book
all I want is a good description of how the connection form is intrinsically defined
@ACuriousMind ?
Can you help me fix the following awesome Lie-representation interpretation of solving the Klein-Gordon equation, based off pages 438 and 441 of Woit
http://www.math.columbia.edu/~woit/QM/qmbook.pdf
?:
@0celo7 Have you read the Wikipedia article for an Ehresmann connection
I give you a manifold (Minkowski spacetime) and it's Lie group (Poincare group). I find it's lie algebra (the Poincare algebra), which I view as a representation of the Lie group, build it's universal enveloping algebra (representation), and construct a Casimir operator (an element of the center of the UEA, i.e. the Simplest Group-Invariant Differential Operator).
@ACuriousMind I think so
I'll read it again
This Casimir is called the Klein-Gordon operator, and setting up an eigenvalue equation for this Casimir give us irreducible representations of the group, which are lie algebra elements. By going to dual space we can turn this Casimir eigenvalue equation into a PDE which can be solved normally, giving us irreducible representations of the dual Poincare group.
Now the functions $\psi(x), \pi(x)$ and their commutation relations provide a unitary representation of a function on the dual poincare group. By combining those $\psi(x), \pi(x)$ representations of the dual Poincare group in a certain way and integrating over all space we find a representation of the original Poincare group (this is the Hamiltonian), and this has something to do with Intertwining operators (i.e. equivalent representations).
Now for some reason the Hamiltonian (representing the original) is in the same space
as the $\psi(x), \pi(x)$ which represent the dual.
It's basically putting formal mathematical words on each step of working out the Free Klein-Gordon in QFT
01:00
@bolbteppa Your terminology seems a bit off - irreducible representations are not "Lie algebra elements" (although they may be classified by their weights, which are duals of the Cartan subalgebra), what is a "unitary representation of a function on the dual poincare group" meant to say, and how "is" a Hamiltonian a representation?
what are those arrows
@0celo7 Something that appears because MathJax doesn't play nice with multi-line posts.
@ACuriousMind well on page 438 Woit says Field operators $\psi(x), \pi(x)$ and commutation relations between them provide a representation of a Heisenberg Lie algebra, a "unitary representation of a Heisenberg Lie algebra on $M \oplus R$, where M is the dual of the space of solutions of the Klein-Gordon equation", so $\psi(x) = \dfrac{1}{(2 \pi)^{3/2}}\int a(p)e^{i \vec{p} \cdot x} + ..$ (p.438, eq. 41.11) is a representation of a function over the dual of the Poincare group right?
Yessss for once in my miserable little life a dual OS install went off without a hitch and for once I have all my data backed up properly.
@NeuroFuzzy ::runs 2 OSs::
welcome to the club
ACM15sok00L is an "excellent" password for 1Password
01:14
I just moved everything to an ssd so my ubuntu boot time is a few seconds :D I should time it.
classes start in a week
omgbbq
I can't wait to make new frieeends aaand play hide n seeek
aaaand eat lots of ice cream
@ACuriousMind that wiki article is familiar to me
is this your final undergrad year?
@skillpatrol yep!
and then I can pursue my career in music
01:18
what I really want is an explicit construction of the connection form I guess
do mathematical music
write a book on GR topology in music
the music of the spheres :P
lol
that might be before his time
I don't get the reference :(
sucks to be you
Have you guys looked at this book amazon.com/The-Topos-Music-Geometric-Performance/dp/3764357312 he does music with topology and sheaves, wtf...
01:20
Musica universalis (lit. universal music, or music of the spheres) or Harmony of the Spheres is an ancient philosophical concept that regards proportions in the movements of celestial bodies—the Sun, Moon, and planets—as a form of musica (the Medieval Latin term for music). This "music" is not usually thought to be literally audible, but a harmonic and/or mathematical and/or religious concept. The idea continued to appeal to thinkers about music until the end of the Renaissance, influencing scholars of many kinds, including humanists. == HistoryEdit == The Music of the Spheres incorporates the...
you should have been no-lifing this place for longer then
:2354266 are you jk or do you not get it either
that's not it
@ACuriousMind can you use your chat memory + 20k powers pretty please??
The topos of music
I want that to be the title of my EP
01:22
track 1, song 1: Simply disconnected
I'll just say most of your favourite albums were done with Pro Tools, fact...
@skillpatrol like a debut album
1
Q: Why are the quantum observables defined on opens sets a presheaf and not a sheaf?

yessIn local quantum field theory one can mathematically describe over each open set $U$ of a spacetime $M$ the quantum observables of the theory. This structure is commonly referred as a presheaf. Why are the states over the open sets not a sheaf structure?

askin the real questions
screw that guy
he could be doing things like curing cancer
physics nerds are so selfish
user54412
01:24
@0celo7 At the largest state schools, how else are they going to teach half the incoming freshmen some basic subject? Often the prof lectures in one hall and they pipe the video into other lecture halls. Of course these days they might just say "don't go to class; download the lecture here."
Feb 21 at 18:08, by ACuriousMind
I want to listen to the hypersphere. That is my goal. We can hear sums. So if I add up every single point on the sphere I can hear it. Those graphs are the 3d one. The equation to number 5 is shown above. — user1698948 21 mins ago
@ChrisWhite but, 5,000 really?
@ChrisWhite I guess my school isn't that big, only 27,000. Aren't some upwards of 100,000?
the army of TAs lol
idk about 100,000
"you take this 100"
01:25
but Arizona state has like 60k
"oh these guys look smart"
5000 in one class?
that is the claim
hmm I thought some were a lot larger than that
@bolbteppa Hm, the $\psi$ is an element of $M\oplus R$, I think. The $M$ is the sum of duals of representations of the Poincare group (since it's the sum of duals of solutions to the KG eq.), so $\psi$ is also an element of the representions, not a function on it, I think, but I would have to read more of that thing to retrace your thoughts. I guess we're quibbling over terminology, anyway.
01:27
Ivy Tech Community College has 100,000
2 hours ago, by Chris White
Perhaps some prof linked to it in a class with 5000 students?
user54412
@HDE226868 I actually haven't heard of it, and it seems to have neither a Wikipedia entry nor any mention in the big orange tome of astrophysics. I'm not saying it's not a thing, just that it's niche.
1,400 pages!
$170!
01:30
What the heck is ivy tech community college?
Not as bad as MTW I guess...
sounds like 3 different colleges in one
@FenderLesPaul Baby Ivy League.
user54412
@HDE226868 What I can tell you is that stellar winds are hard. You can come up with basic models that get basic results (for instance predicting velocities at infinity to be about the escape velocity from the photosphere, in I think the optically thick limit), but reality is much more complicated.
Reality is an enigma
01:31
Enigma is reality
user54412
@HDE226868 One of the biggest problems is that many winds are line-driven -- the momentum transfer is due pretty much entirely to resonant absorption by e.g. iron species. Such winds depend sensitively on the strengths of the lines and the ionization state of the gas, in addition to the usual Rayleigh-Taylor instability and other (M)HD concerns.
you sure do know a lot about this space stuff
Does anyone here use a password manager to generate and store secure passwords?
My father uses 1Password.
user54412
My critical passwords are known only in my head. Everything else is written in plaintext files or on scattered pieces of paper.
user54412
(don't tell the IT people who run the clusters)
01:40
::closes gmail::
Ok...
@ACuriousMind I'd like to quibble over terminology haha, and I'm most likely wrong, but what I mean by calling it a function is that it looks like the Fourier expansion of a function, i.e. a continuous linear combination of irreducible representations of the dual, in the way that functions $e^{ax}$ form a representation of the additive group of the reals & a fourier integral takes the place of this integral, e.g. $\mathbb{R}$ is the space of solutions $e^{iax}$ just like $M$ is for Poincare
user54412
Haha -- unless you sent from a known address and encrypted the email with a trusted public scheme, they probably wouldn't read it anyway. Those people live in their own little world.
They won't read an email from a univ.edu account? Really?
For real, how do sheaves and presheaves help anybody?
@bolbteppa Hmmm...I guess $\psi$ is a "function" (or rather, distribution) on $M\oplus R$, since the $a(p)$ and $a^\dagger(p)$ are elements of $M$, if I understand this correctly, and they are Fourier transformed here. As a distribution, it acts linearly on $M\oplus R$, and hence from the commutation relation it is the representation of an element of a Heisenberg algebra acting on a representation of duals of the Poinvare group.
bah, distributions
01:51
Sheaves are things you do algebraic geometry on just like manifolds are things you do differential geometry on, but on a manifold you define a global manifold and then go down to local open sets, while sheaves let you define local stuff and string it all together to define something global, e.g. analytic continuation and stuff
@0celo7 They are a very general way to understand what "attaching local information to a space" means, and they appear naturally in almost every geometrical setting (e.g. every complex manifold carries the sheaf of analytic functions), and many statements about "finding a global section" or somesuch can be seen to be cohomological statements through them.
In a way, they also are the proper generalization of line and vector bundles
what is "somesuch" in German?
I've never seen anyone besides you use that.
ahnliches?
with umlaut
@0celo7 derartiges, I think
It's an English word, though: en.wiktionary.org/wiki/somesuch
I know it is, but not common.
indeed you are the most frequent user in this chat
@ACuriousMind thanks yeah I think that's what's going on, will think about this overnight, thanks for the help
user54412
01:56
@ACuriousMind Those definitions! "some such thing" "some such"
It seems somesuch surged in popularity in 1986 and is now declining.
"some such" was on a plateau but is now declining.
It's a cool word.
user54412
pics or it didn't happen
user54412
(yes I'm that lazy)
@ChrisWhite Yeah, I don't know what that page gives someone who didn't know the meaning beforehand
01:58
Overall, @ACuriousMind's spelling is disfavored.
user54412
"somesuch" sounds like "an sich" to me -- does anyone use the latter outside discussing Kant?
"an sich" is German, no?
Er hat das an sich getan.
Yes, an sich is German, and used outside of discussing Kant here :P
@0celo7 That doesn't make sense
@ACuriousMind well it does to me
go play with sheaves
"he did that to himself"
@0celo7 That'd be: Das hat er sich selbst angetan. "An sich" roughly means "on its own" as in "That wouldn't be bad on its own, but..." - Das wäre an sich nicht schlecht, aber...
02:03
hmm, that makes sense to me too
I think your German is rusty, pal.
user54412
@ACuriousMind I'd prefer to keep my image of Germans always discussing Kant in every conversation.
@ChrisWhite Sometimes we also talk about Nietzsche ;)
wow a 16 digit alphanumeric caps varying password is "good"
I'd hate to see a great password
user54412
You probably need non-printable characters for that.
correcthorsebatterystaple
02:08
^
well I put in a 16 digit password and I get the error "password must be at least 8 digits"
notallclosedformsareexact
@ACuriousMind would appreciate that one
user54412
Have you ever seen the security.SE folks get up in arms about passwords that can only draw from 60 or 70 characters? It's kind of cute how little they understand.
how little they understand?
user54412
@0celo7 they think that by denying them obscure characters, you deny them strong passwords, regardless of how long you let them make the passwords
02:10
@ChrisWhite Unicode for all!
@ChrisWhite seriously?
@0celo7 I'd rephrase that as NonVanishingDeRhamCohomology ;)
@ACuriousMind yourfaceisanonvanishingderhamcohomologyalsoscrewsheaves
@0celo7 Screwing sheaves could be fun, there even are perverse sheaves
lol
02:14
for pervs :P
I remember HE saying something about certain spacetimes being perverse
they're just so not nice
user54412
Are perverse spacetimes correlated with naked singularities?
Good one.
Perhaps they also have weird bound states.
@ChrisWhite warning: do not google that
02:17
Wow, you really are into BDSM.
Well if you've got an 8 character long password generated from only letter, there are 26^8 possibilities. If you allow alphanumeric characters, it jump to 36^8; adding case sensitivity, you get 62^8; adding punctuation, you get 94^8 possibilities
is that a lot
So there is, in some sense, an argument that increasing the allowed character set would increase password possibilities
just use 16 characters then
@0celo7 I just enjoy making your mental image of me as weird as possible
02:19
@obe do you need anything before I either sleep or read or etc.
But $94^8\sim6\cdot10^{15}$, which probably can be broken in a few minutes
@ACuriousMind I'll be in the Heidelberg area in December!
(if that)
Maybe.
I'll be in Germany and Germany is not that big.
@0celo7 Around Christmas or earlier? (Since I'll not be here around Christmas)
02:21
@KyleKanos How do you break a password like that if you get locked out after 3 tries?
There is that
@ACuriousMind I'm going for 2-3 weeks, Christmas included.
I'm not a white hat, so I cannot say
Where are you going over Christmas?
Nor am I a black hat, for that matter
02:21
It might be closer to where I am.
@0celo7 Home, to Wuppertal
I'll be 18 so I can drive? Maybe?
IDK about German tourism driving laws.
user54412
@KyleKanos I blame the 8-character limit. No reason to restructure every level of an application just so you can include 0x07 as input or something bizarre.
@ACuriousMind I thought that l was a ! and you were really enthusiastic about Wupperta :D
Oh I'm not driving 3 hours to see you.
On the other hand Hberg is an hour, that I'd do.
Yeah, it's a bit far from Heidelberg (which is why I don't go home all that often)
02:24
I'll be in the K-town area.
Also Stuttgart, Munich.
user54412
Europeans' conception of "a bit far"...
Then excursions into Czechia, Slovakia.
Oh, Munich, you could see @Danu!
Austria.
Frankfurt.
We have fam/friends all over.
Frankfurt is 2:30 to Wuppertal.
So if we fly in to Frankfurt, then I could drive up...that's a bit much.
@ChrisWhite We've settled our land a bit more densely than you have :P
02:28
My dad lived 4 hours away from home while in college. Every weekend he made the trip back so he could work on the farm.
@ACuriousMind you're a slacker.
Always have been :)
Always will be?
Probably
@ACuriousMind When are you going to go home? If I have exact dates, I will seriously try to make it work.
Heidelberg can be a cool day trip from K-town.
@0celo7 Hmm...probably on the 20th of December, no earlier, perhaps one or two days later
02:30
ooh, that might work if we come in via Frankfurt.
I should be writing by master's thesis by then, no idea what kind of presence here that will require
But if we fly in via Vienna, we'll do the Munich, former bloc stuff first.
The problem is that you'd probably intentionally mismatch your socks.
1
Q: Distinction between state space and space of functions

user1620696In Quantum Mechanics a particle is described by its wave function $\Psi : \mathbb{R}^3\times \mathbb{R}\to \mathbb{C}$. In that sense, the state of a particle at time $t_0$ is characterized by a function $\Psi(\cdot, t_0) : \mathbb{R}^3\to \mathbb{C}$. The space of all functions like that which a...

@ACuriousMind I want to say Stone's theorem has something to do with it.
@0celo7 Well, one could talk about Stone-von Neumann here to say that it really doesn't matter which space you take as long as it has the canonical commutation relations on it
Like the space of $\Psi$ is constructed by doing $\langle x|\psi\rangle$ as usual and this is an isomorphism because $\langle x|$ is unique?
wtf
I did not see those answers
that's majic
@0celo7 I rather think it is an isomorphism because it identifies $\langle x_0 \rvert$ with $\delta(x - x_0)$, so it is an isomorphism between the larger parts of the Gel'fand triple simply because it sends a generalized orthonormal basis to a generalized orthonormal basis
02:39
Ok, whatever that means :P
I need functional analysis
@ACuriousMind huh, so this happened in the centipede building
obe
obe
@0celo7 Right now no, thanks though.
I'm reading QM, and it's straightforward.
I'll look at that section in Straumann later today.
later today?
you need to sleep!
$\text{no sleep}:\text{person}\mapsto\text{vampire with mismatched socks}$
obe
obe
I will eat insomnia cookies until I finish QM.
actually scratch de Felice from the list
his approach is really complicated
obe
obe
Ok.
02:56
also extend the chapters of Gougoulhon to 1, 14, 15, 16
obe
obe
@0celo7 Frames or 3+1?
Frames

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