8:02 PM
@AccidentalFourierTransform I wish you'd comment this
It's a perfect representation of the problem

I have an excellent physics puzzle.
Suppose I you have a cone. Over the top of the cone you place a loop of frictionless, massless string. At one point on the string, you attach a mass.
At what opening angle of the cone will the mass cause the loop of string to be pulled up over the top of the cone and fall off?
The general relativity people should have an advantage on this one.
@JohnRennie

@SirCumference we are wasting our time

@AccidentalFourierTransform Yeah, I've given up

@AccidentalFourierTransform @SirCumference what's up?

0

When someone deletes an account, the upvotes they left on comments are not removed. When they create another account, they can upvote these comments once again. This is a problem, as you can give a comment as many upvotes as you want by: Upvoting the comment Deleting your account Repeating 1 a...

I dont get people

8:11 PM
@0celouvskyopoulo7 I'm not sure I follow your argument or where I'm supposed to help :D
@DanielSank I think I don't understand what situation you're describing
That is, in no scenario I picture after that description can I imagine the loop of string being pulled over at all, so I must be picturing it wrongly

You'll see it then

^^^
I thought that I got the picture, but the answer was obviously "the string is not gonna be pulled over at any angle"

Suppose the cone is very, very shallow.
E.g. 1 degree shy of being a flat disk.
You don't think the string will be pulled off?

@DanielSank Halp

In the picture in my head I don't see how it could be "pulled off" without ripping

8:21 PM
@AccidentalFourierTransform What do you not understand? They don't want the devs to spend time on this because they think it's not important.
Perhaps you disagree, but surely you see the logic.
@BernardoMeurer 'sup?

the sky, still

@DanielSank I need basic logic help because I'm being silly

@ACuriousMind Well, suppose the cone is really flat, i.e. a disk. Then surely I can move the loop around without it ripping, yeah?

haven't managed to knock it down yet

Suppose I have a CSV with 7 fields

8:22 PM
@ArtOfCode Keep trying.

and n entries
(i.e. a nx7 matrix)

uh huh
Use numpy.

No, no, ged demmit
This is C
I'm writing a parser

use numpy anyway ;)

8:23 PM
Hear me out

@DanielSank I agree with that, but that's not really their attitude towards the post. They are trying to argue that there is no problem at all, not that the problem is not important enough

Wait
No

there is a problem

I got this

8:23 PM
and I might agree that it is not really that urgent

@AccidentalFourierTransform Uh... I read the first few comments etc. and the feeling I got was pretty much "yeah, but this isn't that big of a deal".
> Deleting an account is not a trivial task, so going through that process over and over for this somewhat minor (and largely inconsequential) benefit would be really difficult to miss — assuming anyone would even bother at all.

> Perhaps if there was more benefit to up-voting a comment, it might warrant additional protections. But this is the kind of hypothetical problem you might want to see demonstrated in actual practice before deeming it a problem at all.

there are a couple of comments that say that

but most of them are just "no, not a problem"

Here's the first line of the second answer:
> While I agree this can be annoying and unfair, I don't think it justifies spending development time.

8:25 PM
I dont know, perhaps Im just focusing on the comments that I found unreasonable

^
Yep.

but, oh well, some of them are unreasonable :-P

ZOMG somebody on the internet is wrong!

my point was essentially that "Huge problem, maybe not. Something for developers to bear in mind, IMHO, yes."

I didn't see anything hostile.
I didn't look that hard though.
The responses all seem to be "yeah, you're right that this could happen, but it's not a big deal".

8:27 PM
whatever, not really that important

yep

back to the angle thingy

yep

I think I got the picture right
it seemed impossible for the string to be pulled over because I was imagining the cone way too narrow
but I agree that if its flat enough, it would probably be pulled over
so, what's the angle...
hmm

@ACuriousMind ^

8:30 PM
I guess that it should be 45º?

Why?

because that's what my brain told me

:P

argh, I think I have the right image in my head, but I cannot put it into words
I imagine the string to be completely horizontal
and moving the mass a bit downwards
and the other end of the string should move upwards
hmmm

@AccidentalFourierTransform I think that reasoning should work, but you have to remember this is a 3D problem.

8:36 PM
and the string is inextensible

@AccidentalFourierTransform correct
@AccidentalFourierTransform That's not what I got. How'd you get that?

The Hitchhiker's Guide to the Galaxy is a comic science fiction series created by Douglas Adams that has become popular among fans of the genre(s) and members of the scientific community. Phrases from it are widely recognised and often used in reference to, but outside the context of, the source material. Many writers on popular science, such as Fred Alan Wolf, Paul Davies and Michio Kaku, have used quotations in their books to illustrate facts about cosmology or philosophy. == Answer to the Ultimate Question of Life, the Universe, and Everything (42) == In the radio series and the first novel...

Ah yes.

How'd you get that?

8:41 PM
so its not that
I mean, it has to be a nice angle. If its not 45º, it has to be pi/6
does it have anything to do with parabolas and hyperbolas?
geodesics?

@ACuriousMind Lol, let me try to rephrase...

@AccidentalFourierTransform Why does it have to be nice?

@BernardoMeurer That's another place where using calloc is a win. Without it you have to worry about garbage present in any possible padding biting you i the ass.

@AccidentalFourierTransform Well, yes, since conic sections are those shapes.
@AccidentalFourierTransform mayyyyyyybe....

@DanielSank there is a policy, you know

8:45 PM
Of course you could also use a #pragma to force tight packing with the attendant disadvantages.

@ACuriousMind I have this function $u\in W^{2,q}_\delta$ satisfying $\Delta u=F\in W^{0,q}_{\delta-2-\tau}$. In this situation $W^{2,q}_{\delta-\tau}\subset W^{2,q}_\delta$. So I want to find a $v\in W^{2,q}_{\delta-\tau}$ such that $\Delta(u-v)=0$ outside of the ball $B_R$. Now, I know that $\Delta :W^{2,q}_{\delta-\tau}\to W^{0,q}_{\delta-2-\tau}$ is Fredholm. So its cokernel is finite-dimensional. That means there are finitely many l. i. functions not in $\Delta(W^{2,q}_{\delta-\tau})$.
Agreed?

@AccidentalFourierTransform good point

@ACuriousMind $\Delta =\sum D_iD_i$, of course.

@DanielSank I dont think I will be able to solve it
and goolge is not helping :-P

Guys, why do we need a spherically or cylindrically symmetric charge to have a net field of zero? Gauss’ law doesn’t care about the geometry of the exterior charge, right? As long as the exterior charge isn’t enclosed by your Gaussian surface, you don’t have to worry about its electric field, right?

8:50 PM
@0celouvskyopoulo7 All these indices on the $W$ hurt my eyes

who was a better mathematician, Gauss or Riemann?

it's not a contest

yes it is

@ACuriousMind Well $q$ is the $2$ in $L^q$, $2$ is the number of weak derivatives, and the other index is, well, something controlling the strength of the weight "at infinity"

@ACuriousMind with respect to what operator is the vacuum cyclic

8:52 PM
oh shit, I can't delete it anymore
I misinterpreted it

and it shall stay there forever

@Slereah I don't know what it means to be cyclic "with respect to an operator"

I guess algebra?

@ACuriousMind So...?
Does the setting make sense now

@0celouvskyopoulo7 Whatever they are, I have a hard time parsing that statement because I have to keep comparing indices :P

8:56 PM
I want a $v\in W^{2,q}_{\delta-\tau}$ with $\Delta v=F$ in $R^n-B_R$.
@ACuriousMind Sorry I've been reading these weighted Sobolev papers for a week now :<

dude, use multi-indices
"let $i$ stand for everything I might need"
@DanielSank so, are you gonna tell me the answer or not?

actually, I still don't understand what they mean. So we have a point and a charge at distance $r$. Is the net field equivalent to the flux through an infinitesimal box around the point? But then outside of every charge the net field would be zero, which can't be correct. But how can I say something about the net field then with Gauss? Gauss only gives flux:(

@Slereah Okay. So I guess you're confused about what cylicity means because in an irreducible representation, every vector is cyclic. It's not a property of the vacuum as such

@ACuriousMind whaaat
Every AQFT paper bangs about how the vacuum is the cyclic vector par excellence

thats because in AQFT there is only one vector, the vacuum $\mathcal H=\{|0\rangle\}$
it makes things really simple

9:03 PM
@AccidentalFourierTransform uhhhh
I don't usually give answers to puzzles.
It's not like you have to solve it right now.

@Slereah Well, I don't know about the sort of AQFT you're reading but it's a general fact that every vector of an irrep is cyclic - if a representation contains a non-cyclic vector, then the action of the algebra on that vector generates a true subrepresentation, contradicting irreducibility

Is that related to the selection rules

according to encyclopediaofmath.org/index.php/Cyclic_vector a vector is cyclic with respect to a certain operator. It cannot be cyclic by itself

9:11 PM
@Slereah Superselection rules are basically physicist speak for "the representation is reducible", yeah

Ah yes

Now, if your representation is reducible, then you can have both cyclic and non-cyclic vectors, and I guess the vacuum should be cyclic so that every state can be "reached" from it

So basically every reducible part needs to have a cyclic vector?
One vacuum per sector

effin communists

9:13 PM
@Slereah Well, given a direct sum of irreps, you get a cyclic vector for the sum by just choosing the direct sum of non-zero vectors from these representations
There are non-cyclic vectors here - those whose projection to at least one irrep is zero, but being cyclic still doesn't strike me as all that special

Lemme see if i can find the wording

@ACuriousMind So the idea is this: Suppose there is no $v\in W^{2,q}_{\delta-\tau}$ with $\Delta v=F$ in $R^n-B_R$. Then I take $F$ and change it in $B_R$, and produce a whole infinite dimensional space of functions that $\Delta$ misses outside of $B_R$. But then the cokernel is no longer finite, so that's a contradiction.
The question is if I can modify $F$ like that and actually stay in the Sobolev space. And I don't know about that...

"The Reeh-Schlieder Theorem shows that the spectrum condition entails that the vacuum vector is cyclic for every local algebra."

It's not the usual Sobolev space after all..

for instance

9:17 PM
I'm afraid I don't see how that question has nothing to do with the Sobolev space or how my algebra can help
@Slereah ohhhhhh

it is but one instance because they talk about $\Omega$ being cyclic all the time

Yeah, the special condition here is that you have one space and there's an infinitude of different algebra representations on it (all of these local algebras are different, after all). The special condition is that it is cyclic for all of these.

Ah yes

Being cyclic for a single representation is not special. Being cyclic for so many is.

"In the early twentieth century, it was thought that the referent of T must be a set of axioms of some formal, preferably first-order, language. It was quickly realized that not many interesting physical theories can be formalized in this way. But in any case, we are no longer in the grip of axiomania, as Feyerabend called it"
But I love axiomania :(
How is AQFT supposed to help, anyway
You have several representations for the same Hilbert space
So you only work with the representation directly
But does that really make it unique?
Are we guaranteed that there will only be one theory for a set of rules in AQFT

9:24 PM
@Slereah Pretty sure that's not its goal.

But then why

It's just supposed to axiomatize QFT in a rigorous manner.

Since it's always presented as the solution to the unitary inequivalence

But there are many QFTs, so why would you expect uniqueness?

But why AQFT and not another theory based on a Hilbert space, then

9:25 PM
@Slereah Oh, it evades Haag's theorem by the axioms being so strong that you don't need the interaction picture

What axiom makes it unecessary?

Well, I don't know which of the axioms you're looking at, but I think it essentially assumes that you "have" all the interacting operators already.

Pretty tall order

It's a formal definition of a QFT
Just like Lagrangian or Hamiltonian mechanics. You shouldn't expect any specifics from it

I wouldn't if it wasn't put on such a pedestal!

9:28 PM
And yes, both AQFT and FQFT as axiomatization suffer from the axioms being so strong that we don't know how to produce SM-like theories that fulfill them

wtf is FQFT
what will they think of next??

Well I've seen a few interacting theories based on AQFT
But they're not pretty to look at
functiorial QFT, @0celouvskyopoulo7

@0celouvskyopoulo7 Functorial QFT, it actually works for TQFT, not so much for actual QFTs :P

I thought AQFT was functorial

Also the interacting AQFTs still need renormalization anyway
it's all a ruse

9:30 PM
@0celouvskyopoulo7 The "functorial" here doesn't mean "functors occur in it", but that you essentially define a QFT to be a certain functor on a cobordism category.

@ACuriousMind Hmmm

Well QFT is also a functor in AQFT
The old presheaf

How about I multiply by $1+\phi_\epsilon$, where $\phi_\epsilon$ is nascent delta
that should work
Moving on

I think it's time 4 bed

I think it's time 4 a snack

9:33 PM
@Slereah oh no
"A smooth $n$-dimensional mainfold $(M,g)$ with Riemannian metric $g\in W^{1,q}_\mathrm{loc}(M)$"
How does life even work when the metric isn't smooth

10:03 PM

10:31 PM

So, funny thing about this. The votes themselves are actually deleted. What we don't do, apparently, is update the denormalized Score stored on the comment itself. Regardless of whether or not this is a practical problem, we should probably do that if only for consistency's sake. Having said that, doing so has a caching implication in the API that I'm not sure how to handle off the top of my head (or if it's a real concern - I'm not familiar with that part of the code that much). Gonna have to think about it some. — Adam Lear ♦ 36 mins ago

There is hope

10:49 PM
I passed graduate E&M! I now never have to do electomagnetism again

@JohnRennie I need your code review skillz when you're available
I did some serious brouhaha

hides in bubble where this is how things work

Also, congrats :^)

@GPhys It's a comfy bubble.

o/
Finally submitted to Bonn. I can rest now.

10:58 PM
Nice, GPhys. Though you might have to work on it again, if you plan to keep doing physics :D

Hi, everybody.
@Danu solve the cone problem.

@DanielSank Rytsas

@Mithrandir24601 What?

@DanielSank High Valyrian equivalent of 'hello'/generic greeting :P

The hell is Valyrian?

11:06 PM
@DanielSank Game of Thrones language

I am confused.
I would have thought Mithrandir24601 would speak Quenya or something.

@nbro Something like
> Given that w know that chat has been headed about 1.8 degrees off of the direct bearing to hell at 72 posts per hour for the last 16 hours, determine how much closed to total burnination is the h bar now that at the beginning of the measurement.
perhaps?

Don't know why I never tried learning that... But I didn't :/
Although it's tempting...
But I have actual work to do over the next few months

Sorry what's up Daniel?
I'm on my phone atm

...why are you using SE chat from your phone at 1:20 am? :D

11:20 PM
Why not?
When playing games it's convenient to do it cuz Skyrim crashes when you tab out

@ACuriousMind rebooting laltkp.
Laptop

lol
@ACuriousMind So I have $\Phi:M-K\to R^n-B_R$ a $C^\infty$ diffeo, where $K$ is compact in $M$. This gives coordinate functions $x^i$ on $M-K$. Am I crazy to think there's no particular reason those functions should extend to all of $M$? $M-K$ isn't closed.

Uh...is the sphere w/ stereographic projection not a counterexample?

@ACuriousMind Mmmm, not quite? Stereographic projection goes to all of $R^n$

@0celouvskyopoulo7 Well...remove both poles (or rather, one pole and a small disk around the other), then the target is $\mathbb{R}^2$ minus a small disk, right?
(You didn't stipulate that $K$ be connected)

11:28 PM
@ACuriousMind Lol...it should be morally connected.
But I see the point.
Can you find a connected counterexample?
I can think of weird situations where there's no extension, sure
But none arising in this context

I'm having trouble thinking of a counterexample where $K$ is connected.

What's the question?
To find maps that DON'T extend?

yep

@Danu A paper says to "take an extension" and I don't think that's always possible.

what is $M$?

11:36 PM
A manifold.

not closed or anything?

Noncompact.

and what should the extension satisfy?

It should just be a smooth extension of the $x^i$, nothing special.

So why don't bump/cutoff functions work?

11:40 PM
My immediate reaction is because $M-K$ is open, but since it's covered by one chart it might work...

always first try cutoff functions
:D

Sure, but it's not obvious to me how to do it. Because $M-K$ is open, the $x^i$ can get very bad as you approach the boundary.
That was ACM's point above.
With the sphere example, the functions blow up near the pole.
So connectedness would have to be necessary, if it's true.
Bump functions let you piece together local extensions into a global extension. But for the sphere example, there are no local extensions. My conclusion is that this guy needs to state his definition more carefully.

Also in your sphere example $M$ is compact

@Danu $M$ is not explicitly noncompact, but in applications it should be

ok
solution strategy one: disregard non-compact manifolds
:D
enough trolling

11:53 PM
@Danu I wish!
Noncompact manifolds are horrible
It's a solid reason to disregard GR