The h Bar - 2016-06-06 (page 3 of 3)

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6:00 PM

But still, in this case, the string still corresponds to a location in $R^3$, no?

@Slereah What do you mean by that?

That for a value of $\sigma$, $X^i(\sigma) \vert \omega \rangle$ is in $R^3$

...no

Why not?

That's an operator applied to some state, not a number

6:02 PM

Well the expectation value

jeez

Don't be obtuse :V

Ah, well, then it's in $\mathbb{R}^{1,25}$.

Since that's the target space of the $X$.

Don't be obtuse and don't forget I'm not talking about string theory :p

Then it's in $\mathbb{R}^{1,d-1}$ :P

There we go

I'm not quite sure out of the box how to turn this into the probability of the string to be restricted to a bounded region of space, but it feels wrong somewhat

On the non-physics related front, I've just put a shiny new SSD into my new laptop and installed a clean copy of Win7 and by God it's fast!

6:06 PM

@JohnRennie Truly a great civilization

I guess it should be something like there's some state such that... $\int_\Omega dx \langle X^i\vert \omega \rangle = 1$???

@BernardMeurer they wore socks with sandals - nothing more needs to be said.

@Slereah What's $\Omega$? Don't forget that $X$ is a function on the worldsheet, so you can't integrate it over regions in $\mathbb{R}^{1,d-1}$.

@JohnRennie Socks with sandals is so comfortable, but it's supposedly a more effective contraceptive than condoms

Hm

6:08 PM

I wouldn't know, I've never worn a sock on my willy

@JohnRennie Haha, I mean socks and sandals on your feet are more effective than a condom on your, ehm, willy

Once measured though, you can assign a position to the string state, no?

I see we are once again using the close reason unclear what you're asking as a proxy for this is a really rubbish question:

-1

Q: How can the Moon's moving away from Earth phenomena explained through gravitational waves?

I'm confused with the connection of Gravitational waves and the phenomenon of the moon moving away 3.8 cms per year. As per the concept of gravitational waves and therefore, the moon orbiting being explained as a phenomenon caused by the displacement of time space caused by earth, doesn't it impl...

orbital-motion gravitational-waves celestial-mechanics moon tidal-effect

And more importantly, assign positions where the string is not

So there should be a procedure to see if the string has a probability of 1 to be within a finite volume, no?

@Slereah I suppose you can measure $X(\sigma,\tau)$ - but what are you to use as "the position" of the state?

6:15 PM

Well I guess that, for the whole span of $\sigma$, the position(s) are the expectation values of $X$ for every value of $\sigma$

I don't know if any string state in the Hilbert space is such that it will be entirely contained in a bounded volume, though

I assume that the eigenstates of $X^i(\sigma)$ are not in the Hilbert space?

Probably not

Not sure if there's a state where you can restrict it to a compact region

@JohnRennie no, not the dogs of war

I was gonna say "Maybe in some $[0,1]$ region?", but then that would violate translation covariance for Malament

what on earth

is @Slereah doing string theory???

6:28 PM

Well it's only chapter 1 of Polchinki

The one I've been on for the past year

@BernardMeurer damn right they are

In non-relativistic QM, are there states where the particle is contained within a bounded region?

With probability 1

cf. Hawking Ellis or BEE for details

but you might need linear algebra for that :p

@Slereah ...any wavefunction with compact support is such a state.

@Slereah can the wave function be a bump function?

6:32 PM

@ACuriousMind Well yes obviously but does it make proper sense in the theory :p

@ACuriousMind but there such wave functions

Maybe something else would forbid it, I dunno

And in this case, is there any reason why there couldn't be such a state in string theory

@Slereah yes

@0celo7 what?

@ACuriousMind does the Schroedinger equation have solutions with compact support

@Slereah I don't see any obvious reason, np

But the value of the $X^\mu$ operator is typically not of interest, anyway :P

6:36 PM

Well it is of interest, here :p

Since that is the crux of the problem

@Slereah what are you trying to do?

2

Q: String quantization and Malament's theorem

Malament's theorem posits that, given a few assumptions on relativistic QM, it is impossible to have localized particles. For $E_\Delta$ the proposition that a particle is certain to be found within a bounded region $\Delta$ of space, the assumptions are Localizability : If $\Delta$ and $\Delt...

quantum-mechanics string-theory

@0celo7 What? I can feed arbitrary states into it as initial condition, so yes.

@ACuriousMind but how do you know the support won't be noncompact for $t>0$

Well that doesn't really matter

The point is that it is of compact support on some spatial slice

6:38 PM

@0celo7 I don't, but that wasn't the claim

grumble

Also I suspect that any reasonable theory would only spread out the compact support, not expand infinitely in finite time

I hate diff geo

@Slereah: I think the issue is really the following: You shouldn't be looking at $X^\mu(\sigma,\tau)$, but at the constant part $x^\mu$ in its mode expansion (because that's the "position" operator that enters into e.g. the generators of the Lorentz transformations)

well does that really change my problem with Malament's theorem, though

6:42 PM

Then you can associate to it its spectral projection on some $\Delta\subset\mathbb{R}^{1,9}$.

Man what is it with you and nine dimensions

The point is that from the viewpoint of the theory, this is really just some operator

Isn't 3 dimensions enough for you

@Slereah Habit, but irrelevant here :D

Also how do you even do measurements on a string if position is so irrelevant

How do you model a detector

6:44 PM

So unless you are claiming that Malament's theorem forbids any operator from having states such that there are states with bounded support for it, then Malament's theorem doesn't hold - it suspect what breaks down are the things assumed about the translation operator

@Slereah lol

you don't

Man string theory sucks

Booo

@ACuriousMind that's some "fundamental" theory you've got there

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

6:59 PM

@dmckee o/

Also I note that a lot of my questions, @ACuriousMind answers with "Well we never discuss that!"

That quantity is not to be discussed!

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Forbidden?

IT IS FORBIDDEN

@Slereah I'm not saying that, I'm stating the fact that it usually just isn't discussed

You're free to consider it

Not a very convincing answer, tho :p

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

7:02 PM

Like the forbidden fruit you are free to try it :P

maybe if chat wasn't filled with hooligans we'd have smart people like Lumo who stop by

Oh man

I just realized that my question is the type that Lumo might answer

What have I done

@Slereah with some non-answer, probably :P

He will yell about how obvious the answer is

he will call you a commie

wonder if he reads conservapedia

#rekt

wait no conservapedia hates Einstein

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ why did you delete that

7:05 PM

@Slereah Well, you still haven't posed a convincing exposition of how you think this $E_\Delta$ violating the theorem arises in string theory, either :P

take pride in your words

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I'm trying to "be nice." @0celo7 :P

Well it's just kind of a hunch?

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I'm always nice and look where it got me

Hopefully a string expert will have the answer in mind

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

7:07 PM

ACM will be a string expert someday :P

...what exactly was that supposed to mean?

what part didn't you get

The one about the nuts :P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

3 mins ago, by 0celo7

take pride in your words

good point @Sᴋᴜʟʟᴘᴇᴛʀᴏʟ

but the trolls might ban me

I fear testicles are offensive to people

7:10 PM

Oct 22 '15 at 1:12, by 0celo7

I'm a hypocrite

uhhhh

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

better point

@ACuriousMind SAVAGE

@ACuriousMind do you have a moment for le diff geo

That depends on your actual question, so just ask it

@ACuriousMind Well it's a doozey. I'm having trouble writing the full proof of Lemma 3 in chap 2 of Milnor. What I have so far is: the inverse image of $(0,\infty)$ under $g$ is open in $M$ so it's a manifold. Thus all that remains is to show that a point in $g^{-1}(0)$ (the boundary) is diffeomorphic to one of the boundary sets in $H^n$

and I assume this is what he means by "it's just like Lemma 1"

But it's different, I think

I can draw a picture of what I want to prove, hold on

user image

see the first picture there

the diff that I want takes the top half of the circle on the manifold to the top half of the circle in $\mathbb R^n$, which is an open set of the half space

I thought it might be the $G$ that I constructed, but that's no good -- it's just the $F$ from the first lemma

7:30 PM

@ACuriousMind Can one argue differently as follows? i.e. is this true in general? If $X\subset \mathbb R^k$ is a manifold, then is $\bar X$ a manifold with boundary $\bar X-X$?

because then it's clear that $g^{-1}(H)=\overline{g^{-1}((0,\infty))}$

but I wouldn't know how to prove that one either

@0celo7 Probably not

I already proved that $g^{-1}(0)$ is a boundaryless manifold

But I don't know how to continue

any thoughts?

@0celo7 ...that's just an application of lemma 1, no?

@ACuriousMind yes

that's why I already proved it :P

He says the proof is "just like" the proof of Lemma 1 so I tried to reproduce it and beat it into a form that works here

but all I keep doing is proving Lemma 1 over and over :P

@ACuriousMind did I type all of that for nothing, are you thinking, ...?

You probably typed that for nothing because I don't see what you should do

7:45 PM

@ACuriousMind I have a suspicion that $G^{-1}:V\cap H^n\to something$ is what I need

@ACuriousMind Proof by picture: $G^{-1}$ is a diffeomorphism and $G^{-1}(V\cap H^n)$ looks like a neighborhood of $x$. QED

*$\mathfrak{something}$

@Slereah how about you help

take your mind of the squid theorem

Do I look like a math guy

yes.

then you are wrong

7:52 PM

@Slereah without this theorem I can't prove the Brouwer fixed point thm :(

I need to prove that $D^n$ is a manifold with boundary

Can't you do that the normal way

Make a map from $\Bbb H^n$ to $D^n$

wtf is an energy-momentum tensor?

@Slereah there's a general theorem that for $g\in C^\infty(M)$, the inverse image of $[0,\infty)$ is a manifold with boundary $g^{-1}(0)$

so take $g(x)=1-||x||$ and you get the disk

@BernardMeurer $\frac{\delta \mathcal L}{\delta g_{\mu\nu}}$

@Slereah you know damn well that doesn't help him

7:55 PM

Well you can define it as the Noether current of the translation group???

and it's the functional derivative of $S$, not $\mathcal L$

@Slereah that's the canonical one, he's probably asking about Belinfante-Rosenfeld

@Slereah That makes 0 sense to me :/

Oh boy

@BernardMeurer lemme figure out this proof then I'll skype, explain what it is and you can fix my computer which has broken

Can you guess who actually answered the Malament question?

7:57 PM

Curious One

It is our hero Valter Moretti!

@0celo7 Deal

@Slereah oh shit

> First of all you stated Malament's theorem hypotheses into a not very precise form

REKT

Yeah it's probably true

@BernardMeurer also I think I crashed an ORNL Linux server cluster

8:00 PM

@0celo7 Unlikely, why do you think that?

@BernardMeurer Because it worked fine but when I shut off my laptop it had crashed

No one could access their data anymore :P

I probably uploaded a virus

I deleted its system 32

He answered a lot of important infos but

Linux has no System32

I'm not sure he actually answered the question???

gladly

8:02 PM

I should read this in detail tomorrow

@ACuriousMind $(A\times B)\cap C=???$

can I split that??

????????

what

How do you want to "split" it?

@ACuriousMind hmm, not sure!

you sure do ask good questions mr. acm

@3075 o/ how are you feeling!

@0celo7 lol

8:05 PM

@skillpatrol recovering from surgery.

@3075 Can you smoke already? :)

-.-

@3075 I'll take that as a no :p

good take

@3075 Yesterday I quit smoking and swore I had enough, Till I smelt it in the club and had to take a puff. - Ludacris.

8:06 PM

On a serious note though, how are you feeling?

Really bad because I have a tube inside of me.

@3075 And I'mma blunt blowin'. - Lil Wayne.

And in a lot of pain

@3075 I'm on that good kush and alcohol. - Future.

@3075 😟 Any ETA on brushing teeth again?

8:08 PM

Cmon 0celo7 people will get the wrong idea

Or getting untubbed?

@3075 There is no wrong idea

I brushed yesterday

proof?

@3075 And leaving the hospital?

8:09 PM

@BernardMeurer soon

less than a few days

@3075 Soon™? Or actually soon?

Ah, then just soon

@0celo7 my teeth are nice.

@BernardMeurer quest?

@3075 Nothing new, updates coming soon™

If you feel better I declined my waterloo physics offer and I'm going to uoft

:O UAI?

8:14 PM

Uai?

Oh

WHY

Because I prefer to be close to my family in case something bad happens to me again

I opened a spot for you so be happy

:)

@ACuriousMind Ok, after some shuffling around definitions I figured it out...it was in fact almost identical to the Lemma 1 proof

@BernardMeurer skype?

@0celo7 Call when you can

8:17 PM

doesn't show you as online

Don't ever play yourself

I know, major key right there

Just call

@ACuriousMind in tex, if I want to make a big chunk of text a comment, how do I do that

% on each line?

or is there a way to % the whole thing in one shot

I basically have a proof in my notes that I don't want to finish

but I don't want to discard it

Have you tried googling that? :P

8:24 PM

yes

And you are telling me you found nothing?

...yes

What did you search for?

how to make a big block of text a comment in latex

much to learn you have, young padawan

8:26 PM

:(

Try tex comment multiple lines

same search results :/

oh there's a package

@ACuriousMind my search result worked

thanks anyway

8:38 PM

Did you hear about hungarian scientists claiming to have discovered a new fundamental force ?

May 25 at 15:53, by vzn

Has a Hungarian physics lab found a fifth force of nature? / Nature

You could say so :P

And, what do you think of it ?

My comment would get me banned @PhysicsGuy

Why ?

Because they are out to get me

8:41 PM

lol

I'll wait for further evidence from other experiments before thinking anything about it.

@ACuriousMind CRAP the Brouwer fixed point theorem needs the Weiserstrass approximation threorem?

into the black box it goes

damn

Is someone interested in Riemann hypothesis ?

Riemann is, probably

he's DEAD

8:46 PM

@0celo7 That's what they want you to believe

who're they

@ACuriousMind my god Milnor is such a dense writer, I'm taking hours per page

Good, I go away visiting 190 year old Riemann.

Dec 2 '15 at 0:44, by Chris White

@ACuriousMind after reading him for a while, you do begin to wonder what everyone else is saying with all those words

What do you think of MOND-Hypothesis, an alternative way to explain behaviour of mass without Dark Matter.

MOND kinda lost support as it didn't really pan out experimentally

8:53 PM

nvm

I realized $[a,b)$ is not compact

indeed not

Mandelbrot set abeian

*Mandelbrot set abelian or not ?

Errr how do you define a set being abelian

@Slereah a set is abelian if it's an abelian group obviously

well then call it the mandelbrot group I dunno

9:04 PM

@Slereah so what is a wormhole anyway

if its commutative ab=ba

@0celo7 I hope you realize that doesn't make much sense

It varies in definition

@PhysicsGuy The Mandelbrot set is not a group, what are you talking about?

If you have a spherically symmetric spacetime, a wormhole is a metric such that the radius has a minimum

9:05 PM

A wormhole is the solution of GR called Kruskal-metric

A wormhole can also be just a spacetime with a non-trivial fundamental group

@ACuriousMind Yeah, my thoughts are tangled, thought about something else

There's several definitions of what a wormhole is but overall it's usually the notion that you have a "smaller" region of the spacetime, so to speak

In between two bigger regions

The smaller region being the throat of the wormhole

I want to say that it's a spacetime with handles, but the standard wormhole is usually the Morris-Thorne wormhole, which is just $S^2\times \Bbb R^2 $

@ACuriousMind I do

That's the distinction between interuniverse wormhole and intrauniverse wormhole

9:11 PM

@Slereah do you understand the calculation on page 69 of HE

heheheh

69

what does "integrating along the flow lines" mean

flow lines of what

the fluid

it's the perfect fluid Lagrangian

I dunno

9:13 PM

@Slereah wtf are you even a physicist

No, I'm a software engineer

9:29 PM

@ACuriousMind What's Lagrangian Field Theory?

@ACuriousMind Plox

VTC chat comment as too broad

@BernardMeurer It's...field theory that uses Lagrangians, I don't know how else to answer that without writing half a textbook

@ACuriousMind Will you dedicate that text book to me?

I'd be honored

Yes, I will dedicate the book that I'll never write to you.

BURN

9:33 PM

@ACuriousMind Thanks man! No one ever did something like this for me before :)

@ACuriousMind can you get in this skype call pls, I'm trying to explain it to him

@0celo7 No, I am currently trying to learn how to pretend that I know how renormalization works ;P

@ACuriousMind I gave up on that years ago

(holy shit it's been years since I've done any QFT)

9:57 PM

@ACuriousMind yay I just did a real analysis proof on my own, that class did do something for me

10:42 PM

Wow @ACuriousMind the lemma on page 21 kinda blew my mind

and the proof is so simple but uses a lot of stuff, love it

now that's the kind of topology that's cool

@Slereah what does "topological Casimir effect in compact spaces" mean

If you take Minkowski space and perform some identifications on the boundaries

the renormalized vacuum acquires a negative energy

similarly to the casimir effect, because the identification cuts off infrared wavelengths

what is the renormalized vacuum

PROOF

pls

Visser has a proof

Also Birrell

eeek

Visser is not with me right now

neither is Wald

I need my Wald

I have only my math books @Slereah

I've turned into a math person

If you take the stress energy tensor of a cylindrical spacetime of period $L$, and you renormalize it by substracting the stress energy tensor of Minkowski space, you get $$\langle :H: \rangle = \langle H \rangle - \lim_{L\rightarrow \infty} \langle H \rangle = -\frac{\pi}{6L^2}$$

Using whatever renormalization technique to substract the divergent terms properly

10:53 PM

I still don't get renormalization

why are all QFT textbooks literally shit

Renormalization is basically hiding infinities under the rug

Weinberg and Zee being shit books are the single greatest reason why I'm not studying physics in college

For instance for the vacuum of a scalar field, you have $$H = \int \frac{dk^3}{(2\pi)^3} \omega_k (a_k^\dagger a_k + \frac{1}{2} \delta(0))$$

Which is p. badly divergent

@ACuriousMind @yuggib heh, just typed up a wall of text for you guys that you would have answered with "use Picardy Lindyhop"

So you sweep that under the rug

10:58 PM

took me waaaaay too long to figure it out

@Slereah I'm aware of this integral

The rug here would be the cosmological constant

you just subtract that delta from the definition of $H$

Since the cosmological constant has an energy of $H = -\Lambda$, you redefine it so that the total energy is bounded

Same process goes with masses and interaction strengths to renormalize various quantities

You just need that the divergence is of the same kind as the term you renormalize

FUCK

@Slereah check do Carmo page 203

I was so proud of my proof of that

The cosmological constant is fine to absorb divergences $\approx g_{\mu\nu}$ in the SET

what is carmo

11:03 PM

that will teach me to not do exercises without reading the chapter :P

@Slereah do Carmo, the Riemannian geometry book

I do not know it

...not possible

it's extremely good

@ACuriousMind Reading Milnor is nothing if not humbling

These proofs are too slick for mortals

11:21 PM

@Slereah yeah I never understood that

Well the basic notion is that the coupling constants in the Lagrangian are not quantities that you can measure directly

I know

So they are fitted by renormalization to give the experimental values that we find

It just happens that this basically means that a lot of them are infinite or infinitesimal

at least under the standard model

@Slereah what's the french word for lemma

lemme

11:32 PM

@ACuriousMind Is the operator $\cap$ a kind of derivation on the algebra of sets with product $\cup$

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