The h Bar - 2016-04-10 (page 1 of 2)

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12:00 AM

kind of more important than stuff like projectile, fundamental if you will

linear algebra is a fundamental way of understanding the information you have in front of you

MonaLisaOverdrive

@user507974: Not really, no. See, I'm…not an official physics student. I'm not even a student.

I have an interest in quantum mechanics thanks to several books of fiction (and some non-fiction) I've read and Quantum Mechanics: The Theoretical Minimum is my first attempt at understanding the mathematics of it.

12:17 AM

@MonaLisaOverdrive quantum mechanics is the linear algebra of hillbert spaces

MonaLisaOverdrive

INteresting…

1 min

may have book

fingers oily

cept pinky

MonaLisaOverdrive

Hilbert spaces…are vector spaces, right?

Space of states

12:37 AM

all of quantum mechanics boils down to this object called the wavefunction

its a mathematical object where by performing operations on it we recover physics

@MonaLisaOverdrive yea, a special type of vector space for complex valued vectors that obeys specific rules in relation to the inner product

you may want to read this book at the same time if you are interested in quantum mechanics specifically amazon.com/Applied-Analysis-Hilbert-Space-Method/dp/0486458016

MonaLisaOverdrive

Interesting use of the word recover

i cant speak to the background level of the book, my copy is currently shipping, but the reviews are glowing

@MonaLisaOverdrive i guess if we were to just sample you opinions of my use of the english language would be rated as interesting =)

MonaLisaOverdrive

@user507974: That book…you realize I have next to no understanding of linear algebra, right?

@MonaLisaOverdrive i think it builds up an understanding

I THINK

youd want to learn the absolute basics online while its shipping

i remember not liking my linear algebra book, so i cant recommend that

MonaLisaOverdrive

Um…I didn't understand more than half of the references in the preface alone...

I think I'll start with linear algebra for dummies and work my way up…

1:12 AM

@Slereah We did that in second/third year.

1:37 AM

@0celo7 i think hes saying basically do rydberg equation like stuff, where you are just told the energy scaling law and calculate the energy transition

if so i did that back in high school technically, for the case of hydrogen (rydberg equation)

1:48 AM

@user507974 I know what he's talking about.

We did that in second/third year.

MonaLisaOverdrive

Later all.

2 hours later…

3:41 AM

@skillpatrol hello

hi

Oh? Not on ignore?

nope

@ACuriousMind With $E_1,E_2$ smooth vector bundles with the same fiber $V$ over $M$, is $\mathrm{Hom}(\Gamma(E_1),\Gamma(E_2))$ the set of endomorphisms of the fibers...but with some position dependence?

Here $\Gamma(E_i)$ is an object in the category of $C^\infty(M)$ modules.

@skillpatrol why

4:50 AM

why not?

@0celo7 :28899318 why not both?

All hail @JohnRennie

Hi. Just dropping in to see if there was any interesting chat last night.

@JohnRennie Did you check out my list of GR books

You should get one of them

@JohnRennie I came across a popular science article I THINK may be completely accurate and about fundamental physics if you find that interesting

5:02 AM

@0celo7 Thanks yes, though I haven't looked through it yet. I actually have quite a lot of GR books (including Wald and MTW). It's just finding the time ...

@user507974 yes ...

motherboard.vice.com/read/…

after the first three paragraphs of trying to mentally prime the reader with useless quoting i dont think I saw anything wrong

just havent read up on holographic universe theories

@0celo7 the Gourgoulhon looks fun. That will definitely be on my to buy list

i was actually going to bring up the discussion on this exact topic of non-continuous space and didn't know that was part of holographic theories

@user507974 See backreaction.blogspot.co.uk/2015/12/…

@JohnRennie It is.

@JohnRennie The notation is a bit meh

But it's a good book, covers a crazy amount of stuff

5:07 AM

@0celo7 £68 on Amazon - good grief!

@JohnRennie I got my copy for $25

I wonder is there's a pdf floating around the darker corners of the Internet :-)

"Searching for a prediction based on a hunch rather than a theory makes it exceedingly unlikely that something will be found." This wasnt true of science for a very long time in its history

@JohnRennie I have it.

Legally.

Seriously, I downloaded it from Springer.

I won't ask you to compromise yourself. I'll go hunt down a copy. Anything that has been released as a PDF is floating around the Internet somewhere.

5:10 AM

@JohnRennie just looked at springer website, i swear publishers are the biggest thieves on this planet

Actually I generally buy a paper copy if I'm serious about learning from a book because you just can't learn hard physics from a pdf. But it's useful to have the pdf first to glance through to see if it's a book you really want.

the ebook prices are kind of expected, but the per chapter is ridiculous

>hard physics

God, what are you, a physicist?

@0celo7 OK, OK :-)

@JohnRennie It also has some amazing diagrams

user image

That's the Lorentz factor for a rotating, accelerating observer

5:12 AM

@user507974 Hossenfelder is being a bit pious in that article, but the point is that the ideas the holometer was designed to explore are a bit niche at the moment. Don't take them too seriously.

An Amazing Diagram:

user image

@0celo7 ah, that's interesting, I've done the calculation for a rotating observer in the static frame, but I've always wondered what the metric was like in the rest frame of the rotating observer. I think this is definitely a book to buy.

He doesn't calculate the metric

Well, maybe calculating $\Gamma$ amounts to doing that

@JohnRennie The book is YUGE

Well actually writing down the metric would be an awful mess because of the reduced symmetry. What would be interesting is to get a general feel for it.

@JohnRennie There's chapters on accelerating observers, angular momentum

All sorts of goodness

Rindler horizon stuff

5:17 AM

Anyway, I should have started work 15 minutes ago, so I have to go now. isn't it bed time on the far side of the Atlantic :-)

@JohnRennie I should have gone to bed 15 minutes ago

gf expects me to be functional at 9

But thanks, I will look out a copy of the Gourgoulhon.

@JohnRennie He even talks about an accelerated ruler!

5:38 AM

Where do I find physics from the horses mouth? I'm searching for decently transcripted versions of Einstein, Schroedinger, Newton and Galileo's papers. My library doesn't have any such resource, is there something online, perhaps?

You get what I mean, right.

@Nick Dirac's GR

He's not exactly the horse but he basically outlined the entire theory and its almost a historical document in its own way

plus its only 69 pages

6:43 AM

07:40 a.m. and I've used all my daily allowance of 25 close votes - a new record, but not a good one.

6:55 AM

Miracles of Moving Heavy Objects in the Past Revealed

3 hours later…

user116211

9:57 AM

Hmmm..... don't heat engines work in cycle?

user116211

Your elegant answer is irrelevant because the user botched up his question Besides this, i do not know anything in earth that works on a cycle except the refrigerator. — veronika 5 mins ago

If there's no cycle it's not much of an engine

user116211

@Slereah So, isn't it always true entropy change of the engine is always 0?

wot

I mean I guess?

user116211

@Slereah Let me google what that means ;P

user116211

10:04 AM

Damn wot- World of Tanks!

It outputs heat and take in fuel

user116211

@Slereah Like Otto cycle?

The entropy outside will increase but the entropy inside on average will be the same

Well there's plenty of engine cycles

user116211

I mean there are two reservoirs and there's the engine.....

user116211

It's that the entropy change of the engine is zero while that of the reservoirs is positive.....

user116211

10:06 AM

that's also what Clausius inequality says....

user116211

@Loong: o/

@MAFIA36790 hi

howdy

user116211

@user507974: o/

user116211

10:21 AM

I really don't know what let her so confidently assert that engines don't work in cycle ;(

pround confidence plus oversight

"in real physics we dont bother with that" mentality maybe

user116211

@user507974 Yes, that could be.

@MAFIA36790 well, if our nuclear power plants do not work in a cycle, we call that a loss of coolant accident.

user116211

@Loong My whole point to OP's question is that Clausius' relation always hold; even if there is any irreversibility during the cycle.

user116211

10:38 AM

By irreversibilty, I mean friction, dissipative activities.

user116211

The latest piece:

user116211

@MAFIA36790. Yes, it is not a thermodynamic cycle. And the thermodynamic efficiency is of very relative importance. — veronika 10 mins ago

user116211

I just want to flag this comment....

10:55 AM

blurb

How to prove that $$\lim_{t \rightarrow \infty} \frac{1}{4\pi^2} \int_m^\infty dE\ \sqrt{E^2 - m^2} e^{-iEt} = e^{-imt}$$

user116211

@Slereah: Have you tried MSE?

Microsoft Security Essentials?

user116211

yes.

user116211

But you can also say

user116211

Mathematics Stack Exchange.

11:08 AM

I never get any answer on the math SE

user116211

@Slereah Yes, that's also true... I stalked your last question.... requested them at their chat... but of no avail.

0

Q: Meta-discussion

A spaceship having 0.9c velocity is moving to a star which is 6 light years away from earth .the time the spaceship observer sees on earth observer clock on reaching the star is 1.27 yrs . How will you explain this ? For the spaceship's frame of reference the star is moving towards it with t...

11:39 AM

Common integrals in quantum field theory

There are common integrals in quantum field theory that appear repeatedly. These integrals are all variations and generalizations of gaussian integrals to the complex plane and to multiple dimensions. Other integrals can be approximated by versions of the gaussian integral. Fourier integrals are also considered. == Variations on a simple gaussian integral == === Gaussian integral === The first integral, with broad application outside of quantum field theory, is the gaussian integral. In physics the factor of 1/2 in the argument of the exponential is common. Note: Thus we obtain === ...

Oh wikipedia

12:02 PM

Where's my Big Book of Integrals

12:26 PM

user image

@Slereah : Remind you of anyone?

let's not go there

Man it's been a while since I've had to do fancy integrals

Fortunately I still got it it seems

as opposed to fractal righteousness...

@skillpatrol Dude

@skillpatrol DUDE

I've started calling more people dude

they say that a lot in cali

12:33 PM

What if they are a PROFESSOR EMERITUS

from mit?

From the university of space

Also I need to buff up on my contour integrals

I kinda forgot about them

what is the harvard of france?

ecole

The fancy schools in France are like

Polytechnique

Ecole normale supérieure

And la Sorbonne

Oh and l'école des mines

Those are the big fancy schools

1:01 PM

Never heard of them.

1:38 PM

In France, kids start school very early : school starts at age 2 (for 52% of children) or 3 (for almost 100%)

1:58 PM

Well it's not school school

user116211

@skillpatrol WTF!

It's kindergarten

Not mandatory

Although most people do it yes

user116211

I also began at 3.

2:46 PM

@user507974 I don't know anything about holography

@Slereah Exactly. The integration measure changes by the Jacobian determinant of the transformation, which is -1 for odd and 1 for even dimensions.

@0celo7 Since $\Gamma(\cdot)$ usually denotes the global section function, that's the $\mathrm{C}^\infty$-linear homomorphisms between the global sections of the bundles, nothing to do with fibers.

@ACuriousMind Function?

Is it not a functor?

It's a functor, dunno why I wrote function

@ACuriousMind Since when do you \mathrm the $C$?

Dunno that, either

o.O

2:54 PM

I just got out of bed, don't be so nitpicky :P

@ACuriousMind Ok, but a section is just a way of picking a point in the fiber at each point of the base space

So shouldn't it be at least isomorphic to $\mathrm{Hom}(E_1,E_2)$, the set of bundle homomorphisms?

Hm, no I don't think so. How do you propose to make a bundle morphism $E_1\to E_2$ out of a morphism $\phi : \Gamma(E_1)\to\Gamma(E_2)$?

@ACuriousMind Wait, is $\mathrm{Hom}(E_1,E_2)$ supposed to be a bundle itself

fiber, not vector.

maybe it is a vector bundle, actually

There are two things denoted by $\mathrm{Hom}(E_1,E_2)$ - the bundle whose global sections are the bundle morphisms $E_1\to E_2$, and the set of bundle morphisms itself.

it's the bundle with fiber $\mathrm{Hom}(E_{1p},E_{2p})$

3:01 PM

Some people write the Hom differently (e.g. as mathscr) when they mean the bundle, some don't.

Oh, maybe it's $\Gamma(\mathrm{Hom}(E_1,E_2)\cong \mathrm{Hom}(\Gamma (E_1),\Gamma(E_2))$

Apparently there's a deeper way of saying that $\nabla_XY(p)$ depends only on $X_p$

and it has to do with hom

No, $\Gamma(\mathrm{Hom}(E_1,E_2)) = \mathrm{Hom}(E_1,E_2)$, where the hom on the left is the bundle hom and the hom on the right is just the vector space of bundle morphisms.

Vector space or module?

The coefficients are smooth functions, that's not a field.

It's a $\mathbb{R}$-vector space and a $C^\infty$-module, I guess.

@ACuriousMind Wait what

3:05 PM

@ACuriousMind currently I am trying to prove more this!

4 hours ago, by Slereah

How to prove that $$\lim_{t \rightarrow \infty} \frac{1}{4\pi^2} \int_m^\infty dE\ \sqrt{E^2 - m^2} e^{-iEt} = e^{-imt}$$

@ACuriousMind Ok, let $s\in \Gamma(\mathrm{Hom}(E_1,E_2)$. Then we can construct a map $\phi:\Gamma(E_1)\to \Gamma(E_2)$ pointwise by $\phi:(\sigma,p)\mapsto s(p)\sigma(p)$ where $\sigma\in\Gamma(E_1), p\in M$.

Then $\Phi:\Gamma(\mathrm{Hom}(E_1,E_2))\to\mathrm{Hom}(\Gamma(E_1),\Gamma(E_2));s \mapsto \phi$

Your $\phi$ is strangely written, but yes - every bundle morphism gives a morphism of the global sections.

Strangely written?

$(\sigma,p)$ is not an element of $\Gamma(E_1)$.

@ACuriousMind I said pointwise

Umm

3:11 PM

Still, why $(\sigma,p)$ on the left?

I'm dumb, how should I have written that?

$\phi(\sigma)(p)=s(p)\sigma(p)$?

@Slereah why do you think that is true my child

Because the Peskin sayeth so

@0celo7 Yeah, that looks better

@ACuriousMind I'm bad at $:\mapsto$ notation

Well, if you want to use $\mapsto$ I would have written it as $\sigma\mapsto (p\mapsto s(p)\sigma(p))$, I think.

3:14 PM

eww

Or just $\sigma\mapsto s\cdot\sigma$ and say in words after that that $\cdot$ is pointwise "multplication".

Yeah, that works.

@ACuriousMind Ok, and what about the other way around?

I don't think there is an other way around.

Oh really?

Oh boy

A thing on contour integrals

web.williams.edu/Mathematics/sjmiller/public_html/302/…

It's

LEVEL 3 MATH

3:16 PM

I'm hunting in a book on categorical diff geo for the answer

level up music

@Slereah That formula doesn't make sense, you can't send $t\to\infty$ and still have $t$ on the r.h.s.

BUT THE PESKIN SAYETH SO

You probably don't mean a limit, but an asymptotic there

I assume it's more that

3:18 PM

@ACuriousMind ofc you can

yeah

asymptotically

Don't get your knickers in a twist over notation :V

Anyway how does one show this

That thing "looks" as if one should be able to apply the saddle point method, but the exponential is just linear, so that doesn't work...

@ACuriousMind It appears to work the other way around as well

@0celo7 How?

@ACuriousMind Given a module hom $\phi:\Gamma(E_1)\to\Gamma(E_2)$, let $s\in\Gamma(Hom(E_1,E_2))$ (please forgive lack of mathrm from here on out) by defined by $s_p(\sigma_p)=\phi(\sigma)(p)$

$\sigma\in\Gamma(E_1)$

3:27 PM

Hm

What are some standard methods for asymptotic expansions

@ACuriousMind Also, this map is well-defined in the sense that if $\sigma_p=0$, then $s_p(\sigma_p)=0$.

Your $\phi(\sigma)(p)$ is not guaranteed to define a section at all.

Giving a value at every stalk is not enough to define a section.

Apparently the result for that integral would be $$(it)^{-\frac{3}{2}} e^{-\frac{1}{2}imt} \Gamma(\frac{3}{2})$$

@0celo7 Also, what is that $s$ supposed to be, anyway?

I didn't catch that earlier, but you seem to be using the wrong notion of "bundle morphism". A bundle morphism is a map $E_1\to E_2$, not a map on sections.

@ACuriousMind what

What else do you need

3:35 PM

An element of $\Gamma(\mathrm{Hom}(E_1,E_2))$ is a map $\phi : E_1\to E_2$ such that $\pi_1 = \pi_2\circ\phi$.

Latest gem from Academia.SE:

6

Q: Students keep farting during my classes,

I am teaching a Calculus II session this semester. The class period is from 2pm to 2:50pm, on Mondays and Wednesdays. The problem is that my students keep farting while I am going over important theorems and homework problems on the chalkboard. How can I do something about this? The smell...

mathematics teaching

Lol

@ACuriousMind uhhh

give me some time to think about this

But unless the problem is Maths specific, I'm slightly curious about the mathematics tag applied to the post!

He farts only in prime numbers

@Slereah ?

3:38 PM

A prime number of farts

At least 2

@ACuriousMind Are you sure

In my example, the vector bundles are over the same manifold and vector space.

@Slereah I'm curious about higher advancements :D

Then $Hom(E_1,E_2)$ is the bundle with typical fiber $End(V)$

@0celo7 Yes. For fibers with extra structure it also has to be structure-preserving on the fibers.

Where $\pi^{-1}(p)\cong V$

3:40 PM

@0celo7 That doesn't define the bundle.

Wait how does $(it)^{-\frac{3}{2}} e^{-\frac{1}{2}imt} \Gamma(\frac{3}{2})$ even asymptotically reduce to $e^{-imt}$ at large t

@ACuriousMind Well, at $p\in M$ we define the fiber by $Hom(\pi^{-1}_1(p),\pi^{-1}_2(p))$, I think.

Valentin Tihomirov

What does it mean that the state of a number of q-bits cannot be distinguished from the others?

@0celo7 Just giving the fibers does not define a bundle.

If it would, how could you have two different bundles with the same fiber?

I know, I know

So wtf is $Hom(E_1,E_2)$

3:47 PM

It's the bundle whose sections over $U\subset M$ are the bundle maps $E_1\rvert_U\to E_2\rvert_U$, I'd say.

Hm.

@0celo7 I guess the "bundle" definition should be to take $\sqcup_{x\in M} \mathrm{Hom}_\mathbb{R}(E_{1,x}, E_{2,x})$, equip it with the obvious projection, then find local coordinates for this (over a trivializing cover for both $E_1,E_2$), and then give the transition functions to construct the bundle.

Eww

I should have bought Michor when it was on sale

@ACuriousMind a trivializing cover?

@0celo7 A cover $\{U_i\}$ such that $E_1\rvert_{U_i}$ and $E_2\rvert_{U_i}$ are trivial for all i.

@ACuriousMind Oh, can we always find such a cover for two bundles?

@0celo7 Sure, the existence of a trivializing cover for one bundle is guaranteed by the definition of a bundle, and then you just intersect the two individual trivializing covers to get one that trivializes both.

@ACuriousMind Ah, intersection, of course.

3:58 PM

yay back in Ithaca

@FenderLesPaul check rd.springer.com/book/10.1007/978-3-319-26654-1 please

you might be able to order it now

@ChrisWhite is this the GR@100++ conference?

@0celo7 dude I can't I checked it

@Danu he said I can definitely join his group

@ACuriousMind How do you find the transition functions

@ACuriousMind Can one construct a bundle from its sections?

Sure

Especially if it's trivial

4:17 PM

@FenderLesPaul Okay, nice.

Hm

In the residue theorem

Is the open set of the complex plane required to be compact

well, finite

obviously not compact

Or can you have an infinite one

If it just says "an open set" then take it to mean "any open set"

sure, but in the residue theorem, you also require a curve going "around it"

closed rectifiable curve

I have no idea what exactly you're reading, so I'm not going to comment on the details of whatever proof you're reading.

just basic stuff on the residue theorem

I ask this tho because Peskin does contour integrals that fuck off to infinity

4:21 PM

@0celo7 Let $\phi_1,\phi_2$ be the transition functions of $E_1,E_2$ on $U\cap V$. Then we define the transition function for $\mathrm{Hom}(E_1,E_2)$ to be $\phi_{UV} : (U\cap V)\times\mathrm{Hom}(V_1,V_2)\to (U\cap V)\times \mathrm{Hom}(V_1,V_2), (x,\psi)\mapsto (x,\phi_2\circ \psi\circ\phi_1)$.

@Danu you did a whole class on vector bundles, right

@0celo7 From all sections, yes, since giving the sections on every open set is exactly the sheaf version of the bundle.

@ACuriousMind Sigh, do I need to learn about sheaves

I don't know, probably not for looking at vector bundles

All I wanted to do was some Riemannian geometry :(

Now you're telling me about sheaves

@ACuriousMind $\psi$?

4:25 PM

@0celo7 An element in $\mathrm{Hom}(V_1,V_2)$.

oh, right

I'm off to lunch, we will resume this later

5:06 PM

@ACuriousMind Ok, let's learn about sheaves.

5:16 PM

Oh god all the commutative diagrams

@ACuriousMind I gotta say, this is a bit too abstract for me

5:30 PM

@3075 I saw that.

@0celo7 saw what?

@barrycarter : sure thing, it's my name (all one word) at btconnect dot com.

O-o

@JohnDuffield Email request w/ my address sent.

@barrycarter : received. Howdy cowboy!

5:37 PM

Cowboy?

Oh, my address.

We've urbanized the place quite a bit :)

@0celo7 log

your memory is fine unless this is a recent occurrence.

I don't remember any of that.

lol

@0celo7 what do you mean by this?

@3075 9 stars.

I didn't star it.

5:51 PM

Proof?

want a ss?

k

@0celo7 link

Looks photoshopped to me

...

want to screenshare on skype?

noi

what should we do then

5:54 PM

Star it.

Then unstar.

k done

did you see the star?

Yes.

unstarred.

It was 8 earlier.

Someone starred it just as you entered the room.

well it wasn't me.

and I wouldn't do something so obvious.

I can't believe there are 9 stars on that though.

you're a cool guy xD

10 now. LOL

(it wasn't me I can star and unstar again if you want)

5:56 PM

Just star and let's see what happens?

10 people hate me

might as well delete my account

k i'll star and unstar. look at it right now.

done.

@0celo7 no :'(

Just delete your account? Last time, you were going to kill yourself.

should I do both?

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Apr '1610

The h Bar

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