The h Bar

0celo7

00:28

@tpg2114 Relativistic mass is not a thing...

NeuroFuzzy

01:41

Yoooo!

ACuriousMind

Yoooo?

NeuroFuzzy

02:17

Wasssaaaap!

hwlau

Yoooo...

NeuroFuzzy

02:30

What's crackalackinnnn

0celo7

@ACuriousMind Yoooo!

NeuroFuzzy

02:57

What's up in da hizzouseee!

0celo7

03:24

I'm up

0celo7

03:50

@ACuriousMind I don't see the issue with the $H$ constructed in the Example on page 218.

Why does $H_q$ have to be both $1$ and $-1$?

Secret

04:26

Knowing the history is one thing that help me to anchor to reality when I am too distracted by the algebra

For example, while the formalism of the electromagnetic field tensor is more natural to work with, it comes later after the experiments and models of Faraday, Oersted, Galvani, Volta, Ampere, Ohm, Henry, etc. and of course Maxwell

Some of my friends said electromagnetism in 3rd year should be taught right away starting from maxwell equations, and then deriving other phenomenon from them, while my professors said it is better to start it by following the history and introducing the co…

Secret

09:25

NbCeID 17
http://www.sciencedaily.com/releases/2015/10/151012122853.htm

Stefan Bischof

10:17

@Akshit
Independant of votes you may accept an answer which helped you most. You get 2 reputation and the question no longer is in the unanswered questions list

Slereah

10:43

Hello

Stefan Bischof

10:58

hi

ACuriousMind

11:45

@0celo7 I think the minus is a typo, the issue is not that it would have to be both.

@StefanBischof Not sure that pinged anyone since "Akshit" does not appear for me in the @-autocompletion. They have to have been in chat the last two days for your message to reach them this way.

yuggib

12:36

a JD type question: anybody who wants to discuss theology today? In particular the ontological proof of Gödel in the context of Leibnitz's theism...

I find it an interesting (and funny) subject

the funniest part is that Gödel never published the proof for he was afraid he would have been seen as a religious person

instead he was interested in a logic argument towards the necessary existence of a being with God-like attributes

Slereah

12:58

@yuggib Godel's ontological proof relies on 5 different axioms, IIRC

Those axioms are not particularly self evident

0celo7

13:19

@ACuriousMind Well what is the issue then?

Does it have to do with mod?

ACuriousMind

@0celo7 The $H = -q$ is not a smooth function on the whole torus.

0celo7

wai

ACuriousMind

@0celo7 look at what happens at the point where the coordinate wraps around. It's the same reason there is no global coordinate chart for the circle.

0celo7

so the mod is the issue

damn, why are all non 1080p monitors so expensive

@ACuriousMind what is a good 1080p monitor

ACuriousMind

@0celo7 Not the slightest idea

0celo7

13:26

so you can overclock your display...

Stefan Bischof

@ACuriousMind

@ACuriousMind Right. This Akshit is not in Chat. MMh, I saw, that he never marked his questions as solved.

A comment didn't solve the issue.

ACuriousMind

@StefanBischof Well, users aren't obligated to accept answers. It's annoying, but if they still don't after a friendly reminder in a comment, there's not much we can do about it.

Stefan Bischof

13:41

Correct. At the end of the day SE will end up with 2% never answered questions.

0celo7

Wow, that low?

Stefan Bischof

just guessing

0celo7

:'D

Stefan Bischof

also uninformed :D

0celo7

@ACuriousMind is 1ms vs. 5ms delay worth a ~$50-100 price jump?

ACuriousMind

13:47

@0celo7 Still not the slightest idea

0celo7

what kind of gamer are you

my first build was in the 1500 range

but I could go for a smaller GPU and cut 300

and there's ways to cut small amounts everywhere

Stefan Bischof

I'm always doing this procedure: What kind of FPS / MMO Game do I want to play? I try to fulfill High or Ultra specs of only this game. Not good for future games, but I End up with < 700 €.

Danu

I've got some major confusion about conformal transformations right now. Any help would be much appreciated.

Here it comes:

yuggib

@Slereah It depends what is self-evident for you...some of them are disputable however, I agree

(I am not convinced by the proof, but I think it is interesting nonetheless)

Danu

Under general diffeomorphisms $\phi: x\mapsto x'$ the metric transforms as:

$$ g'_{\mu\nu}(x')=\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^\nu}g_{\alpha\beta}(x)$$

0celo7

13:56

@Danu MAJOR

Danu

Under conformal transformations we have
$$ g'_{\mu\nu}(x')=\Omega^2(x)\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^\nu}g_{\alpha\beta}(x)$$

This is the definition my textbooks all give.

Does that mean that conformal transformations enormously generalize diffeomorphisms?

0celo7

Yes.

Danu

No, right? Then why does the diffeo law seem like a special case.

@0celo7 Do you have sources?

2

Q: Are diffeomorphisms a proper subgroup of conformal transformations?

The title sums it pretty much. Are all diffeomorphism transformations also conformal transformations? If the answer is that they are not, what are called the set of diffeomorphisms that are not conformal? General Relativity is invariant under diffeomorphisms, but it certainly is not invariant u...

These answers say not

0celo7

Well, then I'm just as confused as you.

Danu

@Muphrid: A coordinate transformation when viewed as an active transformation is a diffeomorphism between the same manifolds $M \to M$, so it is necessarily an isometry. So, a conformal transformation is more general than the diffeomorphism or coordinate transformations we encounter in GR. — user7757 Jun 18 '14 at 4:51

This does seem to support my intuitive view (and your view)

0celo7

14:00

The way you've written it makes it seem diffs are conformal trafos with $\Omega=1$.

Danu

Yes.

0celo7

Great, Outlook decided to break itself.

Danu

Perhaps my first equation is not actually the transformation law under all diffeos?

0celo7

Oh

Danu

Just under coordinate transformations? What are the differences between the two? No idea

0celo7

14:02

Well looks like I can't use Outlook any more

Danu

Is the following statement correct: Conformal transformations actually transform the manifold itself, i.e. do not merely amount to coordinate transformations?

The usage of the term "conformally flat" certainly suggests this.

@yuggib It's Leibniz, btw ;-)

yuggib

@Danu :-D you're in a pedantic mood... :P

Slereah

@Danu That's not conformal, though

That's conformal + diffeomorphism

Danu

@Slereah This is literally the definition in three separate textbooks that I've looked at.

Slereah

Hm

Weird

Conformal is just $g \rightarrow \Omega^2(x) g(x)$

Danu

14:13

You're not being precise enough to resolve any of my confusion---I know that statement already and it doesn't help.

Slereah

Although I guess that you could say that two spacetimes differentiated by a coordinate transform still belong to the same conformal class

Since they're the same spacetime

0celo7

@Slereah that's a conformal isometry

Danu

Also, Di Francesco's book has a different definition, namely $g'(x')=\Lambda(x)g(x)$ which translates to
$$ \frac{\partial x}{\partial x'}\frac{\partial x}{\partial x'}g(x)=\Lambda g(x)$$

which is very different from the others, because it seems to be a special case of the usual transformation law.

ACuriousMind

@Danu: The conformal transformations are those diffeomorphisms which change the metric only by that overall conformal factor $\Omega(x)^2$.

Danu

@ACuriousMind This corresponds to Di Francesco's book.

The other definitions are different (do you agree on that?).

The other definitions are from (1) Blumenhagen (2) Schottenloher (3) Ammon & Erdmenger

ACuriousMind

14:21

@Danu If the other ones are different, they're wrong.

Danu

@ACuriousMind I doubt that.

ACuriousMind

But perhaps you are misreading them, let me check Schottenloher.

Danu

Maybe you could have a look at Schottenloher---I trust him most.

@ACuriousMind heh.

Stefan Bischof

Baam! This is coincidence.

ACuriousMind

@Danu Schottenloher's definition is the standard one except that he says it doesn't necessarily have to be a bijection

However, since he directly afterwards mentions that they are locally invertible, they are at least local diffeomorphisms

@Danu This is also what Blumenhagen writes in eq. 2.1, I'm not sure why you say Francesco's is different.

Danu

14:29

The equation I wrote for Di Francesco is different in having $g(x)$ on both sides, while the others don't.

@ACuriousMind I know that---and we can also take the inverse of those derivatives of $\phi$, put them on the other side of the equation and obtain what I wrote earlier.

32 mins ago, by Danu

Under conformal transformations we have
$$ g'_{\mu\nu}(x')=\Omega^2(x)\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^\nu}g_{\alpha\beta}(x)$$

ACuriousMind

Ah, I am beginning to see where the confusion lies

2

Danu

where $\phi=x'$

I suspect there are two different notions floating around that may go under the same name for some people

Also, of course Blumenhagen's equation is different than Di Francesco's, at least visually...

ACuriousMind

@Danu No, I'm pretty sure all those notions are the same, they just are expressed differently.

Danu

In particular, one obtains Di Francesco's equation by discarding the partial derivatives from Blumenhagen's .

I for the life of me cannot see how one would say they're the same, to be honest.

ACuriousMind

First, notice they all state $\phi^* g' = \Lambda(x)^2 g$ in one form or another.

Danu

14:35

@ACuriousMind which is Blumenhagen's equation---agreed?

And it's also the one I just quoted (because of invertibility we can put the derivatives on the r.h.s.), yes?

ACuriousMind

@Danu Yep.

It means the conformal transformations are those diffeomorphisms where $\frac{\partial x^\alpha}{\partial x'^\mu}\frac{\partial x^\beta}{\partial x'^\nu}$ does not actually depend on the indices, but is the same in every direction.

I'm currently not finding the equation in Francesco

Danu

4.1, p.95

@ACuriousMind "It" what? $\phi^*g'=\Lambda g$?

ACuriousMind

@Danu $\phi^* g' = \Lambda(x)^2 g$.

John Duffield

@ArtOfCode : yes I do want to contest it. But not now. I have to go in less than an hour, and meanwhile I'd like to talk physics.

ACuriousMind

@Danu Well, that's the same. You claimed Franceso's definition was different

Ah

I see

Danu

14:40

@ACuriousMind lol

ACuriousMind

He's not a mathematician and considers a mapping from a conformal manifold to itself

There, $g'$ is the metric $g$ after transformation, i.e. it includes the partial derivatives

Danu

@ACuriousMind So there are two different things going on here...

Ah, are you saying that for $\phi: M\to M$ Blumenhagen (et al...)'s equation reduces to
$$\phi^*g=\Lambda g$$ (note the equal $g$'s)?

Okay, I guess that's right.

ACuriousMind

@Danu Yes, and Francesco writes $\phi^* g = g'$ in the physicist's habit to prime quantities after transformation

Danu

@ACuriousMind I think this is the essence.

ACuriousMind

The different usage of the prime is what threw you off

Slereah

14:44

@ACuriousMind But some conformal transformations are not diffeomorphisms!

ACuriousMind

@Slereah Example.

Slereah

Hm, what metrics are conformal to Minkowski again

0celo7

FLRW

Slereah

Yeah

Danu

@ACuriousMind Yes.

Slereah

14:46

And those are not flat space

ACuriousMind

@Slereah So what?

Danu

So, how to explain the terminology "conformally flat"? The spaces are related by a conformal transformation but they are not isomorphic as Riemannian manifolds, because the curvature is not zero in one of the two.

Or do you reserve the term "diffeomorphism" for the (not necessarily Riemannian) differential structure?

ACuriousMind

A diffeomorphism is a smooth bijective map whose inverse is also smooth.

If it preserves the Riemannian structure, it's an isometry

Danu

Right, we've got the word isometry for that.

Slereah

Oh

I guess physicists tend to use the word diffeomorphism a bit more narrowly in GR usually

Danu

14:50

I hate physicists.

7

0celo7

this place

Slereah

Everybody hates everybody

0celo7

@Slereah diffeomorphism in GR is a bit more narrow, I agree

Slereah

the physicists, the math people, the engineers, the experimentalists, the theorists

0celo7

why do people hate engineers

seriously

Slereah

14:52

We don't hate them

We just feel sorry for them

0celo7

that doesn't even make sense

because they feel sorry for you

Danu

@ACuriousMind By the way, in the equation right about (1.1) in Schottenloher, when he writes $(\phi^*g')(a)$ he means $a\in U$ and so he really means $(\phi^*g)\big|_{T_aM}$, right?

ACuriousMind

@Danu Yes

@Slereah That may be even worse than confusing groups and algebras :P

Danu

@ACuriousMind Seriously... I didn't realize how bad it was :\

0celo7

damn, 1TB SSD for $340!

Danu

14:54

@ACuriousMind I think that that's disgracefully bad in terms of notation, too.

Also $T\phi$ for $d\phi$ is just inexplicable to me

Slereah

@ACuriousMind Algebras are just vectors of groups, obviously :p

0celo7

@ACuriousMind I don't think there's a difference

Danu

@0celo7 wow, prices went down that fast?

Only a year ago I bought a 250 GB for around 200...

0celo7

amazon.com/Samsung-2-5-Inch-Internal-MZ-75E120B-AM/dp/…

click over to 1TB

Danu

Wow

That's the exact one I bought a year ago (well 840 but okay)

I love it! :D

SSD prices can't get low enough IMO

0celo7

14:56

that's...tempting to put on the list

ACuriousMind

@Danu Well, it's the map on the tangent space :P

Danu

@ACuriousMind Blurgh

0celo7

but it's still way more expensive than a 7200RPM drive

ACuriousMind

It's better than having the $d$ being both exterior derivative and differential, imo

Danu

@0celo7 have yo uever booted up your computer with an SSD? :)

0celo7

14:57

fucking notation

Danu

@ACuriousMind Really? I disagree

I reserve $\mathrm{d}$ for the exterior derivative.

0celo7

@Danu ::looks at computer with 1TB SSD::

every time

Danu

They only coincide on functions

0celo7

literally

Danu

and there you can just use both

@0celo7 Then don't switch back---you'll be disgusted by the slowness.

0celo7

14:58

@Danu my plan was to get a 120GB SSD for OS and one or two games

and get a 1TB HDD for storage

Danu

I have a 250GB for stuff and then external HDDs for storage.

0celo7

external?

I'd just shove them into the tower

but same concept

Danu

I deal with a laptop mostly :P

0celo7

the 250 is only $85...

@Danu me too

Slereah

The tangent space is obviously just a copy of $R^2$ that you balance at the point

In a space of higher dimension

0celo7

15:02

yeah, simple :)

Slereah

and the metric tensor is just the dot product of the tangent vectors of that space!

0celo7

YES

finally someone puts geometry in words mere mortals can understand

Slereah

(that is how I learned GR)

it's not a bad way to do it really

0celo7

Zee doesn't even do that

the metric is just this thing

Slereah

Yeah most books go straight to MANIFOLDS

0celo7

15:03

that exists

Slereah

They don't really do the whole

Historical path

0celo7

Zee does not even mention tangent space

Slereah

Which is weird because every other physics books do it for everything

0celo7

are you on mobile?

Slereah

No

0celo7

15:06

hory shite a 3TB drive is only 85 bucks

Slereah

The Future!

@ACuriousMind : Are great diffeomorphisms also diffeomorphisms btw

Like Dehn twists and such

0celo7

great diffeomorphisms?

ACuriousMind

@Slereah I know the terminology of "large" and "small" transformations, where the small ones are the ones homotopic to the identity and the large ones are not.

0celo7

I'll twist your Dehn :)

Slereah

@0celo7 : Please don't

It sounds painful

(It involves cutting)

Dehn twist is cutting a torus, twisting it 360° and stitching back up

0celo7

15:15

you thought I didn't know that

sad face

Slereah

I did

Because Dehn twists are pretty rare in physics

Hm

Watching Breaking Bad, I am wondering

Is there any drug people on chemistry SE

0celo7

16:44

You killed the chat, @Slereah

vzn

@dmckee wikipedias thorough pg doesnt say anything about "conspiracy" :P

Loopholes in Bell test experiments

In Bell test experiments, there may be problems of experimental design or set-up that affect the validity of the experimental findings. These problems are often referred to as "loopholes". See the article on Bell's theorem for the theoretical background to these experimental efforts (see also J.S. Bell). The purpose of the experiment is to test whether nature is best described using a local hidden variable theory or by the quantum entanglement theory of quantum mechanics. The "detection efficiency", or "fair sampling" problem is the most prevalent loophole in optical experiments. Another loophole...

dmckee

That was my word, but I stand by it.

vzn

↑ long list!

actually the word "conspirational" is used in some mainstream refs. found a great book on that subj yrs ago with it in the title.

dmckee

Note also arxiv.org/abs/1508.05949

vzn

oops my memory is slightly off it was called "the infamous boundary" by Wick.

but he does talk about the "conspiracy" inside the book. it may have originated with commentators eg mermin or maybe bell himself...?

DS recently hooked me up to an amazing recent ref he coauthored re bells test vs weak measurements.

0celo7

16:54

Methinks quantum mechanics is itself a consipracy to cover up mistakes in the Matrix. In particular, $\hbar$ is just a rounding error.

vzn

yes the Hensen et al work is a tour de force. however my opinion/ thinking/ feeling is the "event ready detectors" construction is just a displacement of the fair sampling/ detector efficiency loophole.

dmckee

> "Experimental loophole-free violation of a Bell inequality using entangled electron spins separated by 1.3 km"

Also, my login state has dropped three times in the past hour, so I may go away at any moment.

0celo7

$[q,p]$ should be zero, but a rounding error makes it not zero.

dmckee

Which won't be a comment on how entertaining y'all are.

@0celo7 That implies a fixed-point system. Very unusual.

0celo7

@dmckee huh?

vzn

16:57

so dmc (playing devils advocate here) presumably you felt the bell testing was pointless after aspect et al in the 1980s?

dmckee

@0celo7 In floating point systems the size of the rounding error depends on the sacle of the measurement. $\hbar$ is consistent across different scales, so the underlying system must use fixed-point arithmetic.

0celo7

uh, sure

the Matrix is mysterious

dmckee

@vzn Of course not, you keep doing it until you have done it right. It's just that once the big gaps are closed you stop expecting anything new.

At this point a change would be a surprise, not continuing to get the same results.

0celo7

internet is slow today

YouTube isn't defaulting to 1080p :/

vzn

Sep 16 at 22:22, by DanielSank

@vzn http://web.physics.ucsb.edu/~martinisgroup/papers/White2015b.pdf

0celo7

17:01

480p! My eyes!

vzn

@dmckee agreed at this point it would be a massive, nearly unthinkable/ inconceivable "upset" if anything anomalous was measured in this area. however think another adj is relevant wrt the long list of loopholes. "suspiciously long".

@0celo7 there are some semi-mainstream experiments now purporting to test "universe as simulation," the idea seems to be gaining increasing credibility/ interest/ traction these days, need to write up/ blog on a batch of links on that.

0celo7

17:18

semi-mainstream?

only a matter of time before the overlords shut them down

those poor guys will have mysteriously shot themselves twice in the heart with a gun that's nowhere to be found

0celo7

18:00

@ACuriousMind Why is $\mathrm{Sp}(2)\cong\mathrm{SL}(2,\mathbb{R})$? If $S$ is a symplectic matrix, I find $\operatorname{det}S=\pm 1$. What is special about $2$ that prevents a negative determinant?

ACuriousMind

@0celo7 For me to answer that you'll have to give your definition of $\mathrm{Sp}(n)$ because people tend to call different things "the" symplectic group.

0celo7

Its elements preserve the skew-scalar product.

ACuriousMind

Over the reals or the complex numbers?

0celo7

Real afaik

Arnold doesn't consider complex numbers atm

wait

doing a calculation

this might hold the answer

ACuriousMind

k, then it's just that the symplectic matrices never have negative determinant, so there's nothing special about 2 ;)

0celo7

18:07

@ACuriousMind wai

an explicit calculation shows the determinant has to equal 1

ACuriousMind

@0celo7 Just look at the Wiki article, the Pfaffian is the elegant way to show this.

0celo7

so why does Sp not always equal SL

if they're just matrices with unit determinant

ACuriousMind

they're not, being symplectic is a stronger condition than having unit determinant

0celo7

Pfaffian...where does that show up in physics?

something something string theory?

ACuriousMind

@0celo7 You might be thinking of Pfaffian forms, which are something different, another name for closed 1-form, iirc.

The matrix Pfaffian might also show up as a Jacobian of some transformation somewhere

0celo7

18:11

Pfaffian shows up in...characteristic classes

so that would show up in the study of anomalies in ST

that identity is suspect

how does one derive it

the one in the symplectic article involving the Pfaffian

ACuriousMind

@0celo7 By direct computation, I fear.

0celo7

I fear it too

well, the exercise only asks for the 2x2 case

which isn't bad at all

so, apparently Sp(2) is homeomorphic to the interior of a solid 3-dimensional torus

wat

@ACuriousMind hint pls

ACuriousMind

@0celo7 Hint for what? Am I supposed to know that homeomorphism off the top of my head?

I've never really looked at symplectic groups, I think

0celo7

first, what does "interior of a solid 3-dimensional torus" mean

how is that set defined

@ACuriousMind yes

ACuriousMind

@0celo7 Hmmmm...I think the "3-dimensional torus" here is just the usual donut.

Since $\mathrm{Sp}(2)$ is three-dimensional, the dimensions match.

And interior presumably means exactly that - the boundary of the torus is not part of the manifold, making it non-compact.

0celo7

18:22

glazed or sprinkled?

ACuriousMind

@0celo7 Whichever you like ;)

0celo7

well, since Sp(2)=SL(2), is SL(2) also homeomorphic to to the donut?

ACuriousMind

It would seem so, but I cannot recall ever having heard that (doesn't mean it isn't true)

0celo7

I don't recall hearing it either

And that really means nothing

looks up group geometry people in math department

I might stop by one of their offices

I don't know where to begin here.

does this have something to do with the fact that the torus is $\mathrm{SL}(2,\mathbb{Z})$, @ACuriousMind?

ACuriousMind

@0celo7 Or you might try googling. Check this out (page 3)

@0celo7 The torus is not $\mathrm{SL}(2,\mathbb{Z})$.

0celo7

18:32

@ACuriousMind uh

you're right

the modular group of the torus is that

or...something

@ACuriousMind damn, you're good at this googling stuff

that seems like a really complicated proof for an exercise

considering Arnold does not talk about that decomposition at all

Slereah

19:46

@0celo7 It's the Moebius transformation group, no?

Or something

Dunno what it is geometrically

Oh wait

$Z$

Not $C$

0celo7

21:32

chat?

chat is kill

obe

21:43

It's lonely here.

0celo7

did you ever do that problem set

NeuroFuzzy

22:11

In need of a sanity check: For atomic wavefunctions, Griffiths just kind of says "Well it's not Hydrogen but we'll just call the electron configurations $|\psi_{n \ell m} \rangle \otimes |\uparrow\rangle$" or something along those lines. But really, since $\hat{H}$, $L^2$, $L_z$ form a complete set of commuting operators

well... does $|\psi_{n\ell m}\rangle$ still have any meaning?

oh... wait $H$ doesn't commute with stuff any more with the interaction terms, huh...

Egh, I'll read Feynman and if that doesn't help bug my profs.

Well, sanity check needed on this statement: Since $H$ is invariant under rotations, at least $L_{z 1}\otimes L_{z 2}\cdots$ commutes with $H$ so the total orbital/spin/orbital+spin values still make sense.

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