The h Bar

0celo7

16:00

@BernardMeurer uh, what

well, I think cigarettes are line bundles

so you could be smoking a hamiltonian system over a circle

Which includes a manifold

So I buy it, very clever.

@ACuriousMind Are $C^k$ manifolds Banach spaces?

ACuriousMind

16:16

@0celo7 Sorry, but that's a stupid question.

0celo7

ACuriousMind

@0celo7 That's not saying the $C^k$-manifold is a Banach space. It's badly expressing that the space of $C^k$-functions is a Banach space.

0celo7

@ACuriousMind Oh, that makes more sense.

(Yes, I know why it was a stupid question; manifolds aren't vector spaces.)

@ACuriousMind Why aren't the smooth functions a Banach space

Inb4 counterexample

Inb4 functional analysis

sharaf zaman

Hello

skill patrol

hello again

sharaf zaman

16:25

oh! i remember you!

still no answer for my question

This proves chemists are not intelligent @skillpatrol

skill patrol

:-/

proof?

and you said you liked maths...

sharaf zaman

yes! lol

Slereah

@0celo7 There are sequences of smooth functions that do not converge to smooth functions

0celo7

@Slereah like what

@sharafzaman that proof was completed long ago

Slereah

Any discontinuous function you can express as a Fourier series, for instance

The series at order $n$ is smooth, but does not converge to a smooth thing

sharaf zaman

16:30

@0celo7 i agree with you but @ACuriousMind doesn't accept

0celo7

@sharafzaman he's just too nice to everyone-{me}

sharaf zaman

@0celo7 hahaha! great use of notation

ACuriousMind

@0celo7 The smooth functions are dense in all $L^p$ and all Sobolev spaces, showing they are not complete.

0celo7

@ACuriousMind ...ok

sure

sharaf zaman

hey if a function is non-differentiable at a point, is this always true that it will have a sharp edge at that point (always?)

ACuriousMind

16:38

@sharafzaman What does "sharp edge" mean?

Slereah

If it's not differentiable and continuous, then yes

sharaf zaman

i mean the graph at that point will be sharp like $|x|$ has at 0

ACuriousMind

I'm not sure I would say that of a fractal like the Weierstraß function

Slereah

Weierstrass is the pointiest function

It has points at all scales

0celo7

@ACuriousMind If $(V,J)$ is a real vector space with complex structure, how does one show that there is an isomorphism $\phi:V\to\mathbb{C}^n$ for which $\phi(Jv)=\mathrm{i}\phi(v)$?

ACuriousMind

16:44

I have the feeling we did pretty much that already when you were proving things about almost complex structures in Arnold.

0celo7

$\phi$ is linear, does one expand in a basis?

ACuriousMind

Also, what have you tried?

@0celo7 Try it!

0celo7

@ACuriousMind I tried expanding in a basis!

But the problem is that we're looking at $V$, not the complexification.

And AFAIK $J$ does nothing special on $V$ itself.

ACuriousMind

What do you mean, "nothing special"? It's a complex structure, that's pretty special

0celo7

@ACuriousMind Perhaps.

@ACuriousMind But the problem is that I have no clue how $J$ acts on a basis of $V$.

I know how to deal with $V^\mathbb{C}$.

@ACuriousMind The standard method is to take a basis $\{e_i,f_i\}$ for which $Je_i=f_i$ and $Jf_i=-e_i$

But one must first show that all vector spaces with complex structure are isomorphic to that one

Then the standard constructions work on all of them

@ACuriousMind Well we know that $V\cong\mathbb{R}^{2n}$ as a vector space, so maybe $\phi$ is the composition of that isomorphism and something that preserves the complex structure?

ACuriousMind

16:55

@0celo7 Try it! I won't tell you what to do. Fail until you succeed.

0celo7

@ACuriousMind Try what?

user116211

Are all linear combination quantum superposition?

ACuriousMind

@0celo7 You say "Maybe $\phi$ is the composition of that isomorphism and something that preserves the complex structure". So try finding that "something".

user116211

2

A: How is a molecular orbital a 'quantum superposition' of the atomic orbitals?

Same story as with the previous question: quantum superposition is always expressed mathematically as a linear combination, but the converse is not necessarily true. Not each and every linear combination expression has something to do with real physical quantum superposition. In many cases it is ...

ACuriousMind

@user36790 What do you mean?

every quantum state can be written as the (quantum) superposition/linear combination of others

sharaf zaman

16:57

0

Q: why do we call lone *pairs*?

According to VSEPR theory molecules adjust their shape to minimize the effect of repulsion. suppose i take the structure of $ClF_3$ i has 3 bond pairs and two lone pairs. So to minimize the repulsion it gets T-shape. the two lone pairs are at corners of a trigonal bipyramidal shape. But my doub...

user116211

> quantum superposition is always expressed mathematically as a linear combination, but the converse is not necessarily true. Not each and every linear combination expression has something to do with real physical quantum superposition. In many cases it is just a mathematical trick.

user116211

@sharafzaman use \mchem

sharaf zaman

ok!

i am new to all languages

ACuriousMind

@user36790 That's wrong. There is no such thing as a "real physical quantum superposition".

user116211

@ACuriousMind @ACuriousMind: I'm coming.....wait.... toilet

ACuriousMind

16:58

@user36790 Uhhhh...what?

sharaf zaman

lol

@ACuriousMind suppose i keep a proton in between two electron will its repulsion decrease or not?

ACuriousMind

@sharafzaman I don't understand the question.

sharaf zaman

@ACuriousMind suppose between two electrons there is a nucleus (electrons are on opposite side) will the repulsion between electron get less or it will be same

ACuriousMind

@sharafzaman Of course a positive charge will attract the electrons. (Note that it doesn't make sense to speak of "opposite sides" on atomic scales, electrons in atoms don't hav definite positions)

sharaf zaman

i see!

what is difference between meta SE and normal SE

John Duffield

17:11

@0celo7 : and you're getting bogged down with mathematical abstraction.

No reply necessary.

0celo7

@JohnDuffield please explain

ACuriousMind

@sharafzaman meta is where you ask questions about the normal SE site, and where discussions about policies happen.

John Duffield

@0celo7 : no. Because you aren't listening.

bolbteppa

@JohnDuffield Why do you hate mathematical abstraction so much when it predicted special/general relativity, positrons and stuff like that?

0celo7

@ACuriousMind Ok, I'm really confused. If I take $\phi=\mathrm{i}\tilde\phi J^{-1}$, where $\tilde\phi$ is any isomorphism between $V$ and $\mathbb{R}^{2n}$, I get a similar equation, namely $\phi(Jv)=\mathrm{i}\tilde\phi(v)$.

John Duffield

17:23

@bolbteppa : I don't hate mathematical abstraction at all. What I dislike is people confusing abstraction with reality.

ACuriousMind

@0celo7 What is $\mathrm{i}\tilde{\phi}$ supposed to mean? there is no multiplication with $\mathrm{i}$ on $\mathbb{R}^{2n}$.

skill patrol

@JohnDuffield why will you not answer my last question?

0celo7

@ACuriousMind Hmm.

And $V$ is not naturally isomorphic to $\mathbb{C}^n$, is it?

John Duffield

@skillpatrol : I've only just got to it.

skill patrol

ok

take your time :)

John Duffield

17:32

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ : can I explain the meaning of that sentence? Not specially. Einstein's writing about simultaneity and time. IMHO the best way to think about simultaneity is to use collisions.

bolbteppa

@JohnDuffield Where exactly has mathematical abstraction led us astray from reality in a negative way?

skill patrol

@JohnDuffield ok, thanks for looking at it

John Duffield

@bolbteppa : people don't understand time because people like John Rennie says it's a coordinate, so they don't understand that the speed of light isn't constant, so they don't understand gravity, so they don't understand electromagnetism, so they don't understand QED, and so on. Then they start rabbiting on about supersymmetry instead of seeing that the Standard Model needs completing.

There's been no significant advances in physics for 50 years, and it's lost its place as the "queen of the sciences". Particle physics in the USA is now history.

user116211

@ACuriousMind So, he is wrong?

user116211

@ACuriousMind: All linear combinations are superposition?

ACuriousMind

17:44

@user36790 I don't know what "superposition" is supposed to mean if not "linear combination"

0celo7

@ACuriousMind Could you please supply a hint?

ACuriousMind

@0celo7 No.

bolbteppa

Is he right that the speed of light is not constant in general relativity? I know the speed of light is constant in an inertial frame right? But in GR can't you always locally transform the metric to be locally the Minkowski metric, so isn't the speed of light constant when you do this?

user116211

@ACuriousMind is superposition only for orthogonal basis or is it applicable for non-orthogonal also?

ACuriousMind

@user36790 I repeat, I don't know what "superposition" is supposed to mean if not "linear combination".

0celo7

17:48

@ACuriousMind because you don't know or because you don't want to

Slereah

@bolbteppa : Define "speed"

user116211

@ACuriousMind yes! got it! So, it's applicable to all basis, right?

ACuriousMind

@0celo7 Because I don't want to, but why does it matter?

bolbteppa

@Slereah 'velocity of propagation of interactions'?

ACuriousMind

@user36790 Yes.

Slereah

17:49

@bolbteppa How do you define velocity when distance and time depend on the coordinate system, though

and you can't compare directly two vectors at two different points

ACuriousMind

@bolbteppa I think Slereah wanted you to define the speed part, not the "of light" part.

Slereah

Velocity is kind of a hard thing to define in GR

Depending on how you define it things can go faster than light, yes

user116211

@ACuriousMind I've read Feynman; he solely discusses about orthogonal basis; can you give me an example of non-orthogonal basis?

ACuriousMind

@user36790 Coherent states for the quantum harmonic oscillator.

Slereah

I already told him that yesterday :V

HE DID NOT LISTEN

user116211

17:52

@Slereah o.O

ACuriousMind

@Slereah That seems to be a theme in this chat lately :P

bolbteppa

"Given this situation, in the presence of more complicated frames and/or gravity, relativity generally relinquishes the whole concept of a distant object having a well-defined speed. As a result, it's often said in relativity that light always has speed c, because only when light is right next to an observer can he measure its speed— which will then be c. When light is far away, its speed becomes ill-defined. "
http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

Slereah

yesterday, by Slereah

Coherent states form a non-orthogonal basis

user116211

@Slereah Have I denied that you didn't say anything to me?

Slereah

Well no, but why would you ask!

bolbteppa

17:54

So clearly it's fine to say that the speed of light does remain constant so long as you consider it's motion relative to a locally-Minkowski coordinate system following the light right? In other words, saying the speed of light changes is incorrect.

John Duffield

@bolbteppa : they end up talking about stuff like closed bosonic strings because they don't understand gamma-gamma pair production, and they don't understand that the electron is a standing wave, or that bosons are waves on an open path whilst fermions are waves on a closed path.

Slereah

In a local frame the speed of light is always the same, yes

user116211

@Slereah I had problem in understanding the quote of the answer at Chem SE; that's why I asked and I don't think it's prohibited, right?

Slereah

It is highly illegal

I am calling the police right now

user116211

@Slereah non-bailable?

bolbteppa

17:57

@JohnDuffield pretty sure your lack of basic linear algebra (diagonalizing a quadratic form) is causing you to make massive errors about the physics of why light is constant.

Slereah

It is immediate death penalty

user116211

@Slereah ＼(◎o◎)／！

user116211

@Slereah It's dictatorship!!!

John Duffield

@bolbteppa : the speed of light varies in the room you're in. The locally-Minkowski coordinate system is of infinitesimal extent, and is nowhere precisely realized in the real world. See the second paragraph here.

user116211

@Slereah If I be hanged, I would be called martyr and revolution will begin!! The Dictator will fall!!

user116211

17:59

The Great Revolution !!!

bolbteppa

@JohnDuffield but that is a completely different issue, you are missing the whole point of physics thinking this way, you may as well deny Galileo's postulates that bodies move in a straight line in the absence of forces because we can never experimentally observe this :\

user116211

in The Periodic Table, 2 days ago, by user36790

7

A: Do we need an orthonormal basis in Quantum Mechanics?

If two states are orthogonal, this means that $\langle \psi | \phi \rangle = 0$. Physically this means that if a system is in state $|\psi\rangle$ then there is no possibility that we will find the system in state $|\phi\rangle$ on measurement, and vis versa. In other words the 2 states in some s...

Slereah

Orthonormal basises are nice because it simplifies calculations

user116211

@ACuriousMind: Is non-orthogonal basis meaningless?

ACuriousMind

@bolbteppa: It's a futile exercise to argue that point with him. Different people have tried it on various occasions in this chat, and on the main site. He never listens, and declares victory when they get tired of the discussion.

Slereah

18:01

No

sharaf zaman

at what age did you guys started quantum mechanics course

Slereah

Isn't there a theorem that states that there's no orthonormal basis for continuous observables or something?

ACuriousMind

@user36790 Well...it's not a basis of states that you could all get as the result of measuring the same operator (eigenbases are orthonormal), but that doesn't mean it's meaningless

bolbteppa

@ACuriousMind sure, just want to make sure I understand the kinds of things people who don't study math make mistakes over

sharaf zaman

@user36790 it means you don't know quantum mechanics

ACuriousMind

18:02

@Slereah Yes, but that's again the issue of the continuous eigenket not lying inside the actual Hilbert space

bolbteppa

@JohnDuffield your little quip about "infinitesimal nature" has just rejected all of calculus and modern science, if only you did a bit of actual study you'd see that

John Duffield

@bolbteppa : no, I'm not making the massive mistake.

@bolbteppa : I have done the study.

user116211

@sharafzaman I'm illiterate.

user116211

@FenderLesPaul: o/

FenderLesPaul

@user36790 sup?

user116211

18:08

@FenderLesPaul what would you do to make 29 Feb memorable ?

user116211

Apart from marriage or divorce or murder?

FenderLesPaul

I'll eat a lot of cereal

user116211

@FenderLesPaul ^_^

bolbteppa

@JohnDuffield are you telling me that, at any point along the path of light moving in a curved gravitational field (giving the appearance of faster/slower than light), when we transform to a locally-Minkowski coordinate system for the light beam, we will measure a speed different than $c$ or no?

user116211

@ACuriousMind: same question

ACuriousMind

18:11

@user36790 Why would I want to make it memorable?

It's just another day

user116211

@ACuriousMind (－‸ლ)

user116211

@ACuriousMind If someone be born in 1972 in 29 Feb, he would be 11 by now.

user116211

marketwatch.com/story/…

user116211

^ 9 reasons to celebrate 29 feb

ACuriousMind

Which are...discounts at various stores. Yay, consumerism!

John Duffield

18:15

@bolbteppa : you always measure a local speed of light to be 299,792,458 m/s because the motion of light defines your second and your metre, which you use to measure the local motion of light. So you measure the local speed of light to be 299,792,458 m/s at the top pair of parallel mirrors, and at the bottom pair.

bolbteppa

Seems to me like, for analogy's sake, he's basically letting friction cause him to deny Newtonian mechanics

user116211

@JohnDuffield pinball?

ACuriousMind

Seriously, you're not going to get me excited about some random date

user116211

@ACuriousMind arg..

John Duffield

See the tautology here: arxiv.org/abs/0705.4507

@user36790 : a depiction of light moving back and forth in two parallel-mirror light clocks at different elevations. The lower clock goes slower when it's lower because light goes slower when it's lower, just like Einstein said.

user116211

18:18

@JohnDuffield O.O

John Duffield

Time for tea.

user116211

@JohnDuffield I prefer coffee.

bolbteppa

@JohnDuffield you claimed that the speed of light is not constant in general relativity, now you are admitting the speed of light is constant in general relativity, but that when you view the light from a different perspective then the speed of light changes :\ (I'm pretty sure!!!) That's also even true in special relativity if you view it from a non-inertial frame, you are not making much sense, even the passage you quoted agrees with what I'm saying not you, did you finish the whole paragraph?

user116211

> The lower clock goes slower when it's lower because light goes slower when it's lower, just like Einstein said.

user116211

When did Einstein say that?

user116211

18:21

@ACuriousMind: Have you any idea when Einstein said the above?

bolbteppa

I think you are letting the language used in the paragraph you posted i.sstatic.net/nI3OV.jpg mislead you, I can see how it would confuse you

ACuriousMind

@user36790 Don't play his word games.

bolbteppa

Okay so maybe all those nobel prize winners were not misleading themselves into thinking they understood relaitivity, qed and stuff like that after all :\

0celo7

@ACuriousMind It matters because I want to make sure the result is correct. Fool's errand and all that.

Slereah

Oh wait

I remember now

sharaf zaman

18:23

@user36790 hey don't lie you are not illertrate

Slereah

even in the simplest case, sometimes energy levels switch due to electron electron interaction

user116211

@sharafzaman I know nothing.

user116211

Doesn't mean I know nothing; it means I'm nothing to all the experts here.

sharaf zaman

@user36790 i agree so do i!

i am also nothing to these experts

user116211

@sharafzaman O.O

sharaf zaman

18:26

@user36790 CHEERS!

user116211

@sharafzaman \o/

sharaf zaman

@user36790 i am nothing compared to u as well, i am just a high school student and you a researchist

user116211

@sharafzaman nay! Who said I'm a researcher? I'm also a high-schooler!

sharaf zaman

@user36790 whats your age?

user116211

@sharafzaman 18

sharaf zaman

18:29

@user36790 GENIUS!!

bolbteppa

You guys should get busy reading en.wikipedia.org/wiki/Course_of_Theoretical_Physics

user116211

@sharafzaman (+_+)

sharaf zaman

@bolbteppa i like lectures by feynman

bolbteppa

To me Feynman is all over the place

FenderLesPaul

I like Feynman by lectures

user116211

18:30

@FenderLesPaul III is the best

user116211

Followed by II

FenderLesPaul

I like II the best

I don't like III

II is quite good though

sharaf zaman

are you saying by volume 2 and three of feynman

user116211

@FenderLesPaul Well, at-least it helped me introducing QM.

user116211

@sharafzaman yep

user116211

18:32

@FenderLesPaul The languages are kinda simple

sharaf zaman

:( i am still at vol. 1 !!! hahahahahahahahahaha

FenderLesPaul

I tend to dislike most books on QM

including Feynman's treatment of it

Feynman knew QM as well as anybody but he tries to teach it in too "intuitive" a manner

and it really just obfuscates the underlying structure of QM

I like Landau's QM the best as far as QM treatments go

sharaf zaman

??

user116211

@FenderLesPaul As I said, you saw the world; have experienced it; I'm just a baby in QM

sharaf zaman

what book i should read for QM

user116211

18:33

@sharafzaman First complete electromagnetism

sharaf zaman

@user36790 u are a baby in QM, i am not even born then.

user116211

@sharafzaman hahahaha

sharaf zaman

@user36790 do u have any chemistry course in ur high school

user116211

@sharafzaman course?

sharaf zaman

@user36790 i mean are studying chemistry or not

user116211

18:36

@FenderLesPaul: Could you understand Feynman's derivation of Liénard and Wiechert potential?

user116211

@sharafzaman Yes

sharaf zaman

@user36790 it is boring or interesting

user116211

@sharafzaman apart from industrial chemistry and metallurgy, I really like chem.

sharaf zaman

then this question is for you

0

Q: my question is about hybridization and a bit quantum mechanics?

I have read in many books that hybridisation is the intermixing of atomic orbitals to form molecular orbital of almost same energy. Hybridisation always takes place when orbitals does not have much energy difference like it can't take place between 2s and 5d Now I know hybridisation can t...

physical-chemistry hybridization valence-bond-theory

bolbteppa

@JohnDuffield en.wikipedia.org/wiki/Course_of_Theoretical_Physics

user116211

18:41

@sharafzaman hybridisation is not intermixing; it's superposition of atomic orbitals as orthogonal basis with definite coefficients that determine the directional character of the wavefunction.

sharaf zaman

@user36790 WOAH!!WOAH!!WOAH.. i am only 15!
tell in layman language

user116211

@sharafzaman okay intermixing.

sharaf zaman

then why 3d not 4s?

user116211

it's like constructive interference

sharaf zaman

if it is constructive it doesn't indicate that the probability of elctron should be in 3d not in 4s

user116211

18:46

@sharafzaman not said in that context; I'm reading your question, wait.

user116211

@sharafzaman Just ask the query in one sentence.

user116211

I'm having difficulty in getting through that body.

sharaf zaman

when electrons of 3s 3p excite why do they jump in 3d not in 4s

(i mean probability)

user116211

@sharafzaman why should they go to 4s?

sharaf zaman

because 4s has less energy "hybridization should have minimum energy difference"

user116211

18:51

@sharafzaman BTW, it's promotion of electron, not hybridisation.

sharaf zaman

what is hybridization of phosphorus in PCl5

user116211

@sharafzaman Hybridisation does not solely depend on energy; it's also dependent on the geometry of the participating wavefunctions.

sharaf zaman

@user36790 say hybridization of phosphorus in PCl5, then i think you will understand my question

user116211

@sharafzaman $\mathrm{ sp^3 d}$

0celo7

@ChrisWhite I'm currently working on a proof of the thing.

It's getting very complicated.

I'm using the topological definition of in/out.

Trying to get a condition on boundary charts.

Then I'll use the uniqueness of the normal vector to get a contradiction. Maybe.

This is hardly "easy to prove".

sharaf zaman

18:58

@user36790 CORRECT! my doubt is why it is sp^3d why not sp^3s

0celo7

Frikken Wald.

user54412

Like most proofs, you'll probably look back and realize there was an easier way.

0celo7

@ChrisWhite yes, I'm trying to formalize this

user116211

@sharafzaman As I said, hybridisation depends on the geometry of the participating orbitals; not on energy; however I would wait to see if someone answers your question at Chem SE

0celo7

One possible strategy is to define $\partial M$ as a level set of some function

then I can use methods on the gradient of that function to do stuff

maybe!

sharaf zaman

19:00

@user36790 Then how did u calculate?

user54412

I have this vague notion that the sign comes about via the chart. Like, in one case we consider integration on the manifold, and on the other we have integration directly on $\mathbb{R}^4$.

user116211

@sharafzaman calculate what?

0celo7

@ChrisWhite I need to do some exercises in Lee

sharaf zaman

hybridization for PCl5

0celo7

He has an analytical condition on a normal vector to be in/outward pointing

But it's an exercise

user116211

19:02

@sharafzaman I know its structure and that predicted what hybridisation $\ce P$ would undergo.

sharaf zaman

oh!

0celo7

Oh, also, I proved that for $\{E_\mu\}$ a vielbein, a volume form $\omega(E_0,...,E_n)=1$ iff $\omega=\sqrt{-g}\,\mathrm{d}x^0\wedge\cdots\wedge\mathrm{d}x^n$ in oriented coordinates $x$.

user116211

@sharafzaman BTW, hybridisation is not real; it's a tactic of VB theory to explain these.

0celo7

Maybe this was clear...but I proved it anyway.

sharaf zaman

i see!

0celo7

19:04

where $g$ is calulated as the determinant of $g(\partial_\mu,\partial_\nu)$

DanielSank

@user507974 It's not an easy problem, is it? ;)

user54412

@0celo7 Not that I want to derail your efforts, but you might try the case I was imagining first -- Minkowski with constant t,x,y,z surfaces bounding a box. We all know what in and out mean there, but if you can see where the sign difference is there, it might be easier to generalize.

0celo7

No! I'm doing this abstractly

NO PICTURES

::erases picture of frame moving along $\partial M$::

DanielSank

@0celo7 Picture shame?

user116211

@ACuriousMind oh! Measuring the operator wouldn't give the element of basis! It must be orthogonal to be an outcome of the measurement!

user116211

19:10

There must be some physical implication of non-orthogonal basis ;/

0celo7

uhhhh

what am I actually trying to prove?

well shit, the normal vector is spacelike for a timelike hypersurface

@ChrisWhite is that right?

@ChrisWhite that's where I was trying to get the sign flip -.-

::sobs quietly::

user54412

@DanielSank Are you using stable 1.x matplotlib or new and fancy 2.0?

user54412

they are changing the default color tables in the new version

user116211

@ACuriousMind: So, if I write the Molecular Orbital of $\ce{H_2O} $, it would look like $$|\psi\rangle = |\psi_{\ce{H_{1s}}}\rangle C_1 + |\psi_{\ce{O_{2s}}}\rangle C_2+ |\psi_{\ce{O_{2p_z}}}\rangle C_3 + |\psi_{\ce{H_{1s}}}\rangle C_4$$

0celo7

@ChrisWhite Oh crap. There's a theorem in Lee that says that if $\omega$ is an oriented volume for $M$ and $N$ is an outward pointing vector field nowhere tangent to the hypersurface $S$, then $i_N\omega$ has the same orientation

no mention of the metric

user116211

19:26

$C_1$ would be $\langle \psi_{\ce{H_{1s}}}|\psi\rangle$ ; $C_2$ would be $\langle \psi_{\ce{O_{2s}}}|\psi\rangle$ etcetera, right? In order to make the energy of the wavefunction lower, we would use variational principle, right?

0celo7

@ChrisWhite Well, I've constructed the outward/inward pointing normal vector to the hypersurface

it's unique

0celo7

19:46

@ChrisWhite right about now is when I would like to listen to the hypersphere

some calm, serene hypersphere music

@ChrisWhite Any ideas? I'm thinking that theorem is wrong. We know that if $N$ is outward pointing, then $\omega$ and $i_N\omega$ are consistently oriented. I have constructed the unique outward pointing $N$ on any hypersurface.

We also know that $i_N\omega$ is the metric volume form on the hypersurface if $N$ is a unit vector!

@ChrisWhite So here is where things could go wrong: if $\Sigma$ is a timelike hypersurface, we cannot find a triad on $\Sigma$ such that one vector is timelike.

But then nothing makes sense

@ChrisWhite mb it's something stupid with how Carroll defined the integration measure

DanielSank

20:05

11

Q: Closing Homework with no attempt as Off-Topic

I find myself all too frequently voting to close a question and having to give a custom reason along the lines of "Homework questions with no attempt at a solution are off-topic". In fact I see a fair few others doing the same thing. Although actually in a few cases (like this one) I see that s...

discussion close-reasons homework

^ Electrical engineering is thinking about homework now.

We may be able to offer advice based on our experience.

@DavidZ

0celo7

I'm probably going about this incredibly wrong

I think the math makes sense, but the GR people are doing something stupid

@ChrisWhite I don't understand these indices

why did anyone think index notation was a good idea

user54412

I'm afraid I lost you

0celo7

@ChrisWhite I'm rambling

if you want to understand me, tell me where you lost me

if you don't care...then whatever

I think you might be right that the GR people are defining the induced volume form in a fucky way

and this is producing a sign

because I've checked the coordinate-free math over and over and over

there's no sign ambiguity

@ChrisWhite hmm, my new idea depends on there being an odd number of spacelike dimensions

so that $(-1)^{odd}=-1$

@ChrisWhite do you have carroll handy

user54412

20:32

yes

0celo7

look at Eq. D.34 on page 448 pls

$\epsilon(n)$ depends linearly on the sign of $n$

but AFAICS $\hat\epsilon$ does not

what gives

dmckee

@Slereah That and the bit where he dies at the end of the first movie.

0celo7

@dmckee people don't die in star wars

they become one with the force

::wiggles fingers mysteriously::

user54412

$\hat\epsilon$ does not?

user54412

if I had the same hypersurface (as a subset) but declared it's normal to be pointing the other way, would it's volume element not change?

0celo7

20:47

@ChrisWhite I'm sure it would

but I don't see that from the equation there

the $\sqrt{\gamma}$ one

user54412

claim: if you change the normal direction you must concurrently do an odd number of the following: (a) change the ordering of wedges of $y^k$, (b) change the overall sign of $\hat\epsilon$, (c) understand that the handedness of your coordinate system has changed

0celo7

@ACuriousMind Is this in the right direction: Let $v\in V$ be expanded as $v_iE_i$. Then $v_i\phi(JE_i)=\phi(Jv)=\phi(J^{-1}JJv)=-v_i\phi(J^{-1}E_i)$. So $\phi(JE_i)=-\phi(J^{-1}E_i)$. This is eerily similar to what $J$ does on $\mathbb{R}^{2n}$.

@ChrisWhite I'm starting to think it's some bullshit surrounding (a)

and that you can compensate for the ordering of the wedges by flipping the sign of the normal vector

FWIW, it's an actual theorem that if $n$ points outward $\Leftrightarrow$ $-n$ points inward

one can prove this using that crazy topological definition I gave earlier

DanielSank

@dmckee Yes that did hurt his popularity campaign, didn't it?

dmckee

He was shaping up to be a good villain, but alas he had no staying power.

0celo7

@dmckee I think he was too British.

DanielSank

20:58

@0celo7 Hey, at least he went full Brit. Leia, on the other hand, wavered between L.A. vulgar English and the Queen's English.

0celo7

LA vulgar English?

DanielSank

@0celo7 I was trying to be funny. I just mean "street English" as opposed to high English.

Anyway, I believe Exhibit A will settle this issue:

Princess Leia's Accent

►

@ChrisWhite Ooooh, does putting "claim" at the beginning give it extra ooomph?

I need to start doing that.

0celo7

claim: this proof about complex structures is hard to prove

claim: english are hard

0celo7

21:44

@yuggib please tell me you know something about complex structures on vector spaces

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