The h Bar

Yashas

12:00

I found a pic

the arrangement is so complicated

but also really cool at the same time

Anonymous

@Yashas fair enough :-)

Yashas

I saw the f orbitals for the first time :O

Anonymous

Really ?

Anonymous

I saw it on many websites before

Yashas

I saw the 3D figure for the first time; we had been taught f orbitals to have the shape of double d orbitals

Anonymous

12:04

Ah, roughly that's correct

Anonymous

@Yashas How was eco ?

Yashas

I got a stats question wrong -_-

Anonymous

@Yashas =P

Yashas

Laspeyres index calculation

6 marks :|

such a simple question

Anonymous

You'll get 94. Chill

Yashas

12:10

I am going to write again in June .-.

Anonymous

@Yashas Noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Anonymous

Not again

Anonymous

You've nothing better to do in life? =D

Yashas

I am stuck at 98 with math

I keep doing some mistake every time

the girls somehow score 100 -,-

Anonymous

Bribe the examiner.

Anonymous

12:12

@Yashas Jealous? =D

Yashas

Not possible. I don't write my name or id in my answer sheet; I write those details in a special sheet.

Only the answer sheets are given to the examiner.

Whoever grades it won't know whose paper they are evaluating.

Slereah

Yashas

12:28

in JEE Preparation, 5 mins ago, by Yashas

Minimum possible atomic number of the atom which can have a 4d orbital is ______

1?

rob

@Yashas Bad question. "Have an orbital" is different from "have that orbital filled when the atom is in the ground state". The second version requires you to know something interesting about how the periodic table is shaped. The first version is an annoying trick question, which you've answered correctly.

0celouvsky

sound about right

@Slereah so what's up?

Slereah

I am thinking about lunch

Probably a steak

0celouvsky

@BalarkaSen So what exactly is $\pi^*E$? I'm having a hard time picturing a pullback bundle over $T'X$.

@Slereah I thought you were poor

Slereah

Well

My parents are alright

I don't have to eat potatoes every day

Does the Lorentz transformation change the metric in GR?

0celouvsky

12:42

what Lorentz transformation?

Slereah

Obviously it's diffeomorphic to, but does the metric change at all

Well, the usual one?

0celouvsky

I don't know of a Lorentz transformation in GR

Slereah

A linear coordinate transformation

0celouvsky

So you mean a local transformation?

Slereah

yes

Hm

0celouvsky

12:44

Then you're going to apply the pushforward metric to the result

But it's not global so I don't know what would change

Slereah

i'm guessing it probably does change the metric, unless boosts are a Killing vector

Well just, you know

The components

0celouvsky

@Slereah you can always go to local inertial frames in GR (?)

that's just normal coordinates

Slereah

Local inertial frames are very local in GR

Like one point

0celouvsky

yeah, so I don't know what your question is

you can't apply a Lorentz transformation outside of a point

and on that point it's just a lorentz transformation

which leaves -+++ metrics invariant

Slereah

Well just

Take the chart

Make some new coordinates that are a boost of the old ones

$x = \gamma(x - t)$, that kind of stuff

Do the components of the metric change

0celouvsky

12:47

Obviously they will change

that's not an interesting question

Slereah

I do realize that i could find out by actually working it out but I am fairly lazy

0celouvsky

components always change under coordinate transformations

Slereah

Yeah I'm guessing so

unless it's a killing vector

0celouvsky

you mean you boost along one?

Slereah

yes

Hence why it doesn't for Minkowski space

0celouvsky

12:48

that doesn't really work because you need your Lorentz transformation to be constant

odds are the Killing vector isn't constant

Slereah

what

0celouvsky

what

Slereah

Nevermind

i'm a bit tired

0celouvsky

do you not know what I'm saying or disagree?

Slereah

Been up since yesterday

I'm not quite sure we're talking about the same thing

I was slightly confused because boosts leave the metric invariant in SR

But that's just because boosts are one of the Killing vector of Minkowski space

Since it is maximally symmetric

was what I'm saying

0celouvsky

12:51

Yes, I buy that.

Lorentz transformations are isometries of the Minkowski metric.

The boost is just a flow along the integral curves of the Killing field.

Slereah

and since the Alcubierre trick for CTCs involves boosting the metric I was a bit confused

0celouvsky

@JohnRennie My book has shipped

Slereah

which

Is it Plato's republic

0celouvsky

"Invariance Theory, Heat Equation, and Index Theory."

Slereah

I hope there's a good modern paper on Feynman's absorber theory

0celouvsky

12:55

what's thta?

@JohnRennie Hello

Slereah

Time symmetric theory

Everything is time symmetric

So when a particle radiates, it radiates both in the past and future lightcone

it requires some weird boundary condition to sort of work

0celouvsky

Would you believe that the Spinor for a neutron precessing in a magnetic field is $$\begin{pmatrix} \cos(\theta/2)\\ \sin(\theta/2)e^{i\omega t}\end{pmatrix}?$$

Slereah

That looks believable

0celouvsky

$\theta$ being the fixed azimuthal coordinate

Slereah

I mean if it was $$e^{e^{e^{\theta}}}$$

I would be skeptical

0celouvsky

12:58

that probably solves a very neat ODE

Slereah

I think there's a double exponential in an integral for the Bessel function

Yashas

in JEE Preparation, 1 min ago, by Yashas

How many electrons in a chromium atom at ground state have $n+l+m = 4$ and clockwise spin?

wth

Slereah

Or at least $e^{\cos(\theta)}$

Yashas

2, 3 and 4 are all correct?

0celouvsky

R A P I D G R O W T H wolframalpha.com/input/?i=e%5E(e%5E(e%5Ex))

Slereah

13:00

it's a fairly steep function

during the brief period where I worked on the topic of my PhD thesis I actually did work on some Pauli equation with magnetic fields

Because in some conditions, it has a supersymmetric partner

Yashas

13:29

in The Periodic Table, Mar 30 at 0:46, by orthocresol

5

Simply Beautiful Art

hahaha

beautiful

Simply Beautiful Art

13:44

Lol, hi @Koolman

Koolman

14:15

@SimplyBeautifulArt Hii

Where is @Kenshin ?????

user228700

14:45

@Koolman Dude, he's probably on vacation.

Koolman

oh god

when will he return @Kaumudi.H

user228700

I have no idea...

Yashas

He is probably busy compiling a Physics SE Q&A book.

Koolman

His site is not working

physics.qandaexchange.com

Yashas

lol

wth

session_start() is undefined

he accidently disabled sessions I think

Koolman

14:54

yes

@Kaumudi.H are you sure

Yashas

she said "probably"

user228700

Yes, I am.

Yashas

:O

user228700

Mar 27 at 8:35, by Kenshin

vacation

user228700

Eh, I'm finding it difficult to find the exact message in which he told that he's going to Japan in April.

Koolman

14:57

Its about two weeks

he was last seen two days ago

Kenshin

3.4k 1 17 40

user228700

I said "probably* because I dunno when he's going (/has gone) and when when he's coming back.

Koolman

okay

John Rennie

@0celouvsky What did it cost you in the end?

0celouvsky

15:44

@JohnRennie that's between me and the devil

John Rennie

@0celouvsky Your soul?

You got a good deal. I'd be lucky to get more than a few copies of New Scientist for mine :-)

user228700

@JohnR: How's the eating? (:-P)

user228700

Ah, OK, you're gone...

ACuriousMind

@JohnRennie Pretty sure he told us already he'S sold that one thrice over

John Rennie

The spirit is willing but the flesh is weak

ACuriousMind

15:54

Or was that @BernardoMeurer

Ah, no, I think Bernardo just sold it for a few gummi bears

user228700

@JohnRennie I keep having to Google things you say .__.

ACuriousMind

Ah, here, no soul for @0celouvsky:

Apr 10 at 1:33, by 0celouvsky

Uh, I lost that when I tried understanding Frechet spaces.

user228700

Do old people speak strictly in proverbs?

0celouvsky

Frechet spaces are the worst

John Rennie

Mar 9 at 18:56, by Bernardo Meurer

I sold my soul for a gummy bear in 6th grade

0celouvsky

15:55

I'm lucky my soul is the only thing I lost

@ACuriousMind Do you know what a Frechet space is?

John Rennie

@Kaumudi.H I wouldn't know, not being old :-)

ACuriousMind

@0celouvsky not really, no

0celouvsky

@ACuriousMind Let $X$ be a manifold. Consider the bundle $\pi:T'X\to X$, where $T'X=T^*X\setminus \zeta$, where $\zeta$ is the zero section of that bundle. Now consider $p:E\to X$ a vector bundle. I understand what $\pi^*E$ is, it's a bundle over $T'X$. But what the hell is it?

This is unrelated to Frechet spaces.

John Rennie

@Kaumudi.H I guess Biblical references are rather better known in the UK than ina predominantly Hindu society :-)

ACuriousMind

@0celouvsky I'd guess it's $E$ with its zero section removed

0celouvsky

15:58

@ACuriousMind But it's pulled back over $T'X$

It's not a bundle on $X$

ACuriousMind

Ohhhh

user228700

@JohnRennie Hmm, yes, I suppose...

0celouvsky

@ACuriousMind it's like a pullback but where $f$ is the projection of a bundle on $X$

ACuriousMind

Hm, then I don't know, looks like a rather odd construction to me :P

0celouvsky

@ACuriousMind Blame analysts who are trying to manifold :P

John Rennie

15:59

English, British English at least, is absolutely littered with Biblical references.

user228700

Oh but that hasn't been my experience watching British vloggers.

0celouvsky

If you don't quote the Bible daily what's the point

Fawad

@0celouvsky what?

user228700

Oh no.

0celouvsky

@Fawad you've got to tell some people what Jesus wants at least once a day

@ACuriousMind It's got to do with this "symbol map" of an elliptic PDE on a manifold

PDE on manifolds are terrible

Who thought any of this was a good idea

@ACuriousMind I guess it's not clear to me what the fibers of a pullback bundle are, exactly

I have a map $p(x,\xi):E_x\to F_x$, where $E,F$ are bundles over $X$, and where $(x,\xi)\in T'X$

So why does that give a map $\pi^*E\to \pi^*F$?

Yashas

16:12

Let $f(x)$ be a continuous and infinitely differentiable function such that:

1. $f^n(a) = 0$

$$f(x) = \frac{f(a)}{0!} + \frac{f'(a)(x -a)}{1!} + \frac{f''(a)(x-a)^2}{2!} + \dots$$

All the terms in the Taylor series except the first go to zero. Therefore, $f(x) = f(a)$.

What is wrong?!

0celouvsky

@Yashas The function doesn't have to equal its Taylor series.

Yashas

It is a real valued function.

0celouvsky

I know that.

Yashas

What condition did I miss then?

0celouvsky

Are you stipulating analyticity?

If it's analytic and all the derivatives vanish, then yeah: it's the constant function.

But take any bump function to see that what you said is not in general true.

Yashas

16:14

It was for a general real valued function.

0celouvsky

Then your equality is wrong.

$f(x)=f(a)+f'(a)(x-a)+\cdots$ is in general not true.

Yashas

aw

If it is not analytic then it can be a straight line locally but not globally.

0celouvsky

Yes.

0celouvsky

16:38

@ACuriousMind Ahhhh, I understand why we want pseudodifferential operators. It's so that the operators and their Green's functions are the same types of objects. Gilkey forgot to mention that part.

And other types of pseudo inverses

0celouvsky

17:02

@BernardoMeurer Michelle hasn't said what labor I have to perform

If she waits too long the tickets will be too expensive

John Doe

Hello all

Bernardo Meurer

@0celouvsky I'll talk to her

Madhuchhanda Mandal

If f'(x) =g(f(x)) and f'(x0)=0. Will the graph of f(x) be constant from x=x0 to x=infinity??

ACuriousMind

@0celouvsky Because you get a map $(\pi^\ast E)_y \cong E_{\pi(y)} \to F_{\pi(y)} \cong (\pi^\ast F)_y$ on each fiber?

John Doe

@ACuriousMind Do you have any idea why a two level system Hamiltonian is given by $\hat{H} = \hbar \frac{\omega_0}{2}\hat{\sigma}_z$ and how $\omega_0$ is determined?

ACuriousMind

17:07

@JohnDoe I don't understand the question, i.e. what you're asking for by "why".

Yashas

@MadhuchhandaMandal u shud show your loop solution so that people here would comment on it

ACuriousMind

That there is a possible Hamiltonian for a two-dimensional system, but without more information, it's only once of many. Without more information, it's also not clear why you'd want to determine $\omega_0$ from anything instead of treating it simply as a constant.

Madhuchhanda Mandal

Okay

0celouvsky

@ACuriousMind so...the pullback bundle fibers are the same as the original fibers?

I clearly do not understand pullbacks

John Doe

@ACuriousMind But what is the motivation for $\hat{H}$ and $\omega_0$? Why is the Hamiltonian defined this way? Not sure why that is a strange question...

ACuriousMind

17:10

@0celouvsky Yes, the fiber of the pullback over a point is the same as the fiber of the original over the image of that point.

John Doe

For a two level system of say spin up an spin down that is the Hamiltonian as far as I know. I'm just intersted in the motivation for why it is given that way...

ACuriousMind

@JohnDoe That's not "the" Hamiltonian. But it is the Hamiltonian for e.g. a spin-1/2 particle in an external magnetic field along the $z$-axis.

John Doe

I understand that the Hamiltonian invloves the spin operator $\hat{H} = \omega_0 \hat{S}_z$

ACuriousMind

The $\omega_0$ is them related to the strength of the magnetic field, since the general interaction term between spin and magnetic fields is $\propto\sigma\cdot B$.

0celouvsky

@ACuriousMind ok. So it remains to check that the induced bundle map is smooth...I hate this

John Doe

17:15

@ACuriousMind What if the two level spin state was not in the presence of an external electric or magnetic field. What would the Hamiltonian be then?

0celouvsky

And independent of choices

Madhuchhanda Mandal

f(x+dx)=f(x) as f'(x)= 0. So f'(x+dx)=f'(x)=0. Now (assuming h to be tending to 0 ) f(x+dx+h)=f(x+dx). (As f'(x+dx)=0). So f(x+dx+h)=f(x+dx)=f(x). Thus f'(x+dx+h)=0. Now if we again consider x+dx+h as our initial point, the same thing loops. So f(x) from x0 to infinity is constant if f'(x)=g(f(x)) and f(x0)=0. Is there anything wrong in the argument?

(Original question : If f'(x) =g(f(x)) and f'(x0)=0. Will f(x) be constant from x=x0 to x=infinity??)

Sergey Dylda

Hello everyone. Can someone explain to me, if there any purely mathematical restrictions that force us to use a real-values as a values of observable quantities? If there is no such restrictions why don't we use some arbitrary abstact fields, or even weaker structure like rings/group to represent our physical quantities?

ACuriousMind

@JohnDoe That's not a well-defined question. If there are no fields the Hamiltonian of something that has only spin degrees of freedom would be just 0, but of course every physical system interacts with something

Yashas

US drops biggest 'mother of all bombs' in Afghanistan: Pentagon

GBU-43/B Massive Ordnance Air Blast

The GBU-43/B Massive Ordnance Air Blast (MOAB pronounced /ˈmoʊ.æb/, commonly known as the Mother of All Bombs) is a large-yield conventional (non-nuclear) bomb, developed for the United States military by Albert L. Weimorts, Jr. of the Air Force Research Laboratory. At the time of development, it was touted as the most powerful non-nuclear weapon ever designed. The bomb was designed to be delivered by a C-130 Hercules, primarily the MC-130E Combat Talon I or MC-130H Combat Talon II variants. Since then, Russia has tested its "Father of All Bombs", which is claimed to be four times as powerful as...

lol

the wikipedia page updated already to include today's launch

DanielSank

17:29

@SergeyDylda Science works like this:

You see something happen. You wonder "why" it happens. You make up a story. Then you see if your story says the right thing about other new stuff you observe. If the story works, then you keep it. If it doesn't, you fix it.

Real numbers work. Why update the theory?

Also, you're making an assumption that's incorrect.

0celouvsky

@DanielSank you didn't explain your measure theory comment.

DanielSank

We don't use real numbers for all observable quantities. The number of apples in my fruit bowl has always been an integer, so I represent the number of apples in my theory with an integer, not a real number.

@0celouvsky So?

I also never explained why I used to call you my "monarch".

I owe you nothing.

0celouvsky

Not even basic decency?

Sergey Dylda

@DanielSank My thought is that everything that can be mathematically generallized should be generalized. It is often that generalization brings new branches of solutions that were not accessible before.

@DanielSank We can redefine integers to be a subset of reals.

Anonymous

@SergeyDylda Don't we already do that?

0celouvsky

17:40

It's more like the integers embed in the reals and we use the same symbols. The concepts are distinct.

Sergey Dylda

@blue I think of integers as quotient set of NxN/~ for certain equivalence relation.

0celouvsky

...why would you think of them like that? Are you a set theorist?

Sergey Dylda

@0celouvsky I just like precise mathematical definitions.

Anonymous

@SergeyDylda I don't know why you are making it way more complicated than necessary...

0celouvsky

Ok buddy.

Anonymous

17:43

@SergeyDylda I see.

loltospoon

Hello

0celouvsky

So you think of real numbers as equivalence classes of Cauchy sequences?

loltospoon

Does the orthonormality of the set of wave functions for the particle in a box also apply when you are using an operator?

i.e.

Sergey Dylda

@0celouvsky As a quotion set of almost-homomorphisms

loltospoon

If $\langle \psi _i | \psi _j \rangle = \delta_{ij}$, then what does that mean for $\langle \psi _i | Q |\psi _j \rangle $ ?

Balarka Sen

17:45

@blue Well, that's exactly what it means.

Sergey Dylda

@0celouvsky But that is not the point, the question was about restrictions on observables.

Anonymous

@BalarkaSen I never claimed that it doesn't.

Balarka Sen

It's not "more complicated". Z is the Grothendieck completion of N :)

0celouvsky

@SergeyDylda wot

Anonymous

@BalarkaSen Ah, you are right. I am not sufficiently acquainted with abstract algebra to make any useful comments. Consider my earlier comment to be a layman's perspective of integers.

Balarka Sen

17:54

rigorously these are fine definitions - technically Z is defined from N by "adjoining the negatives". what is a valid quibble is that these are a tad tedious in actually anything other than foundations in math

I almost never think about set theory and ZF when doing the stuff I do eg

0celouvsky

That was my point.

No one actually thinks of negative numbers as equivalence classes of ordered pairs. That's silly.

Balarka Sen

it's a good perspective

John Doe

@ACuriousMind That is exactly the type of answer I was looking for.

@ACuriousMind Would it be a reasonable assumption to consider that a two level system alwalys sits in a magentic field?

DanielSank

@JohnDoe no

Not all to level systems are magnetic spins.

That said, all two level systems can be thought of as spins in a magnetic field because the Hamiltonians have the same form.

Madhuchhanda Mandal

Can I ask a Calculus related problem?

John Doe

18:02

@DanielSank What Hamiltonians have the same form? All two level systems in a magnetic field?

Madhuchhanda Mandal

@DanielSank Can I ask a Calculus related problem?

Anonymous

@MadhuchhandaMandal Why not ask on the main mathematics site?

Madhuchhanda Mandal

Ok. What's that?

Anonymous

$Mathematics$

Madhuchhanda Mandal

Can you send the link?

Anonymous

18:07

Got it?

Anonymous

<https://math.stackexchange.com/>

Anonymous

math.stackexchange.com/questions/ask

Madhuchhanda Mandal

Is any chatroom available?

Of Mathematics?

Anonymous

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

7

6

Anonymous

Yes

Madhuchhanda Mandal

18:08

Thanks

DanielSank

@MadhuchhandaMandal rule #1 of chat: do not ask for permission to ask a question. Just ask the question.

@JohnDoe the Hamiltonians is two level systems are rather restricted, i.e. 2x2 Hermitian matrices.

So it's easy to think of any two level system in terms of your favorite particular case, which could be a spin in a magnetic field.

John Doe

@DanielSank So the form of the Hamiltonian of a spin up and spin down two level system in a magnetic field is $\hat{H} = \frac{\hbar \omega_0}{2}\sigma_z$, what other two level system has this type of Hamiltonian?

Koolman

A man becomes famous ... only when famous people wait to meet him

0celouvsky

@ACuriousMind My thesis topic just did a 180

@Slereah I am going to be working on static space times with compactly supported perfect fluid tensors

There's a classification out there but people think it can be greatly improved

It involves spinors

Cows

18:32

omg guys I went over some topology today. I am happy to report I finally really know why open set and closed set are not opposites. hehehe

yay

I am sooooooooooooooooooooooooooooooo excited ,

Swapnil Das

Congo :)

Cows

especially about topology

yeeeeeeeyaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

yeeeeeeeeeeehaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Swapnil Das

lol you're too much excited.

Cows

Me so happy I can make my own topologies now bitches!

hehehe

math is epic!

Swapnil Das

It is. But when understood.

Cows

18:34

@SwapnilDas hehe

@SwapnilDas yes I am now just discovering how awesome it is

Swapnil Das

Good luck for further intellectual adventures. Are you a math undergrad?

Cows

neh physics

Swapnil Das

Aha great.

Secret

I don't think anyone can understand time travel without non hausedoff or even non T1 topologies

(unless you live in a spacetime cylinder)

Swapnil Das

Hi! @Secret

Physics Meta

18:44

0

Q: Long answers and stackexchange

Is it true that long answers to questions attract more upvotes than shorter precise answers.I have seen many beautiful short answers but they are neglected and overshadowed by the longer ones

discussion support voting

Jaime Gallego

theverge.com/2017/4/12/15259400/…

They wanted to attract attention, and they did. But that campaign is obnoxious

The company itself edited Wikipedia to influence Google Home answers for good publicity

DanielSank

@JohnDoe Well, that's the Hamiltonian if you only have one dc field.

But anyway, yes, there are other systems with that Hamiltonian.

John Doe

@DanielSank What's a dc field?

DanielSank

For example, the superconducting qubits I work with have exactly that Hamiltonian (when we're not also driving them with control signals).

@JohnDoe Constant in time.

dc = direct current. It's an electrical engineering term but it gets used to mean "constant in time".

Electricity where the voltage/current is constant in time is called "direct current". This is the case for batteries.

If the voltage/current is changing in time, then it's called alternating current, or ac.

John Doe

@DanielSank Oh okay. So is it correct to state that a two level system in the absence of any external influence has zero Hamiltonian?

Secret

18:56

@JaimeGallego Luckily, google and wikipedia contained it shortly after

DanielSank

@JohnDoe Ummmm, I wouldn't quite say that.

Zero Hamiltonian means the two states of the system have equal energy.

That's actually fairly uncommon.

Jaime Gallego

@Secret They tweaked the ad's audio to get it working again.

John Doe

@DanielSank I'm asking because I was told that the Hamiltonian I gave is always the Hamiltonian of a two level system on it's own. But the reasoning could be that the two level system is usually influenced by something and as you say since the Hamiltonians are very restricted we could always consider it as what I wrote. Does that make sense?

DanielSank

Yes, it makes sense.

In particular, $\omega_0$ could be zero.

:-)

Mathematics

Transcript for

$Mathematics$ Mathematics