The h Bar

user218912

12:18 AM

star wall is messed up.

12:30 AM

@BernardMeurer Dude, I wouldn't give you enough sympathy on Google hangouts so now you're trying to get it here from ACM? :P

0celo7

Google hangouts @DanielSank ?

@BernardMeurer I sympathize with you

For some reason @ACuriousMind and other think linear algebra is easy

dmckee

@gonenc I had in mind Valter Moretti when I wrote that, but now that I look in his profile I see that he is actually employed as a math professor; albeit specializing in the math of some physical theories.

Most of the others I have seen exhibit fly-by participation rather than anything on-going.

Bernard Meurer

12:46 AM

@DanielSank I had hopes he had some hack or something

@0celo7 This stuff is hard man

vzn

1:46 AM

@rob agreed! hey huge congrats on your election, top competition there. btw still open to a guest session? would be great opportunity to meet you as mod & discuss your bkg more, & maybe if you do one the other "draftees/ maybes/ TBDs" might fall in line also :| ... nov1 is open!

rob

2:16 AM

@vzn The idea is growing on me. However I have a standing midday Tuesday commitment through December and I can't make the regular times. (alternate Tuesdays, yes?)

Alfred Centauri

2:26 AM

@dmckee or worser, vector spherical harmonics

Swapnil Das

Salutations!

DanielSank

@BernardMeurer but all your problems were just arithmetic mistakes!

That's not linear algebra's fault.

user116211

3:40 AM

@0celo7 :((

user116211

Dull day; nothing interesting to read; same old Royden and other stuffs T__T

DanielSank

@MAFIA36790 o/

user116211

@DanielSank \o

Swapnil Das

I'm finding PAT and MAT papers.

user228700

4:45 AM

Hi everyone :-)

user228700

Wrong chatroom, sorry :/

user228700

@JohnRennie: Sir, good morning! :-) Quick question?

John Rennie

Hi Kaumudi :-)

What do you want to know?

user228700

Are u familiar w/ hybridization of atomic orbitals, sir?

John Rennie

Yes

That was an easy question :-)

user228700

4:54 AM

Okay. I was learning about this and in my textbook, it's written that "Maximum overlap of atomic orbitals will result in the strongest bond. Consequently, the bond formed by a given orbital will tend to lie in the direction in which the orbital has maximum value."

user228700

What does the statement in bold mean?

John Rennie

OK, yes, that's a common way of describing what happens ...

user228700

@JohnRennie Bad luck :-P

0celo7

@JohnRennie Hey let's read a book together

John Rennie

@Kaumudi If you consider a hydrogen $p$ orbital then it has an obvious direction. Are you comfortable with that statement?

user228700

4:58 AM

@JohnRennie A hydrogen $p$ orbital? Why are we considering the $p$ orbital of hydrogen, when it has just the one electron sitting in the $s$ orbital?

John Rennie

The atomic orbitals in many electron atoms are roughly the same shape as for hydrogen, so I'm just using hydrogen as a starting point. I tend to do this because the real world of electronic structure is far murkier than degree level textbooks would have you believe.

But hydrogen, having just one electron, is a reliable starting point.

But my statement is about $p$ orbitals in general i.e. that a $p$ orbital has an obvious direction.

So are you happy with the notion that a $p$ orbital in general has an obvious direction?

user228700

Riight...

user228700

@JohnRennie Yes.

user228700

(Sorry about that-my mother called me)

John Rennie

And likewise an $sp^3$ or $sp^2$ orbital has an obvious direction. Yes?

user228700

5:06 AM

@JohnRennie Well, I've only just started learning about this but my textbook says that to have max. overlapping with an approaching orbital, the hybridized orbitals have some direction...

0celo7

@JohnRennie :(

Not an advanced math book

John Rennie

@0celo7 Ah, OK, what book then?

@Kaumudi A quick Google found thi image of $sp^3$ hybridisation:

There are four obvious directions in which the lobes point. Yes?

user228700

@JohnRennie Ooh, OK, tetrahedral.

user228700

@JohnRennie Yup.

John Rennie

Suppose this is a carbon atom and you bring in a hydrogen atom (with the eventual intent of forming methane CH4)

user228700

5:09 AM

OK...

John Rennie

There is an approximation for the process of bond formation called LCAO

Linear Combination of Atomic Orbitals I think.

user228700

Hang on, u mean this:

0celo7

@JohnRennie Weinberg's old GR book

John Rennie

@0celo7 that book has an awful reputation ...

user228700

John Rennie

5:11 AM

... as an attempt to do GR without using geometry.

@Kaumudi Yes

user228700

OK...

John Rennie

The strength of the interaction between the atoms depends on how much their two orbitals overlap.

user228700

Right...

0celo7

@JohnRennie yes I've read the first half

John Rennie

You'll learn how to make this quantitative at university, but for now if you imaging the orbitals as fuzzy clouds then the overlap is literally how much the clouds overalp.

0celo7

5:13 AM

I don't particularly care for geometry so it's ok

user228700

@JohnRennie OK...

John Rennie

@Kaumudi So if you take that picture and imagine a fuzzy ball of hydrogen atom coming in, the biggest overlap will happen when the hydrogen is in one of the four lobes.

user228700

@JohnRennie OK...

John Rennie

So the eventual C-H bond is going to line up along the same line as the direction of the $sp^3$ lobe.

i.e. when we've added all four hydrogens they'll be at the four corners of a tetrahedron.

J. Pak

@JohnRennie From the last 50 posted questions which one is more interesting?

user228700

5:16 AM

@JohnRennie Riight, but what do they mean by "orbital has maximum value"?

John Rennie

I think that just means greatest electron density

user228700

What is this value of orbital they're speaking of?

user228700

OK, alright. So, this hybridization happens only when a reaction actually happens, correct..?

J. Pak

OK, I go to read it!

John Rennie

Remember that an orbital is a wavefunction that is a function of position $\psi(x, y, z)$, with $|\psi|^2$ giving the electron density.

The function $\psi$ has a maximum value that is somewhere in the centre of the lobe.

user228700

5:19 AM

@JohnRennie Yes, right, OK...

John Rennie

And that's all the phrase maximum value means.

user228700

@JohnRennie OK. Hybridization takes place only during (or right before) a reaction, correct?

John Rennie

Hmm, do I try and explain what hybridisation really means ...

user228700

@JohnRennie Uhh, I dunno any QM, essentially, so...

John Rennie

Hybridisation is a convenient fiction that works well for explaining the shape of simple molecules.

I'd be inclined to leave it at that and not probe too deeply.

When you ask Hybridization takes place only during (or right before) a reaction? you're pushing the metaphor beyond its limits.

user228700

5:24 AM

@JohnRennie What I'm trying to do is understand how the hybridized orbitals will be oriented; whether they will align so as to have max. overlap with the incoming orbitals or whether the incoming orbitals will do the aligning.

John Rennie

If you take a hydrogen atom (hydrogen again!) then its electron moves in a spherically symmetric potential. For spherically symmetric potentials we can describe the wavefunction using functions called the spherical harmonics, and those functions give us the $s$, $p$, $d$, etc orbitals. Does this make sense so far?

user228700

Yep.

John Rennie

If you consider methane then the electrons no longer move in a spherically symmetric potential because now you have five nuclei attracting them from different directions. Yes?

user228700

Well, yeah, I guess...

John Rennie

So the electrons in a methane molecule no longer have $s$, $p$, $d$, etc orbitals. For the innermost electrons ($s^2$ in methane) it's not a bad approximation to treat them as unperturbed and still in the hydrogenic orbitals.

user228700

5:30 AM

OK...

John Rennie

However the outer electrons will interact strongly with the other nuclei so the $2s$ and $2p$ orbitals no longer exist. Instead the electrons move in orbitals that depend on the geometry.

user228700

@JohnRennie "Orbitals that depend on their geometry"?

John Rennie

@Kaumudi depend on the positions of all the nuclei present. By geometry I mean the arrangement of the nuclei e.g. at the corners of a tetrahedron in methane.

user228700

@JohnRennie The nuclei are present at the corners?!

John Rennie

An atom is a nucleus surronded by electrons. Yes?

user228700

5:34 AM

:-P Yessir.

John Rennie

So the position of the atom is basically the position of the nucleus.

user228700

Right.

John Rennie

Conceptually I am positioning the nuclei then trying to understand how the electrons will arrange themselves around the nuclei to form the molecular bonds.

user228700

OK...

John Rennie

Anyhow if we take a carbon atom and impose a tetrahedral symmetry then we can no longer describe the outer electons as $2s^22p^2$ because the symmtry is no longer spherical.

user228700

5:36 AM

Right.

John Rennie

But we can construct $sp^3$ orbitals. These are the wavefunctions in a tetrahedral symmetry.

user228700

@JohnRennie Hm, OK...

John Rennie

Suppose the take a methane molecule and change the C-H bond lengs. Start with a very long bond length so the hydrogen atoms are a long way away.

user228700

OK..?

John Rennie

In that case the C atom is largely unchanged and would be $1s^22s^22p^2$.

user228700

5:38 AM

Yes, right.

John Rennie

Now imagine reducing the ond length to bring in the hydrogen atoms.

user228700

Okay...

John Rennie

As you do this the $s$ and $p$ orbitals become increasingly poor approximations for the electron wavefunctions.

user228700

Right, 'cause of these other nuclei.

John Rennie

And the $sp^3$ orbitals become increasingly good approximations.

@Kaumudi Yes.

user228700

5:40 AM

@JohnRennie OK...

user228700

And that's all there is to hybridization?

John Rennie

Basically yes.

user228700

Cool :-) Thank you!

John Rennie

We call the tetrahedral orbitals hybrids because we can write them as a sum of the original $s$ and $p$ orbitals.

There is a sense in which the change in symmetry causes the $s$ and $p$ orbitals to get mixed up with each other.

user228700

@JohnRennie And when u say sum, u mean something too complicated for my brain to comprehend just now, correct? (:-P)

John Rennie

5:44 AM

I mean an equation like: $$\psi_{sp^3} = a\psi_{2s} + b\psi_{2p_x} + c\psi_{2p_y} + d\psi_{2p_z}$$ for some constants $a, b, c, d$

user228700

@JohnRennie OK...

John Rennie

The name $sp^3$ is used because the tetrahedral orbitals can be expressed as a sum of the $s$ and the 3 $p$ orbitals.

user228700

Oh how I wish I'd had teachers like u back in high school. It's two years of my life I'll never get back, sigh.

user228700

@JohnRennie Yes, OK.

John Rennie

Now suppose we had only three hydrogen atoms, not four, and we arrange them in a triangle in a plane around the C atom.

user228700

5:47 AM

@JohnRennie OK..?

John Rennie

Or instead we could consider a boron atom with three hydrogen atoms ina plane

user228700

Yes, this makes sense. OK...

John Rennie

The geometry is not tetrahedral but instead it is triangular.

And because we impose a different geometry we will get different changes to the original $s$ and $p$ orbitals.

So in this case we wouldn't expect to get the same $sp^3$ orbitals that we get for methane.

Instead we would get ... ?

user228700

$sp^2$.

John Rennie

Correct! :-)

$sp^2$ hybridisation is the same process but occurring in response to a different geometry.

user228700

5:50 AM

Yes, OK...

John Rennie

If we go back to methane for a moment ...

user228700

OK, this answers all my questions! :-)

user228700

@JohnRennie OK..?

John Rennie

I started with the assumption that the hydrogen atoms would be at the corners of a tetrahedron

user228700

Yeah...

John Rennie

5:51 AM

But you could ask why can't then be arranged ina square around the carbon atom?

And they could, and this square symmetry would give a different hybrisation.

So why doesn't this happen?

user228700

@JohnRennie I think I know the answer to this; it has something to do with these orbitals wanting to be as close as possible without interpenetration or something like that..?

John Rennie

No, there isn't a simple answer.

user228700

Oh. I read that somewhere tho :/

John Rennie

The square arrangement turns out to have a higher energy than the tetrhedral arrangement, so the molecule forms a tetrahedron.

user228700

@JohnRennie Alright...

John Rennie

5:54 AM

The point is that the books will tell you the carbon atom foms $sp^3$ hybrid orbitals and that's wht we get a tetrahedral molecule.

But that isn't true.

If the geometry is tetrahedral we'll get $sp^3$ hybrid orbitals, so the cause and effect is really the other way around.

user228700

Hm. It forms a tetrahedral molecule 'cause the orbitals are arranged like that to minimize energy?

John Rennie

Yes, at the end of the day whichever arrangement has the lowest energy will be formed.

user228700

OK!

John Rennie

My point is to be a bit cautious about the logic that:
1. carbon forms $sp^3$ hybrid orbitals
2. therefore we get a tetrahedral molecule

Life is rather more complicated than that ...

user228700

OK, yeah, this is awesome. I need only give a quick read through my book now; it's fairly simple. Thanks for coming to the rescue of the student in distress again! :-D

user228700

5:57 AM

@JohnRennie Yes, I understand ur point :-)

John Rennie

The trouble is that I've ventured into territory far beyond what you need for the entrance exam.

user116211

This guy has done some serious efforts to format the matrices:

user116211

-4

Q: Can the value of concurrence be non-real?

I have a problem as below. Given a density matrix $8x8$ for three qubit, $$B =\begin{pmatrix} \frac{1}2 & 0 & \frac{\sqrt{3}}8 & \frac{-3}8 & 0 & 0 & 0 & 0\\0 & \frac{1}2 & \frac{3}8 & \frac{\sqrt{3}}8 & 0 & 0 & 0 & 0\\\frac{-\sqrt3}8 & \frac{-3}8 & \frac{1}8 & 0 & 0 & 0 & 0 & 0\ \\\frac{3}8 &...

user116211

Unfortunately, they are blunt H/W.

John Rennie

but hopefully knowing a bit more about it helps you understand the university entrance level description a bit better.

user228700

5:59 AM

@JohnRennie Perhaps. Although, I'm fairly certain that at least now I'm confident in this topic so I will perform much better than I would've, having had only a very basic and kind of fragile understanding of all this.

user116211

@JohnRennie; why are some items not reviewable? I thought it was due to poor internet connectivity. But this seems to happen even when I have good connectivity.

user228700

@JohnRennie BTW, most of my syllabus is of university level! That's why I almost always benefit from these discussions :-)

John Rennie

@MAFIA36790 don't know.

user116211

ohh; okay.

John Rennie

It's not a problem I've ever encountered.

user116211

6:11 AM

hmm.

user116211

John Rennie

That normally means you've already reviewed it.

user116211

Check that; only Jon reviewed it.

user116211

Or if I reviewed, why is my username not there?

Qmechanic

6:29 AM

0

Q: Does it cost more energy to turn on a CFL globe than it is to leave them on?

Say there is a 15 W CFL globe. When turned on for an hour, it consumes electricity worth $10 in terms of energy consumption (made up cost for simplicity). If the occupants of the room periodically leave the room for say one minute - about 10 times in the given hour. So for a total of 50 minut...

electricity energy-conservation everyday-life estimation electrical-engineering

Off-topic?

user116211

@JohnRennie Okay checked; it seems I've indeed reviewed that.

user116211

But the message should be You have already reviewed this item.

user116211

@Qmechanic What is the conceptual physics query?

user116211

For me, it's off topic.

Secret

7:41 AM

Black hole information paradox

The black hole information paradox is a puzzle resulting from the combination of quantum mechanics and general relativity. Calculations suggest that physical information could permanently disappear in a black hole, allowing many physical states to devolve into the same state. This is controversial because it violates a commonly assumed tenet of science—that in principle complete information about a physical system at one point in time should determine its state at any other time. A fundamental postulate of quantum mechanics is that complete information about a system is encoded in its wave function...

> In 1975, Stephen Hawking and Jacob Bekenstein showed that black holes should slowly radiate away energy, which poses a problem. From the no-hair theorem, one would expect the Hawking radiation to be completely independent of the material entering the black hole. Nevertheless, if the material entering the black hole were a pure quantum state, the transformation of that state into the mixed state of Hawking radiation would destroy information about the original quantum state. This violates Liouville's theorem and presents a physical paradox.[citation needed]

So that means, in order to preserve unitary, the information that is released due to the pure state transformed into a mixed state has to be going somewhere in the black hole...

$\rightarrow$ hence the start of all this debate

user116211

8:39 AM

@yuggib, o/

user116211

what to do with this:

user116211

0

Q: Finding subject for proposal in cosmology and particle physics

I'm master student in particle physics and cosmology field. Right now I'm supposed to pick a specific area to write my proposal and I really don't no which branch of combination of particle physics and cosmology have bright future in research and also we can get the result by data soon.

particle-physics cosmology education

user116211

I've tagged it with education.

user116211

Frankly speaking, it appears to be off-topic y at the very first sight.

user116211

I voted it to close as too broad as we don't know enough about the educational background of OP.

user116211

8:42 AM

Somewhere in my mind, I don't like such questions in Physics.

John Rennie

This is the sort of thing that is ideally discussed in the chat, but users need 20 rep to join the chat.

user116211

But let community decide.

user116211

@JohnRennie yes, that's why it's out of discussion now since OP is new.

user228700

@MAFIA36790: Ever used Fevicol?

user116211

?

user116211

8:44 AM

Yes.

user228700

U know how u apply it and then it sort of becomes cold to touch?

user116211

Didn't notice.

user228700

Hm. Was wondering why :/ Dude, it becomes really cold!

user228700

Observe it the next time.

user116211

I have seen Feviquick becoming hot after application.

user228700

8:47 AM

@MAFIA36790 No, not super glue. Fevicol, the normal one.

user116211

yeh, yeh; got that.

user228700

Hey, are u at all familiar with the concept of steric number in hybridization?

user116211

Read it; but have to re-read it again to refresh it.

ACuriousMind

@MAFIA36790 I think you voted to close that question from outside the review queue. Obviously, you can't review items on which you have already voted by other means.

user228700

OK...can I ask u a question regarding the concept?

user116211

9:01 AM

@ACuriousMind How does that happen? I've not edited it to remove this from the queue.

ACuriousMind

@MAFIA36790 Not, it's still in the queue. "This item is not reviewable" means it's not reviewable for you

user116211

Okay.

user116211

But that is not the general message given.

ACuriousMind

And it's not reviewable for you because you have already cast an unclear what you're asking vote on it

yuggib

@MAFIA36790 \o

John Rennie

9:03 AM

@Kaumudi steric number?

user116211

@ACuriousMind The general message is this:

user116211

> You have already reviewed this item. It needs more reviews from other users to be completed.

user116211

and my username exists if I reviewed.

ACuriousMind

That is the general message if you have already reviewed it. What I'm saying is that you haven't reviewed that item, you just voted to close it without using a review queue.

user116211

But here the case was different even though I reviewed.

user116211

9:04 AM

And voted to close.

user228700

@JohnRennie Erm, yes. Steric number of an atom in a molecule=number of atoms bonded with that atom+number of lone pair(s) left on that atom.

user116211

@mahla: You need at least 20rep to participate here.

user228700

@JohnRennie: It serves as a tool to identify the type of hybridization in a given molecule.

user116211

@ACuriousMind okay. This is weird for I came across the post only in the queue and voted it to close. The vote was counted but not the review.

user116211

Anyways, thanks ACM for the response.

ACuriousMind

9:07 AM

The timeline clearly shows me that you voted on it outside the review queue. The only vote from review on that post is from Jon Custer. Maybe you opened the post in a new tab for some reason and then used the close vote there instead of clicking the "close" in the review queue?

John Rennie

@Kaumudi OK, what's the question?

user116211

No, no it didn't happen.

user228700

@JohnRennie Trying to figure out the steric number for $d^2sp^3$.

user116211

The only thing that happened was my connectivity got cut at that review. After I reloaded the page when the connection came, I saw that message.

user228700

9:09 AM

Dunno if this rule always works :/

ACuriousMind

@MAFIA36790 Ah, now then all bets are off as to what happened here - if only part of your data got sent, anything might have happened internally. Just forget about it.

user228700

Plus, there's seemingly no way to distinguish b/w $sp^3d^2$ and $d^2sp^3$.

user116211

@ACuriousMind sure.

John Rennie

@Kaumudi Once the $d$ orbitals get involved things get complicated because not all the electrons may be involved in bonding. However suppose they are involved. Then we have six bonding electrons, 4 in the $sp^3$ and 2 in the $d$.

So we're going to end up with a molecule $XY_6$, which will have a steric number of six.

user228700

...which is the same as the other. Okay, then I gotta come up with some way to distinguish b/w the two.

John Rennie

9:12 AM

I don't think there is a difference btween $sp^3d^2$ and $d^2sp^3$. It's just notation.

user228700

@JohnRennie Actually...there is a difference. In the first case, the d electrons of the outer orbital is involved where as in the second case, those belonging to the inner orbital is involved.

John Rennie

If you go back to what I was saying earlier, with an octahedral symmetry the electron wavefunctions can be written as a sum of the $s$, $p$ and $d$ orbitals.

We refer to these wavefunctions as $sp^3d^2$

It's always the outermost electrons affected

But, as I said, tread carefully when the $d$ orbitals get involved. And once the $f$ orbitals get involved (i.e. the lanthanides) all hell breaks loose :-)

user228700

Uhhh. Okay...

user228700

Thanks :-)

John Rennie

-1

A: What force particle mediates electric fields and magnetic fields?

So a Photon at up to infinite distance from its source can tell a charged particle it encounters whether it is attracted or repelled? An this occurs because the 'virtual particle' appears and disappears so rapidly that it does not breach energy conservation rules. How does this occur without brea...

Some mornings I find my patience with these sort of posts is in dramatically short supply

Wow it's convenient having a moderator on tap :-)

ACuriousMind

9:22 AM

I really saw no point in waiting for the review queues to deal with that one

Secret

[Sharing] The microscopic word is pretty

I wonder if we can adapt this single line art in physics context. The sketchy (insert word) style might be useful for something

Slereah

10:11 AM

"You don't want non-measurable sets? This means that you can partition the real numbers into strictly more parts than real numbers, but no part is empty. Yes, let that sink for a moment: you can partition the reals into more parts than numbers!"

@0celo7

Is that the madness you want

Without the axiom of choice

Secret

aren't you need the axiom of choice to prove that non measurable sets exists?

Slereah

Could you rewrite that sentence into a coherent one

ACuriousMind

@Slereah I think I already confronted him with that. He was...a bit shaken, but still wouldn't start believing in Choice.

Slereah

Well

I think most weirdness goes away if you take constructive logical axioms

But then it is much harder to prove anything

Secret

Leasbegue measurability cannot be proven without invoking axiom of choice (hence the notion on whethere a set is non measurable). But perhaps I should rest my case as nearly everything I knew about axiom of choice came from reading wikipedia

ACuriousMind

10:22 AM

@Secret Did you mean to write "Don't you need the axiom of choice to prove non-Lebesgue-measureable subsets of the reals exist?" In that case, yes, with the absense of choice you can construct something called Solovay's model. Don't be misled by all the Maimon posts praising it, it's not really useful.

Secret

ok noted

Slereah

Isn't Maimon a biblical demon

Or is it Mammon

ACuriousMind

It's Mammon

And it's become a poetic (and derogatory) name for money in German

And apparently it already was a name for money/profit orginally, not a demon

Unless you want to imbue money with enough agency of its own to call it a demon, which apparently several people later on in the Middle Ages did

Secret

NB I am actually fine with axiom of choice, because of the weird and interesting results it brought with it, and I like to study anything weird. I am however less familiar with set theory with axiom of choice thrown away.
*anyway, back to study group theory*

Slereah

it is the root of all evil

Secret

10:32 AM

[Unrelated monologue] Sometimes, when I reflected how I learn maths outside of classrooms, it strongly reminds of this:

Chinese room

The Chinese room argument holds that a program cannot give a computer a "mind", "understanding" or "consciousness", regardless of how intelligently or human-like the program may make the computer behave. The argument was first presented by philosopher John Searle in his paper, "Minds, Brains, and Programs", published in Behavioral and Brain Sciences in 1980. It has been widely discussed in the years since. The centerpiece of the argument is a thought experiment known as the Chinese room (described in detail below). The argument is directed against the philosophical positions of functionalism and...

For example, wiki walking and book reading allow me to learn a lot of theorems, and in some cases know where to use it, but in fact, I don't understand the theorem itself

bolbteppa

'Levi decomposition is the Lie theoretic analogue of the Jordan–Holder Theorem', make sense?

Secret

I think I vaguely read about Jordan-Holder Theorem as saying that the normal series of a given group is unique up to an isomorphism of its factors, I am not famialr with Levi decomposition

bolbteppa

Levi decomposition

In Lie theory and representation theory, the Levi decomposition, conjectured by Killing and Cartan and proved by Eugenio Elia Levi (1905), states that any finite-dimensional real Lie algebra g is the semidirect product of a solvable ideal and a semisimple subalgebra. One is its radical, a maximal solvable ideal, and the other is a semisimple subalgebra, called a Levi subalgebra. The Levi decomposition implies that any finite-dimensional Lie algebra is a semidirect product of a solvable Lie algebra and a semisimple Lie algebra. When viewed as a factor-algebra of g, this semisimple Lie algebra is...

It's so weird, Bourbaki seem to link Lie's theorem in lie algebras to Jordan-Holder as well

It's all just JH apparently :\

Secret

Well, the wiki page does say that two levi subalgebras are related by an auomorphism of their largest nilpotent ideal

so perhaps it is related (though I don't see it formign something like a composition series unless it is basically a composition series of length = 1

(I am guessing...)

Secret

11:11 AM

Last night, I dreamt about something called the "untrusted subsemigroup" with the following word salad like definition:

An untrusted subsemigroup is a subsemigroup such that it has so little useful information such that the properties of the semigroup cannot be determined by its characteristics/character. That is, an untrusted subsemigroup is such that the property of the semigroup inferred from this subsemigroup does not agree with that deduced by considering the characters of the semigroup in question

Otherwise, similar to 0celo7's maths solving dreams, I am testing something related to Molfaung loops with left zeros such as the following expression $$a * (b*(c*d))$$ where a is a left absorber

It is perhaps the first maths dream I had where I am solving maths that makes sense (only the Molfaung loop bit)

heather

12:03 PM

what is a "base representation of digits"?

as an aside, why does my chat profile show me as having 6.1k rep when really on physics I only have 3.7k rep? Does it have to do with my rep on other sites?

Secret

I believe base representation of digits is what number system you use for a number

e.g. things like binary , decimal etc.

user228700

@heather The chat rep. depends on the total number of lines u've sent in the chatroom or something like that, if I'm not wrong...

user228700

@heather: Omg, I kept thinking that ur avatar is a small lamp until I actually checked it out just now!

user228700

Kinda like this one:

user228700

ACuriousMind

12:21 PM

@Kaumudi cc @heather The chat rep is the sum total of your rep across all sites.

0celo7

Hello

@ACuriousMind So far I'm not liking FNV as much as F4

The world is strictly worse

Jim

@0celo7 story is better. Sniping is better, rest of actual gameplay is worse

ACuriousMind

@0celo7 The world is strictly better since all the factions have actual goals and motivations instead of the incoherent messes that pass for "goals" in F4. I'll concede that the gameplay is worse.

Jim

^ that

0celo7

The gameplay is fine

But there's nothing to do in FNV

ACuriousMind

12:32 PM

Have you reached Vegas yet?

user228700

@ACuriousMind Ohh. Alright, thanks.

0celo7

@ACuriousMind I need 2000 to get in

I just stocked up on ammo so I'm poor :(

ACuriousMind

@0celo7 No you don't. There's another way that costs less and if you do quests for the NCR they'll also get you in. You just have to find the things to do, they're not pointless quests thrown at you but part of storylines.

0celo7

I've done quests for the NCR

I'm "accepted"

How do they get me in @ACuriousMind

ACuriousMind

If you want to make money try working for the Silver Rush. You can also do the King's story line and demand he gets you in

0celo7

12:38 PM

Wtf how do I join the Kings

They ghoul told me about them

But where do I find King

ACuriousMind

It appears that you haven't explored Freeside yet properly :P

Sir Cumference

TIL I learned that the Newtonian Mechanics equivalent of black holes is called a "dark star"

Dark star (Newtonian mechanics)

A dark star is a theoretical object compatible with Newtonian mechanics that, due to its large mass, has a surface escape velocity that equals or exceeds the speed of light. Whether light is affected by gravity under Newtonian mechanics is questionable but if it were accelerated the same way as projectiles, any light emitted at the surface of a dark star would be trapped by the star's gravity, rendering it dark, hence the name. == Dark star theory history == === John Michell and dark stars === During 1783 geologist John Michell wrote a letter to Henry Cavendish outlining the expected pr...

0celo7

@ACuriousMind I don't tend to explore hives of degeneracy

Also I got there at 1am and decided I should maybe sleep :P

Sir Cumference

@0celo7 Where are you that it's 1:00am?

Oh wait, nvm

0celo7

@ACuriousMind I'm expecting something amazing. Because right now F4 is the far better game

Sir Cumference

12:49 PM

@0celo7 Uh, did you say you stopped playing games?

Jim