The h Bar

Anonymous

7:00 PM

@Curio I gave you a link. Read it

Anonymous

Can a Vampire's bite deal extra damag…

Can a Vampire's bite deal extra damage due to HP drain?

Anonymous

Also what is going on here ^ :P

Loong

Maybe change the host of that room to rpg.stackexchange.com

Slereah

You don't know, it might be a biophysics question

Anonymous

Vampire biophysics :)

Curio

7:12 PM

@Blue I read it but it only says that those elements are irregular

Anonymous

@Curio Valence electron is not a well defined term for transition metals. Normally the electrons which can be most easily removed are called valence electrons...

Slereah

Aren't valence electrons the ones in partially filled bands

Anonymous

@Slereah We're talking about individual atoms here...

Slereah

Oh

Aren't valence electrons the ones in partially filled shells

Anonymous

No, not for transition metals

Anonymous

7:15 PM

For other elements, yes

Slereah

Damn those transition metals

Anonymous

By shell I suppose you mean n=1,n=2, etc...

Anonymous

(principal quantum number)

Anonymous

At least physicists can be proud that there are more handwavy subjects than physics :P

Slereah

Like sociology

Anonymous

7:18 PM

lolol

Slereah

By shells I mean both the principal and the orbital quantum number

Anonymous

@Slereah Okay, then for a certain $n$, some electrons are considered valence and some aren't

Anonymous

For transition metals

Anonymous

The rules aren't clearly stated anywhere

Slereah

Is it that weird thing where some energy levels are switched

Anonymous

7:20 PM

Something like that. I don't remember the details :/

Anonymous

Since @Loong is here...he might be able to say!

Anonymous

2

A: How many valence electrons do elements in the d block have?

The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell. But it is wrong to apply the concept to ...

Anonymous

I don't know if this answer is fully correct

Anonymous

Someone has to verify

Anonymous

Firstly, it depends on what you count as "valence electrons". If you say that, for all the d-block metals, the ns and (n-1)d electrons count as "valence electrons", then the answer is to just look at the group number. However, that obviously doesn't work for Zn, which effectively only has 2 valence electrons. That's really an extreme case though. It is sometimes said that the d electrons transition from being valence electrons in the early d-block (hence Sc only forms $\ce{Sc^3+}$), to being core electrons in the late d-block (hence Zn only forms $\ce{Zn^2+}$). — orthocresol ♦ Oct 20 '15 at 16:28

ooolb

7:30 PM

@BalarkaSen youtube.com/watch?v=NKcPiF3JEcM

dope track for commuting

Emilio Pisanty

7:41 PM

Hey @vzn, this comment is out of line. You don't get to accuse this community of being unscientific without solid evidence - and you definitely don't get to raise that kind of accusation when you systematically refuse to engage in earnest debate the minute any serious objections are raised to your proposals.

Felix.C

Hi guys does someone know by chance if the critical points of the lagrange function are already ensured to satisfy the imposed constraints? or are they merely candidates?

Anonymous

@EmilioPisanty I'm afraid that trying to raise this conversation with vzn will probably be an exercise in vain. He seems convinced that PSE is a close minded community. So, I'd say it would be better to ignore.

Anonymous

ACM tried earlier...

Emilio Pisanty

@Blue Indeed, I have lost pretty much all hope of vzn engaging in a serious debate

but that doesn't change the fact that that comment was out of line

Anonymous

Just flag it as obsolete and move on...

0celo7

7:46 PM

@EmilioPisanty what's the second derivative test in German?

Emilio Pisanty

@0celo7 the what?

0celo7

@EmilioPisanty Hess f is negative definite at a max, etc.

DanielSank

Southern California is still on fire.

0celo7

@DanielSank were you evacuated?

DanielSank

@0celo7 Nope.

Anonymous

7:48 PM

Heard about CA. Sad :/

DanielSank

Still some distance between me a fire.

CooperCape

Ohai

Emilio Pisanty

@0celo7 you want a translation into German of the phrase "the second derivative test"?

0celo7

@EmilioPisanty not a translation, but what do they call it

math can never be directly translated

Emilio Pisanty

sorry, mate, my German is sufficient for grocery shopping, not for scientific inquiry

0celo7

7:52 PM

same

@ACuriousMind Ok I have two doubts for you now

Emilio Pisanty

@0celo7 maybe try and get a German version of Spivak or Courant and see what they say?

there should be a section with that title in the English version, and that'll be easier to trace

ACuriousMind

@0celo7 Why have I been summoned?

Slereah

For QM

Apparently

Emilio Pisanty

@ACuriousMind 0celo7 wants to know how "the second derivative test" is called in German

not sure what the other doubt is

ACuriousMind

I don't think we have a literal translation, we'd use something like "sufficient criterion for extrema" instead.

Emilio Pisanty

8:06 PM

@ACuriousMind seriously?

that's bizarre

ACuriousMind

Yes. I mean, the literal translation would be like Prüfung der zweiten Ableitung, but I don't think anyone who doesn't recognize it as the translation of "second derivative test" would know what you mean

As supplementary evidence, consider that derivative test on Wikipedia does not have a German version

0celo7

huh

@ACuriousMind so what is $\langle x|\psi\rangle$

that's the other doubt

$\psi(x)$ has no meaning for an $L^2$ function

ACuriousMind

When I think back to using that test in school to identify extrema, we always called it hinreichendes Kriterium für Extrema.

@0celo7 It's nothing, a figment of the physicist's imagination :P

0celo7

@ACuriousMind right, but even if you do that rigged Hilbert crap

I'll assume the physicist really loves Dirac notation and defines $|x\rangle$

what is $\langle x|\psi\rangle$ then?

Slereah

@0celo7 what

doesn't it

0celo7

8:11 PM

@Slereah $\{x\}$ is a set of measure zero

vzn

@EmilioPisanty the comment was not out of line but some nameless mod has already deleted it. congratulations

Slereah

Well yes but the function itself is like

ACuriousMind

@0celo7 Well, it's in the dual of Schwarz functions, right, and so you can only take that expression when $\lvert \psi \rangle$ is in the Schwarz space, and the value of the Schwarz function should be well-defined, right? (there are many other functions in the $L^2$ class, but no other smooth Schwarz function).

Slereah

A function of $\Bbb R^3$

You can evaluate it at a point

Emilio Pisanty

@vzn here it is for the record

> sorry, too creative/ imaginative/ unconventional to be acceptable as a Physics (SE) question, but try chat instead esp if you have serious interest/ dedication/ endurance eg chat.stackexchange.com/rooms/9446/theory-salon – vzn 15 hours ago

I'm not sure how you read that in any way other than an accusation of unscientific behaviour

0celo7

8:12 PM

@ACuriousMind well then you're restricted to Schwarz functions, which is also bad

Emilio Pisanty

It is well out of line

vzn

@EmilioPisanty lol so now you defy mod "editorialship" (merely by reposting it)? :P ... thx so much for that :)

Emilio Pisanty

@vzn No, that's very much not what I'm doing. I will very much thank you not to put words in my mouth or otherwise state assumptions about my frame of mind.

ACuriousMind

@EmilioPisanty Wow. Yes, @vzn, indeed it is out of line to imply that your personal chatroom is a better place for "serious interest" than the physics.SE site. If you think that, why participate here at all? (that's a rhetorical question)

@0celo7 Sure, but that's how the rigged space thingy works

vzn

@EmilioPisanty why not go further and flag the question and vote to kill it?

Emilio Pisanty

8:14 PM

@vzn why would I flag the question when my answer is 80% complete?

vzn

@ACuriousMind not in the mood for rhetorical questions. maybe it implies you dont really care about the answer(s) and or contrary pov(s). btw obviously there is no such thing as a "personal chat room" on SE... except those that limit participants, of which the one started by me does not...

Emilio Pisanty

@vzn That's your second accusation of unscientific behaviour today.

ACuriousMind

I indeed do not care about contrary POVs that have repeatedly failed to produce any evidence for their claims at all.

Emilio Pisanty

care to back it up?

vzn

@EmilioPisanty ?!? huh? what "answer" are you referring to? the question stands with no answers currently

Emilio Pisanty

8:17 PM

@vzn the one you haven't seen because I have yet to finish it because it is currently 80% complete

vzn

@EmilioPisanty lol on the edge of my seat, emilio :P

Keep these mind

Which question are you discussing?

Emilio Pisanty

@Keepthesemind this one

John Rennie

@EmilioPisanty I'll be interested to read your answer. I can't see any reason why quantum computing should show up any unexpected behaviour that the millions of other QM based expts haven't. But maybe there are aspects of QC that would be especially sensitive - or maybe not ...

Emilio Pisanty

@vzn The degree to which I care whether you're on the edge of your seat or not is zero.

ACuriousMind

8:18 PM

@vzn I have repeatedly told you that ridiculing others by replying to earnest attempts at conversation with "lol" is irritating. Please stop it.

2

vzn

lol bye

Emilio Pisanty

@vzn ^ I'll tack on to that request

ACuriousMind

@Blue That's an RPG.SE conversation that ended up associated to physics because the room owner has their chat parent site set to physics :D

Emilio Pisanty

@vzn .... aaaaaaand, here we go, people try to engage in a serious debate, and you abruptly refuse to engage.

Keep these mind

@EmilioPisanty Thanks. Is this (the question, not necessarily the answer) related to the 't Hooft ideas?

Emilio Pisanty

8:22 PM

@Keepthesemind I guess, yes

0celo7

@ACuriousMind so it's not a replacement of the L^2 formalism in any sense?

ACuriousMind

@0celo7 Not a replacement - the $L^2$ space is still there as the "middle" space of the Gel'fand triple

it's more an extension that allows you to talk about where the $\lvert x\rangle$ live

But as you have repeatedly pointed out, I don't know anyone who actually uses that extension to any practical end :P

0celo7

@ACuriousMind yeah, but I've gone down a rabbit hole of PSE posts that keep telling me the Gelfand stuff is useful!

and people telling me that $\psi$ and $|\psi\rangle$ are different things. I know they're wrong, but maybe they're right :p

ACuriousMind

I think it depends on what $\psi$ is for you

Keep these mind

If ideas similar to 't Hooft's ideas are (currently) indistinguishable in consequences from the 'mainstream' philosophy of QM, then why not entertain the idea?

Feynman: Knowing versus Understanding

►

ooolb

8:26 PM

if y'all play pokemon you should use the chat version of "pursuit" on vzn.

ACuriousMind

If it's an abstract equivalence class in $L^2$, I think it might be the same. But when a physicist writes $\psi$ (or $\psi(x)$, really), they mean a function, not an equivalence class

Emilio Pisanty

@JohnRennie I was going to pass. But OP does a lot of grandstanding based on crude misconceptions about QM, and I guess I just couldn't stop myself from biting.

answer is up now.

ACuriousMind

I.e. it's more a map between the equivalence class and a particular representant of it

0celo7

@ACuriousMind Ah yes, the greatest notational abuse in math

confusing $L^2$ functions with equivalence classes

@ACuriousMind have you watched the Sopranos

ACuriousMind

Nope

vzn

8:34 PM

@Keepthesemind do you have some interest in 't hooft?

ACuriousMind

@0celo7 Should I?

0celo7

@ACuriousMind so the conclusion is that one knows where $|x\rangle$ lives, but it's in a pretty poor space and it isn't really useful

ACuriousMind

@0celo7 I think so, yes

Keep these mind

@vzn Not really.

0celo7

@ACuriousMind It's interesting

Was wondering if you had an opinion on the grandmother

@ACuriousMind did you kick him?

ACuriousMind

8:37 PM

@0celo7 Well, what other series I might have seen could you compare it to? I don't really know anything about the series

@0celo7 Yes.

0celo7

@ACuriousMind I'm thinking about that. It's really about family drama

I think it turns into a real crime show later, they're hinting at the police becoming important

Anonymous

Since when are you into family dramas?

Keep these mind

@vzn Although, if I needed to place a bet, it would be that some classical system can replicate any QM one. Maybe that is obvious. I'm not informed enough to judge whether computational efficiency is a deciding factor.

0celo7

@Blue since always?

Anonymous

@0celo7 Alright, didn't know that

Emilio Pisanty

8:44 PM

@Keepthesemind my take on it is that I feel that with superdeterminism the cure is worse than the disease.

I've yet to see something that gives me a good reason to care in depth about those theories.

Anonymous

On the other hand I haven't watched TV Series in the last few years

Anonymous

Not even Web Series

Emilio Pisanty

i.e. something that will make it clear that the statistical theories don't actually make things worse.

Anonymous

Just youtube and movies

Sid

@0celo7 if they are hinting that the police is going to be important, then, if they are a clever bunch, it will remain a family drama and not turn into a crime show

Curio

8:46 PM

<< The simplest answer is: d-block elements have number of valence electrons equal to their group number, which is equal to the number of electrons in the "valence shell". This works if you are using the definition of valence shell to be the outermost shell. >>

Anonymous

If you want to watch family drama just watch Indian soaps...you'll be drowned in drama XD

Curio

Palladium ends in 4s2 4p6 4d10. So where's the problem?

Where is the valence shell here?

Sid

@Blue 0celo wouldn't be at the pinnacle of logic anymore if he starts watching that

Anonymous

@Curio n=4 can be called the valence shell....but not all of the electrons in n=4 are valence electrons...

Anonymous

@Sid ikr x'd

Curio

8:49 PM

In A groups this works

So where are the valence electrons? D?

Sid

@Curio normal rules don't apply to transition metals. Having said that, Palladium is a weird case.

Anonymous

@Curio In 4d

Curio

But they aren't 10 then

Anonymous

Yeah, not all of them are valence electrons in 4d

Anonymous

I already mentioned that

Curio

8:53 PM

Yes

Keep these mind

@EmilioPisanty I suppose the main reason to care would be curiosity. If there is only a tiny speck of uncertainty regarding QM, why not completely focus on that speck? That is: if you care about the 'underlying truth'. Then again, I tend to think that they are potentially equivalent; as in: nature isn't quantum or (adapted) classical, it can be described by both (and I don't know if it even matters what nature's hardware is).

Curio

So how many electrons are electrons of valence?

Anonymous

@Curio Equal to group number probably

Sid

@BalarkaSen if you have watched A Clockwork Orange, do tell me if you recommend it

Curio

8 then

Anonymous

8:56 PM

@Curio Roughly...it could vary...

Anonymous

dl.clackamas.edu/ch104-06/valence_electrons.htm

Anonymous

Like here...it's probably a range

Anonymous

And not well defined

Anonymous

Ask some quantum chemist

Curio

Okay thanks

Slereah

9:02 PM

im a quantum sociologist

0celo7

I study Babylonian quantum mysticism

Anonymous

Who's the quantum troll?

0celo7

@BalarkaSen

Keep these mind

@Slereah I assume you just made that up. But it exists.

Slereah

Does it

Is that like """quantum""" economics

just a fancy word for stochastic calculus

Anonymous

9:04 PM

arxiv.org/ftp/quant-ph/papers/0303/0303162.pdf

Anonymous

goddamit

Slereah

Oh man

Immediatly I'm suspisicous

I don't want to sound prejudiced, but

Poor layout is the first redflag of a crank

It's like the paper was made in Word

Anonymous

arxiv is the new vixra ;p

Avantgarde

That's a damn good paper

Slereah

"Members of societies are "atoms" and "individuals" literally, that is, from one
person or one electron it is impossible to make two smaller persons or two smaller
electrons."

I thought a person making a smaller person was pretty common

Anonymous

9:06 PM

lol :'D

Keep these mind

Quantum game theory

Quantum game theory is an extension of classical game theory to the quantum domain. It differs from classical game theory in three primary ways: Superposed initial states, Quantum entanglement of initial states, Superposition of strategies to be used on the initial states. This theory is based on the physics of information much like quantum computing. == Superposed initial states == The information transfer that occurs during a game can be viewed as a physical process. In the simplest case of a classical game between two players with two strategies each, both the players can use a bit (a '0' or...

0celo7

lmao

quantum poker

Slereah

This will be great when they make poker machines with quantum computers

Anonymous

Quantum placed before anything sounds cool...

0celo7

you roll a die and everyone rotates cards

Slereah

9:07 PM

Quantum dickballs

Avantgarde

quantum food. Sometimes it's in your mouth, and sometimes it's somewhere else.

0celo7

@Slereah b&

ACuriousMind

@0celo7 Perhaps you'd enjoy topological games more :P

Avantgarde

Quantum blue

Slereah

@0celo7 If talking about quantum dickballs is a crime then let history remember me as a criminal

Anonymous

9:08 PM

@Avantgarde lel...that could be my new name

0celo7

@ACuriousMind lol

Avantgarde

@Blue if you don't want to be taken seriously

Anonymous

Thank god cranks don't understand GR well enough. Or else it would be as misused as QM

Anonymous

@Avantgarde Heh. I could trade being taken seriously for a cool name ;)

Keep these mind

Quantum game theory is a serious enterprise. You can make fun of it, but that's not very unlike making fun of QM itself. (Unless it is the 'game' part that you're making fun of. In that case, you're at least 70 years behind.)

Avantgarde

9:11 PM

@Blue What about the fables of black holes, wormholes, time travel?

Anonymous

@Avantgarde I've not seen much of those articles claim that they're deducing those stuff using GR

Anonymous

Unlike: According to QM....blah blah blah

Avantgarde

Quantum is more mysterious.

You should try quantum healing.

It cures common cold by inducing loop diagrams in the virus

Anonymous

I sure should :D

Anonymous

LOL

Avantgarde

9:20 PM

the poor virus goes off-shell

It's a ~~classic~~ quantum computation

sup felix

Anonymous

Quantum healing

Quantum healing is a pseudo-scientific mixture of ideas drawing on quantum mechanics, psychology, philosophy, and neurophysiology that purports that quantum phenomena are responsible for health and wellbeing. There are a number of different versions, which draw on various quantum ideas including wave particle duality and virtual particles, and more generally energy and vibrations. Quantum healing is a form of alternative medicine. == Etymology == The term 'quantum healing' was coined by Deepak Chopra, a holistic guru who has more than 65 published works. It invokes the idea of a "healing" process...

Anonymous

Deepak Chopra

Deepak Chopra (born October 22, 1946) is an American author, public speaker, alternative medicine advocate, and a prominent figure in the New Age movement. Through his books and videos, he has become one of the best-known and wealthiest figures in alternative medicine. Chopra studied medicine in India before emigrating to the United States in 1970 where he completed residencies in internal medicine and endocrinology. As a licensed physician, he became chief of staff at the New England Memorial Hospital (NEMH) in 1980. He met Maharishi Mahesh Yogi in 1985 and became involved with the Transcendental...

Anonymous

"Deepak Chopra (born October 22, 1946) is an American author, public speaker, alternative medicine advocate, and a prominent figure in the New Age movement.[3][4][5] Through his books and videos, he has become one of the best-known and wealthiest figures in alternative medicine.[6]"

Avantgarde

What's amazing is that he is actually a doctor.

Anonymous

We should take some #getrichfast tips from him

Avantgarde

9:24 PM

lol

Reminds me of this

Bill Burr - Late Night

►

Start watching at 0:42

vzn

@Keepthesemind would 2nd you on that wager but bells thm is the main evidence (decades est) against the idea. there are glimmers of contrary povs/ new lines of thinking/ evidence within last few yrs & have been assiduously collecting those refs (in my blog).

Anonymous

@Avantgarde wtf is that...

Avantgarde

@Blue a joke. Comedy. Not a fan?

Anonymous

@Avantgarde LOL. My speakers were muted

Anonymous

Was only watching the video

Avantgarde

9:27 PM

F.a.c.e.p.a.l.m.

Emilio Pisanty

@Keepthesemind Oh, I care very much about whether there's something underlying QM or not. But there's very many possible extensions, and superdeterminism is just worse than QM-taken-neat, I think. It would have to show it is exceptionally good to be a better alternative, and thus far it hasn't shown that.

if and when it does, I'll be interested.

Anonymous

Heh, cool :D

Keep these mind

@vzn OK, if you say so. But superdeterminism is not countered by Bell. Explicitly so. From the beginning.

Anonymous

@Avantgarde

vzn

@Keepthesemind focusing on the "speck" that is anomalous wrt current theory. sounds like nature of scientific progress/ advance/ evolution to me. acc to kuhn! :) ... "why not?" but obv the copenhagen interpretation rejects your pov here... hence have been reexamining that myself these days...

Avantgarde

9:29 PM

:)

Keep these mind

@EmilioPisanty OK. But somebody has to do the work.

vzn

@Keepthesemind have not heard much about superdeterminism. dont think it is nec the answer to the (key) question but do think its a step in the right direction. have heard of maybe "superclassical" or something like that... btw do you have physics degree?

Keep these mind

@vzn On your last question: certainly not.

Superdeterminism

In quantum mechanics, superdeterminism is a hypothetical class of theories that evade Bell's theorem by virtue of being completely deterministic. Bell's theorem depends on the assumption of "free will", which does not apply to deterministic theories. It is conceivable that someone could exploit this loophole to construct a local hidden variable theory that reproduces the predictions of quantum mechanics. Superdeterminists do not recognize the existence of genuine chances or possibilities anywhere in the cosmos. Bell's theorem assumes that the types of measurements performed at each detector can...

NB: I'm not a believer. I can merely follow the argument.

Anonymous

@Avantgarde Sorry, I normally mute my speakers before clicking on links. Balarka nearly got me scolded by my mom one day (linked to some batshit crazy loud video). Should keep headphones handy

vzn

↑ nice :)

Avantgarde

9:33 PM

@Blue hahahahaha

Well, fair enough

vzn

@Keepthesemind do admit the arguments are extremely subtle. do not really think QM is "wrong" in some major sense, quite to the contrary (constantly have to reiterate that in this room/ among this audience)... but do "believe" its a statistical approximation of a deeper deterministic theory.

Keep these mind

@vzn Well, I tend to think that both of them are right. :)

vzn

@Keepthesemind both? superdeterminism + copenhagen view?

Keep these mind

@vzn I think the difference between QM (not going into interpretations) and superdeterminism is essentially indistinguishable from within.

0celo7

@Slereah yo

have you ever heard of normal neighborhoods called regular neighborhoods?

@ACuriousMind also asking the German

vzn

9:40 PM

@Keepthesemind ok. thinking a bit )( further, actually do have a feeling that there might be a very subtle indistinguishability yet to be uncovered. believe it is on the horizon. say, within a decade. maybe glimmers of it already in current experiments.

Keep these mind

@vzn I couldn't possibly comment.

ACuriousMind

@0celo7 I don't think I have ever actually encountered "normal neighbourhoods" except when you talk about them ;P

vzn

@Keepthesemind ok. so if not a physics degree, something else?

Keep these mind

@vzn Certainly. Got to go.

vzn

@Keepthesemind ok nice chatting thx for sticking up for the slightly )( non status quo. :)

Slereah

9:47 PM

@0celo7 I did not

Although

"regular" is one of Those Words

0celo7

@ACuriousMind I would ask how you do anything in Riemannian geometry without them, but I think I already know your answer.

Slereah

That people shove in anything

ACuriousMind

@0celo7 "I don't" :)

0celo7

Yeah yeah

Slereah

I still don't know what a "regular" function is

0celo7

9:51 PM

that depends

Slereah

exactly

But sometimes they don't bother explaining what they mean

Felix.C

@Avantgarde good evening

0celo7

@Slereah clear from context

Slereah

I usually assume they mean smooth

0celo7

nah

could mean anything

Bradley G. Cooper

9:53 PM

Hello all, my first time in the chat room, glad to see this, is it new?

0celo7

been here for years

Slereah

Any specific meaning when they talk about PDE solutions?

0celo7

@Slereah depends

Anonymous

@BradleyG.Cooper Hi. No it's not new.

0celo7

Sobolev, Holder, something else...

Slereah

9:53 PM

So in other word I'll never know

Sad!

Bradley G. Cooper

I just posted a question, any answers would be appreciated! thanks physics.stackexchange.com/questions/373701/…

Still having a hard time conceptualizing what exists (or doesn't) beyond the visible universe

0celo7

@Slereah I mean say you have a solution in $H^1$. Then a "regularity result" means that it's in $H^2$, say.

That's the context

Bradley G. Cooper

seems like it would be easier if the universe was in very early stages

Slereah

I guess i should read more what's his name

Bradley G. Cooper

(easier to understand, anyway)

Slereah

9:55 PM

Post modern analysis

0celo7

yes, you should.

Anonymous

@BradleyG.Cooper First of all that is an Astronomy question, more suited for Astronomy Stack Exchange and secondly the title of your post does not match the questions asked in the body.

Bradley G. Cooper

i did not know there was an astronomy stack exchange

but the question does match, just additional questions that are related

Anonymous

However, as this a bit of a cosmology question....it might be okay to keep it here

Bradley G. Cooper

there are lots of questions about universe boundary if physcis

seems like a comsomology question

Slereah

9:58 PM

I'd do more astronomy but

big city life