The h Bar

3:00 AM

@loltospoon Or maybe their test function is zero when acted on by part of the full Hamiltonian. Lots of things could be happening.

ArtOfCode

→ 37 messages moved to Trashcan

loltospoon

@0celo7 oh good point

ArtOfCode

physics is a nice topic let's talk about physics

I did a physic once

0celo7

@ArtOfCode I prefer math.

ArtOfCode

I did a math once too

0celo7

3:01 AM

Do you want to hear a crazy math?

ArtOfCode

it was a good math

I remember it fondly

@0celo7 math away

0celo7

@ArtOfCode There's a strictly increasing function on [0,1] that has zero derivative almost everywhere

ArtOfCode

aaand you've lost me :)

0celo7

i.e. the derivative is zero except for on a set with zero length

crap

@ArtOfCode the square of a negative number is positive

ArtOfCode

I know what these words mean individually :P

@0celo7 ooh, ooh, I understand this bit!

0celo7

3:03 AM

Beats me why that's true.

HDE 226868

@0celo7 Is that the Cantor function?

0celo7

It follows from the ordered field axioms but I never really understood why (-1)^2=1.

skill patrol

Btw @ThomasWard nice work on the smoke detector project :-)

HDE 226868

Or am I confusing that with another one?

0celo7

@HDE226868 It's a modification of that to be strictly increasing.

HDE 226868

3:05 AM

@0celo7 Oh, strictly increasing.

ACuriousMind

I'd be willing to talk about the quantum moduli space of asymptotically conical $G_2$ compactifications, but I don't think that's of interest to anyone else :P

Thomas Ward

@skillpatrol Thank you! Myself and others contribute to make it all better too :)

ArtOfCode

@0celo7 I always justified it with the fact that -2*2 is obviously -4, so -2^2 must be the negative of that, which ends up positive

Thomas Ward

→ 2 messages moved to Trashcan

0celo7

@ArtOfCode But why is -(-a)=a?

Again, not using ring axioms.

What does - even mean?

@ACuriousMind You know I'd be willing but you never explain things enough :P

ArtOfCode

3:06 AM

@0celo7 negative, or a subtraction. Yeah, can you tell I'm not a mathematician yet?

David Z

@loltospoon might be a different way of writing the same thing

ArtOfCode

Similar thing. Way I look at it is that 0-a is -a, so 0-(-a) must be the negation of that.

0celo7

@ArtOfCode Yeah, that's using the ring axioms

ACuriousMind

@0celo7 Well, I took a year of string theory classes to be able to understand this stuff, so excuse me for not being able to communicate it within chat messages :P

0celo7

I mean, how did they explain it to us in 4th grade

ArtOfCode

3:08 AM

At a basic level, I guess you can look at it as two different directions on a number line - positive one way, negative the other, and putting a negative sign in front of something changes the way you're going.

which probably is how they explained it to us in elementary, to be honest

skill patrol

^exactly

0celo7

I guess

@ArtOfCode You can derive the hawking temperature of a black hole by supposing that time is imaginary and cyclic.

ArtOfCode

that sounds complicated

ACuriousMind

My goal of explaining the whole shebang clearly in a single document hit an early snag when I realized that no one knows what Majorana spinors are.

Secret

That,is a lot of (super)earths

0celo7

3:11 AM

@ACuriousMind Why is a -21 answer not deleted?

Secret

So that's what kenshin told us about that NASA announcement

HDE 226868

@Secret 2017 isn't doing too terribly in science so far.

ACuriousMind

@0celo7 Because we do not delete answers for being wrong.

loltospoon

@0celo7 ok so I just tried this problem using a simplified Hamiltonian and it doesn't work. The integral doesn't converge. I used $H=T+V=-\frac{\hbar ^2}{2m}\frac{d^2}{dr^2} + \frac{2}{r} d/dr) -\frac{e^2}{r}$

0celo7

@ACuriousMind I see.

@loltospoon where's your momentum

loltospoon

3:13 AM

Let me make it pretty hang on

0celo7

which integral?

what's your test function

Secret

@HDE226868 well, hopefully they will get alien life forms in a few months. The odds of 7 to 1, sounds a bit too unlikely to be all voided, right...?

0celo7

@ACuriousMind If you learned any cool $G_2$ analysis I could be convinced to listen

although I'm trying to learn some algebra right now

ACuriousMind

@0celo7 I think there's plenty of cool analysis in Joyce's and Kovalev's constructions but I'm treating those as black boxes for now

HDE 226868

@Secret There are a few obstacles there. Possible tidal locking, freaky seasons, irradiation. . . But yeah, the odds aren't terrible.

loltospoon

3:15 AM

Wow that's annoying - can't edit after some time has passed

Secret

Also, that's one crowded system

ACuriousMind

Understanding that part of the analysis is sadly unlikely to bring me closer to the goal of making physically interesting statements about $G_2$ compactifications

skill patrol

@Secret The pats are 7 to 1 to repeat

HDE 226868

@Secret We just had a question about that on Astronomy.

$a_1+a_2$ is tiny. Tiny tiny tiny.

loltospoon

$H=T+V=\frac{-\hbar ^2}{2m}(\frac{d^2}{dr^2}+\frac{2}{r} \frac{d}{dr})-\frac{e^2}{r}$ and the test function is $\psi = C(1-\frac{r}{\alpha})$

ACuriousMind

3:17 AM

@loltospoon 2 mins edit period in chat, gotta get used to that

skill patrol

Mods are unlimited, right?

0celo7

Mod = god

@ACuriousMind For the record, I don't think you're a horrible person

@loltospoon Hmm, yeah. I'd believe that doesn't converge because it's linear.

Integrating linear things never goes well.

loltospoon

@0celo7 then I don't understand how to use the variational principle here. The hardest part is to just figure out which Hamiltonian to use. I know the principle, but which $H$ do I use??

0celo7

You use should be using the full one.

Does your book not have examples?

Secret

Hypothesis: Any article who quoted E=mc2 more than 10 times is likely a crank

0celo7

3:24 AM

@Secret Unless it's a nuclear engineering paper

they use $E=mc^2$ to calculate things in reactors

loltospoon

@0celo7 not for hydrogen atom, no

0celo7

@loltospoon let me check my boosk

Secret

I see

0celo7

@loltospoon get Shankar

look at page 431

loltospoon

@0celo7 I'm using Griffiths

0celo7

3:26 AM

many people think that's a bad book

loltospoon

I've started to see why, yea

@0celo7 which edition?

0celo7

I own the second

you can get a pdf from your school's springer access I bet

link.springer.com/book/10.1007/978-1-4757-0576-8

access that on school wifi or through the school proxy

the equivalence principle??

skill patrol

nvm

loltospoon

@0celo7 got the book. Ok I see the Hamiltonian they use. I'll try it out

0celo7

@loltospoon ok, let me know if it works

skill patrol

3:41 AM

chat.stackexchange.com/faq#nice

That^ needs to be elaborated upon.

loltospoon

@0celo7 didn't work. Same problem.

0celo7

@loltospoon hmmmm

@loltospoon why are you using that function?

loltospoon

It's given

Wait a minute.......

0celo7

It doesn't make much sense.

The state you have is not a bound state

So why would you estimate the energy of a bound state with it

loltospoon

@0celo7 I totally forgot something - it is bounded. The wavefunction is zero when $r>\alpha$ and non-zero when $r \leq \alpha $

Crap.

So does that mean that the upper bound is simply $\alpha$?

0celo7

3:46 AM

-.-

yes

loltospoon

Sorry, still learning.

@0celo7 sigh....looks like even with this change the inegral doesn't converge

I get $\ln(0)=-\infty$.

0celo7

HMMMMMMMMM

ask your prof at this point

I have to do my homework :P

loltospoon

@0celo7 lol he won't be awake

0celo7

I mean tomorrow, obviously

loltospoon

@0celo7 currently studying to prepare for an exam :/

0celo7

3:51 AM

so?

will he not help for that?

loltospoon

@0celo7 not if he's unconscious, no

0celo7

is your exam tomorrow?

loltospoon

Yep

Well, in 12 hours

0celo7

oh, well

does Griffiths have a solution manual?

Skyler

@DanielSank the response: "In this case, the issue you've described is actually just intended functionality "

loltospoon

3:57 AM

Yes but this problem is just a practice problem I found online with no solution

0celo7

oh jesus christ :P

where did you find it?

skill patrol

JEE?

0celo7

JEE doesn't have quantum mechanics

plus, he seems to be American

loltospoon

@0celo7 probably from some abyss

dont remember where

skill patrol

@0celo7 Not yet.

0celo7

4:01 AM

@loltospoon I wouldn't bother with it then

get some sleep instead

and pray

loltospoon

Lol no way. Let's just put it this way - if I can't figure this one out then I wont be able to figure out anything on his exam

If I have to get 0 sleep then I will

@0celo7 any help with this one? physics.stackexchange.com/questions/314124/…

0celo7

@loltospoon idk physics, sorry

you shouldn't lose sleep over something from the abyss

it might be wrong

loltospoon

@0celo7 sigh....I found another mistake I made.

What a wonderful life.

0celo7

what was it?

DanielSank

@Skyler I feel that I'm missing some context here.

Skyler

4:14 AM

@DanielSank From facebook

DanielSank

@Skyler I figured as much.

That's why I amended my statement from saying that their security model is broken to saying that it's bonkers.

Skyler

They also called it a "theoretical attack"

DanielSank

I'm sure they did it that way so that granny can easily get into her account when she gets a notification without remembering how to sign in.

Allright, I'll file a complaint.

How do I do it?

0celo7

What did FaceBook do to you?

DanielSank

@0celo7 I received a notification email that a group I subscribe to is planning an event.

It's an event I thought skyler would also find interesting, so I forwarded him the email.

0celo7

4:16 AM

@DanielSank could you look at a circuit for me please?

Skyler

@0celo7 there we go, the bait and hook

=P

DanielSank

Unfortunately, it turns out that Facebook's emails contain login tokens, so forwarding the email gave Skyler access to my account.

0celo7

Or is that asking to ask

DanielSank

@0celo7 Just post the damn circuit. You know how this works.

0celo7

@DanielSank Is $i_1(t)$ always $-0.5$A regardless of the switch and inductor status?

Sure looks like like it should be

loltospoon

4:21 AM

@0celo7 I forgot this was a 3-D integral, so I casually added the $4\pi$ in front while forgetting that with the 3-D integral you also multiply the integrand by $r^2$....

0celo7

noooo

I told you about the $r^2$ earlier!

loltospoon

Wait really?

DanielSank

@0celo7 Yes that's what a current source means.

0celo7

2 hours ago, by 0celo7

@loltospoon If you're integrating over $\Bbb R^3$ you'd do the usual thing, with measure $r^2drd\Omega$ and $0\le r< \infty$.

loltospoon

@0celo7 that "2 hours ago" physically hurts me to look at...

haha, thank you though

0celo7

4:24 AM

np

Skyler

lets see, i submitted it to their security team: Whitehat Report #1278680065558432 .
You should probably be posting on https://www.facebook.com/help/hacked since i was going through their whitehat portal. I left a post on your account saying "Proof" so if they want to try and backtrack to IP address and everything for submission that should be pretty easy

DanielSank

@Skyler I filed a privacy complaint.

I said that I understand that it's intentional behavior, but that I find it unexpected behavior, as evidenced by the fact that I accidentally leaked access to you.

Skyler

@DanielSank and I'm pretty sure others have too

honestly itd be easy for them to fix too

just make it so you have to put in your password the first time you are on a computer without the account having accessed it before

DanielSank

@Skyler Yes the problem is simply that they're including authentication tokens in the emails.

It's perfectly possible to persist a login session by other means, but apparently they want people to be able to log in without ever entering their credentials, which IMHO is not a good idea.

Skyler

@DanielSank exactly, thats ridiculous, I want to test this a bit more and see if the token expires after some time

if not then I'll tear them a new one

DanielSank

4:31 AM

@Skyler They're not going to care.

The only way to make them care is to get this into view.

i.e. make a Youtube video or something.

0celo7

tell the hacker 4chan about it

Skyler

@DanielSank well I'll just post about the exploit on a forum some whitehat forum

DanielSank

@0celo7 OMG

Skyler

@0celo7 and that

DanielSank

@0celo7 That is a fantastic idea.

0celo7

4:33 AM

@DanielSank Hello? I only have fantastic ideas.

Skyler

i was typing 4chan and backspaced to whitehat forum because i was thinking that might be a dick move

but its for the better imo

0celo7

I only have to type the 4 and it fills in the rest

which board is my secret

Skyler

i also want to try and google search for tokens and see if I can find them readily available online

since all you need to do is find the URL general case before the special token and see if thats readily found online

big middle finger to Xander and his "theoretical attack" crap

0celo7

@DanielSank In that same circuit, suppose I close the switch. Then $R_2$ is shorted and $i_3(t)=0$. Will $i_L$ be the usual RL current?

Skyler

hey dan, is there any way to wildcard in google search?

DanielSank

4:37 AM

@Skyler I have no idea. Ask Google.

@0celo7 I think so. In general you can't have shorts to ground like that, but in this case it works because the reactive elements are all separated from the short by resistors.

bye, everybody

0celo7

bye'

skill patrol

Cya

dmckee

Question that I think is smarter than it looks, but it isn't getting any love on the site:

3

Q: Why is there a spike in the heat capacity of a diatomic gas, at around the rotational temperature of the molecule?

While studying for my Statistical thermodynamics test, I encountered this graph Source: https://www.physicsforums.com/threads/variation-of-specific-heat-with-temperature.399514/ I know this isn't the best graph you'll ever see, but the "bump" was present on several other graphs as well. So ju...

Skyler

@dmckee reading the question literally gives you the answer

at least 85% the way to the answer

this guy has some studying to do

dmckee

@Skyler No it doesn't. He isn't asking around the plateaus. He's asking about the little excursion above 5/2R at the left end of the rotational plateau.

Which is not something trivially explained with equipartition.

Skyler

4:50 AM

sorry

i was being unclear

reading the question title without regard for the details of the question made me feel like that

upon clicking the question became WAY more interesting

dmckee

We all do that sometimes.

And even knowing that I do it doesn't let me rein it in every time. Despite my fondest wishes.

skill patrol

@ACMisbadguy who are you?

You want to see a "bad guy" @ACMisbadguy?

3 hours ago, by Shog9

@skullpetrol you have the dubious distinction of being in a single-digit position among the top most-suspended chat users.

DanielSank

Hi, everybody.

skill patrol

Welcome back :-)

Skyler

@DanielSank hay saxay

DanielSank

5:03 AM

-_-

Skyler

i feel like i solved that bump problem before at @dmckee but i cant remember anything about it for the life of me

OH, I think I've got it

DanielSank

Do tell.

Skyler

wait no

maybe....

so right as you saturate the DOF, wont those modes still have coherences that disappear as you bump up the temperature and drown it out with noise

rob

This figure rkt.chem.ox.ac.uk/tutorials/statmech/hydrogen.jpg suggests two things. That it is a real phenomena and that it is related to spin degrees of freedom (which would be why it is associated with the rotational turn-on but not with the vibrational turn-on). No time to follow up now, but it promises to be a very interesting question indeed. — dmckee ♦ 8 hours ago

I wonder whether the ortho and para curves cross, or merge? That would be informative.

skullpetrol

5:19 AM

I would assume they are asymptotic, no?

Just a guess.

rob

@skullpetrol In the very long run, yes. I'm wondering whether ortho also has a "bump" at 500K or so, analogous to the para bump around 150 K, before they merge.

Skyler

@rob if we talk in terms of electron clouds, whats happening with the ortho case when it starts vibrating

rob

@Skyler Electron clouds, nothing: the two electrons are sharing a sigma bond. (warning: not a physical chemist.) The degrees of freedom at these temperatures are the two nuclei acting like a rigid rotor with energy $E_J = \frac{E_1}{2} J(J+1)$, where the multiplicity of each level is $2J+1$.

Para is even-$J$, ortho is odd-$J$.

For hydrogen $E_1 \approx 15\,\text{milli-eV}$

@Skyler Oh, you said vibrating. I don't remember the energy for the vibrational modes, but it's much higher than this.

Skyler

@rob wait, which one is the "para" then?

rob

@Skyler Para are the even-$J$ rotor states; ortho is odd-$J$.

Skyler

5:32 AM

also notice HD has a bump

rob

Or is it $L$ instead of $J$? I have to re-derive this whole mess every few years.

@Skyler Yes, smaller and cooler than the para-H2 bump.

Skyler

that quickly is surpressed at the same temperature that H2 para modes start to show up

rob

@Skyler Yes, but HD doesn't have the rigid coupling between $L$ and $S$ quantum numbers because the nuclei aren't identical, and the rotor mass and the bond length are different. That's what moves the turn-on temperature lower.

N2 (but not O2 -- no spin) also has the ortho-para separation, but only at temperatures so low that the N2 not a gas.

Secret

6:22 AM

Anyone knew about the current status of superfluid vacuum theory? Are there recent observations that support refute it. Google so far is not really helping on shedding light on this issue

skill patrol

@JohnRennie have you ever thought of starting a second career as a Lecturer?

John Rennie

@skillpatrol I have toyed with the idea of doing some form of tutoring, but I don't need the money and my time is already fairly full.

skill patrol

How about writing a textbook?

John Rennie

6:44 AM

@Kaumudi.H I finally got around to trying the parathas.

And they were very good.

user228700

@JohnRennie Ooh, they look nice. What was the side-dish?

John Rennie

I just cooked a couple of the parathas to see how they would turn out, so it was more of a test than a meal. Though I did eat them :-)

user228700

So u ate them on their own?

John Rennie

Yes. They are thinner than I expected. When my mother makes them she makes them considerably thicker.

They weren't much thicker than a chapati.

user228700

I...see. From the picture, however, they look considerably thicker than a chapati.

user228700

6:47 AM

My mother's chapatis are incredibly thin (and soft <3)

John Rennie

Ah, maybe I'm used to unnaturally thick chapatis :-)

Anyhow the parathas had a lovely flaky texture and tasted delicious. So now I have to decide how to use them in a proper meal.

user228700

It's a shame that u wasted 3 of 'em on their own :-|

John Rennie

Two of them (one torn in half) and I can assure you they weren't wasted :-)

I have loads left - the pack I bought has a dozen in.

user228700

OK :-) Like I said before, try some of 'em with raita.

John Rennie

I was thinking I might cook half a dozen of them and make a range of side dishes.

user228700

6:51 AM

Make a range of 'em? Are u hosting a party of sorts? :-P

John Rennie

All for me! :-)

user228700

:-) OK. Dammit, my hunger grows by the minute but lunch isn't for another 8 minutes!

John Rennie

8 minutes? You could starve in that time!!!

user228700

:-) 6 now...

John Rennie

My next meal isn't for 7 hours yet ...

user228700

6:55 AM

Lol, that's ur own choice--nobody else I know eats just once a day.

skullpetrol

I do too.

user228700

Is it a diet of some sort?

skullpetrol

nope

user228700

:-| How the heck dyou guys do it..?

John Rennie

It's a man thing.

user228700

6:57 AM

-___-

skullpetrol

^

user228700

Alright, I'm off, lunch awaits!

skullpetrol

Cya

@JohnRennie do you take afternoon tea?

Kenshin

7:50 AM

Hello

skullpetrol

8:31 AM

Hi.

Moses

Does anyone know of a good basic book or set of notes to learn about Einstein coefficients A and B?

Kenshin

@Moses wikipedia is probably the best

Einstein coefficients

Einstein coefficients are mathematical quantities which are a measure of the probability of absorption or emission of light by an atom or molecule. The Einstein A coefficient is related to the rate of spontaneous emission of light and the Einstein B coefficients are related to the absorption and stimulated emission of light. == Spectral lines == In physics, one thinks of a spectral line from two viewpoints. An emission line is formed when an atom or molecule makes a transition from a particular discrete energy level E2 of an atom, to a lower energy level E1, emitting a photon of a particular energy...

Slereah

8:52 AM

Welp

did my usual cocktail of meds

Hopefully this cold will pass swiftly

skullpetrol

Antibiotics?

Slereah

Nah

as u may be aware most colds are not bacterial in origin

Just stuff to manage the symptoms

skullpetrol

Right right.

Slereah

that's what you get for being in contact with people

They are full of germs

skullpetrol

Yup, especially foreigners :P

Slereah

8:57 AM

The Germans are the germiest ones

skullpetrol

Germ, germer, germiest

Virus, viruser, virusiest.

Kenshin

vitamin C has been known to reduce the duration of colds

but not severity

skullpetrol

Works for me.

Secret

"In other words, I claim that successfully written science-fiction will trick the reader into temporarily extending the conceptual representation of the things and relations in the world that she has formed in functioning in a cultural reality, so as to include new “rules,” like those that govern time-travel in the science-fiction text."

No, I want to go one step further

skullpetrol

9:13 AM

A flu is a different story.

Secret

I want to write a scifi where the stories themselves are the characters, and the reality as we knew it is also a character

But in order to do so, the 4th wall must be broken to an extent never reached before

Reality is a notoriously difficult character to work with, especially if you have plot elements that are possibly fringe

skullpetrol

Dubious.

Secret

This, I argue, is really one of the two major drive of my curiousity (the other being intrinsic)

I suspect if one ever want to correctly include reality as a character in their stories, it will be the equivalent of knowing everything about reality

You cannot write a good character without a good understanding of your character and how it weaves into the plot

Meanwhile, the main plot is still a mystery...

Part of the challenges is I need to figure out a way to make stories alive, so even after they are wrote down, they are dynamic structures that continues to interact and evolve with time

Kenshin

What is quantum physics?

is that a term that would encompass both QM and QFT?

or is it a term associated with the mysticism?

Slereah

Quantum physics encompasses both, yes

even larger theories

String theory is still a quantum theory

Kenshin

9:27 AM

@Slereah I've heard people say "quantum physics" when they are talking about the "law of attraction" and "the secret"

Slereah

Those people are plebs

Secret

Quantum mystism is a misuse of quantum mechanics, and I consider it absurd even outside of science and in the realm of esoterics

Kenshin

but I want to use the term "quantum physics" as a title in the PHYSICS book to encompass QM and QFT

and I guess string theory too

Secret

Point is, quantum theory simply cannot achieve what the esoterists claimed. Their requirement is something deterministic but quantum is probablistic

So yeah "those people are plebs"

Kenshin

@Secret if you believe in many worlds theory, then there is always a world where they do achieve what they claim :)

Secret

9:32 AM

Well, thatvworld cannot interact with ours, thus the conclusion does not mean anything

Kenshin

@Secret perhaps the esoterists are able to move their consciousness into the realsm where they achieve success

Secret

Recall that in many worlds, unlike their scifi counterpart, each branch are decohered from each other.

So by definition there cannot be any interaction

Kenshin

@Secret I'm not talking about interaction

i'm talking about, suppose in the future my current path splits into two worlds

one is harmful one is good

an esoterist could claim that they can force their consciouness to go into the good path

while a "zombie" takes over the bad path

a wild claim, but i'm just giving thme th ebenefit of the doubt

Secret

Then that is deterministic. I don't think it will work. Actually I am not sure how memory is decided on which branch they will end up, other than probablistically. There is also that philosophical issue of identity. Which memory belongs to you?

(Still typing...)

skullpetrol

(...we know...)

Secret

9:51 AM

Ok I guess a major problem will be, if the esoterists claimed they can literally force a given state (because everything is a quantum state within the framework of a universal wave function) to a certain end state out of the many possible ones, it means the portion of this wavefunction corresponding to those paths are basically something approaching a Dirac delta distribution. What kind of Hamiltonian will allow that unless it is some highly nonlocal Hamiltonian that evolves that portion of the

universal wavefunction so that the probability amplitude take a delta distribution there some later time. If they claimed that they can locally produce such nonlocal Hamiltonian, then they will violate no communication theorem and said interaction will be non unitary

The thing is, if I understood correctly, in order to reproduce the effects claimed by esoterists, even without quantum theory, consciousness/the mind will be some highly nonlocal entity

Transcript for