The h Bar

DanielSank

17:00

I doubt we have to do things as precisely as LIGO though.

FenderLesPaul

@DanielSank did you happen to take GR at UCSB?

DanielSank

@FenderLesPaul Yeah, one quarter of it. Why?

FenderLesPaul

@DanielSank was it taught by Gary or Don?

0celo7

>Gary >Don

you're on a first name basis with them?

FenderLesPaul

@0celo7 All my professors call them by their first names

so I just got used to it

DanielSank

17:01

@FenderLesPaul I had Horowitz, but the quarter after that was taught be Berenstein.

FenderLesPaul

@DanielSank did you like it?

DanielSank

@FenderLesPaul What is the purpose of this question?

0celo7

@DanielSank so you know GR

FenderLesPaul

@DanielSank just want to know what my fate will be

0celo7

please explain the orientation of the normal vector

DanielSank

17:02

@FenderLesPaul Your fate will be nothing similar to mine.

I was not focused on GR at all.

0celo7

@DanielSank and your work is useful

DanielSank

@0celo7 "Know". Yeah..... not really.

FenderLesPaul

@0celo7 fite me bruh

DanielSank

@0celo7 What are you on about?

ACuriousMind

@DanielSank That's a suspicious way to respond to that question ;)

0celo7

17:03

@FenderLesPaul Do you know why the orientation of the normal vector when doing the divergence theorem on a spacelike hypersurface has to be reversed

to point inwards

I've traced a line of like 6 theorems in Lee and I find no reason for it

so it has to be something retarded with the way Carroll/Wald/HE define volume elements

@DanielSank I'm saying quantum computing is more useful than GR/QG

FenderLesPaul

@0celo7 what signature does Lee use?

0celo7

No signature, Riemannian

I've checked every proof

DanielSank

@ACuriousMind I just fail to understand the purpose of asking this. @FenderLesPaul likely will have a different prof and likely a different book. Aside from all of that, the only thing that actually matters in grad school courses is the other students.

0celo7

Nowhere does it matter

I know the typical physicist will say you just flip signs and shit when doing Lorentzian geometry

But I see no reason to.

FenderLesPaul

Wald/Carroll/HE define volume elements with a certain orientation so that timelike boundaries give a sensible result

DanielSank

17:06

Well, ok that's not entirely true. The prof. does matter in that they set the tone and assign either good or bad homework.

FenderLesPaul

Lee probably defines it with only space like surfaces in mind

0celo7

@FenderLesPaul they never make it clear how they orient stuff.

if $\mu$ is the volume element on $M$

DanielSank

@0celo7 How is that relevant to anything else we were talking about? I am confuse.

0celo7

then $i_X\omega$ has the same orientation iff $X$ points out

this is a topological thing

and there's another theorem which says that if $N$ is normal to $\partial M$, then $i_N\omega$ is a Riemannian volume form on $\partial M$

and the only thing that's different for -+++ is a minus in a square root

@DanielSank I was taking a jab at FLP, ignore it

user116211

@ACuriousMind: o/

DanielSank

17:09

Anyone know the reason for this:

2

Q: Why doesn't $σ_x*σ_p$ change with the width of the well in the infinite square well problem (intuition)?

I calculated that the product of the uncertainty in position $\sigma_x$ with the uncertainty in the momentum $\sigma_p$ does not change with respect to the width of the well in the infinite square well problem in quantum mechanics. Why does this happen intuitively?

user116211

So, $S= k\ln \Omega$ only applicable for equilibrium?

ACuriousMind

@DanielSank I don't understand the question. That the product of uncertainties is constant is a property of the Fourier transform. I don't see what it would have to do with the width of the well.

@user36790 Yes

user116211

@ACuriousMind Boltzmann entropy only for equilibrium?

user116211

@ACuriousMind And for general case?

0celo7

@ACuriousMind once again, you're too smart

DanielSank

17:13

@ACuriousMind If the fact that $\sigma_x \sigma_p$ is independent of well width can be traced to a property of the Fourier transform, then that would make an answer.

0celo7

OP wants to see that the change in $\sigma_x$ cancels that of $\sigma_p$

DanielSank

Off the cuff I don't see it, or at least it's not totally trivial (I can see how I might try to go about proving it).

ACuriousMind

@user36790 Yes. Suppose we have $N$ microstates with the $i$-th microstate occuring with probability $p_i$. The statistical entropy $-\sum_i p_i\ln(p_i)$ only becomes $\ln(N)$ if $p_i = 1/N$ is uniform, and uniform distributions are those of equilibrium.

0celo7

Uncertainty principle

In quantum mechanics, the uncertainty principle, also known as Heisenberg's uncertainty principle, is any of a variety of mathematical inequalities asserting a fundamental limit to the precision with which certain pairs of physical properties of a particle, known as complementary variables, such as position x and momentum p, can be known simultaneously. Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the...

@DanielSank See the linked proof.

user116211

@ACuriousMind Oh!! That's Gibbs entropy!

user116211

17:15

@ACuriousMind At equilibrium, all microstates are equally likely; got it.

0celo7

Are they

What about the Microstate on the moon

user116211

@ACuriousMind Should I delete my answer? Waste of time (._.)

DanielSank

@BernardMeurer Isn't that what StackOverflow is for?

I mean, I'm hesitant to give you the email of a colleague if you don't have a specific question.

Is the question of the form "what set of tools should I use to not hate my life?"?

0celo7

@DanielSank I ask myself a similar question every day ;_;

DanielSank

@0celo7 Ok, well, consider using typescript for your js.

ACuriousMind

17:20

@user36790 Well, I don't really know what you're doing in that answer, anyway. There's a lot of words and formulas, but addivity of $\ln(\Omega) = \ln(\Omega_1)+\ln(\Omega_2)$ is really just observing that if I got $\Omega_1$ possible states for one system and $\Omega_2$ possible systems for another, I have $\Omega_1\cdot\Omega_2$ for the combined system.

DanielSank

Typescript is better than js, but compiles to (readable) js.

For web server, this is going to depend a lot on what your project needs to do and where you want to host it.

If you want to host on Google App Engine, you probably want Django.

(since that's what everyone else does)

If you want to use some other hosting service that supports Rails, then you have that option as well.

0celo7

@DanielSank what

DanielSank

@0celo7 wat?

0celo7

@DanielSank that was a good movie

@DanielSank something about js, what are you talking about

DanielSank

@0celo7 This is why I ignore you some times.

@0celo7 javascript

user116211

17:23

@ACuriousMind Well I wanted to shew the entropy formula from the definition of $\beta$ parameter; and using the $\log$ relation, I computed the additivity in equilibrium.

DanielSank

@0celo7 If you follow the chat log this would be obvious.

0celo7

@DanielSank I have no clue what you're talking about

@DanielSank :/

@DanielSank I don't know anything about JavaScript

DanielSank

27 mins ago, by Bernard Meurer

@DanielSank Fundamentally I'm just looking for someone who is a better/more experienced dev than I am so that I can ask a couple of tips about web dev

0celo7

Oh

DanielSank

"web dev"

0celo7

17:24

I was talking about not hating myself :/

Or my life

Whatever you said

ACuriousMind

@user36790 Well...why? The question isn't about deriving the formula for entropy, it asks why it is additive.

DanielSank

@0celo7 Ahahahahahaha I responded to you thinking you were @BernardMeurer.

I'm SO bad at chat.

Dammit, @0celo7!

I cannot handle your random quips!

0celo7

:(

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@DanielSank like everything else it takes practice :)

DanielSank

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Grah! @0celo7 piped in at a weird time and I had a bunch of messages meant for @BernardMeurer pinged at 0celo7 instead.

0celo7

17:27

:(

I'm going back on the ignore list

Bernard Meurer

@DanielSank Should I take you mistaking me for @0celo7 as an insult? :p

0celo7

Because DS can't read :(

ACuriousMind

Best clickbait I've seen in a while on the HNQs: Can't find clit

4

0celo7

@BernardMeurer at least one of us can do basic linear algebra.

So I'd watch your mouth

DanielSank

@ACuriousMind Wow.

Bernard Meurer

17:28

@0celo7 I could go after answering that, but I happen to value our friendship more than that

DanielSank

A problem as old as humanity and it shows up in Linux.

3

0celo7

@BernardMeurer what

DanielSank

@0celo7 You know @BernardMeurer just finished high school, right? How should he know linear algebra?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

AP?

0celo7

@DanielSank because I told him to read the same book I did as a Junior

DanielSank

17:31

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I dunno, we didn't have lin alg in high school.

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ no

DanielSank

@0celo7 Right, and everyone just does what you say.

@0celo7 At least he knows basic programming, so I'd watch your mouth.

0celo7

@DanielSank we were literally discussing exercises in it over Skype

@DanielSank Point taken

But why so hostile

Bernard Meurer

@DanielSank The questions is on the lines of what's the best tool for me to work on the project without having to burn in hell afterwards. Because a dude over at the makerspace is telling us to use Node, but I can smell the hipster in him so I just don't trust him. I was going to say Django but then I wasn't 100% sure on that and thought I was better off finding someone who was more experienced

DanielSank

@0celo7 Hostile? I copy/pasted your own post.

ACuriousMind

17:32

> I can smell the hipster in him so I just don't trust him

0celo7

@DanielSank before that

@ACuriousMind is that referring to you

DanielSank

@BernardMeurer Node is popular so you won't burn in hell. That said, I wouldn't use it because I think it's a foolish idea from the start.

ACuriousMind

@0celo7 I am many things, but I am certainly not a hipster

Bernard Meurer

@ACuriousMind It's a valid concern is it not?

0celo7

@ACuriousMind really?

Bernard Meurer

17:33

@DanielSank Exactly! I just dislike Node, I find it bloated

0celo7

You have long hair, listen to weird music and have a weird life path

ACuriousMind

@BernardMeurer I have not yet decided whether I find that valid or just funny

DanielSank

@0celo7 What was hostile, when I politely noted that @BernardMeurer isn't in college yet? I was defending someone against what was either an insult or a tease (hard to tell difference on chat) from you.

There's no way to construe that as hostile.

Bernard Meurer

@0celo7 Don't get so worked up on it, it wasn't even in italics

ACuriousMind

@0celo7 "Weird life path"? I'm a student, my life path is literally only one step so far.

Bernard Meurer

17:36

And thanks for trying to make him understand the obvious @DanielSank :)

0celo7

@ACuriousMind theoretical physics is pretty weird

DanielSank

@BernardMeurer Anyway, the only reason people invented/use node is to not have to learn multiple languages.

Bernard Meurer

@ACuriousMind You're a german student, that's an entirely different thing

0celo7

And you're a savant

DanielSank

They essentially said "Browsers use js, so let's use js on the server too so we only need one language."

0celo7

17:37

Hipster confirmed

Not a bat thing

DanielSank

It's a good idea in principle, but js sucks.

0celo7

Or bad

ACuriousMind

@0celo7 But I want to be Batman!

Awesome typo

DanielSank

@BernardMeurer Now, I have not actually used it, so I can't tell you from experience whether or not it's good. It is popular though, so it probably works fine for small projects.

Bernard Meurer

@DanielSank That was exactly my thought. I'm just shoving Django down everyone's throat then :)

#pythonmasterrace

DanielSank

17:38

@BernardMeurer mehhhhhh Django has issues too, but yeah whatever.

Bernard Meurer

@DanielSank And so does Ruby :p

DanielSank

@BernardMeurer #scalamasterrace you scrub ;)

Bernard Meurer

@DanielSank Hahaha, praise scala!

0celo7

@DanielSank I was teasing him because he was struggling with linear algebra exercises a few days ago; chill bruh

@ACuriousMind you're too scrawny for that :P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Stop kicking around sand you bully :P

DanielSank

17:41

@0celo7 I like how I make a harmless comment, and then you construe it as hostile and tell me to chill... :-|

0celo7

@DanielSank CHILL

Wow all that rage

ACuriousMind

@DanielSank He's paranoid, that's well-known

2

0celo7

@ACuriousMind well known by whom

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Reverse psychology @DanielSank

0celo7

Who's saying that about me

ACuriousMind

17:42

The hivemind of this chat

DanielSank

@ACuriousMind Oh yeah, I met the Hivemind once. Weird guy.

0celo7

Ah, the puppet master Daniel Sank

ACuriousMind

@DanielSank lol

0celo7

Why is that funny

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Why ask why?

0celo7

17:45

Because I want to know

That's not a good question

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Says you.

0celo7

But I'm always right

DanielSank

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Why ask (why ask why)?

ACuriousMind

@0celo7 A hivemind is not a person you can meet. Subverting expectations is one of the bases of humor :P

DanielSank

@ACuriousMind Holy crap... I real answer... wow.

@ACuriousMind I'm somewhat convinced that the only 100% common element in humor is subverting expectations in a way that is "obvious" in retrospect.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

17:47

@DanielSank toché

0celo7

@ACuriousMind dunno, you can meet the gravemind in Halo, you can talk to Legion in ME

@DanielSank what are you confused about

John Doe

In Griffiths Quantum Mechanics book he describes a bound state (in quantum mechanics) as $E < 0$ . But what kind of physical system has negative total energy? I'm just starting to learn about QM so this might be trivial.

ACuriousMind

@0celo7 You must be fun at parties.

0celo7

@DanielSank correct

@ACuriousMind huh?

ACuriousMind

@0celo7 No one likes people killing jokes.

Bernard Meurer

17:48

@ACuriousMind He just passes out at parties

ACuriousMind

@JohnDoe The zero of energy is arbitrary

0celo7

@ACuriousMind You can't kill a bad joke

John Doe

@ACuriousMind How is it arbitrary?

0celo7

@BernardMeurer for the last time, I was not passed out

I just somehow ended up on the lawn, laying down

ACuriousMind

@JohnDoe You can add any constant to the potential (which is the potential energy) without changing the physics since the force is the derivative of the potential and the derivative of a constant is zero.

Bernard Meurer

17:51

@DanielSank Turns out to get that clit you need to have calibre

sharaf zaman

hello

John Doe

@ACuriousMind Okay I understand that. But how is the zero of $E$ arbitrary? For example if you have $E=0$ then that implies that the system has zero total energy...Are you saying that this system is identical as $E = C$ for some positive constant $C$?

ACuriousMind

@JohnDoe Energy is the sum of kinetic and potential energy. If I can shift the potential energy however I like, I can put the zero of energy whereever I like. What matters is energy difference between systems, not the absolute value of the energy.

John Doe

@ACuriousMind Oh okay I think I understand...What is important is $E-V$? Rather than just $E$?

ACuriousMind

@JohnDoe Not really. It's rather that what is important is the energy difference $E_1 - E_2$ between two different states.

$E-V$ would be just the kinetic energy

Not that that doesn't matter

John Doe

18:02

@ACuriousMind I'm just trying to get an idea of why the classifications for bound states are $E < 0$ and scattered states are $E > 0$. You have given a good reason why $E < 0$ is not a problem. So all I want to confirm is that the importance of the classification. My idea is that the classification is important based on $E < V$ or $E > V$ at infinity at -infinity. That's what I meant. What do you think?

ACuriousMind

@JohnDoe Ah, yes, that's true. The zero of energy is chosen such in this case that a classical particle at infinity with zero kinetic energy would have zero energy. So $E<0$ means you have not enough energy to sit at infinity, i.e. you're bound, and $E>0$ means you are not only unbound, but are also moving.

DanielSank

@BernardMeurer True dat.

ACuriousMind

It is very common to choose the zero of the potential such that it lies at infinity

John Doe

@ACuriousMind For clarity, when you say the 'zero of energy'...does that mean the value of $E$ which corresponds to no total energy.

ACuriousMind

@JohnDoe I mean the zero of the potential.

John Doe

18:17

@ACuriousMind Great thanks.

ACuriousMind

@JohnDoe Yes

hubot

Hi!

0celo7

@ACuriousMind what does that mean

ACuriousMind

18:48

@0celo7 All the information to figure that out is there

0celo7

@ACuriousMind why are you acknowledging his thank you

you never do that for me :(

ACuriousMind

@0celo7 I didn't do that.

0celo7

@ACuriousMind you said "yes" to his "great thanks"

ACuriousMind

No, I didn't :)

0celo7

what

I'm looking at it right now

I hover on the arrow

and his "great thanks" lights up

ACuriousMind

18:52

Yeah, I realize that, but it is not what it looks like

0celo7

best defense ever

I'll remember that for when I get in trouble

ACuriousMind

And I can prove it to you, but I will keep you in suspense until I get back from shopping

0celo7

stackexchange deleted my started post on that stokes theorem question

sigh

it was probably garbage anyway

doing things correctly is apparently not right in GR

0celo7

19:05

@ChrisWhite on the top of page 432 in Wald, should there be wedges in the formula for $\tilde\epsilon$?

I'm checking everything he says now

user54412

don't have books in my office

user54412

also apparently someone stole half the textbooks from the one bookshelf in the department

0celo7

lol

@ChrisWhite you don't have a PDF of Wald?

user54412

it's called laziness

0celo7

@ChrisWhite what's that

also I think Eq. B.2.14 contains a typo

John Doe

19:12

@ACuriousMind For the free particle where $V(x) = 0$, it states that this is always a scattered state. Is the reason for this that that if we consider $E \leq V(x) = 0$ then we get stationary states whose general solution (linear combinations of stationary states) is not normalizable. Hence we consider only the free particle for $E \geq 0$ since even though the stationary states are not normalizable, the general solution is. Is this correct?

0celo7

@ChrisWhite I'm looking up formulas for Levi-Civita symbol contractions

god help me

hubot

I've begun read Landau & Lifshitz book.

ACuriousMind

@JohnDoe It's the same in classical mechanics: Consider the classical Kepler problem and put the zero of the potential to infinity (that's what the usual $1/r$ potential without any constant does, anyway). If the total energy of the particle in this central potential is smaller than zero, then it is in a orbit. If it is larger than zero, it's on an escape trajectory to infinity. Nothing to do with normalizability

2

It is in general the case that the free/scattered energy eigenstates are not normalizable in quantum mechanics, but that's really more an "accident" of the formalism, imo

0celo7

@ChrisWhite here's a crazy idea, what if we go to Riemann normal coordinates and work in Minkowski space

ACuriousMind

@0celo7 Look at the edit history of the post I replied to, then you'll see I originally responded to a different message.

0celo7

19:22

the choice of "in vs. out" should not depend on the coordinates

@ACuriousMind why was this starred?

ACuriousMind

How am I to know? I certainly didn't star it.

I guess someone found it insightful :P

0celo7

@ACuriousMind we star stupid/witty shit here

not insightful

@ACuriousMind Ah.

Didn't notice that.

John Doe

@ACuriousMind Okay I see but what I'm basically asking is why can't we have $E \leq 0$ for a free particle where $V(x) = 0$? If we did then we would have a bound state. Why is it given that $E \geq 0$ for the free particle and hence it is in a scattered state. My guess was that because for $E \leq 0$ we get a general solution which is not normalizable and hence not physically interesting.

ACuriousMind

@JohnDoe For a free particle, the Hamiltonian is just the square of $p$. That can't be negative to begin with.

user54412

@0celo7 genius. wish I thought of that, like, say, yesterday

ACuriousMind

19:30

yesterday, by Chris White

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

John Doe

@ACuriousMind
In this case the Hamiltonian would be $-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$, why can't you get a negative eigenvalue?

0celo7

@ACuriousMind I know he said that :P

@ChrisWhite How about this, what if I pick coordinates so that $n_a\propto \mathrm{d}x^1_a$

now ACM wonders why my basis forms have two indices

ACuriousMind

@JohnDoe Why the $-$? The Hamiltonian of a free particle is $H = \frac{p^2}{2m}$, and in the position basis it's still just $\frac{1}{2m}\frac{\partial^2}{\partial x^2}$.

0celo7

@ACuriousMind o.o

ACuriousMind

@0celo7 Hmmm?

0celo7

19:35

There's a minus from the i^2

ACuriousMind

Ach, damn

Sorry, @JohnDoe, here's the correct answer:

bolbteppa

The speed of light is "variable" in GR because the space changes, not the light right? If you go to a locally Minkowski coordinate system in a curved space the light beam will be the same constant value because the space is locally flat, so saying the speed of light is not constant in general relativity seems very misleading and ill-stated right?

ACuriousMind

The operator $p = \mathrm{i}\partial_x$ is self-adjoint and its spectrum is the entire real line (every possible real value can appear as the momentum. When you square it, the spectrum of the square is just the square of the spectrum, i.e. the non-negative real line.

And now I have to make food before I starve.

John Doe

@ACuriousMind That's a good answer, thanks. Eat your food...

0celo7

@ACuriousMind so Deutsch

@ACuriousMind $p=-\mathrm{i}\nabla$ ;)

@ACuriousMind I don't think that example works @ChrisWhite

because the box will not not have a spacelike/timelike boundary everywhere

user507974

19:57

@ACuriousMind tell us what you end up making

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