The h Bar

Giskard42

16:00

I usually think of charge in a gas as ions, but I haven't put enough energy into the plate to ionize it

G. Bergeron

@bolbteppa ...meh yea-ish

Giskard42

@G.Bergeron Say I induction furnace it, so there's no contact. I'm just boiling it slowly away entirely

G. Bergeron

@bolbteppa It's not meant to act on anything

@Giskard42 There could be enough energy to ionize

@Giskard42 but what will happen is that you will end up with charged gas

that will collect on the other plate

or more like soot particles

Giskard42

@G.Bergeron Sweet, so even though the gas isn't ionized, it'll still be charged?

The electrons will be distributed throughout the gas?

G. Bergeron

@Giskard42 Well, yes. But it will be ionized

Really mildly

bolbteppa

16:02

I know, but in the classical case, when you integrate $\int \rho d^s x d^s p$ you get infinity because of the $2s - 1$ constraints, i.e. in phase space it makes perfect sense for constant energy to eliminate a variable and make this integral bad, but in the quantum case, it's analogue, $\mathrm{tr}(\rho)$, should also become infinite due to conservation of energy, so you write $\mathrm{tr}(\rho) = \sum_n <n | \rho | n>$ or something and conservation of energy should motivate infinity right?

Giskard42

@G.Bergeron Ah, okay.

@G.Bergeron Thanks a lot for your help!

Sorry to interrupt

bolbteppa

Hmm

G. Bergeron

@bolbteppa If the spectrum is continuous then it's like the classical space in that respect

bolbteppa

I think I got it

G. Bergeron

your concentrating a probability measure on a set of measure zero with respect to the previous measure... It has to be infinite

@Giskard42 np

So if it is continuous it won't be a sum but rather an integral

bolbteppa

16:06

If $\rho$ was a constant number, then you have $\mathrm{tr}(\rho) = \sum_n <n|\rho|n> = C \sum_n <n|n> = \infty$, but the ability of writing $\ln(\rho_n) = \alpha + \beta E_n$ allows $\rho_n$ to be a constant since $E_n$ is constant

G. Bergeron

$Tr(\rho)=\int\limits_e \langle e | \rho | e \rangle$

@bolbteppa But that would mean constant on all the spectrum so only one energy state (modulo degenarcies)

bolbteppa

Still making sense of the number formalism tbh, is $\sum_n <n | n> = \infty$ even true?

G. Bergeron

@bolbteppa If $n$ run to $\infty$, yes but it may not

e.g. spins in magnetism

Your sum is $\sum_n 1$

The point is not that the delta appears like that

You put it in

as a constraint

bolbteppa

hmm

G. Bergeron

energy would be a dimension like in the classical case

bolbteppa

16:12

So my infinite trace, if it were true, would justify the delta function, it exactly copies the classical case, the question is justifying that infinite trace, it seems like he might be waving his hands and saying, because we have so many particles, and because energy levels are so close, we can treat it as infinite

G. Bergeron

and you evaluate a non-vanishing distribution on that N-1 dimensional object. So the distribution on the whole space has to be a delta function with respect to the observables (Hamiltonian) associated to the constraint (energy)

@bolbteppa You don't do that in the classical case either

bolbteppa

Landau does do this in his book though

Let me get it

G. Bergeron

You put it in when you assume that energy is conserved and the system is closed

Maybe not explicitly but it appears nevertheless

bolbteppa

archive.org/stream/…

In between equations 4.3 and 4.4

G. Bergeron

Exactly he says "because of the constraint, we'll have to put in a delta function"

essentially

bolbteppa

16:17

You seem to be saying, we add the constraint of a delta function to $\ln (\rho_n) = \alpha + \beta E_n$, but he is literally saying $\rho$ becomes a delta function without working out a Lagrange multiplier problem, it's automatic just from analyzing the integral $\int \rho d^s p d^s q$

G. Bergeron

The real notion is "concentrating a distribution on a set of measure zero"

forget about the log thing

you shove the delta in the integral not the log

bolbteppa

The log thing is literally the most important thing from this perspective :p

G. Bergeron

He says it becomes a delta BECAUSE of the reduced dimensionality of the manifold, another way of saying constraints on the manifold

bolbteppa

Ok he says the integral $\int \rho d^s p d^s q$ is zero, not infinity :\

G. Bergeron

@bolbteppa Not at all on this delta thing

@bolbteppa zero outside of the reduced manifold

where the constraint is violated

Look, it's like if I tell you "I have a point somewhere in a plane"

Now, I tell you I know it lies on a line

The distribution will become a delta function on the line as a distribution on the whole plane

it needs to be infinite because the probability if picking a point on this exact line is zero for any bounded distribution

It's a properties of continuums

The point about the log is that just by stating that your system is invariant under rotations, translations and isolated, which are all extensive quantity, then you have to have the microcanonical ensemble if there are no other conserved quantity

Translation invariance = $\rho$ is not a function of space, Rotation invariance = not much because $\rho$ is a scalar, time invariance = energy is conserved and fixed so it is a parameter of $\rho$.

bolbteppa

16:29

So he seems to be saying, classically, that - you have conjugate variables $(p_i,q_i)$ okay? Because we're in the Hamiltonian formalism, we can treat one of these variables as, say, $q_2$, as the energy. But then $dq_2 = dE = 0$ because energy is conserved. However, the integral $\int \rho d^s p d^s q$ as an integral over $2s$ coordinates has to hold, and so $\int \rho d^s p d^s q = \int \rho d^{s-1} q d^s p dE = \int \rho d^{s-1} q d^s p \cdot 0 = 0$.

The only way to make this integral non-zero somewhere, is if $\rho$ is a distribution, in this case a delta function is chosen so that energy is constant on our hypersurface. That seems to be the logic of why it becomes a delta function. Maybe you can write it as a Lagrange multiplier problem, delta function seems scary but okay, but same outcome right?

The same logic, as Landau seems to indicate, carries over to the quantum case, no Lagrange multipliers necessary, though it may be equivalent. The question is the quantum case, my trace should be zero apparently :p

G. Bergeron

@bolbteppa Lagrange multiplier is not for the per se delta function, it is to obtain the distribution directly

bolbteppa

Yes I agree with what you've written

G. Bergeron

@bolbteppa This is mostly it

bolbteppa

So why is my trace zero

haha

G. Bergeron

@bolbteppa same outcome

it's not

it's only gonna be a delta function if the trace is an integral and then it is the same reasonning

@bolbteppa Anyway, I need to go

bolbteppa

16:33

Alright thanks a lot for the help

G. Bergeron

np

bolbteppa

Oh so $\int_e <e |\rho | e> = C \int_e <e | e> = 0/\infty$?

Top of the page here archive.org/stream/… he seems to justify writing this integral

MetaEd

20 hours ago, by MetaEd

Do I need an add-on in chat that interprets these markups correctly?

bolbteppa

If $\mathrm{tr} (\rho) = \int_e <e | \rho |e> = C \int_e <e|e> = C \int_e 1 = \infty$ then we have a function that is infinite on the energy surface, as a delta function is, motivating writing it as a delta function maybe?

(So close!)

DanielSank

@MetaEd Yes, chatjax.

See link on top-right of this page.

MetaEd

16:43

the MathJax link?

DanielSank

yep

MetaEd

Thank you.

DanielSank

javascript:(function(){if(window.MathJax===undefined){var script = document.createElement("script");script.type = "text/javascript";script.src = "https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML";

var config = %27MathJax.Hub.Config({%27 + %27extensions: ["tex2jax.js"],%27 + %27tex2jax: { inlineMath: [["$","$"],["\\\\\\\\\\\\(","\\\\\\\\\\\\)"]], displayMath: [["$$","$$"],["\\\\[","\\\\]"]], processEscapes: true },%27 + %27jax: ["input/TeX","output/HTML-CSS"]%27 + %27});%27 + %27MathJax.Hub.Startup.onload();%27;if (window.opera) {script.innerHTML = config} else {script.text = config} document.getElementsByTagName("head")[0].appendChild(script);(doChatJax=function()

{window.setTimeout(doChatJax,1000);MathJax.Hub.Queue(["Typeset",MathJax.Hub]);})();}else{MathJax.Hub.Queue(["Typeset",MathJax.Hub]);}})();

@MetaEd, alternatively, copy that text into a bookmark.

Take all three messages I posted there and concatenate them in a text editor, and then make them the contents of a bookmark.

MetaEd

Oooh. Look at that.

Hey-men-whatsup

G. Bergeron

16:52

@bolbteppa Not quite that

Giskard42

@G.Bergeron Sorry to bother, but can I pick your brain again a bit more?

G. Bergeron

:34112266

@bolbteppa This argument could prove alot of things that shouldn't are infinite

@bolbteppa The first and second integral are not on the same space

@Giskard42 yes

Giskard42

@G.Bergeron Is there any way for me to figure out what percentage of the gas would be charged?

G. Bergeron

It depends on the gas

you need to figure out the possible ionization and the associated energy

Giskard42

I guess it would just be the number of gas particles / q, right?

G. Bergeron

16:56

@Giskard42 not quite

ionization is a quantum process

all or nothing

Giskard42

@G.Bergeron I guess I keep thinking about this in terms of metal charge

But you're right of course

G. Bergeron

@Giskard42 It is essentially the same

@Giskard42 They are just spread out

@Giskard42 First approximation: as many particles as needed to cover the total charge in terms of elementary charge unit

Then you could consider the average thermal energy per molecule to calculate the odds of having doubly ionized molecule

But it won't be that relevant

This is because there are no fractional charge (ok, yes there are in exotic condensed matter system)

(but that is an emergent effect)

Giskard42

17:17

@G.Bergeron Thanks for the outline!

Argon

Why does spin 1 have 3 states? Griffiths seems to think its obvious.

user246160

@JohnRennie Does that imply that this answer is somehow wrong physics.stackexchange.com/a/160130/135951 ? I read your points about boiling and they seem reasonable but on the other hand I am not able to connect you conclusion with this answer. Could you clarify that a bit, please ? (You mention that closed systems have no boiling point defined but the answer i linked says the opposite)

John Rennie

@Doraemonドラえもん a liquid in a sealed vessel will boil i.e. it will turn into vapour. But it does so over a temperature range. There isn't a boiling point in the sense that the temperature stays constant while the liquid is boiling away.

If you boil water in an open pan then the temperature rises to 100C then stays at 100C while the water is boiling. That's what we mean by a boiling point. If you put the water in a sealed vessel then the water changes to steam over a continuous range of temperatures.

There is no distinct boiling point as you get in an open pan.

user246160

17:33

@JohnRennie Okay I get your point. But in that case is calling that phenomenon "boiling" justified ? Shouldn't it be just called vaporization ? Because boiling occurs at a fixed boiling point and till the whole liquid vaporizes it continues. But here effectively the vaporization is taking place over a range of temperature, in a state of dynamic equilibrium.

user246160

Or is boiling point defined slightly differently in a closed system? The answer I linked says that boiling starts at point 2 as in the diagram..

user246160

John Rennie

I'm not sure how precisely the term boiling is defined. I agree with you that in a sealed container it's probably better to use the term vaporisation instead, but does it really matter?

user246160

@JohnRennie That is true though. It doesn't matter if we know what is practically happening. But there is (apparently) a lot of confusion about how people really define boiling :P

DanielSank

This is incredible

12

AccidentalFourierTransform

17:41

ewwwww

DanielSank

18:00

@heather did some work on my lin alg book last night.

I have a good plan for progress now.

I need to focus on getting one particular section done, and then the rest of the work is revising material that's already written!

user246160

@DanielSank You are writing a book on linear algebra ?

DanielSank

@Doraemonドラえもん I think that is relatively obvious from my comment :P

Here it is

user246160

Oh...are you planning to publish hard copies ? :-D or ebook ? and how is it different from other books on linear algebra ? :)

user246160

I am checking the link

DanielSank

@Doraemonドラえもん It's different from other books in a few ways:

1) It assumes you already know the basics.

2) It focuses on developing a very solid/deep understanding of the difference between (vectors and linear transformations) representations of those things using columns of numbers (for vectors) and matrices (for linear transformations).

3) After spending the first chapter making sure we understand what matrices are etc., the second chapter introduces the Fourier transform as a basis transformation.

The idea is to make the Fourier transform "obvious".

It is introduced as a way of diagonalizing the derivative, which is very explicitly pointed out to be a linear transformation.

I am doing this because I presented this work to an undergraduate journal club many summers ago, and was met with overwhelming positive feedback.

user246160

18:09

Wow...the link is impressive indeed. I would love to read it fully. I haven't taken any linear algebra courses yet but i am comfortable with basic matrices and determinants. It might be helpful for me next year :-D

DanielSank

@Doraemonドラえもん I have one request:

user246160

@DanielSank Indeed. You should go ahead with this book for sure. It might become a classic in the future. Good luck. :)

user246160

@DanielSank yes tell

DanielSank

As you can see, the book is currently publicly available. This work represents an enormous amount of time and effort on my part. All I ask in return from those who read it now, is that they provide feedback.

The work is on github, and it is trivial to file issues. If you read it and you find a typo, or you have a feeling that something could be more clear, I ask that in return for the book being public, you please tell me via github issues what your thoughts are.

user246160

Sure....I will read it and give you feedback ! It looks interesting. But I don't have a github account...so is it ok to use hbar for that purpose ? And have you written any chapter on matrices and determinants yet ?

DanielSank

18:13

@Doraemonドラえもん Hbar is fine.

The first chapter discusses matrices. There is nothing on determinants.

There is something I would like to write about determinants, but it's not directly related to the goal of this book.

AccidentalFourierTransform

determinants suck

DanielSank

WAT!?

Determinants are awesome.

There is a simple geometrical interpretation: the determinant is the volume of the space cut out by the matrix acted on a set of orthonormal basis vectors.

AccidentalFourierTransform

Down with Determinants!

user246160

@DanielSank BTW in github how to view the math properly? It is showing just the code... (pardon me I never used github)

DanielSank

@Doraemonドラえもん You have to clone the repository and run LaTeX on the code!

@AccidentalFourierTransform Yeah fine, maybe we shouldn't use them for some proofs, but that's a far cry from "down with determinants".

user246160

18:16

@DanielSank how to clone repository ?

Physics Guy

@AccidentalFourierTransform Hehe

DanielSank

@Doraemonドラえもん What operating system are you using?

user246160

@DanielSank windows 7

AccidentalFourierTransform

@DanielSank I know I know, I was just horsing around ;-)

DanielSank

@Doraemonドラえもん Click the green "clone or download" button.

Then "download zip".

Then unzip it.

Then read the instructions for how to build it.

user246160

18:20

Ok doing it

ACuriousMind

@AccidentalFourierTransform While some proofs avoiding it are actually better, I always thought it's silly to say that determinants are "non-intuitive" or "without motivation". They are simply the factor by which a matrix scales the volume of a cube.

AccidentalFourierTransform

and we all love Jacobians, I know

DanielSank

@ACuriousMind Yes. That fact is oddly under-appreciated in school.

ACuriousMind

@DanielSank It's how I was taught them :P

AccidentalFourierTransform

@DanielSank I was taught determinants that way I think

I think thats how its done nowadays

vzn

18:24

@DanielSank saw that, enjoyed it, & was gonna post but couldnt find the right link to it (aaronson gave the main site addr & its gonna scroll off soon). seems like nobody ever talks about aaronson in here... :| what do ppl think of him anyway?

DanielSank

@vzn No idea.

vzn

@Secret plz let me know if you can figure out his name, maybe worth looking into :)

@ACuriousMind all scientific fields, being engaged in by humans, have a political component... it varies to the degree its "swept under the rug" :|

bolbteppa

From $\mathrm{tr}(\rho) = \sum_n <n | \hat{\rho} | n>$ we get $\mathrm{tr}(\rho) = \int <n|\hat{\rho}|n> dn = \int \rho_n(E) \frac{dn}{dE}dE $ and now :(

user246160

@DanielSank I downloaded it

user246160

18:30

Now what to do ?

user246160

How to open the files ?

user246160

Adobe not working

user246160

@DanielSank Where are the instructions ?

AccidentalFourierTransform

@Doraemonドラえもん have you ever used Latex?

user246160

@AccidentalFourierTransform Other stack exchange ?

user246160

18:32

I only used latex to type in SE

AccidentalFourierTransform

almost

user246160

(I never took a course in coding...so bear with me :P)

AccidentalFourierTransform

you used mathjax

user246160

yeah..i used mathjax

AccidentalFourierTransform

it is related to latex, but it is not the same thing

well, today youre going to learn about one of the most important tools for a scientist

vzn

18:33

@Secret would like to analyze the article/ topic more at length with you sometime if you can drop by another chat room sometime, this one tends to be kind of noisy. chat.stackexchange.com/rooms/9446/theory-salon as for "real but nonphysical" my thought is maybe what is meant more is "weakly physical" or "weakly coupled/ interacting". there are many examples of the latter from classical physics... have been developing analogies along those lines for many yrs...

user246160

@AccidentalFourierTransform I know that..but how should i proceed after downloading that folder ? :)

user246160

@AccidentalFourierTransform oh really ? :-D

user246160

I am excited

user246160

:)

AccidentalFourierTransform

@Doraemonドラえもん yes :-) just download latex

you'll see ;-)

user246160

18:34

@AccidentalFourierTransform from where ?

AccidentalFourierTransform

it is a compiler that allows you to typeset professional documents way better than word

and its free!

let me find a good source

do you use windows?

user246160

@AccidentalFourierTransform can you give the link for downloading latex ?

AccidentalFourierTransform

google

user246160

yeah windows

AccidentalFourierTransform

@Doraemonドラえもん yes. Do you use windows?

bolbteppa

18:35

Get texmaker

user246160

@AccidentalFourierTransform yes windows 7

bolbteppa

It is the simplest

and best

AccidentalFourierTransform

I use texworks ;-)

bolbteppa

Has shortcuts for common commands if you're new

Spent months using something awful compared to this :p

user246160

@bolbteppa how to get it ? link please!

AccidentalFourierTransform

18:36

@Doraemonドラえもん see here

bolbteppa

xm1math.net/texmaker

There is a youtube demo etc

user246160

Ok so I basically download that texmaker and then I can view Daniel's files ?

vzn

@DanielSank aka "parallelpiped"

bolbteppa

idk about the files, but figuring out latex will make your life far easier :p

user246160

@bolbteppa wait..wait...is that texmaker enough to read those files ? Or I have to download anything else?

user246160

18:39

I mean these files

user246160

bolbteppa

Oh

Yeah

It should be fine

user246160

oh thanks

Skyler

hey guys

user246160

i will try

Skyler

18:40

is there something defined like a periodic erf function

AccidentalFourierTransform

erf(sin(x))

ur welcome

bolbteppa

Error function

In mathematics, the error function (also called the Gauss error function) is a special function (non-elementary) of sigmoid shape that occurs in probability, statistics, and partial differential equations describing diffusion. It is defined as: erf ⁡ ( x ) = 1 π ...

imaginary error function on that page?

Jesus Christ

They Burmann expand the error function on this wikipedia as if the Burmann expansion is something you should know, the only place I've ever seen it is in Whittaker, my god

user246160

@bolbteppa Fun Fact: I downgraded from windows 10 to windows 7 just because of its simplicity :D

AccidentalFourierTransform

Fun fact: you should be using neither :D

bolbteppa

haha

user246160

18:43

@AccidentalFourierTransform haw ?

bolbteppa

First tex, then linux

AccidentalFourierTransform

its all about those "x"'s

bolbteppa

It's up to you to solve for x (i.e. figure out tex/linux) in each case apparently

user246160

You mean I should first use linux ? :P

user246160

I anyway just use my computer only for serving the net. When I take up computer science as one of my subjects in college next year maybe i will have to use those Operating Systems (like unix,linux, ubuntu) :)

user246160

18:49

BTW thanks a lot guys for the help

user246160

@bolbteppa

user246160

@AccidentalFourierTransform

bolbteppa

Cool

AccidentalFourierTransform

@Doraemonドラえもん cheers

user246160

Bye ! Gotta go!

DanielSank

18:57

@Doraemonドラえもん See the README.

You have to do two little things: convert the svg files to pdf, and then run latex.

bolbteppa

@G.Bergeron How about this: From $\mathrm{tr}(\rho) = \sum_n <n | \hat{\rho} | n>$ we get $\mathrm{tr}(\rho) = \int <n|\hat{\rho}|n> dn = \int \rho_n(E) dn = \int \rho_n(E) \frac{dn}{dE}dE $ but because energy is conserved, $dE = 0$ and so $\mathrm{tr}(\rho) = 0$. Thus the only way to make this work is for $\rho$ to be a delta function as before?

DanielSank

gtg

heather

21:23

hello

Sanya

hey heather

DanielSank

ohai

Sanya

@DanielSank the smbc comic is awesome :)

DanielSank

Yeah, it's good.

heather

21:41

@DanielSank, yes, it's the best =)

dmckee

@DanielSank I've just been doing issues like this with my modern physics class in the last couple of weeks. The timing is perfect.

DanielSank

22:08

@dmckee Heh.

Skyler

23:00

@DanielSank, sent you an email with resources that may be useful for @heather

DanielSank

@Skyler Why not just send them to heather?

Skyler

@DanielSank its probably better for you to structure the approach (also dont have her email)

DanielSank

ah

Skyler

this is all the material from a junior level lin alg of QM course

so if its ever of relevance you can pull problems from it, and if heather ever wants it all she can ask either of us now

also a good reference for your book

heather

hello =)

@Skyler, thanks so much! QM, linear algebra, =D

=D =D =D

@Skyler, I could give you my email, though I'd rather not in the hbar

Skyler

23:10

dw about it, for now dan should have a look at it though to see what parts can help for what you want to do

heather

okay

thanks again

Skyler

this class is in many ways for people who are already done with lin alg and calculus and ODE

your welcome

heather

oh geesh

i really do need to learn multivariable calc

and ODEs

DanielSank

23:35

@heather, let's just do basic ODE right now.

It's easy.

$$\dot{x}(t) = t$$

What's the solution?

Let's just forget all the formal mumbo-jumbo and learn how it works!

heather

what am I solving for?

DanielSank

@heather $x(t)$

heather

$x(t)=t$?

isn't it already solved then?

and is the dot on the $x$ just a notation thing, or does it mean something?

DanielSank

Oh, sorry...

$$\frac{dx}{dt} = t$$

$\dot{x} \equiv dx/dt$

Standard physics notation.

heather

oh, okay

so I'm solving for x(t) where dx/dt = t

okay

DanielSank

23:39

correct

Hint: To solve differential equations, guess.

Then, check if the guess was right.

heather

okay

could you just say x(t) = dx/dt? or am i not thinking straight?

DanielSank

@heather You're not thinking straight.

heather

okay

DanielSank

Let's use better notation. It will help.

Consider a function $f$.

heather

$f$, okay

DanielSank

23:42

Suppose I tell you that $(df/dt)(t) = t$.

Now I ask you to find $f$.

What does $f(t)$ have to be to make $(df/dt)(t) = t$ true?

heather

well, you can divide out the t, right, so df/dt = 1...?

so the derivative of f = 1, so you can take the integral of 1 to get x+c...?

am i still not thinking straight?

DanielSank

You're thinking fine, but you're confused by notation.

$(df/dt)$ is a function.

$(df/dt)(t)$ means the value of that function when you input $t$.

It's not multiplication.

heather

oh, okay

brb i have to eat dinner

DanielSank

k

Transcript for