The h Bar

01:15

@TobiasFünke That's not true; if the system microstate multiplicity increases sharply enough with energy, then there can be macrostate probabilities above ground state that is higher than ground state.

Allie

we were talking about non-degenerate

for distinguishable independent

naturallyInconsistent

If you have more than one particle, it will be difficult to keep the non-degeneracy

That's the point

Allie

well doesnt that depend on what the conditions are

how many energy states are available, what temp were at, etc

my textbook said in many cases it can be a reasonable assumption that there will be enough energy states available that at a high enough temperature the overlap is minimal

naturallyInconsistent

Yes, but that is not a disagreement: it said "in many cases"

Allie

meow

i get it

and i think that's where you get fermi-dirac and bose-einstein statistics, right?

FD for fermions, BE for bosons

naturallyInconsistent

01:25

It is important to understand the difference here because it will be the important difference between microstates and macrostates multiplicity and how canonical ensembles give rise to the seeming stability of macrostates

FD and BE stats are coming from symmetry of the wavefunctions as required by CCR/CAR. It is slightly different from this point.

Allie

i was literally about to say how convenient, FD for fermion and BE for boson. then i realized fermion comes from fermi and boson comes from bose

facepalm

naturallyInconsistent

lol

Legend goes that Bose was trying to lecture about the catastrophe when he made a mistake in the lecture and ended up giving the correct result.

And then he couldnt get it published and had to bring it to Einstein's attention before Einstein helped get it published.

Not sure how true this legend is

Allie

the catastrophe?

naturallyInconsistent

UV catastrophe

Allie

ah

naturallyInconsistent

01:31

At least this story is on Wiki

Allie

ive neevr heard it referred to as just the catastrophe but i guess im a FAKE physicist

(theoretical chemist)

i like stat mech so far though

its cool ive always wanted ot get into this im finally dipping my toes

im done being trapped at 0K

naturallyInconsistent

@ACuriousMind Im not sure how much of that we can agree upon. Nobody complains that positions and vector components depend upon the coördinates chosen, and that it takes some effort to focus strictly upon actually invariant physical quantities, yet somehow it is different when we come to QFT renormalisation.

@Allie mewth

Silly Goose

01:48

Is this a typo in (14)?

Should it be $e^{-i\vec{k} \cdot \vec{n}}$?

Silly Goose

02:21

separately, what is a physical reason for why space translation symmetry should imply that observables evaluated at opposite momenta are correlated?

naturallyInconsistent

03:43

You seriously have to work on context. I can see where it will be obviously true, but contextless it is also obviously false.

Allie

are you talking to me?

naturallyInconsistent

04:00

No

mew mew

Allie

meow

good night

naturallyInconsistent

05:01

mew mew good night

Ryder Rude

06:27

hi

ACuriousMind

08:33

@naturallyInconsistent Fair point, but coordinates are also treated confusingly and strangely when they get into focus ;) (see all the confusion about the nature of GR)

imbAF

@TobiasFünke In the topic of directional derivatives of functions and functionals, that we talked about. I have a question, which is mostly related to math, but I'd like to know the answer to it for personal reasons. I believe I can consider the total differential of a function f(x) as

the directional derivative in some arbitrary direction of an infinitesimal vector dx ? If the total derivative is taken to be zero, that would imply that the gradient of the function and the vector dx are perpendicular to each other. But at the same time the gradient points at the steepest change and the differential df(x)=0 gives the value, for which f(x) has an extrema. So how is it that they are perpendicular (gradient and infinitesimal vector dx) and not parallel?

Tobias Fünke

09:16

Sorry, I do not have time right now to think about this. Perhaps someone else can help

imbAF

I was able to figure it out

Ryder Rude

@imbAF hi. $\nabla f \cdot \vec{v}=0$ implies either perpedicularity or that either of $\nabla f$ and $\vec{v}$ are 0

the extremum stuff is when $\nabla f$ is 0

imbAF

I was able to figure it out. If you are in a level contour and the infinitesimal vector dx is tangent to it, then the total derivative is zero, because the gradient is perpendicular to dx. If you are in an extremum point in that case df(x)=0 because the gradient itself is zero

And one cannot talk about projection

because in the infinitesimal region close to the extream point, there is flatness, and as such the gradient is zero

@RyderRude yes

Then, RR one question. When we jump to physics

naturallyInconsistent

@ACuriousMind yes, but that only strengthens myow point. People freak out over the myriad manifestations in inconsistent manners, so what are we supposed to do?

imbAF

And the functional we consider the the action. When we say that we want the first variation of it to be zero and at the same time the first derivate as such, you'd get an extrema. The function that corresponds to the extrema value of the functional is found out by solving the EL equation @RyderRude right?

Ryder Rude

09:25

yes

imbAF

If it is so, then, why is it that the extrema of the functional for when $\delta S=0$, happens to correspond to the shortest path?

or is not the case?

Ryder Rude

it need not be the shortest path. the configuration space need not even be a metric space, so u have no notion of "shortest'

for the free particle, it happens to be the shortest path

imbAF

Ok, so then Hamilton's principle, how is it different than the math we talked about

In the sense, what is it saying

Ryder Rude

please elaborate

imbAF

I am trying

Ryder Rude

09:28

Hamiltons principle is saying that the actual path is the extremum of the action

it doesnt say anything else

imbAF

But nothing about the type of extrema, and no information about the path characteristics, whatever those may be i.e length or whatever else

?

Ryder Rude

no ... its main purpose is to give u the EL eqn (which u can call a path characteristic)

"length" is usually not well defined for paths in the configuration space

but in theories like free particle GR, the action turns out to be equal to the path length

this is just a co incidence

imbAF

Ok

Ryder Rude

note that u would also be using Hamilton's principle for "path" taken by fields, like $f(x,y,z,t)$

clearly there is no meaning of the length of this function

imbAF

I was under the false idea that the action says that the system chooses the shortest path, because in the example that we were given on the topic, it was that of a ball going from a to b, and the fact that the path taken was the shortest. So I related $\delta S=0$ with the shortest path. Wrongfully of course

@RyderRude You are right.

@RyderRude One more thing I want to ask, because of something you said, which was:
""length" is usually not well defined for paths in the configuration space"
But isn't the 3D Euclidian space such a space? Don't we talk about paths in this space?

Ryder Rude

09:36

@imbAF yes. it is just a co incidence

imbAF

@RyderRude Right

Ryder Rude

@imbAF the 3D space is the physical space. for $n$ particles, u need 3n dimensional configuration space. this is the space (+ time) on which the path of Hamilton's principle lies

e.g. for two particles, u would consider a path in 7D space (6 dof for the particles and 1 time)

imbAF

I see

Ryder Rude

for fields, the configuration space is even crazier. it is an infinite dimensional manifold

imbAF

I understand now fully

@RyderRude I am working towards manifolds

But if I may say, configuration space, seems like the phase space, it resembles it

minus the fact that you can no generalized momenta

but time

Ryder Rude

09:44

@imbAF the configuration space is usually defined without time

paths of the systems lie on $\text{configuration space}\times \text{R}$

@imbAF the phase space usually has twice the dimension of configuration space

this is because the configuration space only encodes positions, while phase space also encodes momenta

u can do a Legendre transform to go from Lagrangian to Hamiltonian mech. this will translate the physics from the configuration space to the phase space @imbAF

imbAF

10:00

@RyderRude Thanks for the overview

ACuriousMind

@naturallyInconsistent Despair :P I think you perhaps read my response to SillyGoose as agreeing/supporting this common view, but I was more just trying to explain what that view is. I'm not really agreeing with it - but as I said I'm not comfortable enough with my own understanding of renormalization to have a strong opinion either way

Slereah

People learn about coordinates in high school when they don't know anything and then they basically never learn it again until GR

Why do people almost never talk about cofibers except for the homotopy cofiber

What's so wrong with the regular cofiber

ACuriousMind

10:23

are you saying people need more cofibers in their diet?

Ryder Rude

descartes had the idea about co ordinate systems before anyone had even formalised real numbers

this is because one gets an almost rigorous idea of real numbers by thinking of them as x.yzwuvabcd......

i dont see anything non rigorous about this tbh

the set is defined as the set of these infinite sequences. and one can define the field operations like how one teaches it to children

i think the only "wrong" thing with this approach is that it takes the base-representation as fundamental

imbAF

10:43

@RyderRude In configuration space of a system, a points represent what/interpreted as what? I am asking this because, I read the following about action functional S :
"Action principles start with an energy function called a Lagrangian describing the physical system. The accumulated value of this energy function between two states of the system is called the action." https://en.wikipedia.org/wiki/Action_principles#:~:text=Action%20principles%20start%20with%20an%20energy%20function%20called%20a%20Lagrangian%20describing%20the%20physical%20system.%20The%20accumulated%20value%20of%20this%20en…

Ryder Rude

10:55

@imbAF what the point represents depends on the system. also, wiki shouldnt call it "energy" to avoid confusion

@imbAF the general idea is that, if $q$ is a point on the configuration space, then $q, \dot {q}$ should correspond to a state of the system

this is analogous to phase space where $(q,p)$ represents the state of the system

@imbAF for particle mechanics, what the point represents is an aggregate of the positions of all the particles, which is why the space is somethign like $R^{3N}$ when we have $N$ particles that each each move around in $R^3$

u might also learn tangent bundle and cotangent bundle. the cotangent bundle is how the phase space relates to the configuration space

as in , the phase space is the cotangent bundle of the configuration space

imbAF

Ok I see

While I have no idea about bundles etc just yet, by the time I reach there, with this in mind, I will be able to understand the connection

imbAF

11:27

@RyderRude For a particle in $R^{2}$ the set of configurations, that need to satisfy mathematical constrains, give a manifold, right? Would this manifold in this case, be a surface in $R^{3}$ ?

naturallyInconsistent

@ACuriousMind now, now, it is all your words against you

lol

qwerty

eeeee I think I have the perfect source for a question someone has. this doesn't happen that much, very exciting

Signor Feynman

11 hours left and I will finally change my username

naturallyInconsistent

so ready to ditch, signore?

Ryder Rude

11:46

@imbAF i dont understand the question...

i think u r saying that, if the configuration space is $R^2$, then the solution to the EL eqn is a surface in $R^3$

imbAF

It is said that: The set of configurations of a system in configuration space, form a manifold

I don't know what that is

but I would assume that geometrically, for 1 particle in $R^3$

the manifold would be a surface in $R^3$ ?

Ryder Rude

i think it is just saying that the configuration space is a manifold?

ooh it is talking about constraints

imbAF

Yeah, I am trying to understand, geometrically what a manifold is, the same way that if I tell you a vector in $R^3$, you imagine a arrow in some direction

@RyderRude Yes, because of constrains, you get a set of configurations, which give the manifold

So, I simply wanted to visualize a manifold, the same way I can, for a vector

And i considered the simple case of 1 particle in $R^3$

Ryder Rude

i think what ur book is saying is that, e.g. if the particle movies in $R^2$, but the particle is confined to a sub-space of $R^2$, then the set of configurations that the particle can take is a submanifold of R^2

imbAF

But I don't want to derail to much. If my intuition about how a manifold looks like is wrong. Maybe in the future I get the right idea

Ryder Rude

11:51

i cant tell without much context

ACuriousMind

@imbAF If you don't know what a manifold is but the text you're reading is using that word as if you should know what it means, you're simply reading the wrong text for your level.

Ryder Rude

what i can tell is that. e.g. say, the particle lives in 3D space. but say, the particle is confined to move in a circle. then the space of configurations becomes a cirlce which is a sub-manifold of $R^3$

qwerty

@imbAF have you looked up a definition?

Ryder Rude

@imbAF u need to learn manifolfds, yes

ACuriousMind

If you want to understand what a manifold is, you should read an actual text on the topic, i.e. either some GR physics texts or math texts on differential geometry. If you want to understand classical mechanics, you should read a classical mechanics textbook. Whatever you're doing currently, these questions reveal it is not working at all.

qwerty

11:53

I thought you were discussing manifolds earlier this week or was I mistaken?

imbAF

I am reading about configuration space. That is a concept present in mechanics. And as to how a set of configurations of the system is defined, it involves the use of manifolds. And I don't know what that is. I simply wanted to know whether my hunch about what it looks like (visually) was correct. But I don't want to invest much on it, since as @ACuriousMind said, that is beyond my level at the moment

But I am taking a course which involves manifolds, topologies etc. So once I reach these concepts, things might be clearer

qwerty

I don't think it is beyond your level...

Ryder Rude

visually, a manifold is just smooth shapes like circles, torus, planes, cylinders, etc. this is what u should be intuiting @imbAF

imbAF

@RyderRude ok

Ryder Rude

but not every manifold is visualisable ofc

qwerty

11:55

mathworld.wolfram.com/Manifold.html

imbAF

@qwerty In my bachelors, we never talked about manifolds, unfortunately

ACuriousMind

@RyderRude the whole point is that they're doing that and this has led them to the misguided idea that $\mathbb{R}^n$ is not a manifold - that's where the strange question about constraints and surfaces above comes from

imbAF

@ACuriousMind Yes, I am attending a differential geometry course

Ryder Rude

@ACuriousMind ooh

ACuriousMind

@imbAF I'm sorry, how are you attending a differential geometry course and don't know what a manifold is? The definition of a manifold should be the first thing you learn in such a course.

imbAF

11:58

@ACuriousMind No, we start by Axiomatic set theory

Signor Feynman

@ACuriousMind ACM suggesting to use a GR book for DG... Is everything okay, man? D:

Ryder Rude

@imbAF did the book mention constraints or did u think the word "mnanifold" had to imply constraints?

qwerty

"this has led them to the misguided idea that $\mathbb{R}^n$ is not a manifold - that's where the strange question about constraints and surfaces above comes from"

this went over my head :S

imbAF

@RyderRude it did

ACuriousMind

@imbAF what

@SignorFeynman no, not really, I'm getting desperate here :P

Ryder Rude

11:59

@imbAF oh. i think ur book is talking about submanifolds of $R^n$

imbAF

@ACuriousMind Logic -> set theory -> topology -> Topological manifolds -> differential manifolds -> Bundles -> Geometry (simple and metric)-> applications in CM,EM,QM,SP ,SR & GR this is the structure of course

ACuriousMind

@imbAF that's like a plan for a two-year lecture series, not a single course

imbAF

Hmmm... The lecturer does mention that we, in physics, won't dive in deeply in the mathematics side

i.e proving something

ACuriousMind

oh, it's a physics course

imbAF

yes

Signor Feynman

12:03

"A manifold is a bunch of coordinates, guys"

ACuriousMind

@qwerty They asked about whether the "configuration space" is a surface in $\mathbb{R}^3$ because they didn't realize that the normal $\mathbb{R}^3$ is a perfectly fine manifold on its own (and hence an admissible configuration space) because the visualization is that we always draw manifolds as these little embedded surfaces/shapes.

imbAF

So it will take a couple of lectures to reach to manifolds

@ACuriousMind Is that a bad thing?

Signor Feynman

I think ACM meant that it's possible, since it's a physics course. In Math, that would be basically almost an entire Bachelor program

Well, actually DG is considered graduate level :P

qwerty

@ACuriousMind ohhh :S I see. I would not have thought this was a pitfall, since you usually say it has n dimensions. Like the way a mathsy book will even say 2-sphere instead of sphere.

ACuriousMind

@imbAF Not necessarily, but when you say "a differential geometry course", that sounded like a math course. If you put a pure math subject next to the word "course", the default assumption is not that it's a "for physicists" version :P

imbAF

12:07

Ahhh, nonono

It's a differential geometry course done from a physics professor, for physicists

It culminates by showing how we use all we learn in CM, SM, etc

ACuriousMind

Yes, so it's not a math course.

I don't know where the communication problem lies this time :P

@qwerty "I would not have thought this was a pitfall" applies to a lot of things I read here :P

imbAF

We are all clear this one time xD

The most interesting thing so far, was to learn that the empty set, is the only set

Ryder Rude

12:33

im having trouble reading things. my head starts to spin

qwerty

Is it normal when you're writing an answer that the preview doesn't show an image you've uploaded, only a link?

Ryder Rude

i feel really bad. i cant study stuff i want to learn

hopefully it gets better

qwerty

@RyderRude are you okay otherwise?

Ryder Rude

no... i have some conditions

it is interfering everywhere

i cant be specific for anonymity

qwerty

I'm sorry to hear. Hope you feel better soon

Ryder Rude

12:37

thanks.. i will try to work on it

it is getting bad. i was able to learn some stuff in the past years

qwerty

this is not a serious solution but I have also thought about what I would do if I couldn't read

Nov 12, 2024 at 19:03, by qwerty

earlier this year I tried to work out what resources there are for learning high level maths with limited eyesight

myria.math.aegean.gr/~atsol/newpage-en/software/braille

Ryder Rude

thanks. it looks like a comprehensive program

@qwerty sorry about ur eyesight problem

Silly Goose

Will figuring out what constants go where in fourier transform haunt me forever?

qwerty

I don't know how much it or braille actually used. I think that just making text bigger on screens is preferred these days

@RyderRude nah i dont have serious problems, just an overactive imagination...

Ryder Rude

@qwerty oh..

i read words but miss most of the meaning cuz i am feeling miserable

qwerty

12:44

:(

Ryder Rude

i will try meditation etc

thanks for support. cya

qwerty

take care

ACuriousMind

@SillyGoose yes

@qwerty not normal normal, but the preview can be weird sometimes

qwerty

ah cool! yeah, it shows the image now and I didn't change anything. weird

User198

13:36

@RyderRude Doing too much philosophy without physical exercise is not a good thing. Mens sana in corpore sano.

DIRAC1930

Stat mech is underrated in its profoundness

User198

@DIRAC1930 Boltzmann would agree

Signor Feynman

13:49

@qwerty Mhhh? What happens when you read math?

User198

When talking about geometric mechanics (1) , symplectic geometry (2), contact geometry (3), Poisson geometry (4) , usage of Lie groups in mech,...

What would be the "roof" term that accompanies all of that? Is it geometric mechanics?

Slereah

I believe the usual umbrella term would be "analytic mechanic"

Silly Goose

@ACuriousMind literally keeping me up at night 😢

@User198 it sounds like geometry

User198

14:12

@Slereah Hm. Thanks. That occured to me, but I wanted to be more specific, otherwise I could just say its "mechanics".

@SillyGoose But that is from maths perspective; differential geometry. I was thinking more about mechanics but described via geometry.

naturallyInconsistent

is there even a roof term? unless, of course "mathematics"

Silly Goose

what is the physical interpretation of (16)?

This is a translationally invariant lattice system and (16) is a two-point correlation function of a field defined on the lattice

i feel there should be some relation to conservation of (quasi-)momentum

but (16) seems weird, to me it says that the field correlations are only nonzero for exactly opposite (quasi-)momentum

ACuriousMind

@User198 It's literally just "mechanics" but done by mathematicians/mathematical physicists :P

If you want to do rigorous formulations of classical mechanics, that's the kind of math you end up with; there no particular "choice" involved to use geometry instead of something else, differential geometry is just the right tool for the job, since the physicists are writing down differential equations on spaces mapped by real coordinates

Slereah

14:28

I believe the nlab occasionally calls it "prequantum geometry" when it's feeling fancy

ACuriousMind

I think the nlab is continually feeling fancy ;)

Slereah

they are fancy lads

ACuriousMind

@SillyGoose How can a lattice system be "translationally invariant"? You're expecting people to infer a lot of context here

Silly Goose

14:42

@ACuriousMind i mean by lattice translations

but i think the form of (16) should also be the case in an actually translation invariant system

ACuriousMind

that's a discrete symmetry, why would it be associated with conservation of anything?

Silly Goose

Well $e^{ik}$ is still conserved in the sense that it commutes with the hamiltonian

ACuriousMind

I mean, in terms of particles all that $\delta$-function is doing is indeed momentum conservation: the ingoing and the outgoing particle (two-point function is "1-to-1 scattering") have the same momentum

Silly Goose

@ACuriousMind hm how is this so, i am seeing that it says they have opposite (in direction) quasi-momentum

ACuriousMind

I'm not sweating a minus sign :P

probably if you want to interpret this in terms of particles there's some convention that gives you a - on the outgoing particle

I just know that in hep-th QFT (which certainly is Poincaré and hence in particular translation invariant!) you always have a $\delta$-function in the momentum-space functions that enforces total momentum conversation between the inputs and outputs

Silly Goose

14:51

Right that i have seen before

ACuriousMind

here, for instance, the difference could come from your position-space version not being time-ordered (because you're not doing particle scattering), so this momentum-space function is not a scattering amplitude

Silly Goose

Hmm i see

Signor Feynman

@SillyGoose at first sight I would say that it's the same thing that happens for the propagator, which in momentum space is diagonal

User198

We can integrate the action of a system over arbitrary small time intervals, and hence get arbitrary small values. But does it physically have any meaning when the action of a system is smaller than $h$?

Slereah

It's usually used as a vague scale for quantum behaviour?

ACuriousMind

14:58

@User198 Why would it matter? What "physical meaning" does the action have anyway?

Slereah

Classical behaviour is usually $S \gg h$

ACuriousMind

also: The Planck units are not actual lower bounds or discretization steps for the values of physical quantities in mainstream physics.

User198

Alright, thanks.

Slereah

h itself isn't really a value with much meaning in itself as far as the action goes

PM 2Ring

15:22

151

A: Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed?

Short answer: nobody knows, but the Planck length is more numerology than physics at this point Long answer: Suppose you are a theoretical physicist. Your work doesn't involve units, just math--you never use the fact that $c = 3 \times 10^8 m/s$, but you probably have $c$ pop up in a few differe...

BTW, Chat is now in the Cloud. meta.stackexchange.com/q/406783/334566

PM 2Ring

15:49

Almost... :)

> Update: Things are taking a little longer than expected, and we're still working on chat.meta.stackexchange.com. Once this one is finished, the other chat servers should still only take up to 15 minutes.

PM 2Ring

16:27

in Tavern on the Meta on Meta Stack Exchange Chat, 8 mins ago, by Spevacus

we're on the cloud now woo

in Tavern on the Meta on Meta Stack Exchange Chat, 5 mins ago, by balpha

@ThomasWard most noticeably, it's much faster now because the database now lives alongside the application

balpha is the dev in charge of Chat. And he originally wrote it.

Ryder Rude

17:23

@User198 ive been doing stretching

Tobias Fünke

17:55

good evening

User198

19:05

Good evening

skullpatrol

hi

User198

19:36

I saw on one page today (not about physics, but a page connected with education in general; some university information posts and some administration): "Schrodinger equation can be regarded as a statement of conservation of energy in quantum mechanics." Idk why they are writing about the Schrodinger equation at all. xD

But how wrong is that?

I think it is wrong, because (I guess) the person writing that thinks that $i\hbar\frac{\partial \psi}{\partial t}\equiv \hat E= \hat H \psi$ so $\hat E= \hat H \psi$ and that is wrong. Should I notify them that that post is wrong?

Tobias Fünke

20:22

@User198 I don't understand what you mean (your equations do not make any sense to me). However, "Schrödinger equation" can mean the time-dependent Schrödinger equation, which for hermitian and time-independent Hamiltonian implies unitary time evolution and that the energy expectation value is constant: $\langle \psi(t), H\psi(t)\rangle= \langle \psi_0, U^*HU\psi_0\rangle=\langle \psi_0,H\psi_0\rangle$.

or any other similar statement regarding the Heisenberg EOM and so on.

That being said, I wouldn't state a a sentence like this (the statement you post).

DIRAC1930

21:08

@SillyGoose I wish I was doing cond-mat theory like you

Well quantum optics anyway

User198

@TobiasFünke Ok thanks.

@TobiasFünke Maybe I overcomplicated, disregard the equations. All in all, saying that "Schrodinger equation is a statement of conservation of energy" is not a correct statement (even meaningless), right?

qwerty

@SignorFeynman oh no, nothing really

morning all

Tobias Fünke

21:39

@User198 Well, I would say it needs further specifications ;)

hey qwerty, how are you doing?

I am currently preparing a talk for tomorrow... I started today :3 heeelp

qwerty

ooooh good luck!!

I'm well, thank you! excited to work on a phys.SE answer later today

Tobias Fünke

niiice :) feel free to post

although I guess I won't understand much

thanks. but I don't need luck: I need time haha

qwerty

aw it's tomorrow, you can do it!! I have known enough academics to know that the art of writing a talk on the plane three hours before a conference is a common one...

Tobias Fünke

22:08

:d yeah I guess so

OK, I gotta go. See you around, and have fun with answering!

Silly Goose

22:39

@DIRAC1930 trying to do CMT XD feeling like i am failing

Signor Feynman

23:06

@TobiasFünke good time, then

2

Break a leg, anyways!

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