The h Bar

Arjun

02:59

@imbAF Hi,they represent generalized coordinates,now a vector might have three independent components in R^3,and if that vector describes the position coordinates of a system each independent component is a q,i.e the vector would be described by three q's

@imbAF by $\veq$ I think you mean the configuration of the system in phase space,in that case qi's are just components,they are not vectors themselves

User1865345

03:52

@Allie are you sure there is one?

Allie

im not....

but i want to read this article :(

User1865345

Exactly. You have to ascertain whether that has been reprinted and translated in another journal

@Allie i can feel you. I have also faced lots of problems because in mathematical statistics too, there come some books or papers that I want to read badly only to see that there is no English translation.

Allie

meow

User1865345

My go to solution is I use a German-Maths dictionary and become familiar with the concepts in hand and try to decipher those in the work.

Allie

yeah but i cant read.... any german

User1865345

03:56

19

A: Book/tutorial recommendations: acquiring math-oriented reading proficiency in German

Landau's Grundlagen der Analysis in the AMS edition comes with a "complete German-English vocabulary", this sounds ideal for your purposes.

@Allie trust me. You don't need to to read articles.

Arjun

@ACuriousMind How did u primarily learn your coursework in physics like classical mechanics,enm,qm etc? You used to study from books or was it just lectures ? I feel so dumb for working like a donkey and still not managing to finish the coursework to a level I would like by the end of the sem lmao,plus do you think it is effective to learn from lectures alone?(Assuming the prof is decent)

User1865345

Of course, it won't be easy. It would take time. But read with a German-English dictionary; it won't be easy but not intimidating.

Allie

meow too much work

User1865345

😅

Also do you have acquaintance with anyone speaking German?

I heard people doing physics at least have a basic knowledge of German and French.

qwerty

@User1865345 this is not true

User1865345

04:02

🙂

In any case, Landau's Grundlagen der Analysis is very helpful. I have picked up few German keywords already.

in Ten fold, Jul 30, 2023 at 8:57, by User1865345 - solidarity Mods

You seemed to hit a dead end when the book you are reading refers to Witting's Mathematische Statistik and you don't know German. It's kind of surprising nobody (even the author himself) thought about translating the book for wider public (English speaking) considering it is referred frequently.

in Ten fold, Jul 30, 2023 at 8:58, by User1865345 - solidarity Mods

Is it that the Soviet books are translated at a greater rate than the German ones? I have hardly come across any German translated mathematical statistics book. I have seen French being translated, though.

I already ranted a year ago.

qwerty

06:04

merry christmas to those that celebrate it :)

6

naturallyInconsistent

merry xmas~

Arjun

06:23

Meryy merry Xmas : )

ZeroTheHero

06:55

Wasn’t there some sort of “Holiday Hats contest” (for lack of a better word) in previous years?

John Rennie

@ZeroTheHero They canned that a few years ago on the grounds it was getting a bit stale.

Which, to be fair, it was.

think_meaning_buildß

07:16

Stale, I am not :-)

Herr Feinmann

08:37

Merry Christmiaos

Meow

GroveRover

Merry Xmas

I'm approaching QIT and I'm confused by LCTP maps. The fact that not all LCPTs have inverse doesn't mean they violate reversibility? The joint evolution operator is unitary, so is its inverse: then it also must give rise to a LCPT.

naturallyInconsistent

M E O W ~

mo-_-

10:14

Merry Christmas 🥳

imbAF

@Arjun if the system is that of N particles in a box or whatever, aren't you having 6N dimensions?

naturallyInconsistent

11:04

@imbAF come on, you know 6N is half for positions and half for momenta.

Arjun

@imbAF If there are no other constraints sure phase space is 6N dimensional

I don't see what's unclear here

ACuriousMind

12:43

@Arjun I have no advice to offer you, I just sat in the lectures and did the exercises and that was mostly enough for me :P

@ZeroTheHero It was called Winterbash and they stopped doing it last year

Arjun

13:33

@ACuriousMind You must've had some good lecturers : )

Prateek Mourya

Wish you all a merry christmas😊

qwerty

@Arjun he was clearly just a good student ;)

Arjun

@qwerty most definitely :p

Herr Feinmann

14:31

@qwerty understatement of the decade :P

Madder

15:43

Hey can someone check the following? i am unsure about the entropy being replaced like this, the authors seem to just give the result. I tried to rationalize it..

\[
Z(N, \beta) = \sum_s e^{-\beta E(s)}.
\]
It ensures proper normalization of the probabilities such that:
\[
\sum_s P[s] = 1.
\]

For its definition, the partition function $Z$ contains all relevant physical quantities of the statistical system at equilibrium.

Not all configurations $s$ have unique energies $E(s)$. Many configurations may share the same energy level $E$. Let $\omega(E)$ represent the number of configurations (degeneracy) with energy $E$. The microstate-level entropy is given by:

In the sum before the final exponential, if i were to write out hte first terms, they would be something of this form right? TS(E_1) - E_1 + ....
because the text just omits the S(E) and writes S.

ZeroTheHero

@ACuriousMind thanks for the link. Not that I was anxiously waiting for it; in fact I had not realized there was none last year.

ACuriousMind

@Madder I'm not sure what your question is - the text you typed out has $S(E)$ everywhere I can see.

Madder

@ACuriousMind No thats what i wrote. the Text does not do that, it skips most of these steps and puts only an $S$

ACuriousMind

16:12

well, there is a way to do this only with the total $S$ and no $S(E)$: You're looking at a probability density $\rho(s) = Z^{-1} \mathrm{e}^{-\beta E(s)}$. The entropy is
$$S = -\sum_s \rho(s) \ln(\rho) = \sum_s \rho(s)(\ln(Z) +\beta E(s)) = \beta \langle E\rangle + \ln(Z)\sum_s \rho(s)$$
which yields $S = \beta \langle E\rangle + \ln(Z)$ and so $F = - \beta^{-1}\ln(Z) = \langle E \rangle- TS$

Leaky Nun

16:28

how fast are other galaxies moving (relative to earth)?

ACuriousMind

@LeakyNun see Hubble's law

Leaky Nun

hmm they're slower than i expected lol

so about 1000 km/s is the fastest galaxy (after adjusting for the expansion of the universe)?

that's like 0.1% c right

i mean i guess that's still really fast

is there anything stopping a galaxy from moving say at 0.5 c?

Madder

@ACuriousMind i will look at your comment later. Thanks for you reply for now.

Leaky Nun

does that mean galaxies at the edge of the observable universe are moving at 3.34 c?

ACuriousMind

16:50

@LeakyNun I don't know about 3.34 but sure, they can appear to move faster than light. The region of the universe in which the galaxies appear to move slower than light is the Hubble sphere

Leaky Nun

i think 3.34 is about right

@ACuriousMind would i be correct in saying that we can't actually observe the galaxies that are moving faster than light (outside the Hubble sphere) so we can't actually know if they move faster than light?

like would i be correct to say that the Hubble's law is only valid up to the Hubble horizon?

Slereah

There is a horizon behind which we will not be able to observe anything yes

Leaky Nun

yeah but i'm saying that "things outside move faster than light" is an interpolation and not an experimentally verifiable result

ACuriousMind

17:07

@LeakyNun Well...careful: We can see the light from the objects beyond the Hubble sphere, because in the past when the light was emitted they still moved slower than light relative to us. These objects don't suddenly disappear for us, their light just becomes increasingly redshifted until we can't see it anymore.

Leaky Nun

@ACuriousMind yes, but we're seeing light emitted from their past, not their present moment

the Hubble constant corrected for the time delay right

ACuriousMind

I'm not sure what time delay you mean or what you mean by "corrected"

Leaky Nun

let's say the Hubble constant is 70 km/s/Mpc, does that mean a galaxy moving at 70km/s is a Mpc away from us now, or in the telescope?

i think it's the "now"

ACuriousMind

As the Wikipedia article explains, the notion of distance in Hubble's law is the proper distance.

which is probably what you mean by the distance "now"

Leaky Nun

yeah

i also meant the light speed delay

if you look into the telescope you aren't seeing into the now, you're seeing into the past

this is independent of the hubble expansion

John Duffield

17:15

@LeakyNun : see arxiv.org/abs/astro-ph/0310808 and note this: "We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light. We explain why this does not violate special relativity and we link these concepts to observational tests". Also check out the ant on the rope.

ACuriousMind

@LeakyNun I think you made up a false dichotomy here: Of course a galaxy with a redshift of 70 km/s in the telescope was 1 MPc away from us when that light was emitted, but of course it's further away "now" and "currently" emitting light we will receive with a bigger redshift in the future.

Leaky Nun

does this cause any discrepancies in the constants?

ACuriousMind

I'm not sure what you mean

Leaky Nun

if i ask about a galaxy 1 MPc away from us "now", and i ask about its velocity "now", will it be significantly different from 70 km/s?

ACuriousMind

In the simple model where the Hubble constant is really constant, no, of course not

in reality the expansion of the universe seems to change over time so the Hubble "constant" isn't really a constant on larger time scales

Leaky Nun

17:23

but if we think about the "ant on the rope", we would get much more distortions that way right?

as in, maybe it's fine for 1 MPc, but not for 5 GPc, right

ACuriousMind

I'm not sure what you mean by "distortions"

nor what you mean by "it's fine" - people didn't just accept Hubble's law because it fit for one particular distance but because they could fit it to a range of distances

Leaky Nun

i'm still referring to the fact that what you're observing in the telescope is in the past, and not now

and as the distance increases, the time delay also increases

ACuriousMind

sure - if the "constant" changes too drastically over time, the expression in there would become some kind of integral over time instead of a simple $H_0$

Leaky Nun

what i meant to ask is, are we talking about the galaxies in the past, or are we talking about the galaxies as they exist in this moment

i think it's the former that we implicitly assume

i think we've implicitly accepted that we're talking about a past lightcone

lemme draw a diagram

ACuriousMind

@LeakyNun when we're just talking about Hubble's law, neither, really

Hubble's law just says that the relative velocity between two objects depends on their proper distance from each other. "Proper distance" is essentially the distance between two objects in the "cosmological frame" at some point in cosmological time.

of course, if you look through a telescope, the redshift you observe indicates the velocity of the object back when the light was emitted, so you need to use the proper distance back at that cosmological time in Hubble's law

Leaky Nun

17:31

yeah that's what i'm asking about

i mean, P.3 of the linked paper also has three diagrams of this sort

with way more details

ACuriousMind

Davis and Lineweaver are explicitly doing a general relativistic analysis of the situation to combat some misunderstandings that have arisen from (over)simplification of the situation in the literature

I would not have linked it to someone first encountering the concept :P

Leaky Nun

well yeah it took quite some time to get you to understand what i meant by "time delay"

> Despite the fact that special relativity incorrectly describes cosmological redshifts it has been used for decades to convert cosmological redshifts into velocity because the special relativistic Doppler shift formula (Eq. 2), shares the same low redshift approximation, v = cz, as Hubble’s Law (Fig. 2).

@ACuriousMind i don't see why this is an "of course", the two different measurements will yield very different results, right

ACuriousMind

@LeakyNun which "two measurements"? One is looking through the telescope, what is the other?

Leaky Nun

@ACuriousMind right, sorry, the other cannot be "measured"

but the other, uh, "value", is the distance and then corrected for light speed and Hubble expansion

i.e. where the green lines cross the black line

ACuriousMind

I'm not sure what you mean by "corrected"

Leaky Nun

17:42

it's where the galaxies are 'now'

that's the correction

ACuriousMind

it's a not a correction, it would just be their proper distance to us now

Leaky Nun

yeah but the two values are different

Silly Goose

how does witten have so many citations? the areas that he works in seemingly are not related (not at the least accessible) to the majority of the physics or math community

ACuriousMind

that's just a completely different value than their proper distance back in the day

one is not a correction of the other, any more than your position today is a "correction" of your position yesterday

Relativisticcucumber

@think_meaning_buildß lol

Leaky Nun

17:43

i don't know why you're fixating on the word "correction", i just meant after adjusting for the light speed

use "adjustment" if you don't like "correction"

Relativisticcucumber

@LeakyNun that is the way of acm

choose ur words wisely or perish

ACuriousMind

I also don't know what you mean by "adjusting" :P

Leaky Nun

i feel like you're ignoring my bigger question and fixating on small details

ACuriousMind

what is your bigger question?

Leaky Nun

the question is is the hubble's law for galaxies now, or for galaxies in the past

ACuriousMind

17:45

I already answered that, it's neither:

14 mins ago, by ACuriousMind

Hubble's law just says that the relative velocity between two objects depends on their proper distance from each other. "Proper distance" is essentially the distance between two objects in the "cosmological frame" at some point in cosmological time.

it's just a relation between relative velocities and proper distances at any time you want (possible with a different value for the Hubble "constant" if the expansion rate changes in your model)

if you want to apply it to what you see in a telescope you have to use it for the distances in the past

Leaky Nun

but we only ever apply it for the galaxies on the red line right

the red line is what we see in a telescope

ACuriousMind

well we don't have a lot of other ways to observe things out there except with a telescope :P

Leaky Nun

which is what i meant by correcting and adjustment

namely we take the observable values and then derive some other values that are not directly observable

such as the galaxies on the horizontal black line

ACuriousMind

@LeakyNun who's deriving any values here? You just plug in the redshift you see in the telescope into Hubble's law and it gives you the distance at the time of emission

2

that's all the law does

no one is computing the "current distance" of these galaxies, at least not in the pure context of Hubble's law

Leaky Nun

i'm just saying that when i first thought about it i thought it would make more sense to have a law about the black line

but now we're clarifying that we're only ever talking about the red line

yeah?

because like when you do newton mechanics you treat space and time separately

so it made sense to have a law for just space

@ACuriousMind yeah that's what i was asking about

ACuriousMind

17:51

I mean there are certainly people who compute a comoving distance (i.e. the proper distance today) for those objects, just that's not what Hubble's law is about

Leaky Nun

whether we're talking about the observed distance or the current distance

again it's because scipop likes to emphasise the difference between the two

they'll say, the observable universe is ... in diameter, but if we're talking about the objects as they are now, they'll be much farther away because the universe has been expanding etc etc

i can't tell you how many times i've heard this line

and also

when u draw a diagram of the galaxy

it does seem like a black line instead of a red line, right?

nobody is also drawing the "time into the past" along the diagram

ACuriousMind

Typically the size of the observable universe is stated in comoving distance, i.e. the distance "now"

@LeakyNun I mean people are reasonably confident that they understand how the galaxy works that they indeed draw the galaxy "as it looks now"

Leaky Nun

i still feel like there are two independent things that are being conflated

ACuriousMind

people are perhaps sometimes not careful about this in science communication

but I assure you everyone who actually works with this stuff is well aware of the distinction :P

Silly Goose

I am confused about why a free QFT is thought of as "many harmonic oscillators". For instance, one contradiction seems to be: The harmonic oscillator in quantum mechanics is not translationally invariant, but the hamiltonian of a free quantum field is translationally invariant.

Leaky Nun

18:02

let's say Adam is moving away from earth at a constant velocity of 0.1 c, and you observe him to be 1 light year away from earth in the telescope. this means that one year ago (!) he was 1 light year away from earth, and now he is 1.1 light year away from earth.

after 1 year, he is now 1.2 light year away from earth, but in your telescope you observe that he is 1.2/1.1 = 1.09 light year away from earth. that means you will calculate his velocity to be (1.09-1) light year / 1 year = 0.09 c

@ACuriousMind this is what i mean by "correction"

ACuriousMind

@SillyGoose It's an analogy, not a formal equivalence. The free QFT has infinitely many c/a operator $a_p,a^\dagger_p$, just like the QHO has a single c/a pair $a,a^\dagger$. Hence a QFT is "infinitely many QHOs".

Leaky Nun

the measured velocity in your telescope is 0.09 c, yet his real velocity is 0.1 c

Silly Goose

but it seems utterly wrong even as an analogy. the properties of the hamiltonian of each theory are utterly different

ACuriousMind

@SillyGoose personally I don't find the analogy particularly helpful, either

but that's what people mean when they say that :P

Silly Goose

sad another misleading statement popularized and promulgated

Allie

18:05

lol

so im reading about spin adapted configurations right now

first of all, do eigenstates of the hamiltonian necessarily need to be eigenstates of the total spin angular momentum operator? because they commute?

Sir Cumference

Happy Chanukah and Merry Christmas

ACuriousMind

@SillyGoose But I would dispute this: There is a straightforward sense in which the free field Hamiltonian is in fact the Hamiltonian of infinitely many oscillators. The QHO Hamiltonian for $a(p)$ is $H_p = \omega(p)a^\dagger_p a + 1/2$ where $\omega(p)$ is the energy of something with momentum $p$ and mass $m$. Then the full Hamiltonian in momentum space is just the sum/integral over all the Hamiltonians for all $p$, i.e. $H = \int H_p \mathrm{d}^n p$.

Leaky Nun

it feels like something has to be wrong with my calculations, because it means that something moving at close to c will appear as if moving at 0.5 c??

ACuriousMind

@Allie No, that two operators commute only means that there is a common eigenbasis, not that every eigenstate has to be an eigenstate of both. As soon as one of the operators is degenerate, there may be eigenstates that are not eigenstates of both

@LeakyNun I don't understand this computation at all. Why would I only observe him 1.09 ly away? If he moves at 0.1c constant, I will see him at 1.1 ly after 1 year.

Allie

huh.

im trying to imagine how to conceptualize this

Leaky Nun

18:12

@ACuriousMind at the initial time, we observe him to be 1 light year away, yet in reality he is 1.1 light year away; this means that "observed distance" : "actual distance" is 1 : 1.1

i've made a diagram to demonstrate this

red line is the universal speed of light, green line is a spaceship travelling near that speed

blue lines are the light signals we receive

orange is the "apparent speed"

after 1 year, the actual distance is 1.2, so the observed distance must be 1.2/1.1 = 1.09

Silly Goose

@ACuriousMind modulo renormalization issues~?

Allie

@ACuriousMind Ok, so I think I can see the reasoning behind this. If operators A and B commute and $A\psi_1 = \epsilon_1 \psi_1$, we have $BA\psi_1 = \epsilon_1 (B \psi_1) = A(B \psi_1)$. Therefore $B\psi_1$ is an eigenfunction of A with eigenvalue $\epsilon_1$. If that eigenvalue is nondegenerate then it follows that $\psi_1$ is also an eigenfunction of B

and I can see how if there is nondegeneracy you could have $B\psi_1$ be the orthogonal eigenfunction of A with eigenvalue $\epsilon_1$ which isn't a constant * $\psi_1$

Leaky Nun

18:56

i feel like people don't emphasise this enough

an object moving at 0.9 c away will be observed as having velocity 0.47 c

nobody ever said this in any scipop!

(and this is not the redshift velocity!)

Silly Goose

19:15

consider a simple model of a lattice of ion cores (ions + bound electrons). can we suppose that acoustic phonons are modeled by a massless spin-1 "quantum field" while optical phonons are modeled by a massive spin-? "quantum field"?

ACuriousMind

@SillyGoose sure, but the "renormalization issue" is just that you have to throw away the 1/2-piece/renormalize it to be vacuum energy

@LeakyNun I think it's not emphazised because it's not really relevant in practice. All the objects that are fast enough that this would matter are so far away that we can't tell they're moving away by any other means than redshift, anyway - all the methods of distance measurement will not be able to tell that some galaxy that's 1 million lys away moved 0.1 lys away from us over 10 years or whatever,

so your "observed distance" doesn't actually exist in practice - only for objects that are near enough and slow enough that this doesn't matter

Leaky Nun

that is a good point, i was just about to say that it doesn't matter because the redshfit velocity doesn't experience this sort of distortion

wait

how do you confirm Hubble's law if you can't measure distance?

oh

Relativisticcucumber

what is meant by the "optical properties" of a material?

Leaky Nun

you use luminosity to measure distance?

ACuriousMind

@LeakyNun I didn't say we can't measure distance

I just said the methods of distance measurement are not precise enough to detect the 0.01% change in distance that you'd be talking about here

Leaky Nun

19:21

it's not 0.01%

ACuriousMind

I just put a random percentage :P

Leaky Nun

i just showed that if an object moves at near +c speed, its observed velocity will be +0.5c

ACuriousMind

@LeakyNun yes, but the objects that move at ~c are like 100 million lys away

so after 1 year, that 100 million becomes 100 million + 0.5

Leaky Nun

ah, ok i understand

ACuriousMind

we can't measure distance that precisely

Leaky Nun

19:22

so how do you measure distance?

Silly Goose

or actually, can I just think of phonons as some quantum field governed by $H = \sum_\sigma \int dp \ \omega_\sigma(p) (a^\dagger_{\sigma}(p) + a_{\sigma}(p)) + h_0$ where the dispersion relation $\omega_\sigma(p)$ is some generically complicated "phonon" dispersion relation?

ACuriousMind

as you said, one way is by the luminosity of standard candles

I'm sure there are others but I'm not a practical astronomer :P

@Relativisticcucumber like refraction index and stuff?

what is this vague question :P

Tobias Fünke

@SillyGoose you can also think of the bosonic Fock space of $d$ modes as the space of $d$ harmonic oscillators $F(\mathbb C^d)\cong \otimes^d L^2(\mathbb R)$.

IIRC, M. Talagrand says one or two things about that in his QFT book.

@Relativisticcucumber what do you think :p?

Leaky Nun

@ACuriousMind and in this case 100 Mly is the "proper distance" and 200 Mly is the "comoving distance"?

ACuriousMind

@LeakyNun To "proper distance" you always need to add the time at which you're talking about it

because the proper distance at present is the comoving distance

Silly Goose

19:26

and where $\omega_\sigma(p) \sim p$ for small $p$ (acoustic) and $\omega_\sigma(p) \sim \sqrt{m^2 - p^2}$ for large $p$ (optical)?

ACuriousMind

but by the 100 million lys I meant the proper distance at the time the light that's now reaching us was emitted

Relativisticcucumber

i have seen this phrase recurring in two places: 1) discussions of how we should classify lattices and 2) literature on high energy density physics. i thought it could refer to permittivity and thus the ability of light to refract, reflect, or transmit through the material but i was unsure @TobiasFünke @ACuriousMind

ACuriousMind

yeah, I would say all of that falls under "optical properties", sure

it's a purposefully vague phrase, not a technical term :P

Tobias Fünke

indeed

Silly Goose

oh wait i wrote the hamiltonian wrong

oopsies

$H = \sum_\sigma \int d\vec{k} \ \omega_\sigma(\vec{k}) [a^\dagger_\sigma(\vec{k})a_\sigma(\vec{k}) + h_0]$

Relativisticcucumber

19:30

@ACuriousMind oh :P darn

Silly Goose

but my question still stands: does $H$ yield the textbook model of phonons? where the acoustic phonons are given by $\omega_\sigma(\vec{k})^2 \sim k^2$ for small $\vec{k}$ (acoustic) and $\omega_\sigma(\vec{k})^2 \sim m^2 - k^2$ for large $\vec{k}$ (optical)?

Leaky Nun

@ACuriousMind how do you express that assuming it always travelled at speed c, its distance now would be 200 Mly?

ACuriousMind

@LeakyNun It wouldn't be, because the universe is expanding - you need to compute the full thing with the time-dependent Hubble constant to figure out how far away it's "now"

Leaky Nun

right, but how would you phrase that distance?

ACuriousMind

that's the comoving distance/proper distance now

Leaky Nun

19:38

then what is the 100 Mly?

ACuriousMind

the proper distance at the time of the emission of the light we receive

Leaky Nun

i see

an unrelated tangent, i've heard it said that the universe isn't actually expanding uniformly everywhere

it's more like only expanding near empty space or something like that

so the atoms in my hands aren't actually getting further apart

ACuriousMind

those are two very different claims :P

Leaky Nun

well empty space as in in deep space, so nowhere near earth

so they're the same claim

oh, you meant the uniformly claim

well are the two claims true then?

Silly Goose

another typo: the minus sign should be a plus sign in the "optical" dispersion relation

ACuriousMind

19:40

the atoms in your hands aren't getting pulled apart because the EM force between your constituent particles is so much stronger than what uniform space expansion would do

Leaky Nun

are we assuming here that the space expansion is caused by a "force" then?

ACuriousMind

and sure, just like the Hubble "constant" isn't uniform in time it's also not exactly uniform in space, either, because the universe at smaller scales isn't the uniform soup the FLRW cosmology treats it as

Leaky Nun

so what actually happens, and how do the "small scales" sum up to get the whole (i'm using reductionism again)

ACuriousMind

@LeakyNun no, not at all - but the distance between your constituent particles is determined by the equilibrum of the repulsive and attractive forces between them. If space expansion pulls them apart a little bit, they just return to that equilibrium

Leaky Nun

i see

ACuriousMind

19:45

@LeakyNun I'm not sure what exactly the question is - in reality you have some $H(t,x)$ but the FLRW assumption is that if you average coarsely enough (on the order of galaxies) you just get a $H(t)$, i.e. the universe is homogeneous and isotropic on large scales

Leaky Nun

i guess the question is, what does the metric look like on earth?

ACuriousMind

I mean, the Earth as a single planet just has a Schwarzschild metric around it

to get "the" metric for all the objects in the solar system at once is probably already an intractable problem computationally

and none of this as anything to do with the large-scale structures like expansion in the FLRW model

Slereah

The gravitation of Earth is a very nuanced thing

Gravity anomaly

The gravity anomaly at a location on the Earth's surface is the difference between the observed value of gravity and the value predicted by a theoretical model. If the Earth were an ideal oblate spheroid of uniform density, then the gravity measured at every point on its surface would be given precisely by a simple algebraic expression. However, the Earth has a rugged surface and non-uniform composition, which distorts its gravitational field. The theoretical value of gravity can be corrected for altitude and the effects of nearby terrain, but it usually still differs slightly from the measured...

many factors

I have a paper somewhere about the GR model for it, lemme see

ACuriousMind

(and also that ^ if you don't approximate the Earth as a spherical homogeneous mass, yes :P)

Relativisticcucumber

Also, returning to a q I asked at the beginning of my CM studies. If this little note that ashmerm make is true, what is going on with the metal staircase? Id expect that to be where things fail moreso than the transition metals?

ACuriousMind

19:50

"ashmerm" lol

2

Slereah

arxiv.org/abs/1708.09456

If you want the full story on Earth and GR

Gravitational anomalies are a fun topic rly

Love the Schiehallion experiment

Those guys spent two years just approximating a whole ass mountain

Tobias Fünke

20:07

@SillyGoose I don't really understand your question(s) about phonons. Have you checked any solid state book on that matter?

they all arrive at the second quantized Hamiltonian form you've written

or do I misunderstand your question?

Silly Goose

@TobiasFünke i don't like starting with classical then quantizing. i am wondering about starting with postulating a quantum model of phonons

i was reading through the volume you cited previously, and i didn't see discussion about the dispersion relation in detail (in particular what in the Hamiltonian encodes optical v. acoustic phonons; or if optical v. acoustic phonons are separate systems, etc.).

and i think ashmerm avoids using the language of quantum field theory (which is another reason i do not like the text)

Tobias Fünke

@SillyGoose you can start by the quantum model "harmonic chain"

and arrive at that Hamiltonian

Relativisticcucumber

@SillyGoose its catching on

Tobias Fünke

20:22

BTW: Is your question with the bands of the free electron gas settled?

Silly Goose

i think yes thank you

Tobias Fünke

@SillyGoose well, you can of course just postulate something. but that's not what is done in textbooks, as far as I know (I've never seen it), because you can actually derive many things

Silly Goose

the number of bands is dependent on the scheme used

but the textbook presentation of phonons (in my opinion) is not a derivation. it is postulating a model, which is the same thing (but messier) than just precisely writing down the hamiltonian to begin with. It is actually worse, since it postulate a classical model and then quantizes it (in some resources).

Tobias Fünke

well, OK, yes and no

they start wit the Hamiltonian of interacting ions

and do a second order expansion of the potential. that should be the textbook treatment

so you start from a general Hamiltonian, apply reasonable approximations, argue why you do so etc., and end up with some Hamiltonian. This then is solved, in the sense that you can re-write it as you've shown, with the well-known solutions.

@SillyGoose ... isn't it the same with e.g. scalar relativistic fields? i.e. they start with some classical field and then quantize?

Silly Goose

@TobiasFünke but that in my opinion is a bad way to present the material

Tobias Fünke

20:27

but I agree that, apart from pedagogical reasons, you can simply start with the final result

why?

much of solid state/condensed matter "lives" through at least some sort of classical intuition, I'd say

Relativisticcucumber

another thing that i have been confused about for awhile, you know how on tv oxygen tanks always blow up? like they are said to be flammable. it seems that the limiting reactant in a combustion reaction is the fuel/the thing you are burning, not the oxygen since there is so much oxygen in the air. so why are these tanks flammable?

Silly Goose

because the deeper perspective (in my opinion) is to first discuss spacetime symmetries (in SR, the Poincaré group), then postulate (following Wigner) some representation theory business, then start talking about what Lagrangians can be built, then probe the theories to see which are realized in the world.

Tobias Fünke

but again, I don't think you need the classical model here

@SillyGoose yes, I agree. But you would never do so in a solid state book (or again, I've never seen this)

Silly Goose

well i think that's the type of CMT/SS book i'm looking for :P

Tobias Fünke

you have much nicer visualization and "real-world" applications in solid state physics, and I think it is completely fine to start like what we've discussed

Silly Goose

20:31

to live through classical intuition seems like a very bad idea in general, though. i mean isn't the lesson learned from heat capacity and etc. that you need to carefully account for what quantum theory brings to the picture

Tobias Fünke

:P tell me when you've found some

@SillyGoose sure! of course

But I can think of Ions moving around like harmonic oscillators, and see why e.g. the Born-Oppenheimer approximation might be reasonable

you don't have to adopt this viewpoint, though

for example, in descriptions of spectroscopic experiments, the language is often very "washed"; on the one hand, this is good, because it gives intuition on what happens and possible suitable approximations; on the other hand, one must acknowledge that the world is not classical, and one has to think about that carefully

Silly Goose

@TobiasFünke I thought the folks at UIUC might have produced something of the sort...but I will look around...I hope such a thing already exists

Tobias Fünke

UIUC?

Silly Goose

illinois urbana-champaign

Tobias Fünke

ah ok

Silly Goose

20:38

i'd like to have a better knowledge of how experimental CM is done :P so that I could reassess my expectations for what CMT can hope for

Relativisticcucumber

@Relativisticcucumber to clarify i already know the oxygen itself isnt flammable -- i mean why would a fire explode when hitting the oxygen tank if the limiting reactant isnt oxygen

Tobias Fünke

@SillyGoose and on a general note, one has to be a bit more careful. I think the classical language and entities, observables etc. are more or less necessary for a suitable description of experiments (which of course are explained/predicted by QM). If I am not mistaken, this is also said somewhere in LL (but I might be wrong)

@SillyGoose well, as I've often said already: CM is a huuuge field, so I don't think one can summarize that easily.

Relativisticcucumber

@SillyGoose lucky for u we r on track to be one CM theorist and one CM experimentalist -- that is if u dont get sucked into the wormhole that tempts u daily

Tobias Fünke

But for example, in spectroscopic experiments, you shoot something (electrons, light) on, say, a solid, and see what comes out (electrons, light or whatever), perhaps with different energy and momentum. this gives you information on the solid itself (e.g. the bandstructure, excitation spectrum)

@Relativisticcucumber haha

Silly Goose

@Relativisticcucumber me "doing CMT" so that I can actually do chern-simons theory ;D

Herr Feinmann

20:41

CMT=CM theory?

ACuriousMind

@Relativisticcucumber the air is just ~20% oxygen. When an oxygen tank explodes in contact with a fire, that blasts a stream of 100% oxygen over whatever is burning, increasing the available oxygen by like an order of magnitude

Tobias Fünke

and as I've said a while ago, the methods and quantities of interest are actually the same as in "standard" HEP: scattering cross sections, Feynman diagrams etc.

@HerrFeinmann yes, presumably

Herr Feinmann

OMG ACM is handwaving

Christmas miracle

ACuriousMind

@HerrFeinmann Coleman-Mandula theorem, clearly :P

Silly Goose

@HerrFeinmann yesh

what is the notation in statements like ER = EPR mean anyways?

is the = meant to be an equivalence?

ACuriousMind

20:43

@SillyGoose it means ER is equal to EPR :)

Herr Feinmann

@ACuriousMind I dislike the word "theorem" in QFT, please tell me you dislike it too

Silly Goose

@TobiasFünke do you have a (relatively recent) favorite experimental CM paper?

ACuriousMind

@HerrFeinmann I mean if people just use it for something they showed with rather questionable rigor that's not good, sure

Tobias Fünke

@SillyGoose uff, let me think about it

ACuriousMind

but e.g. Haag's theorem and Coleman-Mandula are, as far as I know, actual theorems in axiomatic Wightman QFT

Herr Feinmann

20:45

Oh I see. Just that CM is quoted in actual QFT books

Relativisticcucumber

@ACuriousMind but what matters is not the amt of oxygen, what matters is what the limiting reactant is, no?

Herr Feinmann

Haag is avoided like pest

Relativisticcucumber

if the fuel was already the limiting reactant, adding oxygen shouldnt help

ACuriousMind

@Relativisticcucumber Yes, but doesn't this just show that in many fires, the limiting reactant is the oxygen?

Silly Goose

I recently emailed a QFTist at my institution about if he worries about Haag's theorem. He said that he didn't, but upon looking around online again, now he potentially worries again ;D.

Relativisticcucumber

20:47

@ACuriousMind but i cant see how that is true. there's so much oxygen, i mean there is so much air and it can equilibrate so fast

Tobias Fünke

@SillyGoose no, sorry. nothing comes to my mind right now :/

Relativisticcucumber

@SillyGoose why r u proud that u spread qft anxiety like a plague

Tobias Fünke

@SillyGoose hahaha

Silly Goose

@TobiasFünke no worries ~

@Relativisticcucumber haagenpest

ACuriousMind

@Relativisticcucumber but the air is 80% nitrogen, so like 80% of the surface of the fuel at any given time just has no chance to react with any oxygen at all regardless of how quickly the air equilibrates back to 20% oxygen

Relativisticcucumber

20:48

@SillyGoose oh boy do i have one

ACuriousMind

@SillyGoose lol

Herr Feinmann

@SillyGoose "No, I live life peacefully"

Silly Goose

also apperently this site computes irreps of crystal space groups, which seems neat: cryst.ehu.es/rep/repres.html

Relativisticcucumber

@ACuriousMind but how can a fuel be more massive than the amt of oxygen

i mean more readily available

Herr Feinmann

The usual response I get is "what is Haag's theorem?"

I swear, I had this question from the same guy three times

Tobias Fünke

20:49

@Relativisticcucumber pls share

ACuriousMind

@Relativisticcucumber (I'm just making this up as I go along) My thinking is that what's available is the entire surface of the fuel. But only 20% of that come in contact with oxygen. So supplying 100% oxygen should speed up the reaction by a factor of 5 even without any other factors

Silly Goose

@Relativisticcucumber lol

Herr Feinmann

@TobiasFünke you can't go back after that

Tobias Fünke

I am lost anyway

Loong

@Relativisticcucumber no

Relativisticcucumber

20:52

@TobiasFünke oh its a meme lol. so right now im doing the like "process" to see if i can join the research group i want. stage 2 is that i read this paper and then discuss it w the potential advisor. i do find it interesting but i am struggling to grasp the details of the setup lol so ive been staring at it for some time DOI: 10.1126/science.1116955

Allie

hi besties

Relativisticcucumber

@ACuriousMind hm so maybe im overestimating the power of equilibration?

Tobias Fünke

oha, good luck with that!

hi Allie

Silly Goose

@Relativisticcucumber i think you need to do an experiment

Allie

im going to ask a question similar to what i asked before

Relativisticcucumber

20:54

me testing these theories

ACuriousMind

@Relativisticcucumber perhaps? but as I said I just made up that explanation, I don't actually know anything here :P

Tobias Fünke

hehe

Relativisticcucumber

@SillyGoose oy!

Allie

So, LITERALLY the next exercise had me show exactly what you said @ACuriousMind

Loong

@Relativisticcucumber The limiting reagent becomes relevant later when the reaction is going to completion. But in an ongoing fire, the reaction rate is given by both concentrations, fuel as well as oxygen.

$v_0=k[\mathrm{A}]^x[\mathrm{B}]^y$

Allie

20:56

and it also said that in the case of degenerate eigenvalues for B, you can always construct a linear combination of B's eigenfunctions that WILL be an eigenfunction of A

Relativisticcucumber

@Loong im guessing k is some constant, $v_0$ is the reaction rate, and $A$ and $B$ are fuel and oxygen, am i right? so what determines $x$ and $y$?

Allie

Now im wondering, what if both operators are degenerate? Say A and B commute and have eigenvalue $\epsilon$ with a multiplicity of 2. In this case instead of a linear combination, is it more like a change of basis to get from A's eigenfunctions to B's?

Relativisticcucumber

@Loong its been a loong time since gen chem sorry

Loong

@Relativisticcucumber the reaction mechanism determines x and y.

ACuriousMind

@Allie The statement is still true - you (hopefully) didn't use non-degeneracy of A to prove the claim about a linear combination of eigenvectors of B being an eigenvector of A

Loong

20:59

For a real fire, it's actually complicated since there are so many reactions involved.

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