The h Bar - 2022-06-10

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12:04 AM

so we have an interesting situation here. This question physics.stackexchange.com/q/711007/36194 has been bountied but the OP is suspended (for a while yet). Can the OP still award the bounty while suspended?

The electromagnetic force and weak force couple to each other at any energy scale(Glashow's theorem), so there must not exist any only weak interaction governing theory at low energies---yet there exists albeit non-renormalizable but effective low energy Fermi interaction theory which describes some stuff satisfactorily---why this is not a contradiction?

@ZeroTheHero No, the bounty will be awarded according to the rules for awarding it automatically when it runs out. (See "what is automatic awarding?" here)

@ManasDogra I have never heard of "Glashow's theorem". Where did you get that from, and what does is say precisely?

I found this in This answer by Cosmas Zachos

The word "theorem" does not appear in that page

@ACuriousMind Thanks. I did not know this!

12:19 AM

@ACuriousMind Okay sorry, I told in a wrong way, I accept...but is the rest of it understandable?

@ManasDogra The point of Cosmas answer (and Glashow's contribution) is to show that there is no separate $\mathrm{SU}(2)$ symmetry

there's just the electroweak $\mathrm{SU}(2)\times\mathrm{U}(1)$ and the electromagnetic $\mathrm{U}(1)$, but never a standalone unbroken $\mathrm{SU}(2)$ for the weak interaction

afaik, there's no such SU(2) in the effective Fermi theory, so there's no contradiction

Ok, thanks...I have another question.

I was wondering why is the electroweak force inherently quantum mechanical, there must be some step while deriving the electroweak symmetry breaking where quantum mechanics is used, but I fail to see it..

I'm not so sure what you mean by "inherently quantum mechanical"

I have seen the claim that there's no classical electroweak field...

depends on what, exactly, you mean by that

I mean, there's no "classical electron field" either, there's just electrons

but still you can think about the classical field theory of a Dirac field or a non-Abelian gauge field, they just don't seem to have much to do with the reality we observe :P

12:33 AM

Is it sensible to ask why does these classical fields don't occur in reality?

the better question would be why you expect them to

really, the only "quantum field" in the SM that has a somewhat direct relation with a "classical field" is the electromagnetic field

Because...one of them does describe nature classically!

so from that perspective, it's the exception that there is a meaningful classical field associated with a quantum field, not the rule

Yes I was talking of the electromagnetic field...why would it be special?

@ManasDogra it's the only field whose associated particles are both massless and uncharged (after symmetry breaking)

(and even treating photons like other particles is notoriously tricky, e.g. they don't really have wavefunctions in the usual sense because of the lack of a relativistic position operator)

so, in a sense, the quantum objects associated with the electromagnetic field are "maximally unlike" classical particles

12:37 AM

OH....that means the reason why we get to see a classical em field is because after sym break the EM part is unbroken?

that's the "massless" part, yes

the "uncharged" part is also important: it means photons actually propagate and don't clump together (contrast with the strong force/gluon confinement, "glueballs", etc.)

but really what you need to be cautious of is that the quantum->classical limit is actually pretty complicated to make sense of in general

you can't just take a quantum field theory and hope there's a corresponding classical field description

Hmmm...nice.. but is there any alternate way to say why weak forces don't occur for low enough energies but em forces do..

after breaking, the W/Z-bosons are massive

you don't get a lot of weak effects at low energy because there's not enough energy to produce them

while you can produce photons of arbitrarily low energy because they're massless

But, there are terms in the electroweak Lagrangian which don't contain the masses of the bosons, won't they contribute at lower energies also?

I mean, the weak interaction doesn't suddenly stop below a certain energy

it just becomes rare/improbable

12:51 AM

That's what I was thinking, but Cosmas in that answer which I linked above seemingly said otherwise in the last sentence

if you write down any Feynman diagram that includes the W/Z-bosons then it will be highly suppressed at low momenta

@ManasDogra you're being too literal, and Cosmas has a love of flowery language

he just means you won't detect any sort of weak effect at low energies

Oh :p

I mean, we observe stuff like the $\beta$-decay

that's a weak interaction at relatively low energy alright

I guess what Cosmas means - note the "classical" in parenthesis - is that there is no "classical weak force"

there's a bunch of effects like $\beta$-decay, but no actual classical force

(I'll never get tired of posting this xkcd in that context.)

@ACuriousMind Oh, I see...

@ACuriousMind hehe

I have another question...There are some particles like the sigma minus which decays much faster than the relatively less massive neutral sigma particle, why does this happen?

I mean, it will depend on the particular decay modes and so on, but why generally does it happen when it happens..

This is the opposite of what I expect naively from the uncertainty principle

2 hours later…

2:49 AM

Any chemists in the house?

Wanted to know how does one model the potential of 2 hydrogen molecules that collide?

(as a function of distance between nuclei)

Do I need DFT?

5 hours later…

8:07 AM

@ManasDogra I think the answer here is to not naively expect things from the uncertainty principle :P

8:31 AM

looks like from a graviton-photon point of view, photons do indeed acquire an effective mass via mass renormalization

9:16 AM

why do the 3 gluon vertex amplitude depend on the term in the QCD Lagrangian with one and only one derivative acting on a gluon field?

9:31 AM

@ManasDogra what other terms with three $A$ in them do you think there are?

9:42 AM

@MoreAnonymous what exactly are you trying to do?

i.e. are you trying to get a accurate potential for a small system, or is this part of a macroscopic molecular dynamics sim?

1 hour later…

10:49 AM

@NiharKarve well I need this result for a 7 page latex idea (not that that's telling much) but I need an order of estimate for the potential slope when both the molecule collide. Don't need it to be too accurate, just as an order estimate will do.

11:09 AM

@ACuriousMind I am sorry, what I meant to ask is if there were terms like $(\partial A_a)(\partial A_b)A_c$ or $A_a A_b A_c$ would they also contribute? or should terms like $(\partial A_a) A_b A_c$ only does?

@ManasDogra Sure, if there were such terms they would contribute

but there are no such terms in $F_{\mu\nu}F^{\mu\nu}$

Hmm..

user image

I tried but I couldn't see where from the metric tensors come in the picture---
The $-ip_\alpha$ comes by fourier transform, but where from the others come?

Tbh I don't know the rules of converting the vertex diagram to amplitudes, because there were no such term in QED...What is the rule of conversion? or Where from the "$g$"s come?

@ManasDogra It's really just the generalization of how vertices work - if you just view every component of $A^\mu$ as a field on its own, that's what you get when you explicitly write out the summation and apply "normal" rules to the resulting vertices like $(\partial_\mu A_1) A_3 A_0$

it's very annoying to spell this out explicitly, but if you don't see that it must work that way, it might be worthwhile if you sit down and do it

1 hour later…

12:47 PM

U can delete physics meta O_O ??

I deleted it because the meta post it advertised was deleted by its author

1:49 PM

@MoreAnonymous would using the Lennard-Jones potential work?

a more complicated method would probably involve a force field, you might use DFT to fit the parameters there

(it's funny that you should bring DFT up because I started to code a DFT solver today)

2:18 PM

Reading Griffiths (Introduction to elementary particles) I found this remark about virtual particles "If a photon is emitted on Alpha Centauri, and
absorbed in your eye, it is technically a virtual photon, I suppose. However, in
general, the farther a virtual particle is from its mass shell the shorter it lives, so
a photon from a distant star would have to be extremely close to its “correct”
mass; it would have to be very close to “real.” "

Does this imply every photon is virtual and off-shell? I thought there was a net distinction between real and virtual...

@NiharKarve I thought of that but I 've read Leonard Jones Potentials dont work well for hydrogen :/

@NiharKarve Any idea how to do this calculator

@Feynman_00 That's a deeply contentious question

I'm firmly convinced this is nonsense, a myth that has infected even otherwise upstanding physicists by its sheer pervasiveness and intuitive appeal, and that "virtual particles" are really just terms in a perturbation series and not at all, well, "particles"

see more of my ranting about that in general e.g in physics.stackexchange.com/a/275099/50583, physics.stackexchange.com/a/700019/50583 and a comment thread specific to your particular question at physics.stackexchange.com/questions/695288/…

@ACuriousMind I think its historically feynman's fault cause he believed he was they were actual particles

I'm not sure he did

could well be a "Hawking radiation" situation

Hawking explicitly says in his original paper that thinking of this radiation as virtual particle pairs being split up by the event horizon is just a silly analogy you shouldn't take too seriously

and yet I get a (small, but constant) stream of people quoting other parts of Hawking at me in order to show he believed that his eponymous radiation was caused by virtual particles and hence virtual particles must be real whenever I bring this up

And Feynman likewise was someone who liked analogies very much, but who I suspect was much more conscious of the limits of these analogies than the people parroting them are

(alas, in contrast to Hawking, I don't have an original quote telling me this explicitly, so I admit this is just wishful speculation)

2:33 PM

@ACuriousMind I think I saw a Sheldon Glashgow or Murray Gellmann interview on this

Im trying to search for it

2:48 PM

Cant find it

I remember someone complaining about Feynman being weird with the dentist or something

@ACuriousMind I find more acceptable and reasonable to consider virtual particles as mathematical tools. As per the quote from Griffiths I posted above, in this perspective, isn't he wrong claiming a photon emitted by a long distance and then absorbed is virtual (and thus off shell)? Or maybe he's trying to oversimplifying the example making it actually more confused.

@Feynman_00 i never truly understood QFT and then I realised the language used in physics class was just super misleading

and thats the state of affairs atm

in fact not even experts can agree upon an exemplary QFT book

@MoreAnonymous Then imagine how I feel without having taken QFT (next year I will) and having experimental courses throw virtual particles at me and explaining them in terms of energy-time uncertainty...

3:09 PM

A 'virtual particle' is the relativistic analog of a non-relativistic 'intermediate state' also called a 'virtual state' when you do time-dependent perturbation theory over a basis of free particle states, in that case you end up summing over energies so the relation $E = p^2/2m$ fails i.e. it's a virtual particle, similarly in the relativistic case one sums over 4-momentum so the relation $p^2=m^2$ fails. In the non-relativistic case one doesn't commonly hear magical language

3:25 PM

(Deleted, maybe the chain rule...)

1 hour later…

4:34 PM

@Feynman_00 I was marked down during my viva cause I told the person who was taking it there is no rigorous version of the time energy uncertainty principle and the vaccum relation u r talking about

4:51 PM

@Feynman_00 Yes (that's what I'm debating in the comment thread I linked last), I think this is a wrong and misleading description

for one, it makes the strange prediction that all "real" particles are, in fact, slightly off-shell

which should mean that off-shell photons don't propagate at the speed of light or something? I really don't think people who claim this have thought this all the way through, they just think about QFT as "Feynman diagrams" (whether as a simplification or not) and then try to explain everything in terms of these diagrams

@ACuriousMind There are plenty of reasons why real photons propagate slightly under the speed of light really :p

I don't think we've ever actually seen anything go at the speed of light

@Slereah I'm sure there are but they're complicated GR reasons and not silly Feynman diagram reasons :P

@ACuriousMind Those reasons probably compound!

also the lack of an actual vacuum

don't forget QFT is SR

do you claim there are good reasons to believe photons don't propagate that the speed of light in SR?

Well, as you said

"real" photons as measured by real apparatuses are off-shell, since they are just exchange photons for very far apart charges

Very close to but not quite on-shell

4:57 PM

nonononono

that's exactly the idea I'm arguing against

Im under the impression @Slereah is taking the mickey?

there's no such thing as an "exchange photon" that you could measure in an apparatus

How would you describe an actual measurement of light in the real world in QFT then

@MoreAnonymous honestly, it's hard to tell :P

@Slereah well, first of all, "actual light" is usually a coherent state that doesn't even have a definite number of photons

@MoreAnonymous I have never heard this expression except in this video : youtube.com/watch?v=7pSmhZFbCy0

4:59 PM

and there's a lot of quantum optics about how the electric/magnetic fields we measure relate to the actual quantum state

@ACuriousMind How on-shell are they?

very

@Slereah I learned it in the UK

And what kind of toons do u watch

they exist, and hence are on-shell

my eyeeeesssss are bleeeddddinngggg

5:00 PM

Sometimes I act the mickey

the only reason the virtual particle can "be off-shell" is because they're just states in a computation that you sum over, not actual states that you claim exist (that's the "off-shell component Fourier discussion" I also linked above

I recall Feynman arguing for real photons being off-shell

idk how accurate that is though

Vageina from Vegeta OMG

but really to me the simplest and obvious reason that all this is nonsense is that virtual particles are a tool in perturbative computations

this is thing should be reported and banned on youtube

5:02 PM

never

you can't even say what they are without doing perturbation theory

Seems hard to tell if actual measurements are on shell or not, though!

@Slereah I was just telling ACM feynman fluked feynman diagrams and now we have this BS language surrounding them

Does light actually go at c (the parameter of SR relating time to space) or under

I'm not even sure how well defined that question is

well, SI defining the speed of light by fiat certainly didn't make that one easier :P

I mean, if you're saying the speed of photons might vary, you're essentially saying our current definitions of time and space units are nonsense

5:05 PM

there are too many things "the speed of light" can refer to really

another problem is that I don't think "photons" really have a speed

massless objects don't have good position operators, wtf does it mean to talk about their speed

what has a speed is a classical EM wave

photons are really much weirder than people give them credit for, even compared to massive quantum particles which people already mistakenly represent as little balls flitting around

and, as I said, real EM waves usually don't consist of a definite number of photons, it's not like an electron beam where you can just think of it as a superposition of the wavefunctions of a bunch of individual particles

@ACuriousMind yea ... I actually tried to model them like that ... and failed :P

neither do the photons have wavefunctions nor is there a definite number of them, and yet all of intro QM severely tempts people to think about light in exactly those terms

5:32 PM

Everything is complicated in qft...

First section of this is great

> in contrast with “textbook” treatments of Yang–Mills theories... which manage to leave, despite disguises of relatively sophisticated language, a strong impression of arbitrariness, Utiyama strenously tried to derive gauge theory from first principles.

1 hour later…

7:02 PM

@MoreAnonymous I had my first QM course this year and at the same time there were the other courses handwaving energy-time uncertainty everywhere. Next year I'll start with my Master's degree in Theoretical Physics, finally

@Feynman_00 i wish u luck ... personally I love the Condenesed matter way of thinking of QFT which gives u a dictionary to QM

@ACuriousMind Ok, I read the comment section again and I'm glad to hear that Griffiths's statement is misleading at best. I think I'll be going back on this during my future studies and think about it.

@MoreAnonymous I'm interested in QFT and Fundamental interactions :P

@Feynman_00 A lot of particle physics stole techniques from condensed matter including the higgs

did youtube update its video algorithm

I don't think it's fruitful to play cond-mat and hep-th against each other as some sort of opposites

Physics is all One beautiful science, that makes it even more interesting

7:13 PM

Im getting fox news recommendations :(((

@ACuriousMind Yups

they're two sides of the same coin/polygon, both using QFT

@ACuriousMind AQM do u lean towards any theory in QG?

(me curious)

AQuantumMind? :P

but no, I don't have a horse in the race

I know string theory best, but that doesn't mean I'm convinced it's more likely to be true

@ACuriousMind I bet no known race horse will win :P

@ACuriousMind Sounds like a good user name :P

Is that the fusion between qmechanic and ACM?

7:24 PM

I'd like to keep on existing separately D:

@ACuriousMind Am I the only one who is imagining some DBZ fusing

?

no

:D

@MoreAnonymous No, you're not. I was thinking about that when I wrote that message

@ACuriousMind I can imagine something like: "the physics problems of the world has become too much we must do something" "Yes lets do some funny poses and fuse!" "I am AQM!"

Don't worry, thanks to retcons in DBS, potara fusion lasts just 1h...

7:29 PM

@Feynman_00 I swear we might be on our way to being banned :P

Yes, I think we've had enough

why would anyone be banned for talking about DBZ?

@Feynman_00 He musn't find our comments rude since he is secretly AQM! ACM would have been offended by now :P

I wish I could witness them doing funny poses

maybe I just lack the DB knowledge to figure out what would be offensive here :P

most of what I know of that show is through cultural osmosis, not because I would've watched it

@ACuriousMind I wish I could un-remember those pokemon names

7:36 PM

nah, pokemon are great

@ACuriousMind really????

First time in the chat here and was wondering: is there any appropriate venue for asking people to take a look at one's questions?

@ACuriousMind and here i thought u were an intellect

:P

I just thought that physics-only discussion were allowed in the chat. I should read the guidelines...

@EE18 already asked it?

7:38 PM

@EE18 if you just want eyes on your question, bounties are the way to go

@ACuriousMind I think he wants feedback

Yes, I already asked it and bountied it. The bounty just expired though it lets me award it for another 17 hours.

like is this is a good question

ah ... i was wrong

just advertising your question isn't something we really want to see here because if everyone did it for their new questions that would drown out everything else

Gotcha, all good! I will refrain from doing that then :)

7:39 PM

By the way, whoever gave the name to this chat is a genius

@Feynman_00 anything goes as long as no one complains and it stays within the CoC

Ok, thanks

@Feynman_00 It was some user named Kevin, but really naming bar-like venues "h bar" is somewhat common among physicists

@Feynman_00 I can't wait for pi day then I can crack a stupid joke: "Where's my share of the pie? Since it's h by 2 pie"

Its the small pleasures in life that make u go on :P

@ACuriousMind It's a new joke to me. I found out that ħ is "h bar" in English just recently while writing in Latex

7:44 PM

yeah, the pun doesn't translate well

@MoreAnonymous what are your issues with L&L's description of time-energy

I either read it in my native language or just read the symbol in my mind

@bolbteppa Show me a rigourous proof

you're probably just talking past each other - there's plenty of rigorous versions of a "time-energy uncertainty principle", see physics.stackexchange.com/a/53804/50583 and the other answers, but none that would justify the common way of talking about virtual particles etc.

@ACuriousMind I remember I was once so disheartened that Josh's proof didnt mean what I thought it meant and then I fought with you ACM. Remember that?

:P

7:46 PM

vaguely

I've had many fights about the uncertainty principle :P

they blend together after a while

@ACuriousMind I can see where the uncertainty stems from :P

I think @MoreAnonymous means that the "time" that appears in the principle is more of a characteristic time of the system, since there is no time operator

@Feynman_00 Let me ask you think. How would you experimentally verify Josh's proof?

The thing is the moment you do a measurement then you cannot employ Josh's proof since it relies on unitary evolution

how do you verify any uncertainty relations?

@ACuriousMind Using an ensemble

7:52 PM

Why can't you do it for the present case?

@MoreAnonymous sure, and the same works here, you start with an ensemble of identically prepared states with some non-zero $\sigma_H$, you measure $\Omega$ on some of them after a time $t_1$, one some after a time $t_2$, etc.

Yes, that's what I would do

then you plot the $\langle\Omega\rangle(t)$ you get from that and show it always obeys this bound (up to experimental error)

Is it me or mathmode doesn't work in the chat?

but this kind of misses the point because uncertainty relations aren't really claims about measurement

@Feynman_00 look int he room description, there'S a link to how to get mathjax to work

7:55 PM

@ACuriousMind But to 3.72 you need to differentiate the unitary operator

right?

3.72???

"One can show that (see eq. 3.72 in Griffiths QM)"

ah

but what's the problem - each state evolves unitarily until it is measured, and I'm not doing anything with the states after I measure them, I just throw them away

@ACuriousMind Yes I now realized that subtlety

This looks worth reading on time-energy

8:01 PM

Did you guys know last year Jackson's third edition was revised and they made an "international adaptation"?

They removed some parts, though

@ACuriousMind But also I find the whole $t = m \langle x \rangle / \langle p \rangle$ just weird

In some sense that's what Josh's answer endorses :/

I have no idea what you mean

@ACuriousMind I'm talking about his definition of $\Delta t$

I've just chosen the position operator

it doesn't imply anything about a "$t$"

the whole point of "time-energy" being subtle is that there's no time operator

@ACuriousMind Seems like a case of bad naming

To ensure confusion

8:07 PM

well, the $\Delta t$ is a time-span

it's just not the standard deviation of an operator

I get PTSD thinking our professor told us that there no time operator like 5 times

that's why Josh wisely uses $\sigma$ to denote actual standard deviations and $\Delta$ for the quantities in the uncertainty principle

@ACuriousMind Then I can choose $ \Omega = x$ and get my expression no?

Even a nice little Pauli time operator argument in that paper

@MoreAnonymous where did the $t$ on the l.h.s. of your expression come from???

it's still a $\Delta t$

also, $\partial_t\langle x\rangle \neq \langle p\rangle$

8:09 PM

Ah sorry I was being sloppy but would u endorse $\Delta t = m \langle x \rangle / \langle p \rangle$

you can't just pull the time derivative into the expectation value

?

@MoreAnonymous if your observable is the position, you get as a characteristic time uncertainty in position divided by "velocity"

@MoreAnonymous the quantity in the denominator is $\partial_t \langle \Omega \rangle$, not $\langle \partial_t \Omega\rangle$.

@ACuriousMind Ah I see

8:11 PM

you have $\partial_t\langle\Omega\rangle = -\mathrm{i}\langle [\Omega, H]\rangle$ by Ehrenfest

that may or may not be equal to $\langle p\rangle$ for $\Omega = x$ depending on $H$

I see

@bolbteppa Does the Pauli argument remain valid in QFT?

I dont think its bounded from below?

@MoreAnonymous the "no position operators" in relativistic QFT includes time as the "0th position".

That's another curveball in all this

You can quantize a relativistic point particle via $[\hat{x}^{\mu},\hat{p}_{\nu}] = i \delta^{\mu}_{\nu}$, notice $\hat{x}^0$ is a 'time' operator, what now...

@ACuriousMind I do know that but I don't know how to prove that

@bolbteppa I don't think thats legal

If it wasn't, every string theory book would be full of nonsense

Time-energy is simply a legitimately debated topic in QM, it's extremely important for setting up QFT and was historically extremely important before any 'rigorous' attempts which themselves are debated

8:20 PM

@bolbteppa no issues, because the claim about "no position operators" is about QFT, not a single particle

single-particle relativistic QM is a hack anyway, no surprise you can have illegal operators there :P

@ACuriousMind Wait then whats this string theory nonsense claim?

@bolbteppa Also string theory is not normal quantization as far as I (don't) understand: physics.stackexchange.com/a/22614/150174

string theory books often contain "the relativistic point particle" as warmup for the actual string theory

and there's really nothing wrong with quantizing the relativistic point particle

it's just not particularly relevant to QFT or the discussion of time/position operators :P

@ACuriousMind how does one argue a QFT with a position operator cannot exist? Excuse my ignorance

@MoreAnonymous Valter Moretti has an excellent answer here

@ACuriousMind The last line invokes Pauli

But we were figuring out why is pauli still valid in QFT if the spectrum is not bounded from below

8:25 PM

Well that is one indication how wrong all this is

@MoreAnonymous What do you mean by that? The vacuum of a well-defined quantum field theory is the lowest-lying energy state, so the Hamiltonian must be bounded from below

@ACuriousMind Im thinking of the dirac sea

This is all simply more complicated, there are really interesting subtleties here

@MoreAnonymous I'm not responsible for your thoughts :P

@ACuriousMind rolf ..... t

8:27 PM

You can apply Pauli's theorem logic in the relativistic case to a proper time operator, it's the same thinking, that is the way out I see

Literally laughing alot

When you promote $\hat{x}^0$ to an operator, the Schrodinger equation has a 'time' derivative with respect to a proper time $\tau$ (and the Hamiltonian is $p^2 + m^2$ i.e. Klein-Gordon)

In other words, choosing a 'gauge' is a crucial fact in this time operator stuff

Another quick q for you all...does phys SE have a venue for asking if people want to self-study a text together? In particular, I'm looking for someone to work thru Ashcroft and mermin with

@ACuriousMind But I'm confused how is it that by a simple redefinition (of the Dirac sea) to a new definition (the vacuum) you got rid of the loophole :P

I mean I swear the vacuum is nothing than infinite particles ... which can be exicited

Time-Energy is just different because the others are two values of an observable at the same instant, while T-E is two values at two different instants, and we really care about values not operators

8:31 PM

@EE18 the best venue would be here or problem solving depending on the level, but I don't think I've ever seen that successfully take off

@MoreAnonymous the vacuum is just the vacuum

I never really saw a difference between vacuum and dirac sea

I don't subscribe to any weird ontologies

maybe there are infinite particles there, maybe not, as long as they can't come out all at once I don't really care

here as in this chat?

yes

I mean the only other venues we have are the main site and meta, and it's definitely off-topic for both of them :P

@ACuriousMind you can never really though reach infinite with finite means though :P

8:33 PM

ah ok. and there's no way to pin a request is there? i'd just have to come back every once in a while and ask if anyone's down to work thru the text together?

thanks for fielding these qs btw

well, if people like it they can star a chat message and it'll stay on the star board to the right for a bit

@EE18 maybe mention the book u wanna go thru in the same message? So I bookmark can it?

the duration depends on how many silly jokes get starred afterwards :P

2

Yup, I want to find a partner to self-study Ashcroft and Mermin's Solid State Physics (pretty much straight thru) this summer :)

3

Thanks very much @ACuriousMind

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