1:01 AM
1) you should tell us the outcome of the earlier problem
2) you should tell us what Lagrangian you are having now
3) why do you have the angles subtracted like that?

1 hour later…
2:24 AM
I am confused about what a "mental copy of a system" is. The same Hamiltonian, but different initial conditions? I am in particular confused because of this: If we start with a system in some initial state and then make a large number of mental copies of this system in an initial state, then each element of our large number of mental copies will behave in exactly the same way (i.e. trace out the same curve in phase space).
Hence, I am led to believe that by specifying a system, we are not specifying an initial state of the system.
@SillyGoose correct. You are reading about statistical thermodynamics now. We have particularly good physical reasons to believe that every microstate within the same macrostate of fixed energy hypersurface has equal probability of being the system we are studying, and thus that if we want to study the macrostates, we should be reasoning with the ensemble average of the microstates in that macrostate.
okay i see
oh i see this section is still on the microcanonical ensemble
yes, which we do not really really believe is sensible. grand canonical is so much more pleasing
2:39 AM
i was confused bc my actual course covers ideal gas law, etc. etc. and then calls that microcanonical ensemble and then moves on to canonical ensemble
hmm, that's sad. One nice thing about Ian Ford is that we begin with basic entropy and ideal gas stuff, and then clearly delimit a stop and restart and reason with microcanonical, and so forth. The clean delineation helps students see where they are
interesting... yes even this pathria text does not make that start stop so clear
the problem is that the ian ford text seems to be too far off from the content that my actual course covers (or at least the ordering is too far)
Pathria assumes that you already understand stat therm. It is not an introduction.
hm wait what is your def of stat therm (versus stat mech)
Sadly, the Feynman lec-Ian Ford-Callen-Kittel&Kroemer-Feynman stats path, just to read the texts alone, is already 3 months long. There is just no way to understand entropy faster than that.
@SillyGoose Stat therm part is the part we can just spam trickery on partition functions and derive most of the interesting stuff about thermodynamical behaviour of systems. Stat mech is really needing to use the statistical distributions from stat therm and derive physical kinetics relations, like what is the rate at which gases leak out of a hole into a vacuum, etc, and obv stat mech computations are orders of magnitude harder
it is not a really meaningful separation, just a matter of difficulty.
2:54 AM
i see
also, an unrelated question: does a system being in equilibrium imply that the number of representatives points in phase space is conserved?
What does the 2nd part of that question even mean?
I think you might be talking about the Liouville theorem on how phase space volume elements are conserved under Hamiltonian flows?
my understanding is it equivalently means the number of elements of one's ensemble (each copy of your system occupying a point in phase space $(q_i, p_i)$
yes
in which case it does not have to be under equilibrium to do that
you just need to have a Hamiltonian
pathria says that the bottom equation is generally true whereas the top equation is only true for systems in a state of equilibrium
and i am unsure where the assumption that the system is in a state of equilibrium was used in deriving said equation
ooohh
oh gosh i read "former" and "latter" backwards oopsies
hahahaha
2:59 AM
hehehe
hm so a system is in a state of equilibrium iff ensemble average of observables are time independent iff $\partial_t \rho = 0$?
You cannot have it as an iff
a system can be in a dynamical steady state, i.e. not in equilibrium, and still have $\partial_t\rho=0$
ah okay
but equilibrium means that systems are going to have a tendency to visibly seem to stay in their current macrostate for almost infinitely long times, and necessarily that means $\partial_t\rho\approx0$ at least
hm so is the import of Liouville's theorem that it at least gives a way to check if a given ensemble described by density function $\rho$ is not in an equilibrium state?
hm or maybe my question is: I don't get why pathria places so much emphasis here on satisfying $\frac{\partial \rho}{\partial t} = 0$ since it is merely a necessary condition for $\rho$ to describe an ensemble of a system occupying a state of equilibrium. (i am perhaps assuming that describing states of equilibrium is the problem of interest of statistical physics)
3:16 AM
I'm not sure if that is how the logic flows, nor that it would be of practical use
it is not that statistical physics concerns itself with equilibrium. It is that non-equilibrium is too difficult, and we are really only having a satisfactory theory of equilibrium stat therm
oh
hm but even if that necessary condition is satisfied, we don't know that the ensemble won't lead to a too-hard to solve non-equilibrium problem, or perhaps sufficient checks come later in the text/a course in stat mech?
Again, I'm not sure that such directions of inquiry will be fruitful. Maybe you would want to stay with something less convoluted?
3:45 AM
@Relativisticcucumber jigglypuff~
4:31 AM
is there a resource somewhere which says more about the appearance of $h$ as being the general quantity which converts phase space volume into # microstates
@SillyGoose it should not exist. It is simply a fact of life. For example, there is no reason why it should not be pockets of $2h=4\pi\hslash$ rather than $h$. It just so happens that when Planck worked out the quantisation rules, this quantity happens to also be correct units of phase space volume.
5:14 AM
i have a question about a technique in srednicki. so he does this thing that i started to discuss yesterday with nI so he takes the hamiltonian to be $H(p,q) - f(t)q(t) - h(t)p(t)$ and then takes a bunch of functional derivatives to bring down $p$ and $q$s, then he sets $f(t) = h(t) = 0$ but i dont see why this is valid. why can we just add these functions arbitrarily and then why we just say that these are zero?
this is what im referring to:
why isnt there the option to downvote a comment :P @ACuriousMind
5:35 AM
@Relativisticcucumber This is standard partition function trickery, and it is surprisingly more valid than you might think.
@naturallyInconsistent but why is it valid :o
@SillyGoose im not particularly versed in the mathematical proofs in the theory of partition functions to answer... I can only tell you how to think about it...
that works ! @naturallyInconsistent
Consider the absolutely convergent series $\frac1{1-x}=1+x+x^2+x^3+\cdots$, absolutely converging for only $|x|<1$. The LHS is a form that is usable $\forall x\neq1$ and converging for almost the entire complex plane. The RHS is close to one of those $\sum_ne^{-nE/k_BT}$ partition function that we see in statistical thermodynamics
Simply by taking derivatives with x, you can derive some nice properties of the sum. You know this series is absolutely converging for |x|<1, this means near the neighbourhood of origin zero, the derivatives' relations are definitely also going to work.
The stuff that is happening in the question posed by the Lorentzian vegetable is really just the same: the free particle with no external source fields is with $f=0=h$, so yes, we get to set these disturbance sources to zero after the derivatives and get results about the free particle with no external source fields. The derivatives must carry the correct information that we want to get out of these partition function manipulations.
The idea is that, while I am not sure what conditions would be needed to make the partition function exist at all, the manipulations on the series form show that, if the partition function exists, then the derivatives on the partition function must satisfy what the series manipulations say they must have, and thus immediately give us the desired statistical sum that the manipulations must have.
If you did some stat therm, this is similar to how, in Maxwell-Boltzmann statistics of the classical ideal gas, we often want to compute the root mean square velocity or mean squared velocity by $\left<v^2\right>=\frac{\int v^2e^{-\frac{\frac12mv^2}{k_BT}}\mathrm dv}{\int e^{-\frac{\frac12mv^2}{k_BT}}\mathrm dv}$
and this turns into a simple computation when you know the partition function, which happens to be easily differentiable. This is one of the earliest places where the Gaußian integrals appear in physics, and is particularly important that the result is some square root.
Maybe not just similar, but a direct consequence. The physics geniuses who did this created a totally new mathematical field of study in statistics, much to the annoyance to the mathematicians of the day.
6:55 AM
gah
@Relativisticcucumber jigglypuff~
7:12 AM
srednicki wants to suffocate me
i will learn qft if its the death of me !!!!!
What is the issue now?
@naturallyInconsistent here we can also note the reason that this works. this is because the derivative of the generating function of a probability distribution at $0$ is equal to the expectation value
7:45 AM
@naturallyInconsistent i am still trying to parse through what we were discussing above and also another result. :P i think my stat mech is not sufficient to draw the analogy explain above. it is not that i do not appreciate your explanations -- they are usually quite helpful and i save many of them in my personal notes, but in that case it did not illuminate anything for me, sadly. i was hoping someone could offer up something more simple on the main site
but i am sorry if i offended you by not even replying and posting it on the main site. i was trying to parse through a lot of things at once and was in the mode of gathering more information
@Relativisticcucumber why would you not be able to do it? It's just a way to write the correlation functions
the other confusion mentioned in the above message is if, in the PI expression, $\langle q'',t'' \vert a',t' \rangle$ we can write $\langle 0 \vert 0 \rangle$ in this form because there is a part of srednicki where he does an elaborate argument to get $\langle 0 \vert 0 \rangle$ in PI form, but i feel we could just use the PI formula directly to do so and i am trying to figure out why that is not valid
And it's essentially the characteristic function of the path integral probability distribution
eek hope there are no hard feelings i really appreciate your answers :( @naturallyInconsistent
@Relativisticcucumber This is yet another reason why I think it is kinda crazy to be teaching random collections of stuff without considering how a student might be confused on the receiving end. The stat therm bits are really important. Do you think you would be able to understand the $\left<v^2\right>$ expression above, if it is cast in partition function form?
@Relativisticcucumber This is one of the places where Srednicki is bending over backwards to draw an analogy in QM that is usually done in QFT, and it is so convoluted as to cause more confusion than anything else. You might want to not scrutinise everything in that chapter.
7:51 AM
@naturallyInconsistent i basically only know what a partition function is but now im thinking i should revisit that
@Relativisticcucumber do you think you would learn it better if you took the mathematical methods book on the topic? fqq's search term "characteristic function" in the probability and statistics part of maths methods book should do you well. Riley Hobson Bence did a really good job talking about it.
@naturallyInconsistent okay i will check
moment generating functions, etc
Imo there's no need to go and study other books, just the wiki article/a refresher of probability is fine
people usually only learn about PDF and maybe a little of CDF; that is a very limited way to look at probability and statistics. These partition function style results are much more useful.
@fqq for them it will not be a refresher. Wiki is rather bad for learning new things. A really well-written book is always a godsend. Wastes far less time.
7:55 AM
but i am getting a bit lost on what we want to compute again. so yesterday, you said ground state to ground state isnt so important, but the expectation of the correlation function is important. i see that because that is what is in LSZ, but what i dont get is why we consider $\langle 0 \vert 0 \rangle$ to begin with? and yesterday you also said GF, CF, and probabilities are difficult to solve for. are propagators able to be solved for? @naturallyInconsistent
@naturallyInconsistent i will look into this -- this is part of probability theory, right?
@fqq hi. i was wondering if it is more appropriate to call it the "generating function of probability amplitude distribution", as the $e^{iS[\phi]}$ cannot be interpreted as the "probability associated to a path"
@Relativisticcucumber IIRC, Srednicki will show that the propagator = GF will equal this CF of $\left<0|q\cdots q|0\right>$ or something like that. As should be obvious, $\left<0|0\right>=1$ is just a silly sad constant that will not be of any importance. But the functional dependence of it on the added zero functions will be such that the functional derivatives are of immense use.
The path integral will just give you the free particle partition function, from which you can differentiate and get all the propagator greens functions you want
man okay i will look more into partition functions
@Relativisticcucumber yes, studied there.
@Relativisticcucumber it will have different names for the different slight variations, moment generating functions, other generating functions, characteristic functions, etc. If you understand any one of them, they will all be similar; and then you can marvel at how nice they are to use, in the applications
@RyderRude yeah I was speaking loosely to get the message across/give googlable terms. It's not a probability amplitude either
8:09 AM
also i have closed the question because i get now that algebraically it is fine to do this manipulation :P
how is a partition function distinct from a PDF? In stat mech (at least to my impression so far) it seems like it lets one describe the distribution of microstates of a specified macrostate (via specifying the probability that the system occupies that microstate)
@naturallyInconsistent okay i will look into it !
@SillyGoose a PDF is, by definition, a probability distribution function. A partition function is actually a moment generating function (or cumulant, I dont remember silly names), and they are different ways to look at probability and statistics.
The microcanonical ensemble will be a bad place to look at these things. At least in canonical ensembles you have a non-trivial probability distribution from which to observe how these things work.
i will read that next week :D ~~ just finished the microcanonical bit of pathria
Did you never do an intro to stat therm before?
Like, if there is a uni that starts students on Pathria without first doing a sensible intro, that is a uni to run away from
I think books like Pathria, which are clearly not meant to be an intro, should not even have an intro. It should not cover microcanonical at all, since we really do not believe in it, and just jump straight to (grand) canonical, assuming that readers have already gotten a nice intro somewhere else.
8:24 AM
my uni is using the god forsaken text Schroeder for stat mech and the prof recommended pathria if one wanted something a bit more thorough
and the course itself follows a combined flow of schroeder/pathria
Sounds like the uni I was teaching at. Burn it
We are ostensibly one of the best uni in the centre of SEA, and ffs when I realised how horrible Schroeder was, I was livid
Also, I was so interested in learning cluster expansions, and then I realised my prof took it out of the syllabus. Turns out there is a problem limiting its applicability, but I forgot what it was about.
8:44 AM
what would you say the problem with pathria is?
it is not problematic. it is just not an introduction
you read it after you finish the basics
it comes near the feynman stat mech textbook: at the end of the chain of books to read, not the front

1 hour later…
9:54 AM
@Relativisticcucumber It shouldn't be necessary because almost all comments you'd want to downvote are also flaggable. Remember that comments are not meant for answers or discussions and most should just be deleted :P
the newest homework question about finding max height reached by ball leaving hand is so atrociously composed that I dont think leaving a comment about how sad it is, is even going to be helpful. The guy didn't even take a sensible picture of the question
10:22 AM
Yet another nobel prize in optics!

4 hours later…
2:10 PM
hi
2:34 PM
i think they over-dramatise this show. it is pretty cringe
2:58 PM
hm
So this is what a Nobel Prize looks like nowadays.
3:10 PM
Yet another year where I get no Nobel prize
2
Pavlov's canines are not salivating
3:34 PM
Got a copy of Landau lifshitz volume 8 from the library just to check a calculation, it has discoloration due to water drops at about 4 th chapter ಠ⁠_⁠ಠ
Blood, sweat and tears?
2
4:36 PM
Hello folks!
I'm about to sit down to write
any requests in particular?

1 hour later…
5:46 PM
A poem?

1 hour later…
6:51 PM
@EmilioPisanty I just saw this question pop up and thought "Hey, that's something for Emilio" but then I realized it's from you =)
7:27 PM
2

The 2023 Nobel Prize in Physics was announced today, and it was awarded to Pierre Agostini, Ferenc Krausz and Anne L’Huillier, for “experimental methods that generate attosecond pulses of light for the study of electron dynamics in matter”. The documents released by the Nobel Foundation along w...

If you can feed the algorithm and make it work for science communication, that'd be awesome =)
8:04 PM
@ACuriousMind i think i get irked because people say stuff thats just senseless but not warranting of moderator attention
not necessarily wrong but just not additive in any way shape or form :P
but c'est la vie i guess
@Relativisticcucumber that's a classic "no longer needed" flag :P
there's no consequences to the user for having comments deleted unless we actively decide to suspend them for being rude or something
we're deleting hundreds of unnecessary comments each month, most people don't even notice
if you're flagging stuff you shouldn't be flagging you'll notice because the flags will get declined

1 hour later…
9:23 PM
@ACuriousMind 🙂🙂🙂🙂🙂
Boy, today has been a bit of a whirlwind