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9:00 AM
@Blue anupam roy is on the TV. why can't this person stop writing songs?
 
Anonymous
@BalarkaSen lol
 
Anonymous
Even I can't tolerate him
 
Anonymous
But he is somehow famous
 
I can't, please, I mean, stop.
 
Anonymous
Oh ****. He is from my uni!
 
9:03 AM
If he's a singer, I'm a promiscuous blue jellyfish.
@Blue Yeah but Suman is from J too
uh, well, I am not sure if that's a compliment anymore
 
Anonymous
Rituporno Ghosh was the only legit arts department person I think
 
Anonymous
Subir Raha was from my dept
 
I think Rituporno is overrated but that's my opinion.
 
Anonymous
Oops. He is not from arts dept. He studied economics here. But whateva :P
 
Anonymous
@BalarkaSen I can agree
 
9:18 AM
Suman is from the arts dep.
 
@BalarkaSen Which university are you planning on joining next year?
 
hopefully CMI
ISI is also an option
 
ah nice
kolkata ISI?
 
perhaps yes
I mostly want to get to CMI
 
Any particular reason for that?
 
Anonymous
9:24 AM
CMI test isn't very tough. Two of my friends went there this year
 
Anonymous
Good luck though
 
@PrathyushPoduval I want to study math :P
 
Did you go to the IMOTC?
 
@BalarkaSen I mean why CMI? I know you like maths :P
why CMI>ISI is what imeant :P
 
9:25 AM
@Blue Huh. Well, I'm still positively scared about it, but let's see.
 
Anonymous
@PrathyushPoduval Balarka hates competitive maths like IMO problems. He is into high level stuff
 
@PrathyushPoduval Ahh, well, the preference for CMI is because it has more academic freedom.
I think it's unfair to say I hate IMO problems
I just can't do them
it's too hard for me
 
@BalarkaSen It's of a different type then what youre used to doing i think
@BalarkaSen In what way?
 
Anonymous
With the amount of knowledge you have I don't think it (IMO) will be tough for you if you practise for it regularly.
 
You plan on staying in india itself, or going outside inia @balaka
 
Anonymous
9:30 AM
Tbh...imo problems are mostly about tricks. Like puzzles.
 
@PrathyushPoduval As in, you can choose your courses freely. For example I wouldn't want to take 1st year calculus, but I would need to take analysis. Also you can take grad courses in early years.
 
@Blue Blasphemy!!:P
@BalarkaSen Ah okay
@Blue many IMO problems have some elegance in their solution
it's like one couldn't come up with it, unless one solves many problems and knows how to think
 
@Blue I doubt. I think it requires one to think in ways which is just not the way I think.
Ah, Prathyush beat me to it.
 
Anonymous
@PrathyushPoduval That's exactly what I'm saying. They don't require very high level knowledge but an acquaintance of similar problems which is developed by practice.
 
but you called em tricks :'(
 
9:34 AM
@Blue I don't think that means they are beneath advanced (whatever that means) mathematics in any sense.
 
Though they are really good puzzles :P
 
Anonymous
I think you took tricks in a different way.
 
After all, if you can't prove anything with all the math you know, what's the point of knowing a shitload of things?
 
@BalarkaSen You can't compare the two
 
Anonymous
@BalarkaSen I never said that...
 
9:34 AM
IMO maths, and advanced maths are 2 different species
 
Anonymous
It depends on person to person
 
Anonymous
I know people who are great at competitive maths but suck at research. Also the converse is true.
 
@BalarkaSen Just the fun in learning? Many things which humans do not have any point. It is done because people like doing it.
 
Anonymous
Some are good at both though :)
 
Anonymous
@PrathyushPoduval I think Balarka supports your point. He is saying that there's no use of learning lot of advanced math if you can't prove much using all the knowledge you have.
 
9:39 AM
Yes, there is no use. But still it is alright to learn it if you like learning, no matter whether its useful or not
 
@PrathyushPoduval This is true. But, uh, I have never been comfortable with "you do it because you like doing it" justification of the things I do. People do many unconstructive (downright deconstructive) things from their own pleasure.
 
Yeah, that justification requires some more thought :P
 
There was a talk of Benson Farb, a good mathematician/topologist, where he shot a jab at people who learn a bunch of technical machinery but proves nothing with them.
That stuck with me. I think it's better to not know things but know how to use those things than knowing things but not knowing how to use those things.
 
@BalarkaSen If you claim to know the topic, then you must know how to use it and apply it
I agree with that completely
 
Quite.
So in that sense there's a certain value of IMO type problems. They tell you how to use elementary, high school math you know to prove interesting facts.
 
9:45 AM
I have come across a lot of pseudo-experts, claiming to know relativity, advanced maths (not you :P) and so on and realized how stupid they were.
 
But it's also not worth spending your whole life on.
 
since then, i don't claim to know much
 
@PrathyushPoduval me?
 
@PrathyushP Yeah
 
@0celóñe7 wtf from where did you come?
Dont worry its not you too :P
 
9:47 AM
My mother
 
Its some of my friends
@BalarkaSen Yeah agree with that
@0celóñe7 Yes your mother too. How did you know?
 
Ok, I have to get some reading done now.
 
Exam tommorow?
 
Yep.
 
I dont think theyre counted for admissions right?
 
9:51 AM
Admission to where? This is pretest, which is not counted for anywhere much in life.
 
Oh I thought they were your midterms
 
Hah, nope.
 
When does the academic year end in your schools?
 
Feb next year is the HS, I think?
School ends much earlier than that; I think test is on Dec this year
 
9:53 AM
Higher Secondary.
 
Oh okay
Enjoy the holidays then!
 
Anonymous
@BalarkaSen i think I should disturb you today as you have a physics test tomorrow. I read up the la chapters you told me to from that book. :)
 
Anonymous
*not
 
Anonymous
We can continue from 22nd if you wish
 
10:20 AM
[Random pseudophilosophical word]
The feeling of timelessness
 
lzy
Hey guys, quick question here: If R is proportional to L/A, does that mean R is proportional to L and it is also proportional to 1/A?
 
10:45 AM
No. For example suppose L and A both double. R remains the same because the ratio L/A remains unchanged.
R is proportional to L if and only if A is constant.
Likewise R is proportional to 1/A if and only if L is constant.
 
Anonymous
11:04 AM
@JohnRennie Any idea what padding means in a Java program?
 
Anonymous
 
For performance reasons it's often good to have variables aligned to certain boundaries e.g. multiples of four bytes or eight bytes.
When you have a composite object like a struct or class this may mean ading extra unused bytes to the object. Those are the padding bytes.
They are there only to ensure the size of the object is appropriate and aren't atually used for anything.
 
Anonymous
@JohnRennie But aren't variables already assigned a memory space like 4 byte, 8 btye, etc by default? For example int type variable is given 4 bits
 
Anonymous
Is it only for composite objects like Class objects?
 
Arrays of char can be any length ...
Yes, padding bytes are generally only relevant to composite objects
 
Anonymous
11:08 AM
char arrays are already allocated a space of 2N+24 bytes. Is that "24" called the padding?
 
A lattice is said to be primitive if the position of a lattice point r' can be reached from a position r, solely by the operation of discrete translation from r to r'. That is, r'=r+T for some particular triad of u1,u2,u3 associated with all r,r' pairs. Where T=u1a1+u2a2+u3a3, where a1,a2,a3 are unit translation vectors and u1, u2 and u3 are integers. Is this correct?
 
Anonymous
 
Is that specific to Java?
 
Anonymous
I think so
 
Anonymous
But I still don't understand how that extra 24 bytes boosts performance :/ It's left empty!
 
11:11 AM
I think in Java the 24 bytes contains info about the array, so they aren't padding bytes in the sense I usually understand the term.
 
Anonymous
@JohnRennie Oh. That's possible. Could you give me a real example of where byte padding occurs?
 
Anonymous
Say if we write ABC obj=new ABC()
 
Anonymous
obj is padded with bytes?
 
Anonymous
My concept is still hazy...
 
Suppose you have a Java object containing just a single boolean variable
 
Anonymous
11:13 AM
Ok. Say: boolean var=true;
 
Anonymous
var is the single boolean object
 
In Java booleans take one byte, so the size of our object is one byte + the header info for the object.
 
Anonymous
@JohnRennie What is "header info" ?
 
A Java object consists of a header that contains info about the object plus the bytes for the data in the object.
25
Q: What is in java object header

alobodzkCould you give me some information on what is exactly stored in object header? I know, that it's probably JVM dependent, but maybe for HotSpot at least? I'm looking for exact description specifically for a first row. I've read several information that I can't verify positively with information I...

You could have Googled that ...
 
Anonymous
@JohnRennie Okay. I get it now. Thanks :D
 
Anonymous
11:21 AM
@JohnRennie Sorry for being lazy. Actually I wouldn't even know what I was looking for :P header info didn't strike me as a formal term used in CS
 
:-)
 
11:41 AM
@Blue I think I am reasonably comfortable with physics, but yeah, 22nd would be best in the sense that we can talk about these at greater length then.
 
lzy
12:29 PM
thanks @JohnRennie for the answer
just another quick question, does the internal resistance of a battery change as the voltage or current of the source is changed? Or is it always the same?
 
12:42 PM
I was so sleepy that h bar momentarily turned into maths chat and then back
 
@Secret better go to sleep then :P
 
not yet, there's one final piece of python code that I need to done (have procrastinated part of that for 2 weeks already) so I can continue my chemistry PhD
 
1:22 PM
@Secret how are you going to write it in such a sleepy mood?
It seems to be quite impossible
 
Well, I had to, as tmr there's driving class thus there isn't much time left. Currently, I only need to figure out how to update the lists after doing the matrix multiplication without messing up the order
It seems to be quite hard witho numpy, but I have not figure out how to install numpy onto the computing cluster the group is using
so I am basically hard coding with the existing libraries in python 2.6.6
 
@ACuriousMind on a boring car ride, need entertainment
 
@Secret Message the administrator and ask
I still find weird that NumPy is not installed in that kind of system
What permissions do you have?
 
O cool, I can do a multi-assignment for the matrix computed results for each of the xyz components of the vectors to update the coordinates one by one, let's see...

Python 2.6.6. might be too old? Interestingly they have python 2.7.13 but I have checked that there is no numpy library on it

I have nearly full read access to every corner of the cluster except the queue folders and some softwares which I don't use, but only the NCI folks have write access to install applications in the root
 
Then pester those folks
It's still 11PM there
Early, at least by Spaniard standards :)
Or do the same first hour in the morning
 
1:35 PM
O wait, they might had, just not in some folder like form...
module unload intel-fc intel-cc
module load python/2.7.11
module load python/2.7.11-matplotlib
module load hdf5/1.8.14

pip install -v --no-binary :all: --user h5py
the hell is this alien language pip command
 
A tool to install Python packages
 
1:47 PM
@BalarkaSen I need to learn some algebra for the GRE
Tips?
I was going to read through Pinter
 
2:25 PM
wait what is $(\delta_{jk})^2$ (it's the Kronecker delta). It's 9 right?
(j,k = 1,2,3)
 
@CooperCape Depends on what you mean by writing the square. Do you mean $\delta_{jk}\delta_{jk}$ summed over $j$ and $k$?
If so, then no, it's not 9.
 
yeah that's what I mean
is it not?
 
Nope. It's 3.
 
I thought $\delta_{jk}$ = 1+1+1?
 
@lzy the internal resistance isn't constant. It does depend on the state of charge of the battery. However I can't remember the details (if I've ever known them).
 
2:32 PM
@CooperCape No, $\delta_{jj} = 1 + 1 +1$.
 
But if (i,j = 1,2,3) aren't they the same?
 
@CooperCape I think you're getting mixed up about the Einstein summation notation ...
 
Veeery likely
hang on so does $\delta_{ij}\delta_{ij}$ cancel out to be either $\delta_{ii}$ or $\delta_{jj}$?
 
@CooperCape Precisely
 
okay that makes sense
I was getting confused because I did the substitution but left the other $\delta$ still there
 
2:38 PM
@0celóñe7 Don't know that book.
Learn classification of groups of a given order (Sylow theory) really well.
You're supposed to know field theory/Galois theory, right? I think Artin is the right book for that (but not for Sylow theory!!)
Dummit-Foote has tons really good exercises, but n e v e r study from that book.
I don't think I can actually give you a comprehensive algebra textbook. Artin is my go-to text but it's leaned towards pictures more.
 
Sid
Does anyone know where the mobile version option of chat is present?
Never mind. Got it.
 
2:59 PM
@BalarkaSen never?
I very thoroughly dislike algebra and never need it
This is going to be hard
 
i mean, as a reference it's good but not as a textbook
 
3:31 PM
@GauthamShankar You are quite correct that the electrical potential generated by a time dependent magnetic field of the type you describe is not conservative and the potential difference between any two points can't be uniquely defined.
 
3:51 PM
@JohnRennie So are the field ' with potential defined' and the one with 'not defined' two different physical quantities?
 
@GauthamShankar No, they are both electric fields in the sense that a test charge put into them will obey $F=qE$.
 
@GauthamShankar Electrodynamics is my second least favourite subject in physics, but I think in situations like that you need to consider the vector four potential.
This gives you a conservative electric potential when only electric charges and/or constant magnetic fields are present. However when varying magnetic fields are present you don't get a conservative potential if you consider only the electric 3 potential.
Or something like that ...
 
@ACuriousMind But how can the same force be both conservative and non conservative ?
 
@GauthamShankar the electric field is not just the grad of a conservative potential, it's given by: $${\bf E} = -\nabla \phi -\frac{\partial {\bf A}}{\partial t}$$
Here $\phi$ is the (conservative) electrical potential, but in the presence of varying magnetic fields there's that extra term $-\frac{\partial {\bf A}}{\partial t}$
 
@GauthamShankar I don't understand what you mean by "how". In some cases (electrostatics) it is conservative, i.e. the electric scalar potential suffices. In other cases (full electrodynamics) you need to use a magnetic vector potential in addition to the electric scalar potential (which combine to form the relativistic 4-vector potential). That's just how electrodynamics works, I'm not sure what kind of explanation you're looking for
 
3:59 PM
And that extra term is not the gradient of a conservative potential
In that equation $\mathbf A$ is the electromagnetic four potential I mentioned above.
 
@JohnRennie No, it isn't, it's the magnetic 3-potential.
 
Oops, yes, its the magnetic vector potential.
Bloody electrodynamics!
@GauthamShankar: anyhow the point is that there are two contributions to the electric field, one from a conservative potential and one not.
 
Well, I haven't learned what these magnetic 3 potential and the related terms mean. Maybe I will learn it and then get a clear understanding. Thanks for the replies!
 
I think I just accidentally gave a guy 4 notifications by accidentally downvoting his question, then upvoting thinking it would cancel, downvoting again then cancelling the downvote by clicking downvote again... :/
 
@GauthamShankar you learn all about this stuff when you study Maxwell's equations. But you don't usually do that until your first year of a physics degree.
(then if you're me you quickly forget it all again :-)
 
4:09 PM
@JohnRennie And these gaps in my concepts make me eager to learn such stuff.!
@JohnRennie BTW, if electrodynamics is your second least favorite subject, which is your least favorite one?
 
@GauthamShankar Fluid mechanics
An invention of the devil designed to make students miserable!
 
@JohnRennie especially turbulence, I think!
 
4:23 PM
@GauthamShankar I've written up the stuff I posted above as an answer just so it's on record. However I'm not sure this really counts as an an intuitive and less mathematical explanation!
 
4:42 PM
@JohnRennie thanks! I would definitely learn the math and come back to it.
 
5:03 PM
@GauthamShankar Maybe this is helpful:
0
Q: Deriving the magnetic flux from the work done by the Lorentz force, using only infinitesimals

spaceThe question might sound a bit strange at first so allow me to give some context. My prof has given a derivation of the magnetic flux where he used both $\Delta$ and $d$ for elementary line and surface elements. I am confused by the simultaneous usage of both infinite and finite elements, and the...

 
5:21 PM
@JohnRennie literally the most useful physics
Tied with thermodynamics
 
i dig thermo
 
5:38 PM
@ACuriousMind holy crap, 34 rep from reversed voting!? o_O
 
some guy get deleted?
 
@DanielSank Yup. You were being serially downvoted.
 
@JohnRennie sounds about right, at least for Astrophysical fluid dynamics... Mind, at least it's not relativistic Astrophysical magnetohydrodynamics :P
 
Damn... I lost 20rep from some guy being deleted lol
 
Anonymous
5:53 PM
@JohnRennie True. Fluid mechanics was much tougher than Electromagnetism or even Modern Physics (Radioactivity, Matter wave, etc) courses I had in high school. I took a lot of time to understand it well. But once I got through, it seemed like it was pure intuition and nothing else. :)
 
Anonymous
Tbh, the different ways of liquid flow still amazes me.
 
does our syllabus even have any fluid mechanics beyond the dull basics
i remember nothing super interesting from 11th
 
Anonymous
@BalarkaSen I had to learn lots of extra stuff for JEE :P
 
Anonymous
Pretty sure I had covered all the BSc level fluid dynamics
 
5:57 PM
fluid dynamics relates to deep mathematics of PDEs
 
Anonymous
The most interesting topic was that of vortices :D
 
whats that
 
Anonymous
@BalarkaSen Yeah, true. Lots of PDE. I remember having to formula quite a few differential equations for finding fluid profile through pipes
 
Anonymous
In fluid dynamics, a vortex is a region in a fluid in which the flow rotates around an axis line, which may be straight or curved. The plural of vortex is either vortices or vortexes. Vortices form in stirred fluids, and may be observed in phenomena such as smoke rings, whirlpools in the wake of boat, or the winds surrounding a tornado or dust devil. Vortices are a major component of turbulent flow. The distribution of velocity, vorticity (the curl of the flow velocity), as well as the concept of circulation are used to characterize vortices. In most vortices, the fluid flow velocity is greatest...
 
@CooperCape Happens from time to time. When a user is removed - either deleted by mods or voluntarily - all their votes are removed from the system, too. I've taken rep hits in the hundreds once or twice from that
 
6:00 PM
i was looking for a one line description than the whole wiki article, really, but yeah ok
 
Anonymous
Anyhow, I think thermodynamics was the toughest course. At the beginning I was almost in tears coz I couldn't understand entropy :'D
 
When I first did my thermmcis course, I spent half a yearuerstanding entropy. After that it bcomes retively straihtforward (at least compared to electromagnetism
 
i still don't get entropy lol
 
@ACuriousMind ouchie! I was alright tho 'cause upvotes countered it overnight...
 
Anonymous
@Secret Me too :P
 
6:04 PM
(o btw forgive those typos, my keyboard seemed to be failing)
 
Anonymous
@BalarkaSen There are a few good thermo books if you wish to understand it. I could name a few if you're interested.
 
Anonymous
But, since you're in class 12 I don't think you'll want to waste time on it :P
 
That'd be nice. Maybe I'll ask you after 22nd.
 
Anonymous
@BalarkaSen Okay, sure
 
Anonymous
Khan Academy is a good start
 
6:06 PM
I guess what I want to know at some point of time on my life is how to formulate thermodynamics as symplectic geometry (over what manifold?)
@0celóñe7 Do you know this shitz?
 
@Blue Khan Academy is terrible in chemistry.
 
@Blue for me, astro fluids felt like that, only replace 'intuition' with 'guesswork that's wrong as much as it's right' :/
 
Anonymous
@Loong I don't remember if it was chemistry or physics. But khan academy lectures on entropy helped me to understand it well (atleast the basics)
 
Anonymous
They dealt with it as the number of possible configurations
 
k log Omega something something
 
Anonymous
6:11 PM
Rather than the stupid "disorder" thingy
 
Anonymous
"disorder" is the worst way to explain entropy.
 
Anonymous
@BalarkaSen yep :)
 
I never really understood why that notion of entropy agrees with $dS = \delta Q/T$
the "infinitisimal/thermodynamics" definition instead of the statistical one
 
Also, you can't understand thermodynamics unless you've been working in the field for decades...
 
@BalarkaSen It's a little easier to start by noting the equivalence of the 'equilibrium energy distribution in a closed compound system finds the maximum statistical entropy' and 'heat flows spontaneously from hot to cold' versions of the second law.
 
Anonymous
6:15 PM
@BalarkaSen That was just made just to make entropy a state function. There is no "derivation" as such. Also to meet a few other necessities for such a definition. That was explained well in the "Collegepedia" series on Youtube.
 
Then connect the heat flow version to the Carnot/Kelvin/Planck group and make the conclusion from there.
But the simple fact is that classical thermodynamics is suble and tricky.
In two distinct ways.
 
@BalarkaSen I'm not sure what you mean - in the "thermo" approach, that is the definition of entropy, and in the statistical approach, that is pretty much the definition of temperature!
 
@BalarkaSen It's contact geometry, so slightly different
 
In two ways. Both conceptually and in terms of the mathematical toolset you need for dealing with it.
 
I have at various times tried to understand it but it's quite boring
 
6:18 PM
@Blue @ACuriousMind You are both misunderstanding my question. I am not asking for a derivation, I'm fine with whatever people defines as long as it has a meaning. What I mean to say is if I define $dS$ as a 1-form given by $\delta Q/T$, why does it's local antiderivative measure the system's disorderness (whatever that means)? What is the geometric meaning of this 1-form?
I am not asking a pedantic question about definitions.
 
Anonymous
Yeah, temperature can be defined in terms of entropy too. Formal stuff which is not so intuitive for beginners. $T=\frac{1}{\frac{\partial S}{\partial U}}$.
 
@dmckee Hmm.
 
Anonymous
@BalarkaSen quora.com/…
 
@0celóñe7 Hmm!
@Blue That does not answer my question. Also, shit notation, $dq$ is not a closed form.
 
Anonymous
I had a good proof for what you're asking. I'll let you know when if I find it. They had a good explanation as to why dQ/T 's integral measures disorderliness of system. But again, "disorderliness" is a very vague term.
 
Anonymous
6:24 PM
14
A: How is $\frac{dQ}{T}$ measure of randomness of system?

ChristophThis answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level. First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$ S=k\ln\Ome...

 
Anonymous
First of all, entropy is not a measure of randomness.
 
Anonymous
14
A: How is $\frac{dQ}{T}$ measure of randomness of system?

NathanielIn my opinion, it isn't strictly correct to say that entropy is "randomness" or "disorder". The entropy is defined in statistical mechanics as $-k_B \sum_i p_i \log p_i$, where $k_B$ is Boltzmann's constant (which is only there to put it into physically convenient units) and $p_i$ is the probabil...

 
Anonymous
In my opinion, it isn't strictly correct to say that entropy is "randomness" or "disorder"
 
Anonymous
7
A: How is $\frac{dQ}{T}$ measure of randomness of system?

JohannesA wealth of meaningful info is contained in the above answers. However, a short and simple intuitive picture still seems missing. The bottom line is that temperature measures the energy per degree of freedom, and hence $\frac{dQ}{T}$ measures nothing more than the number of degrees of freedom o...

 
Anonymous
**
The bottom line is that temperature measures the energy per degree of freedom, and hence dQ/T measures nothing more than the number of degrees of freedom over which the energy has spread. The number of degrees of freedom describes the microscopic complexity (as others have remarked, the term 'randomness' many consider less appropriate) of the system - the amount of information needed to specify the system down to all its microscopic details. This quantity is known as the (statistical) entropy.
 
Anonymous
6:27 PM
@BalarkaSen Actually, temperature is defined as a function of entropy and not the other way round.
 
"temperature measures the energy per degree of freedom, and hence dQ/T measures nothing more than the number of degrees of freedom" - greatest troll explanation of the week
 
Anonymous
It is almost meaningless to ask why is $dQ/T$ measures "disorder" of the system.
 
@Blue OK. Let me reparse. What the fuck dQ/T measures?
:P
I think that sums up my question, pretty much.
 
Anonymous
@BalarkaSen dS
 
Anonymous
In reality you should be asking why $T=dQ/dS$
 
6:29 PM
Sure, that's all juggling with definitions without any inkling of an answer.
 
Anonymous
I can define entropy and then define temperature with that. Not the other way round, even though some people pretend so.
 
I mean, I am not even sure why $\delta Q/dS$ is well defined because it's ratio of two 1-forms and not every 1-form is a multiple of a given 1-form (the space of 1-forms is not usually 1 dimensional).
But even if it is, the question still stands: Why should $\delta Q/dS$ (where $dS$ is defined statistically like you said) agree with our usual notion of temperature in any sense?
 
Because it works @BalarkaSen
 
@BalarkaSen I think that if you start with knowing $\delta Q$ and $T$, then it really is a surprise that the state function $S$ it defines agrees with Boltzmann's $S = k_B\ln(W)$ that counts the number of microstates available. It's a mystery, a marvelous coincidence, that hints that maybe one should start with $\ln(W)$ as the fundamental quantity and view temperature and heat as derived concepts because then you do not encounter such inexplicable coincidences.
 
Stop trying to derive physics
 
6:37 PM
That the two concepts - the thermo and the statistical notion of entropy - agree is a matter of fact: The whole reason Boltzmann wrote down $S = k_B \ln(W)$ is because after computing $W$ he noticed this produced precisely the correct expression for the entropy in the case of a gas.
 
@ACuriousMind Ok, I'm glad someone understood my question. That I can accept, perhaps. Can you actually compute Boltzmann's entropy for a given physical state?
 
Anonymous
8
A: What's the most fundamental definition of temperature?

robIt's the differential relationship between internal energy and entropy: \begin{align} dU &= T\,dS + \cdots \\ \frac{\partial S}{\partial U} &= \frac 1T \end{align} As energy is added to a system, its internal entropy changes. Remember that the (total) entropy is $$ S = k \ln\Omega, $$ where $\Ome...

 
Anonymous
@BalarkaSen This is a great answer. Do have a look. ^ I think this atleast partially answers your question.
 
@ACuriousMind I see.
 
@BalarkaSen Yes, as I just said, Boltzmann did precisely that for the ideal gas and noted it reproduced the correct expression, see chapter 6 in this paper by Jaynes
 
Anonymous
6:40 PM
@BalarkaSen I agree with @ACuriousMind. The definition of temperature in terms of entropy was a type of "reverse engineering".
 
Anonymous
Most things in physics are like that though :P
 
Anonymous
One more thing. $dQ$ and $dS$ are both entensive variables. A ratio of them is intensive aka temperature.
 
@ACuriousMind @Blue I'll bookmark those, sounds interesting.
@Blue yeah, I guess that means T is a function on the space of states. After all dQ, dS, etc are differential 1-forms on the space of states
One thing that confused me for a while is that dS = dQ/T is an exact differential even though dQ is not. That's why I object to the notation "dQ".
dQ is not really a closed 1-form.
 
Anonymous
That sounds like math lingo. Maybe ACM can answer that
 
@BalarkaSen That's why careful people write $\delta Q$. Also, it's important to note that that equality holds only along reversible/"quasi-static" processes, not always.
 
Anonymous
6:49 PM
Yep. That ^
 
Anonymous
"reversible" processes is the important word
 
@BalarkaSen Indeed, it's just a 1-form, but not closed nor exact.
 
@ACuriousMind Actually I meant to say "exact", but yes, it's not even a closed form.
Ah yeah I agree about the reversibility condition.
I am really switching the words "exact" and "closed" around today aren't I
 
Anonymous
I have a statistical and thermal mechanics course this year again. This chapter doesn't seem to leave me anytime soon :P
 
@ACuriousMind 1-form on what
 
6:59 PM
@0celóñe7 Thermodynamical state space
 
@0celóñe7 "space of states"
 
Which is...?
 

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