04:00 - 20:0020:00 - 00:00

4:53 AM
Hey all whats the rate of say a pressure fluctation?

5 hours later…
10:18 AM
@MoreAnonymous Presumably the fluctuations have a frequency spectrum.
That is, if you measure the pressure at a fixed point as a function of time and Fourier transform you get a frequency spectrum. If there is a characteristic frequency or some sort of average frequency you could take this as the rate.
Or I guess calculate the autocorrelation time, then the rate would be the reciprocal of this.

I don't think it's useful to comment every line of a function, but at least specify what every variable is for, their structure, and what functions do

10:41 AM
@JohnRennie Then whats the frequency spectrum of a system in thermal equillibrium?

You'd get a thermal distribution I guess.
Higher frequencies would have higher energies for a given amplitude, so the amplitude would decrease with frequency.

depends on your ensemble
e.g. the microcanonical ensemble should have constant pressure, no fluctuations
the very definition of equilibrium is that some things don't fluctuate

They do in thermal equilibria

@JohnRennie Im confused ... are we saying microcanonical ensemble cannot be in thermal equillibrium? Or are you disagreeing with Acuriousmind

There's a video somewhere on YouTube where Sean Carroll waxes lyrical about the difference between a thermal equilibrium and an eigenfunction in QM. His point being that the former fluctuates and the latter doesn't.
Ah, it's still in my history:

10:54 AM
that's a bit too vague to say anything - of course some things "fluctuate" (have non-zero standard deviation?) in a mixed state that don't in a pure state
that doesn't say anything about pressure in particular

Hi All...
$\tau = r \times F = I \alpha$. How $r \times F$ take care of bodies rotational inertia?

@ACuriousMind what about fluctation theorem?
The fluctuation theorem (FT), which originated from statistical mechanics, deals with the relative probability that the entropy of a system which is currently away from thermodynamic equilibrium (i.e., maximum entropy) will increase or decrease over a given amount of time. While the second law of thermodynamics predicts that the entropy of an isolated system should tend to increase until it reaches equilibrium, it became apparent after the discovery of statistical mechanics that the second law is only a statistical one, suggesting that there should always be some nonzero probability that the entropy...

@MoreAnonymous the very first sentence of that article says it's about systems away from equilibrium...

The FT is one of the few expressions in non-equilibrium statistical mechanics that is valid far from equilibrium.
It also says^

sure
but you said you wanted to talk about systems in thermal equilibrium
so I'm confused why the fluctuation theorem would matter

11:00 AM
Sorry I crtl + C 'ed the wrong line :P
the FT gives a precise mathematical expression for the probability that entropy will flow in a direction opposite to that dictated by the second law of thermodynamics.

@123 it's not clear what you're asking, are you asking where $I$ comes from in the derivation of $\tau = I \alpha$. The wikipedia shows two expressions for $\mathbf{L}$, $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ and $\mathbf{L} = I \vec{\omega}$. By computing $\frac{d \mathbf{L}}{dt}$ for both expressions you get $\tau = I \alpha$ (the wiki gives something slightly more general)
In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically...

So if Im in max thermal equilibrium I can deviate away from it over time and return aka a fluctuation ... right?

@bolbteppa I have read this wikipedia article.

@MoreAnonymous no
again, the theorem is about systems in non-equlibrium

You're repeating yourself now and ignoring specific questions, come on

11:02 AM
an equilibrium state is by definition invariant under time-evolution, it doesn't change in time

My confusion is that, if we use torque formula $r \times F$. This formula does not take care of rotational inertia of body. It means this formula produce same amount of torque no matter how large or small body is.

@ACuriousMind I thought thermal equilibrium was simply maximum entropy ... no?
From what I remember of my stat mech

@MoreAnonymous well, but why is equilbrium maximal entropy?

Again, what does that mean. The wikipedia derives $\tau = r \times F$ from $\frac{dL}{dt} = \frac{d}{dt} (r \times p) = r \times F$, it also equates this to $I \alpha$ because $L = I \omega$ implies $\frac{dL}{dt} = \frac{d}{dt}(I \omega) = I \alpha$. What is wrong with this, it's right there in the wiki article

because thermodynamical statistical systems move essentially along the gradient of entropy, and so the equilbrium point (i.e. a state that doesn't move) must sit at a stationary point of entropy
turns out that stationary point is a maximum under reasonable assumptions

11:05 AM
How it is possible we can have same amount of torque with different size and size of bodies. Requirement of this formula just require perpendicular distance from axis of rotation to the line of action of force and force itself.

but you don't define equilibrium by maximal entropy - what would be the point of that? Why would you care about the state of maximal entropy if there wasn't something special about it, namely that it is invariant under time evolution?
@123 If I exert a certain force on a body via e.g. a spring, that force doesn't care how massive the body I exert it on it, it's always the same force
the mass just determines how strongly (acceleration) the body reacts to the force, just like the moments of inertia determine how strongly (rotational acceleration) a body reacts to torques

If the same $\tau$ is applied to larger and larger bodies, it means the $I$ in $\tau = I \alpha$ gets bigger and bigger, thus the $\alpha$ will get smaller and smaller to compensate. The value of $\tau$ being the same in all these cases is because the same $F$ is applied in each case and $r$ is the same, i.e. $\tau = r \times F$, that doesn't mean the resulting accelerations in each cases will be the same

@ACuriousMind What if body has two different moment of inertia. How $r \times F$ can produce same torque on each body separately.

I don't understand the question. Compare: "What if two bodies have different masses? How can I exert the same force on each with the same spring?"

You're basically forgetting that the $\alpha$ will have to change for two different $I$'s when $\tau = I \alpha$ is the same for two different bodies

11:10 AM
@ACuriousMind No no... Let me explain you. I have two bodies having different moment of inertia but same shape, size and same reference point on each body. I apply same force on each body separately at same distance from reference point.

Then you get the same torque, but different rotational accelerations. Where's the problem?

How both bodies can have same amount of torque in above case. Because there is no parameter for moment of inertia in formula $r \times F$

@ACuriousMind Is the wiki wrong then
?
In statistical mechanics, thermal fluctuations are random deviations of a system from its average state, that occur in a system at equilibrium. All thermal fluctuations become larger and more frequent as the temperature increases, and likewise they decrease as temperature approaches absolute zero. Thermal fluctuations are a basic manifestation of the temperature of systems: A system at nonzero temperature does not stay in its equilibrium microscopic state, but instead randomly samples all possible states, with probabilities given by the Boltzmann distribution. Thermal fluctuations generally affect...

@ACuriousMind Ooooooooooooooooooooooooohhhhhhhh.... I seeeeeeeeee...

@MoreAnonymous Right above the table: "The expressions given below are for systems that are close to equilibrium" [emphasis mine]
stop quoting non-equilibrium expressions :P

11:13 AM
The question of stationary states in statistical mechanics is a big deal, technically you can't set up a stationary state for a statistical system, you unavoidably need to use density matrices

@ACuriousMind Thank you. It means the value of torque does not mean body have same rotational acceleration. This is what i was confused.

@bolbteppa the density matrix in equilibrium is a stationary state
it's just not a pure state
that is evident when you write the thermal state as an exponential in the Hamiltonian $\rho\propto\mathrm{e}^{\beta H}$, because then $[\rho,H] = 0$ since operators commute with functions of themselves

@ACuriousMind isnt that because its fluctuating? I mean it says clearly: "In statistical mechanics, thermal fluctuations are random deviations of a system from its average state, that occur in a system at equilibrium."

@MoreAnonymous the intro of the article does not apply to that latter section
if you read carefully, the intro also says the "control variables" like $V$ do not fluctuate

I've never seen it called a stationary state, but if you mean 'stationary state' in that sense then maybe it makes sense

11:16 AM
and then your table at the end gives non-zero fluctuations for $V$
it's about systems close to equilbrium, not systems in equilibrium

@ACuriousMind Yes but we were talking about $P$

But in terms of the multi-particle system being described as a stationary state solution of the time-independent Schrodinger equation, technically one can't even set up such a problem even if one tries (which one should really try to do) even for a closed isolated system, for important reasons

@MoreAnonymous pressure is a function of the control variables in the mic.can. ensemble
it's just the value of $\partial_V S$
@bolbteppa this pressure isn't external forces, it's the pressure your closed isolated system would exert on its enclosure

Wait no, more complicated...

@ACuriousMind I thought it was particle number $N$, $E$ energy and $V$ volume

11:21 AM
@MoreAnonymous sure, and entropy $S$ is a function of those, so $p = \partial_V S(N,E,V)$ is a function of the control variables, too

Also why does wiki write pressure as the average of a quantity?
In statistical mechanics, the microcanonical ensemble is a statistical ensemble that represents the possible states of a mechanical system whose total energy is exactly specified. The system is assumed to be isolated in the sense that it cannot exchange energy or particles with its environment, so that (by conservation of energy) the energy of the system does not change with time. The primary macroscopic variables of the microcanonical ensemble are the total number of particles in the system (symbol: N), the system's volume (symbol: V), as well as the total energy in the system (symbol: E). Each...
Yes entropy fluctates too (at least thats what I thought)

how would it? An equilibrium system sits at maximal entropy, as you yourself said
if it "fluctuated away" from that, it would no longer be in equilibrium
I think you're just confusing a setting where you think about systems close to equilibrium with actual equilibrium
in many use cases you'll have to think about the former, so that's not inherently wrong or anything, but you need to know that you're not talking about actual equilibrium

Hm
If I have some function $\phi$ and I have a set of points $(x_i)$ where we have the value of the function at that point $(\phi_i)$
And given some bound on the derivatives of $\phi$
Can we define a subset of the function space where that function may be

@ACuriousMind Can the experimentalist set up a system in thermal equilibrium of one near thermal equilibrium ?

depends on the experimentalist :P

11:28 AM
@ACuriousMind I was hoping for a serrious answer :P

@Slereah sounds like a job for Sobolev norms (::runs away screaming in horror::)
@MoreAnonymous that is a serious answer

@ACuriousMind I'm pretty sure it is, but is there a specific name for this

sometimes equilibrium is simple to set up, sometimes not, depends on your system

@ACuriousMind A gas?

I'm pretty sure the answer I'm looking for is like "bla bla bla Sobolev space bla bla bla compact open topology"

11:29 AM
@MoreAnonymous you're thinking too general

I need a little ball in $H^k(\mathbb{R}^n)$ or something

If you have a function you know it's values everywhere right, if you don't have the function and just have information at a finite number of points and a bound on the derivatives at those points, plenty of functions could satisfy those conditions no

like, suppose I want to equilibrate some block of material at constant temperature $T_0$

@ACuriousMind A gas in a box?
with a piston

the first issue is whether I have anything that can supply that constant temperature
@MoreAnonymous no, what I mean is that this depends very much on your specific experimental setup

11:31 AM
@ACuriousMind hmmm ... I see
Lemme assimilate todays discussion

i.e. what sort of instruments you have and how "far from equilibrium" you're still willing to count as equilibrium
I mean, equilibrium is a point in thermodynamic state space, you'll never hit that point, just like you'll never perfectly place a pebble at the top of a smooth dome
the question is whether for your purposes it is useful to model the system as being in equilibrium, or whether you're interested in phenomena that make the "close to equilibrium" approach more relevant

I see .. thanks
@ACuriousMind also can I ask for your opinion on something I wrote (it requires statistical mechanics and GR knowledge)? Its like 6-7 latex pages?
(no worries if your not up for it)

@MoreAnonymous sorry, I'm not reading 6 pages

@ACuriousMind No worries
Also has anyone read tristan needham differential geometry? Its quite nice and visual
So far

11:47 AM
The complex one is great to skim now and again

@bolbteppa I'd recommend the differential geometry one as well :D

1 hour later…
1:08 PM
Typography question
What should I call a temporal function
$t$ is a bit overloaded already, $T$ feels a bit weird, $\tau$ is for my proper time, $\mathcal{T}$ also looks weird, $\mathfrak{t}$ is more of an algebra thing
I could call it $\theta$ but that's already angles and the solder form
Maybe I could call it $\pi_t$ since I am gonna treat it like a bundle anyway

1:31 PM
$t_{temp}$

2:29 PM
oh great
Another conference monograph that's impossible to find
It is technically here, but I'm not sure I ever got research gate to work for anything
hm, that one does download without asking for anything else
Praise Urania

It's an accident when you find anything useful on there

ResearchGates is at the bottom of the list for my "places to find a paper"
It's barely above viXra
The hierarchy is something like ArXiv > Journals > Euclid Project > Personal sites from physicists > Internet Archive > those weird NASA papers > the INA archives > the very bad old timey journal websites for some french journals you sometimes find > Research Gate > Vixra

@ACuriousMind

There's been barely any unified theories on there the past few months :\

2:44 PM
@JackRod yes?
@Slereah just use $\vartheta$ :P

hi is it real?
some one shared with me

Nothing is real

I don't know how you expect me to figure that out

use your mind powers

I don't know anyone at CERN and I'm not in academia, I have no secret knowledge of what sort of things CERN might offer or not

2:46 PM
actually these kind of queries may be raised by someone
on the site or in chat and you are mostly present in the room+you are phy interested guy

loong is from nuclear science?? @ACuriousMind

It's listed on there

I'm sorry, I don't understand what you want from me

@ACuriousMind nothing I was just replying for the last comment you made
@bolbteppa thanks
first time when I heard about cern was in 2015 about there discovery of a new particle

2:52 PM

It does look interesting
I will give it a look, thanks

"If you are applying in the fields of experimental or theoretical physics, you must hold a PhD (or be about to finish your thesis)."...probably not good for you, right?

That's only for the 'senior' one

CERN researchers find flaw in faster-than-light measurement

but Slereah has an MSc iirc, so wouldn't that disqualify him from the junior profile?

2:56 PM
1 min ago, by Jack Rod
CERN researchers find flaw in faster-than-light measurement

"a “Senior Fellow” profile, with a MSc level diploma or PhD and between 4 and 10 years’ relevant experience after completion of your MSc."
I don't have 4 years of experience after

they could not proved it

junior could work, though

there result were wrong

Not 100% sure that this program applies to me but I can give it a shot
It is also a very generous salary that they offer

3:08 PM
Yeah now I'm a bit confused on the wording of it, I'm not sure what's going on with this thing and when you click apply it gives no better info, I'd say this is the kind of thing one could enquire about and find out what's going on with this 'offering projects' aspect of it and the msc vs bsc aspect of it, e.g. what are they offering to say biology bsc's that a tp msc is excluded from

Ah well
We'll see
plus it's probably not a bad thing for me to train to write a cover letter

I think this is just a few examples of how it goes

2 hours later…
5:18 PM
For anyone interested, because you can write $C M C^{-1} = M^*$ with $C$ also being an element of the representation (along with the property $C=-C^{-1}$), the quantity $(C\psi^*)_\alpha$ transforms exactly like $\psi_\alpha$. Therefore, you can relate those two representations together allowing you to write $\epsilon_{\alpha \beta} (C \psi^*)^\alpha \psi^\beta$ and finding that $\psi^1 \psi^{*1} +\psi^2 \psi^{*2}$ is invariant.
In this way $\epsilon_{\alpha \beta}$ transforms as $V^* \times V^*$

5:44 PM
Hi all

6:03 PM
Question: if g = -9.8, by the second derivative test, wouldn't that mean the distance is always decreasing?
What does that even mean?

It means that things fall down

OK, but what about a ball on a straight plane?
It has no change vertically
g is still -9.8, but $d_y$ doesn't decrease.

If a ball is on a straight plane, there is an equivalent reaction force in the opposite direction
The actual acceleration is therefore zero

Ah
Thanks

Also keep in mind that the second derivative test is at critical points
You can throw a ball up, but it will only start going down once its speed reaches zero
at the apex of the trajectory

6:16 PM
Why do the Poisson brackets encode in a classical theory?
Is it the classical equivalent of the time evolution for Heisenberg fields in quantum theory?

it's the other way around, the quantum commutator is the quantum equivalent ;)
but yes, the Poisson bracket with the Hamiltonian encodes the time evolution of observables: $\dot{f} = \{f,H\}$

I'm just reading the part of Dirac's book and it's crazy lol

I've been saying for years in here that I wished people learned classical Hamiltonian dynamics properly before learning QM
because the way it is often taught people often get very confused about what's quantum and what's just Hamiltonian mechanics

So the quantum Poisson Brackets just emerge from preserving the order of the dynamical variables i.e. converting them to operators?

I don't know what a quantum Poisson Bracket is :P
"naive" quantization is "replace $\{-,-\}$ by $\mathrm{i}[-,-]$"
it doesn't really work, but it works often enough

6:29 PM
It's what Dirac calls $\imath [ -, -]$ lol
He even says it doesn't always work

yeah, that's just the commutator

It's amazing how his book still holds up

@DIRAC1930 you can show that is only true for the $\mathrm{SU}(2)$ subgroup of $\mathrm{SL}(2,\mathbb{C})$

@bolbteppa Yeah, with $SL(2,\mathbb{C})$ in general it won't work

Dirac has some crazy argument for why Poisson brackets become commutators, Landau has a very simple argument in chapter 2 (in a footnote...)

6:39 PM
Does the same thing happen with all symmetry groups?

does what happen?

I'm not sure entirely
how to even ask the question
Elements of representations turning to operators doesn't really change anything does it because the elements were matrix operators to begin with right?

I'm not sure what you mean
classically, you don't necessarily have "representations" in the sense of linear algebra - the phase space does not (need to) have a vector space structure

If I have a classical field, it will always transform under the Lorentz group right?

ah, we're suddenly doing field theory, okay :P
sure, the fields all take values in linear representations of your symmetry groups, and these representations will be compatible with the representations on the quantum space of states just like the reps of the Lorentz group are

6:51 PM
So nothing changes except the normal representation becomes the projective representation?

depends on what you mean by "nothing changes"
in the classical theory, you just have the representation on the target space of the fields - finite-dimensional, no requirement to be unitary
quantumly, you still have the representation on the target space, but you also have the unitary representation on the space of states
and you always have a compatibility condition like $\rho_\text{cl}(g)\phi = U(g)\phi U(g)^\dagger$ that says these two representations play nice with each other

For every $\rho$ there will only be one $U$ right?

depends on what you mean by "one" (I know I'm being annoying :P)
there is "one" representations on the quantum space of states for each of your symmetry groups, but it doesn't need to be irreducible

I don't see why we look for an exact explicit form of the unitary representations. Won't it just have all the exact information that $\rho$ has because it's just a representation of that?

in a way, yes, but if you only know $\rho$ you don't know how the states transform in isolation
If you know $U$, then you can just write $\lvert \psi\rangle\mapsto U(g)\lvert \psi\rangle$
if you only know $\rho$, then you need to first figure out how to write $\lvert \psi\rangle$ as being created from the invariant vacuum like $\lvert \psi\rangle = \phi(x)\lvert 0\rangle$, transform that to $(\rho(g)\phi(x))\lvert \psi\rangle$ and then figure out what the latter means as a state
and if $\lvert \psi\rangle$ cannot be so simply expressed in terms of the action of $\phi$ on the vacuum, this is very cumbersome/intractable

7:01 PM
Can a spinor take any values for it's components? i.e. could I have $\psi^1 =100$, $\psi^2 = i1000$ or something?

I don't see what that has to do with anything above, but sure
it's just a vector (in the linear algebra sense, not the "transforms like a vector" physics sense)

So why isnt a vector of $SO(2)$ the same as a spinor of $SU(2)$?

I don't understand the question at all.
What's a "vector of SO(2)"?
what does it mean for it to be "the same" as a spinor?

Well there doesn't seem to be any constraints on what the components of each of them can be
Just the way they transform

why "just"?
when you're talking about vectors, spinors and whatever, their transformation behaviour is what you're interested in!

7:05 PM
But don't we have all the information just from the elements of the representation?

once again, I don't know what you mean :P

If I just had a 2 component matrix on it's own, theres no way for you to tell if it's a spinor or a vector

correct, that's because it "is" neither

It seems arbitrary what group it transforms under
It seems like a bookkeeping device

you need to think about vectors in a representation always as "a vector with its transformation information attached"
if you just give me a list of numbers, that's just not enough information

7:10 PM
So it is kind of a bookkeeping device?

what is?

Nevermind
This is a point that's confused me since I started learning representation theory

7:32 PM
@ACuriousMind Hi man, I wanted to ask you something in statistical mechanics ofc

7:53 PM
Can classical theories have no internal symmetries?

@ACuriousMind What of $\{ e \}$
Also $\mathbb{Z^2}$, quite possibly