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12:56 AM
@Mr.Feynman That would be both awkward and inconvenient. The coupling constant should be the fine structure constant and its cognates. The electric charge should be whatever is quantised and conserved, and Noether charge is far more suitable. I swept all the equations clean and found out that the only constant that I have appear in EM is the fine structure constant. speed of light and electric charge went to the unit choice.
 
 
2 hours later…
2:36 AM
@ACuriousMind mindblown. It is fascinating. Also, in Joking, Feynman recounted that he gave up learning Japanese because of all these confusing variations, so it would be fun to ask him what he thought of old English Herself. Am also sad that the modernist commie revolution wiped away most of these nuances in modern spoken Chinese. (It still exists when we want to give some respectful flair, though, miehehehe)
@Mr.Feynman Reminds of the Japanese seiza. Oh, miao miao is such an otaku
weeaboo
I had a prof who insisted we call him by his given name. It sounded sooooo rude. This is where pronouncing it in a cute way is the appropriate thing to do miaaaaa
 
2:59 AM
@Odestheory12 It is better now, but in a sense, we can only know how to help you if we solve the problem itself. The result that you are trying to derive is kinda obvious by physics intuition, but to get the precise mathematical reason why, we would have to think about it. You would have a far better chance of getting it solved by asking in here than on the site.
 
3:10 AM
@SillyGoose i mean the extra term is U(1) invariant so yes
 
meow
 
3:24 AM
M I A O
 
3:48 AM
hermitian conjugating commutes with partial trace, right? to clarify, $\rho$ is not necessarily hermitian above
 
Why would you have a density operator that isn't Hermitian?
 
i decomposed a density matrix into a sum of its matrix components
to let me more easily compute this quantity
 
@SillyGoose Kaluza-Klein theory unites GR with electromagnetism by introducing an extra compactified space dimension. In that theory, momentum in the extra dimension equates to electric charge. So the U(1) symmetry of charge becomes a literal spatial U(1) symmetry.
 
@SillyGoose This is irrelevant to what I was asking you. As long as you have $\rho=\sum_ip_i\left|\psi_i\right>\left<\psi_i\right|$ then it has to be Hermitian.
 
4:07 AM
Yes the entire rho will still be hermitian but i am trying to find relationships between the parts i decomposed the matrix into
and so the rho is a sum of partial traces of non hermitian matrices some terms of which i am trying to establish a relationship with other terms
 
 
2 hours later…
123
6:24 AM
Hello Everyone...
 
6:46 AM
@naturallyInconsistent I learned a new word :D
 
@Mr.Feynman sit that way today~
 
What is today?
I had a bad night
1.5h of fighting against a mosquito
Little bastard was massless
 
a point that has no size
 
My disappointment is Lebesgue-immeasurable and my day is ruined
 
7:17 AM
A Dirac delta distribution whose position is wildly varying is very day-ruining indeed.
 
7:46 AM
It was massless
Bro was lightspeed 💀
 
123
@naturallyInconsistent Hello... Why N3L tells about forces are equal and opposite not pressure or stress?
 
@123 Sigh, don't tag meow
 
123
Okay . . Why you always get angry..
 
This tagging, if I do not open it fast enough, will go to the message queue on the main site. It is very annoying. I use that queue for finding stuff I want to pay attention to, and this is something I do not want to pay attention to.
 
123
I tagged you because you gave me most satisfactory answers.
 
7:53 AM
And I told you that stress as a concept came long after pressure. The concept of pressure came after forces. Forces is simpler to understand than either of pressure or stress. Why is it a problem to consider forces before pressure or stress?
 
123
I think your knowledge is helpful to peoples.
@naturallyInconsistent Aaaah... I see
It means pressure and stresses are derived from force. And force is the basic quantity.
 
When you have two bodies colliding, it is not necessary that the collision be a pressure. Pressure means that it is the same pushing in all directions. For all you know, you think that the collision is pushing perpendicular to the area of contact. That is a stress of some form, but it might not be pressure.
In fact, when you throw a ball and make it spin very quickly, when it collides with a wall or floor, it can bounce back differently. The difference from just dropping a ball shows that there can be non-perpendicular stresses. It is not just pressure that you have to deal with.
 
123
Yes this is what i am thinking. Because during collision stresses between objects should not be equal. because both bodies can be different material , shape, etc..
 
But if you think about it from forces, you can argue the same thing without caring which kind of stress it is.
 
123
Oooh i see..
 
7:57 AM
No, during a collision the stresses must be the same. You agreed that the area is the same and the forces are the same, so the stresses must also be the same. The fact of the matter is that during the collision, we DON'T CARE what the area is.
And the reason why we don't care, is that, underneath, the important quantity we actually care about, is how it will move later on, i.e. its momentum. We don't really even care about forces.
 
123
It means during collision forces, pressure and stresses are same for both bodies?
 
In fact, N3L needing to be written down, is proof that forces are also the wrong way to look at it. No other conservation law is written down like N3L does, because it is just mathematically silly to do that. Just look carefully at momentum, and you can work out mechanics properly.
 
123
@naturallyInconsistent Oooooh okay..
 
Also, the forces during a collision would be a sudden spike. If we don't know how long it takes for the collision, and how specifically the forces spike during the collision, we cannot do much. However, any collision has a well-defined momentum transfer, and we only need that to get the long-time behaviour handled correctly. This is what we actually really care about.
Why would you want to pull your hair out with pressures and stresses when simple arguments will work, I do not know, and it is clear with your many days asking this here and nobody else answering, that nobody else wants to touch this question either.
 
123
It means in collision we care about how objects move after collision not the effect of forces on the bodies like pressure or stress. If we deal the effect of force on bodies it will be study in properties of matter.
 
8:03 AM
@123 It is about making things simple. Of course, if you want to study how the body deforms and turns and what other horrible things, we would then need to study that. But beginners should only deal with the centre of mass moving here and there, and the centre of mass only ever responds to the momentum transfer. It does not care about the particular details. It is VERY EASY to work with. Physics is difficult enough. Do easy stuff first.
 
Let's leave that to engineers
 
123
Aaah.. Thanks for your answer.
 
I'm fine with rigid bodies :P
 
It is not that what I am saying is new or insightful or what. If you study things properly, doing the easy stuff before moving on, you would have understood these things yourself. Most of the other people here, they know what I am saying is true. It is just that they know that trying to not get into a long-winded discussion.
@Mr.Feynman Special Theory of Relativity is not fine with rigid bodies
 
123
@Mr.Feynman I am also working with lab equipment's since 15years. troubleshooting maintenance and application.
 
8:08 AM
@naturallyInconsistent Oh, me knows that
Cool lab @123
 
123
I am working on these equipment's. Rheometers, density meter, refractometer, polarimeters, viscometers, ion chromatography, HPLC, GC, FTIR, UV-Vis, etc..
 
@Mr.Feynman rigid bodies between legs, however, miehehehe
 
123
@Mr.Feynman It is a very very huge lab with too many equipment's .
 
@123 How are you using these things and yet asking basic questions on Newtonian mechanics?
 
123
@naturallyInconsistent I am PostGraduate in Industrial Chemistry since 8 years back.
But i am having interest in physics since 20years when i was in 10th standard. Physics always bothered me in understanding the basic ideas. And we don't have such a good teachers in my country. Also there is no satisfactory explanations available in youtube, quora and stackexchange as well as in books. That's why i asked question here.
 
8:13 AM
@123 does that mean that you are familiar with quantum theory?
 
123
@naturallyInconsistent Yes i am familiar with QM and QFT. But not in terms of math directly. but i understand the ideas. My study is going on . Hopefully i will reach to that fully conceptual in 3 to 5 years.
 
@123 No. While I have my own annoyances with how things are taught, there are still many satisfactory explanations for the basic stuff you are asking. It is just that you have a pathological approach to learning stuff, and that is blocking you from seeing the answers for what they are.
@123 That's good. I'm more willing to help you if you had told us your situation earlier.
 
123
I also animate physics and math problems in softwares.
@naturallyInconsistent Thank you for your support and this community as always.
These days i am in Austria for the research purpose. I am heavily engaged but my curiosity engaged my with this community.
pls see the link above i have created projectile motion simulation in polar and cartesian coordinates.
 
@123 my friend is being bullied in Leoben and in Vienna. I'm not very happy with Austrians.
 
123
8:32 AM
@naturallyInconsistent :P . But i always found Austrians are very humble. Hopefully i will not face that
 
8:53 AM
@Mr.Feynman is it non measurable or it has infinite measure?
 
Unfortunately the quote goes "immeasurable"
It's like a joke I used back in the day when I was studying analysis. "My rage has no limits"
It would be more appropriate to say it diverges but it would break the joke
 
9:24 AM
Why not "My rage is unbounded"
 
123
Pls confirm that during collision between two objects forces, pressure and stress are all always equal?
 
We cannot be sure that the pressure or stresses will be equal. The sum or decomposition of the same force over same area can be into different combinations of pressure and stress.
 
123
Also N2L can also be explained in a situation for example i have system of three noninteracting bodies and external force 1 act only on body 1 and external force 2 act on body 1 and body 2 and there is no external force act on body 3. These situation can also handle by N2L?
 
Under Newtonian mechanics N2L is required to be true always.
 
123
I asked this question because knk always talk about interacting system when applying N2L and this system as a whole act by external force at COM.
@naturallyInconsistent Great answer. Now i understand it better.
 
9:31 AM
K&K is correctly emphasising that if you have a motion problem, you should be considering N2L.
 
123
One last question. knk said motion can always be decompose into linear and rotational motion. And i saw the equations in books where angular momentum of COM can be decompose into two parts. First part is angular momentum of COM about some fixed intertial frame (which he says orbital angular momentum) and second part is angular momentum about its COM (which he says spin angular momentum).
 
@Mr.Feynman ;)
 
123
what i expect when motion can be describe by linear and rotational . The linear part should only have linear momentum as we study in linear dynamics but the equation is angular momentum of COM not linear momentum of COM about some fixed inertial frame. Is there any case at which we describe motion by linear momentum (for linear motion) + angular momentum (for rotational motion)
 
@Relativisticcucumber B A H
 
@123 tagging when asking a question is also just not really allowed in this chatroom as far as i understand
@naturallyInconsistent BAH
 
9:45 AM
@Relativisticcucumber you people are dirty minded
I like that
 
@123 Motion can always be decomposed into linear, rotational, and whatever is leftover. If you say orbital angular momentum and spin angular momentum, you will disagree with quantum theory, so avoid it unless you understand it to be in agreement with quantum theory
@Mr.Feynman smirks
@123 What do you mean "Is there any case"? That is the usual way we do or think about it.
 
10:01 AM
@naturallyInconsistent Hello, thanks. What's the physics intuition behind it?
 
10:15 AM
@Odestheory12 Consider just one sphere. Then it will definitely roll down. Consider two spheres but the bottom one is very heavy and the top one is very light. Then the light one can never provide sufficient friction at the sphere-sphere contact point to pull the bottom sphere back. So it will also definitely roll down. So the only possible case in which it doesnt roll is the $m_1<m_2$ even without calculations.
 
123
@Relativisticcucumber Now i also understand... :-)
 
Oh wait, is there friction between the spheres?...
 
@Odestheory12 without that you definitely cannot solve the problem
 
I was assuming there is not since the problem didnt state it
 
123
But the community is always supportive and helpful
 
10:17 AM
Well that was it, that's what i was missing. Sigh.
With that condition we can prove it mathematically with what I wrote in the post.
 
The person who set the question is thinking what I am thinking, purely intuitively, and just didn't even bother to set the details correctly.
 
I see, it wasn't meant to be solved analytically
 
If you want to calculate the thing in the easiest way possible, pick your origin to be on the inclined plane, but also the tangent from both spheres touching. That unique point will be the most convenient.
And obviously decompose your problem's forces along and perpendicular to the inclined plane
And you are supposed to use torques.
It is not just force balance
 
Uhm I see. In the site I wrote the relation between masses as:
$$ m_1 = m_2\frac{\mu_2\cos{\theta}-\sin{\theta}}{\sin{\theta}-\mu_1\cos{\theta}} $$
If $m_1 < m_2$ then $${\mu_2\cos{\theta}-\sin{\theta}}<{\sin{\theta}-\mu_1\cos{\theta}} \to (\mu_2+\mu_1)\cos{\theta} < 2\sin{\theta} < 2$$
Since $(\mu_1 + \mu_2 )\cos{\theta}< \mu_1 + \mu_2 $ we have that $\mu_1 + \mu_2 < 2$ which is true
 
 
5 hours later…
3:19 PM
@SirCumference thanks
@SirCumference i have trouble manipulating tensors
 
 
3 hours later…
6:12 PM
Dies anyone have an introduction resource to topological qfts
 
6:41 PM
I am looking at Atiyah's paper and so far it is sounding like a modified homology/cohomology theory. is my impression a problematic way of thinking about TQFTs?
 
7:03 PM
what are the first principles that boltzmann statistics is based on?
like the fundamental assertions
 
@SillyGoose Atiyah explicitly discusses the similarities and differences with ordinary (co)homology theories in the paper, what exactly is the question?
 
@ACuriousMind i am wondering if that is just a heuristic or if it is just like another homology/cohomology theory
 
I mean...it isn't a (co)homology theory
 
@Obliv what do you mean by boltzmann statistics
 
it's just a functor on cobordisms that looks a bit like (co)homology if you squint
 
7:11 PM
okay i see
 
Can anyone help me understand bound and stationary states and also whether we can have time dependent bound states?
What is the difference between bound states and stationary states?
Also if we consider a confined system in a region where a time dependent potential is present, would this mean, that the solutions to the TDSE, are time dependent bound states ?
 
when the Hamiltonian is explicitly time-independent, a "stationary state" is literally just an eigenstate of the Hamiltonian. It is called a "stationary state" because it evolves under the Hamiltonian by picking up a phase. Hence, if you were only evolving one such eigenstate of the Hamiltonian, it would not be evolving at all as time moves forward (since states that only differ by a global phase are physically equivalent).
 
@SillyGoose mostly referring to this but idk what the context is
I don't have to worry about what they mean by quantum effects etc because I haven't studied non-classical physics yet but yeah besides "every microstate is equally probable", idk what other assumptions are used
also what "distribution" means
but i'll keep reading
Actually, I think I have the gist of it.
 
7:27 PM
@SillyGoose I am familiar with what you are saying. Basically multiplying with a complex time dependent exponential the solutions of the TISE, does not bring any physical change, as the norm is conserved. But what about the rest of the things I asked
 
this formula looks incorrect :P but i might not be understanding it
 
what do these terms represent in the generalized distribution $$Pr(\omega)\propto \text{exp}\left[\sum\limits_{\eta=1}^n\frac{X_{\eta}x_{\eta}^{(\omega)}}{k_BT}-\frac{E^{(\omega)}}{k_BT}\right]$$
 
what is $X_\eta x_\eta$?
 
I have no clue, the wiki page doesn't elaborate.. and googling the generalised boltzmann dist. doesn't really return much lol
 
if it is chemical potential for a particle times the number of such particles, then that statement is describing the grand canonical ensemble
 
7:30 PM
$\omega$ is a particular state?
 
$\omega$ labels a particular microstate specified by number of particles $n_\omega$ with energy $E_\omega$
im not really sure about this maxwell-boltzmann stuff, but it literally just looks like using the grand canonical/canonical ensemble to compute the expectation value of number of particles that have certain energies
if u are familiar with the grand and canonical ensembles
 
reading it now, thank you :)
 
the microcanonical, canonical, and grand canonical ensembles all are derived from the basic postulates of statistical mechanics + specifying how you want to specify your macrostate: $(N, V, E), \ (N, V, T), \ (\mu, V, T)$, respectively. How you specify your macrostate determines what possible microstates your system can actually occupy and with what probability. The Maxwell-Boltzmann distribution tells you how these microstates are distributed (classically) for the grand canonical ensemble.
 
I have to relearn thermodynamic equilibrium lol, I thought it meant state of maximum entropy
but it's not quite that.
 
It should
a system in equilibrium
is characterized by maximal entropy
the state
 
7:36 PM
this is my idiosyncratic definition :P
 
well like using the thermodynamic identity $dU = TdS - PdV + \mu dN$ max entropy is dependent on other state variables
macrocanonical ensemble depends on more than just energy i think unlike in the specific boltzmann distribution
in my book derived for 1 atom and a reservoir, $dN = 0$ and $PdV$ is negligible compared to $dU$
(not explained ofc, just says refer to appendix A for the QM stuff lol)
 
you can look at pathria's text on statistical mechanics if you would like a detailed derivation of each of the three textbook ensembles
 
what is there besides micro and macro?
I just wanna derive the one that is most generalized for the basic thermodynamic identity
if that's the grand/macrocanonical one I'll do that
 
Grand canonical
which additionally to the energy it exchanges matter as well with the environment
in difference from the canonical ensemble that only exchanges energy
 
I think we only did canonical in my book then
 
7:46 PM
so microcanonical = isolated system. No exchange of energy or mass.
Canonical = closed. It exchanges heat/energy
Grand canonical = open. it exchanges heat and mass
I don't think there's a generalized one
it depends on the case
It depends what is being exchanged
 
seems like microcanonical one is used for relatively small systems (shocker)
 
You can have more than those three ensembles. You can exchange volume for instance in which the quantity that is equilibrized is pressure
 
i guess you need QM for that one
 
I don't know whether it's possible to transition, mathematically, from the solutions of the grand canonical ensmble to those of the mce, since they represent two different situations. But perhaps I am wrong, and maybe someone who knows more. i,e ACmind can help
 
@Obliv it is used for isolated systems in general
 
7:49 PM
@SillyGoose that's kinda cool.. so the individual volumes of the particles change the pressure in the whole system
 
What I personally don't know, is how the phase space of those 3 ensembles differ from each other. What is a "visible" difference between the three.
In my course we only considered the phase space of MCE and that was it
 
Well you only really need to know the microcanonical ensemble to do all others
that is how you can derive the others
 
phase space?
 
because a system coupled to a temperature reservoir if taken as a whole is just an isolated system
well if you are saying that you described in general the phase space of an isolated system, then tou would be able to describe the phase space of what i just described
 
you mean system + environment = one big isolated system ?
 
7:52 PM
yes
 
then, would it be correct to say that in the phase space of CE or GCE, you'd have sources and sinks ?
 
if they're all describing the system at thermal equilibrium, they should all describe the same phase space afaik
 
This question is related, to the idea that the phase space of a MCE, can be interpret as an incompressible fluid. @SillyGoose
 
Im not too familiar with the phase space picture
 
@imbaF Liouville's theorem is probably related.
 
7:54 PM
i ignored the quasi classical business in pathria :P
 
@Obliv I am not sure, as in GCE, locally you are not having conservation of mass
overall yes, but locally I don't know
of course, I am always considering equilibrium state
 
good point
but overall, I think they should be the same/conserve all the same stuff
otherwise that's weird for classical theories
 
Because one more thing in statistical mechanics, is the volume one consideres for when a system is in equilibrium. FOr example in MCE, at equilibrium, the volume, which includes all the points in phase space, which are all the microstates characterizing the macrostate, which happens to be the state of equilibirum, the volume, while it doesn't change in value, it does change in shape
In equilibrium always
I don't know how this volume behaves prior to the system being in equilibrium, in either of the ensembles
 
@imbAF we would reset N to be constant and E to be constant in which case we would obtain the equiprobability distribution, ie microcanonical ensemble.
(If we started with the grand canonical ensemble for a single particle species)
 
Oh yeah
the volume changes in shape, because a microstate changes over time. While the energy it represents does not, the position and momentum of the particles of the system does change, so this change must be showed in the phase space. This is showed, by having a point in the phase space of a system, which represents a microstate, moving/traveling in phase space, implying different
positions and momenta for the constituents. Now, if you extrapolate this behavior for all the points included inside the volume, that volume will change shape, but not its value
@Obliv
 
8:07 PM
@imbAF hmm, that's weird I gotta let that one sit lol.
yeah that makes sense.
I wonder what pressure is then, if it depends on volumes occupied by individual particles.
like if it depends on the shape of the volume or just the total volume
i mean locally it should depend on the direction/shape (i.e measuring pressure on one microscopic section of a wall)
idunno
 
pressure?
 
yeah if we take pressure to be the equilibrized quantity like sillygoose mentioned
exchanging volume between particles instead of energy for instance
 
Ok, pressure from what I remember, is the result of the collision of the particles with the boundary
boundary of the regions in which they are confined
When they are not, perhaps is the result of the collision of particles
 
maybe it's some divergence theorem thingy
 
For some reason 'bos has given up his blog, mostly given up physics debates/questions online, and is now mainly posting about Czech stuff on quora...
 
8:15 PM
Maybe
@Obliv The pressure can be calculated statistically using the virial theorem or obtained from the equation of state of the system.
 
@imbaF the issue is theres like no symmetry whatsoever lol in both the div. and the flux on the surface area.. like change in volume isn't constant
 
in all 3 ensembles the volume, physical volume, is kept constant
 
There is only one density matrix in quantum mechanics, i.e. the canonical density matrix, it will behave differently in the case that particle number is fixed vs when it is variable ('grand canonical'). However, when we're talking about a closed system, things are so simple that we can approximately describe the density matrix with a simpler distribution, the microcanonical distribution, that's all that's going on
 
9:05 PM
@Mr.Feynman Oh wow, I just noticed you changed your name back
 
did he change it again lol
@SirCumference can I call you Cumference, for short? Or are we not that well acquainted yet..
 
well "cumference" does sound a little weird lol
I guess SC would avoid any strange innuendos
 
lol jk :P
 
i might've mentioned it here but i have had a friend completely misunderstand what my name was supposed to mean lol
granted english wasn't his first language
 
Lmao, I didn't know that.
that was your original alias afaik too
true OG
 
9:13 PM
I think I was originally Pies when I first came here
changed it for the pun but kept the same profile pic I guess
 
oh that would explain the pie
always looked like a delicious pie
 
i suppose it does still work as a $\pi$ joke for circumference
might be stretching it a bit
 

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