@psa iirc, the issue is that the thermodynamic identity $dU=TdS-PdV$ is for systems undergoing microscopic changes in its state variables while in thermal equilibrium (or quasi-equilibrium? i forgete)
but the free expansion of a gas into vacuum isn't an equilibrium process
also, I see no reason to take the box as the volume. it's the gas which expands and the gas whose internal energy you're interested in
(you should still not get any work, since there's no pressure to resist said expansion)
(right, free expansion isn't quasi-static. that's what I was forgetting)
the basic issue is that, since it's not quasi-static, you can have an increase in entropy without having heat transfer
free expansion of a gas is an adiabatic process (no heat flow) but not reversible
@DarkRunner Is there anything in particular that's confusing? It's mostly just talking about acceleration and $\mathbf{T}, \mathbf{N}, \mathbf{B}$ as a basis for the space that the particle is moving in.
That's a very math-y way of talking about kinematics though. It's usually only used in vector calculus classes.
I think I'm understanding! Basically, we have 3 vectors that describe the motion of any object in 3d space: The Tangent and Normal vector, of course, but also this new "Binormal" vector
@DarkRunner yeah. you of course need 3 vectors to form a basis of $\mathbb{R}^3$. the binormal vector serves that purpose. I think the physical intuition as far as what it describes is torsion.
so you've got velocity which is related to the tangent vector, curvature which is related to acceleration (and the normal vector) and torsion which is like the "3d-ness" of the curve.
@Knight I...don't really consider 26 "small". It's the "intended" age at which one would be expected to hold a university degree or otherwise have been working for a few years around here.
And I've been self-dependent ever since I've moved out from my parents years ago! :P
@ACuriousMindi don't know the exact date so thought to wish you earlier. If I go by the Hindu tradition then my DOB is not fixed in Gregorian calender!
@Knight Federally funded scholarship, but even if that hadn't been there, I'd have received a guaranteed federal loan like every other German citizen whose parents earn less than a certain threshhold.
I'm having serious trouble understanding this. If a fuse burns out when too much current is flowing through it, then how does turning on all knobs to their full setting burn out the wall-plug?
Shouldn't turning on all knobs increase resistance as you're adding more components to the circuit? Can...
@JohanLiebert it's asking about the design of ovens. I did hesitate before voting to close, but ultimately I don't think it's a good question for the site.
@ACuriousMind I want to know if anyone who does Master’s degree 📜 become eligible of understanding all the researches going on or being done?
I mean if you look me, I cannot understand the discoveries in Quantum Theories because I don’t know Hilbert Space, I cannot understand the stuffs in General Relativity because I don’t know higher dimensional Tensor Analysis. So, does everyone become that much knowledgeable by the end of Masters so to understand almost anything in Science.
If you cannot tell about everyone, then please tell me about yourself. Is there anything which YOU cannot understand? If it is there then why you are unable to understand it.
@Knight There is vastly more material having been created and being created than you could ever even read, let aloneunderstand, and understanding one tiny subfield will not make understand an unrelated subfield much easier.
Subfield here is not "QM" or "GR", it's stuff like "low-temperature condensed matter theory", "high-energy particle accelerator physics", "quantum field lattice simulations", to name just three that are sort-of "QM" and there's dozens if not hundreds more
I wanna know, if you look me now it’s the intellectual incapacity that prevents me from understanding high level things. What happened at you people’s level?
and that course would have to be specific to the subfield I work on (namely, the theory of strong-field light-matter interaction in atoms and molecules)
if you had someone who did an MSc "in QM" but it was geared towards e.g. condensed matter, high-energy physics, quantum chaos, cold atoms, etc., then they wouldn't really be able to follow along a detailed discussion
there's just too much material to read up on
@Knight that type of information is in my profile
@Knight and I'm definitely not going to answer that.
@ACuriousMind Try to understand me, if I were to read some political philosophy book I might not be able to understand it because it requires so so much contemplation but I cannot understand GR because my mathematics and few things in between are absent. Reasons are different, what is the problems we get when we older at your age or Emilio’s
@Knight The problem stays the same, always: That in order to understand something you need to understand all its prerequisites.
There is no magical level of math and physics knowledge where "all research" suddenly becomes clear to you. Each subfield still has a mountain of prerequisite knowledge unique to it that you must work through.
Typically, if somebody who just completed an MSc that was specifically constructed to teach a given subfield (i.e. a field of study as specific as 'theory of quantum simulators using BECs in optical lattices), then you'd expect them to be able to follow along with material from any cutting-edge research in that field, so long as it's presented in a way that's general to the field
i.e., the sort of presentation you'd have in a conference on the topic
but they would still need one or two years of dedicated work to really understand enough of the details to start making new contributions
@EmilioPisanty Does that type of course given in your country to anyone who wants it? As you must have been aware that in my country you require money and reputation for getting that kind of goal oriented MS
@Knight It's not that you're asking irrelevant things, it's that you don't seem interested about the actual answers and you're pretty dismissive of the things I'm saying, so I don't see much point in continuing to say them.
@Knight You simply seem to reply to things that we're not actually saying (cf "old man", asking for specific prerequisites when you've been told it varies highly between fields).
If you don't listen, don't be surprised that people stop talking.
There is no magical level of math and physics knowledge where "all research" suddenly becomes clear to you. Each subfield still has a mountain of prerequisite knowledge unique to it that you must work through.
The prerequisites there are all the important work that's been done in that field since its inception.
You know same happened to me when I read Bertrand Russell’s “Introduction to Mathematical Philosophy”
ACM and Loong I need your advice in one thing. I was thinking of learning German because many important Physics and Mathematics Text are written in German and I don’t why but I feel like the English translation somewhat takes away the real feelings of the original author.
That may be, but I neither see why you'd need "feelings" for a physics text nor why you'd read Heisenberg instead of a modern exposition of QM except for historical curiosity.
A nice sentiment, but as an introduction I doubt the usefulness - usually our understanding will have evolved in the intervening years and the old account will of course not reflect how people today look at it.
I've heard that Dirac's book on QM holds up nicely but I've never heard the same about Heisenberg's, probably for a reason :P
The bar is closing and I have to change locations, so I'll post my question anyways. I'm reading a proof showing the tensor product representation of an orthogonal or Lorentz group $G$ with metric $g$ on antisymmetric 2nd rank tensors coincides with the adjoint representation on $\mathfrak{g}$. $L$ is defined in the usual way as the intertwiner between $\mathbb{R}^n$ and $\mathbb{R}^{n*}: v\mapsto g(v,\cdot),$ and the intertwiner $\phi:\Lambda^2\mathbb{R}^n\rightarrow\mathfrak{g}$ is defined as
@dsm I don't understand your definition of $\phi$. $\phi(T)$ is an element of some algebra $\mathfrak{g}$ - how am I supposed to apply that to a a pair of co-vectors ($v$ and $f$)?
@dsm given a Lie algebra $\mathfrak{g}$, you do not have in general that its elements are (1,1)-tensors on something. And even more, if you define $\mathfrak{so}(n)$ purely abstractly by its commutation relations, you also don't get that
You only get that if you realize that $\mathfrak{so}(n)$ "are" n-by-n matrices on $\mathbb{R}^n$ and therefore act as (1,1)-tensors on it
Hmm, how do you interpret the following sentence in my text before defining this intertwiner: "To define $\phi$ abstractly we interpret $\mathfrak{g}$ not as a space of matrices but rather as linear operators on $\mathbb{R}^n.$"
Whatever distinction that sentence is trying to make is lost on me, except maybe that they're saying they don't want to pick a basis, i.e. talk about matrix components in a particular basis.
@dsm Maybe this helps: $g(v,w) = L(w)(v)$ and $L(w)(\phi(T)v) = \phi(T)(v,L(w))$.
The first is "by definition of $L$" indeed, the second is more or less "by definition of a (1,1)-tensor", together these are all you need to get from your l.h.s. to the r.h.s.
Ok, I can see that. And to clarify, you are pulling out the $\phi (T)$ from linearity of the inner product, and then denoting the abstract dual with $L(w)$? Otherwise it's kind of weird notationally: $(v,L(w))=(v,(w,\cdot))$
No, I'm saying that applying the covector $L(w)$ to the vector $\phi(T)v$ is the same as applying the (1,1)-tensor $\phi(T)$ to the pair of vector/covector $(v,L(w))$
How could someone justify the assumption "The interaction takes place at the nuclear surface only" for Surface Delta Interaction schematic interaction in nuclear physics?
Because the vector $\phi(T)v$ is by definition the vector defined by $\phi(T)(v,\cdot)$ ($\cdot$ being a covector slot) and applying the covector $L(w)$ to that vector is by definition just slotting the $L(w)$ in there.
So changing the direction of your slots to represent a vector as a $(0,1)$ tensor, is the following chain correct: $L(w)(\phi (T)v)=L(w)\phi(T) (\cdot,v)=\phi (T)(L(w),v) = \phi (T)(v,L(w))$?
I see you're jumping to it because it's probably clear, but the steps aren't yet obvious to me
Ah, I see, oops. I guess I'm confused on how you immediately know that applying $L(w)$ to $\phi (T)v$ is the same as applying $\phi (T)$ to $(v,L(w)).$ Is that obvious?
For any vector $x$, we can think of it as such a map $x(\cdot)$. Now, applying a covector $y$ to $x$ as $y(x)$ is the same as $x(y)$, we're just switching the notation around.
Just ask your questions, if someone can and wants to answer them, they will (but I'll say right off the bat that I have no idea what "Newtonian World Index Algebra" is)
They're quoting me now and claiming that im laughably wrong, the quote they're ridiculing right now is "Astrophysics is a branch of physics, it is not particle physics nor is it theoretical physics, because they're all concerned with different things" lol
altho tbh it does seem a bit reductive in hindsight idk how to phrase it better tho
Why was the particle smaller than a grain of sand at the Big Bang and not something very massive that left all this dust and stuff out there.
