but the actually important thing in relativity is to have a reference velocity, and this is what the ether would have provided
the orientation of the axes or the origin don't really matter that much
note that you've already been thinking like this - in your example with the person and the car, you talked about "the car's reference frame" when in fact there are infinitely many different reference frames in which the car is stationary and non-rotating - take one frame in which the car is, and just rotate its axes or put the origin elsewhere, and it's still a frame in which the car is stationary and non-rotating
it's just that this sort of difference where the frame just differs by the position of the origin by a constant vector or a constant rotation of the axes is not all that relevant/interesting
a cute observation. suppose your spinor is normalized, i.e., $\langle \psi|\psi\rangle=|\psi^1|^2+|\psi^2|^2=1$. then $P=|\psi\rangle\langle \psi|$ is projection onto $|\psi\rangle$ and $\sigma=P-1/2=1/2(2P-1)=1/2[P-(1-P)]$
that makes it explicit that $\sigma$ is reflection around the $|\psi\rangle$-ray up to a factor of $1/2$
A thin oblate spheroid with sizes 100X100X1 collide with small sphere radius 1. that move with velocity v and hit the spheroid 99 unit from its center(near the edge). The spheroid is bended and get rotation and velocity after the collision. find the mass moment of inertia of the bended spheroid a...
@NiharKarve yeah I've been using obsidian.md for a year or so and am a big fan. I used roam before that, but it was expensive and had a lot of problems. Notion also seems pretty popular, but didn't fit my use case
there's a ton on youtube about each of those, though the videos can get a bit fanatical at times
@ACuriousMind do you recall when we discussed about the phase space, and you told me that $\rho(\vec x,t)$ which is the pdf in phase space, can be interpret as the macrostate, and can be thought of cluster of points, each representing a microstate of the system?
it's just a basic linear algebra fact - it follows directly from "cyclicity" $\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$ and basis changes acting as $A\mapsto PAP^{-1}$ on matrices.
@imbAF what do you mean by "space of eigenstates"?
also, what are "eigenstates of a system"? Do you mean eigenstates of the Hamiltonian of the system?
We describe a physical system in quantum mechanics by a) a Hilbert space that tells us what the possible states for the system are, b) a Hamiltonian operator on that space that tells us how states evolve in time according to the Schrödinger equation, c) a bunch of other operators that represent physically interesting observables of the system in question
Nothing about this changes when we do statistical mechanics.
sure, elements of the classical phase space represent the state of the system classically, and elements of the quantum Hilbert space represent the state of the system in quantum mechanics.
and it's a fact that for any operator that's not a multiple of the identity, not all vectors in a vector space are eigenvectors, because you can always add two eigenvectors with different eigenvalues to get a vector that can't be an eigenvector
if you have two different operators, which do not commute , can you still say that :"you can look for the few vectors in this hilbert space that are eigenvectors"?
if you're lucky, you can find some vectors that are eigenvectors of both, but you'll never find a basis of such shared eigenvectors if the operators don't commute
We have system K and operator A and B. Operator A has it's own eigenstates (Like the hamilton operator with $|n\rangle$. These eigenstates belong to a hilbert space and we can express them via ket vectors, who are orthonormal and therefore can be used to express every other state that the system can be in, physically as a superposition of the eigenstates, and mathematically as a linear combination of the ket vectors multiplied with some constant.
The same applies to the B operator. And if they do not commute, then each operator has it's own basis ket, which abstractly/physically is a hilbert space containing the eigenstates and every possible state that is a superposition of them.
whether operator A has it's eigenstates or the system, idk which is the correct way to say it
I was trying to find the classical partition function for the 1d harmonic oscillator, and while I got the same result in the solutions, I did something for which I have no explanation as to why I should. When I integrated I took $-$ and $+$ infinity as my boundaries.
While in phase space that makes sense,
physically how do I justify this?
The harmonic oscillator cannot have infinite displacement or momentum
@ACuriousMind One question. In the case of particles in box, the condition for classical interpretation of the situation is when $\frac V N >> \lambda$. For the classic harmonic oscillator I found for the entropy $S= kln(\frac{kT}{\hbar \omega}) + k$ and it is said that this makes physically sense if $kt >> \hbar \omega$
Why it makes sense only in this case?
and why is this the classical limes for the oscillator
Ofc if $kt<< \hbar \omega$ we have a negative value, but that doesn't mean that the whole entropy is negative, which is not supposed to happen
the $\hbar \omega$ essentially is the "size" of one coarse-grained microstate here (remember, entropy is "supposed to be" the logarithm of the number of states)
and if your $kT$ is not much larger than that, then this approximative coarse-graining fails - e.g. you'd be saying there's "half a state" for $kT = \frac{1}{2}\hbar\omega$, which is absurd
One more thing, but this is a mathematical one that has been bugging me for quite sometime
When we consider the quantum 1d harmonic oscillation, with the Hamilton operator etc. When we want to find the quantum canonical partition function, we use the following formula (which was never explained how do we get it, but it's not important rn)
Man, the more I study physics, the more insecure I become about my choice. I have began to think that only gifted people should study this. Others are simply wasting their time trying to hard and achieving nothing
one way to define this is by saying "multiplying $H$ with itself $n$ times is defined, so we'll define this as what happens when you plug $H$ into the Taylor series of the exponential", the other is to say "it's the operator whose eigenstates are the same as those of $H$ but each eigenvalue is $\mathrm{e}^{E_n}$"
the latter way is nicer, because mathematically you'd actually have to wonder about what convergence of a series of operators means in the first case, etc.
@imbAF to be honest, I have the impression you're "running before you can walk" - you're doing quantum statistical mechanics but you seem to be lacking proper training in both classical and quantum mechanics without the added complication of statistics. This makes this much more difficult than it needs to be
when we took the hilbert space we said something about a bunch of functions that are square intregrate-able
and a bunch of other nonsense
And because work + university is hard, I had to leave some exams, and one of those is classic electrodynamics, which has nothing to do with current quantum physics that we do, so I am forced to spend time on both fronts, for two upcoming exams
one in classical electrodynamics and one in quantum statisticalmechanics part 1, cuz I tookl part 2 without giving one :D
I wasn't trying to blame you - I don't know your life, and I know there are many places where this sort of curriculum that sends you running before you can walk is just the way it's done
But it's not that I am lacking because I have all the nice explanation and I fail to understand. It's the contrary, we are given bits of information, and then for the rest : "Search on x y z book"
That is time consuming and many times useless, because one concept or a topic that is explained in the lecture, which is hard to understand in the first place, is treated or explained differently in different books, and now you are forced to adjust the explanation in the lecture to that of what you read, while at the same time having minimum understanding of what's going on
I understand, but this won't ever change again - the more advanced your topics get, the less there will be the "one textbook version" and the more there will be different accounts of the same topic that might seem to say wildly different things at first glance
^that. 100% understanding at first try is simply unrealistic in most cases. Living with uncertainty is a useful skill also outside of quantum mechanics ;)
I do that rarely, and hope that that what I don't understand it's not in the exam
yolo
@ACuriousMind I have done a small "project" of mine, trying to explain the canonical / grand canonical ensemble, schematically with ofc mathematic explanation attached to it. When I finish, could you have a look at it?