@RyderRude the point of bounties is to draw attention. If it worked like that we'd be packed with bountied questions and old bounties wouldn't even come up at the top
@Mr.Feynman this is y i proposed a secondary bounty page where u dump the expired bounties. these will no longer get attention but the answeres cant still.be rewarded
it's like when u put a bounty on someone's head, it is for forever, can be collected by anyone anytime
@ACuriousMind @SillyGoose I just checked the Italian wiki page of American Coffee (in the sense you mean) and at the beginning of the article there is:
> Nella lingua italiana con "caffè americano" (talvolta indicato come caffè all'americana) si può però intendere anche il caffè come nelle preparazioni americane, cioè un caffè filtro lungo.
> In Italian language "American coffee" (sometimes called "American-style coffee") may also mean drip coffee, though.
@Slereah I've seen a definition of the EH as the boundary of the closure of the chronological past of future null infinity. Most sources use instead the causal past, anything to say?
i also hav an idea to model Aristotle's physics. u define the geometry of spacetime to be a 0 connection. so the geodesic eqn is $\frac{d x^{\mu}}{d\lambda}= v^{\mu} (x^{\mu})$. then we can define a sort of covariant derivative $\nabla=\frac{d}{d\lambda} - v^{\mu}$. then non geodesics r given by $\nabla (x^{\mu}(\lambda))=q^{\mu}$
to make particles come to rest in absence of interactions, u define the 0-connection to be a time-like vector field in some preferred rest frame
@SillyGoose I was wanting to give some trivial arguments. Your question is actually somewhat ambiguous. There are some processes that violate superselection rules, and then they cannot happen. That is one reading of your question. If your process can happen under SM, then there must be a (not necessarily unique) simplest diagram. This is because all predictions of SM come out of S-matrix, and Feynman diagrams are how they get in the S-matrix in the first place. The series is ordered in terms of
number of vertices, and the complexity of diagrams rises dramatically with number of vertices; so if your process obeys all the conservation and all the superselection rules, it will be utterly insane if it cannot happen.
In section 2.3 of abatanasov.com/Files/Supersymmetry.pdf, how is the 2nd anitcommutation relation consistent? If bosonic operators $B$ are in $g_0$ and fermionic operators $F$ in $g_1$, surely this means that $\{F_i, F_j\} \in g_2$ since $1+1=2$ however $g_2$ doesn't even exist
unless we have i+j modulo 2 or something but surely if that were the case it would be stated everywhere eg. on wikipedia en.wikipedia.org/wiki/Graded_Lie_algebra etc
@naturallyInconsistent They weren't asking "For every process allowed by the SM, is there a diagram?", they were asking "For every input/output combination that satisfies all conservation laws, is there a diagram in the SM that mediates that process?"
presumably this is an attempt to determine how seriously Gell-Mann's totalitarian principle should be taken
@DIRAC1930 It is stated everywhere
what do you think the algebra being $\mathbb{Z}/2\mathbb{Z}$-graded means?
So is this $i+j$ meant to be the composition of the $\mathbb{Z}_2$ elements, i.e. $1$ is bosonic, and $-1$ is fermionic, therfore $[X,X] \in g_1$, $[X,Y] \in g_{-1}$, $\{Y,Y\} \in g_{1}$?
Yes, when reading the definition, I didnt realise that the $i+j$ was reference to $Z_2$. I just thought it was a general expression and I didn't realise the $0$ and $1$ refered to elements of $Z_2$
I have a feeling the way the quantum chemists do QM must be quite different from the way physicists do it. I haven't studied "physics" side of QM, but I've come at it from a chemistry angle
"Votive offerings in the Greek world as in other ancient cultures could naturally take the form of intellectual offerings representative of the intellectual performance of the dedicat. [...] But this was a different case from ours, because Ptolemy did not dedicate the astronomical instruments through which he achieved those results, but the (shortened) text itself where the theory is explained. This sort of offering is indeed rare outside from mathematics."
Offering my thesis to the gods
"Xenagoras is said to have inscribed his calculation of the height of Mount Olympus at the Pythium there. We read in Porphyry's Life of Pythagoras that Pythagoras' son Arimnestus dedicated a bronze tablet containing "seven knowledges" to Hera in his temple at Samos."
@DIRAC1930 have you considered that you only know the "old ones" that left a lasting impression? I'll bet that if you go at the actual list of physics nobelists from the early days there'll be a bunch of names there you aren't thinking of with your "different caliber" claim
people in the past weren't different, we simply merely remember fewer of them
Dalén, "for his invention of automatic valves designed to be used in combination with gas accumulators in lighthouses and buoys" - man, that sure is a level of physics no one else reaches today
@RyderRude I don't think it's just due to their being drastically new physics at the time. If you read some of the textbooks from Pauli or Schrodinger, their knowledge of theoretical physics from areas that had nothing to do with what they won their nobel prize for is crazy
@DIRAC1930 I'm confident there's plenty of physicists today with similar breadth of knowledge, but the field is simply orders of magnitudes bigger today than it used to be, so individuals stand out less
It just irks me that Dirac did so much of Feynmans work (that he essentially didn't really add much to) and that Feynman didn't credit Stuckelberg for a lot of things he stole
> After Albert Einstein’s paper had been published, rumor has it that Sir Arthur Eddington, one of the propagators of the theory, was one of the only three persons in the world to understand it. When asked about the rumor during a casual conversation, Eddington allegedly paused for a moment, then replied: “I’m trying to think who the third person is.”
Although in a sense I am astonished. Today Physics is so broad that unless you do on purpose, you don't end up reading the same guys' papers all the time
When you look in detail at a lot of theories that are famously hard you start to see how a lot of them were constructed with pretty well established ideas
the trace of the composition of two linear maps $A$, $B$ can be computed taking the trace of the product of two matrices representing $A$ and $B$ also in different basis?
it seems so from the cyclicity of the trace but it also seems a bit weird
@ekardnam_ I think you should probably try writing out whatever proof you have in mind explicitly, because this is definitely not true: take the matrix $A$ that has a 1 in the upper left corner and zeroes everywhere else. it is its own square, so the trace of $A^2 = A$ is 1. If you permute the basis vectors for the second $A$ (which is an allowed change of basis), the product of the two matrices will be zero, which does not have trace 1.
@Mr.Feynman but say $A_1$ is the matrix representing $A$ in a basis $b_1$ and $B_2$ represents $B$ in the basis $b_2$ then $B_1 = P_{12} B_2 P_{12}^{-1}$ represents $B$ in the basis $b_1$ but then $tr(A_1 B_1) = tr(A_1 P_{12} B_2 P_{12}^{-1}) = tr(A_1 B_2)$
guys in C-T it is stated that if I consider (single-particle) $\mathbf{J}^2,J_z$ and I take a pair of eigenvalues $m,j$ than to this pair we can associate an eigensubspace whose dimension is denoted is $g(j,m)$. Do you agree with this statement? I thought that once we specified $j$ then the correspondent eigensubspace is of dimension $2j+1$, but I don't quite get why there would be ulterior degeneracy
I dislike the usage of "well-defined" in this context. It's not helpful imo. Well definedness has a meaning in math but in the context of quasi-static volume change I have no idea what it means.
this notion of pressure as a property of the system would be ill-defined if the pressure was a function of position
@ClaudioMenchinelli It just depends on that system you're looking at - there is no guarantee that every system has only a single copy of the irrep for each $j$ value.
sure, for a single particle there's only one $2j+1$ subspace for each integer $j$, but e.g. the 3d rigid rotor has higher degeneracies
Can't pressure be an emergent property from knowing everything about the phase space, so we consider the pressure on one side of a container to be the total momentum of the particles in contact with that side at any given time? I don't think it makes sense in a continuous way though
@ekardnam_ i also have a really nice thought experiment about free will. suppose there is a deterministic block universe with time travel such that past cant be changed. and u know that something really bad might have happened yesterday. do u choose to time travel to make sure it doesnt happen?
this thought experiment reveals free will even in this universe
@Obliv in thermodynamics, over a macroscopic range, they assume pressure is constant and so on. This can be refuted ofcourse, if gravity is in the pic, then there will be a pressure gradient- but still, over reasonable ranges, like 1 m^3, its constant. its one of those equilibrium things.
@Obliv this is what is done in kinetic theory of gases. assume every gas molecule is a point mass, that have no internal interactions except collisions, and each gas molecule on avg traverses vast distances before it hits another one, and so on. you can derive an expression for pressure then, through a time average
you will need to invoke "randomness" conditions though
@Obliv usually what they do is, given a speed range $v$ to $v+dv$, what is the probability that some molecule is moving in the direction of $\theta-\theta +d\theta$, and $\phi-\phi +d\phi$, and they say it is $\frac{d\Omega}{4\pi}$, the solid angle of a small patch, over the total solid angle
I can't understand this: I wanna add to angular momenta $j_1,j_2$ (my initial kets, wrt the following set of commuting operators $\{J_1^2,J_2^2,J_{1z},J_{2z}\}$ is $|j_1,j_2,m_1,m_2\rangle$
Now I wanna switch to this set $\{J_1^2,J_2^2,J^2,J_z\}$ (the first two operators will be omitted to obtain a ket of the form $|J,M\rangle$). If I fix $j_1,j_2$ then I know that $M = m_1+m_2 \in \{j_1+j_2, j_1+j_2-1, \cdots, -j_1-j_2\} $
The problem is: how do I know what values can $J$assume given $j_1,j_2$?
I mean, I'm pretty sad: basically this proof is the only problematic thing, but it's necessary, in fact if you know the values taken by $J,M$ once you fix $j_1,j_2$ you can prove that there's a one-to-one correspondence between a pair $$ (J,M)$$ and a vector belonging to the subspace $\mathcal{E}(j_1,j_2)$
@ClaudioMenchinelli The answer is that the possible $J$ lie between $\lvert j_1 - j_2\rvert$ and $j_1 + j_2$, and at least the upper bound is what is stated in the paragraph you marked red
I know what the permitted values are, as I went on with the theory leaving this behind. I can't understand the strategy he uses to obtain the eigenvalues of $\mathbf{J}^2$. It's pretty complicated, and I'm searching for a different "proof"
@ClaudioMenchinelli The simplest way is a dimensionality argument: You know that there is $J = j_1 + j_2$ in there, and you can now apply lowering operators to this. The question is when this terminates. Let $J_\text{min}$ be the lowest possible value for $J$, then the dimensionality is $\sum_{J = J_\text{min}}^{j_1 + j_2} 2J + 1 = (J_\text{max} + J_\text{min})(J_\text{max}-J_\text{min} + 1) + (J_\text{max}-J_\text{min} + 1)$.
Now equate that to $(2j_1 + 1)(2j_2 + 1)$, plug in $J_\text{max} = j_1 + j_2$ and solve for $J_\text{min}$
eh, I think if you carefully argue why this goes down in integer steps you actually directly get the result anyway
You start at one state with $J = j_1 + j_2 = m$, this is $\lvert j_1, m_1 = j_1\rangle\otimes \lvert j_2, m_2 = j_2\rangle$. Now you get a two-dimensional space of states with $m = j_1 + j_2 - 1$ (you can either lower the $m$ of the first or of the second vector). One of these dimensions is the $J = j_1 + j_2, M = j_1 + j_2 - 1$ obtained from the top state via lowering, but the second dimension needs to be something else,
and so it needs to be the top vector of its own subrepresentation with $J = j_1 + j_2 - 1$
you can either explicitly work this out iteratively, or you can then write down the sum I wrote above and solve for $J_\text{min}$
@ClaudioMenchinelli It is just saying that if $M=j_1+j_2$ then you know immediately that $J=j_1+j_2$ because there is only one way to get that. But if $M=j_1+j_2-1$, then it can either come from $J=j_1+j_2$ or from $J=j_1+j_2-1$
Ah ok, I see now.There's two possibilities: the max value taken by $M$ in the subrepresentation where $J = j_1+j_2-1$ or the the subrepresentation where $J = j_1+j_2$ but the value of $M$ just one unity lower
@ACuriousMind @naturallyInconsistent thanks those are illuminating hints haha
Of course I'm being so reductionist I don't want to bother much with Riemannian geometry as it's only a special case of DG. I'm hating myself for that lol
and so the idea here is we want to conjecture some maximal set of mutually commuting observables. based on the data we have, it looks like $\{H, L^2, L_i\}$ for any $i \in {1, 2, 3}$ must at the very least be part of this maximal set of commuting observables. And conventionally, we just pick $i = 3$ and correspond it to the $+z$ direction in space?
and we want to look for a maximal set of mutually commuting observables $\{C_i: \mathcal{H} \rightarrow \mathcal{H}\}$ because it is a linear algebraic fact that $[C_1, C_2] = 0 \iff C_1 \text{ and } C_2 \text{ are simultaneously diagonalizable}$. Hence, such a maximal set yields a basis labeled with the most information (corresponding to quantities of interest) possible?
i guess I am not sure about why we want a maximal set other than for the purpose of judging how much information we can know all at once
as in upon taking a measurement of one of the observables in the maximal set, you can measure the most other distinct observables if you pick from the maximal set
maximality of the CSCO just guarantees that the labels are non-degenerate, i.e. specifying the eigenvalue for every operator in the CSCO uniquely specifies the state
in that case, it's a mathematical fact that the spherical harmonics are a basis of $L^2(S^2)$ and for them the $\ell, m$ label is non-degenerate, and you can also show that the radial functions $\psi_n(r)$ form a basis of $L^2(\mathbb{R})$, so what you have altogether is a basis of $L^2(\mathbb{R})\otimes L^2(S^2) \cong L^2(\mathbb{R}^3 -\{0\})$ labeled by non-degenerate labels $n\ell m$
so something like prove for arbitrary $n,l,m$ that $\lvert nlm\rangle - \lvert n'l'm'\rangle = 0$ where $n = n', l=l', m=m'?
I am trying to understand 3.267 and 3.268 in a rep theoretic framework. in particular, i am trying to understand why the structure of the eigenvalues is the same no matter the particular system. I think this is because the irreps of $SO(3)$ over $\mathbb{R^3}$ don't care about the Hamiltonian or about the specifics of the space. So long as we are dealing with a vector space isomorphic to $\mathbb{R}^3$ we have the same structure of eigenvalues for generators/($L^2$ is a casimir inv?) of $SO(3)$?
i guess this is the same as in the finite dimensional case with spin, but I have a more concrete belief in this fact for the finite case because i know that projective irreps of $SO(3)$ are characterized by dimension of the space, not by the space itself in any way. and I do not know the analogous result for the infinite dimensional irreps
@SillyGoose you don't need to know anything about infinite-dimensional reps here
SO(3) is compact, all its irreps are finite-dimensional
The spherical harmonics $Y_{\ell m}$ are precisely the decomposition of $L^2(S^2)$ into the finite dimensional irreps of total angular momentum $\ell$ and dimension $2\ell + 1$
Hm well, I guess I am confused. I have only seen results for finite dimensional representation theory, and I am not sure what would change when considering infinite dimensional representations. But you say that irreps of $SO(3)$ are finite dimensional, okay. Then, the procedure to classify the irreps of $SO(3)$ is the same whether talking about finite dimensional systems or infinite dimensional?
the representation of SO(3) on $L^2(S^2)$ is just the one inherited from the fundamental rep (any time you have a representation on a space $X$, you also get representations on the function spaces on $X$), so you know this is a proper linear representation of SO(3) and not just a projective one
@SillyGoose I don't understand the question - as I just said, the irrep are finite-dimensional, so why would the system being infinite-dimensional matter?
where the procedure I have in mind is the classification of irreps of $\mathfrak{sl}(2;\mathbb{C})$, which is equivalent by various results to irreps of projective irreps of $SO(3)$
well I guess I am confused because if we account for the fact that we can have projective irreps, the procedure to classify projective irreps of $SO(3)$ results in finding projective irreps of $SO(3)$
so that is the reason why I am wondering why we are disregarding these projective irreps (the ones that do not lift to normal irreps)
