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00:05
@Mr.Feynman oh hm i guess i didn't read far enough
 
2 hours later…
02:06
consider a degenerate subspace $D$ of $\mathcal{H}$. Suppose we're solving a perturbation theory problem. If degeneracy is not broken after a first order correction, then is the problem that we cannot apply nondegenerate perturbation to obtain second order corrections? Or is there a different problem that occurs if degeneracy is not broken?
I am thinking that the problem is that degenerate perturbation theory says: 1) choose 0th order states so as to avoid singularities in inverting an operator in schroedinger to obtain first order corrections to eigenstates, 2) usually degeneracy will now be broken, so just apply nondegenerate perturbation t heory
so when sakurai says "this procedure works provided that there is no degeneracy in the (first order corrections)", is it referring to the fact that (2) can no longer be done?
(i don't wish for the detailed answer to what needs to be modified since I wish to work this out myself, just if i am identifying the problem correctly or not)
02:27
@SillyGoose Sakurai later provides an example of when a degeneracy is partially lifted and that there are still other degeneracies left.
@SillyGoose Nobody does things that way. Everybody does the simplest case first, and then handle the complications that might arise that the simplest case explicitly asks students to ignore. Especially when things are difficult, this is the only order of presentation in which students can hope to keep up with.
Not to mention that the perturbation theory you are learning is actually going to give you asymptotic series. Renormalisation and grappling with infinities are everywhere in physics!
@ekardnam_ I would point out a totally different viewpoint. Both bras and kets are abstract, things living in the real world, that exist independent of representations. A bit like points on a manifold, vectors in a tangent bundle of said manifold, they are abstract stuff that exist in a world we cannot do much with until we introduce some representation to manipulate them. Abstractness is enforced so that we can assert that, for correctness and internal self-consistency, they obey
certain contravariant or covariant transformation laws when the representations change.
Kets $|\psi\rangle$ are abstract vectors living in some Hilbert space. Bras $\langle\phi|$ are abstract linear functionals living in the dual Hilbert space to that. There is no need to state which representations you are working with, when you state these facts. However, whenever we actually want to do something, we have to choose representations.
02:42
@naturallyInconsistent but i think this way of presenting perturbation theory makes it confusing to see which steps in the procedure are generally true and which break down due to degeneracy. it also masks the fact that in both cases we are actually doing exactly the same thing. In nondegenerate, you disregard the 0th order term of the ket youre solving for, in degenerate, you disregard the 0th order degenerate subspace of the ket youre solvingg for
where the degenerate subspace is the literal generalization of the trivial eigenspace that you are disregarding in nondegenerate p theory
What that corresponds to, is the choice of the decomposition of the identity operator. In particular, we get to choose $\int|x\rangle\mathrm dx\langle x|=\mathbb I=\int|p\rangle\frac{\mathrm dp}{2\pi}\langle p|$ and it is this choice that fixes which Heisenberg algebra you are supposed to be using. In particular, note that $|\psi\rangle=\int|x\rangle\mathrm dx\langle x|\psi\rangle$ is still an abstract vector living in the Hilbert space, whereas old notation $\psi(x)=\left<x|\psi\right>$
is a pure complex number function that is what lives in $L^2(R)$ of the Hilbert space. Because of the limitations of the old notation, it becomes extremely difficult to sort these concepts out, and Dirac notation helps us conceptualise these things in a far better manner.
@SillyGoose If the subject matter is simple enough, sure, I am generally a proponent of treating the most difficult cases directly. However, perturbation theory is difficult enough and full of pesky details enough that this is pretty much impossible to perform coherently in a pedagogical setting. If you manage to unify the treatment and go straight to the final result, sure. However, prior to any actual educational experimental verification that this could even be done, I wont be entertaining it.
@ekardnam_ note that while ACM is here telling you that $|x\rangle$ and $|p\rangle$ actually lives in rigged Hilbert spaces, those Gel'fand triple that you said you are not too familiar with, ACM also often points out that the actually rigorous treatment of QM is not using rigged Hilbert spaces, but rather uses the standard Hilbert space mathematical machinery and proving enough stuff so that the limits leading up to $\langle x|\psi\rangle$ and $\left<p|\psi\right>$ are well-defined.
@naturallyInconsistent i am attempting to write up the presentation i described let's see how it turns out :P
That is, you might want to avoid wasting too much time on rigged Hilbert spaces, and instead focus on those advanced Hilbert space machinery if you really want to hammer all these details down.
@SillyGoose If you need help on nice notation, I could make some suggestions.
oh yeah the notation i was writing in yesterday was just a bad decision LOL
i have switched to sakurai's notation but what notation do you use? (I don't think I'll change it in my note since I have written up a lot of it, but for future reference)
But first, I need to rush to the doctors before they close to rest. I'm coughing my lungs out. It's my 6th moderna and I'm sure I just have a bacterial sore throat. Sooo bad.
02:54
oh good god
i hope you feel better soon :P
@SillyGoose To minimise the amount of weird extra stuff, $(H_0-E_n\mathbb I)\left|n\right>=0$ and $(H_0+\lambda V-E_{n,\lambda}\mathbb I)\left|n,\lambda\right>=0$ where the stuff with $\lambda$ are exact quantities. The expansions are $\lambda\Delta_n=E_{n,\lambda}-E_n=\sum_{r=1}^\infty\lambda^r\Delta_n^r$ and $\left|n,\lambda\right>=\left|n\right>+\sum_{s=1}^\infty\lambda^s\left|n_s\right>$
note that my $\Delta$ has one extra factor of $\lambda$ factorised out, so that you can extract it out along with the $\lambda$ that goes with V, and that makes for slight notational wins. Not very much, but still.
 
3 hours later…
05:56
@naturallyInconsistent if you're interested i uploaded the note here: scribd.com/document/700337213/tipt
06:24
Is there some way to see it without downloading or something? Scribd is needing an account to download
07:06
oh
okay here's a link to the onenote 1drv.ms/u/s!At9Qgpejg8l2gyg_eavGhcNTs7xa
 
1 hour later…
08:39
@Minsky Carroll thinks that it's an Occam's razor situation where we are just adding redundant things on top of the laws of laws
He misses the nuances of a topic like this. i think he does not have much experience in philosophy
@Minsky i agree with Philip's philosophy, although he seems to have problems conveying it. he is talking about this philosophy plato.stanford.edu/entries/structural-realism
08:53
Well, Pofessor Carroll did start off by saying "I don't care about consciousness."
*Professor
maybe he shudnt talk for 3 hours about consciousness if he doesnt even care to get informed about the topic
Philip is informed and knows the subtleties of the topic
Philip correctly replaced eliminative materialism.with the mathematical universe hypothesis. Then Carroll started talking about the naive mathematical universe hypothesis, when that wasn't even Philip's point
and both parties should've defined what they meant by "behavior". The precise term should be "measurements". a lot of confusions would go away if we bother to make things precise
09:13
I think the topic definitely has a deeply philosophic question at its core having to do with your perception of your "mind."
09:24
Comparing a couple of GR books I think Rovelli's General Relativity: The Essentials is one of my favorites. Very introductory but in so few pages he touches many interesting things
 
1 hour later…
10:24
@naturallyInconsistent I see would definitely need to read more about this
@ekardnam_ about which thing?
@naturallyInconsistent about the actually rigorous treatment of QM
@naturallyInconsistent im at the end of my masters in theoretical physics and i still dont know so much about the rigorous treatment of QM mathematically, yet I still seem to know more than many other people in my same situation
10:33
@ekardnam_ That's because you are doing theoretical physics and not mathematical physics.
There is a chasm between those
@naturallyInconsistent yes and no. My thesis would definitely belong in between hep-th and math-ph, at least for where most of the papers im reading come from
@ekardnam_ yeah, there are some brave souls in the chasm...
@naturallyInconsistent many papers labelled as math-ph are not even so mathematically rigorous to be fair
i do like the mathematical rigor though. at least as far as it does not make everything more complicated than necessary
Oh, it will be way more complicated than necessary
@naturallyInconsistent sometimes it is an acceptable level which actually makes stuff clearer imho
10:47
@ekardnam_ Yes, and when that is the case, I generally would support that, and advocate for teaching using that level of rigour.
11:06
anyone knows of a good pedagogic introduction to instanton counting in $\mathcal{N} = 2$ gauge theories? I tried reading so many times the paper by Nekrasov but I would have to read so many other references and a self-contained explanation would be of help
11:26
@ekardnam_ well, hello :P
With that in mind I took measure theory this semester. When I have more time I'll start reading Conway's functional analysis and maybe finally learn wth is wrong with infinite dimensional spaces
11:43
@Mr.Feynman okay when you learn it lmk
It's far down in my list though
In contrast to differential geometry, I find functional analysis kind of boring
It's too technical
There are technical lemmas all over the place
Zee says"For example, when a physicist thinks of
a rotation, he or she typically has in mind a 3-by-3 matrix. A mathematician, in contrast,
might think of rotations as abstract entities living in some abstract space, defined entirely
by how they multiply together."
It is apparently a physicist practice to conflate groups and matrices
physicists define abstract entities to be concrete entities, which is convenient but bad
12:00
@RyderRude I think it's the opposite
Mathematicians often define the rotation group $\mathrm{SO}(3)$ as the matrix groups of isometries of $\mathbb{R}^3$
I've seen this discussion about the "abstract" group more by physicists than by math folks
@RyderRude That said, this is not conflating anything. The rotation group is a matrix group regardless
Where by matrix Lie group I mean isomorphic (in the sense of Lie groups) to a closed subgroup of the general linear group
@Mr.Feynman me too in fact I do understand differential geometry way better than functional analysis
@Mr.Feynman yeah
@Mr.Feynman i wud take the definition of the rotation grp as an abstract set with a composition operation which is isomorphic to the isometry composition on R^3
I don't know which I understand better but I can certainly say that geometry lingers in my mind, while all the technicalities of analysts don't last much
12:06
this definition does not favor the 3d rep over the 5d,7d,9d .... reps
becuz they r all reps under this definition
but in the physicist definition, the 3d one is the group itself while 5d, 7d are more like reps
what do u think about this viewpoint? @Mr.Feynman
@RyderRude I think talking about isometries is already enough. Isometries are not matrices per se, it's as abstract as you want. Then, the point is that in euclidean space there is a natural isomorphism between linear maps and matrices (as the choice of the basis falls on the canonical basis of $\mathbb{R}^n$) and this is where you transition to matrix, if you want to be pedantic
yeah. talking about isometries is abstract.
@RyderRude I think it is worth questioning it (I did that too a couple of months ago and you may find that in the chat) but not so worth looking for alternative formulations
but i guess u r still favoring the 3 dimensions, but it's subjective
beucz the isometris are on R^3
After all, it's not by chance that is it called also the defining representation
Now it's lunch time :P
12:09
alternatively, i cud define SO(3) as the 5d rep modulo choice of basis. becuz of this modulo, the thing becomes an abstract group on a 5d space
and this grp is identical to the abstract grp of isometries on R^3
and both r equally abstract becuz we modulo-d out the basis
so i think, on top of modulo-ing out the basis, we shud also modulo out the dimensions
and all we r left with is the manifold structure, and the generator algebra which is common to all reps
@Mr.Feynman i think there is barely any difference between these formulations anyway. we dont lose anything by defining the abstract thing to be a concrete thing
so im.not really against it
i guess it's just against the spirit of abstraction.
for e.g. in QM, we can define the abstract wavefunctions to be L^2(R) and the abstract X to be x, but this gives an arbitrary preference to a particular basis
12:50
@Mr.Feynman I think of the rotation group as the geometric algebra rotors that have half-angles...
13:07
@SillyGoose your notes are not doing Sakurai at all. It seems to be an amalgamation of other texts on the subject. I am also not able to fully decipher if you are treating degenerate perturbation theory correctly either.
13:28
lets say we hav two reps of a group : n dimensional and m dimensional. from these we can construct an n+m dimensional rep by doing a direct sum
and we can construct an nxm dimensional rep by doing a direct product. $D^n (g)$ and $D^m(g)$ get mapped to $D^n(g) \otimes I + I \otimes D^m (g)$
this is on the space $V^{(n)}\otimes V^{(m)}$
@RyderRude that's a tensor product, not a direct priduct, and your way of summing the two representations is wrong - that's how Lie algebra representations combine on the tensor product, not group representations.
oh yeah... this is how the $X$ , $P$ operators are mapped on the multi particle space
the grp members are $e^{iaX}$, etc
so... my question is only the former is called a reducible rep, the latter shud also be called "reducible rep -type B"?
becuz the concept of reducible reps is when we cheat to produce more reps from smaller reps
tensor product is also a way of cheat. the other rep decomposes into a tensor product of reps
is "reducible rep type -B" not a useful concept?
both of these representations are generally reducible/decomposable
I have no idea what you mean by "type -B"
13:36
the former is so by definition. the latter is becuz ... it also happens to decompose becuz of some theorem?
I don't know what you mean
a representation is reducible if it has a non-trivial subrepresentation
the book says it's reducible if it's a direct sum of sub reps
lemme see
that's the wrong definition, that's decomposable
but for nice groups decomposable and reducible mean the same
oh. the book is kinda misleading here. it is written in a casual way
Direct sum of representation sounds very decomposable :p
13:38
it gives decomposable reps and says "these are reducible reps"
A representation is decomposable when it can be written as a direct sum of two or more other representations. A representation is reducible when it has a non-trivial subrepresentation. These two notions are not, in general, equivalent.
oh
this book is misleading. maybe theyll clear it up later
Both the direct sum and the tensor product you talked about are usually (for most nice groups we look at) decomposable (and hence in particular reducible) - the direct sum always by construction, the tensor product more or less because of the way irreps work
yeah. the tensor products decompose for angular momentum
im familiar with this
i love this book but theyre giving wrong stuff i feel
the proof of Cayley's theorem was also incomplete
they just said "the multi table is a permutation, so proved"
just don't learn group theory from physicists :P
13:43
i bought it but it later lead to contradictions
@ACuriousMind but mathematicians focus on physically irrelevant things
"oh no, I might learn more than I strictly wanted to know!"
@RyderRude because group theory is... Math?
yeah... but mathematicians focus on weird things
physics study of groups is entirely based on reps
How dare you pollute math with physically motivated understanding :P
Physicists just butcher Lie theory :(
i still love this book. and reducible almost equal to decomposable, so it's fine
theyre making things simpler intentionally
@Mr.Feynman i will learn diff geo from mathematicians
13:54
Speaking more seriously. Of course everyone is free to choose whether to learn something from the mathematicians or the physicists for their own reasons but then you have be coherent with your choice. It appears to me that you want to stick to physics-oriented treatment of math, complaining about things that would be clear in a more standard math class, which you want to avoid
Don't act like a cat :P
Are you even surprised that he would be like this?
@Mr.Feynman im.not complaining anymore tho. the definition they gave made me ask a weird question
14:31
under a translation on a 2 particle hilbert space : $|x_1, x_2\rangle \rightarrow $|x_1+a, x_2 +a\rangle$. for each $|x_1, x_2\rangle$, the set $|x_1+a, x_2+a\rangle$, $a \in R$ is an invariant subspace
so this means this is reducible by the mathematician definition
im not sure if this is decomposable
i guess u cud decompose it into an infinite number of invariant sub-spaces
@RyderRude I'm here to help if I can, not to be surprised :P
so each invariant sub space is characterized by the difference $q=x_2-x_1$
so this decomposes into $\oplus _{q} V^{(q)}$
is this correct @Mr.Feynman
$V^{(q)}$ is the set of points (x_1,x_2) such that (x_1-x_2)=q
Once again: $\lvert x_1,x_2\rangle$ are not elements of the actual Hilbert space
@Mr.Feynman i will give up on the book if it weirds me out again
yeah. so i shud look for wavefunctions. i think wavefunctions defined on $V^{(q)}$ will also not do
becuz they r sort of discontinuous on R^2 becuz theyre 0 except on $V^{(q)}$
ooh but i cud look for wavefunctions defined on some continuous region which is a direct sum of $V^{(q)}$
this region is an invariant space for these wavefunctions?
14:49
but the action in terms of functions is straightforward: Choose coordinates $x_\pm = x_1 \pm x_2$, then translation acts as $f(x_+,x_-)\mapsto f(x_+ - 2a,x_-)$. Choose any Hilbert basis of $L^2(\mathbb{R})$ as $\psi_i(x_-)$. Then $L^2(\mathbb{R}^2) \cong \bigoplus_i (L^2(\mathbb{R}) \otimes \mathrm{span}(\psi_i))$, where the first factor is functions of $x_+$ is a decomposition into subrepresentations of translation.
it's really clever to use those co ordinates
thanks
If you want to keep the Jacobian trivial, the choice is $\frac{x_1+x_2}2$ and $x_2-x_1$
 
4 hours later…
19:25
@user85795 he certainly cares, and wrote more than one paper on it. So i think he says that to feel more comfortable.
@RyderRude interesting link, will read it
 
1 hour later…
20:53
Can physics explain consciousness ?
what does that even mean
21:13
Could anybody confirm this is ok? I just want to be sure
1,2 refers to two distinguishable particles
to which I assigned their own subspace
@ClaudioMenchinelli In that case, what even is there to prove? $[S_{1i},S_{2j}] = 0$ should be true by however you defined these operators, so $[S_1^2,S_2^2] = 0$ follows immediately.
Yeah that’s what I wrote indeed
The term is equal to zero, I wrote it under immediately
the other commutator involves operators in $\mathcal{E}_1$ only instead
so why is there six lines of computation and some argument about antisymmetric terms here? Am I missing something?
@ACuriousMind it means if you think that the current physics theories would find plausible that a piece of matter has inner/private experiences
apart from behaviours, which can be explained (in theory.)
@Minsky I do not know any mainstream physics theory that has anything to do with "experiences" of matter
21:21
why do you think that there aren't @ACuriousMind ?
i.e. this isn't the sort of things physics theories talk about at all
@ACuriousMind I checked and what I wrote seems reasonable still, I might be overlooking something haha
@Minsky uh...because I know a lot and none of them are about that kind of thing? Physical theories are generally concerned with predicting the time evolution of some dynamical variables (like an electric field, or a current, or the position of particles). Where is "consciousness" supposed to enter here?
Because many hypothesise that it's a property of physical systems, maybe ? @ACuriousMind
@Minsky I don't know what that is supposed to mean
21:26
That if you create a human, like a baby, at some point it starts having a representation of the world, like a visual experience. @ACuriousMind
most physicalists tend to believe that it's something in the structure of our brains that has to generate whatever we call consciousness
that's the realm of biology and neuroscience, not physics
Well, but it should be expected then, and we could ask what is it made of, for example @ACuriousMind
@naturallyInconsistent oh I did not intend to follow Sakurai's presentation (just his notation)
(Life is about biology, but schrodinger wrote interesting insights from physics, as well)
@Minsky A computer is made of metal and silicone; the physical description of what the currents in the CPU do will not help you understand anything about the computations it's doing
21:28
Well, it should to some extent. For example, what is the screen for ? and also, is there anything non-physical in the computer ?
No, so, what is consciousness made of ?
don't be a silly reductionist; just because in a physicalist worldview everything is made of "physical stuff" that doesn't mean physics as a natural science has something to contribute to explain every phenomenon we observe
@Minsky What is the addition that a computer can do "made of"?
Well, it must hold in every phenomenon, especially in the ultimate nature of it, unless you think there are non-physical things
demanding that there is something emergent phenomena are "made of" is a category error
My point with the computer is the following: if I know it has leds, I know there will be an image
You cannot measure any consciousness but your own
21:30
so you can certainly say some stuff
It makes the notion of applyinh physics to it pretty difficult
@Slereah Yes, but why ? how does this relate to the physical stuff ?
@Minsky Computer screens are usually not made of LEDs
Because it is part of the scientific method that physics relates to measured phenomenon
@ACuriousMind great point
21:32
my point is: The computational capacity of a computer is an emergent property of the way its components interact; the physical description of these interactions is secondary and only very few characteristics of it actually matter in this emergent system
there is nothing this capacity is "made of", it's a property of the system as a whole - and that's something we fully understand
@Slereah that sounds to me like being lazy and not really interested in the question, which is fine.
so why should there be something consciousness is "made of"
I mean why not apply physics to love by that argument
Maybe we are just being too lazy to do it
i'm not talking about the computational capacity, but of the stuff that a computer is made of @ACuriousMind
@Minsky What do that have to do with anything?
the computer is made of metal and silicone, a human is made of water and carbon and hundreds of trace elements
21:34
It does have imho
this tells you nothing about the computer's capacity to do computations, just like it tells you nothing about the human's capacity to have subjective experience
@Slereah let me put it into other terms. If we assume that consciousness (and assume you know what is meant by it) appears in certain physical systems, then is that experience expected in physical terms ?
I think it's a fair question if you put minimum will to read it and not dismiss it straight away
I mean since we cannot measure it, we don't know if it appears in certain physical systems
We can't assume that it exists in other people or doesn't exist in rocks
Well, you do though
To be honest, I've heard this kind of question dozens of times and I really think most people are not actually asking any kind of coherent question when they talk about consciousness
21:37
unless you aren't conscious
Do I?
I am a unique case
Pretty difficult to generalize
@ACuriousMind Yes, me too, and I think people don't want to think anything other than book stories
@Slereah well, it is, yes, but you may as well assume others have
Tons of people have tried to solve that problem for... about 5000 years, at least
And people still are
I'm aware but just dismissing it isn't a good idea
Slereah's point is that precisely because consciousness is subjective, there is nothing that distinguishes a physical system "with consciousness" from one "without consciousness" from the viewpoint of experimental physics
so how could physics possibly have anything to say about it?
21:40
what do you mean by subjective ? not interesting for physicists ?
@Minsky I cannot do anything kind of experiment to determine whether you are conscious or not.
this aren't good reasons imho
you know you have experience
Not everyone even agrees on this
most do though
if you say: can you imagine a rose ? they say yes. ok you can distrust, but you have your imaginary rose at least.
@Minsky so what? Physics isn't about what I feel, it is about generalizable, repeatable experimental results
21:42
but that's a crazy property of matter
@Minsky But then all you observe is the behaviour, not the consciousness
and i doubt one could not be interested
Unless you're a behaviourist and you say the two are the same
@Slereah that's why i said 'you have your rose'
@Minsky No one in this conversation has said it is not an interesting question
we're saying it's not a question that lies within the realm of physics
21:43
i disagree and have shown you why, it's a property of matter
and you seem to agree that it exists
no, you haven't shown that at all
it is consistent with physics to not be a physicalist
well, do you think it is not a property of matter ?
@Minsky I'm not taking a stance either way
of course, i mean humans
then that doesn't mean it's non-physical
especially if you agree in the existence of NCCs
And again, X being a "property of matter" in this very general sense doesn't mean physics has to be the science to explain it. All the stuff happening with plants and animals is surely "properties of matter", too, yet the science dealing with that is biology, not physics.
21:49
@Slereah that has an interesting caveat: if someone is in a body that can not move, then behaviourists would say it's not conscious, seems quite wrong.
no one expects physics to explain the mating behavior of penguins
In quantum field theory, penguin diagrams are a class of Feynman diagrams which are important for understanding CP violating processes in the standard model. They refer to one-loop processes in which a quark temporarily changes flavor (via a W or Z loop), and the flavor-changed quark engages in some tree interaction, typically a strong one. For the interactions where some quark flavors (e.g., very heavy ones) have much higher interaction amplitudes than others, such as CP-violating or Higgs interactions, these penguin processes may have amplitudes comparable to or even greater than those of the...
no but you could say they are made of matter right ? @ACuriousMind
and they move due to matter gradients and electric signals right ?
so, it has things to say
@Slereah you're all npcs, I'm the mc
It's the beetle box problem all over again
21:55
@Slereah yeah, yesterday I found out that an acquaintance of mine (physics layman) has a Dirac equation tattoo
At least the equation was written properly
@Minsky but again: The science that deals with that is called neuroscience, not physics
no, the question does not require to focus in neuroscience, but nevermind, i am tired
sure, there's an intersection with physics when it comes to some aspects of electric signals and so on, but there's no "physical theory" here that would directly imply anything about the behaviour of the animals
And the neuroscience mostly treats mental states as being identical to neuron states
If you are hoping for spirit atoms or etheric consciousness you are in the wrong field
22:02
no, i'm just no so blind as to not worry about an interesting phenomenon @Slereah
You should read theosophy if that's what you want
i think you need it lol
maybe, arealist
Isis Unveiled: A Master-Key to the Mysteries of Ancient and Modern Science and Theology, published in 1877, is a book of esoteric philosophy and Helena Petrovna Blavatsky's first major work and a key text in her Theosophical movement. The work has often been criticized as a plagiarized occult work, with scholars noting how Blavatsky extensively copied from many sources popular among occultists at the time. Isis Unveiled is nevertheless also understood by modern scholars to be a milestone in the history of Western esotericism. == Overview == The work was originally entitled The Veil of Isis, a title...
All the mysteries of the spirit
imho is quite disrespectful to accuse someone like you are doing
i wonder what's more on topic in this forum
Physics
22:03
so maybe you could avoid accusations like that one ?
@ACuriousMind but biology is a trivial consequence of ph-AGGDKFHRVAJRVOSGDIDVRISBRBRHDH
Never
damn ACM, why did you slam my head into the keyboard?
Feynmann was a lot more open minded lol
I guess it got old
Go take it up with Feynman then
22:05
well, Feynmann attacked first
Feynmann might have been, but I think Feynman would align with ACM
but obviously, you will here lose out of group behaviour
yes, he would
I'm not sure. What are we talking about right now?
@Mr.Feynman 'twasn't me, it was the model of me that lives in your head
Come to think of it, there is an ACM in my head
22:08
I know :)
Anyone know if the 45 degree angle part matters in the context of this problem? Not sure if it's just extra unneeded info
people never realise that group behaviour feels quite bad, at some point, you'll get attacked by the ones that know, just because you are curious.
Like $F = \frac{\Delta p}{\Delta t}$
The pressure on the pane will only involve the normal component of the force
you should believe @Slereah, he knows the truth. /s
22:11
It's true
@ACuriousMind More specifically. I realized that in my first times hanging out in the chat I used to interpret many of your messages with an argumentative tone. Over time I realized that it's the same way I speak in real life and so now I read your messages as calm and chill as a human being can possibly be :P
oh so it's just the force $\cdot \cos(45)$
not cod45
@Mr.Feynman :P
@Mr.Feynman i vaguely recall him having more flair in those days. I feel like the site in general had a lot of know it alls when I used to frequent it in like 2016.
but overtime, dealing with so many different people I don't know if arguing with everyone is worthwhile lol
@Obliv I got on the hbar around 2022ish
22:17
In my experience it is not :p
idk it could be the same as it was back then, I don't frequent the main site much anymore.
I'm sorry, @Minsky what "group" are you talking about? ACM has been discussing with you politely and Slereah replied a couple of times but he was not fooling you
It was not a productive dzbate
I mean I've been around here for almost 10 years now, of course there's going to be some wear and tear :P
22:18
I chuckled lmao
@Slereah I read the title and got hyped. I read the description and took my hype back
How often do you have to ban people @ACuriousMind ? I remember people would get hit with the hammer frequently back then.
@Obliv you probably remember kicks, not bans
there was a rowdy time where kicks were somewhat frequent
actual bans for an extended time are very rare
I will not name names to protect the guilty, and so should you :P
@Slereah Dude, I remember playing an old fps shooter (quake live) in like 2019 or later.. the game is practically dead but this one guy with a name similar to "Michelson morley test on ISS" or some "Einsteins relativity is incorrect" was on every single night, not sure if he even played the game or just idled. He even hosted a server under similar names. I tried asking about it and he started to explain the issues with the MM experiment. shrug
Right, I do remember kicks. I think there were a couple people with permas, but yeah I don't want to say the names :P
Just spreading the truth
The Quake people must know
Although if you're going to slander Einstein why not go with Red Alert
A game he is actually in
22:24
I do give credit to crackpots that actually familiarize themselves with science to some degree. I was listening to some Alex jones rant and I was just amazed. can't tell if he's just manipulating people for profit or he's actually delusional.
Lol i did not know that. I played red alert 2 a bit, i think in our high school computer labs lol.
@ACuriousMind why do you want them to be anonymous? Not that I care about who they are, just asking
@Slereah quick ACM translate what he says
Did Einstein just vaporize hitler with a handshake
@Mr.Feynman it's been years and they've served their time and not re-offended, so it doesn't have any benefit bringing it up again
Sorry for intruding, about earlier, I got to a different expression when computing $[\mathbf{S}_1 \cdot \mathbf{S}_2, S_{1z}+S_{2z}]$ which led to $$ i\hbar\epsilon_{i3k}(S_{1k}S_{2i} + S_{1i}S_{2k})$$ and I believe that in this particular case-scenario the antisymmetric-object and symmetric-object wrt to $i \leftrightarrow k$ is the only argument to get zero.
Uh, I see
22:28
this is also the general philosophy behind suspensions on the main site: after they've expired there is no public record of them - we want to avoid some kind of eternal public pillory
MQ dopo le 23 è proibita, Claudio :P
But yeah, as long as that trick serves right what's missing?
@MrFeynman just checking
@Mr.Feynman my professor basically said: "this is a complete set of commuting observables, if you don't believe it, go and prove it yourselves hahahah"
@ACuriousMind it just reminded me of a guy on tiktok saying that being in jail is the greatest honor and hated police lol. SE>real world
@ACuriousMind do you think abstract algebra is cool? I know you took a lot of math but I'm having mixed feelings. It might not help that I accidentally spent 8 hours studying the wrong textbook and trying graduate level math problems yesterday.
Maybe I'm just burned out after that but I feel like physics is just more fun to learn.
@Obliv I do!
22:41
It was hungerford's algebra btw. I for some reason didn't check the cover very well just used the first pdf I found :\
the undergrad text was soo much more forgiving and nice, for what it's worth. But I kind of psyched myself out I think, trying to engage with a grad lvl algebra text.
Like, how do people get to that level.. and why
because playing with structures is fun :)
to me, much of modern abstract algebra is driven by the realization that stuff like groups, rings and algebras appears everywhere in math, and so investigating their properties very generally divorced from any specific application will be useful in many different contexts
Right, I just gotta breathe through it and take it as it comes. I guess it's not terrible if you learn it at a reasonable pace, varying things in your life along the way.
@ACuriousMind Ah, I see. Thank you

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