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Obliv
Obliv
1:12 AM
@Mr.Feynman jon snow was a hero!
as plato or socrates once said: i don't know what I don't know
humility in a truth seeker leads to proper self-awareness :)
maybe an unpopular/useless opinion but "understanding" isn't encompassed fully in analysis (probably obvious?) and while I don't think one or the other is better for physics (phenomenological or mathematical), I think to truly "connect" with the world we have to look at things for what they are and not what we think they should be.
i.e, while it might sound nice to describe things as "laws" of physics, at the end of the day (I think) nature doesn't care how we describe it.
In introductory thermodynamics & classical mech we often talk about things with broad strokes (I guess necessarily) but it's slightly irritating that we describe nature with such things as law of energy conservation etc.
It's like stating a fact about a mathematical structure without detailing a proof. I'd rather read a proof than just be told "that's how it is"
but in the context of physics there aren't proofs like in mathematics. We kind of just have to believe that things make sense and that we can pick apart why things behave a certain way.
@naturallyInconsistent meow?
 
naturallyInconsistent
naturallyInconsistent
1:38 AM
meow~
 
Relativisticcucumber
Relativisticcucumber
2:35 AM
HONK
@Obliv meow.
@SillyGoose LOL
 
user 85795
user 85795
honk
meow
 
naturallyInconsistent
naturallyInconsistent
3:16 AM
H O N K
 
 
3 hours later…
Relativisticcucumber
Relativisticcucumber
6:16 AM
 
Silly Goose
Silly Goose
tell me what becomes of them :0 @Relativisticcucumber
 
Relativisticcucumber
Relativisticcucumber
@SillyGoose i think they die 0.0 isnt it referring to curiosity killed the cat XD
acm is in danger
2
 
user 85795
user 85795
My guess is they end up in Wonderland.
 
Relativisticcucumber
Relativisticcucumber
@user85795 R U A SWIFTIE
 
user 85795
user 85795
Nope.
 
Relativisticcucumber
Relativisticcucumber
6:26 AM
D,:
 
user 85795
user 85795
:D
 
ACuriousMind
ACuriousMind
@Relativisticcucumber D:
 
user 85795
user 85795
@ACuriousMind r u a swiftie
 
ACuriousMind
ACuriousMind
@Obliv the quote is: "I know that I know nothing" (cf. en.wikipedia.org/wiki/I_know_that_I_know_nothing)
 
user 85795
user 85795
6:42 AM
Sounds like a paradoxical statement to me.
cf. I think, therefore I am.
 
nickbros123
nickbros123
All these issues can be resolved by not thinking, I think
Oh wait
 
Ryder Rude
Ryder Rude
7:01 AM
@user85795 exactly. Socrates's sentence is paradoxical. the correct way to interpret it is that it's a proof by contradiction that we do know something
 
user 85795
user 85795
Chernobyl will be habitable again in about 20,000 years due to the long-lasting effects of ground absorption of radiation. TIL
 
Ryder Rude
Ryder Rude
1. If we know nothing, we cannot know that. 2. Therefore, we know that if we knew nothing, we couldnt know that. 3. Therefore, we know something
but the previous proof is based on rules rules of logic which are assumed to be known
 
Mr. Feynman
Mr. Feynman
 
 
1 hour later…
123
123
8:18 AM
Hello Everyone...
What is the criteria to take axis of rotation in rotational motion? Is it always around at which the body rotate ?
 
Slereah
Slereah
If it's a rigid rotation sure
 
123
123
@Slereah Hello. Can we take arbitrary axis of rotation other than axis at which body rotates. Same as we take origin any where to calculate angular momentum and torque?
I am reading KnK they take origin always along the axis of rotation. It is always necessary to take origin on axis of rotation line or we take it anywhere?
I am confused about axis of rotation and origin in rotational motion.
 
Sanjana
Sanjana
8:38 AM
How do we derive the matrix form of non-Cartan generators of some Lie algebra given its roots and weights in any rep and hence the algebra in Cartan-Weyl basis?
I could derive the Cartan generators by simply using the weights of fundamental reps because they are diagonal. I am trying to work out the $SU(3) \sim A_2$ example for now.
 
Sir Cumference
Sir Cumference
I just realized, was there no winter bash for 2023?
ah dang, they got rid of it
 
naturallyInconsistent
naturallyInconsistent
9:18 AM
@Relativisticcucumber miao am alive miahahahaha
 
Ryder Rude
Ryder Rude
this shows that earth is extremely hostile to life
there r mechanisms in place to brutally species-cide everyone every few million years
 
DIRAC1930
DIRAC1930
9:57 AM
@Sanjana I am interested in this also
 
Mr. Feynman
Mr. Feynman
10:21 AM
WTH is Tong doing here? First he considers a snapshot but the whole thing is integrated...?
Of course it is the Nambu Goto action $$S=-T\int d^2\sigma\sqrt{(\dot{X}\cdot X')^2-\dot{X}^2 X'^2}$$
I mean, I can see that in that gauge, at an instant in which there is no kinetic energy, the lagrangian is $R\sqrt{(d\vec{x}/d\sigma)^2}$
I can't see what is going on with that time integral
 
Sanjana
Sanjana
10:46 AM
@Mr.Feynman He is considering a static string whose spatial embedding coordinates $X^i(\tau, \sigma)$ are independent of worldsheet time $\tau$.
The time is evolving but the string is static.
 
DIRAC1930
DIRAC1930
@Sanjana Isn't this equivilant to finding the matrix form of operators. I.e. if we know the weights of all the $k$ basis states, say $|s,s^z_k \rangle>$ and the action of operators $\hat{S}^z, \hat{S}^+, \hat{S}^-$ on these states, cant we just do $S^+_{ij} = \langle s,s^z_i | \hat{S}^+ |s,s^z_j \rangle>$
 
Sanjana
Sanjana
@DIRAC1930 Yes but I am starting from the Cartan matrix.
I don't know what $S^\pm$ looks like: that's the question
 
DIRAC1930
DIRAC1930
We know the action of these just from the roots right?
 
Mr. Feynman
Mr. Feynman
@Sanjana why does it call it "snapshot"?
 
Sanjana
Sanjana
@DIRAC1930 Yes...I know $S^z=\text{diag} (1,-1)$ and $[S^+,S^-]=S^z, [S_+,S_{-}]=S^z$ from the root $\alpha$.
 
DIRAC1930
DIRAC1930
11:00 AM
So why can't you generalize to spin 1, spin 3/2... irreps?
If you know the weights of all states, you have the diagonal generators, and then find the other ones by studying the action of the raising and lowering operators and transfer to matrix form
Which is given from the roots
 
Sanjana
Sanjana
@Mr.Feynman Maybe he wanted to just mean "consider a static string", but instead of that he said "consider a snapshot of a moving string" which is kinda equivalent to a static string? I am not sure about this though. I read this stuff from Zwiebach...He gives excessive details on this, and I relooked into it right now. He is indeed considering a static string too
If u have it, its in section 6.7 of the 2nd edition
 
Mr. Feynman
Mr. Feynman
@Sanjana I'll have it in five seconds ;)
Because I'm close to the library of course!
 
user 85795
user 85795
of course, we all are ;)
 
Sanjana
Sanjana
@Mr.Feynman I thought the library name ends with "gen" :p
 
Mr. Feynman
Mr. Feynman
Yeah nowadays there are libraries around any corner, y'know
Uh, Zwiebach looks more complete than Tong's notes
 
Relativisticcucumber
Relativisticcucumber
11:10 AM
@naturallyInconsistent i never want any cats too die -- they are too precious. only of old age and in their sleep
 
Mr. Feynman
Mr. Feynman
(Thr Librarian was damn quick)
@Relativisticcucumber wym cats don't die
They're just MIA
 
Relativisticcucumber
Relativisticcucumber
fsdlfjdslfjslk i love cats so much eek
one time i went to a cat cafe and they all snubbed me XD
wait have you seen lil dumb dumb? @Mr.Feynman
its my grandmas cat but she invaded my life during covid. she would pounce onto my shoulder and chair during online classes
 
Mr. Feynman
Mr. Feynman
@Relativisticcucumber No, I haven't
 
Relativisticcucumber
Relativisticcucumber
she is v mischievous
user image
2
 
Mr. Feynman
Mr. Feynman
11:27 AM
I starred only one but really the star holds for all of them
 
Mr. Feynman
Mr. Feynman
12:09 PM
I was comparing GSW to Tong's notes and I noticed that (unless I skimmed too fast) they don't use the standard names (Nambu-Goto, Polyakov etc.). Why?
 
Ryder Rude
Ryder Rude
12:44 PM
what is everyones favorite extinct animal
mammoths mustve been cute bundles of joy
 
bolbteppa
bolbteppa
1:04 PM
@Sanjana If there was an easy answer to this you'd see matrix descriptions of E8 etc like nobodies business, instead I don't think I've ever seen them
@Mr.Feynman They do for NG, for Polyakov it's not always called that they reference papers by 5 other people for the origin of that, which some people even call it by...
Polyakov action
In physics, the Polyakov action is an action of the two-dimensional conformal field theory describing the worldsheet of a string in string theory. It was introduced by Stanley Deser and Bruno Zumino and independently by L. Brink, P. Di Vecchia and P. S. Howe in 1976, and has become associated with Alexander Polyakov after he made use of it in quantizing the string in 1981. The action reads: S=T2∫d2σ−hhabgμν(X)∂aXμ(σ)∂bXν(σ),{\displaystyle {\mathcal {S}}={\frac {T}{2}}\int \mathrm {d} ^{2}\sigma \,{\sqrt {-h}}\,h^{ab}g_{\mu \nu }(X)\partial _{a}X^{\mu }(\sigma )\partial _{b}X^{\nu }(\sigma ),}where...
 
Mr. Feynman
Mr. Feynman
@bolbteppa I see
 
DIRAC1930
DIRAC1930
1:20 PM
@bolbteppa Shouldn't this be simple for the SU(3) case since the generators are Hermitian therefore the eigenvectors are orthogonal therefore expressions like (for $su(2)$ case $S^+_{ij}=\langle s, s^z_i | \hat{S}^+ | s, s^z_j \rangle$ should be relativiely straight forward to obtain for different irreps right?
Probably wouldn't take long in Python or Mathematica or something
 
bolbteppa
bolbteppa
I don't know
 
DIRAC1930
DIRAC1930
All the diagonal elements will be $0$ for $S^+$ and $S^-$ without even calculating anything
from the orthogonal condition
since $S^+$ will transform a state from $|s, s^z_i \rangle$ to $|s, s^z_j \rangle$ where $i\neq j$
Hmm maybe it's not so simple for anything other than $\mathfrak{su}(2)$
 
Sanjana
Sanjana
2:00 PM
@DIRAC1930 The problem isn't in the computational time. I don't understand how to do it even in principle. And I agree that I haven't seen any $E_8$ generator matrix stuff but yes, $SU(3)$ laddering operator matrix elements are commonly found as certain combinations of Gell Mann matrices.
@DIRAC1930 I understand what you are saying formally...But I am looking for a method to calculate an explicit expression
 
Sanjana
Sanjana
2:21 PM
@RyderRude Ryder's questions are more unexpected than quantum mechanics :p
 
DIRAC1930
DIRAC1930
2:32 PM
This detailes a general method physics.stackexchange.com/a/590766/310229
 
Sanjana
Sanjana
2:48 PM
@DIRAC1930 This is from $\mathfrak{su}(2) \sim A_1$
The last two equations for Adam's answer when generalized to any simple Lie algebra is supposed to give what we are looking for.
 
DIRAC1930
DIRAC1930
Yes, won't $\mathfrak{su}(3)$ be extremelly similar just with $2$ diagonal generators
 
Sanjana
Sanjana
@DIRAC1930 "similar" is relative :) For me, Gell Mann's matrices is not at all similar to Pauli matrices. I mean...I can't just guess Gell Mann's from Pauli's...Yeah the first three look similar cz $\mathfrak{su}(2)$ is a subalgebra but still, atleast I don't know how this can be done
 
DIRAC1930
DIRAC1930
They should bring out a book that just gives explicit representations lol
I would definitely buy it
 
Sanjana
Sanjana
The farthest I could get into was $$R(H_i) \left(R(E_a)| \mu\rangle \right)=(\alpha_i+\mu_i) \left( R(E_a))|\mu \rangle\right)$$ but this is by simple application of commutators
$H$ are the Cartan generators, $E$ are the non-commuting generators. $\mu$ is a weight and $\alpha$s are roots. $R$ denotes a representation
The $a$s should be $\alpha$s
@DIRAC1930 I am pretty sure this is known...but this is not that insightful maybe that's why its not done explicitly in books...
 
DIRAC1930
DIRAC1930
There's another way but it is probably worse. You can just tensor two representations and symeterize and antisymmeterize it accordingly. This will give you a reducible representation which you will have to manually make it irreducible
I do not recommend this at all since doing the first part is easy in Python or Mathematica but doing the 2nd part is not
 
Sanjana
Sanjana
3:05 PM
@DIRAC1930 How can this give me the laddering operator matrices? I spend last few days learning rules on how to work these out in pen and paper...Verified quite a few from Slansky and found LieART on Mathematica to do this for me, and wrote a short code to generate this using 1. Dynkin labels 2. Young Tableaux (whenever it is a $A_n$ rep exercise) 3. Dimensions (most common form)
Here's how it looks...
The first three lines are the only inputs...
 
DIRAC1930
DIRAC1930
Well for $\mathfrak{su}(2)$, I would have $\mathcal{P}^{\pm}S^+ \otimes S^+\mathcal{P}^{\pm}$. $+$ will give the spin-1 representation and $-$ will give the spin-0 representations (here $\mathcal{P}^{\pm}$ is the symmeterizer ($+$) and antisymmeterizer ($-$). However these representations will be reducible since they will be $2\times 2$ dimensional
However I do not recommend doing what I wrote above
This is exactly what we are doing in qm. The state space is the tensor product of single particle Hilbert spaces which each transform under 1 copy of $\mathfrak{su}(2)$ spin 1/2. Therefore, 2 particles will transform under the tensor product of these and we just project out the spin-1 part and the spin-0 part
 
Obliv
Obliv
3:22 PM
for the van der waals state equation $(P+\frac{aN^2}{V^2})(V-Nb)=NkT$ I don't get the first term added to pressure
I get that the density of a gas, $\frac{N}{V}$, is directly proportional to the potential energy due to attractive forces. So $\frac{aN^2}{V}$ is the effect of the attractive forces for all the molecules on the potential energy.
with some constant $a$ related to the specific molecules
this is the section
Oh I guess it's just adding this contribution nvm i get it
nvm it's covered in 7.4 of schroeder
@Relativisticcucumber since ACM wishes he were a cat, we can surmise that they are not yet one so the danger is minimal.
@naturallyInconsistent on the other hand, I'm not entirely sure is safe.
 
DIRAC1930
DIRAC1930
4:34 PM
Things seem quieter on here than usual
 
 
1 hour later…
DIRAC1930
DIRAC1930
5:54 PM
Where is everyone?
 
Mr. Feynman
Mr. Feynman
On Earth
No ACM no party :(
Uh wait, today is the 27th
@lucabtz We want updates
Ok, ACM is offline and Luca is graduating. That's enough evidence to infer ACM is in the crowd
3
 
John Zimmerman
John Zimmerman
6:30 PM
Is there a theorem that relates a discrete symmetry group to a continusou symmetry group?
I recall noethers' theorem, but
 
ACuriousMind
ACuriousMind
@JohnZimmerman in what sense?
@Mr.Feynman hi :P
 
Obliv
Obliv
what's a continuous symmetry group?
 
ACuriousMind
ACuriousMind
a Lie group (at least in physics that's what we usually mean)
 
Obliv
Obliv
"However, a discrete symmetry can always be reinterpreted as a subset of some higher-dimensional continuous symmetry" off wiki
no source though unfortunately
 
John Zimmerman
John Zimmerman
@ACuriousMind well I was thinking that a mathematical object could admit both global discrete symmetry group as well as a continous symmetry group and I wondered if these two would be independent or mutually exclusive from one another
@Obliv interesting
 
ACuriousMind
ACuriousMind
6:35 PM
@JohnZimmerman there's in general no relation at all, you can have a discrete symmetry that's completely unrelated to the continuous symmetry, or you can have something like rotations and reflections, which you initially might frame as invariance under continuous rotations $\mathrm{SO}(3)$ and discrete reflections $\mathbb{Z}_2$ but which combine into the disconnected continuous $\mathrm{O}(3)$ of rotations and reflections
 
John Zimmerman
John Zimmerman
hmm that's what I was sort of thinking (that there's not a relation)
 
Sanjana
Sanjana
Oh ACM...you arrived :)
Can you pleeeeeeeeeeease say something on this?
 
Obliv
Obliv
What about a lie algebra that consists of an infinite number of discrete symmetry groups lol
 
ACuriousMind
ACuriousMind
@Obliv are you just putting random words together? :P
 
Obliv
Obliv
YES
well i read this off reddit (lol) : "A discrete group has elements which can be numbered. The crystal structure groups are a good example eg the operations you can perform on a cube to get back the same cube (rotations by multiples of pi/2, reflections etc.).

Continuous groups have elements labelled by a continuous parameter. Free rotations space are an example of this. These groups all contain an infinite number of elements in the same way as there are an infinitely many decimal numbers between two other numbers.
 
ACuriousMind
ACuriousMind
6:41 PM
@Sanjana I don't understand the question - what do you mean by "matrix form"? There is no distinguished basis for an arbitrary representation, and matrices are relative to a fixed basis
 
Sanjana
Sanjana
@ACuriousMind If I fix the representation to be the $2$ dimensional defining representation for $su(2)$ and I just want to derive the usual $2 \times 2$ $J^\pm$ matrices where I already have $J^3=\text{diag}(1,-1)$ then?
I am fixing the basis to be the one in which the Cartan generators are diagonal and of a given form. Will that be enough?
 
John Zimmerman
John Zimmerman
@Obliv I haven't made myself clear - but that's because I don't know if the idea is of any merit or exists at all
but thanks
 
ACuriousMind
ACuriousMind
@Sanjana Why would that fix a basis? Matrices can be diagonal in more than one basis.
 
Obliv
Obliv
would there be another noether charge associated with magnetic "charge" if monopoles exist
like gauge invariance for electric charge
 
ACuriousMind
ACuriousMind
@Sanjana the Pauli matrices are special because of the low dimension - the commutation relations yield enough independent equations to have a unique solution in 2-by-2 matrices, this does not hold for higher dimensions
 
Sanjana
Sanjana
6:51 PM
@ACuriousMind Ok...is there no way to derive Gell Mann matrices naturally starting from Cartan matrix information?
@ACuriousMind In some books like Georgi it is given that $E_{(1,0)}=\frac{1}{\sqrt{2}}(T_1+i T_2)$ etc for $su(3)$ and $(1,0)$ is one of the root vectors. If I assume the form of Gell Mann matrices which are $2 T_1,2 T_2$, then by finding the eigenvalues I can see that $(1,0)$ should indeed be the corresponding root vector...but is there no way to do the converse i.e. given this and other root vectors and hence the commutation relations, getting the form of the $E_{(1,0)}$ matrix?
 
ACuriousMind
ACuriousMind
@Sanjana again, you have not, in general, fixed a basis by any of the conditions you've stated
 
John Zimmerman
John Zimmerman
ohhhh
 
ACuriousMind
ACuriousMind
now since the GM matrices are for the fundamental representation, I think once again they might be fixed because of low dimension by the commutation relations
 
John Zimmerman
John Zimmerman
I forgot about D-branes
it stands for dirichlet membrane
 
ACuriousMind
ACuriousMind
@Obliv that depends on your exact formulation in which you model those monopoles, however in the usual formulation their charges will be of a topological nature rather than a Noether charge
 
Obliv
Obliv
7:00 PM
so you'd use lie groups/algebras?
 
Sanjana
Sanjana
@ACuriousMind What about how we do it in the case of $su(2)$ algebra. We have some $J^{\pm}|j m \rangle =\sqrt{j(j+1)-m(m \pm 1)}|j, m \pm 1 \rangle$ for $su(2)$. Is this and its generalization to other algebras not derivable via generic weight theory?
 
ACuriousMind
ACuriousMind
@Sanjana Now there you've done something different: You've fixed a particular basis, namely the $\lvert m = -j\rangle,\dots, \lvert m = j\rangle$
that fixes a unique basis, that's something we can work with
 
Sanjana
Sanjana
@ACuriousMind But this is just the weight system for a particular irrep. Is this also fixing the basis?
 
John Zimmerman
John Zimmerman
Is anyone here interested in p-adic d-brane theory?
 
Sanjana
Sanjana
@ACuriousMind Can I do this for $\mathfrak{su}(3)$ and other algebras...I wrote an almost similar relation here
 
ACuriousMind
ACuriousMind
7:09 PM
@Sanjana how is that similar? you have the unknown $R(E_a)$ in there! The point of your relation for $\mathfrak{su}(2)$ is that you don't have the $J^\pm$ appear on the r.h.s., so this is just an equation that gives you the matrix elements of $J^\pm$ in the $\lvert jm\rangle$ basis
you should write the r.h.s. as $R(E_\alpha)\lvert \mu\rangle = \lvert \mu + \alpha\rangle$
however this doesn't define a basis when your weight spaces are not one-dimensional as in the case of $\mathfrak{su}(2)$
it's just a vector of weight $\mu$ mapped onto a vector with weight $\mu + \alpha$
 
Sanjana
Sanjana
This is weird...We can derive matrices for the Cartan generators but not the non-Cartan ones?
One article even says "Knowing this one can find the representation of $H_\alpha$ in the $T_3, T_8$ basis. Since the main
goal of this section is to express the roots and weights in the T-basis, one does not need to find the representation of the $E_{\pm \alpha_i}$ in this basis. The additional computations can be done abstractly."
"need to"
 
ACuriousMind
ACuriousMind
I'm not excluding that there is a way but I don't think it's something you need often :P
for anything but low-dimensional stuff like the literally two examples you've given, why would you torture yourself with an explicit matrix representation :P
 
Sanjana
Sanjana
nooooooo dont give up :)
 
ACuriousMind
ACuriousMind
and in both your examples I don't think anyone derives those via Lie algebra theory, those are just very natural bases for 2 and 3 dimensional matrices to choose
 
John Zimmerman
John Zimmerman
don't give up 🙌🏼
 
Sanjana
Sanjana
7:22 PM
@ACuriousMind My actual target is to derive for $E_8$ but then I noticed I dont know how to do this even for simpler cases....
 
bolbteppa
bolbteppa
@Sanjana these notes might help, e.g. it shows the $E_{\alpha}$ as the usual matrices for $SU(3)$, a big problem with all this is that if you start with the Cartan-Weyl basis you've still got to deal with all the $[E_{\alpha},E_{\beta}]$ commutators, i.e. all the $(\mathrm{ad} E_a)^{1-A_{ab}} (E_b) = 0$'s
Nobody ever writes anything like this out, it's not going to be useful, e.g. in $E_8$ are they going to be $248 \times 248$ matrices or half that or something?
 
ACuriousMind
ACuriousMind
@Sanjana see maths.usyd.edu.au/u/bobh/hrt.pdf for the actual algorithmic computation of matrices of $E_8$; you should note that while the algorithm indeed uses the Cartan-Weyl basis it is far from straightforward
 
Silly Goose
Silly Goose
Does anyone know of a nice way to algorithmically organize the generalized gellmann matrices?
each package i use has a different order for them :P
Oh wait i think i know a way
 
ACuriousMind
ACuriousMind
why is this chat so obsessed with matrices :P
 
John Zimmerman
John Zimmerman
I hate matrices
but they do serve a purpose
and i respect that
 
Silly Goose
Silly Goose
7:34 PM
I would try to do coordinate free optimization if i could :P
but the manifold optimization packages i have seen don't seem to support optimizing over SU(N) sadly
 
Sanjana
Sanjana
@bolbteppa They just put the matrices there adding the Gell Mann matrices up or something :)
I wanted to "derive" it somehow directly from the Cartan Matrix. I understand that ofcourse I have to choose a basis. I already chose to express the roots and weights in terms of $e_x,e_y$ i.e. the standard Euclidean/Cartesian orthogonal basis. I didn't think that I would have to do anything more
@bolbteppa Hmm...but it bothers me that it is not doable even in principle...
@ACuriousMind Ok...it is doable
 
ACuriousMind
ACuriousMind
note that they only do it for the adjoint, where the matrix is indeed fixed by the structure constants, I didn't really understand when skimming it whether this generalizes to all representations
 
bolbteppa
bolbteppa
What you're trying to do is derive matrices for every generator in the algebra just from the Cartan matrix which is expressed only in terms of the simple roots, but the Cartan matrix wont do that, you've got all these $[E_{\alpha},[E_{\alpha},E_{\beta}]]$-type commutators to worry about which produce the remaining elements/roots of the Lie algebra
 
Sanjana
Sanjana
@bolbteppa But I can produce the full set of roots from Cartan matrix...either by highest weight theory or by doing some root string computations.
In that sense I thought that the full Lie algebra is encoded within the Cartan matrix
I am able to reproduce the algebra in Cartan Weyl basis using the roots...I could get the Cartan generators in matrix form too. Getting the $E_\alpha$s would give it a finishing touchhh
 
bolbteppa
bolbteppa
As far as I can tell, the best all this is going to give you matrices with a single $1$ above of below the diagonal to signify a raising/lowering operator, maybe you could then put your massive 248 or whatever matrices into Generalized Gell-mann form I'm not sure
 
Sanjana
Sanjana
7:54 PM
I think the point(for me) not getting the result but knowing the method how to do it, and the simplicity of the result increasing its tantalizing appeal more...
 
bolbteppa
bolbteppa
As far as I can tell, writing the $E_{\alpha}$ matrices as matrices with a single $1$ off-diagonal, and then using the Gell-mann basis, is what you wanted above
Good luck doing this for $E_8$
 
Silly Goose
Silly Goose
this talk is interesting (Four Ways of Thinking: Statistical, Interactive, Chaotic and Complex - David Sumpter
)[https://www.youtube.com/watch?v=PPCfDe8TfJQ]
oh well about hyperlinking it XD
 
Sanjana
Sanjana
@bolbteppa That's the result that I wanted, yes. But I wanted to know how/whether one can arrive at this form of $E_\alpha$s with some knowledge of the roots from which I could derive the Cartan generators $\lambda_3,\lambda_8$ explicitly. I think, I missaid that I wanted it for $E_8$...I wanted to know how one can do it for any algebra, and a small non-trivial example would satisfy me
 
bolbteppa
bolbteppa
As far as I can tell what I said will work for any algebra that has this Cartan-Weyl decomposition, presumably the ordering of the matrices corresponds with the ordering of positive/negative roots
 
Sanjana
Sanjana
@bolbteppa I think your result is indeed correct.
But I dont understand the method
 
bolbteppa
bolbteppa
8:00 PM
I don't know about the $H$ matrices in general
 
Mr. Feynman
Mr. Feynman
8:11 PM
Now I understand why Thorgott from the MSE room claimed to hate root systems
 
Sanjana
Sanjana
The $H$s are always going to be given with the help of roots themselves, I think.
They are just diagonal with the diagonal elements given by $a^{(i)}_j$ for the $i$th Cartan generator Cartan generator $H^i$ where $a^{(i)}_j$ is the $j$th component of the $i$th root vector and $i=1,...,\text{rank}(\mathfrak{g})$ and $j=1,...,\text{dim(R)}$.
 
bolbteppa
bolbteppa
Is that going to reproduce $[H_a,E_b] = A_{ba} E_b$ and $[E_a,E_{-b}] = \delta_{ab} H_{a}$
 
Sanjana
Sanjana
It gives the correct 3rd and 8th Gell Mann matrices and the correct 3rd Pauli matrix
@bolbteppa So assuming the other matrices, this is indeed reproduced
 
bolbteppa
bolbteppa
But what if you did this for $E_8$
 
Sanjana
Sanjana
I don't know the $E$s for $E_8$ for sure :p
@bolbteppa I will try it sometime using your proposed $E$s
But some of the $E$s in $Sp(2N)$ have some $-1$s here and there :(
Omg how did they derive this ever!
The notes you sent was from lectures by Van Nieuwenhuizen...I heard he is a good teacher
 
bolbteppa
bolbteppa
8:24 PM
I have no idea why those $C_n$ matrices are different, likely the above is wrong and things are just an even worse mess
This matrix approach is not going to be very useful so nobody discusses it but yeah...
 
Sanjana
Sanjana
Hmm...u r right
 
bolbteppa
bolbteppa
Now I'd like to know the answer to this, why can't they also just be $E_{ij} = \delta_{ij}$ matrices :\ Maybe it's due to a choice they make I don't know...
 
Sanjana
Sanjana
@bolbteppa If they are identity matrices, won't they become part of the CSA?
 
bolbteppa
bolbteppa
That's supposed to mean a $1$ in the $ij$ entry, $i \neq j$
 
lucabtz
lucabtz
9:31 PM
@ACuriousMind eh physicists
Anyhow I'm officially unemployed
Graduated today
10
 
ACuriousMind
ACuriousMind
congrats
3
 
lucabtz
lucabtz
@ACuriousMind thanks
 
bolbteppa
bolbteppa
Congrats, are you going to stay on with this stuff
 
lucabtz
lucabtz
10:09 PM
@bolbteppa yeah that's the plan
 
123
123
In the above example of Knk, When they are say taking origin B. Then why diagram on the right side coordinate system origin is not shifted above to take origin B.
I am trying to understand because angular momentum and torque depend on the choice of origin. But book taking different origin without changing coordinate system origin. Why?
 
DIRAC1930
DIRAC1930
11:00 PM
This should be doable but a labourious task for $\mathfrak{su}(3)$
But I really doubt you will gain anything from doing this for $\mathfrak{su}(3)$
Even doing it for $\mathfrak{su}(2)$ isn't particularly insightful
 

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