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Silly Goose
Silly Goose
8:33 AM
shall we start the TPS party once more
@ACuriousMind i read a tiny snippet of category theory today :P and it suggests itself in this TPS context i feel
can I think of $U$ here as a map between isomorphisms?
or like
a map between sets of isomorphisms
or harness some greater unifying power to organize this information more clearly XD
i badly want to think of $U$ as some analogue to a right action on a set of isomorphisms
but we're mapping between sets of isomorphisms defined on potentially different domains, codomains
and i feel like thinking about it just as something acting on a set washes out all the interesting structure that we actually want to pay attention to :P so maybe that route is invalid
 
ACuriousMind
ACuriousMind
8:52 AM
@SillyGoose sure: Any homomorphism $f : X\to Y$ induces - via composition of functions - a map from homomorphism $\mathrm{Hom}(Y,A)$ to homomorphisms $\mathrm{Hom}(X,A)$ by $- \circ f: \mathrm{Hom}(Y,A)\to \mathrm{Hom}(X,A), a \mapsto a\circ f$
I don't particularly like the usage of the same symbol for this induced map as in your screenshot, but it's common enough
also I would write eq. (1.10) as $U([T']) = [T]$ :P
 
Silly Goose
Silly Goose
is it not an equivalent perspective to say that the unitary includes what you called before the fixed isomorphism
also so it lasted for a day XD
 
ACuriousMind
ACuriousMind
@SillyGoose that's too vague for me to say whether it's right or wrong :P
 
Silly Goose
Silly Goose
previously there was a perceived problem in comparing $\bigotimes_i \mathcal{H}_i$ with $\bigotimes_i \mathcal{H}_i'$. the proposed solution was to recognize that these tensor product factorizations must be isomorphic to begin with. hence for the equivalence of TPSes definition to be satisfied, there exists an isomorphism between such tensor product factorizations.
oh wait nvm i recall the problem
what is the benefit of writing 1.10 in that way? @ACuriousMind
hm well idk... i know that for the equiv class equality condition to be satisfied, the equiv classes must be related by a bijective map. i am struggling to build up the additional structure of the map between the two equiv classes required.
okay okay XD
maybe ill justt be content with this
hm well that's not quite right. it is too strict a requirement. bleh i shall be back after getting more thoguhts togehter :P
 
ACuriousMind
ACuriousMind
9:14 AM
@SillyGoose I thought we were in a setting here where we already have fixed an iso $\mathcal{H}_i\cong\mathcal{H}_{j_i}'$ so that both the $T$s and the $T'$ have the same target $\otimes_i \mathcal{H}_i$
because otherwise it makes no sense to talk about $U : \mathcal{H}\to\mathcal{H'}$ mapping a $T'$ to a $T$
 
Silly Goose
Silly Goose
hm well i am trying to build up the motivation behind fixing such an iso. i feel like the reason for having an iso should fall out of the definition of a TPS (singular, not equivalence between)
because if we are saying $T \in \mathcal{T}$ and $T'U \in \mathcal{T}$ and nothing else, then we should be able to say that okay well if $T'U$ has the target $\bigotimes_i \mathcal{H}_i'$ then this $\mathcal{H}_i$ is constrained to be equivalent TPS-wise to $\mathcal{H}_i$? But by definition of TPS, not anything else. idk if that makes sense
or is that what you mean also?
 
ACuriousMind
ACuriousMind
@SillyGoose your definition of a TPS $\mathcal{T}$ fixes $\otimes_i \mathcal{H}_i$ as the target of all $T\in\mathcal{T}$.
you can't even say $T'U\in\mathcal{T}$ if $T'U$ isn't a map with target $\otimes_i \mathcal{H}_i$
 
Silly Goose
Silly Goose
ah i see. so that is where you are saying the implcit iso is coming from
yesyes i see it now :D
ah i am confused. because i feel like $\otimes_i \mathcal{H}_i$ is a representative of its equiv class. so that $T'U$ could have other targets (as in have other representatives of the equiv class in question as targets). but the otimes notation suggests invariance of $\sigma$-permutations and change of bases....
the last point being confusing because how would you even write the target being any other representative? it seems like we are notationally working on a zoomed out level so the invariant details are washed away (bases are not indicated, $\sigma$-permutations have no notational effect)
so then it is also confusing because then $\otimes_i \mathcal{H}_i$ isn't a representative...it is like all the equiv class all at once :P
 
ACuriousMind
ACuriousMind
let's look at your definition:
This fixed the target of all $T$ in the same equivalence class to be $\mathcal{H}_1\otimes\mathcal{H}_2\otimes\cdots$, there is no indication here that the target can change
 
Silly Goose
Silly Goose
9:30 AM
i thought the "and permutations of subsystems" indicates that the target can change
 
ACuriousMind
ACuriousMind
@SillyGoose ah, no
 
Silly Goose
Silly Goose
D: oh dear
 
ACuriousMind
ACuriousMind
by this definition $T_1 T_2^{-1}$ is a map $\otimes_i \mathcal{H}_i \to \otimes_i\mathcal{H}_i$
the "permutations of subsystems" clause means in technical terms the following: If I have an isomorphism $f_{jk} : \mathcal{H}_j\to\mathcal{H}_k$, then this induces a "permutation isomorphism" $\sigma_{jk} : \otimes_i \mathcal{H}_i \to \otimes_i \mathcal{H}_i , \dots\otimes v_j\otimes\dots\otimes v_k\dots \mapsto \dots\otimes f^{-1}(v_k)\otimes\dots\otimes f(v_j)\dots$
the definition says that $T_1T_2^{-1} = \prod_i \sigma_i \circ ( \otimes_i U_i)$ where $\sigma_i$ are such "permutation isomorphisms" and the $U_i : \mathcal{H}_i \to \mathcal{H}_i$ are local unitaries
 
Silly Goose
Silly Goose
what are the $v_i$ here?
 
ACuriousMind
ACuriousMind
just elements of $\mathcal{H}_i$
I'm saying the permutation isomorphism is defined on simple tensors by taking the factor $v_j$ from the $\mathcal{H}_j$ and replacing the $v_k$ factor by $f(v_j)$, while putting $f^{-1}(v_k)$ into the slot that is now free
this is just the annoying formalization of what "the tensor product is commutative" means for two isomorphic factors
 
Silly Goose
Silly Goose
9:48 AM
so what you were saying the other day is that this manifests in terms of these $\sigma$-permutations of isomorphic tensor factors
so in that sense the targets are the same because if you swap two notationally identical factors you end up with the same target
or is that still inaccurate :P
 
ACuriousMind
ACuriousMind
I mean the targets are isomorphic
but the thing is that in the abstract setting $\mathcal{H}_j$ and $\mathcal{H}_k$ are not the same thing even if they're isomorphic
in a concrete example you might just write $\mathcal{H}_j = \mathcal{H}_k = \mathbb{C}^n$ and this whole "permutation isomorphism" is literally just the swap of the $j$-th and $k$-th factor and $f$ is the identity
but in the abstract you can't just say "swap the $j$-th and $k$-th factor" because elements of $\mathcal{H}_k$ are not elements of $\mathcal{H}_j$ without the isomorphism $f$
 
Silly Goose
Silly Goose
hm okay i suppose that makes sense.
wait so what do you mean by fixed when you say the target is fixed
 
ACuriousMind
ACuriousMind
10:11 AM
@SillyGoose I mean it is always the same space, $\otimes_i \mathcal{H}_i$ with the same $\mathcal{H}_i$
 
 
5 hours later…
Amit
Amit
3:16 PM
I'm trying to understand a bit better what you can / can't do with vectors on a manifold's tangent spaces. So I was wondering, suppose I have some $V\in{T_pM}$ on a manifold with connection. Can I write a "small" change in this vector in the following way: $\Delta{V}=(\nabla_a{V})(\Delta{X}^a)$ where $\Delta{X}^a$ is a small displacement in the $a$ direction. In components that would be $\Delta{V^n}=(\partial_a{V^n}+V^m\Gamma^n_{ma})\Delta{X^a}$
Even though it's clear that it's completely chart dependent, I wanted to know if this is even well defined on a specific chart. Because it looks like I am getting a "distance" without a metric, but assuming I only use it as part of assuming $\Delta{X}^a \rightarrow 0$ would it be ok?
 
ACuriousMind
ACuriousMind
3:35 PM
@Amit What does $\nabla_a V$ mean if $V\in T_p M$
you need a vector field for the derivative to make sense, not just a single tangent vector
 
Amit
Amit
Yes, you're right, I'm assuming also it is a vector field. I mentioned only $p$ to mean I am deriving at that point
Should have written $V(p) \in T_{p}M$
 
ACuriousMind
ACuriousMind
also, what sort of object are the $\Delta X^a$ supposed to be
 
Amit
Amit
Chart objects, that is, I have coordinates $X^1,..,X^n$ and I define a small increment directly by using them
 
Obliv
Obliv
@ACuriousMind how come GR doesnt work in quantized for
form*
Like i mean curvature of spacetime in the realm of small particles
and following geodesics
is it because spacetime isnt locally flat in the quantum lvl
 
ACuriousMind
ACuriousMind
@Amit and what meaning do you think is there to multiply some small coordinate increment with a vector at the point $p$?
What you can do is talk about the vector-valued 1-form $(\nabla_\mu V)\mathrm{d}x^\mu$
 
Amit
Amit
3:43 PM
With the derivative of a vector... well the idea was actually to see if I can use it to derive the Riemann curvature tensor
 
ACuriousMind
ACuriousMind
which, as $V = V^\mu \partial_\mu$ is equivalently a mixed tensor $\nabla_\mu V^\nu (\partial_\nu\otimes\mathrm{d}x^\mu)$
 
Amit
Amit
But can you write the $\otimes$ with the contraction like that?
I mean what sort of object would you get? On the one hand it should be a scalar, on the other hand the $\otimes$ means it can't be?
 
ACuriousMind
ACuriousMind
@Amit the relation between covariant derivatives around a loop and the curvature tensor is called the Ambrose-Singer theorem
@Amit It's a (1,1)-tensor
you probably just aren't used to my notation :P
 
Amit
Amit
Ah I see, because you're writing it "post-fix" so it remains a tensor
But are you meaning to say that this tensor, given a "small vector" and a "small coordinate change" will do what I intended?
 
bolbteppa
bolbteppa
In other words, from $\partial_{\mu} \mathbf{A} = \partial_{\mu} (A^{\nu} \mathbf{e}_{\nu}) = (\partial_{\mu} A^{\nu} + \Gamma^{\nu}_{\mu \rho} A^{\rho}) \mathbf{e}_{\nu} = \nabla_{\mu} A^{\nu} \mathbf{e}_{\nu}$ you have
$$d \mathbf{A} = (\partial_{\mu} A^{\nu} + \Gamma^{\nu}_{\mu \rho} A^{\rho}) dx^{\mu} \mathbf{e}_{\nu} = \nabla_{\mu} A^{\nu} \mathbf{e}_{\nu} dx^{\mu} = \frac{\partial \mathbf{A}}{\partial x^{\mu}} dx^{\mu}$$
 
Amit
Amit
3:48 PM
is the $\nu$ redundant on the LHS?
 
Obliv
Obliv
Wait are tiny blackholes made at the LHC often?
 
Amit
Amit
ok :)
 
bolbteppa
bolbteppa
Typo
 
Obliv
Obliv
Or are they not really blackholes but mathematical blackholes
like high enough energy in small enough space
 
Amit
Amit
but how can you "make a number" from $dA$
 
Obliv
Obliv
3:51 PM
referring to "This creates a black hole. If you scatter them at even higher energy, you would make an even bigger black hole, because the Schwarzschild radius grows with mass. So the harder you try to study shorter distances, the worse off you are: you make black holes that are bigger and bigger and swallow up ever-larger distances."
Im not sure if hes talking out of experience or experiment or theoretically
tiny blackholes would indicate gravity exists at that lvl at least
 
Amit
Amit
@bolbteppa I mean, I know this exterior derivative is a one form, one form can operate on vectors to produce scalars. So would there be a significance to the scalar produced by the operation $dA(\partial_{a})$ for some coordinate direction $a$?
In other words, is it true in some sense that $dA(\partial_{a}) = \Delta{A^a}$ ?
Looks wrong I know...
lol it's not clear why you would star that but that's funny
 
ACuriousMind
ACuriousMind
@Obliv there is no experimental evidence for micro black holes that I know of
 
bolbteppa
bolbteppa
Don't know what you're asking, $d \mathbf{A} = \frac{\partial \mathbf{A}}{\partial x^{\mu}} dx^{\mu}$ is technically a vector-valued 1-form, it spits out vectors when you use the $dx^{\mu}$ to 'eat' vectors, maybe you mean the coefficients $\Delta A^{\nu}$ in $d \mathbf{A} = \Delta A^{\nu} \mathbf{e}_{\nu} = (\nabla_{\mu} A^{\nu} dx^{\mu}) \mathbf{e}_{\nu}$
 
ACuriousMind
ACuriousMind
the point of the answer isn't so much that this actually happens, but that a naive approach to quantizing gravity would predict these large block holes at high energy, but that that isn't how quantum field theories work, so unlike other non-renormalizable theories, people do not expect to just find some "more fundamental QFT" that solves the problem
 
Amit
Amit
spits out scalars right?
 
ACuriousMind
ACuriousMind
4:03 PM
no, when you feed this a vector $\partial_\nu$ the result is the vector $\partial_\nu \mathbf{A}$
 
Amit
Amit
Oh, right.
So these coefficients , are just the components of the covariant derivative
 
ACuriousMind
ACuriousMind
sure
 
Amit
Amit
They certainly don't have any "length" units, which was what I was trying to get to
 
ACuriousMind
ACuriousMind
I'm not really sure what you're trying to do
 
Amit
Amit
Basically trying to understand how formally correct is that mess I've written here: https://math.stackexchange.com/questions/4649152/is-this-a-correct-way-to-derive-the-riemann-curvature-tensor
:)
 
ACuriousMind
ACuriousMind
4:07 PM
I think you're just trying to get at the Ambrose-Singer theorem I linked but you're using very strange notation :P
 
Amit
Amit
BTW I will definitely give the Ambrose-Singer theorem a look @ACuriousMind , thank you for that
 
bolbteppa
bolbteppa
You can derive the Riemann Curvature from e.g. $\partial_{\mu} \partial_{\nu} \mathbf{A} = [\partial_{\mu} , \partial_{\nu}] \mathbf{A} + \partial_{\nu} \partial_{\mu} \mathbf{A}$
 
Amit
Amit
lol, yes I am definitely using strange notation. I am only trying to see whether it's only strange or also wrong lol
@bolbteppa That's kind of the definition. I wanted to verify that the definition comes from this "closed loop" idea
 
ACuriousMind
ACuriousMind
when you look at how a vector changes when you parallel transport it around a parallelogram and then shrink the parallelogram to zero, you get that the change in the vector is the commutator of covariant derivatives, i.e. the curvature tensor
 
Amit
Amit
Okay that's part of my confusion, doesn't parallel transport mean the derivative vanishes?
 
ACuriousMind
ACuriousMind
4:10 PM
a formal presentation of this idea int he specific context of Riemannian geometry is here
@Amit the crucial thing is what derivative
when you transport along the $X$ direction, the covariant derivative with respect to $X$ vanishes, but not the one with respect to $Y$
and vice versa
 
Amit
Amit
Oh, right.. that makes sense... so in the definition of the curvature tensor we always derive in a direction which is not the one we parallel transport by
And the tensor itself is just the information about how such combination of parallel transport in direction $a$ & derivative wrt $b$ relate to one another?
Anyway, thank you I think that's probably clear enough, I just need to give it a bit more time to sink in :)
 
ACuriousMind
ACuriousMind
@Amit it tells you how much parallel transporting the vector around an infinitesimal parallelogram spanned by $a$ and $b$ changes the vector
 
Amit
Amit
When I can't understand something 100% I settle for 90% understanding and 10% dogma. Don't judge me!! :)
@ACuriousMind I see. Makes a lot of sense. Thank you, I think it really clicked when you said that the derivative only vanishes in the direction of parallel transport
 
ACuriousMind
ACuriousMind
a nicer version of the derivation than the one on Wiki (which I think also contains at least one typo) is here
 
Slereah
Slereah
the Riemann tensor basically tells you how much the space differs from Euclidian space
 
Amit
Amit
4:21 PM
@ACuriousMind Thanks!
 
Slereah
Slereah
In Euclidian space, parallel transport around a parallelogram in Cartesian coordinates will not move the vector's components at all
The curvature tells you precisely how much that differs here
If you have 2 dimensions it's 4 components, one to tell you how the x component is affected in the x direction, the x component in the y direction, etc etc
Well that's the Ricci tensor
 
Obliv
Obliv
i wonder why ive never heard of Dr. William Gilbert in any of my classes at school. Dude basically pioneered EM magcraft.com/william-gilbert
he da goat
 
Slereah
Slereah
Riemann will tell you about the transformation of a vector along those two displacements
 
Amit
Amit
Can you reconstruct Riemann from Ricci?
Because I know Riemann has a lot less independent components than overall components
 
ACuriousMind
ACuriousMind
no, there are manifolds that are Ricci-flat but not flat
 
Amit
Amit
4:24 PM
Oh, cool.
 
Slereah
Slereah
You can split the Riemann tensor into Ricci parts and non-Ricci parts
 
Amit
Amit
Ricci and Mannci?
 
Slereah
Slereah
Ricci decomposition
In the mathematical fields of Riemannian and pseudo-Riemannian geometry, the Ricci decomposition is a way of breaking up the Riemann curvature tensor of a Riemannian or pseudo-Riemannian manifold into pieces with special algebraic properties. This decomposition is of fundamental importance in Riemannian and pseudo-Riemannian geometry. == Definition of the decomposition == Let (M,g) be a Riemannian or pseudo-Riemannian n-manifold. Consider its Riemann curvature, as a (0,4)-tensor field. This article will follow the sign convention R i...
 
Amit
Amit
@Slereah Cool, I'll get there in due time :)
 
ACuriousMind
ACuriousMind
@Obliv no he didn't
he thought electricity and magnetism were fundamentally different things :P
 
Slereah
Slereah
4:27 PM
He also thought that there was a giant magnetic island at the north pole
 
ACuriousMind
ACuriousMind
Physics classes are confusing enough as it is, you usually don't learn about phlogiston theorists either even though they often contributed considerably to our understanding of fire
 
Slereah
Slereah
you don't even really learn about anything pre-Galileo
 
ACuriousMind
ACuriousMind
you learn about Maxwell because Maxwell essentially figured out EM as we know it today
 
Amit
Amit
Faraday was probably the most underrated EM guy ever :) He was considered an inventor and not as a scientist
 
Obliv
Obliv
Well we also get virtual experiments which are literally pointless, i want to learn physics not recite my textbooks
 
ACuriousMind
ACuriousMind
4:29 PM
I don't know what a virtual experiment is
 
Slereah
Slereah
just head to the tower of Pisa with a bowling bowl
 
Obliv
Obliv
Its a sad attempt at doing a physics experiment online for when covid hit
I just want to be endowed with a grant from queen elizabeth I like dr gilbert and do physics at my liesure
:(
 
ACuriousMind
ACuriousMind
I mean most scientists pre-late modernity were independently wealthy people
not a lot of opportunities to become a scientist when you were a factory worker or farmer
 
Obliv
Obliv
Yup, its the same today. You have more resources but experiment is still limited by monies
 
ACuriousMind
ACuriousMind
@Obliv no, not at all the same as today
 
Obliv
Obliv
4:34 PM
You can still be born to a poor family in a 3rd world country is my point
but definitely near impossible back then
 
ACuriousMind
ACuriousMind
yes, but just because we haven't eliminated this kind of barriers doesn't mean it's the "same"
 
Obliv
Obliv
What you said some time ago about how humans have existed in their current form for tens of thousands of yrs but modern civillization only existed for ~5000 yrs just goes to show how difficult it is to get to this point
 
ACuriousMind
ACuriousMind
and in particular it is not the case today that you need independent wealth to become a scientist - science today is mostly funded by governments and companies
 
Amit
Amit
 
Obliv
Obliv
@ACuriousMind i meant like having wealth today translates to better opportunities for research
 
Amit
Amit
4:37 PM
The situation today?
 
ACuriousMind
ACuriousMind
@Obliv how
in what way could a scientist do better research if they were a millionaire
 
Obliv
Obliv
I was more thinking a billionaire
because you can buy the equipment and stuff that the govt wont fund you
Basically that @Amit
 
Amit
Amit
I mean you probably can't generalize, I'm sure some scientists are well aligned with the research interests of their advisors, research funds, etc... but I don't know what's the statistics on that. My guess is that most of them have to compromise on some level and a lot more interesting things would be done if they were given more freedom. But I hope I'm wrong :)
 
bolbteppa
bolbteppa
How about this for a quick derivation in terms of closed loops
\begin{align}
\oint_{\partial S} d \mathbf{A} &= \oint_{\partial S} dx^{\mu} \partial_{\mu} \mathbf{A} = \iint_S dx^{\mu} d(\partial_{\mu} \mathbf{A}) = \iint_S dx^{\mu} \wedge dx^{\nu} \partial_{\nu} \partial_{\mu} \mathbf{A} = \frac{1}{2} \iint_S dx^{\mu} \wedge dx^{\nu} [\partial_{\nu} , \partial_{\mu} ]\mathbf{A} \\
&= \frac{1}{2} \iint_S dx^{\mu} \wedge dx^{\nu} R^{\rho} _{ \ \ \sigma \nu \mu} A^{\sigma} \mathbf{e} _{\rho}
\end{align}
 
Amit
Amit
Borderline witchcraft!!! :) Actually I don't know enough about integration on manifolds to understand this. But I see you're switching from a boundary to a surface integral basically, this makes sense I guess
 
bolbteppa
bolbteppa
4:51 PM
I don't know how to integrate on manifolds either yet I just did it
Technically I just integrated a vector-valued one-form on a curved manifold using baby calculus
 
Amit
Amit
So @ACuriousMind can perhaps comment if what you did without knowing is unknowingly correct
baby calculus, is that a moral type of calculus?
or is that a teeth problem?
It also reminds me that I was wondering whether the "good old" curl from vector calculus is related to the Riemann curvature tensor. It seems like it has something to do with it... perhaps even its generalization?
baby curl
 
Obliv
Obliv
How does a permanent magnet work i thought magnetic fields were produced by moving charges
 
bolbteppa
bolbteppa
Here you are using stokes' theorem for differential forms, $\int_{\partial S} \omega = \int_S d \omega$, so differential forms are the 'generalization' in this situation
Generalized Stokes theorem
In vector calculus and differential geometry the generalized Stokes theorem (sometimes with apostrophe as Stokes' theorem or Stokes's theorem), also called the Stokes–Cartan theorem, is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus. In particular, the fundamental theorem of calculus is the special case where the manifold is a line segment, Green’s theorem and Stokes' theorem are the cases of a surface in R 2...
 
Amit
Amit
Yes, I know that much. But I don't feel comfortable enough with this stuff yet
@Obliv I think it's related to the QM "spin" of electrons
 
Obliv
Obliv
Ohh thats why its called spin
but im told its not to be thought of actually spinning, why not
 
Amit
Amit
5:00 PM
The word is probably where the similarity ends
Read about the Stern Gerlach experiment, I think it will give you an idea why that is a peculiar kind of spin
 
ACuriousMind
ACuriousMind
5:20 PM
@Amit you don't need spin necessarily
just the orbital angular momentum of the electrons around the atoms is enough to get magnetic fields
it's true that spin contributes to that because it is also a kind of angular momentum but for the basic idea it isn't necessary
 
Amit
Amit
which one is the greater contribution in a permanent magnet? or does it depend on the specific material?
 
ACuriousMind
ACuriousMind
I'm not sure you can so cleanly separate the contributions
because the spin and orbital angular momenta couple in non-trivial ways to give the total angular momenta
 
Amit
Amit
interesting, thanks I didn't know that
 
Obliv
Obliv
Regardless, that raised some more questions like could the SG experiment be done then with hydrogen atoms?
Any electrically neutral atom* so hydrogen as an example
 
ACuriousMind
ACuriousMind
but there are certainly cases where the spin doesn't contribute at all because if the atom has all the orbitals filled in a way that every spin-up is paired with a spin-down
 
Amit
Amit
5:24 PM
oh, yeah that will cancel the overall spin, cool
 
ACuriousMind
ACuriousMind
 
Obliv
Obliv
Cool
would electron orbital angular momentum cause deflection of the atoms along the inhomogeneous mag field
seems like it could contribute to the deflection
 
ACuriousMind
ACuriousMind
yes, but again only unpaired ones
silver has all shells filled completely and then just a single electron in the outermost shell, which has 0 orbital angular momentum
so there's no orbital angular momentum contribution there
 
Obliv
Obliv
If total angular momenta is quantized then would both spin and electron orbital angular momentum be as well?
 
ACuriousMind
ACuriousMind
sure
 
Obliv
Obliv
5:30 PM
Would Hydrogen atoms have 0 orbital angular momentum?
since its 1 in the shell
 
ACuriousMind
ACuriousMind
in the ground state, yes
 
Obliv
Obliv
In general when two or more electrons are in the outer shell it will have orbital angular momentum
because the charges repel each other? or up until it is filled
so it will have some stability
 
ACuriousMind
ACuriousMind
at this point you probably should just learn the shell model of the atom :P
 
Slereah
Slereah
like an egg?
 
Obliv
Obliv
Why is Leibniz's Monad description a contradiction to anything? It seems like hand wavy metaphysics
Or rather how can metaphysics in general contradict logic/math if its construction isnt rigorous itself
contradiction has a well defined meaning in logic
 
ACuriousMind
ACuriousMind
5:44 PM
what
 
Obliv
Obliv
in the wiki page "history of subatomic particles" one excerpt "In contrast, certain ideas of Gottfried Wilhelm Leibniz (see Monadology) contradict to almost everything known in modern physics"
it seems like a weird thing to say
thats like saying religion contradicts science
when religion isnt even the same language as math/science
wait maybe a bad example
 
ACuriousMind
ACuriousMind
@Obliv that particular view is known as NOMA and not universally shared - plenty of both religious and non-religious people disagree with this assertion
 
Slereah
Slereah
physics is just natural philosophy
 
ACuriousMind
ACuriousMind
@Obliv I don't really know anything specific about Leibniz' monads but the claim here is simply that Leibniz thought that there was a single fundamental "stuff" everything else was made out of and our modern understanding is that there are several different kinds of fundamental particles that constitute everything else
 
Obliv
Obliv
Oh I see. I dont know much about it either I just thought how could something metaphysical like that contradict physics if its not even a real theory
maybe too many fundamental particles ;)
 
Slereah
Slereah
5:56 PM
are strings monads
 
Obliv
Obliv
If a fundamental particle is anything like prime numbers, then im afraid we will need a larger standard model
or a generalized one
 
Amit
Amit
I think Leibniz was greatly motivated by the desire to refute Newton... and he was brilliant enough that if differential geometry was evolved enough at his time, perhaps he could do it. He complained to a certain princess that Newton is an occultist because his theory has action at a distance lol
 
Obliv
Obliv
@Slereah Im reading it now and he uses monads to explain the "soul" as one dominant monad commands the other monads in the body. Do strings have a similar property ;P
 
ACuriousMind
ACuriousMind
@Obliv careful: there are different ways in which "metaphysics" is used
 
Slereah
Slereah
@Obliv maybe
 
ACuriousMind
ACuriousMind
6:01 PM
I personally often use it in the literal sense of something that comes "after physics"
but as a branch of philosophy, metaphysics is not so much separate from physics
 
Slereah
Slereah
I only do hypophysics
 
Obliv
Obliv
@Amit he definitely would have been team einstein for sure.
actually maybe not since einstein cant explain the soul like monads
So like metaphysics pertains more to our minds and the non physical, after we use physics to get info from the physical @ACuriousMind
 
Amit
Amit
I think this monad idea partly dates back to Aristotle's "The Unity of Knower and Known"
I don't know if that's how he puts it but it was one of his ideas
 
Obliv
Obliv
As an attempt to unite the mind and body
and soul if your into that
 
Amit
Amit
It's a mystical concept as much as I can see, that the knower and the object known are united in the knowing. It sounds like a word salad but a lot of mysticism does :)
 
ACuriousMind
ACuriousMind
6:10 PM
@Obliv not really, no
 
Slereah
Slereah
The idea that all is one substance is very old
At least back to Thales
 
ACuriousMind
ACuriousMind
it's might discuss notions that are prior to being able to perform science, asking questions like "What is existence?" or what are cause and effect
 
Amit
Amit
Plato's forms are also very close, I think. I didn't read any of these original works admittedly
Wittgenstein, it is said, was the one who terrorized metaphysics out of the map. At least on the face of it... instead of metaphysics we talk about language, and we shy away from calling "Philosophy of science" by its real name "Metaphysics" :)
@Slereah Yes, very old and expressed in different ways, sometimes leading to different lines of thinking
 
Obliv
Obliv
@Acuriousmind admittedly this answer has caused me to ask these questions and my brain temporarily broke.
Without order and structure nothing is, and there would be no questions to ask
 
Amit
Amit
It must be some kind of a manifold, existence
 
Obliv
Obliv
6:17 PM
So we exist to ask such questions in a way
 
ACuriousMind
ACuriousMind
@Obliv what answer
 
Obliv
Obliv
@ACuriousMind this one
 
ACuriousMind
ACuriousMind
did you never ponder the nature of reality before
look into the abyss that is our utter inability to comprehend the marvel of our own existence
 
Slereah
Slereah
or even just look at a movie
They're not exactly obscure topics
 
Obliv
Obliv
I have its just that every time I do it makes no sense
its like a failure(or success) in my programming
 
Slereah
Slereah
6:20 PM
Have u seen the Matrix
 
Amit
Amit
I remember the very first thing that didn't make sense to me in that regard. It didn't make sense to me as a kid, that everyone is having the same kind of a separate existence, it just looked impossible for some reason. And I wondered if tomorrow I wake up as someone else, and someone else wakes up in my place, would we know lol
 
ACuriousMind
ACuriousMind
@Slereah do kids these days even watch that
 
Obliv
Obliv
@Slereah yes ive also dropped acid and listened to pink floyd
 
Slereah
Slereah
idk what's the modern equivalent
Have you ever watched the 1976 student film Dark Star
 
Obliv
Obliv
Im fairly certain im barely younger than you both
 
Amit
Amit
6:22 PM
The sequel to Matrix should have been called Tensor
 
Obliv
Obliv
@Amit i remember as a kid thinking death made no sense because we're here now and if I can remember my life how could I just not exist
back to work ttyl
 
Amit
Amit
have fun
 
Slereah
Slereah
what if all of this is a dream
Like the hack plot of so many episodes
 
Amit
Amit
It is a dream in many ways, because we are conscious of such a tiny fraction what is actually going on
And even that is screened through senses, mental conditioning, language, culture, etc.
 
ACuriousMind
ACuriousMind
@Slereah I mean what if all my half-remembered dreams are actual lives :P
 
Amit
Amit
6:25 PM
Culture is a kind of a dream machine
 
Physics Meta
Physics Meta
0
Q: Question of sending internet packets through time closed for "off topic"

LalalaMy question about sending internet packets through time was closed as "off-topic" as it was "non-mainstream-physics". This seems like mainstream physics to me, so it should fit on the site.

discussion closed-questions reopen
 
Amit
Amit
You know about the aboriginal tribe "dream time"?
 
Slereah
Slereah
I have the dream time every night
I just go honk mi mi mi
 
Amit
Amit
lol
 
user4539917
user4539917
merrily, merrily, life is but a dream
albeit, a persistent one
:P
 
Silly Goose
Silly Goose
6:28 PM
Hm the exact same spaces or isomorphic spaces is what is confusing me @ACuriousMind because
 
Slereah
Slereah
I have seen Star Trek 5 yes
 
Amit
Amit
said the Cheshire cat
right before he quantumly teleported
 
Slereah
Slereah
 
Loong
Loong
@Slereah Go to sleep, Spock.
 
Silly Goose
Silly Goose
I never remember dreams:P
 
Amit
Amit
6:31 PM
Birds singing in the binary trees, dream a little dream of me
 
ACuriousMind
ACuriousMind
@SillyGoose the same
@SillyGoose I rarely do, but when I do they're either nightmares or were so good I hate that I woke up, I'd prefer not remembering :P
 
Slereah
Slereah
I remember a fair number of dreams
although many of them are boring
 
Silly Goose
Silly Goose
Ack hm I guess I need to understand the permutation isomorphisms better…
 
Slereah
Slereah
quite a lot of them involve commuting
As above so below
 
Silly Goose
Silly Goose
@Relativisticcucumber’s dreams are full length movies that I would like to see produced XD
I have strange spooky kafkaesque dreams when i am stressed i think XD
 
ACuriousMind
ACuriousMind
6:36 PM
@Slereah mine feature a lot of zombies
which is odd because I'm not really a zombie or horror fan
 
Amit
Amit
Would Shakespeare agree with this notation? $<\text{sleep}|\text{dream}>^2 = perchance(rub)$
 
Slereah
Slereah
I had a dream where I went to ask a robot at a desk if they had a time machine I could use and he answered that ah, no, otherwise they'd never run out of "You've been warned!" jokes
Then he flickered as if disappearing from existence and another future version of him flickered in yelling YOU'VE BEEN WARNED
before disappearing
What did he mean by this
 
ACuriousMind
ACuriousMind
lol
he meant you've been warned
 
Slereah
Slereah
Probably my funniest dream
 
ACuriousMind
ACuriousMind
better heed the warning
 
Slereah
Slereah
6:39 PM
that's some Red Dwarf nonsense
 
Amit
Amit
He meant that in the future (now) you're gonna tell this dream and it will be funny (hence the joke), clearly he knew that because he used the time machine, that's why he couldn't allow you to use it
Full disclosure the robot was not me
 
Slereah
Slereah
The robot had a more MST3K look
Crow looking dude
 
Amit
Amit
I imagined a Dalek the moment you wrote it said "ah,"
 
Slereah
Slereah
 
Amit
Amit
Wow the resolution is low in the future
 
Slereah
Slereah
6:43 PM
it has all the pixels it needs
 
Amit
Amit
lol, yes because the vision must be enhanced. and beings aren't so material as they are now
 
Slereah
Slereah
 
Amit
Amit
I won't be surprised if we are within a robot's dream
pigments of imagination (no typo)
 
Mr. Feynman
Mr. Feynman
Meow
 
Amit
Amit
Cheshire cat?
 
Mr. Feynman
Mr. Feynman
6:55 PM
Ordinary cat
 
Amit
Amit
Ordinary cats don't compute lie brackets
Most don't even lie
Is this a turing-purring test?
 
Obliv
Obliv
that reminds me i wanted to start dr who
 
Amit
Amit
Start from the beginning?
 
Slereah
Slereah
Something's not right
 
Amit
Amit
lol, difficult to pinpoint
 
Obliv
Obliv
7:03 PM
i saw a clip of the show where there is a void ship which was a cool concept
totally nonsensical but ya know
 
Amit
Amit
I like the Tom Baker seasons, and most of the modern ones
 
bolbteppa
bolbteppa
What does DAN cat say though
 
Mr. Feynman
Mr. Feynman
@Amit who says I can compute Lie brackets?! Meow
 
Amit
Amit
lol, It's an educated guess, and I didn't even ask Cat-GPT
 
Mr. Feynman
Mr. Feynman
Cats don't need Lie algebras
 
Amit
Amit
7:11 PM
Let's not be splitting fur
 
Slereah
Slereah
7:37 PM
what about
 
Amit
Amit
lol
Axiomatic cat theory
 
user4539917
user4539917
7:55 PM
"What is the most important thing a scientist should cultivate in himself?
One should get rid of excessive ambition. One should not think that only a
genius can be happy. One must learn to appreciate even a small achievement,
to rejoice in it, and never overestimate oneself. One has to cultivate a love
for work. One has to understand and cultivate the joy of learning, which
is almost the same as the joy of life. Happiness is when your life’s work is
needed."
Sobolev, Sergei Lvovich
 
Silly Goose
Silly Goose
can't one be ambitious as well as content with the process of learning :)
 
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