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12:34 AM
@SillyGoose what does this even mean? That Newtonian mechanics describes very well what happens when you fire projectiles on earth will never change by demonstrations that Newtonian mechanics cannot describe some other situation; there is nothing gained from understanding QFT if you just want to understand where a shot-put ball thrown by an athlete will land
 
12:50 AM
chat.stackexchange.com/transcript/message/64831054#64831054 The Cauchy foliation solves a heat equation on Euclidean space, therefore if you want to investigate it in that space you are no longer dealing with a Lorentzian manifold nor a Cauchy foliation
I think I figured it out
 
1:39 AM
@ACuriousMind if you have two facts $F_1$ and $F_2$, then they could be independent. That is, there does not exist an underlying theory which gives both $F_1$ and $F_2$ as consequences. Or, they could be dependent. That is, there exists an underlying theory which gives both $F_1$ and $F_2$ as consequences.
Under this vocabulary, we would gain something from understanding a theory $T$ which implies a modern theory of physics which itself implies Newtonian mechanics. In particular, we would gain an "explanation" of from what Newtonian mechanics comes from
 
@ACuriousMind someone sounds triggered
 
The alternative is that there does not exist any theory $T$ which implies a physical theory $P_1$ which itself implies a known physical theory $P_2$. which would mean that the two physical theories $P_1$ and $P_2$ are independent and "in pieces"
 
2:07 AM
@Mr.Feynman lmao
 
 
4 hours later…
5:38 AM
tong points out in his notes that the equation for a classical field and for a one particle wave function are the same. is there anything to this..?
 
 
2 hours later…
7:31 AM
@SillyGoose H O N K
@SillyGoose The fact of the matter is that we now have the SM, which unifies 3 of the 4 fundamental forces in one cohesive framework. Only gravity, dark matter and dark energy are missing, as far as hoomanity can tell. That is not a "forever remain in pieces"
@Relativisticcucumber It is the same, by construction, by picking the same PDE for both, so that we can study the quantum system with an approximate classical picture. There is not much to it other than deliberately doing so for the analogy.
jigglypuff
bah
 
@SillyGoose Wigner mentions in his paper that reductionism cant be taken for granted. we can doubt if physics derives chemistry or biology (but this doubt cant be cleared imo because the complexity is too high to do the derivation)
@naturallyInconsistent we dont know the derivation of classical physics from the standard model yet
classical physics and quantum physics are very much pieces @naturallyInconsistent
 
7:51 AM
@RyderRude Why must you comment on things when you don't know it? The process is long and convoluted, but we know how to map that stuff. Not everything has to be mathematically exactly pinned down.
And that is not an invitation for you to continue the conversation.
 
There are hints of the map, but we dont know the map even in a mathematically non rigorous sense
the problem isnt of mathematical rigor here. the problem is the measurement problem
or rather, we know several maps depending on the interpretation
we dont know the correct one
 
 
2 hours later…
9:33 AM
are phd people better physicists than masters people or do they just know one area of physics better?
 
9:52 AM
i want to retract my statement. the measurement problem does not show up in the emergence of classical mechanics
 
10:19 AM
@RyderRude derivation is an unreasonable ask, but there's a lot of qualitative , and sometimes quantitative reasoning borrowed from physics that applies well to chem and bio which predict things
 
@nickbros123 i think so too. but Wigner thought he had reasons to believe that biology is non-derivable
that possibility is not worth thinking over
 
10:33 AM
philosophy.stackexchange.com/a/106053/27700 In this post, people are taking Descartes's demon to ridiculous limits
claims like the demon may have fooled us our entire lives into calculating 2+2 wrong
 
In representation theory terms, why is that we can classify irreps using Casimir operators? I'm not sure Schur's lemma alone suffices, does it?
 
 
1 hour later…
11:50 AM
Why do CY compactifications have a trivial bundle structure? I mean why do people consider the global product space $\mathbb{R}^{3,1} \times X^6$ rather than considering product structure to hold locally only?
$X^6$ is a Calabi-Yau 3-fold.
 
@Mr.Feynman The formal statement is the Harish-Chandra theorem which shows that the highest weights of two representations with the same Casimirs lie in the same Weyl orbit together with the theorem of the highest weight that says both those weights are dominant integral.
There's only one dominant integral weight in each Weyl orbit, so the two representations have the same highest weight and are hence isomorphic.
 
I asked the above question because the real world is locally flat, not globally...So why would people want to consider such a compactification for "realistic" model building?
 
@SillyGoose Yeah, I don't think this kind of formal reductionism is even coherent: What exactly is "a physical theory" here? I can make a Frankenstein physical theory by stapling together two previously disjoint theories ("For more than 100 particles, use theory A which explains F_1, for less, use theory B which explains F_2"), call it ACM's monster theory. This theory then explains both F_1 and F_2.
 
12:07 PM
Trying to read the Grassmann book that nlab think is cool but apparently it is already pretty steep in the weird lingo of the era
idk what you're supposed to read to get used to this, are you supposed to read like Fichte or something
 
@Slereah the Ausdehnungslehre?
 
> The opposition between the discrete and the continuous is (as with all true oppositions) fluid, since the discrete can also be regarded as continuous, and the continuous as discrete. The discrete may be regarded as continuous if that conjoined is itself again regarded as given, and the act of conjunction as a moment of becoming.
And the continuous can be regarded as discrete if every moment of becoming is regarded as a mere conjunctive act, and that so conjoined as a given for the conjunction.
@ACuriousMind that's the one
 
note that even the people of the era thought it was incomprehensible :P
 
I'm guessing it's just the terminology german idealists used in general but none of them seem to bother to explain it
Yeah I'm guessing there's a reason why it never became that popular
But I assume that there's some original book(s) where this is defined
I'm guessing maybe Fichte's book "The doctrine of science" but idk
As a German full of idealism what do you think
Is there a dictionary of german idealism
 
The only idealist I have a good grasp on is Kant
the others like Fichte and Hegel are mysteries to me, too :P
 
12:14 PM
I briefly looked into Fichte's book and it seemed to at least use example so I'm thinking maybe but otherwise who knows
 
@Slereah maybe this is more comprehensible in the original German, from which part of the book is that sentence
 
Section B of the introduction (DERIVATION OF THE CONCEPT OF EXTENSION
THEORY), at the send of part 4
 
nope, I'm afraid it's just as strange in the original
oh I think the problem is actually the meaning of "discrete" and "continuous"
he defines at the beginning of part 4 what he means by that and it's not what we mean today :P
 
that definition isn't entirely clear either I fear
Apparently I'm not alone
> Now, you have to be careful here, as I once met a Kant scholar who told me he started studying Kant in order to understand Hegel and never finished the "studying Kant" part and therefore became a Kant scholar.
 
@ACuriousMind thanks!
 
12:22 PM
@Slereah I think if you cut through the weird language he's saying that something "continuous" is just some indivisible unit that is created by writing down an axiom, while the "discrete" comes from taking some given things and connecting them (i.e. saying "hey, here I have a set and an operation, let's put them together and call the pair a vector space")
and the passage you quote is just the rather trivial fact that you can choose to "forget" that something discrete was made out of individual parts, and view it as something indivisble and given on its own
 
I s'ppose
I guess I can say that next time someone asks me if mathematical objects are truly made of sets
Apparently this is the dictionary
I guess the category people are truly the heirs of the german idealists in their refusal to explain anything
 
I just skipped the whole philosophical waffle and this actually becomes an interesting text about constructing something like a vector space
 
12:41 PM
I am told he tries to build a theory of intensive and extensive variables in there
 
12:52 PM
@Slereah I mean, Ausdehnungslehre means "theory of extension"
and I just looked into his revision of the book and apparently he thought better of the philosophical stuff at the beginning of the first version and so the second just starts with a formal definition of vectors
 
Is it just the first chapter?
 
@Slereah why r u learning the philosophy of discrete and continunous
 
why not
 
it's great. does it help with physics or math too?
 
@Slereah The first chapter is these mathematical generalities where he explains why what follows is about "extensive variable" but I don't really see a "theory of intensive and extensive variables" in there, it's as you already saw mostly a bunch of idealist epistemology
 
1:02 PM
I mean is the first chapter where the philosophical faffing concentrated
 
guess I can skip that :p
 
I skimmed through the rest and it becomes much more "mathy"
and really the revision cuts out almost all this philosophy, so perhaps you should read that
 
oh so u r tolerating the philosophy amidst the math
 
but then it's just a text about algebras, really
 
1:03 PM
@ACuriousMind there's only one English translation as far as I know
and it contains the everything
 
in mathematics, the continuum is just the power set of the discrete, together with topology given by Euclidean metric
i read there was another way to define the continuum, but using field stuff
let me check
this other definition uses rationals and Dedekind cuts to define reals
 
Are doubly warped product metrics used as much as warped product metrics in GR?
 
I can't think of much of them?
 
@Slereah Okay, I skimmed through enough of the revision to say that his notion of "extensive quantity" is just what today we'd call a vector or an algebra element
 
Is that the origin of vector spaces
 
1:12 PM
looks like it
the revision reads almost like a modern textbook on linear algebra except for a few weird terminology choices
I think people are right when they say Graßmann was ahead of his time, but it was in this presentation of the math, not in that incomprehensible philosophizing about being and becoming :P
 
I think we can really say in the end that Hegel invented the vector space
 
did Grassman like Hegel
it seems like Hegel has inspired math somehow
u keep discussing it
 
According to the editor this had apparently no influence on the actual invention of vectors which were based on the work of Hamilton instead
weird how things shake out
 
@Mr.Feynman because Casimir's commute with every element of the Lie algebra, they can be simultaneously diagonalized with the elements from the Lie algebra that you're diagonalizing (i.e. elements of the Cartan subalgebra in a semi-simple Lie algebra) you don't really even need to invoke Schur
 
@Slereah Not sure this is the right way to state what happened - apparently Hamilton knew and appreciated Graßmann's work
 
1:21 PM
the concept of the arrow vector may have been invented before the concept of a vector space
 
@ACuriousMind I'd have to look into it I guess
 
the German wiki article on Graßmann claims, among other things: 1. Hamilton explicitly mentioned Graßmann in his "Lectures on Quaternions" 2. Famous mathematicians who highly valued Graßmann (although only decades after the first Ausdehnungslehre) were: Klein, Clebsch, Lie
like, yes, Graßmann was ignored for a few decades, but he was still alive when people started to understand the significance of his work
 
Apparently he weighed in on the topic himself in a letter
"The position of the Hamiltonian quaternions in extension theory"
 
quaternions are more properly a field in the vector space context
but they are also a vector space with some extra structure
there is some very interesting structure of math that i encountered
let's say we start with groups as a primitive object (as it barely has any structure). when u represent it, u need a vector space which needs a field. but a field is itself an abelian group with some extra structure
so the concepts all circle back on themselves
 
1:40 PM
@ACuriousMind The constraint question has turned into a saga
 
the way i think about this structure is : the simpler groups can be directly represented as a division algebra. and the more complex groups are represented as matrices built on top of the simpler groups (which r the division algebras)
and ofc there r even more complex groups which cant b represented as matrices
so representation theory is combining the simpler groups with matrix multiplication to build more complex groups
the simpler groups being the fields
 
@ACuriousMind But they only understood that the work is significant, but did not understand the work itself. Otherwise, we would not have had to have vector algebra wars and just get straight to GA from the get go.
Of course, that is not to diminish nor deny that Graßmann's extremely weird writing is a huge hindrance to the adoption and understanding of his work
 
1:58 PM
For having read some of it I can tell it probably didn't do him any favours
 
@naturallyInconsistent I'm not sure what you mean by that; Graßmann himself finds (what we today call) the exterior product much more interesting than the "geometric product", that's why the exterior algebra is also called the Graßmann algebra, and the Clifford algebra the Clifford algebra
 
He goes on about conjunctions of objects for a while without defining what that could be
 
and "geometric algebra" is really just the theory of Clifford algebras in a trenchcoat that for some reason has a bunch of people acting as if it was some obscure concept
 
@ACuriousMind Not disagreeing with this, but that was pretty ignored in the physics side as opposed to the maths side. The successful rebranding and bringing into the attention of half-fringe physics side is still noteworthy
 
Hestenes says it makes problem solving easier
 
2:04 PM
@Slereah I'd use it and stab at those who want to introduce only rigorous maths first before teaching any physics, but really, it is not anything to do with rigorous maths, and instead everything to do with abstract philosophising before giving concrete examples.
 
There are places where understanding how to do something in multiple ways is very useful, however re-doing everything that involves vectors with Clifford algebras is not one of them
 
yeah, physicists would maybe experiment with it if they find anything lacking in the current math
but they r too busy with research to unnecessarily change their toolbox
 
If either of the commenters above actually saw how much shorter the solutions that Hestenes uses in solving problems are, they would not be making those comments.
 
if it's clearly demonstrated, it would slowly spread
just because some problems become easier doesn't mean all of them do
 
Oh, how innocent, the naïveté
 
2:17 PM
if it's worth it for those problems, it would slowly spread
it's free market for math :P
 
I have never seen the $\mathcal{N}=4$ SYM action before...Here it is and it contains a CG coeff! How to make sense of the CG coefficient term in the action? Can somebody give another simpler example where such a thing arises?
I can make some sense of it that we are tensoring two spinors $\psi_a$ and $\psi_b$ to build a vector $\phi^i$(spacetime scalar but $SO(6)$ vector), but why is the $i$ index and the $\phi^i$ there?
Reminds me of Yukawa coupling terms, but it wasn't a commutator...Why is there a commutator? To antisymmetrize because we want $SU(4)$ covariance?
 
2:40 PM
@Sanjana note that it is a summed-over-square of a commutator. Again, the kind of problems you are running into is so difficult that I cannot hope to aid you.
 
Yeah ofcourse...the Lagrangian density has to be a scalar afterall...
 
It arises from reducing the spinor terms from 10 to 4 dimensions, and involves and $\eta^a \gamma_i \eta^b$ for the $\eta^b$ a basis of $SO(6) = SU(4)$ spinors
 
jigglypuff!!!
bah
 
@Sanjana Don't forget that all the fields here are adjoint-valued, and that the $X^i$ come from the now scalar 6 compactified components of the gauge field. The "4-photon" terms of the gauge field in $F\wedge F$ does look like $[A^\mu,A^\nu]^2$ when you actually write it out, and this is just $A_\mu \mapsto X^i$
 
2:55 PM
@ACuriousMind Yeah, I understood the last term...I was talking about the 2nd last term. I got bolbteppa's point that I have to actually do the KK reduction...but I can't even think why would a term such as the 2nd last one appear? And how are these CG coeff terms supposed to be in a generic field theory?...i mean tell me something/anything about these...
 
@Sanjana likewise, for the "Clebsch-Gordan coefficients": The original term in the reduced action ends up as $\bar{\psi} \gamma^i [X_i, \psi]$ with the $\psi$ still the 10d fermion, and one can see this as a fermion bilinear $\langle \psi, \gamma^i [X_i,\psi]\rangle$. The right side is in $\mathbf{6}\otimes \mathbf{4}$, and the only non-zero contributions to a pairing with another $\mathbf{4}$ can come from non-zero $\mathbf{4}$ in the CG decomposition of that, so:
$\gamma^i [X_i,\psi] = \gamma^i c^{a} [X_i,\psi_a] = C^{iab} [X_i, \psi_a ] \psi_b$
and it's the $C^{iab}$ from this decomposition that appear in the Lagrangian
they're not really pure CG coefficients, but the also contain the effect of the "lost $\gamma$ matrices" from the dimensional reduction
 
3:16 PM
@ACuriousMind How did you know that? $6 \otimes 4=4 \oplus 20$, and then did you do $6 \otimes 4 \otimes 4$ or something?
under $SO(6)$
 
how did I know what?
 
"the only non-zero contributions to a pairing with another $4$ can come from non-zero $4$ in the CG decomposition of that"
@ACuriousMind And what did you do in the 2nd step? is there a typo: I see three "$i$"s
 
@Sanjana a linear map that produces a scalar (i.e. something the trivial $\mathbf{1}$) from a $\mathbf{4}$ is by definition in the dual of $\mathbf{4}$
i.e. the $\langle \psi,-\rangle$ is in $\mathbf{4}^\ast$, and the thing in the $-$ position has to be in $\mathbf{4}$ for this to yield a scalar
so we have to decompose the $\gamma^i[X_i,\psi]$ which is in $\mathbf{6}\otimes\mathbf{4}$ into $\mathbf{4}$s
 
3:32 PM
Wait a sec how is the RHS in $\mathbf{6} \otimes \mathbf{4}$? The $\psi$ is the 10 d fermion right? The reduced fermion $\psi_a$ is $\mathbf{4}$
 
The $\gamma^i[X_i,-]$ is in $\mathbf{6}$, it acts on the $\psi$ in $\mathbf{4}$, the result is in $\mathbf{6}\otimes \mathbf{4}$
 
I don't get how the $\psi$ is in $\mathbf{4}$...It is a 10 D fermion. It transforms as $\mathbf{16}$ of $SO(9,1)$. But how do you say that it is in $\mathbf{4}$?
The $\psi_a$ is in $\mathbf{4}$, but how is the 10 D fermion also in $\mathbf{4}$??
 
$\psi = c^a\psi_a$
it's just the vector of the 4 4d fermions
@Sanjana $\mathbf{16}_{9,1} = \mathbf{4}_{3,1}\oplus \mathbf{4}_{3,1} \oplus \mathbf{4}_{3,1} \oplus \mathbf{4}_{3,1}$, the $\mathrm{SO}(6)$ mixes these 4 representations of $\mathrm{SO}(3,1)$, so this is simltaneously (4 copies of) a $\mathbf{4}_{6,0}$
 
This is under $SO(1,9) \to SO(1,3) \times SO(6)$ right?
 
sure
 
3:44 PM
Now how is $\gamma^i[X_i,-]$ in $\mathbf{6}$? I thought $X^i$ itself is the $SO(6)$ vector
 
I think the indices are throwing you off here
the $X^i$ is one component of a $\mathbf{6}$ vector $X$
well, it's not really the $\gamma^i[X_i,-]$ that's in $\mathbf{6}$, actually
it's in $\mathbf{1}$
but the $[X_i,-]$ alone is in $\mathbf{6}$
 
YES...that's what I thought!
 
and we want to turn this into some expression that no longer contains the $\gamma^i$
because we don't want any higher-dimensional gamma matrices left in our 4d theory
so you look at the $[X_i,\psi_a]$ as something in $\mathbf{6}\otimes \mathbf{4}$, and the $\gamma^i$ act on stuff that's in $\mathbf{4}$, so you decompose the $[X_i,\psi_a] = c^{iab} \psi_b$ to get $\gamma^i c^{iab} \psi_b$ as the result, let the $\gamma^i$ act on the $\psi_b$ to get some $\gamma^i \psi_b = g^{ibc} \psi_c$ and the $g^{ibc}$ and the $c^{iab}$ combine into a decomposition $\gamma^i[X_i,\psi_a] = C^{iab}\psi_b$
some messed up positions there but there's like 10 different ways of writing down what's happening here
it's just something you need to either accept blindly or work through very carefully yourself :P
 
Got it, thanks, is there a place where the complete dimensional reduction is done? Let alone this, I see no place where any reduction explicitly done...
 
it's in a bunch of the original papers but yeah, most sources just state the result and leave the details as an exercise
 
3:55 PM
There is a video reviewing it somewhere if I find it I'll send it
 
@ACuriousMind :( Every single paper I saw says this
@bolbteppa yey...thanks
 
 
1 hour later…
5:09 PM
This explains it partially, this partially does it (as does his book), this fully does it, this talks about the 4's, 6 and CB's.
CG's*
 
Learning about lattices is annoying since joins and meets are essentially equivalent and I'm starting to suspect that not all books use the same conventions
 
5:53 PM
Does stronger coupling (bonding) strength between atoms imply higher spin-spin correlation?
 
how would you be measuring this?
 
@bolbteppa Thank you so much!
 
6:36 PM
@naturallyInconsistent Yeah, I am not sure. But my intuition (is it wrong?) was that the magnetic interactions between neighboring, non-equivalent NMR-active nuclei, known as spin-spin coupling, are stronger when the atoms are more closely bonded.
 
@Keane I am not disagreeing with you, but you would have to have a measure from which to make that argument. I know at least one way to make you correct, but I think this is mostly going to be confusing than helpful, so you should specify what it is you want
 
@naturallyInconsistent Sure, you're definitely right. To give you a more clear idea, I recently thought of imagining classical weighted graph as atoms and their coupling (bonding strength) between them, and then I got curious whether the shortest path might correspond to the lattice sites at which the correlation function of their spins $G_z(i_1, \cdots i_n) = \langle T\bigr\{\sigma_{i_1, z}(t_1)\sigma_{i_2, z}(t_2),\cdots,\sigma_{i_n, z}(t_n)\bigr\}$ might be minimal/maximal.
 
you need to surround the LaTeX with $
Anyway, I have no idea what "classical weighted graph" is, and what "atoms and their coupling" would have to do with that, and why there is any "path" being involved in any of these things.
 
Atoms and their couplings have to do with approaching classical optimization problem by using quantum mechanics, just like what finding ground-state energy of Ising Hamiltonian does. As I said, it's probably flawed, but the intuition was that higher bonding strength could imply higher correlation between spins of atoms, and so one could look at the correlation function of their spins rather than finding ground-state energy of Ising Hamiltonian.
 
6:53 PM
No, because, say, consider the diamond giant molecular structure. The covalent bonds in it is incredibly strong, yet you know that it is not magnetic. It is not even anti-ferromagnetic, just non-magnetic. Similarly, the similar bonds between some magnetic materials would sometimes present as ferromagnetic, and sometimes as anti-ferromagnetic. The correlation functions of these spins are varying wildly, but might be similar in bonding coupling.
 
@ACuriousMind :o you ask questions :oooo
 
$$2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)}$$

Can someone walk me through the steps in taking the Mellin transform of this equation? (w.r.t. the $x$ variable)
In mathematics, the Mellin transform is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series, and is often used in number theory, mathematical statistics, and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform, and the theory of the gamma function and allied special functions. The Mellin transform of a function f is {...
 
@Relativisticcucumber you could have discovered that at any time by looking at my profile :P
 
I'm reading the section (linked) but I don't understand why
 
@naturallyInconsistent That makes sense, yes.
 
7:07 PM
my attempt is that the LHS remains the same and that the RHS becomes something, I don't know how to deal with the square root of the differential
 
7:21 PM
@Relativisticcucumber One of ACM's questions (the one about Majorana spinors IIRC) got Mike Stone to write a 40ish pages paper
At this point I'm an ACM lore expert
 
7:46 PM
@naturallyInconsistent Apologies if I am being imprecise again, however: obviously we can measure the least possible energy (bond entalphy) to break one mole of a bond between atoms of a molecule. Similarly, does the bond entalphy of multiple bonds correspond to the bond entalphy of individual bonds in a molecule?
 
What is the mathematical object corresponding to spacetime defined defined to be in GR? If there is a definition
Is it merely “a set of points”
My question is that someone might say that “the Einstein field equations relate geometry of spacetime with energy”.
Previously i thought the definition of spacetime included the metric. As in “Minkowski spacetime”
But the metric which gives “rise to the geometry” is what is solved for using EFE, which itself is defined already in reference to a spacetime
So it seems like the metric cannot be part of the definition of spacetime, and i am confused
 
@SillyGoose it's just a pseudo-Riemannian manifold
and yes, it includes the metric
you have the space of all pseudo-Riemannian manifolds $(M,g)$, and then you can think about the EFE as being an equation for such manifolds together with a stress-energy tensor $T$ on them. Some triples $(M,g,T)$ are solutions to the EFE, others are not
 
@SillyGoose This seems related.
 
Is T a “nice way of relating the various metrics g”? Also is g there a usual metric of real analysis? Or is metric in this context always a metric tensor that assigns a metric to each point on the manifold
Lol releasing
 
8:01 PM
metric in this context always means metric tensor
a Riemannian metric would induce a metric in the sense of metric spaces via path length, but a pseudo-Riemannian does not
I don't understand what you mean by "relating"
 
8:20 PM
From what I understand, in Maldacena's original conjecture $\mathcal{N}=4$ SYM lives on the worldvolume of the $N$ coincident D3-branes, and the brane system itself produces a $AdS \times S^5$ geometry...You can get rid of the $S^5$ by reduction. But what makes people say the CFT lives on the conformal boundary?
I understand that the theory is defined in one lesser dimension, but I don't get the geometry...
If I look at the global AdS cylinder, where and how exactly are the branes oriented?
 
8:40 PM
Okay nevermind my question. So then if one wanted to be clear, is it advised to say that PR manifolds are pre-spacetimes and solutions to EFE are spacetimes
 
no, not really
 
What do you mean by PR
 
I'm guessing pseudo-Riemannian
 
the standard terminology is that spacetime is just a pseudo Riemannian manifold of the appropriate signature
They're not required to be "on shell"
Sometimes people will throw in a few other conditions but nothing so bold as requiring it to be on shell
 
@Slereah yes pseudo-riemannian
 
8:44 PM
like, you don't call all possible paths in classical mechanics "pre-trajectories" and then the solutions to the equations of motion "trajectories", they're all just trajectories, some of which solve the equations of motion and some don't
 
hm well in that context trajectory is a concept defined without respect to physics
but i can see what you mean
 
It's pretty common to not even worry about what the spacetime obeys as far as an equation of motion
Sometimes you just consider the metric by itself
 
i think it is confusing because a spacetime that lives in the space of pseudo-riemannian manifolds does not have the same properties as a spacetime that solves the EFE. So to call them the same object is leaving out the particular properties that "solving the EFE" endows
 
After all there are several theories of gravity with different EoM but you can look at the properties of a spacetime independently of this
 
@SillyGoose what do you mean "it does not have the same properties"?
 
8:48 PM
Also like every spacetime is on-shell for some stress energy tensor T
 
I'm pretty sure you can find some $T$ for any $(M,g)$ such that $(M,g,T)$ solves the EFE
the point of the EFE is not really to select specific manifolds
remember that physics is chiefly a science concerned with prediction - the point of the EFE isn't to find the right manifold or whatever, it's to start with initial data for $g$ and $T$ on a Cauchy surface and solve them for the whole manifold
 
hm okay
 
It's pretty common that we don't even know the details of the stress energy tensor and we just kind of guess the spacetime from other features
 
hm okay so at a conceptual level. we posit that things exist in "spacetime", which is modeled as a differentiable manifold ($\mapsto$ M). we then posit that the presence of things in spacetime determines the geometry of spacetime ($\mapsto g$). The pair $(M, g)$ must be pseudo-riemannian. We refer to both the raw spacetime manifold $M$ and the pseudo-riemannian manifold $(M, g)$ as spacetime
@Slereah does on shell here mean a solution to the EFE?
 
@SillyGoose $M$ by itself should not called spacetime
 
9:00 PM
yes
@Sanjana I'd call it a spacetime
 
@Slereah hm? without the metric, too?
okay it's a word afterall...
 
sure
 
well okay how about this: does the EFE only solve for $g$ or does it also solve for $M$?
as in do we say given an $M$ solve the EFE
or do we say let's look for solutions $(M, g)$ to the EFE
 
@SillyGoose it's a local equation so it does not solve for $M$ at all
 
okay then this is my confusion
 
9:03 PM
since it's a local equation you can look at any open subset $U\subset M$ of some solution $(M,g,T)$ and then $(U,g\vert_U,T\vert_U)$ is also a solution
 
@ACuriousMind You too? I mean, sure when we are not being careful about the terminology then it's okay but in most places they atleast include the metric and call the full tuple a spacetime...
 
@ACuriousMind I mean it can constrain M in some ways
 
@Sanjana there is a difference between being careful about terminology and being pedantic about it :P
I see no point in chastising peoiple for just calling the manifold "spacetime", since for many notions the metric doesn't matter at all (e.g. for "a function on spacetime" or "a vector field on spacetime")
 
Fact 1) things exist in "something". Often, this "something" is called spacetime. Fact 2) Spacetime, seemingly, corresponds to a pseudo-riemannian manifold $(M, g)$. Fact 3) The EFE require a specification of $M$ to solve for a $g$. We say that the EFE solves for the geometry of spacetime, not spacetime itself. Fact 4) The EFE is written in terms of objects (energy tensor, metric) defined on $M$.
 
I wouldn't say that the EFE "solves for the geometry of spacetime"
 
9:07 PM
so my confusion is in calling $M$ spacetime, which we do when saying "the metric is defined over "something", i.e. spacetime. and in calling $(M, g)$ spacetime like in the above conversation.
 
the $T$ is a function of all the other dynamical fields of the theory, and you're solving for them as well (together with their associated E-L equations)
 
@ACuriousMind hm i see
 
@ACuriousMind Sure thing :p
 
well then my question can simply be: physically, what is the metric tensor defined over?
 
really, if we take the principle of least action seriously, the EFE is just one of the E-L equations for the Einstein-Hilbert action + some matter action
 
9:08 PM
don't bring out the pedestal ! hehe jk
 
The way that you would normally "solve" a spacetime is to consider space at some moment $t$, the geometry of that space at that moment, and the matter content
And then you look at what it is in the future
The geometry at the initial time isn't determined by the matter content, it's the initial conditions
 
yeah, that's what I meant when I brought up Cauchy surfaces
 
i thought the answer to my above question is: spacetime. but spacetime is defined to be a tuple $(M, g)$. so that can't be right. unless one is using this extended definition of spacetime to refer to both or either of $M$ and $(M,g)$.
 
but I feel @SillyGoose is once again trying to over-abstract a physical theory before they've seen how it's going to be used :P
 
what a silly goose
 
9:10 PM
@SillyGoose didn't we just say that you can call either the manifold itself or the pseudo-Riemannian manifold "spacetime"?
 
well i am just confused about how the EFE is a coherent statement given how it is colloquially stated
@ACuriousMind Sanjana disagreed
 
the EFE is just the E-L equation for the Einstein-Hilbert + matter action on some manifold
it's not magic, it's literally just another equation of motion
 
If you already have $(M,g)$ you don't need to solve the EFEs
The stress energy tensor can be deduced from $g$ in this case
 
@SillyGoose Believe me goosie...Although what ACM and Slereah says is 100% correct and that's the way most people use it when talking about stuff...the way books define it in the beginning is that they call the tuple spacetime, not just the $M$.
 
@ACuriousMind i'd just like to get the terminology straight :P. it seems like abuse of language similar in horridness to that used in representation theory in physics. and this abuse of language is unfriendly in my opinion to someone learning the material for the first time
 
9:13 PM
@SillyGoose I just don't see what this distinction matters for
why does it matter whether "spacetime" really refers to just the manifold or the manifold with the metric tensor?
 
for stating the EFE without making it sound inconsistent (or incoherent, not sure what the right word here is)
 
I mean, as ACuriousMind said, Einstein field equations is just a differential equation $R_{\mu\nu}[g] - \frac{1}{2} g_{\mu\nu}R[g] = T[g]$ involving a metric g, and for every g, there exists a stress-energy tensor T[g] such that it solves the EFE. This equation can be derived from Einstein-Hilbert Lagrangian, just like any other theory
But obviously a (pseudo)-Riemannian metric won't uniquely determine M.
 
Positing that an object lives in spacetime and defining spacetime to be a tuple $(M, g)$ makes the EFE not understandable to me, for example.
 
@SillyGoose You have some manifold $M$, and the EFE are equations for a metric $g$ and other fields $\phi_i$ on it, which have a stress energy tensor $T(g,\phi_i)$
why does it matter which part of this we call "spacetime"?
 
@SillyGoose It doesn't "define" it in the sense that it wouldn't exist if it wasn't there
It just determines its evolution
 
9:16 PM
Yeah^^
 
@Slereah hm so you are saying that an $(M_0, g_0)$ would be (part of?) an initial condition for instance
 
Same way that charges do with the electromagnetic field
 
@SillyGoose no, an initial condition is data for the $g$ and the $\phi_i$ on a Cauchy surface in $M$
 
i think it makes more sense considering that bit then
 
you can even try to ask the question that if you just have the data of the Cauchy surface itself and no $M$ given, which possible $M$s it might embed into
 
9:18 PM
@ACuriousMind Isn't the point of EFE to prescribe time-evolution of g under a particular stress-energy tensor? From what I know, it doesn't attempt to uniquely determine M
 
so then, if i understand correctly, as slereah emphasized: the EFE solves for the dynamics of spacetime, not "a spacetime", "the spacetime", etc.
 
@Keane did I say anywhere that this yields a unique solution for $M$?
and yes, what I'm saying is that it's just a time-evolution equation - a Cauchy surface is a "surface of simultaneity", i.e. represents an instant of time in a generalized sense (it might not be covered by a single coordinate patch so just saying it's a surface of $t=\text{const}$ in specific coordinates doesn't cut it)
 
Thinking that there would be no spacetime without matter is more of a Machian thing than GR
 
okay well if it is just a dynamics relation like any other then it makes a lot more sense :P. for some reason I thought that the EFE are supposed to solve for spacetime given the existence of objects (vice versa), which did not make sense to me.
(because then one has to ask how can an object exist if not embedded already in some spacetime)
well maybe to someone that question has an answer or makes sense. but i would not imagine in the sphere of physics
 
@ACuriousMind I didn't say that you did. However, I'm unable to follow the point of bringing Cauchy surface, although it looks interesting to me
 
9:24 PM
@Slereah what do you mean by this? that it is a historical opinion or that it is a perhaps reasonable line of thought that only a subset of relativists believe?
 
@SillyGoose That was some older theory that influenced GR
But it is a bit more extreme
I'm told that it still exists today in relativistic form under this name : en.wikipedia.org/wiki/Shape_dynamics
 
"Shape dynamics is not formulated as an implementation of spacetime diffeomorphism invariance" ::eye twitches::
 
@ACuriousMind Surely you should enjoy that fact
 
In some sense it is true: GR is definitely different from other theories.
In other theories, the spacetime is given apriori, but in GR you dynamically determine the spacetime.
 
I'm not sure Machian mechanics makes much sense especially in modern time tho
 
9:27 PM
good god
 
It's kind of too much structure missing to make anything sensical
 
the starving artist of our generation
 
Also me
 
hehe
 
Machian theories are kind of weird because they don't have kinetic terms
Since there's no sense in which a single object can have a meaningful velocity
 
9:29 PM
@Slereah I would enjoy it if I could go a single day without encountering the phrase "diffeomorphism invariance" :P
 
Would you prefer the theories varied with diffeomorphisms
 
@ACuriousMind well now that we're on the subject...
 
Who doesn't prefer diffeomorphism invariance? :)
 
what makes someone think that space and time is real at all. or, how can a physicist have an informed notion of what it means for something to be real
 
@SillyGoose That's a question best left to the philosophers
I don't know if cars are real but I still look both ways before crossing the streets
 
9:32 PM
measurements honk
 
If you want to know more about such nonsense you can look at uuuuh
I dunno something in there
 
i agree with splitting thought into i) the everyday and into ii) the encompassing of all. but beyond that i would be interested if a physicist had an informed notion of what it means for something to be real. maybe ACM :-)
 
The notion of realness in physics is typically taken to be something that can be measured
Although there's no lack of people arguing against that
 
They haven't been in a lab
taking measurements
 
It's a thorny issue according to philosophers but then again I've never seen a philosophy of science book saying an issue wasn't thorny
Every philosophy of science book says that science is in crisis
 
9:35 PM
A theoretical physicist's most useful measurement tool is his pen, I would argue.
 
It's a perpetual crisis
 
@SillyGoose I don't care about whether things are "real", I care about whether our models are useful.
people get too hung up on ontology
 
@ACuriousMind How useful has your QFT knowledge been so far
 
@Slereah It has provided me with a great source of the absurd :P
 
Beautifully absurd though :D
 
9:39 PM
I believe spacetime is real
I see it
 
I'm reminded of this recent comment:
"That's like denying the existence of "angular momentum" because "angular momentum" is just a "model" with math and the actual existence or proof of it is "just philosophy."" - You have summarized my epistemological position correctly. — ACuriousMind ♦ Nov 21 at 14:21
 
@ACuriousMind What was your motivation for formally learning philosophy?
 
@Slereah There's probably no way to make sure of the existence of spacetime, as it is just a mathematical model/abstraction to bring an explanation to the phenomena we observe in nature. Just like how numbers are used to compute the trajectory of a falling ball under the influence of gravity, yet these numbers do not exist.
 
I believe in numbers
 
9:42 PM
In physics and cosmology, the mathematical universe hypothesis (MUH), also known as the ultimate ensemble theory, is a speculative "theory of everything" (TOE) proposed by cosmologist Max Tegmark. == Description == Tegmark's MUH is the hypothesis that our external physical reality is a mathematical structure. That is, the physical universe is not merely described by mathematics, but is mathematics — specifically, a mathematical structure. Mathematical existence equals physical existence, and all structures that exist mathematically exist physically as well. Observers, including humans, are...
 
Here they are
 
@Sanjana While I think a lot of philosophy goes down strange and useless paths, the basic idea of Kant's three questions resonates with me: What can I know? What should I do? What can I hope for? These are essential questions worth trying to answer.
6
 
@Slereah Yeah ;P
 
@ACuriousMind Kant's answers was only more philosophy for all three
 
What can you hope for? 10 stars on this message.
Come on people, let's do it!
 
9:46 PM
What can you hope for? A Stack Exchange profile with 100k reputation.
Let's try our chance!
 
@Slereah As you might have gathered over the years, I do in fact think Kant's moral philosophy has a lot of merit to it
 
@ACuriousMind Are you gonna rat out the jews to the Gestapo
 
@ACuriousMind Is there a philosopher whose ideas you're interested in most, like Kant?
 
@Slereah no, I said it has a lot of merit to it, not that I believe it's entirely correct :P
 
I'm an anti-Kantian
I believe in the categorical imperative to do evil
 
9:48 PM
it's not the specific implications of the categorical imperative I find valuable, but the idea of reason as the unifying trait of moral agents and the idea that there is such a thing as a rational duty
 
@ACuriousMind Does state-operator correspondence in a CFT hold for SCFT, too? This is probably a trivial question, as SCFT is essentially a CFT but with supersymmetry
 
@Keane yes, since as you say an SCFT is in particular a CFT
 
Yep, right
 

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