also, the algebraic formalism is more general than both the Hilbert space and the projective Hilbert space, since if there is more than one GNS irrep then there are algebraic states which cannot be simultaneously expressed as vectors or density matrices on the same irrep

@ACuriousMind i don't see it as a desire for elegance; although, i will reflect on that. i just saw it as a logical consequence sort of thing. namely, that a physical theory should talk about transitions from physical states to other physical states.

@SillyGoose But that's what the ordinary Hilbert space formulation also does, just that your states are not points in the Hilbert space but rays

@ACuriousMind There are papers on this, but I have never had the chance to read them yet. However, the question was about substituting H as the space of pure states with P(H), and is possible. The space of mixed states is not H but a convex set, and represents a distribution of P(H) states, so there should be no direct reference to H

we are aware that "the state" is really a ray and that the vectors we pick to do calculations with are merely "representants" but it would be so cumbersome to always phrase it in those (doubtless technically more correct terms!) that no one does it except when it becomes relevant (for instance in the context of talking about projective representations)

Also, it may be noted that the distinction between H and P(H) is weak: H contains P(H) and P(H) has a natural tautological bundle with fiber H

@GroveRover That's my question: Is it really a "distribution of P(H) states"? Because as you say the set of states is convex, meaning the mixed states are sums of the pure states (a density matrix can be represented a linear combination of projectors onto pure states), but you cannot "sum" points in projective space.

@ACuriousMind I've seen this objection in another form, but it's not clear to me. Sums aren't observable, what is observable are distributions

Regarding my last comment maybe the inequivalence is in the hermitian form, which is lower dimensional on P(H) and can't be lifted to H

@GroveRover But what is this "distribution", mathematically? If I give you a density matrix $\rho$, how do you express it in terms of the projective space?

@ACuriousMind A collection of points and scalars?

@ACuriousMind but that's the thing the operators we use (in particular annihilation or even simply $\sigma_-$) maps from certain rays into not a ray

even in the conventional quantum framework where we start to work with density matrices, we have $\overline{\mathcal{B}}^+(\mathcal{H}$ as the space of states. Not talking about any projective business. Yet we will still have that an operator like $\sigma_-$ maps out of this space of states.

@SillyGoose but the annihilation operator is not a "transition" (by which I assumed you mean time evolution or at least some kind of physical operation), it's just a linear operator on H

@ACuriousMind That's my point, you can't observe the action of the operator, you just see expectation values, well defined for P(H)

@GroveRover I don't know what that means - a pure state $\lvert \psi\rangle$ goes to its corresponding point in the projective space. An observable $A$ goes to a function $\langle \psi\vert A\vert \psi\rangle$ on P(H) if I understand you correctly (and I'm not 100% convinced this does not lose information, either). What does a mixed state $\rho$ go to?

@ACuriousMind I think you can use the polarization identity to not loose information

I don't know how people treat mixed states, as I said before, but on the top of my mind you could think of them as pairs of points of P(H) and scalars that sum to 1. The scalars are summed, not the points.

Also as I said before, I am not sure this is relevant to the original question, see above for the reasons

@GroveRover but how do you identify those pairs of points? An arbitrary density matrix can be expressed as the sum of pure projectors in more than one way, meaning there are different sets of points and weights that correspond to the same density matrix, i.e. you're re-introducing a redundancy that this formulation is supposed to avoid!

@ACuriousMind You can just take an equivalence class

obviously, but then SillyGoose will complain again that we really should be looking at the space of those equivalence classes and not projective space :P

@ACuriousMind and I'd think that's a fair point

But again, I'm concerned about the distinctions between these spaces and H

To define this equivalence class you need H first, just as for P(H)

The only thing you have too much information of is the hermitian form since it descent to a lower dimensional one

In any case, this discussion is more about philosophy of physics than physics itself

@SillyGoose I think maybe the issue here is really that you imbue this problem with the "missing zero" with more relevance than it deserves: Consider a classical system that exists in $\mathbb{R}^n\setminus\{0\}$. If we wanted to describe how rotations work for this, everyone would obviously (at least to me) just describe them exactly the same as rotations on $\mathbb{R}^n$, and you can write down the generators of rotation which will generically have zero eigevalues,

i.e. "map some states out of the space of states". Still there is much value to be found in discussing them, and the linear algebraic structure contains a lot of information (especially if you know all your state spaces are of the form "vector space minus the zero"). If you insisted that we must not talk about the generators of rotation because they sometimes map a vector to zero, would you really think that's a sensible position to take?

I was indeed trying to think about implications on the representation theoretic structure. I mean at the least I think we are fine at the group level (since we conventionally use projective representations).

but i suppose there is a blurry line between "necessary" for a physical theory and "convenient". At this point it seems that symmetries (in all senses of the word), which heavily use Lie algebraic machinery are somewhat crucial for physical theories even though they can also be seen as merely convenient.

@SillyGoose Wait they're not only convenient, they're needed

H = P^2/2m is a group theory consequence

@ACuriousMind but i mean as we see with Leinaas and Myrheim such a seemingly irrelevant removal can have drastic effects (if we are to take the framework of L&M as being reasonable and serious).

@GroveRover do you mean characterizing a unique translation + rotation invariant operator?

@SillyGoose to characterize the galileo's generator of time translations

@SillyGoose I would suggest that at least for a subset of physicists there is no difference at all: The purpose of physics is the convenient description of nature, and so what is convenient is necessary.

hm i am unfamiliar with the galilean group and its representation theory. (more familiar with the restricted lorentz group and its projective unitary infinite dimensional representation theory and finite dimensional projective representation theory).

@SillyGoose And I mean there are also "effects" in QM, such as projective representations!

it's not as if people ignore the nature of states as rays

@ACuriousMind dropping them may be beneficial in various aspects, and has been in the past

@GroveRover dropping what?

@ACuriousMind convenience tools assumed as necessary

@ACuriousMind well i meant to additionally point out the fact that the exact classical analogue of what we’re discussing can be interpreted as having a drastic effect

@SillyGoose I'm confused how it's relevant to my analogy - my point is that we don't stop using linear algebra just because the zero is not a state, so demanding that we stop using linear algebra in QM because zero is not a state seems like a non-sequitur to me

again, no one is pretending the zero vector is a state

@ACuriousMind I never had the impression that he meant we should stop using linear algebra, that would be insanity

I think He meant it's not the more fundamental and naked way to see the theory

And indeed it is not. For example the parallelism between classical and quantum is completely obscure on H but evident on P(H)

@ACuriousMind Concerning my diffeomorphism problem, I am not sure what Deane is trying to say. He says we're imposing a gauge condition on f_T and right after he states that we're imposing it on d/dt (f_t^* g).

Indeed as Grove says I am not claiming to not use linear algebra or the Hilbert space machinery or to use any other convenient machinery.

It may be helpful to consider a simpler example of projective geometry. We can do 3D computer graphics by working with $\mathbb{R}^4\setminus\{0\}$ and using homogeneous coordinates. This permits us to use matrix multiplication for both rotation and translation, and also other useful transformations, like scaling & shearing. And perspective projection.

@GroveRover They just mean that $f_T$ is a diffeomorphism such that the resulting $f_T^\ast g$ fulfills whatever gauge condition you want

Well i think my point is i think most people nod their heads to the inclusion of the zero vector as just a convenience. But i have not yet seen what a calculation that relies on the zero vector looks like in projective space, which I imagine would clear my conscience

@SillyGoose I'm really not sure what you mean by "just a convenience" here

@ACuriousMind I understood that, then He proceeds to say we're imposing this condition on d/dt (f_t^* g)

@SillyGoose As I showed you in my answer, there is no need for such calculation

@GroveRover well i guess in principle it can be circumvented

@ACuriousMind i guess by convenience i mean is not actually necessary to do a computation or general computations

@SillyGoose the fact that you can circumvent them is the very reason you can use just P(H) without H, since those operators are not defined on P(H)

This also makes me recall that seemingly stating entanglement in terms of tensor products is insufficiently abstract

@SillyGoose I don't know what that means either, really - obviously if someone asks you to compute what $a\lvert 0\rangle$ is you need the zero :P

@GroveRover well i see this for amplitudes indeed. But for building a theory like QFT for instance which relies so heavily on c/a operators seems nontrivial even if “in principle” it can be done

@SillyGoose I am just starting to learn QFT, so I won't comment on that

@ACuriousMind but i think as Grove points out and I have remarked before it should never be necessary for such a thing to appear in any quantum mechanical probability amplitude computation

@ACuriousMind I want to stress that the question would be physically meaningless

The fact that you can apply operators to states and stop there is a (very useful) artifact of the theory

@GroveRover I think the problem is that you didn't show us what your professor did and it's not clear from your post. Let's just collect the facts: You have a metric $g$ and a family of diffeomorphisms $f_t$. The generating vector field $X$ of that family is what physicists call the "infinitesimal diffeomorphism", and to first order in $t$, we have that $f_t^\ast g = g + L_X g$ for $L_X$ the Lie derivative. What exactly is the question?

By the way @GroveRover are you a maths student?

@SillyGoose Why not? The intermediate steps of computations are not "real"! Personally I would love it if coordinates were never necessary in differential geometry as nothing should ever depend on coordinate choice, but sometimes you just have to roll up your sleeves and prove some statement in Riemann normal coordinates or whatever because no other proof is known.

likewise, I see no reason to demand that the zero should not appear "in computations" except for aesthetics, precisely the same kind of aesthetic preference I have against coordinates

as long as it drops out at the end, there's no problem, just as there's no problem with using coordinates if the end result is not dependent on any specific choice

@ACuriousMind I thought I cleared in the edit: He treats the lie derivative as if it was a diffeomorphism that returns f^* g = g + L_\xi g. Then uses the new metric as a valid solution. This is present also in some general relativity book

He first states this in general, then He applies it to fix the GW gauge and to prove the conservation of the stress-energy tensor by substituting the variation of the metric in the action

@GroveRover it's just standard physicist behaviour of dropping terms beyond first order

@ACuriousMind You can do that under assumptions, you can't just drop orders as you please

He treats it as a general fact

in the more careful notation you have something like $f_t^\ast g = g + L_{X} g t + \mathcal{O}(t^2)$ (Taylor expansion) and because we said the word "infinitesimal" we have $t^n = 0$ for all $n>1$

I spent most of my education approximating to the first order, I am familiar with the procedure

@SillyGoose formally physics, but on the mathematical physics side. I'm somewhat in the middle

@ACuriousMind Again, no issue with that. But you can't just use it in generic contexts as you please. You can in certain approximate situations. The conservation of T^{mu \nu} is definitely not a approximate situation

I personally believe this is just plain wrong and He doesn't know what He's doing

This happens often

There is no reason from this alone to assume that this is the only condition on $T$ so that $U$ is unitary, but...it turns out it is, and physicists often do this without necessarily proving that it's enough.

@GroveRover Are you sure this isn't a cases of the magic of the Lie correspondence? For instance, we might start with a unitary operator $U^\dagger U = 1$. Now I expand infinitesimally $U = 1 + \mathrm{i}T$, dropping all terms higher-order in $T$: $(1 - \mathrm{i}T^\dagger)(1 + \mathrm{i}T) = 1 + \mathrm{i}(T - T^\dagger) = 1\implies T = T^\dagger$ and so I've "proven" that the infinitesimal generator of $U$ is a self-adjoint $T$.

@ACuriousMind God I hate physicists so much

Likewise, your prof is starting with some condition on $f^\ast g$, expanding infinitesimally to derive some condition on $X$ and implicitly acting as if $X$ fulfilling that condition is enough for the finite $f^\ast g$ to fulfill the original condition. It's not at all obvious this is true, but with much more effort akin to the finite-dimensional Lie correspondence you could probably show it is.

@GroveRover I have purposefully written this in a way maximally upsetting to mathematicians :)

@ACuriousMind However it can't be the case since with the lie correspondence you get an operator, here we don't get a diffeomorphism so in no way the result can be right

@ACuriousMind My education is in physics, I just hate them when they do random things. Physics could have gone so much quicker and deeper if they just weren't scared of mathematics

@GroveRover I would be careful with that statement - vector fields and diffeomorphisms correspond to each other in a manner very much like finite-dimensional Lie groups and their algebras, and exactly this works to show Killing fields generate isometries: An isometry has $f^\ast g = g$. Infinitesimally: $g + L_X g = g\implies L_X g = 0$, so vector fields with $L_X g = 0$ generate isometries :)

@ACuriousMind yes but because the derivative is null. If it's not, your first-order expansion is not a diffeomorphism (it's a sum, how could it be). The stuff with U works only because the sum of an operator is still an operator

@GroveRover since you haven't written out the actual condition your prof used anywhere it's impossible for me to tell what exactly he might have meant

@ACuriousMind The point is that there is no such thing

He just stated "diffeomorphism invariance" and "infinitesimal translation" and started writing symbols

He does not even know what a pullback is

but then I would assume all he was doing was showing that the infinitesimal version of a diffeomorphism is $g\mapsto g+L_Xg$

@ACuriousMind yes but then using it as if it was an actual diffeomorphism

in what way did he "use" it like that? you seem to avoid spelling that out

i told you before

"He first states this in general, then He applies it to fix the GW gauge and to prove the conservation of the stress-energy tensor by substituting the variation of the metric in the action"

The conservation of T^{\mu \nu} in the general case, not in linearized gravity

ah, it's the derivation of the covariant stress-energy conservation

Maybe in both cases you can resort to first order, in linearized gravity you could in some way use the smallness of h and for T^{\mu \nu} you could use the fact that you're making a variation

but then it's just again my infinitesimal argument from above: you start from $S' = S$, where $S'$ is the action transformed under a finite diffeomorphism, you do $S' = S + \delta S$ and the $\delta S$ is the infinitesimal change under diffeomorphism

Or it's possible that if you compute the riemann tensor of the new metric it is still the same, however that would be problematic for me because it would mean you have more degrees of freedom than just those from diffeomorphisms

this answer probably does a computation similar to what you saw

@ACuriousMind That's what I meant, it is a random variation of the metric for which the action must also be stationary, so you derive and set the variation to 0

I think I wrote something similar earlier today

@GroveRover but then what's the problem?

@ACuriousMind Yes exactly

@ACuriousMind That he stated this as invariance under diffeomorphisms, but has nothing to do with that

Invariance under diffeomorphisms has a precise meaning which is different

This is only the stationarity of the action under arbitrary variatios

And the variation generated by a random flow simplifies the calculation

@GroveRover I'm not following you: Invariance under "diffeomorphisms" is $S' = S$. The infinitesimal version is $\delta S = 0$ where the $\delta S$ is the infinitesimal change of $S$ along the flow of the vector field $X$, which results in a bunch of Lie dervatives etc.

@GroveRover No, careful: At the point where you impose $\delta S = 0$ here you are not assuming the dynamical variables fulfill the equations of motion

this is not "the variation of the action is zero on-shell for all variations"

this is "the variation of the action is zero off-shell for symmetries"

On-shell means only satisfying the motionequation? Not familiar with this terminology

Yesh

on-shell/off-shell = with the dynamical variables satisfying the equations of motion/without the dynamical variables satisfying the equations of motion

Ok I think I finally got it

@GroveRover how nice. i am doing physics, but i like to dabble in some (algebraic) math when i can :P.

Thank you very much @ACuriousMind

I will write this down tomorrow hoping everything adds up conceptually speaking

the lennard jones potential is given by $2*\varepsilon [A_{12}(\frac{\sigma}{r})^{12} + A_{6}(\frac{\sigma}{r})^6]$ -- can i say that $\sigma$ is the same between similar atom types -- e.g. noble gases

i think no but i pray yes

And I'll find where is the approximation for which this works also as a gauge choice in linearized gravity

this kind of confusion usually first pops up in the context of Noether's theorem, when we derive conservation laws from the variation of the action being zero under the infinitesimal symmetry, and then at least one person goes "but isn't the variation of the action zero under all infinitesimal variation?"

classic

(al mechanics)

@ACuriousMind Unfortunately I've seen the canonical conservation without the action, so I just used the equation of motion + explicit independence from the position

@GroveRover You're doing GR but you've never seen a derivation of Noether's theorem for symmetries more general than position invariance?

@ACuriousMind both in my GR and QFT classes we derived it this way

Is it that uncommon?

I would insist that a classical mechanics course should have taught Noether's theorem long before that but I fear I already know the truth :'(

I believe Schwartz only derives Noether's for the energy momentum tensor

@ACuriousMind We don't have classical field theory here unfortunately, you just see it at the start of QFT

Noether's theorem and variations of the action etc. are not a field theory concept (but again every time I talk about classical mechanics in here I realize how exceptional our classical mechanics courses were)

where my condensed matter fam at

XD

@ACuriousMind of course they're not, but the energy-momentum tensor yes. The point particle version I've seen does not lead to a tensor, rather vector quantities conserved by lagrangian invariance

@ACuriousMind OT are you still publishing?

my classical mech class was 1. tides and two-body problem blah blah 2. smol lagrangian mechanics 3. coupled oscillators. in essence, i think

martha blah blah

@GroveRover publishing what?

lol

@ACuriousMind papers lol

I haven't published a paper in my life and I'm no longer in academia :P

@ACuriousMind I know you're not in academia but there are people who still publish outside

One I met at a quite important conference

@SillyGoose you have to discover geometry then

You'll find out that's the thing you always missed but never knew

BTW this chat is trash, you can't see the message I'm actually replying to and I look insane

@GroveRover You can - there are little arrows to the left of the @ that jump to the message you reply to when you click them

(unless you're on the inferior mobile view :P)

@ACuriousMind I know but it's not evident - also I can't reply to myself

@ACuriousMind replying to myself

@GroveRover im marking a spot on my bingo card

@ACuriousMind HOW

@SillyGoose Are you in the course part or the research part of your PhD?

@ACuriousMind Nice thank you

@GroveRover i am in the course part. i am a first year

I'll use it right now

@GroveRover this I am sure because I spent one day trying to find a canonical tensor in very different ways for point particle before understanding it is not possible

i might be able to do my research in topology related physics muahaha

@SillyGoose Oh here we had one of the main contributors to QTFT

Which field more specifically?

@GroveRover are u a phd student?

@SillyGoose I'm a weird student

Taking a gap year between Bachelor and Master but still taking Master classes, doing research and going to schools

i was suggested by a potential advisor to start learning about/looking into topological defects/goldstone bosons. i suppose they both fall under spontaneous symmetry breaking, so that is what I have been recently (past week) doing. but now i need to finish a final project and study for EM final, sadly.

@SillyGoose EM final? In your PhD?

QED maybe?

naw just classical electromagnetism

What

I had that at my second year

update: i have found in the literature the sigmas are indeed different for the noble gases

sad times

at least in the states i think generally the core curriculum in physics phd is QM, stat mech, EM

@GroveRover what content did your course cover?

@SillyGoose the first I had in the third year, the other two in my second

@SillyGoose All of griffiths, some of Jackson

oh in what country are you based?

Here the first courses of the masters are QFT and GR and then you specialize

@SillyGoose Italy

It's quite different here, we don't get introductory classes

my undergrad (which is probably far below average for physics) had EM based out of griffiths. and i think we only covered about half. the current EM course i am taking is based off jackson, and we are just about finishing waves in media i think (it is end of semester now).

e.g. we see no QM until the third year, but then you take a class in functional theory and straight up eat the Ballentine

interesting

well i notice the theory papers i read out of institutions in italy are quite nice (and plentiful)

@SillyGoose You need also to consider that we have way more hours than you

You can take only classes in physics and math here

ah i see

i went to a liberal arts college, so i think maybe 1/4 or so of my courses were unrelated to physics/math.

@SillyGoose here people are the most knowledgable in the world but they lack any other skill

For example we usually don't do research until master's thesis

No conferences

No schools

No internships

No contact with professors (classes of about 200 people and 1 professor)

in the states it is seemingly the opposite. people trying to get first authors before they apply to grad school :P.

I know, there should be an equilibrium

@GroveRover does everyone really take functional analysis at your uni?

@ACuriousMind what about the german system

(every physics student)

@SillyGoose It's compulsory

interesting. have you found that beneficial to your studies so far?

@SillyGoose not really. In QM everyone pretends things are well defined so you don't get to see what's under the hood

my undergrad advisor, for instance, said that he actually did not really use the differential geometry that he learned as a student (and he did his phd in geometric Langlands).

Actually the first part, the one about hilbert spaces yes

But not the one on operators

Also functional were useful for when I took distribution theory (also compulsory)

@SillyGoose What

Lmao

I think the thing i use the most is differential geometry

maybe he was understating something

GR, QFT, QM, Quantum Gravity

Everything is geometrical if you look deep

i still do not myself really have a strong grasp of differential geometry. i only know the basic constructions. in my undergrad, i took linear algebra, probability, abstract algebra, real analysis, algebraic topology, topology, and lie theory and some calculus course.

@SillyGoose algebraic topology?

Interesting but weird

i actually have been using algebraic topology lately

since i have started looking into topological defects (which are classified by homotopy groups)

@SillyGoose what do you mean by defects?

Like monopoles?

I've seen the most basic type of monopole and it was brutal. Had to go through almost a semester of principal bundles

this is my understanding of a topological defect. you have some field $f$ and the presence of a topological defect in $f$ is signified by having some homotopically non-trivial loop in the target space. i think.

@SillyGoose I am not familiar with this honestly

It seems some sort of principal bundle in disguise but f has the opposite direction

I guess one should be able to rewrite this sort of thing in the language of bundles

but the simpler examples just live on some nice base space like $\mathbb{R}^2$

The only one I've seen was ( R^3 \ {0} ) x R and was already hard

I imagine doing this things on non trivial spaces

i think monopoles are considered an example of a topological defect. or can be subsumed under such a class of phenomena

Anyway I think it's a formative area to start working in

That said, I'm going to sleep

see ya

If you want to learn more about P(H) I'll send you some reference

sure that sounds nice

(the formulation of QM not P(H) itself lol)

@SillyGoose write a comment under my answer then and tomorrow I'll send you

See yea

The basic example of a projective Hilbert space is the Bloch sphere

No examples

😓

important math guys

@Slereah what am I looking at

physics

Noice

can't be physics. the important guys are maths guys.

morning

@SillyGoose I mean, geometric Langlands is mostly algebraic geometry and number theory, I'm not surprised that an algebraic geometer doesn't have a lot of use for differential geometry :P

unless you're a complex algebraic geometer and can use GAGA, the two geometries don't really touch each other a lot

complex algebraic geometry? gaga indeed

@qwerty It's the established name for the correspondence between complex analytic and complex algebraic geometry!

hehehe

hi

So what about the discord?

@GroveRover what subjects do you study

i meant to ask : what subjects are you interested in

@RyderRude Fundamental physics and geometry

What about you?

@GroveRover i am interested in fundamental physics and philosophy of science and consciousness

@RyderRude Will you be at the European conference in fundamental physics?

@GroveRover idk what that is... is it an annual event

@GroveRover are you interested in the measurement problem

i am thinking that the measurement problem and quantum gravity are the same problem

most physicists seem to think that

@RyderRude This event, it's organized by the philosophy of physics society

@GroveRover oh

@RyderRude I am not sure most physicists think that

A part for sure, there are collapse model

@GroveRover i mean except string theorists maybe

Also Everettians

but string theorists aren't sure what their principles are

@GroveRover yeah.... forgot about them :P

@GroveRover yes

A lot of them also doesn't care at all

i meant most physicists working in foundations

other physicists may be working on more practical stuff

Probably, not too much into the field yet

Met both Diosi and Penrose in September tho

oh

@GroveRover nice

Penrose supports consciousness based theories

i am currently agnostic toward consciousness based theories. probably more pragmatic approaches are needed

@RyderRude Not familiar with that, i just heard about microtubules

@GroveRover yeah..

I'd like to read more about it but time is always tight

@GroveRover what areas do you study in fundamental physics

i hope to learn measurement problem physics and GR and qft

Right now I'm still a student so I'm taking classes in black hole theory, quantum information, QFTCS etc etc. In a few weeks, I'm starting some research on the geometric aspects of quantum mechanics

i only some QFT so far and have read some basic measurement problem physics

What do you mean by "measurement problem physics"? Like collapse models?

@GroveRover exciting topics

Decoherence?

@GroveRover yes

i don't study interpretations all that much cuz I think they r all wrong

but i will give QBism a try

@RyderRude I mean, how can you know if you don't study them hahaha

@GroveRover i know their basic ideas... they r wrong at the basics :P

plus my credence is 99.99% that measurement problem and quantum gravity are the same problem

and no interpretation is convincing. maybe some ideas of QBism are right tho

@RyderRude Hopefully we'll see

Are you a PhD student?

illbe back

@GroveRover i do physics as hobby

@RyderRude Got it

in the second half of the video, there is a conference of measurement problem community

these people are agonistic towards the final answer, but they do think that objective reality is incorrect

@RyderRude oh it's the IQOQI, I'm going there in April

I'll watch some sections when I find the time

@RyderRude I would not take seriously anyone saying they have the have a solution to the problem

This is one of the hardest tasks in physics and we should not be carried away by hype like we did in quantum gravity

@RyderRude I am once again suggesting you stop wasting your time watching these hours-long clickbait videos and start learning proper physics. Or at least not post those videos every time you fall for the bait.

@ACuriousMind Not familiar with this content but I've seen that the second half of this video is actual lectures at the IQOQI

The first half seems to be a round table with the 3 recepients

The title is cringe I'll give you that

@GroveRover The channel didn't record the lectures, they're just using them. The video is infested with ads (without adblockers I can't skip around even 5 minutes without getting an ad). If someone wanted to debate the content of the lectures, they should just link the original recordings.

The channel itself (as one can find out by looking around a bit) are people actively trying to promote "non-materialist metaphysics". This is one step above the pseudo-scientific grifters RR usually links, but it's still not a neutral source of information to direct attention/traffic to.

@ACuriousMind wtf lmao

@ACuriousMind ok ok got it

wait. is this some philosophy faction?

@ACuriousMind the title may be click bait, but it is an interview with genuine researchers who were awarded by the measurement problem research community. in the second half, there is the conference of the community

@ACuriousMind I'm dying "Essentia Foundation is a registered non-profit committed to making its content as accessible as possible and without advertisements. Therefore, we depend on contributions from people like you to continue to do our work. There are many ways to contribute."

@ACuriousMind what is wrong with non materialist metaphysics??

it is a genuine philosophy

but i haven't watched the channel's specific videos on non materialist metaphysics

as I was asking the other day, is there some philosophy crackpot index and are these people scoring on it?

I recognize bad practices but I'm sure you can consider the round table with the recipients as a good source of informations

@ACuriousMind also, I'm not saying that these videos are a substitute for learning. i only watch them to know the opinion of physicists

Don't know what RR usually links tho so I can't comment on that

@GroveRover feel free to use the search box lol

@GroveRover i almost never link psuedoscience

i accidentally linked a pseudoscientific video but that was months ago

@qwerty I've looked at the first few but they seem to be interviews with well-established physicists

🧶👕🙀🍚

@ACuriousMind a small curiosity, can you formalize GR as a gauge theory on the GL^+ (1,3)-principal frame bundle?

yes

Although with some complications

@Slereah Interesting, is there any (at least conceptual) advantages?

Well you get to use the various tools of gauge theory

It's not typically that useful except for very specific applications though, hence why it isn't that commonly done

@Slereah such as?

@GroveRover Have you seen physics.stackexchange.com/a/706483/50583?

I mean, I know the most basic constructions but I don't see them as convenience tools but rather the simpler way to formulate field theories

And in the case of GR you already have a simpler way

wow the hbar is v active lately yay

@GroveRover Like constructing the actual configuration space of GR

Which is nice to have if you want to quantize it, for instance

@Relativisticcucumber I did not check the screen for 10 or 20 mins... then 30+ messages lol

@SillyGoose ...string tee cat rice?

@ACuriousMind I was at the conference for his 90th birthday and had no idea of this series of papers. Might deeper look into it. Seems very interesting.

Trautman did a lot of cool papers

Reading your answer I couldn't help but ask myself why you are not publishing

@Slereah Right, I have not came in contact with quantum gravity/formal quantization yet so I didn't think about it

Like if you really want to be fancy with GR you could add all the ghosts and whatnot

Many ghosts

spooky

@Slereah you website has certificate issues

does it

and I can't understand if the CV has a broken button or you linked your StackExchange profile as your CV

I don't remember

lol

Website is as it is, if you want more feel free to pay me to write it :p

@Slereah can I pay you to take it down?

Sure, but it will be the same price

You're a software engineer too t.t how is it possible

I talked about principal bundles 4 times in the last two months and 3 of them were with software engineers

@GroveRover Trautman has a lot of interesting papers if you're interested in mathematical physics and geometric viewpoints

@GroveRover it pays the bills

@ACuriousMind still you're way more educated in geometry than the vast majority of people whose bills are paid by physics departments

ahhh to be young