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peterh
peterh
2:49 AM
Mods are revolting the SE. Last time it was not really effective, as I can remember - Monica did not get her diamonds back. The never before seen SE "lead", who expelled Monica, has left the company to the linkedin not so much later. Part of the mods came back, part of them did not. SE generated yet another bunch of documents, regulations.
 
 
3 hours later…
More Anonymous
More Anonymous
5:31 AM
If I have an isometry in my geodesic deviation vector? Does it mean my geodesics are parallel?
I think this is wron?^
*wrong
 
Silly Goose
Silly Goose
is 2.143 the last line a correct usage of the tensor product symbol?
I would expect composition not tensor product
 
Silly Goose
Silly Goose
6:23 AM
well i guess i would think that the usage of tensor product symbols seems dubious in 2.143 overall
 
 
1 hour later…
Slereah
Slereah
7:37 AM
The most terrifying thing about reading old secret documents is thinking that people this stupid are given that much government power
Reminds me of that Feynman anecdote when some general asks him if he could make a tank that runs on sand
 
 
2 hours later…
Slereah
Slereah
9:29 AM
The wiki article on Klein geometry doesn't even link the original Klein article
 
 
2 hours later…
John Rennie
John Rennie
11:35 AM
This is in a couple of days. Hopefully it will be interesting for those of us who have some QFT but know nothing about string theory.
 
ACuriousMind
ACuriousMind
Most physicists need to know exactly nothing about string theory :P
 
Slereah
Slereah
Many physicists know nothing of string theory and yet still feel like having opinions
 
user 858770
user 858770
11:51 AM
Does this push the boundaries of what it means to be a physicist.
 
ZeroTheHero
ZeroTheHero
12:22 PM
Interesting post from another site regarding AI-generated contents: academia.meta.stackexchange.com/q/5286/90441
2
 
ACuriousMind
ACuriousMind
12:33 PM
@ZeroTheHero also pay attention to the edit history: The VP of Community does not have time to actually address any of the concerns users have, but he does have time to personally remove the featured label from unwanted meta posts
 
ZeroTheHero
ZeroTheHero
@ACuriousMind Thanks for highlighting this small but revealing point.
 
user 858770
user 858770
Sounds like a power trip.
Revealing a hidden agenda to "clean things up."
 
ACuriousMind
ACuriousMind
what on earth are you talking about?
 
user 858770
user 858770
Your revealing point.
 
ACuriousMind
ACuriousMind
Yes, what does that have to do with a "hidden agenda to clean things up" or a "power trip"?
 
user 858770
user 858770
12:41 PM
A judicious use of time management by the VP.
 
rob
rob
12:56 PM
The removed featured might be a test of the "strike," since restoring that tag is a moderation action, which the moderators have said they won't engage in.
 
user 858770
user 858770
Classic union flick.
✊
 
ZeroTheHero
ZeroTheHero
1:12 PM
@rob This whole story is just sad. Nothing was broken (at least in PSE) so why force a fix?
 
user 726941
user 726941
284
Q: Moderation Strike: Stack Overflow, Inc. cannot consistently ignore, mistreat, and malign its volunteers

Mithical Introduction As of today, June 5th, 2023, a large number of moderators, curators, contributors, and users from around Stack Overflow and the Stack Exchange network are initiating a general moderation strike. This strike is in protest of recent and upcoming changes to policy and the platform that...

discussion community company
 
Amit
Amit
I have a title for the next episode
TSSB - The Stack Strikes Back
 
rob
rob
2:16 PM
@ZeroTheHero The claim was a that (a) software GPT detectors have a pretty high error rate, that (b) the errors discriminate against users who aren't native writers of English, and that (c) the geography of GPT-related suspensions suggests those biases apply on the SE network. I think those are legitimate concerns. But the company essentially forbade most GPT-related removals and suspensions, based on their internal analysis, and doubled down when the moderators were unconvinced.
If they had asked the moderator corps for more data before (or instead of) issuing orders to their volunteers, the whole shitstorm might have been avoided.
 
Slereah
Slereah
what if the post say "As an artificial intelligence, ..."
as it often does
 
rob
rob
@Slereah I don't know how much of the private moderator guidance I can share publicly.
But posts whose copy-paste includes the "regenerate response" prompt, and similar, are taken as an "admission" of chatbot usage.
 
Slereah
Slereah
You can tell that Urs Schreiber is a true mathematician because his powerpoints are horrendous
 
आर्यभट्ट
आर्यभट्ट
@rob agree.
What u said to zero the hero.
 
Relativisticcucumber
Relativisticcucumber
2:36 PM
@Slereah @SillyGoose ur idol
hello - I have a question about the metric in GR. so if a metric tensor is defined to be a bilinear map from two tangent vectors to a real number, does this mean the components of the metric tensor are the "real number" in discussion here? if the metric tensor can be written as $g = g_{\mu,\nu}dx^{\mu}\otimes dx^{\nu}$, I am a bit confused on how these two tangent vectors serve as the inputs to the metric tensor
 
Slereah
Slereah
It is the same thing as every tensor, your actual tensor is $g$, which you can decompose in a basis as $g = g_{\mu\nu} dx^\mu \otimes dx^\nu$, and then the application of the two vectors is done by $$g(X,Y) = g_{\mu\nu} dx^\mu(X) \otimes dx^\nu(Y)$$
The dual basis projects your vector to their components via $$dx^\mu(X) = dx^\mu(X^\nu \partial_\nu) = X^\nu dx^\mu(\partial_\nu) = X^\nu \delta^\mu_\nu = X^\mu$$
Alternatively you can consider $g_{\mu\nu} = g(\partial_\mu, \partial_\nu)$
The components are the inner products of the basis vectors
 
Relativisticcucumber
Relativisticcucumber
if $\partial_{\nu}$ and $dx^{\mu}$ are different bases, why is their product delta ?
 
Slereah
Slereah
The dual basis is generally defined to be specifically the basis such that $dx^\mu(\partial_\nu) = \delta^\mu_\nu$
You can show that there always exists such a basis of the cotangent bundle
 
Relativisticcucumber
Relativisticcucumber
ok i see
and one more question: so if we have this criteria that a metric tensor must map two tangent vectors to a real number, i see that $g$ satisfies that, but is it necessary that the metric tensor should be invariant ? im not seeing how this necessarily comes from the definition, but isn't that the essential point of the metric tensor?
 
ACuriousMind
ACuriousMind
2:57 PM
"invariant" under what?
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind coordinate transformation, right?
 
ACuriousMind
ACuriousMind
You used the word, you tell me what you mean by it :P
 
Relativisticcucumber
Relativisticcucumber
oh no i dont understand
or, i dont understand the question i mean
 
ACuriousMind
ACuriousMind
What do you mean by the metric tensor being "invariant under coordinate transformation"?
And in what way do you think that property - whatever exactly you mean by it - is not just part of the general definition of what a well-defined tensor is?
 
Relativisticcucumber
Relativisticcucumber
ok i think my confusion is maybe in muddling the line element and metric tensor. i see that tensors are defined by obeying the tensor transformation laws, so obviously they are invariant under transformation. then, i know the line element also is a quantity that is invariant under coordinate transformation, so i guess i need to answer the question of how the definition of the metric tensor gets us to the line element because the line element is a scalar and the metric tensor is a tensor
but somehow i feel they are the same thing so i must be muddling something
they = metric tensor and line element
 
ACuriousMind
ACuriousMind
3:10 PM
For any curve $\gamma : I\to M$, you get its "length" according to $\int_\gamma \sqrt{g(\dot{\gamma},\dot{\gamma})}$
the "line element" notation $\mathrm{d}s^2 = \dots$ is just a mnemonic, you shouldn't consider $\mathrm{d}s$ as an object that exists independently of the curve whose length you are interested in
 
Relativisticcucumber
Relativisticcucumber
also i feel that obeying the tensor transformation law does not follow from mapping two tangent vectors to a real number. i interpret it to be the case that a tensor is a quantity that can map from two tangent vectors to a real number, right? so then it's not necessarily the case that any entity that obeys this map should be conserved under coordinate transformation? or am i misunderstanding?
 
Amit
Amit
It may be that the line element is a slightly loaded term, it has all kinds of associations that you need when doing integrals for example. But I think to understand the metric tensor more simply you can just think of it as the map that defines an inner product between a pair of vectors
 
ACuriousMind
ACuriousMind
@Relativisticcucumber you're misunderstanding, the transformation laws of the components follows directly from defining the tensor to be a bilinear map on the tangent space at every point
it's just linear algebra: AT each $p\in M$, $T_p M$ is a vector space, and so bilinears on it are elements of the vector space $T^\ast_p M\otimes T^\ast_p M$. $\mathrm{d}x^\mu$ are a basis for $T^\ast_p M$, and so you can expand any such bilinear tensor $T$ as $T_{\mu\nu}\mathrm{d}x^\mu\otimes\mathrm{d}x^\nu$
 
naturallyInconsistent
naturallyInconsistent
@Relativisticcucumber There are many types of tensors...
 
ACuriousMind
ACuriousMind
from the definition of $T$ not being coordinate dependent and the known transformation of $\mathrm{d}x^\mu$ via chain rule, the transformation behaviour of $T_{\mu\nu}$ follows without further assumptions
 
Relativisticcucumber
Relativisticcucumber
3:16 PM
@ACuriousMind but the derivation that i saw for the transformation law is to say "tensors are conserved" and "chain rule exists" and then to derive the law, so this seems circular?
 
naturallyInconsistent
naturallyInconsistent
@Relativisticcucumber Most new theories sound circular in its postulates.
 
Relativisticcucumber
Relativisticcucumber
not conserved -- invariant , sorry
@naturallyInconsistent but surely it's not actually ?
 
naturallyInconsistent
naturallyInconsistent
I actually think it should be
Münchhausen trilemma
The only one physics and maths may use is the circular argument, hence we have to axiomatise / postulate, and then work from there
 
ACuriousMind
ACuriousMind
@Relativisticcucumber I don't understand where the circularity is supposed to be
 
ZeroTheHero
ZeroTheHero
@rob The points you reported actually muddle the issue, which is how AI will affect the quality of the questions and answers. There may or may not be a true geographical bias but if there is then origin information should be masked, and if this is not possible immediately then a fix should be in the works.
I don't follow closely the related closures on PSE but your post on our Meta suggests the actions taken were appropriate, so I still wonder why this has to be a system-wide requirement. If there are local abuses deal with those locally.
 
Relativisticcucumber
Relativisticcucumber
3:22 PM
@ACuriousMind if we say "this entity takes two tangent vectors to a real number", does this statement alone mean "this entity" is necessarily a tensor? i think this might clear my confusion up perhaps
 
Amit
Amit
Add multi-linearly and yes
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind wait this means yes to my question i think
 
Amit
Amit
Also symmetrically and non degenerately :)
 
Relativisticcucumber
Relativisticcucumber
i see -- so these two definitions of tensors are equivalent it would seem
 
Amit
Amit
Defining a tensor by the transformation law is afaik a heuristic, maybe it can be shown to be equivalent to the more axiomatic approach but idk
 
ZeroTheHero
ZeroTheHero
3:27 PM
@rob I mean don't get me wrong: Mother SE is trying to muddle the issue with bias stuff. It may very well a legitimate concern but the solution offered is not a solution.
 
ACuriousMind
ACuriousMind
@ZeroTheHero We don't understand it either! And SE refuses to explain it in a way that makes sense
it seems they've gotten some data that GPT detectors are unreliable (they are!) and immediately concluded that since moderators make their decisions mostly based on such detectors (they don't!), most of the suspensions issued must be invalid, but they did not present any reasoning to actually support that jump in logic
and instead of asking moderators how we actually moderate, they just dropped that policy on us
 
Slereah
Slereah
Imagine that you got it wrong and instead of removing an AI you removed a post so badly written you confused it with one
 
Amit
Amit
@Amit correction about that -- these two additional conditions apply only for an inner product, not a general tensor
 
ACuriousMind
ACuriousMind
@Slereah I am beginning to suspect that people who communicate all day in corporate speak may actually not understand how different GPT output is from how normal people would answer an SE question :P
 
ZeroTheHero
ZeroTheHero
 
Amit
Amit
3:38 PM
Lol
 
ZeroTheHero
ZeroTheHero
how many of those before they reached their decision...
 
Amit
Amit
Did things start going south when SE was acquired in 2021?
 
ACuriousMind
ACuriousMind
@Amit no
 
Amit
Amit
Ok
Interesting, thanks
 
Slereah
Slereah
The site is run by the robots now
 
bolbteppa
bolbteppa
4:02 PM
@JohnRennie It will likely be an updated version of this:
 
Amit
Amit
@ZeroTheHero @Slereah I knew that was agent Smith at the board! 😂
Nah I hope it's allll a big misunderstanding and everyone will come to a good agreement, no need to be mean 😁
 
Slereah
Slereah
4:26 PM
Some articles of nlab are smaller than others
 
Amit
Amit
Lol, can be read instantoneously
 
Relativisticcucumber
Relativisticcucumber
4:41 PM
bah i have another question -- so it says in Carroll that the partials $\{ \partial_{\mu} \}$ are a basis for the tangent space, but in another book i am reading, it defines the basis vectors for the tangent space as $e_{\mu} = \frac{\partial \mathbf{r}}{\partial x^{\mu}}$ -- are these contradictory ?
here $\mathbf{r}$ is the position vector
 
ACuriousMind
ACuriousMind
@Relativisticcucumber those are "intrinsic" vs. "extrinsic" definitions
the second definition makes only sense when your manifold is embedded into an $\mathbb{R}^n$ so that $\mathbf{r}$ makes sense
under the embedding of the manifold into that $\mathbb{R}^n$ the $\partial_\mu$ are sent to the $\frac{\partial\mathbf{r}}{\partial x^\mu}$
 
Relativisticcucumber
Relativisticcucumber
okay, so in the first definition, how does one obtain the components of the metric?
 
bolbteppa
bolbteppa
Don't do GR without the position vector $\mathbb{r}$, you will get lost in formal abstraction whereas using the position vector everything is just basic calculus
 
ACuriousMind
ACuriousMind
@Relativisticcucumber didn't we already do that? You define $\mathrm{d}x^\mu$ as the dual basis to $\partial_\mu$, then expand $g$ in the basis $\mathrm{d}x^\mu\otimes\mathrm{d}x^\nu$ of $T^\ast M \otimes T^\ast M$
 
Relativisticcucumber
Relativisticcucumber
i guess what im not seeing is how this gets us $g$ explicitly as the second definition does because im seeing we have a manifold and then we know we have some $g$ that satisfies the bilinear relation mentioned above and that we can write this using the expression you mentioned but none of this seems to arrive explicitly at the components of the metric which i think needs to stem from the coordinates, right?
@bolbteppa bah
 
bolbteppa
bolbteppa
4:48 PM
Even defining something as simple as a derivative (covariant derivative) becomes a massive chore involving magical thinking about parallel transport which is really just the special case of the derivative of a constant vector field turned into some formal definition to satisfy pedants that it's actually more general instead of a special case
 
ACuriousMind
ACuriousMind
@Relativisticcucumber I'm not sure what you mean by "arrive explicitly at the components".
In practice, you will always be "given" a metric by someone specifying the $g_{\mu\nu}$ in one particular coordinate system
and if you want to know how these look in another coordinate system, you just use the transformation behavior of the $g_{\mu\nu}$
while the abstract approach insists that the idea of the coordinate-independent $g$ is important, you effectively almost never have any way of specifying a specific $g$ if not by its components
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind if i use the second definition i can literally get the metric elements. for example i can write down $x = r\text{cos}\theta$ $y = r\text{sin}\theta$, take the partials of $\mathbf{r}$ and take the inner products to get the elements of the metric arriving at a matrix that is $\text{diag}(1,r^2)$
 
ACuriousMind
ACuriousMind
@Relativisticcucumber sure, but the second approach has an embedding into $\mathbb{R}^n$
that embedding is additional data, in particular it is an isometry
in general, such isometric embeddings always exist, but may be very hard to find and the $n$ in the $\mathbb{R}^n$ may be larger than you like
(e.g. you can't embed a Klein bottle in $\mathbb{R}^3$)
 
Relativisticcucumber
Relativisticcucumber
so are you saying in the first approach we cannot write down the metric components explicitly? we need to be provided a metric? but this metric must come from somewhere, right?
 
ACuriousMind
ACuriousMind
@Relativisticcucumber A metric is additional data for an abstract manifold
when you have an embedding of manifold into another (pseudo-)Riemannian manifold, then you get an induced metric on it (that's what's happening with the embedded versions)
but if you just have the standalone manifold, there just is no metric on it
it's additional data you need to specify, you cannot compute it from somewhere
 
naturallyInconsistent
naturallyInconsistent
4:57 PM
@ACuriousMind not if you are trying to integrate EFE themselves, e.g. numerical gravity
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind :o
 
naturallyInconsistent
naturallyInconsistent
@Relativisticcucumber I would have to point out that the moment you think that position is a vector, you are in a nice situation. In general, it is not a vector.
 
ACuriousMind
ACuriousMind
for instance, there's plenty of different metrics you can choose on a torus - there's the standard Riemannian metric on it you get from its embedding into Euclidean $\mathbb{R}^2$ but there's lot's of Lorentzian tori, e.g. flat ones or the Clifton-Pohl torus
all these are different (i.e. non-isometric) as pseudo-Riemannian manifolds, but they are diffeomorphic as smooth manifolds - the underlying manifold is just the 2-torus
that's what I mean when I say the metric is additional data and not some intrinsic property of a manifold
@naturallyInconsistent well, you'd start at least with initial data on a Cauchy surface, no?
 
naturallyInconsistent
naturallyInconsistent
@ACuriousMind I'm not sure, actually. Can we not do it in terms of Maxwellian local volume elements?
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind im so jumbled now. wait so does this mean that since we model spacetime as a manifold and we talk about all these metrics, we are assuming some embedding in some higher dimensional manifold?
 
ACuriousMind
ACuriousMind
5:01 PM
@Relativisticcucumber no, what did I say to imply that?
as I said, the metric is just additional data
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind probably nothing i have a defective brain
let me reread ur messages
 
naturallyInconsistent
naturallyInconsistent
@Relativisticcucumber I would point out that the phase space, symplectic manifolds that we talk about, are all the same, except the specific Hamiltonian and the specific metric that corresponds to a specific system is different.
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind i think this ? not that u implied what i said but just that i guess im misunderstanding this
 
ACuriousMind
ACuriousMind
@Relativisticcucumber I'm saying that a standalone manifold has no natural metric on it
but that instead the metric is a choice of additional data (precisely the choice of a bilinear form on every tangent space at every point)
 
bolbteppa
bolbteppa
@Relativisticcucumber It's not you it's the nature of having left the simple realm of the position vector, this is only the beginning you've been warned
 
ACuriousMind
ACuriousMind
5:03 PM
you can make many different such choices, in general
 
Relativisticcucumber
Relativisticcucumber
@bolbteppa LOL
 
bolbteppa
bolbteppa
It's a big price to pay just to be able to do GR on Klein bottles
 
naturallyInconsistent
naturallyInconsistent
@bolbteppa Exactly. The niceness of the Cartesian coördinate system that allows us to have trivial parallel transport and position vectors, is very difficult to pry ourselves away from
 
bolbteppa
bolbteppa
I'm still stuck on simple things like deriving the Kerr solution which is insane if one looks into it
 
Amit
Amit
But the position vector is just a helpful conceptual device right? Imagining that every tangent space for example is actually defined by an absolute position within an embedding space. But it doesn't really simplify any calculation, does it? Just makes stuff easier to visualize?
 
Relativisticcucumber
Relativisticcucumber
5:07 PM
@ACuriousMind but i thought this was the definition of a metric
ah sorry if im making u repeat urself. i have become very tangled with these concepts
can you explain what you mean by "additional data" and "natural metric"
 
bolbteppa
bolbteppa
I have never in my life needed to picture multiple tangent spaces, it's all a big waste of time, you will get to the depths of string theory without any of that nonsense, it's really all just basic calculus applied to vector fields where you allow the basis vectors to vary, next one moves on to the physis
 
naturallyInconsistent
naturallyInconsistent
@Amit Not just that! I mean, how would you express the force of one hemisphere of a charged conductor on the other hemisphere? You usually express the force using spherical parametrisation but using Cartesian vectors, and then integrate. How are you going to express that dayum integral without the Cartesian trivial parallel transport?
 
Relativisticcucumber
Relativisticcucumber
@bolbteppa so you think there is no need for formalism #1 in general?
if i dont learn it i need to change my name to pseudorelativisticcucumber
 
bolbteppa
bolbteppa
Well I can give you books where you need to do that, but it doesn't mean it's actually fundamental to the ideas at all
 
Amit
Amit
Hmm by choosing an appropriate chart? But I think I see what you mean, that charts become less arbitrary looking when you think in terms of a position vector?
 
naturallyInconsistent
naturallyInconsistent
5:11 PM
@Amit How would you express "same direction" using merely charts?
 
ACuriousMind
ACuriousMind
@Relativisticcucumber Maybe let's take a step back and consider just the linear algebra notion of an inner product on a vector space $V$
 
Amit
Amit
Oriented atlas? :) idk, as you said I've usually done this kind of integrals assuming a transformation to cartesian space...
 
ACuriousMind
ACuriousMind
if I just give you a vector space, there's no inner product on it, right? I need to specify an inner product, e.g. by choosing a basis $e_i$ and then declaring what $\langle e_i,e_j\rangle$ for each pairing of the $e_i$ is.
the metric on a manifold is just a generalization of this concept where we need to provide such a specification on every tangent space simultaneously
and just like there can be many different inner products on a vector space, there can be many different inner products on a manifold
 
Amit
Amit
But I am starting to see that yes, more calculational machinery is apparently involved when working totally "intrinsically"
 
ACuriousMind
ACuriousMind
but when the vector space $V\subset W$ is a subspace of a vector space $W$ on which I already have an inner product, then $V$ inherits an inner product (just the restriction of the general inner product to the subspace)
 
Amit
Amit
5:16 PM
Or at least more formalism is required
 
ACuriousMind
ACuriousMind
the same happens for a manifold embedded into an $\mathbb{R}^n$ where we use the Euclidean (or Minkowski) metric on $\mathbb{R}^n$ - it inherits a metric (the "induced" metric) - and that's what you're computing with the derivatives of the $\mathbf{r}$
 
naturallyInconsistent
naturallyInconsistent
@Amit It is fundamentally much worse than that. Once spacetime is curved, "same direction" is path dependent, even if you have well-defined parallel transport
 
Amit
Amit
Interesting, I guess it's not something I ran into yet
Not a GR veteran :)
 
Slereah
Slereah
If it's not path dependent it is literally flat space
 
bolbteppa
bolbteppa
If someone thinks GR is too easy using the position vector, I'll give you something too difficult ;)
BKL singularity
A Belinski–Khalatnikov–Lifshitz (BKL) singularity is a model of the dynamic evolution of the universe near the initial gravitational singularity, described by an anisotropic, chaotic solution of the Einstein field equation of gravitation. According to this model, the universe is chaotically oscillating around a gravitational singularity in which time and space become equal to zero or, equivalently, the spacetime curvature becomes infinitely big. This singularity is physically real in the sense that it is a necessary property of the solution, and will appear also in the exact solution of tho...
This is a god damn rabbit hole
 
naturallyInconsistent
naturallyInconsistent
5:24 PM
You dont have to be fully into GR to know this. Just do any sphere shit and you should see it.

Imagine you have a vector on the North pole, pointing towards the South pole passing through x=0 line. If you move this vector down to the equator, it will be pointing the straight towards the South pole. Then you parallel transport it to the y=0 line via the equator, it will be pointing still straight towards the South pole. Now you transport it back up to the North pole, and you realise you have a 90 degree rotation of the original vector
 
bolbteppa
bolbteppa
@Slereah is that stuff going in your GR book
 
Slereah
Slereah
One thing you have to keep in mind is that it is a fundamentally weird idea to try to cling to flat space in GR
Like Euclidian geometry is just an idea
It's not a particularly obvious fact
@bolbteppa Not so far no
My book is more focus on like math
If I have to include all the cosmology and all the astrophysics it's gonna go from 100 chapters to 10,000
 
naturallyInconsistent
naturallyInconsistent
Our relationship with Euclidean geometry is actually rather nuanced. We utilise knowledge of it to construct all the curved stuff we want to talk about, but it is therefore not obvious to know where it starts and where it ends.
Much worse than a stereotypical love-hate relationship
It is more like "You cannot know that you do NOT live in a Euclidean geometry until you go out and look!"
 
Slereah
Slereah
Also there are like
Equivalence things
Like part of the challenge of physical geometry is deciding what a "straight line" is
In mathematical geometry we just give a bunch of properties for lines, but we don't know what lines are
In physical geometry you have to decide what is or isn't a straight line
and that is one of those bootstrap problem
 
Amit
Amit
@naturallyInconsistent oh ofc I know this classic example that demonstrates what curvature does to a parallely transported vector in a closed loop trip. I am just not making the leap where calculationally it is easier to do with the extrinsic formalism, it's clearly easier to visualize extrinisically :)
 
naturallyInconsistent
naturallyInconsistent
5:35 PM
@Amit Well, again, how would you even get to express the integral we want to calculate? This is one of those hangups that I really want to solve, so that then we can migrate fully to post-last-last-century maths in physics.
It is incredibly infuriating that when we want to actually compute some integrals, it is often the case we have to break out of the nice maths tools we have gotten, tensors, diff geo, etc, and go back to basically outdated maths.
 
Slereah
Slereah
You can use fancy maths for anything in GR
But most people don't because it would be a huge pain
 
Amit
Amit
If you have the metric for this sphere, no matter how you obtained it, you can integrate on any path can't you? Oh yeah and the LC connection I guess :)
Or maybe you don't need it for the integration.. hmm I don't remember it comes up explicitly at least
 
Relativisticcucumber
Relativisticcucumber
@ACuriousMind ok ok that makes more sense
 
Amit
Amit
I do remember there is some problematic point in defining orientability, if you define the integral naively it doesn't work.. and it requires to loan some methods from measure theory, to define the tensor density correctly? This is all vague memory now, from watching a lecture by Schuller on integration on manifolds in GR
 
Slereah
Slereah
You can also just define your volume form up to a sign
It's not like the sign matters that much*
 
Amit
Amit
5:43 PM
Lol
 
Slereah
Slereah
That's essentially what doing it with fancy measures does anyway
 
Amit
Amit
Yes, I just liked the asterisk
Will there be measure theory in the book? At least an appendix or two?
 
naturallyInconsistent
naturallyInconsistent
@Amit Setting up the framework to allow for integration is easy. How specifically to get the alignment of the directions so that you can express it as a specific integral is the difficult part.
 
Slereah
Slereah
If you're bothered by orientation for integrals just use the double cover and divide by two
 
Slereah
Slereah
6:50 PM
@ACuriousMind If a space is connected, does there exist like some bunch of overlapping open sets connecting any two points
 
Slereah
Slereah
7:00 PM
probably wouldn't work as a definition otoh since the open sets also need to be connected
 
Silly Goose
Silly Goose
@Relativisticcucumber honk
 
PM 2Ring
PM 2Ring
7:15 PM
The mod strike post on MSE is currently at 42k views. meta.stackexchange.com/q/389811/334566
 
Physics Meta
Physics Meta
7:43 PM
0
Q: Is there somewhere that we could officially recommend "companion" stack exchanges?

Logan J. FisherSome stack exchanges are clearly closely related to each other. For example, "companions" of the physics stack exchange might include: History of Science and Mathematics Mathematics TeX Electrical Engineering Chemistry Quantum Computing Academia And others... While we don't want to discourage p...

discussion feature-request scope questions stackexchange
 
 
1 hour later…
ACuriousMind
ACuriousMind
8:45 PM
@Slereah path-connected definitely, merely connected I'd have to think about
 
Slereah
Slereah
Picturing this
But open sets
 
naturallyInconsistent
naturallyInconsistent
@Slereah ewwww
 
Slereah
Slereah
wot
 
Amit
Amit
decorative manifold!!
o manifold, o manifold, how lovely are thy fibers
 
Elements in Space
Elements in Space
9:39 PM
I don't see a post about the strike on Physics Meta. Are ya'll on board with the strike?
473
Q: Moderation Strike: Stack Overflow, Inc. cannot consistently ignore, mistreat, and malign its volunteers

Mithical Introduction As of today, June 5th, 2023, a large number of moderators, curators, contributors, and users from around Stack Overflow and the Stack Exchange network are initiating a general moderation strike. This strike is in protest of recent and upcoming changes to policy and the platform that...

discussion status-review community company
 
Amit
Amit
@ElementsinSpace Did you see this one?
 
ACuriousMind
ACuriousMind
@ElementsinSpace I personally support the goals of the strike, but we have laid out our reaction to the new policy here - for the time being we will continue to moderate as we have before.
 
Elements in Space
Elements in Space
@Amit I did, that post doesn't mention the word "strike" (though it does appear in the comments).
 
Amit
Amit
9:54 PM
Yeah, I consider the Physics.SE stance even better... If indeed the mods are not following the new policy, I suppose "civil disobedience" is in some ways even better than a strike isn't it? :)
 
user4539917
user4539917
I wonder which sites are not striking?
(or are in full agreement with the policy)
 
Elements in Space
Elements in Space
10:11 PM
It makes more sense to me to try to actually try to get the network policy changed, rather than just ignoring it.
 
Amit
Amit
I'm sure it's not just a passive act of ignoring, but idk what is happening behind the curtains of mods - SE management interactions...
I would find it remarkable if there isn't any dialogue resulting from all this
 
ACuriousMind
ACuriousMind
10:49 PM
@Amit There is a lot of communication going on behind the scenes. I wish more of it were public (and I think some it would make SE look better if it were public), but that's up to SE to decide.
 
Amit
Amit
mm, makes sense, thanks for sharing (what little you can)
 
PM 2Ring
PM 2Ring
The analysis has been provided to moderators, now, so we're mostly in the loop. We can discuss things more meaningfully with the CMs. (Can't share details, but wanted to let people know about the positive development.) — wizzwizz4 2 hours ago
wizzwizz4 is a mod on Retrocomputing
Currently, 616 people have signed the strike letter, including 4 former SE staff: Shog9, Tim Post, Jon Ericson & hairboat. And the strike announcement on MSE has 51k views.
 
Amit
Amit
11:20 PM
@PM2Ring The information must flow! :)
 
peterh
peterh
@user4539917 Roughly all of the company statements show, they do not know, what to do with the SE site network, they consider themselves as Stack Overflow. I think, probably the face loss is their major reason to not sell or close the whole site network et al.
 
PM 2Ring
PM 2Ring
11:57 PM
@Amit The Physics.SE attitude towards GPT has always been a little different to the SO policy.
Dec 6, 2022 at 17:39, by ACuriousMind
oh, sure, but if their posting volume is high and the answers are garbage we can already suspend people for "consistent low-quality contributions"
Dec 6, 2022 at 17:40, by ACuriousMind
I don't really want to have to make an argument whether the content is generated or not in that case
 

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