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4:45 AM
Now this is an interesting twitter account
I'm so glad I found this account
 
 
3 hours later…
7:50 AM
Kerr seems to have recently given an example of a singularity free Kerr black hole in classical GR! What do you guys think 'bout it?
 
5
Q: Why does Roy Kerr claim that the Kerr black hole does not contain a singularity?

noir1993In a preprint posted on a popular forum, Roy Kerr claims that there is a widespread misunderstanding related to the singularity inside the black hole that bears his name. Can anyone explain his argument in "simple terms"? Are there any pitfalls in his reasoning? Here's a link to the preprint:...

 
8:07 AM
That account is driving people crazy
 
 
2 hours later…
10:21 AM
hi
 
@Sanjana From what I can tell he basically found quasi-regular singularities in basic black hole solutions, which casts into doubts what the singularity theorem actually says?
Since quasiregular singularities are the boring ones
fucked up if true and nobody noticed so far
 
10:57 AM
Why on earth is this on ResearchGate and not arXiv
 
Kerr is a million years old, maybe he is having trouble with it
 
and there's no one who could've helped him with it?
 
Who knows
 
is it necessary for quasistatic states to be in equilibrium?
or is it necessary for it to be slow only?
 
One of the bibliography element is just "[2] R. Penrose,"
Good thing this is just a preprint :p
 
11:02 AM
I'm just saying that "very old physicist claims breakthrough idea, then publishes it through non-standard channels" is a kind of pattern that should make people suspicious
 
Well it's Kerr and I've seen it posted thrice in different communities so far
If there's something wrong I'm sure we'll hear about it soon
From a casual reading I'm not seeing any wild claim?
He's just distinguishing the notion from a curvature singularity to a general b-boundary point
I didn't check his math so idk if it's correct
 
@Slereah Not sure - did people ever really rebuke Atiyah's "proof" of the Riemann hypothesis? From my memory they just let it more or less slip into silence and everyone agreed not to talk about it anymore.
 
I'm not even sure if this is a novel claim really, nitpicking on the notion of singularity in the singularity theorem isn't a super new thing
It's just a pretty niche topic
 
and honestly the abstract does not inspire confidence:
> Roger Penrose[1, 2] claimed sixty years ago that trapped surfaces inevitably lead to light rays of finite affine length (FALL’s). Penrose and Stephen Hawking[3] then asserted that these must end in actual singularities. When they could not prove this they decreed it to be self evident.
that's not a professional way to start your paper
 
Yeah he is being a bit confrontational :p
That guy thinks he invented the Kerr metric or something
2
Also he seems to personally believe that there shouldn't be curvature singularities
at least for stellar black holes
 
11:13 AM
@ACuriousMind I think you're right. Out of respect, people just left it as an exercise for the interested reader(s).
 
11:24 AM
I don't know if his paper is correct but it certainly does him no favor to be that agressive about it
 
I think aggressive and competitive come as a package.
 
It's not a competition :p
 
The internet has made it seem to be one...
 
also apparently Penrose was aware of this issue even back then so idk if it's as novel as he'd like to think
 
12:37 PM
iep.utm.edu/hume-ima Hume is saying that consciousness is based on a copying mechanism
according to Hume, first person experiences get copied and these copies are what ideas are
what do u think about this?
i dont think this mechanism is correct. the correct mechanism i feel is where the input from input organs comes first. this input gets translated into both a first person experience and a copied memory.
i think the copying happens before the experience. the input signal first gets copied onto ur neurons which u then experience. this is like a cache memory copy
and then the post-experience again gets copied into ROM memory
this model also explains hallucinations. it is when the cache memory copy is a bady copy, or is unrelated to the physical signal
 
 
2 hours later…
2:21 PM
The real reason for the Kerr paper
Those damn kids
 
3:10 PM
::curses on their cursory glances::
Nice find, btw. 👍🏻
💯%
 
 
5 hours later…
8:39 PM
@ACuriousMind Fair call. But (as far as I can tell) he isn't proposing a breakthrough idea. He's just saying that the singularity theorem doesn't actually prove that singularities must exist in a (ideal) Kerr spacetime. And that there's never been a good reason to assume that a singularity must exist in a BH formed by collapse.
 
Remember how things went when Atiyah started making big claims about deriving the fine structure constant etc
 
@Slereah Please consider writing an answer for that Kerr question. Currently, the only answer comes from safesphere, who freely admits that he didn't actually read the paper.
 
yeah his answer seemed a little off
 
@Slereah study it all for 2 days then write an informed answer
He might actually read your answer, that would be something
 
Sure, as safesphere says, you can avoid singularities in the extended solution (with the white hole, etc). But Kerr says that stuff's not really relevant for a BH formed from a collapsing star.
 
8:47 PM
@PM2Ring His abstract is essentially saying that Penrose and Hawking were wrong (and dismissively saying they just assumed something they could not prove). This is an extraordinary claim, especially in this tone of combativeness.
If the stuff he's actually saying in the paper is more tame, fine, but then don't market this as some grand rebuke to the singularity theorems
 
8:59 PM
I don't have the differential geometry or GR skills to fully understand this stuff. It seems reasonable to claim that light rays of finite affine length must end in a singularity, even if one cannot prove it rigorously. To me, it seems like Kerr (a mathematician) is rebuking the GR community for being insufficiently rigorous on an issue that needs rigour because it cannot be investigated empirically.
 
is there a stack post somewhere that explains in what situations the metric space completeness of Hilbert space is used in quantum mechanics?
I see answers which say roughly: it allows us to do calculus in Hilbert space, but I haven't seen a more descriptive answer
 
@SillyGoose the whole notion of a Hilbert basis relies on completeness
 
Even when the underlying vector space is finite dimensional?
 
finite-dimensional vector spaces are automatically complete
 
9:04 PM
there's no non-complete finite-dimensional vector space
 
@PM2Ring His point is that it doesn't have to
Although he's using "singularity" as in "curvature singularity", here
which whatever I guess
Although he hasn't proven [I think?] that the examples he gives can be done without singularities at all
He showed that you can have curve of finite affine parameters in Schwarzschild and Kerr that don't end on the central singularity
 
Yeah. It seems like he's saying that the light-like curves can have finite length by asymptotically approaching a curvature singularity without actually terminating at a singularity. Or something. ;)
 
I guess his idea is that since you can have curves like that in between the two horizons of the Kerr metric, maybe you can make a solution without singularity on the inner metric without violating Penrose's theorem?
But I don't think he gives an actual example of that
 
@ACuriousMind hm so is guaranteeing a basis the primary implication of (metric space) completeness?
 
just that it's not out of the picture
 
9:10 PM
@SillyGoose "Completeness" is the statement that all limits that should exist actually exist (inside the space)
what I mean is that it guarantees that for any basis $\psi_i$ and any square-summable sequence $c_i$, $\sum_i c_i \psi_i$ exists
 
hm and "linear combinations" of states with components corresponding to a square-summable sequence are precisely the square integrable functions?
 
it's the statement that the space isn't "missing points": If you have a Cauchy sequence of vectors $\psi_i$ (i.e. $\lvert \lvert \psi_i - \psi_j\rvert\vert \to 0$ as $i,j\to \infty$) that get ever closer together, they should be converging to something. Completeness says: Yes, they do!
@SillyGoose I don't understand the question
I didn't talk about functions at all
 
I am confused about the role of square-summable sequences because it seems to be a sort of constraint on what states we can have
 
it's because you need the norm of the linear combination to be finite
$\langle \psi,\psi\rangle = \sum_{i,j} \langle c_i\psi_i,c_j\psi_j\rangle = \sum_{i,j} c^\ast_i c_j \delta_{ij} = \sum_i \lvert c_i\rvert^2$, which is finite iff $c_i$ is a square-summable sequence
 
i see
@ACuriousMind this is a consequence of the fact that every finite sequence converges?
 
9:25 PM
@SillyGoose no, it's that any two norms on a finite-dimensional vector space are equivalent, and the standard Euclidean norm is complete
 
oh mayn
 
the proof of this is not "obvious", it usually involves Bolzano-Weierstraß
 
is this sort of content covered in a linear algebra course?
 
however, it's easy to understand why Bolzano-Weierstraß fails in the infinite-dimensional case: The sequence $\psi_i$ of infinitely many normalized basis vectors does not converge (then it wouldn't be a basis) but it is bounded (because it is normalized)
in some sense, the fact that the unit sphere of an infinite-dimensional vector space is not compact is the root of all the subtleties compared to the finite-dimensional case
 
I think I learned B-W in my real analysis course, but we didn't talk about Hilbert spaces (only metric spaces generally)
 
9:30 PM
@SillyGoose Given that I have heard exactly 1 linear algebra course in my life, I couldn't say :P
We did cover normed vector spaces and the equivalence of norms
your course might not
 
i think we covered the definition of isometric but went no further
but im not sure if that is the same concept as equivalence of norms
since it sounds more like equivalence of metric spsaces
 
isometry is a stronger condition than metric equivalence
it is true that all finite-dimensional normed vector spaces of the same dimension have equivalent norms (which in particular means that any sequence that converges in one of them converges in the other)
it is false that all finite-dimensional normed vector spaces of the same dimension are isometric
for instance $\mathbb{R}^n$ with the $p$-norm is an inner product (Hilbert) space only for $p=2$, so it cannot be isometric to the non-Hilbert spaces for $p\neq 2$
but if you're interested in these subtleties you should find a course or text on linear algebra that discusses them
 
i would like to understand the subtleties eventually...for if i ever get to teach QM :D since googling around for "why use a Hilbert space for quantum mechanics" does not yield very lucid answers
 
@SillyGoose I do not recommend reading them at your current level (because he assumes you already know how quantum theory works), but a significant fraction of Valter Moretti's work has been spent on this question, and his book Fundamental Structures of Quantum Mechanics is a good starting point
the basic idea is that, under reasonable assumptions, one can prove that the lattice of observables (as in this answer) is necessarily the lattice of projectors of an algebra of operators in a Hilbert spaces
@SillyGoose but at a much more basic level, the practical reason is just that we want to do linear algebra, and Hilbert spaces are the infinite-dimensional vector spaces which are the closest to finite-dimensional vector spaces and usual linear algebra
(precisely because they have an inner product and are complete)
 
9:48 PM
hm i see. what sorts of resources would you recommend to see how quantum theory works :0
 
uh, the usual textbooks?
It's not just that you should "know" how it works but be familiar with it
where I mean the kind of familiarity which comes from solving exercises or actually using QM to solve a problem
I quote from Moretti's introduction: "This book should appeal to a dual readership: On one hand mathematicians who wish to acquire the tools that unlock the physical aspects of quantum theories and on the other physicists eager to solidify their understanding of the mathematical scaffolding of quantum theories"
I guess what I really mean is that you should know how e.g. the hydrogen atom or the QHO works by heart before you start trying to engage with these abstract mathematical structures
not exactly because it's a prerequisite, but because it will make you go "oh, so that's why it works" instead of being puzzled what all this abstract structure has to do with physics :P
 
10:04 PM
i do have a problem with doing problems :P i should take a look at some...
 
 
1 hour later…
11:16 PM
He also wants to reform Physics education.
> What is the peer instruction method?
> In peer instruction, the instructor poses a question with discrete options and gives students the chance to consider and record their answers individually, often by voting using clickers. Students then discuss their answers with neighbors, explaining their reasoning, before being given a chance to vote again.
 
11:42 PM
good god i hate peer instruction
 
honk
 
honk
 
like any education method, you need to know how to use it
there's a scene from "Experimental Physics 2" burned into my brain where the lecturer strung up a giant tiger plushie on a wall, aimed a crossbow at it and told us everything was rigged so that the plushie would start falling down at the time he pulled the trigger. The question was: Where did he need to aim to hit the plushie with the bolt?
 
did you all vote with clickers
 

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