The h Bar

Slereah

7:00 PM

If your equations and initial conditions have a bad regularity, you lose determinism

Obliv

I end up with $\dfrac{2m+1}{2m - m}$

in the limit

ACuriousMind

@Slereah What's imo troubling about Norton's dome is that "bad regularity" here is still a perfectly reasonable situation - it goes against all intuition that we lose/gain determinism when just varying exactly how a dome is curved.

Slereah

good thing we got rid of classical mechanics then!

Except now our theories are even worse with those types of scenarios!

Obliv

@acuriousmind I don't think I can solve the limit as it is actually. The original ratio of the expressions in the limit depend on the value of $i$. I have to take an infinite sum of each of the limits. Do you know how I could do that?

bolbteppa

Mach's Principle - Sixty Symbols

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Maybe this has the final answer please please

Anonymous

7:05 PM

@Obliv Your question looks very vague tbh. Maybe state the whole problem you're trying to solve

Slereah

The GR people just try to pretend really hard that we have proven that naked singularities can't form but it's pretty vague :p

and pretty bad if they can

ACuriousMind

@Obliv I'm not sure I understand your problem. Your limit is now $\lim_{m\to\infty} \frac{2m + 1}{2^i m - m} = \lim_{m\to\infty} \frac{2 + 1/m}{2^i - 1} = \frac{2}{2^i - 1}$. And the infinite series of summing these up seems to converge nicely.

Obliv

@acuriousmind Somehow when I do it, the $i$ term is gone and I have $\frac{2m+1}{2m - m}$

maybe i'll write the steps so I can see where I go wrong

$\dfrac{2m+1}{\dfrac{(2^i-1)2^i m}{2^i}}$ becomes $\dfrac{2m + 1}{\dfrac{2^{i+1}m - 2^i m}{2^i}}$

Anonymous

m missing in denominator's first term

Obliv

thanks

ACuriousMind

7:15 PM

...$2^i \cdot 2^i \neq 2^{i+1}$, and why don't you directly cancel the $2^i$ from the numerator against hte $2^i$ in the denominator? You don'T need to multiply the bracket out.

Obliv

wait really?

Anonymous

$(2^{i})^2=2^{2i}$

Obliv

oh man hold on

oooooooh

Anonymous

$2^2\times 2^2=2^{2+2}=2^4$

Obliv

it's been so long yeah you can directly cancel that out too

so I get $\dfrac{2m+1}{2^i m - m}$

and you multiplied all the terms by 1/m i'm guessing

thanks @acuriousmind @blue

bolbteppa

7:22 PM

"and it's no coincidence, in my opinion, that the cross section of a black hole coordinate singularity perimeter physical singularity origin is identical to the polar view of a pseudo-sphere, which is the inversion of a spherical body" ohhh

Emilio Pisanty

@JohnRennie This library github.com/valandil/wignerSymbols was throwing up gfortran errors

¬¬

John Duffield

Hey, who wants to talk about black holes?

Emilio Pisanty

@Blue I am working as a physicist, and I'm happy to discuss cutting-edge research. But I know nothing about entropic gravity.

John Duffield

And did I see Eddington-Finkelstein coordinates?

ACuriousMind

@Obliv np

Semiclassical

7:25 PM

The only bit of black hole Physics I know is that conversations about entropic gravity usually turn into black holes

Anonymous

@EmilioPisanty I meant what's your feeling on being called an engineer by @0celo7 ? :P

Anonymous

I was clearly kidding there

Semiclassical

@EmilioPisanty there’s a lot of edges to cut with :p

bolbteppa

you just need to use Eddington-Finkelstein coordinates to transform out of that black hole quick before the singularity arises from the pseudo-sphere

Obliv

@acuriousmind I'm curious though, how do you get the convergence of the infinite sum?

John Duffield

7:27 PM

Eddington–Finkelstein coordinates

In general relativity, Eddington–Finkelstein coordinates are a pair of coordinate systems for a Schwarzschild geometry (i.e. a spherically symmetric black hole) which are adapted to radial null geodesics. Null geodesics are the worldlines of photons; radial ones are those that are moving directly towards or away from the central mass. They are named for Arthur Stanley Eddington and David Finkelstein. Although they appear to have inspired the idea, neither ever wrote down these coordinates or the metric in these coordinates. Roger Penrose seems to have been the first to write down the null form...

ACuriousMind

@Obliv Well, it's pretty much of the form $\sum_i 1/2^i$, which has the well-known value of 2.

I didn't think very hard about what the actual value of your series would be, but that^ is why I'm certain it converges.

John Duffield

It would seem that Eddington-Finkelstein coordinates were invented by Roger Penrose, and popularised by Misner/Thorne/Wheeler. IMHO they contain a "schoolboy error".

Obliv

.

bolbteppa

This is becoming hilarious now

John Duffield

"When some probe (such as a light ray or an observer) approaches a black hole event horizon, its Schwarzschild time coordinate grows infinite. The outgoing null rays in this coordinate system have an infinite change in t on travelling out from the horizon. The tortoise coordinate is intended to grow infinite at the appropriate rate such as to cancel out this singular behaviour in coordinate systems constructed from it."

ACuriousMind

7:30 PM

@Obliv That doesn't make much sense - there no $m$ in the formula you're talking the limit of. :P

Slereah

@ACuriousMind well then it's trivial!

easy as pie!

ACuriousMind

But in any case, anything that qualitatively goes as $1/2^i$ will converge.

Obliv

er

bolbteppa

I bet if they weren't called Tortoise coordinates, nobody who doesn't know calculus would have no idea what they are

Anonymous

@Slereah Unless $i$ depends on $m$ :p

Anonymous

7:32 PM

@bolbteppa Triple negative. I like that ;)

Slereah

@ACuriousMind [1] Ζήνων ὁ Ἐλεάτης, Φυσικὴ ἀκρόασις , 400 BCE

John Duffield

It takes forever for the downward-moving light beam to cross the event horizon, so it never ever does. But Eddington-Finkelstein coordinates does away with that by saying in essence that we use bigger and bigger seconds.

ACuriousMind

@Slereah Heh

Slereah

Although I guess technically Zeno thought that was a problem

and that it should diverge

Obliv

okay written correctly I think,$$\sum_{i=1}^{\infty} \dfrac{1}{2}\dfrac{2}{2^i -1} - \dfrac{1}{2^i}$$

ACuriousMind

7:33 PM

@Slereah Yes, I'm under the impression that Zenon specifically thought that limit didn't exist :P

bolbteppa

@Obliv you should use brackets

John Duffield

It's similar for Kruskal-Szekeres coordinates‌.

ACuriousMind

Or rather, that he had no conception of limits at all.

Slereah

IIRC he thought that applying such notions to the real world was nonsense and that is why he came up with that example

Obliv

@bolbteppa how do you write those?

@acuriousmind do you know how to find the sum since it converges? Sorry for all the questions xD

bolbteppa

7:35 PM

If Alexander runs a half mile, and the tortoise runs half an inch, iterate, they never cross, therefore tortoise coordinates do not exist

Slereah

although looking into it, that might be an interpretation

Since apparently we don't know all that much about Zeno

Not a lot of writings about him

Anonymous

It was a long time back

Anonymous

Zeno's paradoxes

Zeno's paradoxes are a set of philosophical problems generally thought to have been devised by Greek philosopher Zeno of Elea (ca. 490–430 BC) to support Parmenides' doctrine that contrary to the evidence of one's senses, the belief in plurality and change is mistaken, and in particular that motion is nothing but an illusion. It is usually assumed, based on Plato's Parmenides (128a–d), that Zeno took on the project of creating these paradoxes because other philosophers had created paradoxes against Parmenides' view. Thus Plato has Zeno say the purpose of the paradoxes "is to show that their...

Slereah

yeah, a lot of philosophers we basically don't know anything about

John Duffield

The infalling elephant does a hop skip and a jump over the end of time and is in two places at once at the same time:

bolbteppa

7:36 PM

@OBliv $\sum_{i=1}^{\infty} (\frac{1}{2} \frac{2}{2^i - 1} - \frac{1}{2^i})$

Slereah

Except through mentions by other people

Because writing things down wasn't that common and it was hard for all of these to survive

Anonymous

Gamow's book had a good discussion about that paradox (iirc)

Anonymous

I didn't know limits back then

Slereah

Your Friendly Liberal Neighborhood Ku Klux Klan

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ACuriousMind

@Obliv Nope. (although $\sum_{i = 1}^\infty 1/2^i = 1$, so you only need to find the first term)

Obliv

7:41 PM

@acuriousmind Hmm okay I'll give it a shot later. do you have any resources on solving a sum of this form? I haven't studied infinite sums for years and I forgot all of the techniques..

anyway, thank you so much for your help you are truly a savior :)

Anonymous

See Paul's online notes

Sir Cumference

Oh hey, @Obliv, you're back!

ACuriousMind

@Obliv Have you forgotten that I'm really bad at recommending resources? ;) Anyway, the main techniques all boil down to educated guesswork - try to reduce it to series you already know, try to sandwich it between two series you already know, etc...

bolbteppa

There are basically no techniques to actually sum a series unless it's a geometric series in some form

Do you need to sum it or just show it converges

There is some crazy residue trick to sum some easy ones

John Duffield

7:56 PM

All you have to do is read what Einstein said to know that there's some significant issues with papers like Penrose's 1964 gravitational collapse and space-time singularities.

Semiclassical

@bolbteppa it’s sorta like counting problems: one doesn’t hope for one approach that does everything so much as a collection of techniques which cover a wide range

Sir Cumference

Right, series, so much fun and guesswork

Semiclassical

(And they’re sorta tied together insofar generating functions are used to solve combinatorial problems)

Anonymous

It's quite a broad topic

bolbteppa

The other technique is to just know it's the series expansion of some nice function :p

Semiclassical

8:02 PM

Integrals are like that as well

Anonymous

@Semiclassical Ah, generating functions are the interesting stuff

Semiclassical

@bolbteppa everyone loves perturbation theory :)

@Blue definitely

And insofar as coefficients of Laurent series link up with contour integrals via Caucht’s formula, it also links up with integral representations

Anonymous

I actually like the applications of generating functions in statistics. There was this book which I started reading (The Problem of Moments: bookstore.ams.org/surv-1), although could only get through a small part of it, due to lack of time. It's amazing how much you can deduce about distributions from the first few moments itself, and how they vary for different kinds of distributions

Semiclassical

Definitely.

What I dig is how the asymptotics of coefficients are related to contour integrals

That lets you do Laplace’s method in order to get estimates

Anonymous

What's Laplace's method?

Anonymous

8:08 PM

Haven't heard of it

Anonymous

Laplace's method

In mathematics, Laplace's method, named after Pierre-Simon Laplace, is a technique used to approximate integrals of the form ∫ a b e M f ( x ) d x {\displaystyle \int _{a}^{b}\!e^{Mf(x)}\,dx} where ƒ(x) is some twice-differentiable function, M is a large number, and the integral endpoints a and b could possibly be infinite....

Anonymous

Ah

Semiclassical

Method of steepest descent , saddle point method, stationary phase approximation

They’re all pretty closely related

So I’m not that careful about distinguishing them

Anonymous

I see