@ArtOfCode I concluded that hours ago

But maybe it's not English at all!

@0celo7 Hmmmm, how might someone named ArtOfCode go about figuring that out...

It is encrypted

Not a Caesar shift.

How do you know?

@ACuriousMind I DON'T KNOW

you're sassy today

@0celo7 Because I've just run all the possible permutations of the cipher, and none of them make sense.

@0celo7 You can't figure out that he/she's writing programs to brute force the solution?

@ACuriousMind oh

how would I know that!

Art. Of. Code.

I can do cute plots in MATLAB

I don't know what's possible with computers!

Stop sassing me!

regarding me being shit at math

Be careful what you wish for...

not this :(

Well how are we supposed to with @ArtOfCode around

Seriously, you're trying to be meaner now :(

I thought we had something special

The only thing that Google comes up with for any of it is knowledge.safe.com/articles/841/…

Y'know, I never said it wasn't random gibberish...

Not saying it is, though B)

can we just ban him

/s

No rules broken :)

what if it's some language he made up for his RPG games

what if he's a modern day Tolkien

Then we're unlikely to solve it without extensive samples of the language.

exactly, it could be hopeless

Quite possibly.

@0celo7 C'mon, at least it's not quadratic equation solving ;)

@ACuriousMind oh for the sake of crap

I can solve god damn quadratic equations

Screw you $\mathfrak{Astronomer}$!

The issue was factoring by inspection

Not solving

@0celo7 factors of $c$ that add to $b$.

Ew Fraktur

I hate it when old papers have Fraktur symbols

@ArtOfCode YES

that's what it was

Simple :)

now how does it work when you have $a\ne 1$

@0celo7 then it gets fun

is it factors of $ac$ that add to $b$?

no, it's factors of $c$ that add to $b$ when one of them is multiplied by $a$.

@0celo7 Now I know what you were talking about! Vieta's formulas!

@ChrisWhite no one gives a shit about the other one :P

@ArtOfCode hmm

for example, $2x^2+11x+12$

A moderator deleted an answer of mine. I updated the answer. The answer is clearly better than the other answers on the page. Can I get a moderator to help me improve the page?

@ACuriousMind Pretty much

So can we please stop this stupid meme that I can't solve quadratics

@user25246 Use the flag link to notify the moderator or drop a link here so some 10k+ users can look at it and vote to undelete if deserved.

factorises to $(2x+3)(x+4)$ - 3 and 4 factors of 12, and 3 and 4 add to 11 when the one opposite $a$, i.e. 4, is multiplied by $a$ - i.e. $8+3=11$ - @0celo7

physics.stackexchange.com/questions/197010/…

thanks

@ArtOfCode yes that's the process I was looking for

thank you

Hm

@ACuriousMind of course thought I couldn't plug into the quadratic equation

I should make that list of CTC spacetimes I was planning to do

takes me a minute to do, just make sure you have the right ones

Might be handy for things

To the Excel!

it's basically brute force in your head

Now analysis, on the other hand, is way too hard

@0celo7 what do you recognise as analysis?

@ArtOfCode epsilon delta proofs, point set topology, convergence

@0celo7 yeah, I have no experience with those words, or the topics they represent.

So I'm going to assume it's complicated :)

@user25246 Oh, sorry, since it was deleted by a mod and not by a review process, normal users can't vote to undelete. You'll have to use the flag link, flag for moderator attention, and just write you've expanded the answer and would like it to be undeleted.

@ArtOfCode Apparently it's super easy and 10 year olds do it in Germany

@ArtOfCode Wait...are you a computer hardware guy?

@0celo7 Yeah, but German 10 year olds don't count.

@0celo7 Not massively.

Damn

I need an expert

More software. I know some hardware, though. Look at what sites I mod... :)

Do you know what NB RAM frequency is

@ACuriousMind Do you mind giving a hint for the cypher?

@ChrisWhite I think so.

@0celo7 IIRC it's basically an overclock frequency. Comes from the northbridge being the performance part of the CPU.

Question: What determines NB freq? Is it a constant?

Constant for constant hardware specs.

You get base frequency, then you get Northbridge frequency which is higher, which is when you're using the northbridge for some task.

@0celo7 For constant specs, yes.

@ArtOfCode Mine varies from 800 to 3300 depending on which way the wind blows. (MHz)

Guys

Is Schwarzschild's topology $S^2 \times S \times R$

@Slereah yes

I forget

@Slereah no

Is it not

uh

wait

hmm, what the heck is it

@0celo7 I've seen that problem before. It's not a problem, that's just what's getting reported. The actual max is constant, the 800-3300 is the reported current-running-at frequency. I guess 3300 is your max.

Oh right

Hm

So $S^2\times\mathbb{R}^2$?

Does anyone know about dynamical systems?

Is it $S^2 \times R^2$ or $S \times S \times R$

@ChrisWhite that works

or what

why $S\times S$

that's a torus

I think Chris is correct.

he is, the origin isn't in the manifold

I mean the maximally extended Schwarzchild

@obe A tiny bit. What's the question?

A really tiny bit, though.

I want to learn about dynamical systems.

and what research is done in that area.

@HDE226868 Why don't extremals cross for small times

books to read in dynamical systems.

@obe Arnold

Wait a moment extremals are geodesics and we can find a convex normal nbd

Maybe

@HDE226868 thanks for the hint

@Slereah books.google.de/…

ok then.

:D

brb

@0celo7 No idea.

Ah yes, it is $S^2 \times R^2$ apparently

@0celo7 Hey, I'm trying to learn set-builder notation.

@HDE226868 I don't even know what that is

so good on you!

@0celo7 Literally just the notation used in set theory.

AHA

Yes!

I got it!

I think I can apply the theorems of Hawking and Ellis to a problem in Arnold!

Maybe!

Question time! Is there anything that determines the global structure of spacetime in usual GR? I.e. that governs whether usual Schwarzschild or the maximally extended version is the actual spacetime manifold?

Do you mean experimentally

No, mathematically

Well you can see this amazing post

Hm

What's the topology of an actual wormhole

that's not quite what I meant

It's $(R^3 + T^3) \times R$ I think

I don't want some generic statements about what kind of manifold spacetime is

I want a statement in the dynamical theory that tells me what specific manifold it is

Well there's nothing preventing spacetime from being such and such manifold

(except the reasons I listed)

Although there is a theorem saying that the spatial hyperslices of the spacetime can't change topology

Because, you see, this is a bit silly, but what the heck does the Lagrangian live on when we start not knowing the manifold before solving the EFEs?

(at least not without troubles)

You can't have an action be a functional of a metric on an object that's not yet determined, can you?

If at a time $t$ the manifold is of topology $\Sigma$ changes are good that it is $\Sigma \times R$

So who tells the Lagrangian what it's being integrated over?

There are a few cases where the topology can change, but that brings a whole lot of troubles

Also in quantum gravity you can try to do a sum over manifolds but that is an extremely bad idea

@ACuriousMind It lives on the initial data.

@ChrisWhite Yes. I have no trouble with that, it's clear what the equation is, it just fails to have a good solution. My problem right now is that there is nothing the action lives on, and hence nothing to obtain the EFE from prior to determining the manifold.

Question time! Arnold states, without proof, that the extremals (of $\int L$) emanating from a point don't cross for $t< t_0$. How does one show this? It just occurred to me that one can use the Maupertuis principle to transform the problem into a geometric one and use the theorem that "geodesics can't cross in a convex normal neighborhood". But I suspect that's overkill...

@ACuriousMind Everything lives on the initial data! I don't see what the issue is.

@0celo7 No, it does not. It is integrated over the entire spacetime, and is formally fed data from certain jet bundles of the spacetime. How can it be fed data from an object we don't yet no

And then you solve the equations to evolve the system in time.

@ACuriousMind Ah, one doesn't solve the EFE using the action.

I am not talking about solving the EFE

Then what are you talking about!

I am talking about obtaining it as a well-defined equation on well-defined objects in the first place!

See, what are the objects in there, formally?

No Godel CTCs are geodesics, right?

The Christoffels are some weird connection on the frame bundle, for instance. But if we don't know yet the manifold we are working with, there is no frame bundle

The manifold is usually assumed to be known

@Slereah But both Minkowski space and the (extended or not, I don't care) Schwarzschild spacetime are "solutions" to it, and they are not topologically equivalent

Well yes

It's not that we obtained different metrics on the same manifold, they are different manifolds

The manifold is kind of implied by the coordinates, usually?

So what did the equation live on prior to us solving it, exactly?

@Slereah That is exactly what a manifold is not.

I suppose, but there it is

The fact that some coordinates are cyclical usually implies what the manifold looks like

the coordinates live on $\mathbb{R}^n$, and you're solving the equation on that

I mean technically it should not

What is the formal procedure to obtain the correct manifold from the local solutions to the equation? Why don't we ever get manifolds that aren#t covered at all by a single coordinate patch instead of the usual cases where we have only single points which aren't covered

But in practice, you never use $R^n$ coordinates

You can't

Local solutions don't depend on the manifold

Exactly, so how is the manifold determined by the theory?

Well as said

It is usually assumed at the beginning