The h Bar

Secret

13:09

@0celo7 Do you have a link to it so I can learn more about the details?

0celo7

@Secret No, ask @Slereah

Slereah

Why me :(

When people deal with CTCs they try too hard to apply science fiction tropes to it

CTCs aren't that hard to deal with

You just treat them using normal mathematics

Don't try to think too hard on the meaning of it

Solve the equation, see what happens

For Minkowski surgery it's very easy if you do it with geometric optics

Secret

Well here we are after all these years, finally isolate concretely why science forbid back to the future style time travel to be a possible reality of time travel (This will surely change my view of time travel stories in general)

->Violation of Maxwell equations

Slereah

It's not because of this.

A field is a function

For every spacetime point, it can only have one value

Even if it has discontinuities, you can't change the past because of this

If you want to allow changing the past you have to change the structure of spacetime quite radically

Secret

What prevents the field to be a function of both spacetime and proper time, is it because it is lorentz invarient thus it must have a unique value in spacetime that agrees to all observers?

ACuriousMind

13:18

@Slereah Locally defined gauge fields can be "multivalued" all right. ;)

Slereah

Proper time is just a specific coordinate system

Try to make your statements coordinate independant

Secret

right...

> If you want to allow changing the past you have to change the structure of spacetime quite radically

such as multivalued spacetime (not a very welcomed concept in spacetime as single valued quantities are easy to handle)?

ACuriousMind

@Secret what is a "multivalued spacetime"? You keep stringing together words in ways that don't really make sense.

Secret

I am not sure, probably a half baked understanding of what Slereah previously shared to me about something branched something spacetimes or something, I don't remember

@Slereah (back to ordinary questions). The general approach in solving them is to compute the curvature tensor and then work out the geodesic equation subjected to the boundary conditions of CTCs?

Slereah

Sure

Secret

13:33

ok

0celo7

@ACuriousMind How good are your psychoanalysis skills

ACuriousMind

Uh, rather non-existent, why?

0celo7

@ACuriousMind because I obviously need some psychoanalysis for something?

ACuriousMind

@0celo7 ...I guess I should've expected that answer :P

Slereah

My diagnosis

You're crazy dude

Secret

13:44

in Mathematics, 2 mins ago, by Secret

Hey guys, I have a question concerning limits that I am not sure how to make it narrow enough so it can fit MSE question guidelines:

I am interested in finding a function $f(x)$ such that its limit

$$\lim_{x\rightarrow 0} f(x)$$

Is hard to compute using known rules of limits, but

$$\left(\lim_{x\rightarrow 0} f(x)\right)^2$$

will allow the limit to be simplified into two bits

$$\lim_{x\rightarrow 0} g(x)+\lim_{x\rightarrow 0} h(x)$$

where the limits of $g(x)$ and $h(x)$ are easy to compute, thus allowing the original limit to be determined by taking the square root of this limit?

It is annoying when I have questions that is literally asking

"What is the question that will satisfy the given conditions"

because I have no idea how to quantify the plausibility of solving a question given a specific method

For example, if I have a question that asks "what is the function $f(x)$ such that its integral is most easily solved by squaring the integral?", we all knew $f(x)=e^{-x^2}$ (for starters)

but limits are not like integrals. Even the rules on solving and approaching limits are more art than solving integrals

ACuriousMind

...you do realize that "easy to solve" is not exactly an objective condition?

Secret

(arghhh, you fill in the blanks for me for what my incoherent mind is trying to say, since you guys are somehow better and grasping what my mind is thinking)

...for the integral example, more like, we most commonly solve it by squaring it

Joke: Basically, I am tempted to prove that the solution to "what is the question that has the working solution as follows <workings>" is the same as "How to reach the tree of knowledge"

I have no idea why most of my questions are equivalent to this question

"How to reach the Tree of Knowledge"

0celo7

@ACuriousMind Yes it is.

What is the average time 1,000 monkeys will take to solve it.

Amazon emails are so helpful

Obliv

lol physics.stackexchange.com/questions/262321/… @johnR getting flagged for spam

rekt

people like this guy should get a temp ban honestly. Just aggressively arguing with people giving constructive feedback. Unscientific and poor humility when asking physicists questions.

0celo7

13:59

>You dont even know what a number is mathematically.

Holy crap

Obliv

poorn jinawee, didn't deserve that savagery

Slereah

Have you even seen a number in the sky

0celo7

I think I have $25 in change at my house

I could buy a Springer book with that...

Obliv

is that a film @slereah

0celo7

let's see what I need...

@Obliv no but it's pretty dramatic

> Why dont you go and write "The right answer" to every question on this site, let us see how clever you are.

Savage!

> Capische?

ACuriousMind

14:03

@0celo7 Uh, that's a pretty standard comment, I've had versions of that adressed to myself a couple of times :P

0celo7

HE WENT ITALIAN ON HIM

> Before closing someone should give a reason, I will take this up with moderators and I hope you will be banned,

Yashas Samaga

1

A: Elemetary Rotations -imagining differential rotations - intitutive proof of such rotations being vectors

All vectors, except $\:\mathbf{r}\:$, are infinitesimals. I wonder if the author (Irodov) makes use of this result anywhere in his textbook.

Can somebody help me with that answer?

"All vectors, except r, are infinitesimals." - this thing is screwing my brain

Obliv

sounds like a job for... @Secret !

0celo7

-1

Q: causal structure

Denition 2.10. A spacetime (M; g) is called globally hyperbolic, if it satises the strong causality condition and for all x; y 2 M the set J M + (x)\J M􀀀 (y) is compact. Can anyone tell me how does second condition implies, that no gravitational singularities without an event horizon exist (als...

general-relativity

Why can't people TeX?

Slereah

Why can't people $\LaTeX$

user116211

14:08

@YashasSamaga Have you asked him?

Yashas Samaga

Yes in the comments but I would be leaving today and wont be there for 2 days. So I thought someone could clear it up for me before I leave. :/

user116211

It is quite clear although.

Yashas Samaga

My understanding of some physical quantity being infinitesimal is some change that is so small that it causes no real change. It only gives a finite value on integration.

user116211

$\bf r$ is the position coordinate and about that the infinitesimal rotations took place; it's explicitly clear from the pic.

0celo7

I know the definition of globally hyperbolic off the top of my head

Kill me now

Yashas Samaga

14:10

He says all vectors except r are infinitesimals.

0celo7

@Slereah oh, I could get Hirsch

Now how to pay Springer with quarters

Slereah

insert them in your floppy reader

0celo7

@ACuriousMind I actually have several issues I want psychoanalyzed

Yashas Samaga

@MAFIA36790 Can you cite some reference on axial vectors? I don't have any intuition yet. I am still staring at that picture since 10 minutes.

Secret

if a vector is infinitesimal, it means these vectors are small enough that we can regard them as basically lying on a plane at that point. Therefore the usual parallelgram rule applies and you can add them as if they live in flat space

user116211

14:13

Have you checked the wikipedia article?

user116211

@Secret And that's why they are infinitesimals ;P

0celo7

@Slereah Hey did you ever write the full proof that compact implies acausal

Obliv

damn @0celo7 going in on the mods AND the OP

user116211

@Obliv what happened?

Obliv

@mafia just some shennanigans physics.stackexchange.com/questions/262321/…

Yashas Samaga

14:14

I have studied about tensors up to order 3 but I have no intuition for axial vectors and I ignored these minuscule points in rotational kinematics in my class.

ACuriousMind

::sigh:

Slereah

Well, I basically have everything except the part where $\exp_p(I^+(0)) = I^+(p)$

I think that's the gauss lemma or close to it

user116211

@Obliv ah!

0celo7

What the hell is $I^+(0)$ anyway

Slereah

The future light cone in the tangent plane

0celo7

14:16

You're gonna need the fact that spacetime has no boundary somewhere

Which means the argument is going to be ugly

You might need a homeo to an open set of $\Bbb R^4$

Slereah

Well it's a spacetime

So obviously no boundary, yes

Secret

@YashasSamaga So that diagram of Frobenius is basically saying at every point along those circles, parallelgram rule is obeyed thus the tangential vectors dr1 and dr2 are indeed vectors

0celo7

@ACuriousMind what's that for

Slereah

And obviously a spacetime with boundaries can be compact and causal

Like a spacetime that is a square

0celo7

correct

Yashas Samaga

14:17

@Secret That diagram shows the solid rotating by nearly 30 degrees. How is that finite? R.I.P my fundamentals are so bad lol

0celo7

What you have to show is this

you need to show that the $I^+(p)$ cover $M$

So how does one actually do this

Well, you have to show that a given point $q$ lies in $I^+(p)$ for some $p$

But how to actually do this is...

I guess set up a normal neighborhood

Oh shite you have to prove that $I^+(p)$ is open

Secret

0celo7

Is that what the trouble is?

Obliv

:'D what kind of response is this kind of question even looking for physics.stackexchange.com/questions/262349/…

Secret

@YashasSamaga (see above) the 3 disks are actually each rotating like this a little bit

0celo7

14:19

please stop being rude to me@ACuriousMind — Aaradhya 2 hours ago

:D @ACuriousMind

You're a freaking bully

Slereah

@0celo7 yeah

If what I said is true (it probably is), then it's the image of an open set

0celo7

@Slereah the existence of normal neighborhoods is not a problem btw

Slereah

So it's all good

0celo7

it's the inverse function theorem

have you proved that $\exp_p$ is open?

Slereah

wot

0celo7

14:22

You said $I^+(p)$ is the image of an open set

I agree with this

But you need to show that $\exp_p$ is an open map

Secret

@Obliv I am guessing the OP is asking whether the big bang is caused by basically a universe filled with dark matter collided with a universe filled with dark energy

0celo7

Otherwise its image need not be open

Slereah

Isn't continuity enough

Yashas Samaga

@Secret I understood now but how are those angular displacements infinitesimally small?

0celo7

No, continuity means that the preimage of an open set is open

Not that the image of an open set is open

Obliv

14:25

@secret that's like asking if we can create new universes by shooting black holes with dark energy bombs. It's so out there lol

Slereah

Hm

0celo7

And you'd have to prove continuity too

Although that might follow from Picardy

Slereah

Well I think if the connection is continuous, the exp map is too

0celo7

> 1+(f/,..II) will be denoted by 1+(f/), and is an open set, since ifp E..II can be reached by a future-directed timelike curve from f/ then there is a small neigh- bourhood ofp which can be so reached.

Proof >:(

Secret

@YashasSamaga The disks are rotated by only a bit, e.g. maybe like 0.001 degrees each all at the same time, thus the sector traced out by that rotation on each disks is roughly a triangle. The diagram exaggerate this in order to make them visible

0celo7

14:26

What do the other texts have to say about it

Slereah

I've got a good feeling about it

0celo7

The Standard References

Slereah

That's all the proof I need baby B)

0celo7

physicists are disgusting

I swore I would never do GR again

Now look at me

Slereah

Maybe you should go back to engineering

Make a really good table

0celo7

14:28

@ACuriousMind Is it curious that I despise point-set topology but seem to have fun with it when it's in GR

Secret

@Obliv Well, it is not entirely outlandish if you knew a bit about brane cosmology (which john or slereah can probably tell you more about it). In brane cosmology, one possible origin of the big bang is when two branes collided

ACuriousMind

@0celo7 That is strange, yes

Slereah

GR topology is all about drawing cones on potatoes

0celo7

sexy potatoes

Secret

I personally have drew many of those back in those time when I work with that model. Drawing cones help visualise the causal structure of the spacetime since you can easily see where they twist and do interesting things

0celo7

14:32

@Slereah Oh lord Sachs-Wu has "show that the vacuum Einstein equations arise as the Euler-Lagrange equations of the functional $\int S\Omega$" as an exercise

Slereah

What's a ess omega

0celo7

S is the scalar curvature, $\Omega$ the metric volume element

Slereah

Is S the Sssscalar curvature

I have to say

Obliv

how is a cartesian product of a group and a set written? $G \times A$ where $G$ is a group and $A$ is a set. Are elements $(g,a)$ with $g \in G$ and $a \in A$?

Slereah

Pretty weak that he didn't even include the boundary term

Straight up amateur hour

0celo7

14:33

Still not sure why one needs that thing

Slereah

Although I'm guessing that the vacuum EFE is always integrable for the action

0celo7

faints

Slereah

Since $S = 0$

0celo7

There's an exercise

To show that the Cauchy problem is well-defined

What is up with these exercises o.O

Slereah

they're not fucking around

No hand holding

You know what you could check for that?

HE.

0celo7

14:36

Ah, it's only for Ricci flat spacetimes

But exercise 8.5.6 asks you to generalize it, lol

Secret

@Obliv A group is a special case of a set, I would expect $G \times A$ be a set in general as closure may be violated in general for the elements in the cartesian product

0celo7

What is the Ambrose-Hicks theorem

Slereah

Well

Hm

Secret

4 mins ago, by Slereah

Is S the Sssscalar curvature

Joke: Creeper alert

0celo7

I don't get it

That theorem might be in Kobayashi-Nomizu

Slereah

14:38

Thinking about it

ACuriousMind

Hint: Prefacing things with "Joke" doesn't make them funny.

Slereah

The Ricci scalar isn't a linear PDE of the metric

So it's gonna be shit to prove

0celo7

@Slereah nope

@Slereah Yes it's pretty terrible

@ACuriousMind What's the hardest exercise you've seen in a book/whatever you read

Slereah

I have it somewhere

ACuriousMind

@Obliv What, exactly, is your question?

Obliv

14:39

the group action is defined strangely in this text.

Secret

@ Acuriousmind, I never said I intend them to be funny. I am like a computer thus I tend to index things given I rarely make jokes anyway

0celo7

I think proving the well-posedness of the GR Cauchy problem is the worst I've seen

ACuriousMind

@0celo7 ...have we forgotten again that I haven't actually read many books? :P

0celo7

@ACuriousMind I EDITED IT

@Secret That's the saddest thing I've read in a while.

Obliv

It's a map $G \times A \to A$ that is associative and has an identity? But it's not defined in terms of ordered pairs

0celo7

14:40

Uhhh

Secret

@ 0celo7 Well, I am just saying the truth of myself

0celo7

@Obliv it just means you associate $b\in A$ to $g\in G,a\in A$

Which is just $(g,a)\mapsto b$

Slereah

That's the one

Welp

Nothing comes to mind

ACuriousMind

@0celo7 I don't think I've ever seen exercises except for the ones from my courses

0celo7

@ACuriousMind Do you agree that proving the GR Cauchy problem is a ridiculous exercise

ACuriousMind

14:41

Which are usually rather tame

Obliv

does it associate it to the group operation between $g,a$ ? That's what it looks like in this text

Slereah

I'd say proving the GR cauchy problem is a ridiculous thesis assignment

ACuriousMind

@0celo7 Yes, I agree that GR is ridiculous :)

0celo7

@ACuriousMind what does that mean

What's wrong with it

hunts theorem in book

Slereah

You can probably prove it with bits from HE and Wald

He discusses the topic

0celo7

14:43

HE contains a full proof

But it requires a lot of functional analysis

Holy shit o.o

That's like a chapter in BEE

How to write a GR book: solve all exercises in Sachs-Wu

Slereah

The first one I think is in the causal hierarchy of spacetime

0celo7

WHAAAAAAT

These are just stupid

Slereah

Ahahah

Obliv

A group action of a group $G$ acting on a set $A$ is a map from $G \times A \to A$ (written as $ga$, for all $g \in G$ and $a \in A$) satisfying: $g_1 (g_2 a) = (g_1 g_2 )a$ ...

Slereah

He asks you to prove the fucking Geroch theorem

0celo7

14:45

Forget Jackson

These are the hardest physics exercises ever

@dmckee Jackson has been dethroned

Slereah

@0celo7 What about the last exercize of Visser

Obliv

can someone explain what associativity is being satisfied here? Is this associativity between $G \times A$ and $G$? (I say that because $ga$ looks to be an element of $G \times A$)

0celo7

@Slereah that's a joke exercise

These are very real

Slereah

Are we sure

ACuriousMind

@Obliv $ga$ is what one writes for the image of $(g,a)$ under the action $G\times A\to A$.

Slereah

14:47

Did someone actually do those exercize

I'm not sure Ellis could do them

ACuriousMind

That is, $ga$ is an element in $A$ - the one you get after acting with $g$ on $a$.

0celo7

Ellis can't even understand shit in his own book

Obliv

thanks @acuriousmind

0celo7

@ACuriousMind look at these exercises

ACuriousMind

@0celo7 Why?

0celo7

14:48

Slereah

That's just the raychaudhuri equation?

0celo7

@Slereah :(

He wants us to prove the singularity theorems

Slereah

The question is, does he give you the elements to solve it?

Or are you supposed to magic them up

0celo7

He gives a hint

Slereah

HINT : Use math

0celo7

14:51

"See Hawking-Ellis"

Slereah

Well why not read HE directly then

0celo7

@ACuriousMind so you can tell me they're hard and I stop feeling small and inadequate

:(

ACuriousMind

But I can't tell how hard they are because I don't actually know any GR :P

0celo7

@ACuriousMind My prof who does Riemannian geometry/geometric analysis would probably just laugh

I mean why even bother making these exercises

Slereah

Well the first one is proving that a nonlinear PDE has a well defined initial value problem

0celo7

14:53

They should have just said CHAPTER 8: FACTS

and referenced each one

@Slereah Does O'Neil prove that $I^+(p)$ is open

Slereah

Lemme check

0celo7

did you buy it?

Slereah

Legally, yes

0celo7

pic?

Slereah

He shows it's open in an open convex set

hm

0celo7

14:58

@Slereah I just posted a pic of that

what is a convex open set

a convex normal neighborhood?

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