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1:09 PM
@0celo7 Do you have a link to it so I can learn more about the details?
 
@Secret No, ask @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
 
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
 
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
 
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?
 
1:18 PM
@Slereah Locally defined gauge fields can be "multivalued" all right. ;)
 
Proper time is just a specific coordinate system
Try to make your statements coordinate independant
 
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)?
 
@Secret what is a "multivalued spacetime"? You keep stringing together words in ways that don't really make sense.
 
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?
 
Sure
 
1:33 PM
ok
 
@ACuriousMind How good are your psychoanalysis skills
 
Uh, rather non-existent, why?
 
@ACuriousMind because I obviously need some psychoanalysis for something?
 
@0celo7 ...I guess I should've expected that answer :P
 
My diagnosis
You're crazy dude
 
1:44 PM
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
 
...you do realize that "easy to solve" is not exactly an objective condition?
 
(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"
 
@ACuriousMind Yes it is.
What is the average time 1,000 monkeys will take to solve it.
Amazon emails are so helpful
 
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.
 
1:59 PM
>You dont even know what a number is mathematically.
Holy crap
 
poorn jinawee, didn't deserve that savagery
 
Have you even seen a number in the sky
 
I think I have $25 in change at my house
I could buy a Springer book with that...
 
is that a film @slereah
 
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?
 
2:03 PM
@0celo7 Uh, that's a pretty standard comment, I've had versions of that adressed to myself a couple of times :P
 
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,
 
1
A: Elemetary Rotations -imagining differential rotations - intitutive proof of such rotations being vectors

Frobenius 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
 
sounds like a job for... @Secret !
 
-1
Q: causal structure

AaradhyaDenition 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 Mô€€€ (y) is compact. Can anyone tell me how does second condition implies, that no gravitational singularities without an event horizon exist (als...

Why can't people TeX?
 
Why can't people $\LaTeX$
 
user116211
2:08 PM
@YashasSamaga Have you asked him?
 
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.
 
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.
 
I know the definition of globally hyperbolic off the top of my head
Kill me now
 
2:10 PM
He says all vectors except r are infinitesimals.
 
@Slereah oh, I could get Hirsch
Now how to pay Springer with quarters
 
insert them in your floppy reader
 
@ACuriousMind I actually have several issues I want psychoanalyzed
 
@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.
 
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
2:13 PM
Have you checked the wikipedia article?
 
user116211
@Secret And that's why they are infinitesimals ;P
 
@Slereah Hey did you ever write the full proof that compact implies acausal
 
damn @0celo7 going in on the mods AND the OP
 
user116211
@Obliv what happened?
 
 
2:14 PM
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.
 
::sigh:
 
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!
 
What the hell is $I^+(0)$ anyway
 
The future light cone in the tangent plane
 
2:16 PM
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 $\Bbb R^4$
 
Well it's a spacetime
So obviously no boundary, yes
 
@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
 
@ACuriousMind what's that for
 
And obviously a spacetime with boundaries can be compact and causal
Like a spacetime that is a square
 
correct
 
2:17 PM
@Secret That diagram shows the solid rotating by nearly 30 degrees. How is that finite? R.I.P my fundamentals are so bad lol
 
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
 
Is that what the trouble is?
 
:'D what kind of response is this kind of question even looking for physics.stackexchange.com/questions/262349/…
 
@YashasSamaga (see above) the 3 disks are actually each rotating like this a little bit
 
2:19 PM
please stop being rude to me@ACuriousMind — Aaradhya 2 hours ago
:D @ACuriousMind
You're a freaking bully
 
@0celo7 yeah
If what I said is true (it probably is), then it's the image of an open set
 
@Slereah the existence of normal neighborhoods is not a problem btw
 
So it's all good
 
it's the inverse function theorem
have you proved that $\exp_p$ is open?
 
wot
 
2:22 PM
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
 
@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
 
Otherwise its image need not be open
 
Isn't continuity enough
 
@Secret I understood now but how are those angular displacements infinitesimally small?
 
No, continuity means that the preimage of an open set is open
Not that the image of an open set is open
 
2:25 PM
@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
 
Hm
 
And you'd have to prove continuity too
Although that might follow from Picardy
 
Well I think if the connection is continuous, the exp map is too
 
> 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 >:(
 
@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
 
2:26 PM
What do the other texts have to say about it
 
I've got a good feeling about it
 
The Standard References
 
That's all the proof I need baby B)
 
physicists are disgusting
I swore I would never do GR again
Now look at me
 
Maybe you should go back to engineering
Make a really good table
 
2:28 PM
@ACuriousMind Is it curious that I despise point-set topology but seem to have fun with it when it's in GR
 
@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
 
@0celo7 That is strange, yes
 
GR topology is all about drawing cones on potatoes
 
sexy potatoes
 
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
 
2:32 PM
@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
 
What's a ess omega
 
S is the scalar curvature, $\Omega$ the metric volume element
 
Is S the Sssscalar curvature
I have to say
 
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 $a \in A$?
 
Pretty weak that he didn't even include the boundary term
Straight up amateur hour
 
2:33 PM
Still not sure why one needs that thing
 
Although I'm guessing that the vacuum EFE is always integrable for the action
 
faints
 
Since $S = 0$
 
There's an exercise
To show that the Cauchy problem is well-defined
What is up with these exercises o.O
 
they're not fucking around
No hand holding
You know what you could check for that?
HE.
 
2:36 PM
Ah, it's only for Ricci flat spacetimes
But exercise 8.5.6 asks you to generalize it, lol
 
@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
 
What is the Ambrose-Hicks theorem
 
Well
Hm
 
4 mins ago, by Slereah
Is S the Sssscalar curvature
Joke: Creeper alert
 
I don't get it
That theorem might be in Kobayashi-Nomizu
 
2:38 PM
Thinking about it
 
Hint: Prefacing things with "Joke" doesn't make them funny.
 
The Ricci scalar isn't a linear PDE of the metric
So it's gonna be shit to prove
 
@Slereah nope
@Slereah Yes it's pretty terrible
@ACuriousMind What's the hardest exercise you've seen in a book/whatever you read
 
I have it somewhere
 
@Obliv What, exactly, is your question?
 
2:39 PM
the group action is defined strangely in this text.
 
@ 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
 
I think proving the well-posedness of the GR Cauchy problem is the worst I've seen
 
@0celo7 ...have we forgotten again that I haven't actually read many books? :P
 
@ACuriousMind I EDITED IT
@Secret That's the saddest thing I've read in a while.
 
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
 
2:40 PM
Uhhh
 
@ 0celo7 Well, I am just saying the truth of myself
 
@Obliv it just means you associate $b\in A$ to $g\in G,a\in A$
Which is just $(g,a)\mapsto b$
 
That's the one
Welp
Nothing comes to mind
 
@0celo7 I don't think I've ever seen exercises except for the ones from my courses
 
@ACuriousMind Do you agree that proving the GR Cauchy problem is a ridiculous exercise
 
2:41 PM
Which are usually rather tame
 
does it associate it to the group operation between $g,a$ ? That's what it looks like in this text
 
I'd say proving the GR cauchy problem is a ridiculous thesis assignment
 
@0celo7 Yes, I agree that GR is ridiculous :)
 
@ACuriousMind what does that mean
What's wrong with it
hunts theorem in book
 
You can probably prove it with bits from HE and Wald
He discusses the topic
 
2:43 PM
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
 
The first one I think is in the causal hierarchy of spacetime
 
WHAAAAAAT
These are just stupid
 
Ahahah
 
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: $g_1 (g_2 a) = (g_1 g_2 )a$ ...
 
He asks you to prove the fucking Geroch theorem
 
2:45 PM
Forget Jackson
These are the hardest physics exercises ever
@dmckee Jackson has been dethroned
 
@0celo7 What about the last exercize of Visser
 
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 $G \times A$)
 
@Slereah that's a joke exercise
These are very real
 
Are we sure
 
@Obliv $ga$ is what one writes for the image of $(g,a)$ under the action $G\times A\to A$.
 
2:47 PM
Did someone actually do those exercize
I'm not sure Ellis could do them
 
That is, $ga$ is an element in $A$ - the one you get after acting with $g$ on $a$.
 
Ellis can't even understand shit in his own book
 
thanks @acuriousmind
 
@ACuriousMind look at these exercises
 
@0celo7 Why?
 
2:48 PM
 
That's just the raychaudhuri equation?
 
@Slereah :(
He wants us to prove the singularity theorems
 
The question is, does he give you the elements to solve it?
Or are you supposed to magic them up
 
He gives a hint
 
HINT : Use math
 
2:51 PM
"See Hawking-Ellis"
 
Well why not read HE directly then
 
@ACuriousMind so you can tell me they're hard and I stop feeling small and inadequate
:(
 
But I can't tell how hard they are because I don't actually know any GR :P
 
@ACuriousMind My prof who does Riemannian geometry/geometric analysis would probably just laugh
I mean why even bother making these exercises
 
Well the first one is proving that a nonlinear PDE has a well defined initial value problem
 
2:53 PM
They should have just said CHAPTER 8: FACTS
and referenced each one
@Slereah Does O'Neil prove that $I^+(p)$ is open
 
Lemme check
 
did you buy it?
 
Legally, yes
 
pic?
 
He shows it's open in an open convex set
hm
 
2:58 PM
@Slereah I just posted a pic of that
what is a convex open set
a convex normal neighborhood?
 

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