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12:00 AM
@yuggib Well, actually, this highlights a subtle feature - the category of topological spaces is wired "the other way around" compared to the usual algebraic example. In the algebraic case, preserving the structure is always taking something in the source and preserving it under the image - the topological case takes something in the target and preserves it under the preimage
 
seriously, I can't come up with an invertible continuous map that isn't bi-continuous
 
one may ask why continuous maps and not open maps are chosen to define the $\mathrm{Top}$ category
 
@0celo7 Seriously, that's one of the absolute standard examples or exercises in any introduction to topology
 
what are the open maps? homeomorphisms?
 
@0celo7 An open map is just a map under which the image of an open set is open.
 
12:01 AM
@ACuriousMind of course
:P
@ACuriousMind seriously, I have no clue
 
Hm
There are some exact results for $\varphi^4$ in 2+1 D, right?
 
@Slereah there are some, yes
 
@0celo7 Seriously, it isn't hard. Just take the identity on some set and equip the source with a different topology from the target.
@Slereah Define "exact result"
 
at least for small values of the coupling constant
 
@ACuriousMind Huh? edit doesn't help
 
I have no clue what you're talking about
 
@ACuriousMind solutions known for two points correlation functions
 
@0celo7 Then...think about it. You really should be able to construct an example (even if it takes you a bit, I think this is worthwhile).
 
a more recent link with a more functional integral approach:
 
12:07 AM
@ACuriousMind How can the identity map take you into a different topology ??
 
to be accompanied with
 
"If in addition each $\alpha_i$ is a polynomial in $g$ (with infinite coefficients) plus a
finite function of $g$ then the problem is said to be Superrenormalizable."
But when is it super duper renormalizable
Really for what I want I don't care too much about renormalization :p
 
@ACuriousMind What if I have a point and a line and consider the map $f:\{0\}\to\mathbb{R},\{0\}\mapsto \{0\}$
 
Since I want to do the energy as $\langle H \rangle_\omega - \langle H \rangle_0$
Pretty brutal renormalization
 
the point is an open set in the first space but not in the second
and by definition continuous
but since going forward it maps a point to a point, it's not open
it's not bijective
back to the drawing board
 
12:14 AM
I think it's written in martian
 
argh this topology is too hard
 
After CTCs wormholes are pretty well behaved, really
They have a foliation and all!
I can just used basic QFT
 
@ACuriousMind are there any functions (squiggly lines on graph paper) that have this property
 
Well...the line I'm thinking of (the identity $\mathbb{R}\to\mathbb{R}$) isn't exactly squiggly...
But I suppose any other bijective map - squiggly if you like - also works.
 
Yeah, I don't get what you're trying to tell me with the identity
 
12:25 AM
You do realize the topology on a set is not unique, right? That there is more than one topology you can choose?
 
of course
even @HDE226868 knows that
 
Then I don't get what you don't get :P
 
The empty set has only one topology :p
And so do all singletons!
 
@0celo7 I have no idea where the conversation's gone to/is going, so I'm not sure whether to feel complemented or insulted.
Though I have my suspicions
 
@HDE226868 well, did you know that :P?
 
12:28 AM
@0celo7 Yes, I did. It sounds familiar, at least.
 
@ACuriousMind right, you just map points in $R$ to points in $R$
and the topology is different in the two
so what?
 
It very easy to choose the topology such that the map is continuous and bijective, but not open.
 
do I have to construct the topologies explicitly?
@ACuriousMind >very easy
I'm assuming I can take the standard topology on the first copy, right?
 
The map is just the identity, what is there to construct! All you have to do is pick two topologies!
It pretty much doesn't matter which two you pick as long as they fulfill a certain property relative to each other.
 
aha!
take the first one to have the discrete topology
and the second one to have the standard one
 
12:35 AM
Yes, that works.
 
what "certain property" were you thinking of
that the open sets of one are open sets of the other
but not vice versa?
 
Yes.
We call the first topology finer than the second one (and the second coarser than the first) in this case.
 
Is there any $f:R\to R$ where both copes of $R$ have the standard topology but $f$ satisfies the aforementioned properties?
 
"In this letter, we argue that in curved spacetime, β is in general not directly related to temperature in the sense of the zeroth law. For in stationary curved spacetimes, tidal forces and the Unruh effect impede this int erpretation of the KMS parameter. "
Oh no
 
@0celo7 Hmmm...if the reals were compact, the answer would be no, but I don't see anything that would forbid it.
 
12:45 AM
@ACuriousMind why if they were compact
 
@0celo7 Because a continuous bijection from a compact space to a Hausdorff space is a homeomorphism :)
 
uh, sure
 
well obviously
 
I don't dare ask for a proof
 
Hm
Is the stress energy tensor easy to define in a scattering, I wonder
Perturbatively, anyway
(not for the asymptotic states, that is)
 
12:50 AM
what happened to CTCs and wormholes and shit
 
CTCs are a heap of trouble
When you try to do QFT without a foliation things get tricky
 
12 hours ago, by Slereah
You may notice I don't have a good attention span :p
 
well according to JD there are no CTCs, right?
 
obviously
 
what is naive set theory
 
12:52 AM
Some people say "Well maybe quantum gravity will save CTCs!" but I don't have a fucking clue on how to do QG with CTCs
Even in the most trivial case
@0celo7 : That's set theory where a set is just "That thing that contains object with such property"
$A = \{ x \vert f(x)\}$
 
as opposed to
 
You can't build any set like that in ZFC
 
what if I do
 
Or other frameworks of set theory
Then you get Russell's paradox
 
$S:=\{x|x\in S\}$
 
12:56 AM
@0celo7 Russell's paradox. We've been over this.
 
@ACuriousMind have we
@Slereah any naive set?
I'm confused now
 
You can build a lot of the same sets in both naive set theory and ZFC
But not all of them
 
Sep 8 at 17:39, by 0celo7
idk about Russell's paradox
 
@ACuriousMind I think my Turkey Day drinkathon killed some brain cells
 
2:28 AM
@0celo7 Wow, that's particularly bad.
 
seems most energy condition papers nowadays use local QFT
 
Sigh...some people.
-8
Q: how this question could be solved?

Mazen Ahmed I want ur help in findin a solution to this question

 
Seems I won't avoid having to learn it
 
Can't even use proper grammar
 
2:54 AM
I have no idea if any of the information herein is accurate, but it's got an awesome animation:
 
3:51 AM
@Slereah what do you know about spin structures
 
They are the even Clifford bundle of the manifold or some shit?
 
@ChrisWhite That Einstein equation question has got me thinking... has anybody defined a... spacetime... Reynolstein number?
 
@Slereah if the momentum of a particle is future directed timelike, is the mass necessarily positive?
 
If things get non-dimensionalized in a "proper" fashion, some sort of non-dimensional number relating "convective" and "diffusive" forces will show up
 
I think so?
Even if it's past directed, really
 
3:59 AM
@tpg2114 What is the Reynolds number, anyway?
@Slereah then the positive energy theorem makes no sense
 
@tpg2114 : I vaguely recall a continuum mechanics analogy of GR, but that kind of stuff isn't really used I think
 
it says that the ADM momentum is future directed timelike/null
 
@0celo7 Ratio of convective to viscous forces in a fluid. $Re = u L / \nu$
Can be thought of many different ways
 
does that imply that $P^0>0$?
 
But basically as Re -> infinity, diffusion is not important
 
4:01 AM
Yes
 
@Slereah you just said no
 
Did I
I said the mass, not the energy
 
maybe
mass energy, E=mc^2 and all that shit
 
$m^2 = p^2$
As usual
 
but
the ADM mass is $P^0$
 
4:03 AM
Different mass definition I s'ppose
There's a bunch of masses for GR
 
according to Straumann $P^0$ is indeed the mass
for Schwarzschild it's the mass parameter
hmm, what's the difference between a future and past directed timelike vector
how does one tell which it is
 
@0celo7 I know nothing about GR, but wouldn't a future/past timelike vector just have a normal of (-1,0,0,0) or (1,0,0,0) in (t,x,y,z) coordinates?
 
@tpg2114 Yes...something like that.
 
Maybe nothing is an exaggeration... I know stuff gets heavier as it goes faster, and it probably gets shorter too.
 
But that sign difference is what is bugging me!
What if that 1 is an m
Then how does one know if it is positive or negative!
 
4:07 AM
@0celo7 You can't!
You have to first define
a TIME ORIENTATION
 
But isn't that the point? A single point in space/time/etc is meaningless
If I stick you in the middle of nowhere without a compass and ask you what way North is, do you know?
 
Two timelike vectors have the same time orientation if $g(x,y) < 0$
 
And while we're at it, does it even matter?
 
(if the signature is (-+++))
 
Vectors only matter if something changes in some direction.
 
4:09 AM
Usually the time orientation is defined by the time coordinate
 
@Slereah ADM assumes a nice spacetime, so this isn't a problem
Well the ADM mass is constant.
But apparently it's future directed and timelike or null!
 
@Slereah But to have any coordinate, you have to have some direction right? Otherwise it's ill-defined. If I stand in my room and look forward and declare that "north", is it?
There isn't really a way to know what the unit vectors are
 
@tpg2114 Time is the direction of increase of entropy.
 
There is actually a coordinate time such that this is the case
It is the York time used in quantum gravity
 
@0celo7 Again, not sure I can comment because I don't do anything remotely relativistic (my thesis is actually on low-Mach number fluids...) -- but isn't entropy dependent on the observer/system?
 
4:13 AM
not really no
 
@Slereah Then you're defining the coordinate time by the time orientation, not the other way around ;)
 
Ah, @ACuriousMind
If the momentum is future directed timelike/null, does this imply that the mass is positive? Or even just that the energy is positive?
I'm not sure what the distinction is.
 
I have no idea at all about "ADM masses" or "positive energy theorems".
 
@ACuriousMind Bah, ignore all of that
 
York time is $\text{Tr}(K) = C$ apparently
 
4:14 AM
Just a general momentum vector
 
That's not something you learn in geometry, and all of my GR ability comes purely from that :P
 
@Slereah trace of the extrinsic curvature?
 
Yep
It is a constant
 
@ACuriousMind all of these proofs are done by mathematicians
 
I thought York time would be the time of the timezone in which York lies :P
 
4:15 AM
it's apparently an interesting application of geometric analysis
 
@Slereah If you read the comment thread here: physics.stackexchange.com/questions/218505/… it appears somebody thinks entropy depends on the observer
 
and who doesn't like spinorial PDEs
 
@0celo7 Yes, but by mathematicians interested in GR :P
 
Again, above my pay grade, but it does seem odd
 
Entropy depends on the observer but not increase of entropy
 
4:16 AM
@ACuriousMind pls answer my basic special relativity question :)
 
@0celo7 I don't even know how the mass could not be positive - $m = \sqrt{-p^2}$ rather by definition - that can't be negative!
 
Energy!
I'm not convinced that these GR math people are making the distinction.
 
Well then say energy, not mass
You dunderhead
 
I've seen it written as "positive mass theorem" and "positive energy theorem"
 
Well...then it depends whether you think $p^0$ or $-p^0$ is the energy :D
 
user54412
4:18 AM
@Slereah Weird stuff that's not really used? Sounds like your thing :p
3
 
@ChrisWhite Yeah
 
I might just have to get this book.
@ACuriousMind AAAAAAA
 
There was a brief passage in a NASA report on advanced propulsion doing an analogy of GR with continuum mechanics
 
I wish there was some way to include alcohol consumption in data.SE... There could be a very strong correlation between propensity to downvote stupid questions and drunken-ness. At least for me.
 
4:19 AM
To give an idea of the rigidity of spacetime
 
@tpg2114 how so
 
@tpg2114 By that logic, I'm drunk most of the time I'm here.
3
 
@0celo7 In the comment thread in question, it seems odd that whether mixing of identical molecules depends on who sees them... If I don't "see" whether they are left or right before I remove the barrier vs. somebody who can see that
If I can't tell left from right, I see no entropy change
But if you can see left and right, you do?
Seems... unsettling.
 
let's ask @ACuriousMind
@ACuriousMind What does "future directed" even mean
 
@ACuriousMind It's a good thing I don't need to explain my downvotes... Otherwise there would be many "-1, are you f'ing serious????" comments
 
4:23 AM
is it just that the time component of the vector is greater than zero
time to get out
 
@0celo7 : What I said
 
@0celo7 With respect to what reference frame?
 
THE BIBLE OF SPECIAL RELATIVITY
 
"Future directed" means that it is in the same direction as another vector which defines an orientation
 
@0celo7 Uh...by definition, your spacetime comes with a time orientation, no? I.e. a smooth partition of timelike tangent vectors into two equivalence classes, which you call "future directed" and "past directed".
 
4:24 AM
In most spacetimes that just means $x_t > 0$
 
In Minkowski, those classes are indeed defined by the sign of the time component, but I dunno about arbitrary charts/spacetimes
 
Side note -- I feel like "with respect to what reference frame?" is a question anybody who knows nothing about physics could ask a physicist in order to sound smart
 
@tpg2114 Well, more often that not, that is actually a good question! ;)
 
@ACuriousMind bah
globally hyperbolic spacetimes are basically Minkowski
 
Kind of like "how does that make you feel?" works for psychologists/psychiatrists
 
4:26 AM
@ACuriousMind it's not
the reference frame is obvious
how about this
 
If your spacetime is globally hyperbolic, there's usually a coordinate system such that the time coordinate defines a proper time function
 
suppose we have some future directed velocity
$v$
and then we have some negative mass $m'=-m,m>0$
 
user54412
@tpg2114 One of my professor's favorites to ask of colloquium speakers: "But what about the magnetic fields?"
 
@0celo7 Wat
 
then the momentum $p=m'v=-mv$ is past directed
from this I infer that positive mass iff future directed momentum
QED
@ACuriousMind what confuses you, my child
 
4:28 AM
Greetings, minions.
 
I'm scared that this might actually be the thinking behind "positive mass/negative mass" :P
 
Eh boss I need some money
Getting hungry over here
 
@DanielSank Hello @dmckee or was it... wait... which sock puppet are you again?
 
@ACuriousMind Huh?
 
@DanielSank Greetings, me.
 
4:29 AM
@tpg2114 This is the main account, i.e. no the sock puppet.
 
If $v$ is future directed then $-v$ is past directed, right?
 
@ACuriousMind Indeed.
 
@0celo7 I guess so, yes
 
@ACuriousMind So $mv$ can only be past directed if $m<0$?
 
@DanielSank Ohhhh! I got lost in the many aliases and puppetry...
 
4:30 AM
I feel like I'm the only one who isn't a puppet
At least, I don't feel someone reaching up my butt to control me.
 
@0celo7 Hm. Well, if $v$ is future-directed, then $mv$ is future-directed iff $g(v,mv) = mg(v,v) > 0$ (in the convention where $g(v,v) > 0$ for timelike), so yes.
@0celo7 It's important that you believe that.
 
@0celo7 But maybe somebody is reaching out of it, with memories of controlling you? After all, the $t$ component of the vector may be negative.
 
@ACuriousMind Hmm, yes. Here's the bible of SR saying the exact same thing.
@tpg2114 sigh
engineers really shouldn't do physics
 
@0celo7 That is 100% the truth
I'm slightly disappointed. Drunken trolling of the physics room has gotten me 0 stars tonight.
 
4:36 AM
I'm waiting for someone to call me out on that
or maybe everyone agrees
 
@tpg2114 I got three, though :)
 
@ACuriousMind Life is so not fair.
Maybe stars depend on the observer too. You see you got 3, but I see that I got them all.
 
why are spin bundles so horrible
now there are p-form spinors
 
spinors are horrible
 
pretty soon there will be p-form-valued spin-valued hilbert space operators
 
4:39 AM
Could be worse
 
that act on a spin hilbert space
 
If you lack orientability then you have to use the Pin group
 
which is a bundle on a Teichmuller universe
 
There are 8 different components of the pin group
 
all in the category $\mathsf{HorrbleMathObjcts}$
 
4:40 AM
@tpg2114 You needed to get drunk to notice that? ;P
 
@0celo7 I did want to respond to your comment on the turbulence question but didn't want to drag out the conversation there -- I'm sure the Einstein equation is well known for people who deal with it, and I'm sure I wouldn't have recognized the notation if I saw it, but I was hoping to see the equation so I could identify the mathematical nature of it. Based on the rest of the comment thread, a second-order PDE that is linear in the 2nd derivative but non-linear in the first
That's the same as Navier-Stokes and so maybe some inter-disciplinary knowledge could happen. It's one thing that's cool about our site
 
@tpg2114 $R_{\mu\nu}-\tfrac{1}{2}Rg_{\mu\nu}=T_{\mu\nu}$
 
@DanielSank Does this mean I get to be yellow, possessed of an unspecified small number of eyes and dedicated to serving evil where-so-ever I can find it?
 
@0celo7 He wants to see the derivatives, silly :P
 
I may not know the physics of the equation, but I could recommend trying to do something we would do for a fluid that might lead to some insight
 
4:42 AM
ah
$R\sim \partial^2 g$
something like that
 
@0celo7 I mean... all of Navier-Stokes is really just F = ma. So we can be really obtuse about it if we want :)
 
@tpg2114 well Einstein's equation is just $\delta S=0$
although if you insist on doing GR on a compact thingie $S$ can get nasty
personally I like my ill-defined integrals over the whole manifold, no annoying boundary terms
 
^ That's why finite volume numerical methods are better than finite difference... conservative, and lower order derivatives to compute
 
wait, why do we do the action over a compact subset anyway, @ACuriousMind
because the action should be finite?
 
@0celo7 So it's guaranteed to be finite?
Seems a bit difficult to do the usual "fields fall off fast enough at infinity" on arbitrary spacetimes.
 
4:47 AM
I've never seen spacetime done as a fluid so far I'm afraid
 
@ACuriousMind literally what I just said...
@tpg2114 what
 
@0celo7 Numerical approaches to equations are much easier by integrating the equations once. You lose a lot of the ambiguity at the boundaries and it becomes easier to get the right answer
 
I've seen it done as continuum mechanics
And optics
 
Fewer "annoying boundary terms" to deal with
 
but it's kind of a fringe formalism
 
4:50 AM
@Slereah In the question about Einstein turbulence, somebody linked to a paper that shows dimension $d+1$ spacetime can be realized in $d$ dimensional Navier-Stokes (fluid)
 
Maybe
There's a lot of GR formalisms
 
But I don't know how reputable that paper is... they cite themselves a lot and it isn't peer-reviewed. I know physics is cool with the whole arxiv thing, but from a fluids perspective that's usually a sign of nonsense
 
I had a reviewer reject a paper because citing an arxiv paper is only a "very small, mostly crackpot, step above citing Wikipedia"
 
@tpg2114 oO
 
4:52 AM
@ACuriousMind engineering :)
 
Hard to say because when you have such marginal theories, you will end up citing yourself a lot
The best way to check, I suppose
Check the math by hand
Check if this is indeed equivalent to GR
 
yes
 
@0celo7 Pretty much -- although, almost all fluid dynamics research happens in peer-reviewed journals. Preprints just aren't a thing we do.
 
calculate the metric by hand
 
Even "physics" and not "engineering" approaches
 
4:53 AM
@tpg2114 I know some of my profs use it
nucl-ex, etc.
 
I'm sure it's very field-specific
 
but there's a lot of specific journals, too
 
My gut reaction when I come across a paper on arxiv is "Okay, find the actual journal paper" and most of the time if I can't, I dig into the arxiv one and find it... lacking.
 
wow
I trust anything on arXiv with my life
"Einstein equations solved explicitly"
Hell yeah!
 
Arxiv is mostly fine
 
4:56 AM
^ field dependent
 
There's a few crazy papers but overall it's not really problematic
 
@Slereah you're into stuff where 75% of the citations are from the author
 
Nah
 
I have not found any paper on arxiv that I would use without finding the corresponding journal paper
Well, I found one but then got destroyed by a reviewer for it
 
@tpg2114 does the reputation of the author count for anything?
@tpg2114 you have tainted Georgia's rep
 
4:57 AM
@0celo7 Possibly? But I haven't found any of the big names in my field on there either
 
@tpg2114 : Are you allowed to reference unpublished papers and private correspondance :p
 
@Slereah WALD
 
@0celo7 Woah woah woah... make sure you have your universities straight. I'm not U[sic]GA here
 
@tpg2114 Georgia Tech then
at a UTK student you're all the same filth to me ;)
 
@Slereah No actually. Unless it is so inconsequential as to not influence anything in the paper and could be safely omitted.
 
4:59 AM
Theoretical physics you basically have
[22] Frie Danfrie, private communication, not to appear (it's private).
 
@tpg2114 can you outsource all of your proofs to Hawking-Ellis and not even bother giving the page number
 
@0celo7 Hey now. Keep your SEC hatred to yourself. The ACC doesn't like any of them.
 
[36] Covariant Formalism, Kneel Mark is rubbish, private thought.
 

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