The h Bar

Slereah

1:00 AM

I found a paper that mentions states for the Thirring model but not Sine Gordon

a bit odd

Oh wait, the states are just for the massless case so far

Which is just the free field

Or maybe not

I dunno

0celo7

2:39 AM

It's generally considered good form to link the abstract page, not the PDF. — 0celo7 13 secs ago

@dmckee @ACuriousMind HAPPY?

Slereah

2:50 AM

"A nonlinear superposition principle by which any two solutions induce a third one. This is called the Bäcklund transformation and is a kind of creation operator for the system"

Oh my

0celo7

3:00 AM

@Slereah Sounds sexy

Slereah

"Creation/annihilation of wormholes supported by the Sine-Gordon phantom (ghost) field"

whaaat

It's written by two Vladimirs

0celo7

what the hell is an infinite subset

Slereah

"WORMHOLES SUPPORTED BY A PHANTOM MEXICAN HAT"

¡Dios mio!

0celo7

what are you reading

vixra?

Slereah

No, this is a real paper

Weirdest part is I wasn't looking for wormhole stuff

I was looking at sine gordon papers

0celo7

3:14 AM

Aha, limit point compactness implies sequential compactness

Very interesting.

No, other way around!

user116211

4:23 AM

0

Q: Show that $(x-iy)g(r)$, $z g(r)$ and $(x+iy)g(r)$ are mutually orthogonal

I want to show that $$\psi_1(x,y,z) = (x-iy)g(r)$$ $$\psi_2(x,y,z) = \sqrt{2}zg(r)$$ $$\psi_3(x,y,z) = -(x+iy)g(r)$$ where $g(r)$ is an arbitrary function of $r = \sqrt{x^2 + y^2 + z^2}$, are mutually orthogonal. I see that $$\int_{-\infty}^{\infty} \psi_1^*\psi_2 d^3r = \sqrt{2}\int_{-\i...

quantum-mechanics

user116211

HW?

Slereah

6:51 AM

Guys

If I add a bump function to a metric, it will cease to have a well defined Cauchy problem, since from the initial slice we can have several developments

But what part of the initial value problem theorem of GR does that violate

Is the metric still a proper spacetime metric, too

0celo7

Fire alarm

Why am I up

@Slereah ...what are you talking about

Physics Meta

7:33 AM

0

Q: Why the cited questions aren't counted active?

The title of the question pretty much tells it but still... A question being cited somewhere on the PSE itself actually stimulates a good amount of activity regarding the cited question. But the question is not showed 'active today' even if it is cited today. Its activity is displayed only on the...

feature-request citation

Slereah

Well posedness of GR says that for some initial slice, there's a maximal globally hyperbolic development that is unique up to diffeo

Are the other possible developments non GH ones

0celo7

2:30 PM

@ACuriousMind The German word for explode sounds made up.

ACuriousMind

@0celo7 Again, all words are made up!

0celo7

2:47 PM

@ACuriousMind Not really

It just sounds like someone tried to say "explode" with a German accent + suffix

Conjugation thingie

Whatever the tail part is called

@ACuriousMind there are definitely real and "fake" sounding words

Most French words sound fake imo

Obliv

@0celo7 all them damn foreigners and their 'fake' soundin' words. despicable

@slereah you hear that? your language is actually fake <|8O)

user54412

3:13 PM

@RobJeffries ok, let the experts vote — Wolphram jonny 22 hours ago

user54412

What kind of troll challenges the stellar physics expert on the site with a popularity contest?

ACuriousMind

@ChrisWhite The kind of troll that also indiscriminately votes "Leave Open" on every close review he does.

Obliv

@acuriousmind hey is this true: $(1~2~3)\circ(1~3~2)$ don't directly commute but if you do conjugation $gxg^{-1} = x$ you get $(1~3~2)\circ(1~2~3)\circ(1~2)(2~3) = (1~2~3)$?

i.e centralizer $\ne$ center of a group?

user54412

@ACuriousMind interesting...

ACuriousMind

@Obliv I don't understand what you mean by "if you do conjugation"

user54412

3:22 PM

nor do I, but in general the centralizer is not the center, if that was the point

ACuriousMind

@Obliv The centralizer of a subset of a group is a generalization of the concept of center, and the centralizer of the entire group is the center.

@ChrisWhite That's one way to put it :P

Obliv

I meant since $C_G((1~2~3)) = \{g \in G\mid g(1~2~3)g^{-1} = (1~2~3) \}$ the condition is called conjugation, right?

on wiki the centralizer was just defined as $C_G(A) = \{g\in G \mid ga = ag, \forall a \in A\}$ which I don't agree with for the symmetry group $S_3$

maybe i'm just doing the operation incorrectly for the elements in $S_3$ though

ACuriousMind

@Obliv "Conjugation by $g$" is the operation $x\mapsto gxg^{-1}$. The condition that $x=gxg^{-1}$ is that $g$ commutes with $x$.

@Obliv How can you "not agree"? $ga=ag\iff gag^{-1} = a$.

Obliv

well I'm probably doing it wrong for $(1~2~3)\circ(1~3~2)$ then

wait

yeah i was doing it wrong. those elements commute

how would lagrange's theorem make finding centralizers & center's of groups $S_3,D_8,Q_8$ easier?

user54412

centralizers (and thus centers) are themselves subgroups

user54412

3:36 PM

since the cardinality of $D_8$ is 8, you know any subgroup must have cardinality 2, 4, or 8

user54412

so if you know of 5 elements, for instance, in a subgroup, the subgroup must be the whole group

Obliv

ohh I see. @chriswhite thanks. Similarly, in $S_3$ if I get $3$ elements in the subgroup, I can stop checking because it won't be larger than $3$

user54412

well, if you know the subgroup is proper

Obliv

what does that mean?

user54412

"proper subgroup" = "subgroup that is not the entire group"

user54412

3:42 PM

there exists a subgroup of $S_3$ that has a full 6 elements, namely $S_3$

user54412

(unless your course is doing some weird thing where all subgroups are considered proper by default)

Obliv

hmm yes okay

Secret

4:03 PM

newscientist.com/article/…

arxiv.org/abs/1505.07800

Actually reading this prompt another question: If not all regions of the universe during its evolution is globally hyperbolic (i.e. no cauchy surface can be defined for that region, thus the evolution there is not necessary uniquely determined by the matter distribution on the spatial hypersurface), then how can we be certain of the whole geometry of the universe in space and time...?

user54412

I'm not sure what the connection is between that paper and that comment, but...

user54412

regarding the comment: any such catastrophic failure of unique evolution spells the end of science as any of us knows it; I for one prefer to assume that science works when doing science

user54412

regarding the paper: just about anything posted to gr-qc and then cross-posted to astro-ph is pseudoscience -- you would do well to avoid it

user54412

In astrophysics we just laugh at the nonsense in that part of arxiv. It's basically people who excelled at doing symbol manipulation in their physics courses, but who never understood what physics was, claiming that enough QFT-like arcana explains all the complex, emergent phenomena in the universe.

Secret

4:19 PM

Re comment 1: O crap, never thought about that... (except that will mean we cannot realy plot out the spacetime manifold because of the nonunique evolution in time of space)

Re comment 2-3: Newscientist is sharing about a couple of groups trying to use spacetime curvature to explain that dark energy might be an illusion due to the average curvature became more negative as the universe expands. That paper quoted is the latest of the chain of studies done by those groups. I unfortunately don't have enough background to recognise pseudoscience in cosmology however, but I do aware that there a…

Having said that, since I don't have the background, I am going to just crossing fingers and see which one is the better description. Based on that paper, the backreaction effect the group claimed seemed to be minor in explaining the observed accelerated expansion of the universe

thus I suspect the dark energy explanation will still be the more accurate one

user54412

I will agree that dark energy is a bit more tenuous. It unfortunately is irrelevant in the early universe, so seeing it in the CMB and BAO is nontrivial. The most direct evidence is from SNe, but I can tell you from first hand experience that SN people tend to throw out any objects that don't fit on the line they want, which is pretty scary.

user54412

Not to mention we don't actually know what Type Ia SNe even are.

Secret

and speaking about curve fitting in cosmology...
I heard my professor and my peers said as in in-joke that the data is so noisy that nearly any curve is a good fit...

user54412

Bonus points if you can identify the science in question

user54412

^ looks like a pretty good fit to me...

Secret

4:25 PM

I wonder if long period gravitational astronomy will provide an extra way to help deconvolute the noise in the data...?

(of course I am not talking about using it in its current highly noisy state, but maybe some time in the future after those other inferometers are built to constraint better the angular dispersion of the data)

user54412

Slereah

Well if spacetime isn't globally hyperbolic it's not the end of the world

Solutions may fail to be unique but they're not random either

user54412

The anisotropy power spectrum is even more telling. Those high-l points have smaller error bars now (wiki's image is 10 years old). But even then, the relative height among the first three peaks tells you right away how much matter is in the universe.

Secret

@ Slereah but then how can we plot out the spacetime manifold if there are more than one solutions in a region. And if it is not random, what determines the evolution? (And I am pretty sure you might not be implying possibly non hausodoff spacetimes here)

@ Chris white I don't quite understand that power spectrum, thus I cannot comment much other than that the matter content as revealed in the spectrum must be underestimating the true amount which is the 20% something of dark matter

Correction: the spectrum must be underestimating the true amount, otherwise we cannot account for the 20% something contributed by dark matter

Slereah

4:44 PM

You can't.

If you drop the assumption of global hyperbolicity, you can add any bullshit topology doodads and gizmos in the future

You can just drop handles and cuts and naked singularities wherever

Secret

that's clearly non hasodoff

Slereah

what

Secret

suppose you begin with a matter distribution A(x,t) in space at some time coordinate t. If the region is globally hyperbolic, then you should only expect a unique matter distribution A(x,t+t0) at the t+t0 coordinate, right?

Slereah

You can also have non unique solutions in a globally hyperbolic spacetime

The Hawking theorem is that a globally hyperbolic spacetime only has unique solutions for field evolution if those fields have unique solutions in minkowski space

Obliv

@slereah what is backreaction?

Slereah

4:51 PM

Backreaction is the effect of matter on a spacetime

Usually used in the context of the test field limit

Obliv

so the paper argues that matter affects spacetime geometry?

Slereah

You calculate the field in a fixed metric

Then you calculate the stress energy tensor of that field

This is the backreaction

Secret

If the region is not globally hyperbolic, then you can have e.g. B(x,t+t0), A(x,t+t0) etc. Since they all shared the same t coordinate but different x coordinates (because the distribution of matter can differ in space because of the nonunique evolution, then how can we plot B(x,t+t0), A(x,t+t0) in the same spacetime manifold as there will be points where there is matter if B is the distribution resulted but there is none if A is the distribution that resulted?

and in the absence of multivalued fields, we want a unique field value for each spacetime point

Slereah

You can apply this method in the hope that the solution converges

Secret

or more simply, a nonunique evolution will mean there are at least two different possible matter distributions hence curvature for the same time coordinate. The spacetime manifold is supposed to have all events in the past present and future all plotted in, but given the above nonunique solutions how do we understand where they are on the spacetime manifold?

except invoking a non hausdoff topology?

Slereah

4:58 PM

what

Secret

ok let me try again. The stress energy tensor T determines the energy momentum for each point in spacetime

Normally if we have unique evolution, then T at some time coordinate t will be some value, and then it wil become some other value at t+dt

If the evolution is nonunique, then there will be more than one possible T at t+dt.

But since spacetime is supposed to have T defined for all t, if we don't allow T to be mutivalued at all x and t, then how can we plot this T at t+dt in (since there will be more than one possible of these because of more than one solution)?

ACuriousMind

@Secret You're not making sense. The spacetime has T defined for all points - it's just that more than one possible T solves the equations of motion.

Secret

and those possible T just happened to have the same value for the points on the hypersurface (hence the nonunique evolution towards the future)?

ACuriousMind

5:16 PM

Yes

0celo7

@ACuriousMind my advisor said I'm "pretty much a physicist"

3

How can I stop people from thinking this

Secret

but in that case, since the T differs for any points outside the hypersurface, this will imply we need to consider more than one possible spacetime manifold if we want to show the matter distribution in the future and past of the hypersurface? and we surely cannot plot all of these onto the same manifold, so how do we analyse them?

do we analyse the curvature induced by each possible T separately?

Slereah

You don't

vzn

@0celo7 lol maybe give up on GR and change the name of your cat... (study GDP instead...?) :P

Slereah

If uniqueness of time evolution breaks down you can't predict it

Secret

5:23 PM

so basically we canot predict it, yet it is still nonrandom because we can determine exactly all the solutions possible for that given info on that hypersurface?

Slereah

Well it will be one of the solutions

vzn

@Slereah so why so into sine gordon lately? like it myself... am personally amazed nobody seems to do soliton computational experiments/ simulations...

Secret

hmm... it must be a semantic issue, but sometimes I don't understand the difference between random and unpredictable.

Because in quantum mechanics, you also have the wavefunction telling you all the possible eigenvalues you can obtain for some given operator, but which eigenvalue you get is randomly determined by what the state is being projected to with a probability.

Is the nonunqiue time evolution discussion above differ from this quantum scenario in that we don't even have probabilities telling which solution we will end up from the hypersurface, other than we only know that "it will …

vzn

@Secret try bohmian mechanics & his philosophy on implicate vs explicate order :)

0celo7

@vzn I'll name my next cat "Milton"

ACuriousMind

5:32 PM

@vzn What do you mean? A particular kind of solitons, the instantons, have been the focus of much non-perturbative research in quantum field theory for decades

@Secret "random" means "according to a probability distribution". "unpredictable" means, well, exactly what it says, that something cannot be predicted. Whether you call random things "unpredictable" is a matter of preference.

vzn

@ACuriousMind what do you mean "non-perturbative research"?

@0celo7 Milton was quite religious. thought you might go for a nihilist philosopher like Nietzche or Camus... how about Kafka? o_O

ACuriousMind

@vzn I mean that this line of study does not use the perturbation series of QFT

And in particular therefore is not limited to the perturbative regime of small coupling

@0celo7 You stop people from thinking you're a physicist by not being one ;)

vzn

@ACuriousMind not familiar with the area; but havent heard of it in computational experiments/ simulations...

ACuriousMind

@vzn I'm pretty sure instantonic effects do occur in the computational approach to lattice gauge theories

vzn

@ACuriousMind hmmm ok sounds interesting

ACuriousMind

5:40 PM

"Experiments" is a bit vague because these are solutions to the Euclidean version of a QFT - they do not correspond to "physical objects" (whatever that is) in a straightforward way

vzn

anyway QFT simulations are limited to a very "small" region of space with not a lot of interaction arent they?

Secret

@ Acuriousmind Hmm, I see. Suddenly I felt quantum mechanics is more predictable than nonunique time evolution because for the former, you still have probabilities evolving deterministically

ACuriousMind

@vzn The lattices are usually rather small indeed, but that doesn't really correspond to a "small region of space" so much as a very coarse graining of the spacetime. I'm not sure what "not a lot of interaction" is supposed to mean - simulating a free theory would be pointless

vzn

interesting wikipedia agrees with you early on re solitons/ QFT/ lattice etc en.wikipedia.org/wiki/Lattice_field_theory

wonder why "numerical simulation gives access to field configurations that are not accessible to perturbation theory, such as solitons."

ACuriousMind

What do you mean "why"? It can access the non-perturbative regime simply because it doesn't have to make the restrictive assumptions that allow perturbation theory to work.

vzn

5:47 PM

ok

Slereah

@vzn I like me some analytic solutions

ACuriousMind

@vzn Perhaps to clarify that quote - the thing that gives access to the non-perturbative regime is the lattice formulation itself, not the numerical simulation. It's simply the case that numerical simulation is often the only way to compute anything in the lattice formulation.

vzn

seems to be focused on quark/ gluon interactions en.wikipedia.org/wiki/Lattice_QCD

Slereah

Also @ACuriousMind I was right there is totally a way to turn classical solutions into quantum states

It's some semiclassical path integral shit

vzn

@Slereah ??? have been looking into that for many yrs, think there is another way

ACuriousMind

5:50 PM

@Slereah I didn't so much say that you were right or wrong as I insisted that you actually state the procedure for that ;P

Slereah

Well I am currently reading Rajamaran's book on solitons for the details

But apparently you can have the whole "particle" spectrum for sine gordon

You can even do a superposition principle, kinda

vzn

@Slereah ofc something to be said for that but dont be afraid of simulations :)

vzn

6:08 PM

you might be interested in this soliton pg by Kasman where he talks about "nonlinear superposition," decomposition of a particle collision into a virtual particle, etc kasmana.people.cofc.edu/SOLITONPICS

0celo7

6:59 PM

@ACuriousMind but I'm not one :((((

Stop bullying me

@ACuriousMind advisor wrote "ups" in an email, should I tell him it's "oops"

(In English, that is)

ACuriousMind

I wouldn't bother

0celo7

7:39 PM

@vzn nihilism is self-detonating because nihilism is irrelevant

Obliv

8:20 PM

@0celo7 do u have any earbud recommendations, mine broke :(

they were default samsung ones

0celo7

why would you ask me

Obliv

you seem like the musical type

0celo7

8:39 PM

@Obliv I hate music

vzn

@0celo7 actually was exaggerating, those are technically considered existentialist philosophers/ writers who bordered on nihilism only occasionally/ at extremes :)

Obliv

@0celo7 i guess you don't consider lil wayne to be music, then. gottem

vzn

← doesnt entirely consider lil wayne to be music either :P

0celo7

@Obliv I don't

Obliv

you consider it gospel truth, then? @0celo7

0celo7

9:05 PM

@Obliv yes

Obliv

ill ask in math chat actually

3075

I'm depressed now.

Obliv

@3075 whys that

3075

the inverse of the reason I was happy a few days ago.

:(

Obliv

what a mathematical approach to happiness :\

right multiply by the happiness reason again

3075

9:17 PM

this is my first time dealing with this kind of occurrence.

and I feel so broken.

Obliv

is it love-related? I've been there.. sort of

3075

yeah.

0celo7

@3075 swag your shit up

3075

lol?

0celo7

I'm talking Js, dope watch, tight ass clothes

New haircut

Don't sag if you're white

If you're not...consider it

3075

9:19 PM

you saw my profile image dude.

of course i'm white, pale even.

and I like how I look right now xD

maybe a haircut but...

0celo7

That reminds me

I need to clean my shoes

Obliv

@3075 if it's rejection, don't worry everyone gets rejected. it's just natural

3075

@Obliv it's worse than that.

I didn't even ask her.

Obliv

oh damn. yeah I almost went that route. ended up asking and got semi-rejected.

well is it too late or something? can't you ask later? @3075

Physics Meta

1

Q: Why close list-based questions?

After seeing some close votes on this question, I'm confused about the rationale behind closing list-based questions. The problem is that almost any question can be rephrased into a list question, and vice versa. Consider the following examples: List: where is technique X applied in physics? N...

discussion scope questions

3075

9:23 PM

@Obliv I found out she has a bf... :( if only I was a few weeks earlier.

and it's too late for now, because schools finished and we're going to different universities in the fall.

Obliv

well I'm not good with what to do in this situation. ask @0celo7 how to swag your way to steal her from him

oh .. well to be truthful, it might not have worked out anyway. I waited till the last 2 months of senior year of hs too and it wouldn't have worked out anyway.

3075

@Obliv :(

0celo7

@Obliv and what did you do in your year of college

Obliv

@0celo7 that's classified information

i gotta go. peace o/

0celo7

@ACuriousMind I THINK I UNDERSTAND IT

3075

9:27 PM

@Obliv bye.

ACuriousMind

@0celo7 :)

0celo7

@ACuriousMind The only problem is I don't know how to phrase it any better either

So maybe I don't understand it

@ACuriousMind in particular, I understand this

ACuriousMind

@0celo7 Well, if that's our standard for understanding, then I evidently don't understand it either...

0celo7

@ACuriousMind My philosophy prof seemed insistent that being able to explain = understanding

But he's a bit of a troll

and likes to get people fired up

@ACuriousMind ugh :(

What

@ACuriousMind A connected component of $f(I)\cap g(J)$ is diff to a connected component of $\Gamma$, right?

ACuriousMind

Yep

0celo7

9:36 PM

which you showed has to be the graph of a closed interval

so the connected component of the graph is diff to this closed interval

and since diffs preserve closedness, the connected component of $f(I)\cap g(J)$ is closed

so that's a contradiction, but how do we get out of this by ending the intervals on the boundary?

ACuriousMind

@0celo7 Ah, that part's a bit fishy - diffs send closed subsets to closed subsets, but what is the diff you are thinking of here?

0celo7

A graph is naturally diffeomorphic to its domain

I'm not sure what your argument is then

(the second half)

ACuriousMind

A component of $\Gamma$ corresponding to closed intervals $[i_0,i_1],[j_0,j_1]$ tells youi that $f([i_0,i_1])$ is closed in $M$ and is a connected component of $f(I)\cap g(J)$, but $f(I)\cap g(J)$ is the intersection of open sets and hence open in $M$, and so it s connected components should also be open.

Hm

Yeah, I'm not explaining this one very well either :/

0celo7

@ACuriousMind Hmm, on $[i_0,i_1]$ isn't $f$ a diffeomorphism

also is $g^{-1}\circ f([i_0,i_1])=[j_0,j_1]$?

ACuriousMind

@0celo7 Yes

I have the sinking feeling that the whole argument only makes sense to me because I'm secretly already imagining $M$ as an interval...

0celo7

9:51 PM

@ACuriousMind Is $\Gamma$ actually diffeomorphic to $X:=f(I)\cap g(J)$

ACuriousMind

@0celo7 Yes, the diffeomorphism is given by $\Gamma\to f(I)\cap g(J), (x,y)\mapsto f(x)=g(y)$.

0celo7

@ACuriousMind Ok so if $\Gamma\approx X$ then connected component of $\Gamma\approx$ connected component of $X$

thus connected component of $X$ is closed

but I don't see how ending $\Gamma$ on the boundary gets us out of this

does this mean that the connected components are no longer closed?

oh they are but they're clopen?

user54412

If you close the question you prove creationism — D J Sims 1 min ago

user54412

Is that a challenge?

ACuriousMind

@0celo7 Ah, see, that's a problem - how did you/we conclude that?

I want to use that it is closed in $M$ to conclude it doesn't exist - but the diff only preserves closedness relative to $X$.

0celo7

10:04 PM

@ACuriousMind The image of a closed set under a continuous map is closed? (that might be wrong)

@ACuriousMind Hmm

ACuriousMind

@0celo7 That's wrong, not every continuous map is closed, but diffs are closed.

0celo7

@ACuriousMind homeomorphisms are closed

and a diff is a homeo

ACuriousMind

Yeah, bit still, I claimed closedness of the components in $M$, which you don't get this way. I want to say that if the interval $[i_0,i_1]$ doesn't involve the boundary of $I$, then you get that $f([i_0,i_1])$ is still closed in $M$

0celo7

Ok, back to more thinking

ACuriousMind

@ChrisWhite I can prove things by closing questions here?

that is a whole new level of power

user54412

10:18 PM

wow that comment thread escalated quickly

user54412

and heretofore that person's only notable activity on the network was a somewhat unhealthy obsession with legs

HDE 226868

@ChrisWhite That was partially my fault. The guy's got a long record of bad behavior, which I didn't recognize because he had a different username.

He does have more notable activity; it's just deleted. Fortunately.

user54412

Remember the dmckee strategy -- just come here to vent about stupid questions

HDE 226868

Ah, I should have copied the text of some of my comments before I got rid of them.

0celo7

@ACuriousMind so...your understanding of the proof is wrong?

3075

10:33 PM

@0celo7 you didn't add my new skype :(

user54412

@3075 What happened to your space pic? It now has... airplanes?

3075

yeah.

it's pearson airport in toronto.

I like going there and looking at planes on weekends by myself. :)

it has a blue vibe to it, idk why.

user54412

I like the black-blue-white color scheme. I want to try it on my visualizations.

0celo7

@ACuriousMind How about this:

we prove that $i_0$ has some neighborhood which must lie in the graph

do you think that's true?

3075

@ChrisWhite thanks, which visualizations?

user54412

10:36 PM

Unfortunately too many people in my field use jet/spectrum/rainbow

0celo7

this would prove that the graph simply cannot end there unless we're at the boundary

ACuriousMind

@0celo7 Yes, that sounds good

3075

@ChrisWhite oh I see.

user54412

@3075 All my simulations of accretion disks around black holes.

3075

yeah I figured.

that's really cool.

computational physics seems really interesting but sadly I'm dumb when it comes to programming.

0celo7

10:37 PM

I'm going to be on a DHS project soon, so there might be spooks here soon

3075

@0celo7 what?

0celo7

@ACuriousMind I think the local diff property of $g^{-1}\circ f$ is important here... but what exactly does he mean by that?

user54412

@3075 Everyone has to start somewhere.

0celo7

@3075 I'll be working for the government

3075

@0celo7 I know but why will spooks be here?

to follow you around?

0celo7

10:39 PM

seeing if I'm divulging secrets I guess

3075

@ChrisWhite I did my fair share of programming, but I'm bad at it. I can barely remember syntax.

@0celo7 divulge them in the form of rap lyrics.

user54412

Actually the nice thing about computation I just realized is this: there are no genius prodigies to contend with. In pure math or linguistics or music there are always those people who were born able to do everything. But they just don't exist in computation. I don't quite know why.

0celo7

@ChrisWhite The really smart people find it boring

3075

Lol

0celo7

Really smart people find engineering boring too

@ACuriousMind is computation boring?

3075

10:41 PM

@0celo7 you find engineering boring?

0celo7

@3075 I'm not really smart

evidence: cannot figure out this proof

3075

smarter than me at least.

user54412

I would think engineering is more exciting than computation. All I can destroy is a small cluster. Engineers can destroy labs, bridges, mountains... probably secret moon bases too.

ACuriousMind

@0celo7 No, although it's not the most exciting thing to me either.

user54412

@ACuriousMind Did you ever get that computation to work?

3075

10:44 PM

@ACuriousMind do you recommend that I finish my bachelors in 3 years and do a masters then ph.d or finish in 4 years and do a ph.d right away?

it all depends on whether I take all the 2nd year courses in my first year.

0celo7

why would you ask the person who's taking 4 years for his masters

3075

he's experienced.

and he's only spending 1 year on his masters, iirc.

ACuriousMind

@ChrisWhite I lean towards thinking it actually worked as intended, and that the expectation that I should see non-trivial results was wrong, but I don't know for sure either way

@3075 I don't feel qualified to answer that at all. Being European the idea of going bachelor -> PhD straight away just seems rather strange to me

3075

yeah.

ACuriousMind

@0celo7 3, actually :P

Junaid Aftab

10:47 PM

Hi,

My professor wants to give me (and another kid) a problem in quantum cosmology. To that end, he asked me to read through his recent paper that appeared in the *Physical Review Letters*. He said that I should be able to go through it since all the paper employs is (quantum) scalar field theory, which you have been doing as part of your studying QFT in the summer.

I, however, upon looking at the paper am clearly stmped as I can't make a word out of it. To me, it seems heavy on GR and a lot of other stuff I haven't done.

user54412

What is it with everyone trying to rush through their undergrad? Done right, it's pretty much the best 4 years of your life. And that's not just for those who party their way through school. Whether you like socialization, academic freedom, intellectual diversity, or whatever, it's all available.

3075

@ACuriousMind 3???

how does that work.

0celo7

I should draw a picture of this situation

3075

@ChrisWhite I guess you're right.

ACuriousMind

@3075 I'm not sure I understand the question.

0celo7

10:49 PM

@ACuriousMind grabs pen Secret diagram incoming

3075

@ACuriousMind I thought masters degrees were 1 year after 3 years of undergrad.

ACuriousMind

What does there have to "work" for this? I simply did not graduate thus far, and it'll be another half year to year till I do

3075

then it's ph.d

ACuriousMind

@3075 oO Bachelor is three years and masters two years, here

3075

oh I see.

makes sense.

0celo7

10:54 PM

@ACuriousMind double checking: the statement that the endpoints of the connected components of $\Gamma$ can be open if it goes to the boundary is the same as saying that $(0,1]$ is closed in $(0,2)$ wrt. the subspace topology?

even though it has a $($

user54412

@JunaidAftab It looks reasonably heavy on the GR side, if you haven't seen any at all before. That said, the GR used in cosmology is almost as trivial as Minkowski spacetime. The issue is the formalism and tools often are far more powerful than that. So for instance a full appreciation of ADM 3+1 stuff is at least a full semester into a GR course, maybe 2, but its use here isn't that deep.

user54412

So I'd cautiously say this might be understandable if you're comfortable with QFT and you have the prof available to answer whatever questions come up.

ACuriousMind

@0celo7 yeah

0celo7

god I hate topology

have I ever told you guys that before?

@ACuriousMind Ok, so what we need to prove is that two of $i_0,i_1,j_0,j_1$ are boundary points?

It's clear from the diagram I drew :P

or is it that one of $i_0,j_0$ must be on the boundary of an interval and the same for $i_1,j_1$?

@ACuriousMind Is each connected component of $\Gamma$ diffeomorphic to the corresponding $[i_0,i_1]$ and $[j_0,j_1]$?

I'm 100-.000000000001% sure

vzn

11:14 PM