Hey anyone here uses mindmaps to make you notes?

@0celo7 But how do I decide which name to take?

I already agonize too much about what to name my own characters!

@ACuriousMind I picked "Drake"

same first letter as your suggestion

naming my Pikachu...uh

Tunechi

sounds cute

@0celo7 That appears to be...some sort of drug?

@ACuriousMind o.o

It's Lil Wayne's nickname

well, one of them

it can be heard here: youtube.com/watch?v=5z25pGEGBM4

@0celo7 Googling it gives me that and that it's a name for "spice" or "synthetic marihuana" (???)

@ACuriousMind huh

well maybe that's where he got it from :P

Admittedly, I don't find anything that looks very reliable :P

I thought it was what his grandma called him

Maybe some people named the spice after him? I don't know

perhaps

Time to kill this eevee

@0celo7 :(

@ACuriousMind it is done.

It's been many years since I've played gen 1

I always wanted to have the choice of having an eevee as my starter, too

@ACuriousMind isn't that what you do in XD?

XD?

pokemon XD

bulbapedia.bulbagarden.net/wiki/…

hmm, now what?

go north?

don't I need a map and pokeballs and shit?

@0celo7 I never had a console, so I never played that

@0celo7 You'll get it. Just walk into the grass to get slaughtered

slaughtered?

I just slaughtered a pidgey if you know what I mean

Oh

right

I think you get the balls only after you've made that first trip?

It's been a while

oh don't I have to get the package?

Oak's package

Yeah, get that package and return it, then he'll give you his balls

Or something

the graphics are shit

oh it's a parcel, not a package

what's the German word for parcel

Paket just like for package, I think

Or Päckchen if it's small

wow we Americans are sophisticated in our nomenclature for boxes that go in the mail

@ACuriousMind I sent a packchen to Germany the other day

I wrote the German customs people a nice note in case they opened it

...what did you send?

I really liked your most recent answer @acuriousmind

I'm pretty surprised someone would go to grad school for theoretical physics and still ask a question like that though

@ACuriousMind samples of various oxides to the ion beam at GSI

@ACuriousMind interesting "delta function" question

why did you think of that?

have you checked e.g. Deligne et al. to see what they say about it?

@0celo7 Because it came up in the "advanced string theory" course I'm taking.

delta functions or the fact that the delta functions don't make sense?

@0celo7 That construction, and the lecturer said "It's a Poincaré dual, it perfectly well-defined" and then I said "That space is not compact" and then he grimaced and said we could still define it but we had to go on with the lecture instead of hashing that out.

@ACuriousMind Can you describe it by the level set $f(x)=c$ some regular function and then $\delta(\Sigma_p)=\delta(f-c)$ (regular delta)?

@0celo7 No, it's meant to be a form

@ACuriousMind Ohhh

@0celo7 I don't see any chapter in Deligne et al. talking about branes

Right, that's why he would say it's the Poincare dual.

@ACuriousMind Both volumes?

Yep

Dang

Oh crap rival battle and I haven't trained at all

The hint in the answer to deRham's book is nice insofar as that seems to be one of the few decent references on currents, but I'm not 100% convinced it's supposed to be a deRham current

Dunno anything about those

Looks like the best possibility currently, though

@0celo7 They're "just" distribution-valued forms (or equivalently distributions on forms)

Don't know anything about distributions either!

aaaaaand I lost a rival battle

noob

:)

@0celo7 which version are you playing?

yellow

I prefer pokemon blue.

@0celo7 Tong's notes skip so many steps...

when proving formulas and stuff

does time dilation also affect mechanical clocks? I know atomic clocks demonstrate the effects of time dilation and there has been experimental evidence of it.

@Obliv it affects all clocks

it's time dilation not atomic clock dilation

Hello

@0celo7 can you look at page 24 here.

I don't understand 2.54.

Why does the second term drop at the end?

It says "by lagrange" but I have no idea what he means.

@3075 there is no term there that drops.

The second-to-last equality is by the Euler-Lagrange equations, and the last is just the product rule.

So

If I want to solve the EFE on a weird manifold

Do I really need to do the full serious manifold apparatus

Do I need to have overlapping coordinate patches, transition maps and all that

Or

@ACuriousMind Oh I see.

Thanks.

Can I get away with just the identification of the edges of the connected sum

@Slereah What?

Which connected sum

And what do you intend to do when you try to "get away with just the identification"?

@ACuriousMind also on page 25 is $\frac{\partial L}{\partial \dot{\mathbf{r}}_i} \cdot \mathbf{n}$ from 2.54?

why is it dotted?

Why is what dotted?

dot product

where does it come from?

is that taylor expansion?

No

It's just what Noether's theorem gives you for a translation in direction $\mathbf{n}$.

I don't know how to realize that by myself.

did he

just use 2.56 in 2.54?

@ACuriousMind Just $\Bbb R^2 \# T^2$

@3075 Because he says "We can compute the conserved quantity associated to translations" directly before that?

what does GR fail to describe ?

My secret plan is to just solve the EFE on the plane - 2 disks, solve it on the cylinder

@ACuriousMind so it's the definition of the conserved quantity associated to translations?

And then say that the metric and its derivatives will be identical along the edges of those disks

Does that sound like a good enough idea

@3075 Full sentences make it easier for me to understand you ;)

Is what the definition of what?

...is 2.57^

I mean how does he know 2.57 works?

The conserved quantity associated to a symmetry $Q(s,t)$ is what 2.54 gives you. 2.57 is the result of plugging the translations as your symmetry into the last line of 2.54

exactly that's what I said I think.

just use 2.56 in 2.54?

is that right?

2.56 is just the statement that the translations are a symmetry. You can't directly "use" it in 2.54, but it is the reason you are allowed to plug in the translations as your $Q$.

I see.

Maybe I should check Visser

That's basically how he builds the Schwarzschild coordinate type wormhole

I suspect that any attempt at what I'm doing will not be very nice

Since it has basically 1 Killing vector

how do cross ratios come up in conformal field theory. I tried following and convincing myself I know what is happening without success. . . a few separate times. Is there a paper, text or something that assumes I am super stupid(almost true) and actually shows the steps to getting this not just stating the results? I have never sat in a CFT class.

i tried asking a question about this but there were some threats to close it so I am asking here

. . . . . can anyone guide me or help me think about this?

I done some googling and looked at some texts but I think I need to talk it out with someone to try to figure out what is going on

hopefully at a super simplified but still technical level

anyone there. . . .

I'm not sure, but don't the standard texts derive this?

which standard text?

Let me check di Francesco

the big yellow book scare me

*scares me

I am checking right now. Let me see if i can follow his treatment of it

Well, but if you want a "super simplified" version, you simply get stuff in terms of cross-ratios because cross-ratios are Möbius-invariants.

So since the correlators on e.g. the Riemann sphere must be PSL(2) invariant, it is clear that they can only depend on Möbius invariants formed out of the coordinates they depend on, and you can form such an invariant simply by forming cross-ratios of the coordinates.

ah I see, this is the link!

Up to 3-point functions, you get the complete form of the correlators just from this PSL(2) invariance.

The 4-point function then is the first that has an undetermined function of the cross-ratio in it, and you need to use more than just invariance to determine it

@ACuriousMind did all the current stuff get worked out?

@MikeMiller Well, let's say it is in progress...I'm not yet sure one can do everything with currents that I've seen my lecturer do with these objects (in particular I'm not sure how to make "$\delta(\Sigma)\wedge\delta(\Sigma) = 0$ in flat space" precise, given that $\delta(x)^2$ is usually ill-defined)

is there any geometrical shape in which changing the scale can create asymmetries and forming a new shape? Or is changing the scale by definition the changing of size while keeping the shape the same?

What is $\delta(\Sigma)$?

@MikeMiller The object with $\int_M \delta(\Sigma)\wedge\omega = \int_\Sigma \omega$ for $\Sigma$ a p-brane.

The physicists seem to usually call it a "Poincaré dual" and leave it at that.

Not sure what a p-brane is. In any case, currents are not enough to take wefhe products (just like you couldn't take wedge products of some sort of Sobolev differential forms and expect the products to be in the right sobolev space). If you want to take wedge products, you should either be wedging with smooth (compactly supported) forms or expandin your differential forms beyond currents.

you're a pea brain

Like, let's just think on the level of $\Omega^0$. Obviously distributions don't suffice or as you say I could say $\delta^2$ was a distribution. Perhaps you could take the stmmetric algebra on the vector space of distributions.

Oh, a p-brane is just a (p+1)-dimensional submanifold of $\mathbb{R}^{1,9}$ in this context

Sorry, sometimes I forget to switch to math when talking to mathematicians ;)

Closed submanifold? Or is $\omega$ compactly supported?

Well, surely this is math, because it couldn't possibly be physics ;-D

@MikeMiller I have not yet discarded the possibility that $\delta\wedge\delta = 0$ is just a dirty physicist trick to make a term vanish that's otherwise very difficult to show to vanish

@MikeMiller non-compact submanifold (you can just think of a p+1 dimensional plane), and $\omega$...may or may not be compactly supported depending on how regular you like your physical fields. I'd go with it not being compactly supported.

I mean, I just don't know how to integrate it. There must be some rapid vanishing assumption and some tameness assumption on $\Sigma$; being a plane would suffice for tameness.

I guess there's probably some artificial way to put a product on distributions.

It's a plane in its "ground state". Supposedly it can move and twist, but I'd be fine with understanding this construction for a $\Sigma$ that doesn't deform for now

@MikeMiller Yeah, $\omega$ probablly is silently assumed to fall off sufficiently fast at infinity for most things to converge

...except where the divergence might be physical. I've never been all too sure which things we must tame and which contain "physics" :P I think there should be nothing divergent here, though

Hm

I guess perhaps

But yeah, I'm not really sure how to have a chain-level wedge product in the theory of currents. I'm not sure if that's in G&H.

I could have a kind-of angular Killing vector

@MikeMiller deRham essentially defines the notation $\int T\wedge \omega$ by $T[\omega]$.

Except it's appolonian circles instead of circles

So it might just be abuse of notation

Well, angular for the wormhole part, appolonian circles for the asymptotic spacetime part

deRham also defines a more general wedge product on currents when the supports overlap in specific ways (and in some other cases)

But I have not yet found something that would allow me to wedge a current to itself

Ah, that makes sense. I agree if you have some conditions on supports.

Yes, I am rather skeptical of that.

Have you considered tackling your lecturer and not moving until he makes it make sense?

annoyingly, the lecturer is at a conference and I can only ask him in two weeks what he meant by that, so I thought I'd try soliciting explanation from elsewhere

@MikeMiller That's plan B

I'm pretty thin, though, so I might starve before he makes sense :P

Sorry I couldn't be of much help. Since most things are salvageable with enough cheating and tears, I suspect this is salvageable. But I also suspect the intended audience was not the weak-willed mathematicians like I.

Nowhere I've ever been.

They say things like "all y'all", though.

"Close" for clothes is pretty common though.

@MAFIA36790 I've been at sea, at a conference ;-)

Hey guys i need urgent help please

Lets say we have a spring and we put a projectile in it

we pull the string by x

and then we release it

What will be the relationship between the distance pulled and the speed of the projectile

i love tal as a chess player btw ^^

@MikhailTal if it obeys hooke's law, $F = -kx$ and $a = -\frac{kx}{m}$ integrate with respect to x, $v\frac{dv}{dx} = \frac{kx}{m} \to v dv = \frac{kx}{m} dx$ integration yields $\frac{v^2}{2} = \frac{kx^2}{2m} \to v = x\sqrt{\frac{k}{m}}$

I love tal too, definentely my favourite player

Lets see here

Is is supposed to show up as a formula?

cause it shows up as $\frac

Rather than as a latex type thing

@obli

you need to use chatjax

@Obliv

meta.stackexchange.com/questions/97938/…

make it a bookmark and just click it once to render latex in chatrooms

@ChrisWhite I've never noticed it. Though admittedly, it is easy for a native speaker to not be consciously aware of subtle things they are doing with language.

hey @dmckee how drastically would the universe change if the speed of light was altered slightly?

Thanks a lot, so it means they are proportional?

yes directly proportional

Awesome thank you so much

I have my ia to hand in

If you know what IB is

but my first one failed and i had to come up with something really quickly

idk what that is

@Obliv IB is the international high school diploma

oh

m is the mass of the projectile right>

@Obliv That's not really the right question to ask, because the speed of light is a scale. Instead ask about dimensionless constants.

Perhaps you are really asking what happens is the fine structure constant changes. In that case chemistry is different.

The simplest effects are linear in the fractional change, but there are thresholds.

Which elements are ferromagnetic and how strongly depends on the fine structure for one thing.

well I mean, if the speed of light changed, all of the equations involving it would change, right?

@mikhail yes

@Obliv Let me try again. Science fiction author have occasionally played around with the idea that they could put their characters in a system where everything is the same except that $c$ has changed, right?

yes

@dmckee I know the magnetic permeability constant would change (?). I'm wondering if it is possible that the speed of light(photons) are geometrically dependent on the curvature of spacetime and that the curvature changed at some point (5 billion years ago) when the change in the expansion of the universe occurred. Just wondering if this thought would be immediately crushed by something intuitive (like if light's speed changed, something dramatic would happen)

So here is the question: how would you tell that this thing had happened?

@Obliv the speed of light is a dimensionful value. It's not clear what it means to say that the speed of light "changes" because you need to fix a unit system to talk about its numerical value and, for instance, our current unit system fixes the speed of light to have a particular numerical value.

if it was dependent on the curvature of space time and we had extremely powerful measurement devices, it would be reasonable to assume we could detect slight deviations from the speed of light 100 years ago or something maybe @dmckee

You'd try to measure the speed of light, right? But doing that relies on tools made of atoms and so you have to assume that the tool are still doing what they did before.

There is a real problem with circularity there that the sf guys simply ignore.

It's better to ask about dimensionless constants. $m_e/m_p$, $\alpha$, and so on.

well it's only dimensionless because we treat it as dimensionless to make calculations easier, right? @acuriousmind but it should have dimensions distance/time,no?

Because they are ratios, and they still have meaning even if your tools have been altered by the change.

OH u said dimensionful

@Obliv ...what? I never said "dimensionless"

I'm not seeing how the tools would also change @dmckee unless the way you measured the speed of light depended on the speed of light somehow

See posts like physics.stackexchange.com/questions/21721/… and physics.stackexchange.com/questions/2725/…

the first post says you can't change newton's constant because of philosophical reasons. I don't understand it. If the gravitational constant changed by 1/10^100 an order of magnitude, how does that screw us fundamentally?

@dmckee Is the reasoning that is shared by all of these constants: our tools of observation are dependent on what we are observing?

@ChrisWhite Nowhere I've ever been.

Actually...I might do it.

But I speak mangled German-English.

@0celo7 Do you? your english seemed fine to me.

really american.

nope

ask @ACuriousMind

@3075 It's true, he sounds like a barbarian

@BernardMeurer quest?

@3075 I was waiting for you to say that :p

@3075 No change, Deny on CS, Application on Physics

looking bleak.

@BernardMeurer didn't you say they will call you in a few days?

@3075 Yeah... I think CS is foobar for good, which was expected; I'll hit Brian an email anyway

o/

@Danu /o/

\o\

@ChrisWhite orange?

@ChrisWhite wut

Meh. Chatjax limitation kill a fine joke.

@dmckee I'm curious. Post a pic?

Something like this:

@dmckee I didn't get it :(

hint: Village People.

YMCA

AAAAAAAAAAAAAAAAH

That's too old for me man

Earworms like that don't have an era: they propagate.

@dmckee Nice, nice.

@dmckee huh?

@ChrisWhite I think that's supposed to be an illustration of the propagation of earworms

I'm not clicking that link, though :P

@ACuriousMind Erdinger is a good beer man

Erdinger Dunkel that is

@ACuriousMind it's a how-to on raising beetles as food for critters

@BernardMeurer It's fine I guess?

@ACuriousMind Yeah I mean like, it's good, not amazing, but it's great value

@ChrisWhite Propagating worms?

I don't know man...

I've been writing out a SUSY computation all day... I don't even know anymore.

@ChrisWhite oh, yeah. I take it you're in town?

@ChrisWhite oh yeah, how's the thesis going?

Haven't asked in a while

@ChrisWhite fun :-P well, if you want to meet up, let me know, we could probably work something out

Man, @ChrisWhite, I must've spent that month in Princeton so close to your office. I was in the basement of Peyton, too.

I was stuffed away all the way in the middle, where nobody ever comes

Next to some kind of experimental room (?!)

It was pretty depressing there.

God, I wish I'd have been less shy, so that I'd have actually talked to people there.

@Danu Because of the experimental room?

Marth's side smash is OP

I don't remember.

@0celo7 No it's not :P It's so slow!

@Danu well you're probably a good player

@Chris y'know you could come out ;-)

I got the full expression for the SUSY action for a general Kaehler potential

PRAAAAISE DA LAWD

I wasn't really planning to be in that part of campus much but I do have some free time

@Danu And now that you have it, what are you going to do with it?

NOTHING :D

It was a terrible exercise

I've been doing this since last afternoon.

(last being the one a few hours ago :P)

Your perseverance is...impressive ;)

I'd have said "fuck this" hours ago :P

Well, it helps that I've got a midterm (?!?!) on this next week, and the TA told me this will be on the exam.

Urgh

Also, "midterm" in Germany?

What sorcery is this?

Hence the (?!?!)

It's a Dutch professor

Oh well, in some sense it's nice; it counts for 50% of the grade.

Takes off some pressure come finals---I'm doing 6 courses now so I'll need all the free time I can get by then.

Taking this SUSY course was probably a bit of a mistake/time-waster; We're doing pure PHENOMENOLOLOLOLO :( Not my type of thing

@Danu The terminology is so ridiculous...how am I supposed to remember what kind of stuff is inside a "hypermultiplet"? They could really have tried to come up with more descriptive names :P

Oh, that ain't shit

This calculation...

I can say, with confidence, that it has been the hardest one in years, in terms of tricky signs etc.

...but yes, everything is bad.

So, so, so many different indices.