The h Bar

Obliv

3:00 PM

@acuriousmind I see. That's if you consider attraction to be positive and repulsion to be negative lol

0celo7

@ACuriousMind Is it still arbitrary when we do weird SUSY shit so that proton charge = - electron charge?

Or is there some absolute definition of charge in SUSY

ACuriousMind

@0celo7 What?

SUSY is even more horrible in the sign conventions you have to make

0celo7

Hmm

ACuriousMind

And yes, the sign stays arbitary.

But I don't see what SUSY would have to do with the proton or electron charge

Slereah

Isn't the sign of the electron charge basically the basis you pick for $U(1)$?

0celo7

3:07 PM

@ACuriousMind Oh, not SUSY

I meant SO(10)

I think.

Maybe SU(5)

One of those grand unified theories.

Slereah

Is SU(5) still going on

ACuriousMind

I think it's quite dead

0celo7

Why?

Slereah

Like the particle spectrum is too light or something

ACuriousMind

We should've seen some things by now that we didn't see, I think

Slereah

3:09 PM

None of those light particles have been produced

Did u know

Emilio Pisanty

@DavidZ so it turns out that this old question

13

Q: What sort of experiment would directly test time reversal invariance?

I guess the title says it all: how could/would you experimentally test whether our universe is truly time reversal invariant, without relying on the CPT theorem? What experiments have been proposed to check this? Have any of them been performed? I know that there are indirect tests of time rever...

Slereah

You could almost do electroweak theory with $U(2)$

Emilio Pisanty

has had a different answer for like two-thirds of the time since it was posted

Slereah

Because it's like $U(2) \approx SU(2) \times U(1) / Z_2$

Emilio Pisanty

Kinda makes you wonder what other old semi-canonical questions need revisiting

0celo7

3:14 PM

@Slereah Shit, it really is $\Bbb R^2\times S^2$

Slereah

It is

0celo7

Unless...

Slereah

It's more complicated for Kerr because the singularity is spacelike

0celo7

@ACuriousMind $\Bbb R^3-\{0\}\approx \Bbb R\times S^2$?

seems reasonable honestly

ACuriousMind

If $\approx$ denotes "homotopy equivalent" then yes.

0celo7

3:16 PM

Isomorphism in the smooth category @ACuriousMind

"diffeomorphism"

ACuriousMind

Oh. Yes, I guess that's also true

0celo7

@Slereah HA

Slereah

Well

PROOF???

0celo7

@ACuriousMind I take it back, you're not terrible at geometry

@Slereah Of what?

ACuriousMind

Since removing the point you have that the polar coordinates from $\mathbb{R}_{>0}\times S^2$ cover it and there is a diffeomorphism $\mathbb{R}_{>0} \to\mathbb{R}$.

0celo7

3:17 PM

@ACuriousMind Yes, I know.

I needed someone with pedigree to confirm.

Slereah

On the other hand

Hm

0celo7

Now, is that good for maximally extended too?

Slereah

It's hard to think about Schwarzschild casually

0celo7

I guess look at a Penrose diagram

Slereah

Because it has that weird thing where $t$ becomes spacelike beyond the horizon

0celo7

3:19 PM

@JohnDuffield

Slereah

Nothing but woo

0celo7

It seems to me that the Penrose diagram is $\approx\Bbb R^2$

Slereah

Yes

They suppress angular coordinates

0celo7

And there's a sphere at each point

so it's definitely $\Bbb R^2\times S^2\approx \Bbb R\times(\Bbb R\times S^2)\approx \Bbb R\times(\Bbb R^3-\{0\})$.

Slereah

Well I suppose it makes sense, from Schwarzschild coordinates

0celo7

3:21 PM

@NeuroFuzzy There you go.

Slereah

I wasn't too sure how much you could say from them since they don't cover everything nicely

0celo7

I just picture the spacetime in my mind

but the factor of $\Bbb R$ isn't the Schwarzschild time

it's the

GEROCH TIME FUNCTION

Slereah

:o

0celo7

from the GEROCH THEOREM

So, let's calculate the topology of Reissner-Nordstrom

Slereah

Not too sure about that one

0celo7

3:24 PM

Slereah

I want to say "same", but I dunno

0celo7

Honestly...that looks like $\Bbb R^2$ as well.

@ACuriousMind what do you think

it's certainly an open set of $\Bbb R^2$...

ACuriousMind

I have no idea what you two are talking about

Balarka Sen

Me neither.

0celo7

Is a simply connected open set of $\Bbb R^2$ diff to $\Bbb R^2$?

Balarka Sen

3:26 PM

Yes.

0celo7

Proof?

Slereah

the reader can show

Balarka Sen

Riemann mapping theorem.

This is hard.

0celo7

Could one use Jordan-Brouwer?

Balarka Sen

No idea how you'd do that.

0celo7

3:27 PM

consider the boundary of the thingie

then it encloses a piece homeomorphic to $\Bbb R^2$...

because it's homeomorphic to a disk

Balarka Sen

Oh, you mean Jordan Schoenflies.

0celo7

We don't even really need smooth stuff here

@BalarkaSen ...maybe?

Balarka Sen

Boundary of a simply connected set in R^2 need not be a circle, though.

0celo7

@BalarkaSen open set

what else could it be

Balarka Sen

Chuck the x-axis out.

0celo7

3:30 PM

ah.

Yes, in my black hole case it's not bounded either

Balarka Sen

Chuck a cantor set out.

0celo7

that set continues on to infinity in both directions

@Slereah Well, it's certainly reasonable that it's $\approx\Bbb R^2$ just by stretching it

Slereah

I would beware

It is full of singularities

0celo7

@SlereahBut doesn't RN have like ring singularity shit or is that Kerr

@Slereah but the singularities are on the boundary

they're not a part of the open set

they would be contained in the closure

Slereah

mb

Kerr has ring singularities

0celo7

3:33 PM

Should I post a thing on it on PSE?

Slereah

Since a point singularity can't carry angular momentum

0celo7

I'd get a nonanswer from Timaeus and a diss by JD

@Slereah what does that even mean

Slereah

Points can't rotate on themselves, yo

0celo7

...electrons

we all know they spin, see Einstein-de Haas

Balarka Sen

Not points.

ACuriousMind

3:35 PM

@0celo7 ::twitches::

0celo7

@ACuriousMind yes?

Obliv

I would have thought @0celo7 was better than that.

Balarka Sen

But I'll duck out since I am not a physicist.

0celo7

I'm genuinely confused

I was trolling but people seem to be confused by my non-trolling too

Slereah

btw

there is a nice little proof that, if you allow for torsion

0celo7

3:36 PM

Proof?

Slereah

A black hole with non-0 spin density has no singularity

0celo7

I'll write a question on PSE

The troll answers will be fun

Slereah

Since a 0, 1 or 2D singularity carries no torsion

0celo7

what does it mean for a singularity to carry torsion

I need HE

@ACuriousMind Dammit, I need to start carrying HE with me again

@Slereah haha, holy shit HE references HE

@Slereah what are the various RN thingies called

maximally extended, ...?

er, no

not maximally extended

there's like $e^2>m^2$, etc.

@Slereah what are those called

@Slereah when do you get a ring singularity for Kerr?

clearly $\Bbb R^3-\text{ring}$ is not $\Bbb R^3-\{0\}$.

@ACuriousMind ...right?

@Slereah Is "ring" just some embedding of $S^1$?

ACuriousMind

3:55 PM

@0celo7 Right. The one with the ring is not simply connected, if I understand what you mean correctly.

0celo7

@ACuriousMind Indeed.

I would ask this question on PSE but I highly doubt I'll get an answer

Physics Meta

0

Q: Problem with questions

About five minutes before I posted this, a question came up in the feed asking users to come up with innovative project ideas for a highschool project with a photoresistor. The question had no research, and it was posted originally without a homework tag. While this question was deleted (maybe a ...

discussion homework policy

John Rennie

@PhysicsMeta personnally I like the idea of an addon that electrocutes the OP just before they post crap.

We could use that in the chat room too, though it might decrease the population a bit too much.

0celo7

@JohnRennie ah, can you answer my question

@JohnRennie savage.

John Rennie

:-)

@0celo7 What question? I almost certainly can't answer, but just for a laugh I can attempt it ...

Slereah

4:09 PM

Extremal, @0celo7

Obliv

@JohnR I find something oddly ironic about a supposed middle-school student complaining about high-school students asking poor homework questions :^)

John Rennie

@Obliv :-) People take things very seriously when they're young. It's only when you get to my advanced age you decide it's much more fun to just bugger about :-)

0celo7

@JohnRennie I have a conjecture

that all black holes are diffeomorphic to $\Bbb R^2\times S^2$.

Obliv

@0celo7 what's $S^2$

John Rennie

@Obliv 2-sphere

@0celo7 silly question, but is Minkowski space diffeomorphic to $\Bbb R^2\times S^2$?

0celo7

4:18 PM

@JohnRennie Don't think so.

because $S^2\not\approx\Bbb R^2$.

John Rennie

Well the four standard black holes are all diffeomorphic to Minkowksi space

0celo7

No.

John Rennie

Why not?

0celo7

you have to remove the singularities from the manifold

John Rennie

Yes, so?

0celo7

4:20 PM

wait wait wait

What is the Penrose diagram of Minkowski spacetime

@JohnRennie fiddling with lasers, give me a minute

Suppose I take $S^2$, and for some reason I remove the north pole

then the resulting space is not longer diffeomorphic to $S^2$

and it's really a myth that Schwarzschild becomes Minkowski in the limit $M\to 0$ @JohnRennie

it becomes Minkowski except for the point where the singularity was

(actually it's a 1D submanifold I think)

John Rennie

@0celo7 leaving aside time for the moment, I can describe space with the usual polar coordinates $r$, $\theta$, $\phi$. And constant $r$ is the submanifold $S^2$. So why can't I foliate all of space as $R \times S^2$?

ACuriousMind

@JohnRennie The origin.

0celo7

Because $r=0$ is inadmissible

it's not a diffeomorphism there

John Rennie

Aaaaaaah. Gosh you mathematicians are sneaky.

0celo7

not even a bijection

I'm not a --- whatever

Slereah

4:27 PM

Penrose for minkowski is a diamond

0celo7

@JohnRennie have you never heard of this before?

the metric in polar coordinates is $dr^2+r^2d\theta^2$

so the volume element is $r$

And at $r=0$ there's a singularity!

But that's just a coordinate singularity

But this also shows that $\Bbb R^2\not\approx \Bbb R\times S^1$.

(for one, they have different fundamental groups)

@ACuriousMind Is their cohomology different too?

de Rham

John Rennie

@Slereah a diamond or a triangle i.e. half a diamond?

0celo7

Well, now I'm not convinced that Penrose diagrams are global charts

Slereah

I don't think it matters all that much, since it's just a conformal diagram, but usually diamond

Penrose diagrams are just a spacetime conformal to the one you got, where you bring infinity to a finite distance

John Rennie

Wouldn't a diamond imply an extension to Minkowski space analogous to the maximal extension of the Schwarzschild metric?

0celo7

4:31 PM

but my argument would then show Minkowski being $\Bbb R^2\times S^2$ as well

which is wrong

Slereah

Not really?

It's flat space

You can't really extend it

all geodesics already go ALL THE WAY

baby

John Rennie

@Slereah That's my point

ACuriousMind

@0celo7 Yes.

0celo7

@ACuriousMind What is the cohomology of $S^1$?

Then you apply that one theorem

...Mayer-Vietoris?

Been a while since I read Lee's cohomology shit

ACuriousMind

@0celo7 Use Poincaré duality and that the homology of spheres is $H_n(S^n) = \mathbb{Z}, H_0(S^n) = \mathbb{Z}$ and zero otherwise. Then use homotopy invariance to contract the $\mathbb{R}$ to a point.

0celo7

4:34 PM

...what

ACuriousMind

I guess you could also use Künneth to compute the cohomology of the product, but that's overkill.

0celo7

Oh, Kunneth

that's what it is

Slereah

Is there a word for "manifold where between two points, there is only one geodesic"

0celo7

@Slereah Hmm

I haven't heard of it

Slereah

Basically axiom 1 of Euclid

I think that means that the normal neighbourhood can cover the whole manifold?

0celo7

4:37 PM

I guess

@ACuriousMind can you please say that slower

Slereah

I think that in those cases, the Hadamard form once renormalized is entirely determined by $\varpi_\gamma(x,y)$

0celo7

I know that all cohomology groups of $\Bbb R^2$ are trivial

ACuriousMind

@0celo7 Nope. The zeroth isn't.

0celo7

@ACuriousMind what is it

ACuriousMind

@0celo7 $\mathbb{Z}$.

Or $\mathbb{R}$, if we're doing real coefficients

John Rennie

4:40 PM

Phew, they've gone off on an extended discussion about cohomology so I'm off the hook :-)

0celo7

@JohnRennie this is how GR gets done

Slereah

@ACuriousMind Winding number?

ACuriousMind

@Slereah ?

0celo7

@ACuriousMind Oh, I'm literally retarded

Slereah

Trying to remember how the cohomology works :p

0celo7

4:42 PM

$H^0$ counts the number of connected components

@ACuriousMind Ok, so let's go over this again

We want to calculate $H^k_{dR}(S^1\times\Bbb R)$.

Now you're saying $S^1$ and $S^1\times\Bbb R$ are homotopy equivalent?

@JohnRennie please unstar

that was in very poor taste

John Rennie

@0celo7 done

0celo7

@Slereah that's the fundamental group...

@ACuriousMind why isn't cohomotopy a thing

Slereah

All topology looks alike to me

Obliv

5:00 PM

nvm

ACuriousMind

@0celo7 Who says it isn't? (Google "cohomotopy" :P)

Slereah

I guess the simplest way to find out what the terms of the Hadamard form are is to plug them in I s'ppose

Fortunately $\Delta = 1$ in Minkowski space

That's a load off my ass

$$\Box_x \frac{1}{4\pi^2} [\frac{1}{|x-y|} + v(x,y) \ln |x-y| + \varpi(x,y)] = \delta(x-y)$$

Things simplify a fair bit

Hm wait

For the wave equation the Green function is $\delta(t-r)/r$

Is the van Vleck determinant really 1 or is it a delta

Obliv

5:17 PM

shouldn't torque on a dipole in a uniform electric field be $\vec{\tau} = 2\vec{p}\times \vec{E}$?

since the positive and negative side feel the same torque?

Balarka Sen

@0celo7 It is a thing. In appropriate context, cohomotopy is also known as cohomology.

@0celo7 Yes, $\Bbb R^3$ minus ring deformation retracts to a torus, whereas $\Bbb R^3 - 0$ deformation retracts to a sphere.

Torus and sphere are different.

vzn

5:37 PM

156

Q: What are some examples of when Mathematics 'accidentally' discovered something about the world?

I do not remember precisely what the equations or who the relevant mathematicians and physicists were, but I recall being told the following story. I apologise in advance if I have misunderstood anything, or just have it plain wrong. The story is as follows. A quantum physicist created some e...

soft-question math-history big-list

↑ via reddit/ hot network question, lots of refs to famous physics cases

...

@0celo7 you might enjoy this, its big on the radio/ top40 right now, found the deconstruction amusing/ deep/ penetrating genius.com/9271412

Balarka Sen

Cellular automatons are great stuff.

vzn

@BalarkaSen =D

ritwik sinha

http://physics.stackexchange.com/questions/266426/what-does-the-moons-orbit-around-the-sun-look-like

So does every other satellite have roughly same orbit ? Also does the moons of other planets have rouhly same orbit ?

user54412

5:56 PM

@ritwiksinha Roughly, but some of them are more exotic. Note the spiraling pattern everyone mentions is not what the Moon looks like. Well, Jupiter's orbital speed is 13 km/s, while its moon Io goes around at 17 km/s. Thus for part of its orbit Io is in fact going backward around the Sun.

0celo7

6:11 PM

@ChrisWhite Do you know what the topology of a generic black hole is?

156

Q: What are some examples of when Mathematics 'accidentally' discovered something about the world?

I do not remember precisely what the equations or who the relevant mathematicians and physicists were, but I recall being told the following story. I apologise in advance if I have misunderstood anything, or just have it plain wrong. The story is as follows. A quantum physicist created some e...

soft-question math-history big-list

Physics is not mathematics.

@ACuriousMind So do you now know something about M theory?

user54412

@0celo7 spherical?

user54412

3

A: Gravitational lensing image from merging black holes

Not even light can pass through a toroidal black hole before the hole closes up, so the question is moot. To set the stage for why toroidal black holes are interesting, I'll start with Hawking's 1972 result that "a stationary black hole must have topologically spherical boundary." That is, once ...

user54412

Unless you're including the whole thing inside the horizon too. In which case, are they supposed to be the same?

ACuriousMind

@0celo7 yes

vzn

@0celo7 did you read any of the answers?

user54412

6:19 PM

@0celo7 Halfway down the page and I haven't found any accidental examples. Apparently mathematicians think anything a physicist does is accidental :/

lucas

@Qmechanic Welcome back! I miss you! :-)

0celo7

@vzn yes

most of them are random physics things

@ACuriousMind so what is it

vzn

@ChrisWhite your use of the word "accidental" seems not to match that used/ interpreted

ACuriousMind

@0celo7 Nobody knows!

vzn

am a bit surprised you guys dont like the question/ answers, its like a big ad for physics on the math site, and youre all kneedeep in math etc, it highlights/ unifies key discoveries etc

0celo7

6:23 PM

@ACuriousMind That's a thoroughly dissatisfying answer and you know it.

ACuriousMind

@0celo7 But it is true

M-theory is the conjectured theory of which all known string theories and 11d SUGRA are particular limits.

But no one knows what that theory is

0celo7

hmm

when I am retired I will revisit this @ACuriousMind

ACuriousMind

Some of its objects - like the M2- and M5-branes - are known by finding their corresponding objects in the limiting cases and noting they are stable against smooth deformations ("BPS states")

Qmechanic

@lucas : Thanks. I'll be officially back on Sunday.

0celo7

BPS state?

ACuriousMind

6:27 PM

@0celo7 That's what one calls these states that are safe against perturbative corrections and hence should still be present in the full non-perturbative theory

They usually fulfill some extremality condition

Kind of like the masslessness of a vector boson in QFT- also an extremal mass condition - inherently protects it against acquiring mass

0celo7

@MAFIA36790 Pokemon GO is the top free app in the app store

But not a single advertisment for it anywhere

quite impressive, really

@ACuriousMind what does that mean

@ACuriousMind I don't know anything about massless vector bosons...

ACuriousMind

@0celo7 I'm not sure how to answer that briefly. There is usually a significant function of mass and charge of an object associated to that object, and there's something special about when this function becomes extremal - for example, the number of the allowed fields living on it change, and this cannot be a smooth transition, so if the object is sitting there at weak coupling it must also be there at strong coupling.

@0celo7 You do not know that all massless vector bosons are gauge bosons and protected from acquiring mass by gauge symmetry? Shame on you! ;P

0celo7

@ACuriousMind No, I'm not a physicist.

All I remember from the QFT days are pain.

QFT was enough to make me hate all of physics, do you really think I know anything about it?

I'll revisit it after my grad course on QM. If it's still beyond me I will conclude I'm not smart enough for it.

Slereah

@ACuriousMind I'm saying it's a path integral

Just to bother all the people who told me UGH STRING THEORY ISN'T AN ACTION THEORY

0celo7

@ACuriousMind Although I will say I understood Timo's notes on it just fine

But as soon as people start talking about irreps my brain aneurysms

Obliv

6:38 PM

@0celo7 is it the math that bothers you?

0celo7

I think his notes were relatively group theory free

@Obliv I do not understand representation theory.

And I've come to despise it so I never want to learn it.

QFT books make me feel like an inferior human being.

Hard math books make me feel like I have a lot to learn.

ACuriousMind

@0celo7 That's because the second-half of QFT2 is missing from them where we discussed anomalies and instantons and symmetry breaking :D

0celo7

Hard physics books just make me feel like shit.

Slereah

Isn't representation theory just a morphism between whatever group and a subset of $M(n)$

ACuriousMind

@Slereah What the heck does that even mean?

Slereah

6:40 PM

I'm saying it's an action theory

0celo7

No, according to Weinberg it's $D_{\sigma\sigma'}\Psi$ shit.

Slereah

Bam

0celo7

Or something.

Slereah

Can't prove me wrong!

Until u prove M theory

0celo7

And Zee says it's something with transformations, I don't even know.

ACuriousMind

6:43 PM

@Slereah What?

0celo7

Am I the only one who feels this way?

ACuriousMind

A single representation is a group morphism from the group to the group of automorphisms of some vector space

If you mean matrices with $M(n)$, then you have excluded all infinite-dimensional representations, in particular the entire quantum theory of the Lorentz group :P

0celo7

@ACuriousMind Why not just say a homomorphism to $\mathfrak{gl}(V)$

ACuriousMind

@0celo7 It would be $\mathrm{GL}(V)$ (we're talking groups not algebras), and how is that different from what I said?

Slereah

@ACuriousMind Operators are nothing but continuous matrices

ACuriousMind

6:46 PM

@0celo7 What way?

Slereah

That's right I said it

ACuriousMind

If you mean feeling like physicists are shit at explaining group theory, then certainly no

0celo7

@ACuriousMind Hard math is motivating but hard physics destroys the soul.

ACuriousMind

@0celo7 I'd guess other mathematicians might also feel that way!

0celo7

I'm not a mathematician.

Slereah

6:48 PM

Hm

Thinking about it

I guess the first two terms of the Hadamard form might describe like

The local behaviour of fields?

ACuriousMind

What are you talking about now?

Slereah

Since the first one is $\approx 1/r$ and the second one is $\approx \ln r$

ACuriousMind

Do I want to know?

Slereah

5 hours ago, by Slereah

So the Green's function in curved spacetime is $$G(x,y) = \sum_\gamma \frac{\Delta_\gamma (x,y)^{\frac{1}{2}}}{4\pi^2} [\frac{1}{\sigma_\gamma(x,y)} + v_\gamma(x,y) \ln |\sigma_\gamma(x,y)| + \varpi_\gamma (x,y)]$$

This

0celo7

Proof?

Slereah

6:50 PM

Well I'm trying to find the proof

It's somewhere in there

archive.org/details/lecturesoncauchy00hadauoft

0celo7

is there no reference in the book you're reading

Slereah

Not in the same notation, no