The h Bar

Steven Sagona

12:09 AM

was my question in the question pool answered already?

the "what circumstances led you here?" question

heather

similar questions were answered throughout the transcript, yes.

Steven Sagona

thanks ill start from the top then

ACuriousMind

@StevenSagona Daniel has bookmarked the AMA as a conversation here in case you don't want to scroll up to get to the top

Steven Sagona

@ACuriousMind thanks but the deed is done

@heather I'm surprised you aren't expressing any boredom for your classes. I was in a similar situation as you when I was your age (but before the internet was as good of a tool for learning) and my math classes drove me insane. The only thing getting ahead really ended up doing for me was giving me an ego that'd I'd annoy the students around me.

anonymous

12:25 AM

@StevenSagona Drove you insane because the maths classes were too easy for you ? Or am I misinterpreting ?

Steven Sagona

@anonymous yeah but mostly because I already learned the material so it was pointless

things are really simple at the beginning - they just make you do algebra over and over again

0celo7

@ACuriousMind You here? I need to bounce ideas off of someone

anonymous

@StevenSagona True. That would be definitely boring but I am not sure it would drive one insane :). For me it would be a good time to sleep in class without getting noticed :P. I've faced such situations before :D

However, they sort of help in revising the concepts....

user228700

@DanielS: About that drink. In the first place, I must warn you that I made it for three people and it was mostly guess-work. I mixed the following ingredients in a bowl using an electric mixer: 3 cups of milk, 1 spoon of instant coffee powder, 5 tablespoons of cocoa powder, 3 "pastries" 5 spoons of sugar and 10 ice cubes. The mixture became quite frothy and warm so I kept it in the freezer to chill for about 10 minutes before gulping it down.

user228700

Pastry: Choco pie--

user228700

12:31 AM

> A choco pie is a snack cake consisting of two small round layers of cake with marshmallow filling with chocolate covering. (@JohnR: See here <---)

anonymous

Who mixes choco pie in a drink ? :P

The marshmallow would affect the flavour

user228700

Me, that's who! :-) It was delicious.

anonymous

LOL :D I will try doing that. Choco-pie+Chocolates+Wafers+Pastries+Fruit Cakes + Coffee + Hot Chocolate... in mixer and make a milk shake out of it :'D

0celo7

that sounds healthy

obe

yum

anonymous

12:43 AM

I think cola and wine might be good addition to your ingredients :') @Kaumudi.H

I've tried coffee with hot chocolate before

But your recipe is a class apart :P

DanielSank

12:56 AM

@Kaumudi.H sounds good

0celo7

@ACuriousMind Suppose I have some group action on a pointed space. Can I decompose the group as $F\rtimes T$, where $T$ acts transitively and $F$ leaves the basepoint fixed?

Er, I mean, where $T$ is any subgroup that acts transitively.

ACuriousMind

@0celo7 What about the standard $\mathbb{Z}_2$ action on the sphere by the antipodal map?

0celo7

@ACuriousMind I want the action of $G$ to be transitive, sorry

ACuriousMind

@0celo7 Then consider any transitively acting $G$ on the sphere and the action of $G\times\mathbb{Z}_2$.

Well, I guess your question trivially has the answer "yes" for $F$ the trivial group and $T$ the original group...

0celo7

$T$ is given

I have some transitive action $T$

$T\subset G$

ACuriousMind

1:03 AM

Then you can't, by my example. $G\times\mathbb{Z}_2$ acts transitively but you can't decompose it into a product with $G$ such that the second factor leaves any point fixed.

0celo7

What is the $G\times\Bbb Z_2$ action?

flip, then apply $G$?

ACuriousMind

Hm, you're right, two separate group actions don't give an action of their product

0celo7

@ACuriousMind The context is that I'm trying to compute isometry groups (rather, follow Wolf's proofs) of Minkowski spaces and the model constant curvature spaces

He writes down $I(\Bbb R_s^n)=L\cdot V^n$, where $V^n$ is the translational group and $L=\{f\in I(\Bbb R_s^n):f(0)=0\}$

he uses $\cdot$ for the semidirect product

So I'm wondering how he got that

A priori, why should the isometry group be of that form?

ACuriousMind

Ah, here's a silly counterexample: Consider the circle, and the action of $\mathrm{U}(1)$ on it, which is clearly transitive. You can also define a transitive $\mathrm{U}(1)\times\mathrm{U}(1)$-action just by composing the two rotations. As everything is commutative, there's no problem here, but clearly you cannot decompose the $\mathrm{U}(1)^2$ in your fashion w.r.t. to the $\mathrm{U}(1)$ because nothing but the identity leaves any point fixed.

@0celo7 Given any isometry $f$, you can compose it with a translation to achieve $f(0) = 0$.

Nothing to do with general group actions, everything to do with your specific situation :P

0celo7

what?

ACuriousMind

1:14 AM

For any $f\in I$, we have that $g(0) = 0$ for $g = t_f \circ f$ where $t_f$ is the translation that shifts $f(0)$ to $0$.

0celo7

Hmm. I guess the semidirect product is needed for if you decompose $f\circ f'$ in that way?

ACuriousMind

Now apply the inverse translation to get $f = t_f^{-1}\circ g$ and therefore the claimed decomposition.

@0celo7 Yep, since the second isometry will act on the translation coming from decomposing the first.

0celo7

Right. So for the dS/AdS type spaces I have a transitive action of $O^s(n)$ by isometries. Pick some basepoint $x$. So given an isometry $f$, I find a $t\in O^s(n)$ that moves $f(x)$ to $x$. Now, the group leaving $x$ invariant is just $O^k(n-1)$, where $k=s$ or $s-1$. But that group is the "maximal group preserving the inner product on $T_x\Sigma$" and I don't know what the significance of that is

I think that means that if $l\in I(\Sigma)$ leaves $x$ invariant, then $dl\in O^k(n-1)$

So if $L$ is the stabilizer of $x$, then...hmm

$L\subset O^k(n-1)$

@ACuriousMind why is algebra so hard

Ah! An isometry is determined by its value and first derivative. But the value is fixed since it's a stablizer

So we're looking at $L\to O^k(n-1)$ and that's injective

So it's a subset in a natural way

Then we get $I(\Sigma)\cong O^k(n-1)\rtimes O^s(n)$

but the action is just by multiplication, so we can absorb that factor

vzn

1:31 AM

in theory salon, 1 min ago, by vzn

Hundreds gather in Copley to ‘stand up for science’ / bostonglobe

in theory salon, 8 mins ago, by vzn

The geeks shall inherit the earth: The smart and funny signs from Boston's Stand Up for Science protest clever enough to confuse even scientists / dailymail

↑ science + politics! aieee o_O

0celo7

2:40 AM

@Danu Getting into the good stuff. If $M^{2n}$ is complete and connected with constant sectional curvature $K>0$, then $M$ is isometric to $S^{2n}$ with radius $1/\sqrt K$ or to $\Bbb RP^{2n}$ with the same radius.

@ACuriousMind hola

@Secret hi

Secret

Hi, I missed the AMA cause Japan time is 8:00, where we are still asleep. Glad to see it went well

0celo7

@SirCumference hello

Sir Cumference

2:59 AM

Howdy

@0celo7 Quick question, this is complete nonsense, right?

0celo7

i think so

Why?

Sir Cumference

@0celo7 It relates back to that paper I'm reading, about pseudo profound bullshit.

Secret

[Attempt at localising unicorns] Suppose there exists something call antiferromagnetic force and suppose it act by enforcing antiferromagnetic order to whatever is being interacted. Now suppose there exists something called iohydrogen. We might expect that it is a state of hydrogen that tend to antialign any hydrogen it encounters in the way. Now suppose negatively curved means concave. There are no known electrostatic configuration that can induce the reversal of electric charge so that

,hydrogen becomes antihydrogen and vise versa. Therefore we concluded iohydrogen does not exist (as charge reversal of hydrogen just turn it into antihydrogen, and it does not preferentially antialign spins of surrounding (anti)hydrogens

0celo7

3:15 AM

@SirCumference I bet I could come up with some pseudo profound bullshit

@SirCumference do you want me to try?

Sir Cumference

@0celo7 I'm open ears. I dealt with those kinds of people hijacking my tours back when I worked in the museum of natural history.

A bit funny sometimes, concerning at others

Secret

Though if we can engineer a system such that it antialign spins wrt itself and the effect is proportional to some function of distance, then such system is very useful in introducing skymyrions

0celo7

@SirCumference If you interpret temperature as imaginary reciprocal cyclic time, then you get get the Hawking temperature and Hawking-Bekenstein entropy of a Schwarzschild black hole without ever mentioning quantum mechanics. Thus there is a strong link between quantization and temperature and time that we don't understand...it's not just a coincidence.

Sir Cumference

@0celo7 I just showed this to my friend, not telling him it's BS. He said "well, hopefully we can figure it out"

Tbh I would've fallen for it too XD

0celo7

@SirCumference The first sentence is correct.

Sir Cumference

3:21 AM

@0celo7 Do I trust that?

Because you might be messing with me

0celo7

You don't, but I can show you.

It's a standard trick in thermal quantum field theory

Secret

how is temperature even cyclic?

0celo7

But you apply it to classical GR and somehow it works

@Secret the time is cyclic

Secret

Yeah I know but a cyclic temperature will mean it is somehow periodic in the system. That cannot happen unless it has suitable boundary conditions?

0celo7

I'm not claiming it makes sense

let me get out my notes

@Secret the proper way is to use boundary conditions, yeah

Sir Cumference

3:24 AM

Brb

0celo7

@Secret do you know about the partition function in quantum stat mech?

Secret

stat mech yes, quantum stat mech no unless you mean something to do with the way entropy of a quantum state is defined to be the trace of some oservable (something something)

0celo7

@Secret it's $Z=\mathrm{tr}\, e^{-\beta H}$, where $\beta= T^{-1}$

Secret

I see. That makes sense as the Hamiltonian is the total energy of the state and we are summing up all these contributions which are the energy of the eigenstates

0celo7

so that's $\sum \langle n|e^{-\beta H}|n\rangle$

Now if you write $\beta=it$, then you have $\sum\langle n|e^{-iHt}|n\rangle$

but that's just a path integral going from $|n\rangle\to|n\rangle$ in time $t$

so the states always end up where the begin (cyclic time)

Now let's put this idea to use. Consider the standard Schwarzschild spacetime
$$g=-\left(1-\frac{r_\text{S}}{r}\right)dt^2+\left(1-\frac{r_\text{S}}{r}\right)^{-1}dr^2+r^2d\Omega^2$$
and expand it near the horizon
$$g\simeq -\left(\frac{r-r_\text{S}}{r_\text{S}}\right)dt^2+\left(\frac{r_\text{S}}{r-r_\text{S}}\right)dr^2+r_\text{S}^2d\Omega^2$$
Change variables to $\rho^2=4r_\text{S}(r-r_\text{S})$. Then $\rho d\rho=2r_\text{S}dr$, so that $\rho^2d\rho^2=4r_\text{S}^2 dr^2$ or $(r-r_\text{S})d\rho^2=r_\text{S}dr^2$. Plugging this into $g$, we find that the spacetime near the horizon is desc…

where in the last step, we set $t=-i\tau$. Formally, this is known as a Wick rotation. If we now change variables again, setting $\tau=2r_\text{S}\psi$, we obtain
$$g\simeq d\rho^2+\rho^2d\psi^2+r_\text{S}^2d\Omega^2$$
This is the metric of a plane in polar coordinates and and a sphere with radius equal to the Schwarzschild radius of the black hole. So near the horizon, we have the homeomorphism
$$\text{Schwarzschild}\cong E^2\times S^2(r_\text{S})$$

One can picture this as a sheet with a black hole-sized sphere at every point. Since $\psi$ is an angular coordinate, it has periodicity $2\pi$. The Euclidean time $\tau=2 r_\text{S}\psi$ has a recurrence period $\beta =2 r_\text{S}(2\pi)=4\pi r_\text{S}$. Thus, according to the previous discussion, the electromagnetic field propagating near the horizon thinks it is living in a heat bath with temperature
$$T_\text{H}=\frac{1}{\beta}=\frac{1}{4\pi r_\text{S}}=\frac{1}{8\pi GM }=\frac{\hbar}{8\pi GM k}$$

@SirCumference

Crap, I meant to put this at the beginning:

Consider a quantum field, be it the field of a photon, an electron, or whatever, propagating in spacetime. Suppose that it discovers that time is actually imaginary and cyclic. The field is fooled into thinking that it is living in a heat bath, with the temperature determined by the inverse of the recurrence period $\beta$ of this bizarre imaginary time.

Secret

3:56 AM

I wonder if the imaginary cyclic time means the expectation value of the energy is due to the amplitude oscillating between an earlier and later time. Also, that the recurrence time is a function of the schwartchild radius, I wonder it has some to do with the space and time axes interchange at the horizon?

0celo7

4:17 AM

@Secret Hmm.

I'm not sure!

jyotishraj thoudam

4:42 AM

Has anyone seen this question?physics.stackexchange.com/questions/313453/…

DHMO

@JohnRennie Please ping me when you come back; I have some questions not related to physics

Sir Cumference

@0celo7 I think we both know that's way above me

Secret

[random question] is it possible to boost to some noninertial frame such that the magnetic field of the spin becomes an electric field in this new frame. What will such electric field look like?

Secret

5:04 AM

6

Q: How do I calculate electric fields due to currents of magnetic dipoles?

Short version of my question: Do dipole currents cause fields? I think currents of aligned magnetic dipoles cause an electric field, but I don't know how to calculate this field except in the simplest of cases. I'd like to know how! Full version of my question: Suppose I have a wire (or pipe) ...

The problem: what if I had a stationary electron, then we cannot model that as dipole currents

user228700

6:22 AM

@JohnR: Morning :-) Mangoes and salt?

user228700

@anonymous Lol, thanks :-P

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