The h Bar - 2017-02-20 (page 3 of 6)

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3:00 PM

@YashasSamaga, yeah

Something which has a higher heat capacity takes in more heat to increases its temperature, so it is suitable for cooling.

@CAF You get a sum over the $i,k$ when you compute the full amplitude since there's no reason to restrict to the diagram where the gluon couples to the blue quark - to get a full amplitude for the process, you must sum over all possible colors of the quarks involved.

@Ramanujan Something which take less heat to increase its temperature is suitable for heating as its temperature goes up very quickly for relatively less amount of heat.

@YashasSamaga, and for healing

@YashasSamaga, then for heating C, is it?

I already gave the reasoning.

Read my previous message.

3:06 PM

We know, Q=msdt, so, Q is directly proportional to s. So C takes less heat..

Keep thinking...

Think for a while, you will realize which one is more suitable for heating. I have already given the answer somewhere above.

@ACuriousMind Yes, that makes sense, thanks! So I guess with the colour analogy, when I consider other diagrams with different quark colours, perhaps when i sum them all up it might be indicative that the exchanged gluon is equivalent to the 'ninth' gluon ie the one that is unphysical?

@CAF I guess you could say so

I'd watch a movie titled The Ninth Gluon.

@YashasSamaga, about third question

That is pretty easy.

I won't answer.

You are going to reason out.

I will only give hints.

3:11 PM

@ACuriousMind How about The Last Sobolev Space?

@Ramanujan You need to start thinking.

@0celo7 ::yawn::

doesn't sound as catchy to me

@ACuriousMind Splitting the Sheaf?

(for some reason Sobolev spaces remind me of the last airbender)

@YashasSamaga, what's hint!!?

I think I have synesthesia

3:13 PM

@Ramanujan clues

@0celo7 Much better, that sounds exciting!

@Ramanujan How are you going to pass the exam if you expect someone to keep answering all your questions? Nobody is going to be there to help you out on the exam day.

@ACuriousMind I rewatched The Matrix on Friday. I still have no clue what "there is no spoon" is supposed to mean.

@YashasSamaga, I will manage in exams

.

.@YashasSamaga

,if I know the answers now

huh, do you have the question paper with you? -.-

3:15 PM

, I will read them for exam

@0celo7 ...that it's not real because it's merely part of the simulation? I think it's a silly line because it could be applied to everything around them.

@YashasSamaga, I only have the old questions

@Ramanujan Use Q=ms (change in temperature) for the third question

@ACuriousMind That part I get. But why do Chow et al. quote it in their Ricci flow book?

I don't see how it applies to Harnack estimates.

...unless they are saying analysis is not real?

@0celo7 How am I supposed to know what those crazy analysts think :P

3:19 PM

@ACuriousMind Because you're crazy too?

@0celo7 I'm a very different kind of crazy :)

.@YashasSamaga

, Object C

Did you guess?

How did you come up with that answer?

@ACuriousMind what's the difference between algebraic and analytical crazy?

@YashasSamaga, Q is directly proportional to s (SHC)

3:24 PM

What does cooling have to do with Q?

@0celo7 they're probably dual to each other, but uncanonically so

Heat will be lost faster?

What about topologically crazy:P ...?

@Ramanujan No, the temperature reduces quickly even for small loss of heat.

"In all these questions, informal descriptions, accompanied by pictures, of the covering maps is sufficient, but describe in enough detail so that a mathematician would be convinced."

@ACuriousMind My topology class has abandoned rigor and reason!

3:28 PM

@YashasSamaga, oo.. However the answer is C

because C takes the least heat to decrease its temperature

which is correct

@0celo7 Completely standard for a topology class

@YashasSamaga, the last one...

@ACuriousMind How am I supposed to draw $\Bbb RP^2\vee \Bbb R P^2$??

@0celo7 Probably in the "gluing polygons together" style.

3:33 PM

"Process philosophers charge the Minkowski universe with being a static one. The philosopher of the manifold denies this charge, saying that a static universe would be one in which all temporal cross sections were exactly similar to one another and in which all particles (considered as four-dimensional objects) lay along parallel lines. The actual universe is not like this, and that it is not static is shown in the Minkowski picture by the dissimilarity of temporal cross sections and the nonparallelism of the world lines of particles. The process philosopher may ... seems to be characterist…

"Philosophers of the manifold also urge that to talk of events becoming (coming into existence) is not easily intelligible. Enduring things and processes, in this view, can come into existence; but this simply means that as four-dimensional solids they have an earliest temporal cross section or time slice."

Well, except this slice is relative to your reference frame, and in extreme spacetimes (those that are nonorientable in time, or in general, (forgot) killing vector (forgot) does not vanish), it is not even well defined.

@ACuriousMind So its fundamenal group is $\Bbb Z_2*\Bbb Z_2$, which I'm assuming is infinite?

@0celo7 Yup, the free product with anything but the trivial group is infinite

Yes

It's also easy to see a copy of $\Bbb Z$ sitting inside it

"easy"

let a generate the first Z/2, b generate the second. ab generates a Z

3:40 PM

that's what Hatcher says

Idk how I'm supposed to find the universal cover of that space without using Cayley complexes

what's the univ cover of RP^2?

maybe we'll talk about them in class

@BalarkaSen infinite sequence of 2-spheres

univ cover of RP^2 is S^2... but you seem to know what the univ cover of RP^2 v RP^2 is already

user image

so why do you complain

3:44 PM

@BalarkaSen because the way Hatcher proves that has not been discussed in class

@0celo7

what was that crazy paper

does he use Cayley complexes?

on the singularity theorem

I can tell you how to do it without those

ptep-online.com/index_files/2006/PP-06-06.PDF

3:44 PM

thx

@BalarkaSen First, I need to find two non-isomorphic 2-sheeted covers. I don't know how I'm supposed to think about this

"Hilbert’s corruption of the spacetime metric"

Very ominous

@0celo7 I'll give you some ideas. Univ cover of RP^2 is S^2, yeah? What does a basepoint of RP^2 lift to? Two points, agree?

Yeah, because the action is $\Bbb Z_2$

Righto. So, based on that idea, can you give me an "obvious" double cover of RP^2 v RP^2?

(Hint: Think of the wedged point as a basepoint)

3:48 PM

Disjoint union of two $S^2\vee S^2$?

not disjoint.

but two of those guys maybe?

So four $S^2$s wedged?

What's the double cover map, though?

Note that you want a Z/2 action on the double cover

yeah, I need it to be regular

double covers are always regular anyway

huh

why?

corresponds to index 2 subgroups downstairs; those are always normal!

3:51 PM

ah, normal in $\pi_1$?

$\pi_1$ of the base, yeah.

What about $S^2$ glued to $S^2$ at two points?

antipodal points

Bingo

What's the map?

$(x,-x)\mapsto x$?

A little unclear what that map means

3:55 PM

@BalarkaSen How would you write the map $S^2\to RP^2$ in terms of elements?

I know that its a quotient

How about $x \mapsto [\pm x]$? :)

jyotishraj thoudam

can anyone help me

@jyotishrajthoudam nope

@BalarkaSen Is that for the two copies?

jyotishraj thoudam

I have a problem in understanding an equation of motion derived in a tutorial video

I could post the link her

*here

It's intuitively clear what the map is, but I'm not able to write it down

jyotishraj thoudam

3:58 PM

please?

@jyotishrajthoudam You can ask any question you want here; whether anyone is available and willing to help you is up to them.

jyotishraj thoudam

0

Q: Equation of motion

I'm not sure this equation is the right one if we invert the direction of positive y-axis. Am I right if we assume the positive direction of the $y-axis$ downwards, the correct equation of motion would be $m\ddot{y}=mg-{\frac{\rho A C}{2}\dot{y}^2}$ as the body is accelerating downwards? Help

homework-and-exercises newtonian-mechanics drag free-body-diagram

@0celo7 Right, it is that for both copies of $S^2$'s. I would write it as $f : S^2 \times \{0, 1\} \to \Bbb{RP}^2 \times \{0,1\}$ by $(x, i) \mapsto ([\pm x], i)$ for all $i = 0, 1$, and then mention $g \circ f$ factors through $X = S^2 \times \{0, 1\} /(p, 0) \sim (p, 1)$ where $p$ are the two poles of $S^2$. $X$ is your space.

"Unless one accepts a process philosophy and thinks of the flow of time as going around and around such closed timelike world lines, it is not necessary to think that Gödel's idea implies eternal recurrence. Events can be arranged in a circle and still occur only once."

Yeah, attempting to naively foliate(?) spacetime with a CTC and one get (for that observer) a bizarre phenomenon where the event split into copies, where one move forward in time and one move backward in time, and then they recombine at some later time. In such case, the event technically occured once only, but will look ve…

NB I am actually not very sure if I can think of the most basic non spacetime cylinder CTC as a simple loop in spacetime. God, my GR background sucks too much!

Where $g$ is the quotient map $\Bbb {RP}^2 \times \{0, 1\} \to \Bbb {RP}^2 \vee \Bbb {RP}^2$ identifying two points of $\Bbb{RP}^2$ in each of the disjoint copies.

And that map $X \to \Bbb{RP}^2 \vee \Bbb{RP}^2$ is precisely your double cover.

It takes a while to translate visual ideas to words, but that's it, yeah. You got it.

@0celo7 Anyway, this is one double cover. But this is the "non-obvious" one, what's the "obvious" one?

4:06 PM

@BalarkaSen Lol, I thought this was pretty obvious

@BalarkaSen I was walking to class :)

Hey guys does anyone of you happen to know a site which does some heavy integration apart from Wolfram(cause I haven't got the paid version yet)

I certainly thought you wouldn't come up with this one!

user image

The best site

You unfold both projective planes @BalarkaSen

Right, good intuition. How about you don't unfold both?

4:07 PM

then make sure the gluing makes sense wrt. the quotient map

so that's what we just did

@BalarkaSen unfold just one?

Or rather, there's two ways to unfold an RP^2. Into an S^2 and into two disjoint RP^2's

Two disjoint $RP^2$'s?

@0celo7 Right

doesn't the covering space have to be connected?

Yeah, just RP^2 x {0, 1} \to RP^2.

You can use this to make a connected covering of RP^2 v RP^2 distinct from the one you constructed!

4:09 PM

Lol

I guess you always have a double cover like that?

Sure. Not just double, but of any number of sheets.

X x {1, 2, ..., k} --> X

right

now if this is regular, doesn't it have to carry a group action?

so what if your fundamental group has less than $k$ elements?

@Slereah thank you

@0celo7 Regular covering spaces are by hypothesis path-connected, actually

@BalarkaSen not according to my notes

4:13 PM

Well that's not right then

E --> X being regular means image of $\pi_1(E)$ in $\pi_1(X)$ is normal.

But for that you need $\pi_1(E)$ to make sense!

@BalarkaSen Hatcher does not define it like that.

Neither do my notes

(We are using hatcher)

Of course you can work on a connected component... but ya know

I forget how Hatcher does it

He says that if $x$ lifts to $\tilde x$ and $\tilde x'$, then there's a deck transformation taking $\tilde x$ into $\tilde x'$

@BalarkaSen he then proves this under the assumption everything is path connected

I see.

Well, then, X x {1, 2, ..., k} --> X is regular.

But the fundamental group issue you speak of is gone to dust because nothing is connected

right

so what is the group action then?

4:19 PM

@Slereah agreed

you need a group action on $RP^2\vee RP^2\sqcup RP^2\vee RP^2$ to quotient out

One minor but definite achievement was getting listed in the acknowledgements

@0celo7 That's not connected. I wanted a connected cover of RP^2 v RP^2.

you think there's another connected double cover?

Yup

Your unwrapping technique works, but you unwrap the two RP^2's in RP^2 v RP^2 differently

4:21 PM

Ok, so I unwrap one of them

Then I have $S^2$ with $RP^2$ attached two $S^2$

aha

One point of $RP^2$ attached to two points of $S^2$

Shouldn't it be the other way? two RP^2's attached to the poles of S^2?

Ah, that's what you mean

@BalarkaSen you could do that too

because what I have is just yours but with the two $RP^2$'s identified

RP^2 v S^2? I don't think that works

4:24 PM

right?

What I have is RP^2 v S^2 v RP^2

"... according to the Copenhagen interpretation of quantum mechanics the state of a microsystem is relative to an experimental arrangement. Thus energy and time are conjugate: no experimental arrangement can determine both simultaneously, for the energy is relative to one experimental arrangement, and the time is relative to another. (Thus, a more relational sense of “time” is suggested.) The states of the experimental arrangement cannot be merely relative to other experimental arrangements, on pain of infinite regress; and so these have to be described by classical physics. (This parasitis…

yes

is that not a three-sheeted cover?

Oh, no! The Z/2 action is by identifying the antipodal points of S^2, and for the two north and south RP^2's identify them

But you would get a three-sheeted cover by unwrapping the $RP^2$'s again!

4:27 PM

Why would you unwrap the RP^2's again? RP^2 v S^2 v RP^2 is exactly your double cover.

You don't do anything to that anymore

@BalarkaSen because the next question is to find a triple cover :P

Ah. Yeah, sounds about right.

No I think you get a 4 sheeted cover

yeah, 4 not 3.

"It is continuity that enables modern mathematics to surmount the paradox of extension framed by the Pre-Socratic Eleatic Zeno—a paradox comprising the question of how a finite interval can be made up of dimensionless points or instants."

A continuum: Every sequence converges to some finite number

:/

There's a little trick involved in constructing a 3 sheeted one

4:31 PM

The problem says it won't be regular

Dear god. Yeah, could be.

"Other criteria perhaps involving other sense modalities might lead to slightly different spans of time, but the interesting point is that, if there is such a specious present, it cannot be explained solely by ordinary memory traces: if one hears a “tick tock” of a clock, the “tick” is not remembered in the way in which a “tick tock” 10 minutes ago is remembered."

Neuroscience suggest the sensation of "now" actually takes 2-3 seconds long

@BalarkaSen assuming I can find it, how does one show it's not regular

"
Two matters discussed earlier in connection with the philosophy of physics have implications for the philosophy of mind: (1) the integration of space and time in the theory of relativity makes it harder to conceive of immaterial minds that exist in time but are not even localizable in space; (2) the statistical explanation of temporal asymmetry explains why the brain has memory traces of the past but not of the future and, hence, helps to explain the unidirectional nature of temporal consciousness. It also gives reasons for skepticism about the claims of parapsychologists to have experime…

So far there is no conclusive evidence of either

"By postulates, atomic time is the uniform time of electromagnetism. Leaving aside relativistic and operational effects, are SI seconds formed at different times truly equal? This question cannot be answered without an invariable time standard for reference, but none exists. The conclusion is that no time scale can be proved to be uniform by measurement. This is of no practical consequence, however, because tests have shown that the atomic clock provides a time scale of very high accuracy."

Well, if that is the case, everything else is scaled in the same way, thus nothing is really affecte…

Bernardo Meurer

4:46 PM

@JohnRennie Are you around?

@0celo7 That's a good question

Bernardo Meurer

Does anyone here have a pair of tungsten electrodes by any chance? In particular with a spiky end

@BernardoMeurer Hi

@BalarkaSen without a legitimate construction and equation for the covering, it seems impossible with any standard of rigor

Even then...

well there may be an algebraic proof that a index 3 subgroup of Z/2 * Z/2 can't be normal.

4:50 PM

@ACuriousMind ::algebraic bat signal::

@BalarkaSen it also says that one can use the three sheeted cover to show that the fundamental group is nonabelian

I suspect it's not hard to construct a degree 3 cover. If X --> RP^2 v RP^2 is degree 3, restrict to RP^2. A degree 3 cover of RP^2 is the trivial cover RP^2 x {1, 2, 3} --> RP^2

Bernardo Meurer

@JohnRennie Do you have tungsten electrodes laying around?

So you have to glue a bunch of RP^2's in a strange way. No S^2 business

Bernardo Meurer

And an old TV?

I need a flyback transformer

@BernardoMeurer Sadly not.

4:54 PM

I gotta get dinner

Bernardo Meurer

@JohnRennie Dangit :P

Bye

Bernardo Meurer

@0celo7 Can you mill me some tungsten electrodes?

@BernardoMeurer dude

You don't mill tungsten.

Bernardo Meurer

@DanielSank Can you mill me tungsten?

I need some

4:59 PM

@BernardoMeurer No way. Tungsten is way too hard for any tools I have.

Bernardo Meurer

:(

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