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4:00 PM
A few months back there was a long argument about how to pronounce Euclid but I can't remember what the conclusion was.
 
everyone is wrong and no one really knows
 
Hello @JohnRennie
 
It's supposed to be eookledes
 
Evening Mr. Einstein.
 
I really think giving Euclid credit for the asymptotic heat kernel on a compact manifold is stretching it a bit
I mean come on
 
4:01 PM
in the original greek
 
He wasn't as good at geometry as his brother Noneuclid.
 
Εὐκλείδης
 
@JohnRennie, How are you?
 
user228700
@JohnR: Ah, saw your mail just now. I've replied.
 
Euclid did come up with Euclidian gravity
pretty good for ancient greece
also euclidian field theory
 
4:04 PM
oh no
OH NO
Chavel has been using a different definition of differentiability on a closed set this whole time
Well, screw all of this
 
lel
 
I'm doing algebra
@ACuriousMind Where do I start?
 
Also annoying : HE and Penrose use a different definition for the Cauchy development
 
What are the definitions?
 
The difference is whether or not the set $S$ for $D(S)$ is allowed to be non-achronal
 
4:09 PM
John Duffield is telling me that "the temperature before the big bang was absolute zero."
 
what does "before the big bang" even mean?
 
If it's not achronal, it makes a difference whether or not you use trips or curves
 
Oh crud, sorry
@0celouvskyopoulo7 shrugs
 
JD is banned, how is he talking to you?
 
4
A: Amount of energy of the Big Bang

Sir CumferenceLet's start by making some points clear: 1. We don't know what the Big Bang was. Rather, we know that the Universe is expanding. If you extrapolate backwards, you'd expect the Universe to be denser and denser. More specifically, we talk about this as a change in the scale factor $a$, and this g...

 
4:10 PM
Lol, did he really quote Hawking?
He hates him.
 
@0celouvskyopoulo7 It is a comment on another SE site, probably astronomy SE.
 
@peterh Probably
 
Is Duffield not banned
I thought he was banned from all SE
 
Guess not
-4
A: How can an infinite universe expand?

John Duffield I understand the expansion of the universe as actually an increase in the ratio of space to matter. Is this a correct understanding? It isn't wrong. The ratio is increasing. But it isn't a "correct understanding". It's merely an observation of one of the results of the expansion of space. ...

 
@SirCumference Here is it.
@0celouvskyopoulo7 Here is it.
 
4:12 PM
If $v\in \bar C^1(\Omega)$, then for any $w\in\Gamma$, we let $\gamma_w(s)$ denote the geodesic emanating from $w$ with initial velocity $\nu(w)$, and define the "normal derivative" of $v$ at $w\in \Gamma$, $(\partial v/\partial\nu)(w)$, by $$\frac{\partial v}{\partial \nu}(w)=\lim_{s\uparrow 0}\langle (\nabla v)(\gamma_w(s)),\gamma'_w(s)\rangle.$$
What?
 
@0celouvskyopoulo7 Btw, his chat ban is a different thing as his PSE site ban, his chat ban will expire in around a half year.
 
@peterh can't wait
he has been gone too long
 
Oye, posted the wrong link
Or guess not...
That's JD's response
 
@0celouvskyopoulo7 Also I miss him.
 
4:15 PM
I couldn't mention the universe being infinite without him yelling "NO PROOF!"
 
@SirCumference He has a point.
There is no proof.
 
@0celouvskyopoulo7 Yeah, but still
I think he said that there is an edge to the universe
 
what happens at that edge
 
@Slereah Too edgy for me
 
4:17 PM
Is there a custom agent?
 
Aug 28 '16 at 16:15, by John Duffield
@SirCumference : yes. IMHO if the universe was truly infinite it could not expand.
Bad understanding of expansion
Aug 28 '16 at 16:01, by John Duffield
@SirCumference : not quite. I've said we have no evidence that the universe is infinite. And that there's a non-sequitur wherein people claim a flat universe is an infinite universe. I've also said that I think that the expansion of the universe is evidence that it is not infinite.
 
How is infinite and non expansion related
 
$$u:(\bar\Omega\times[0,\infty))\setminus(\Gamma\times\{0\})\to\Bbb R\in \bar C^1\cap C^2(\Omega\times(0,\infty))$$
Christ
 
@Secret He probably thinks "expansion" means the universe is a bubble that's growing
 
Is that not what it means?
 
4:18 PM
 
@SirCumference Can you perhaps not pass judgement on chat-suspended users here where they cannot respond? Just let it rest.
 
I think the Universe is not flat, instead it has a small constant positive curvature (below measurement limit) and thus a spherical topology. But I can't prove it.
 
The edge of the universe
@ACuriousMind that is not judgement
It seems fairly objective
 
@ACuriousMind I'm not attacking him, I'm mentioning the comments he's left on my posts
@peterh Why positive?
 
@SirCumference But...why? Let's just not talk about him at all. (That's not a moderatorial instruction, but I'd greatly prefer we just ignored him)
 
4:20 PM
@SirCumference So it is not infinite
 
@SirCumference Still that does not stop something infinite from expanding. sure it will look like noting had happened, but that is far from not actually expanding at all given even in infinite volume, the spacing betwen galaxies are still increasing, and that is finite
 
@SirCumference Because spherical topology has a constant positive curvature and it is cyclic.
 
Jesus one person at a time
@0celouvskyopoulo7 Infinite is good
 
$$\textrm{I} ♡\infty$$
 
@peterh Why do you want a cyclic universe?
 
4:22 PM
@SirCumference its't the only physically realistic one
this is clear
 
is it
 
@SirCumference Constant negative or zero curvature would close out a spherical topology, although I think it could be still cyclic, but not so naturally. I think the spherical Universe is coming from 3 things: 1) isotropy 2) homogenity 3) Aristotle's ultimate argument: everything should be spherical, because the sphere is the most perfect thing :-)
 
@Slereah I haven't seen an infinity before, and neither have you
 
@peterh a positively curved universe is necessarily cyclic? That's a bold claim.
 
4:22 PM
I haven't seen a spherical universe either
 
@JohnRennie Yes, but I didn't said that (probably)
 
@JohnRennie Positive curvature implies the spatial slices are quotients of spheres
 
It would be cooler if the entire universe is one gigantic CTC atvthe comiving direction
 
As @JohnRennie said, that's only if $\ddot{a} < 0$
@peterh I don't think any of these 3 things are related to the curvature being positive...
 
this is well known...
 
4:23 PM
@0celouvskyopoulo7 No it doesn't
 
@JohnRennie For example, a spring has a constant positive curvature but it is not cyclic.
 
@Slereah Proof?
 
Jesus, one step at a time
 
A complete 3-manifold with constant positive curvature is a quotient of $S^3$.
I'm quite sure that's correct.
 
@peterh We've measured $\ddot{a}$ to be greater than 0. Our universe won't collapse.
 
4:25 PM
@SirCumference There is also a fourth, this time from Marx: qualitative changes always transform to quantitative changes after a point. It closes out the infinite universe being the same everywhere. If I add these 4 argument together, the result is a spherical Universe.
 
@SirCumference well, that's true if dark energy is constant. Since we have no idea what dark energy is we can't tell how it will behave in the future.
 
@0celouvskyopoulo7 Take $\Bbb R^4$ with the metric $$ds^2 = \frac{\eta_{\mu\nu} dx^\mu dx^\nu}{(1 + \frac{1}{4} \eta_{\mu\nu} x^\mu x^\nu)^2}$$
Curvature of $1$ everywhere
Not a sphere at all
(Though geodesically incomplete)
 
Even if it does collapse, there's many problems with the cyclic model
 
@Slereah Riemannian you genius.
Curvature in cosmology refers to the spatial slices.
 
Also works for a Riemannian manifold
 
4:27 PM
1
Q: Experimental test for a universal scale closed timelike curve?

SecretSo recently, in chat, I was interested in about what if our universe contains only one CTC (thus making it unlike the Gödel metric, van Stockum dust where there can be more than one CTCs found) and that is the time dimension of the whole universe wrapped back onto itself (meanwhile the spatial di...

I am currently fixing this question
 
just remove a point from the sphere
 
@Slereah It's not complete
 
@peterh I'll let Baum & Frampton (2006) explain the problem with an oscillating universe:
> ...one principal obstacle is the second law of thermodynamics which dictates that the entropy increases from cycle to cycle. If the cycles thereby become longer, extrapolation into the past will lead back to an initial singularity again, thus removing the motivation to consider an oscillatory universe in the first place.
 
Well yeah
 
We assume completeness
 
4:27 PM
But why would the universe be complete
 
@SirCumference A recollapsing universe ends in a singularity (classically at least) and the equations of motion can't be integrated beyond that point. To claim cyclic behaviour seems to me at best a supposition.
 
Why would the universe be homogeneous and isotropic?
 
What if there's a timelike singularity in the universe?
 
@SirCumference I think gravity can somehow change the direction of the entropy. I suspect, if a quantum gravity will work, it will also find ways to reverse the direction of the entropy. This is my another thing what I suspect but I can't prove.
 
@JohnRennie Isn't one of the principle things about dark energy that its density remains constant as space expands?
@peterh Wait what
 
4:29 PM
@SirCumference depends
there are various sources for dark energy
 
@SirCumference that's true iif dark energy is a cosmological constant.
 
@Slereah My observation is that almost everyone in the lay public has room for only one notion of "particle physics" in their head at a time, and they focus on "most powerful" as the only figure of merit.
 
KEK is cool, but it's not "most powerful" so it doesn't make an impression.
 
Why nobody seemed to interested in the time topology of the universe?
 
4:30 PM
@JohnRennie Ok, either I haven't gotten to that or I need to re-read my notes/textbook
I'll ask the prof
 
Because there's no time topology
 
@peterh Why would gravity reverse entropy?
 
@SirCumference Look at the black holes, they give out energy and they will be hotter. Look a simple gas cloud, or any star, the denser components will concentrate around its center. If it is about gravity, there are always things in thermodynamics which somehow doesn't work as we accustomed to it, producing thermodynamical formulas based on hot gases on different cylinders.
@SirCumference I only suspect, somehow it can.
 
> they give out energy
You mean Hawking radiation?
 
It seems it could be reasonable that there could be a CTC along the comoving direction that is so large that the fields don't have time to accumulate enough to cause a divergence?
 
4:32 PM
what
 
What is happening
 
@SirCumference Yes. Smaller black hole has higher temperature. As the BH loses energy by HR, its mass shrinks, and its temperature grows. On the end stage, it explodes light a big, big nuclear bomb.
 
Guys, I’ve read Newton’s Third Law described in this way: "If two objects, sufficiently isolated from interactions with other objects, are observed in a local inertial frame, then their accelerations will be opposite in direction, and the ratio of their accelerations will be constant."
and the mass defined as such
However, I’m slightly confused by the definition of mass. What if we measure two electrons and how they repulse each other? We would then measure a huge mass. So should we only consider neutral objects then?
 
@peterh I don't know too much about HR, but I don't see how that relates to the second law of thermodynamics
 
@ShaVuklia That sentence is valid only if the two bodies have the same mass.
 
4:34 PM
Suppose the universe has a CTC spanning billions of light years in diameter and for all observers in the comoving frame, they will agree the CTC is pointing along the time direction with no spatial components, then how would that differ from a cyclic universe?
 
@peterh wrong
 
@ShaVuklia I don't quite follow - why do you think we will measure a huge mass?
 
CTCs only cause divergences if they're compactly generated
 
@ShaVuklia Sigh, misread, forget this
Sigh, i gtg
 
@ShaVuklia: mass has dimensions so it's absolute value depends on the units. We could use kg or pounds or stones and the number we assign to the mass would be different with the different units.
 
4:36 PM
@SirCumference HR is black body radiation. Calculating the temperature which gives the same black body radiation as a given black hole, you can order a temperature to the black hole. In this formula, the temperature of the black hole is reversely proportional to its mass. Thus, smaller BH is hotter.
 
@ShaVuklia: the point of the article is that the ratio of the accelerations gives the ratio of the masses and that's true regardless of what units we choose for mass.
 
It doesn't matter of what nature the force is - as long as it's not one of the situations for which Newton's third law doesn't work, that sentence holds true, regardless of whether there is charge or not.
 
@JohnRennie If I jump, I give for example around $1\frac{m}{s^2}$ acceleration to my body, while I give only a negligible acceleration to the Earth.
 
@ACuriousMind I'm not entirely sure why I am thinking this, but I understand that if we are only dealing with the gravitational force, we can measure the mass. But what if we're only dealing with electrical force? This force is 24 digits larger, so wouldn't we get a huge mass?
 
@peterh Sure. The point of the quoted answer is that no matter how hard you jump, the ratio between your acceleration and the acceleration of the earth will always be the same.
2
 
4:38 PM
@John yea I'm not bothered by the units, but I'm rather bothered by the nature of the force.
but maybe I'm confused because this law is presented before newton's second law
 
@ShaVuklia No, and I don't understand why you would think so. The force between the two particles is much larger, but the ratio between the accelerations stays the same.
 
(in the answer on stack)
So the third law technically states that whatever interaction two objects have, if you place them together, their ratio of acceleration will always be the same?
 
@SirCumference The calculation of the Hawking Radiation is hard for commoners, but the results aren't. They are simple high school math formulas. Some constants, somewhere an $\frac{1}{M^3}$ or similar in them.
@SirCumference Playing with these formulas is very funny (for example, you can calculate that around Moon-sized black hole had been evaporated since the Big Bang).
 
@ShaVuklia yes, that's one way of phrasing it and arguably an elegant way of phrasing it. (ratio as measured in the centre of mass frame of course)
 
@ShaVuklia Yes, and that's precisely how the answer phrases it
 
4:42 PM
@ACuriousMind Right, I didn't read it enough carefully, sorry.
 
But how do we know then for sure that we will always calculate the same mass when dealing with different forces? So say we have objects $X$ and $Y$, and $A$ and $B$, and an object $I$, that we consider to have unit mass. Say the force between $X$ and $Y$ is electrical of nature, and for $A$ and $B$ gravitational. And say our object $I$ has no charge. What happens between $X$ and $I$, and $Y$ and $I$ would then be gravitational I think, and between $X$ and $Y$ electrical.
So because it is always the same for $X$ and $Y$ by the law, we know that we should be able to get the ratio between the two, because it's also the same between $X$ and $I$, and $Y$ and $I$? Hmmm, I'm just really confused. I don't "see it" why we can always measure the mass this way.
 
Facebook is such a useful resource. It's just informed me that Saturday is World Naked Gardening Day. It's a bit nippy in the UK though ...
 
Couldn't we have stated that the interaction between one object and all other objects will always be the same (i.e., constant acceleration)? And then deduce from that that the interactions between all those other objects are the same?
I'm just worried that if we fix all accelerations at once, that there could be contradictions, no?
Because in theory, there could be "random" accelerations
 
I'm still not following you
 
@ShaVuklia: the point is that regardless of the nature and magnitude of the force the ratio of the accelerations will always be the same.
And therefore the ratio of the masses will always be the same.
 
4:50 PM
You have a ratio $c_{Ix}$ for all objects $x$ with the reference object $I$. Its mass is $m_x = -c_{Ix}$, if we define $m_I = 1$ to be our "unit" of mass. What is the problem?
 
But with electrical forces, the force doesn't even depend on the mass.
So how can we deduce the mass from this interaction then?
 
Sorry, still not following you. Do you deny that the Coulomb force charged particles exert on each other obeys the third law?
The "nature" of the force really doesn't play a role here
 
okay, but say that hypothetically nature assigned random constants of accelerations between each possible pair of objects in the universe
how could we have ever deduced something from this?
because this law doesn't tell us yet that the acceleration is linked to a force.
this law is placed before newton's second law (in the post)
so technically, we have no idea of forces yet.
 
Right. But the definition of $m_x = -c_{Ix}$ doesn't depend on a notion of force - it just needs the fact that there are constant accelerations, which is the content of the third law, and a choice of reference object
I don't get what the problem is supposed to be
 
sure, there's nothing wrong with that definition, but
we can't deduce from two other objects (that are not unitary) their ratio of masses from the ratio of acceleration
 
4:56 PM
Oh, are you worried about transitivity? that we might have $c_{Ix},c_{xy}$ and $c_{yI}$ given, but not behaving as $c_{yI}c_{Ix} = c_{yx}$?
 
EXACTLY. :)
in case the ratio is assigned randomly, there might be no transitivity
and then our definition is useless
as is the concept of mass
 
I suppose that you technically need to add that assumption to this variant of the third law. I.e. extend it as "the ratios of accelerations are constant, and for every triple of objects $A,B,C$ we have $c_{AB}c_{BC} = c_{AC}$".
 
yea, that's quite a big deal
 
what is the purpose of \usepackage[utf8]{inputenc}?
 
alright thanks
 
4:59 PM
I see it used rather often
 
@AccidentalFourierTransform It allows me to use things like ä without having to type \"a
 
huh
interesting, it probably works for á,ñ,etc too
 
In more sophisticated terms, I think it allows you to put utf8-encoded files instead of mere ASCII into pdfTeX, and therefore you can use unicode characters instead of having to jump through hoops
You don't need it with XeLaTeX or LuaLaTeX, iirc
But you should probably ask an actual TeXpert about that :P
 
so its for lazy people
I get it
 
@JohnRennie I think something in my computer's video pipeline might be failing
May be my drivers though
 
5:06 PM
So if I used the halloween Latex package, what would the symbols be used for
The ghost is for ghost fields, obviously
what is the witch
and the pumpkin
 
Does anyone here know how to clean blood off of surfaces?
 
That's...a suspicious question.
Is the next question how to get rid of a corpse?
 
No, I know that part
@JohnRennie You're a colloid scientist, how do I clean blood?
5
 
last time I suggested that we needed blood cleansing ACM scolded me
6
 
@BernardoMeurer you mean clean blood off something?
 
5:13 PM
@JohnRennie Yes
 
you need to denature it
do you have any kind of strong acid around?
 
If it's a porous surface like a fabric or carpet it's virtually impossible as blood contains cationic polymers that will bond strongly to the substrate. Use cold water and clean up before the blood starts clotting. But you'll never clean it well enough to escape detection.
If it's a non-porous surface like a ceramic use cold water then bleach.
Right, I'm off to drink beer and read about archaeology in the New World.
 
@JohnRennie It was last night, so it's dry
And it's a wall
White wall
@AccidentalFourierTransform I have bleach
 
what the hell happened?
I hope there are no other body fluids involved
 
Nope
Just blood
Not much though
 
5:24 PM
ehem
it is yours, right?
 
5:42 PM
I'm trying to find the magic trick in Peskin's treatment of the interaction picture
But he does that thing where he doesn't mark clearly what field operator is free and which one is $\phi^4$
so it's a bit confusing
 
ah yeah, good ol "let $\phi_I=U\phi U^\dagger$, and let us drop the $I$"
 
Still better than "Let $\varphi = \langle \phi \rangle$"
 
@AccidentalFourierTransform Aha! So you do read Cyanide and Happiness! :D
 
It seems a bit clearer in Schwartz
He just calls the free operator $\phi_0$
 
5:47 PM
@paracetamol I never said I did not (but I do not)
 
:'(
 
although the problem with Schwartz is that he does it a bit out of order
 
Behold! L'entropie
 
I'm trying to find the "correct" explanation of the interaction picture, but it's not easy
They mention that we can totally decompose the field variable $\phi$ into ladder operators
which I am sure is true, but I'm guessing those ladder operators have nothing to do with the ones of the free theory
Let's see Haag on the topic
"We have presented the magic formula without giving any reasons. Let us try to justify it."
 
5:55 PM
at any fixed time, you can always (technically) write $\phi\sim\int a\ \mathrm e^{ikx}+\text{h.c.}$
its just a Fourier expansion
 
Yes, but the ladder operators will just be random operators
 
if $\phi$ is interacting, $a$ depends on time
yes, well, they still satisfy $[a,a^*]\sim 1$
 
"We start from the assumption that at a fixed time, the fields $\Phi$ and their canonical conjugate momenta satisfy the same (canonical) commutation relations as the free fields $\Phi^0$ and that we may identify $\Phi$ and $\Phi^0$ as well as the respective canonically conjugate momenta at this time, say at $t = 0$"
Ah, that's the trick
We just ASSUME THEY'RE THE SAME
Dang it
that's not as clever a trick as I thought
Is there any motivation for this outside of "well at some point the particles will be far apart"
 
you need some algebra
why not the one we use for free fields?
 

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