The h Bar

Nital normally uses quite dilute acid. Nitric acid is only really dangerous when it's concentrated, because conc HNO3 is such a strong oxidising agent.

0celo7

4:58 AM

@JohnRennie I'll look at it in the morning, I have to sleep now

Night

John Rennie

Actually the methanol is probably more dangerous. Don't breathe the vapour!!!

0celo7

What happens if you do?

What happens if you breathe ethanol vapor btw

John Rennie

You go blind.

0celo7

...yo like how much

I totally took a whiff yesterday

John Rennie

Methanol is metabolised to formaldehyde, and formaldehyde kills tissue like crazy.

0celo7

4:59 AM

not super high concentration

No, ethanol

John Rennie

@0celo7 A brief whiff will have done you no harm.

0celo7

We don't use methanol regularly.

John Rennie

You can sniff ethanol as much as you want - just don't drive afterwards :-)

0celo7

Ah, good.

It smelled good ;)

John Rennie

As generations of homo sapiens have noticed :-)

0celo7

5:01 AM

cheerio

John Rennie

G'night

Secret

nautil.us/issue/30/identity/…

I have a different way to distinguish them, by asking them each with the same question:

Q: Is A and -A known as a contradiction? Answer every second for 100 seconds

Assuming ja means yes
Then the true god will answer with a stream of ja
The false god will answer with a stream of da
and the random god will answer with a stream of random ja and das

One can say I kinda cheated, by technically splitting 3 questions into 300 questions, and exploiting that a random outcome depends on the trial number and the elaspe of time

However, if someone ask this true random and false god question and insist that no logical trappign questions can be asked , then this will be a solution to this more generalised version of the puzzle

----wait a minute, I only exposed who is random, back to the drawing board!

The Dark Side

5:54 AM

Trilogy complete:

ConfusinglyCuriousTheThird, San Francisco, CA, United States

872 1 3 11

:D

Secret

Acuriousmind CuriousOne and now ConfusingCuriousTheThird

Only a Curious Mind

213 1 2 8

particle-physics photons quantum-mechanics energy gravity mathematics molecules

and 80+ more curiouses

The Dark Side

6:10 AM

Curiosity is a meme on Physics.SE!

John Rennie

You have three beings called ACuriousMind, CuriousOne and ConfusingCuriousTheThird. One always answers questions truthfully, one never answers questions and one is unknown. You have to determine which is which, but:
- you are not allowed to ask about cohomology or German goth metal bands
- comments don't count

5

The Dark Side

@JohnRennie "Comments don't count" makes this a trivial puzzle.

Meanwhile, on SuperUser:

134

Q: Will it damage my MacBook if I put it in the fridge to cool it down?

I've a got longstanding problem with laptops overheating (MacBook Air/Pro) and it's not only related to one machine. The laptops are overheating especially during hot days (summer). I've found that keeping them in the fridge for half an hour makes a dramatic difference in their performance. Howe...

laptop macbook cooling overheating case

Followup question: "How do I get the garlic smell off my MacBook after keeping it in the fridge?" — svidgen 2 days ago

:D

user54412

7:39 AM

Public service announcement: Please don't go looking to apply close reasons at all costs. Just because a reason exists doesn't mean there is a reward for whomever can tag the most questions with it.

user54412

Just because a question mentions the free will theorem doesn't mean it has anything to do with free will in philosophy, and certainly it is a mainstream result. Also, just because a question mentions a device does not mean it is asking how to engineer such a device.

2

Physics Meta

8:38 AM

0

Q: Can a question be offered for bounty twice by a same user?

If a user offers some bounty for his own question and awards it, can he offer bounty for the same question again?

support

user116211

10:32 AM

Well, Curious es and Jim s are popular here ;))

user685252

will the MAFIA ever be as popular? @MAFIA36790

Domenico Bagnato

Given a continuous function f in R, and which satisfies the relation: 4(f(x))^2 - 4 f(x) + 1 > 0, for each x in R show that if a point x0 exists such that f(x0) = 0, then the function has an upper limit . .

what can the proof of this be?

yeah

Balarka Sen

10:47 AM

@user685252 Is that skull?

user116211

11:15 AM

Breaking News: Borris Johnson will not be contesting for the post of PM.

user116211

@JohnRennie: Like it? ^^^

John Rennie

@MAFIA36790 Thank God for that. Johnson is a very intelligent man, so he differs from Trump in that respect, however he is just as politically irresponsible and would have been a dangerous leader.

I suspect that more or less hands the contest to Theresa May as Michael Gove is allegedly viewed by many of us as a weasally piece of crap.

user116211

Meanwhile, everyone is asking for Corbyn to resign...

user116211

But he seems to be too adamant.

John Rennie

That's a long and (if you're not a Labour supporter) entertaining story

0celo7

11:31 AM

@JohnRennie Please no insults.

user685252

Trump has danger written all over him.

Balarka Sen

What's the state of things about the election, btw? Is Trump winning?

user116211

@BalarkaSen NO>

Balarka Sen

Too bad.

0celo7

No one knows who will win

Balarka Sen

11:45 AM

So there's still hope.

user685252

Big money always wins.

user116211

@0celo7 Whom do you want to win?

0celo7

@MAFIA36790 Trump.

user116211

....

0celo7

@yuggib You reading Tao's new paper?

user685252

11:55 AM

No supporter of Trump lost his life because he was absorbed in the contemplation of a mathematical diagram.

Balarka Sen

There needs to be another war to shake people out of their apathy and wrong-headed decisions, and I can see it coming fast - with all the worldwide terrorism involving all the countries and the bad takes on handling them.

0celo7

@user685252 What?

@yuggib What on Earth is his $\mathcal M$ supposed to be?

Why is he attaching a torus

user685252

so that the next war can be fought with sticks and stones?

Balarka Sen

@0celo7 Apparently $\Bbb R^m \times (S^1)^{d-m}$

0celo7

@BalarkaSen Yes, which is $\Bbb R^m$ with a torus attached at each point.

Balarka Sen

12:05 PM

Sure, why not.

0celo7

@BalarkaSen Or maybe it's a torus with $\Bbb R^m$ attached at each point!

Balarka Sen

Both descriptions are equivalent.

But perhaps not so useful, depending on context.

0celo7

Perhaps.

user685252

have you started an exercise routine yet? @BalarkaSen

Balarka Sen

Exercise routine?

0celo7

12:09 PM

@user685252 Why do you keep changing names?

Only an old woman would ask that question.

user685252

:(

Balarka Sen

Oh, I literally thought "exercise" as in math.

user685252

for your health

Balarka Sen

I realized, yeah. Well, I'm still ill.

user685252

plan to start something as soon as you get better

0celo7

12:14 PM

Why is my coffee machine making less coffee?

It's not filling the cup up all the way anymore.

user685252

perhaps it's a message to cut back

0celo7

I drink one cup in the morning on an irregular basis.

Balarka Sen

I like tea more than coffee.

user685252

me too

0celo7

I don't.

Slereah

12:21 PM

I'm an energy drink man

0celo7

@Slereah Should I get a 2L soda today

Slereah

Possibly me

@0celo7 Isn't it mandatory in America

0celo7

@Slereah who is that dude anyway

Slereah

That is the Spurdo

0celo7

Not the fucking bear

Slereah

12:25 PM

knowyourmeme.com/memes/people/ainsley-harriott

0celo7

I know the meme just not the dude's name.

Slereah

the name is included

0celo7

Your face is included.

user685252

why is he pouring it into a frying pan?

ACuriousMind

@JohnRennie This is getting ridiculous :D

Slereah

12:28 PM

How else are you supposed to oil your pan

user685252

i see

ACuriousMind

Maybe I should really get a new name

0celo7

Bajoran!

user685252

Rigor!

or Rigour ;)

0celo7

Honestly ACM is not that rigorous.

ACuriousMind

12:30 PM

@0celo7 I actually like the sound of that

@0celo7 Not compared to true mathematicians, no

0celo7

@ACuriousMind I would be honored to get to name you, mother duck.

ACuriousMind

Hmm..."Bajoran the mother duck"

That would probably weird people out :D

0celo7

More like mother f

:)

user685252

nice

0celo7

@ACuriousMind You are literally a troll right now.

ACuriousMind

12:32 PM

@0celo7 Orc, not troll.

Get your imaginary species right!

0celo7

@ACuriousMind Did you know that any closed set of $\Bbb R^n$ is the zero set of a smooth function?

user685252

Jun 25 at 19:41, by ACuriousMind

::quacks angrily::

0celo7

I still can't believe it!

user685252

Oct 14 '15 at 21:05, by ACuriousMind

::quacks protectively::

ACuriousMind

@0celo7 oO

0celo7

12:33 PM

ACM is a bona-fide duck

ACuriousMind

I'd like to see the function for the Cantor set!

Balarka Sen

@0celo7 Well, any closed of R^n is zero set of some continuous function. And then, like, smooth it out.

0celo7

@BalarkaSen No, that's not how you do it :P

@ACuriousMind Ok, lemme calculate it

Balarka Sen

@0celo7 I think it can be done like that.

0celo7

@BalarkaSen But the continuous function is easy: the distance to the set

Balarka Sen

12:35 PM

Yes, of course.

I want to smooth that out by bump function techniques.

user685252

@0celo7 he could be a goose :P

0celo7

Ok, let $h:\Bbb R^n\to \Bbb R$ be a smooth bump function equal to 1 on $\overline{B_{1/2}(0)}$ with support in $B_1(0)$. Let $K$ be the closed set, then $\Bbb R-K$ is open and for $x\in\Bbb R^n-K,\exists r\le 1:B_r(x)\subset\Bbb R^n-K$. By second countability, we can cover $\Bbb R^n-K$ with a countable collection $B_{r_i}(x_i)$ os such balls

user685252

@0celo7 the goose that laid the golden eggs

ACuriousMind

Is there really no requirement for the zero set of a smooth function to be mesaurable?

I have a feeling there should be

Slereah

Ducks aren't geese

0celo7

12:38 PM

For each integer $i$, let $C_i\ge 1$ be a constant that bounds the abs val of $h$ and all its partial derivatives through order $i$

Then the desired function is $$f(x)=\sum_{i=1}^\infty \frac{(r_i)^i}{2^iC_i}h(\frac{x-x_i}{r_i})$$

You use the M-test to prove it's smooth

and for each $x\in K$, you're outside of the support of all of the bump functions

@ACuriousMind Convinced?

ACuriousMind

Hm, yes, it appears you use nothing but closedness.

0celo7

@BalarkaSen I would like to see your smoothing argument

@ACuriousMind More generally, this is true on any smooth manifold.

user685252

@Slereah Ok, how about the million dollar duck?

0celo7

@ACuriousMind And it is furthermore true that given any closed $K\subset\Bbb R^{n-1}$, there is a properly embedded manifold in $\Bbb R^n$ such that its intersection with the hyperplane $\Bbb R^{n-1}\times\{0\}$ is $K$.

ACuriousMind

@0celo7 Okay, that's even weirder.

Balarka Sen

12:45 PM

@ACuriousMind I think I have an explicit construction for the cantor set. Like, take $[0, 1]$, remove a middle third. Consider the bump function $f_1$ which is $0$ only at the bd of the removed interval, and zero elsewhere (so graph looks like a bump on $(1/3, 2/3)$ and zero elsewhere. Now remove middle thirds from the remaining components, and consider $f_2$ the graph of which looks the same as $f_1$ except one has two extra bumps of smaller height at $(1/9, 2/9)$ and $(7/9, 8/9)$. And so on.

0celo7

@ACuriousMind The proof was one of my diff top homework questions :)

Balarka Sen

I think these $f_k$'s should converge uniformly.

0celo7

Take the manifold to be the graph of the smooth function whose zero set is $K$.

Balarka Sen

The resulting function would then be a smooth function vanishing at precisely the cantor set.

If the heights of the successive bumps are chosen appropriately, I think it's believable that it should converge uniformly.

@0celo7 It was just an idea, though. I'll think about it and let you know if I get something.

0celo7

@BalarkaSen Oh, btw

Turns out I did not need that elaborate generalized tube lemma

There was a weaker version in the appendix

it was sufficient

Balarka Sen

12:49 PM

Ah. What was the weaker version?

0celo7

@BalarkaSen Er, do you have GP handy?

Balarka Sen

Yep, I do.

0celo7

lemma 2 on page 204

But even then I'm not sure I needed that either

I wanted to have one open set $O$ that works for all $t$, but actually I did not need that for the next proof

And I misunderstood what homotopy stability of local properties is anyways

@ACuriousMind Where does one actually use the Whitney embedding theorem?

Balarka Sen

@0celo7 Thanks.

0celo7

It seems like a neat fact, but so what?

The embedding theorem as shown in GP

where they already defined manifolds as being embedded

ACuriousMind

12:55 PM

I'm not sure whether there's any "use" for it besides people thinking the question of how small the surrounding space can be made is interesting in itself.

Balarka Sen

Don't think one uses the embedding theorem on a regular basis.

ACuriousMind

For abstract manifolds, it's of course very relevant to know that their theory is equivalent to the theory of embedded manifolds.

0celo7

@ACuriousMind Of course.

Speaking of abstract manifolds, is it possible to define $S^n$ intrinsically?

Balarka Sen

Depends on what "intrinsically" means.

0celo7

SO(n+1)/SO(n) is not an answer

Balarka Sen

12:57 PM

I can glue two disks by the identity map along the boundary.

I mean, given any smooth manifold, you can use the information about the atlas to reconstruct the manifold from the charts and the transition maps.

0celo7

Yes...

yuggib

@0celo7 it is a space where some directions are eventually bounded (with "periodic" boundary conditions)

essentially a generalized cylinder

user685252

\o @yuggib

yuggib

o/

ACuriousMind

@0celo7 Define $S^0$ as two points, then define $S^n = S(S^{n-1})$ where $S(-)$ is suspension.

0celo7

1:15 PM

@ACuriousMind You absolute madman.

ACuriousMind

What? Suspension, along with cylinders and cones, is a wonderful method to create new spaces

0celo7

Never heard of it

Balarka Sen

It's a rather homotopy theoretic construction tho

Suspensions are rarely manifolds.

0celo7

@BalarkaSen That would explain why it's not in Lee

ACuriousMind

Well, you didn't specify that I need to construct $S^n$ in the smooth category! :P

Physics Meta

1:23 PM

2

Q: Does the resource catalogue question still get updated?

I have been looking at Book recommendations which is a catalogue of links to resource recommendation questions. For me it has been (in combination with the individual resource recommendations that are linked in the wiki) one of the most useful things on physics.SE. So thank you to the creators! I...

discussion resource-recommentation

0celo7

@ACuriousMind I study geometry, not your algebraic nonsense

I haven't seen an algebra in the night sky

user685252

you haven't look hard enough :P

ACuriousMind

@0celo7 Geometry != smooth. Saw a torus fibration in string theory yesterday where the fiber degenerates to a "pinched torus" at one point. Still clearly a geometric construction (to me), but far from being a manifold.

user685252

2 hours ago, by user685252

No supporter of Trump lost his life because he was absorbed in the contemplation of a mathematical diagram.

Balarka Sen

I find algebra and geometry thoroughly correlated in many contexts.

0celo7

1:31 PM

@ACuriousMind That string theory sounds a lot better than the group theoretic hell of BBS and the CFT hell of BLT

ACuriousMind

@0celo7 I'm liking it a lot better, too. It's almost looking deceptively simple, but I guess that's becauase we a) always look at the nice examples and b) currently only work at a rather qualitative level

All that CFT and group theory still lurks beneath the surface

And occasionally rears its head in things like "Well, now we need to find a 496-dimensional group that satisfies some weird formula for the relation of traces of various representations"

I really need to learn things about that $E_8$ thing, it occurs suspiciously often.

user685252

how many courses are you taking?

0celo7

@ACuriousMind I still don't know what that is

I pretty much gave up on string theory when BBS started talking about $E_8\times E_8$ heterotic and I had no clue what $E_8$ is and no one seems to care

And I'm always skeptical of string theory questions on PSE, because I cannot believe anyone legitimately understands that shit :P

Balarka Sen

I vaguely recall a s.c. 4-manifold with intersection form E8 is not smoothable.

0celo7

1:47 PM

s.c.?

Balarka Sen

Simply connected.

0celo7

ah

I thought the intersection form was a map like $\Gamma: H^2\times H^2\to \Bbb Z$

Balarka Sen

Yeah.

0celo7

Where does $E_8$ appear

user685252

you'll need to trade some of your internet time for physical activity @BalarkaSen

Balarka Sen

1:48 PM

It's a quadratic form on H^2.

0celo7

@user685252 shoo lady

user685252

that's old lady :P

Balarka Sen

@0celo7 Well, a quadratic form can be represented by a symmetric matrix.

0celo7

Yeah?

I thought $E_8$ was not a matrix Lie group

Balarka Sen

And there's a matrix called "E8".

Hmm, really?

ACuriousMind

1:49 PM

@0celo7 It's a weird Lie group that arises from a remarkably unique lattice in 8 dimensions. I've never seen a physicist write down anything explicit about it, but someone needs to have done it because people determined that it was one of only two groups that can yield an anomaly-free SUGRA of IIb type (I think, might've been IIa or I).

0celo7

@ACuriousMind See, bullshit statements like that are why I hate physics :P

Where on Earth would you even find that proof

ACuriousMind

I think I have the reference somewhere in my notes

0celo7

@ACuriousMind Have you even proved that $E_8$ is a Lie group?

What's the manifold like

What's the Lie algebra

ACuriousMind

@0celo7 Well, there's nothing to prove if you define it as the Lie group belonging to the $E_8$ lattice. Then what you have to prove is that the $E_8$ lattice is an admissible root system, but that's just a bit tedious, not hard.

0celo7

@BalarkaSen Are the exercises in Hirsch considered hard?

ACuriousMind

1:53 PM

(You need Lie's theorem about algebras integrating to groups, though)

0celo7

@ACuriousMind That is one of the few things I know about Lie groups

Next summer I should read Helgason

Balarka Sen

@0celo7 I don't know what's the connection with the E8 Lie group is, but some googling tells the E8 matrix is the one with 2's along the diagonal, 1's along the super diagonal except the first entry which is 0, 1's along the subdiagonal except the first which is also 0.

Perhaps you'd find that familiar.

0celo7

@BalarkaSen I don't know shit about $E_8$, so no :P

Balarka Sen

@0celo7 Yes, so I have heard.

Me neither.

0celo7

@ACuriousMind Can I apply Taylor's theorem to a differentiable function or must it be $C^1$?

Balarka Sen

1:56 PM

Don't you need your function to be C^\infty before you try to prove it's analytic?

0celo7

Who said anything about analyticity?

I want to write $f(x+h)=f(x)+g(h)$ with $g(h)$ some error term

And the error term should be like...

$O(h)$

ACuriousMind

@0celo7 I think being differentiable is enough.

Balarka Sen

@0celo7 To me Taylor's theorem explicitly says the series expansion of an analytic function at a point.

0celo7

@BalarkaSen No, Taylor's theorem is $f(x)=f(x_0)+derivatives+error term$

and if I want $k$ derivative terms, do I need $C^k$ or $k$-fold differentiability

I'm working on an exercise in Hirsch

Balarka Sen

I see, you're thinking about the approximation.

0celo7

2:00 PM

I basically need to bound the image of a compact set under a differentiable map

Balarka Sen

You don't need C^k, I think.

0celo7

I don't like books with hard exercises, it makes me think I missed something in the text

Balarka Sen

I usually complement the book with a book with easier exercises then.

user116211

Don't know why I'm getting this ;\

user116211

2:07 PM

Saw the extra white bar? Also the grey background is incomplete ;(

user116211

Why is this happening?

user116211

I reloaded the page; but of no avail T__T

Emilio Pisanty

2:29 PM

Man

scifi.stackexchange.com/users/28603/kyle-hale?tab=topactivity

Anyone notice anything funny there?

user116211

@EmilioPisanty ?

Obliv

@Emilio there was a 3150 rep bounty on that question for whatever reason.

ACuriousMind

@EmilioPisanty You mean a user answering a sum total of one question getting access to essentially all the relevant moderation tools like editing, closing and reopening?

@Obliv There wasn't.

Obliv

@acuriousmind I guess my eyes deceive me scifi.stackexchange.com/questions/2937/…

0celo7

Looks like 3150 to me!

ACuriousMind

2:37 PM

@Obliv Hover over the bounty text. That wasn't a single bounty of 3150.

Obliv

really ._. @acuriousmind you are the master of semantics sometimes

t-minus 3 hrs until my calc final

guess i better work through some more algebra in preparation

0celo7

@ACuriousMind What?

@Obliv are you in summer school?

I can't hover over it

Emilio Pisanty

@ACuriousMind Yeah. That's sixteen separate bounties on that question.

ACuriousMind

@0celo7 Put your mouse over the blue box and wait a few seconds.

Obliv

i'm taking a couple of summer classes so I don't fall behind. I missed some deadlines and couldn't take some crucial classes for my major ^^

0celo7

2:41 PM

@ACuriousMind Yes.

That doesn't work.

@Obliv Which calc

Obliv

2 lol

ACuriousMind

You should see a tooltip listing all the people who have awarded a bounty to the answer

0celo7

@ACuriousMind I know what I should see!

ACuriousMind

@EmilioPisanty Well, if you look in the comments, it appears that question was somewhat "famous" on SciFi.SE. We have a similar phenomenon (though without bounties) with this user on our site.

Emilio Pisanty

@ACuriousMind Yeah, I remember that

0celo7

2:43 PM

@ACuriousMind lol that -1 on his answer

Emilio Pisanty

It was the xkcd effect, though

(From a what-if post if memory serves)

I think our closest equivalent to the question itself is this one, though

101

Q: Superfields and the Inconsistency of regularization by dimensional reduction

Question: How can you show the inconsistency of regularization by dimensional reduction in the $\mathcal{N}=1$ superfield approach (without reducing to components)? Background and some references: Regularization by dimensional reduction (DRed) was introduced by Siegel in 1979 and was shortly...