The h Bar

WillO

12:00 AM

it's not injective on every fiber, but i believe it's a monomorphism

0celo7

Well I've seen the result in a few places

ACuriousMind

I think you need to be precise about whether that thing is a monomorphism in the category of bundles (locally free sheaves) or whether it is actually injective.

WillO

yes exactly. if it's actually injective, i believe it.

ACuriousMind

If it is actually injective, you can define the quotient locally and glue that together again

WillO

right.

0celo7

12:02 AM

@ACuriousMind Huh?

WillO

ocelo: he's just saying what we agreed on before. define B/A fiberwise, and patch.

0celo7

I understand that I can define the quotient, but how do I prove uniqueness up to isomorphism

@WillO yeah, I figured

ACuriousMind

If it is just a mono, the result is false - the quotient of two locally free sheaves is rather irritatingly not always locally free

WillO

but: the category of vector bundles on X is equivalent to the category of projective modules on C(X), and clearly there are monomorphisms of projective modules whose quotient is not projective.

0celo7

@ACuriousMind Ok you know I don't understand that

Ok now you two are showing off

ACuriousMind

12:03 AM

@0celo7 The quotient of two vector bundles does not always exist (as a vector bundle)

0celo7

@ACuriousMind Yeah, sure

I'm not making that claim

ACuriousMind

Just being a monomorphism in the category of bundles is not enough to guarantee the existence.

WillO

you are claiming that for certain sorts of maps A-> B, the quotient exists. this is true if "certain sorts" means injective, false if it means mon

mono

0celo7

Are you telling me a monomorphism is not an injective homomorphism

WillO

yes.

or more precisely, the other way around.

not every monic is injective.

0celo7

12:07 AM

I mean an injective bundle homomorphism.

WillO

then the result is true.

0celo7

I know that it's true, I don't know how to prove it.

WillO

we've just sketched the proof three times.

0celo7

Of uniqueness? Not that I saw.

WillO

uniqueness is just the five lemma.

0celo7

12:08 AM

The fuck?

WillO

If E and F are both quotients, then by definition you have maps E-> F and F-> E

check these are iso by just chasing the diagram

0celo7

...I do?

WillO

Lift from E to B, map from B to F, check the composition is well defined (follows immediately by exactness)

0celo7

Hmm, using the five lemma there is interesting

and ingenious

Does the 5-lemma work when $\alpha,\gamma,\delta,\epsilon$ are isomorphisms?

i.e. does this imply $\beta$ is one, too?

WillO

that is very easy to check for yourself!

0celo7

12:17 AM

I don't find diagram chasing very easy or fun

@WillO A yes/no would be nice so I know if I'm chasing my tail or not

WillO

i'm pretty sure that's a no

0celo7

Ok, then I don't see how the five lemma applies after all

ACuriousMind

@0celo7 Fill with zeros to one side to position the map you want to show as iso in the middle

0celo7

Because you'd get a diagram like the above, where $D$ and $D'$ are the two quotients

WillO

your rows are 0->A->B->E->0 and 0->A->B->F->0

0celo7

12:20 AM

and $\alpha,\beta,\gamma,\epsilon$ are isos

@ACuriousMind ...I'm sorry, what?

@ACuriousMind Oh

Ohhh

DJ Khaled - You Smart, You Loyal

►

@ACuriousMind ^

@WillO thanks too

ACuriousMind

:)

vzn

in theory salon, 1 hour ago, by vzn

↑ havent read this yet/ much too excited, even trouble sleeping this wk! not sure if they lay this out. re spacetime fabric, extending analogy of EM waves to seismic waves it seems clear the E/B EM components are like s-waves and it appears gravitational waves are like the p-waves!

0celo7

12:36 AM

@ACuriousMind oh no that's the wrong one!

DJ Khaled You Smart, You A Genius

►

@ACuriousMind there lol

@ACuriousMind and:

Dj Khaled Celebrates "Hold You Down" Having SEX! YOU SMART YOU LOYAL

►

lol that video title

vzn

@ACuriousMind there can be a subtle interplay/ difference between "unintuitive" and "cognitive dissonance" & think some of the two are being mixed up wrt physics knowledge... (maybe akin to yin-yang)

@Bass agreed and it is now shown in new/ ingenious experiments that even animals "understand" (in a behavioral sense) classical physics. its the "physics of everyday experience/ life" that is also directly/ inextricably tied in with evolutionary survival... in other words classical physics is even tied up with eg human psychology in various ways...

peterh

12:55 AM

@ACuriousMind It would be so good to understand, what are you doing and why. Unfortunately, your meta posts don't contain enough details, and here you seem to ignore me. This essentially limits the cooperative discussion. Did you ever thought to be in the center of an AMA event?

Physics Guy

What is the meaning of these videos ?

vzn

@peterh 2nd would like to see ACM at one

0celo7

ACM isn't a researcher

peterh

@0celo7 Yet.

vzn

have noticed that myself in more ways than 1 but think hes close )( :P

0celo7

1:14 AM

)(?

bolbteppa

1:25 AM

@0celo7 $a^2 + b^2 = 1$ is the 'implicit equation' representation of a circle, and it can be parametrized by $a = \cos(\lambda \theta), b = \sin(\lambda \theta)$. If you want to rigorously prove this, Bloch 'Real numbers' chapter 7 derives trig from basic calculus without assuming even $\pi$.

If you want to link $SO(2)$ to this, start from the equation of a circle $x^2 + y^2 = 1 = \vec{v}^t \vec{v}$, orthogonal symmetries of this are $R$'s s.t. $1 = (Rv)^t(Rv) = v^tR^tRv = v^t v$, $R^tR = I$, then do what Zee does in his group book, expand $R = I + \omega$, find $\omega$ explicitly by solving $\omega^t = - \omega$, exponentiate this and you'll explicitly construct the rotation matrix which describes the elements of $SO(2)$.

Power series trig is linked to geometric trig by the Taylor series man, the book you were reading in your post is just so insane compared to these questions tbh

vzn

@0celo7 wheres slereah when we need him o_O

May 31 at 16:26, by Slereah

246

Q: How do I draw a pair of buttocks?

I'm trying to develop a function which 3D plot would have a buttocks like shape. Several days of searching the web and a dozen my of own attempts to solve the issue have brought nothing but two pitiful formulas below. They have some resemblance to the shape I want, though not quite. Could you...

ps have you thought about hanging out in Mathematics chat instead? it might be more apropos yaknow

0celo7

2:07 AM

@bolbteppa hmm

@bolbteppa what

@vzn what's that in reference to

what what

In the butt

In mathematics, a rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the origin is kept fixed and the x' and y' axes are obtained by rotating the x and y axes counterclockwise through an angle θ {\displaystyle \theta } . A point P has coordinates (x, y) with respect to the original system and coordinates (x', y') with respect to the new system. In the new coordinate system, the point P will appear to have been rotated in the opposite direction, that is, clockwise through...

0celo7

3:33 AM

@bolbteppa too hard

bolbteppa

3:56 AM

@0celo7 P71 - 74 books.google.fr/…

Landau is literally the best writer in existence, Pauli-Lubanski out of thin air

0celo7

4:09 AM

google books stucks

bolbteppa

Thanks for the thank you

0celo7

You linked the section that's also in his GR book, it seems like

bolbteppa

The only $2x2$ matrix, up to a scalar, satisfying $\omega^t = - \omega$ is $\begin{bmatrix} 0 & 1 \\ - 1 & 0 \end{bmatrix}$, so now you have $R = I + \theta \omega$ and $R = e^{\theta \omega} = \sum_{n=0}^{\infty} \frac{\theta^n \omega^n}{n!} = ?$

Since $\omega^2 = - I$ that breaks into the series expansion of sin and cos separately

$R = e^{\theta \omega} = \cos(\theta) I + \sin(\theta) \omega$

0celo7

Yeah, so?

I've read his book

bolbteppa

Circle group

In mathematics, the circle group, denoted by T, is the multiplicative group of all complex numbers with absolute value 1, i.e., the unit circle in the complex plane or simply the unit complex numbers T = { z ∈ C : | z | = 1 } . {\displaystyle \mathbb {T} =\{z\in \mathbb {C} :|z|=1\}.} The circle group forms a subgroup of C×, the multiplicative group of all nonzero complex...

You just constructed elements of the circle group $SO(2)$

0celo7

4:19 AM

So?

bolbteppa

You wanted to know whether $a^2 + b^2 = 1 (= v^t v)$ would allow you to write $a = cos(\theta), b = \sin(\theta)$, this is a proof chat.stackexchange.com/transcript/message/31674840#31674840

0celo7

How expensive is Landau

@bolbteppa Oh, I figured it out hours ago.

Thanks, but I know Zee's method and that was exactly not what I was looking for...

bolbteppa

If you knew it, why did you ask whether you could suppose this was true, when this is a proof

and why did you say it was too hard a few minutes ago

0celo7

Well wait now

You did not prove that if $a^2+b^2=1$ implies $a=\cos\theta,b=\sin\theta$, I don't think.

What you should do is write $\theta=\arctan b/a$

bolbteppa

Well there we go, you don't get it yet, think about it, I put the vectors in brackets for a reason

0celo7

4:23 AM

I'm not terribly interested in cryptic comments...

The future of gaming

nigga you up late

Wow

Early

I'm listening to conservative Christian heavy metal

wtf

it's so good

Slereah

4:25 AM

Can we make a script that bans any posts with the words "relativistic mass"

bolbteppa

Ah, conservatism explains the thanklessness

0celo7

@bolbteppa You're being cryptic, as usual

I'm sorry I'm not thankful for that

so good

bolbteppa

No I'm not, I explained it and you don't understand it but you said you did so I said think about it, bit of a contradiction there...

0celo7

@bolbteppa the $a=\cos\theta$ is just polar coordinates

what's there to understand

bolbteppa

Well that's what I said to myself when I read you confused about it in the first place, but you were unsure if you could assume this, so I gave you a proof...

If you want to get technical, where do polar coordinates even come from

0celo7

4:29 AM

what I want to know is why $\sin\theta=opp/hyp=\sum_{i=1}^\infty stuff$.

bolbteppa

Do you know what a taylor series is

0celo7

No

bolbteppa

Really?

Seriously?

0celo7

If you know the answer, why did you ask

bolbteppa

Are you serious or joking?

0celo7

4:31 AM

probably joking

bolbteppa

Wow

Alright, very quickly, you know $f(x) = f(a) + f'(c)(x - a)$ is a linear approximation right?

Could you prove that $f(x) = f(a) + f'(c)(x - a) + \frac{1}{2}f''(c)(x - a)^2$ is a quadratic approximation, i.e. approximating $f$ by a parabola?

How about cubics etc...

Why can't you keep going until you completely generate the curve near $a$? That is the jist of a Taylor series

Proving it can be done in a few ways

Integration by parts is the best most constructive non-assuming way I know

0celo7

why are you telling me about Taylor series

bolbteppa

Because you said you don't know what they are

Dude

0celo7

I said I was joking

bolbteppa

Why are you asking why $\sin(\theta) = \sum_{i=1}^{\infty} stuff$ then, and why did you ask it earlier, if you know the answer?

0celo7

4:35 AM

I don't know the answer, clearly

I guess I should compute the derivative of sine and cosine

Then it should be trivial

deja vu, I feel like I figured this out a few hours ago

bolbteppa

Yeah, that is the Taylor expansion of sin and cos

0celo7

I know

bolbteppa

...

This is getting a bit ridiculous man, one minute you don't know the answer, the next you knew it all along

0celo7

No

Slereah

that is how @0celo7 works

0celo7

4:37 AM

I don't know how to compute $\sin'x=\cos x$

Difference quotient, yeah

bolbteppa

haha

0celo7

but how

bolbteppa

Ok

Alright

0celo7

although I know that's in my analysis notes...

if you use an angle sum formula I will leave

you must prove that result first

bolbteppa

The awful way to do it is with a difference quotient, then expand $\sin(x + dx)$ using $\sin(A + B) = \sin(A) \cos(B) + \cos(A) \sin(B)$, but I always forget that and hate that proof

There is a cool picture

0celo7

4:39 AM

picture != proof

bolbteppa

20

A: Intuitive understanding of the derivatives of $\sin x$ and $\cos x$

I agree with David (+1), this is the pertinent graph, and it works for me: From here. Updated (added brief explanation to make this self-contained): The main right triangle (in blue) gives $\cos \theta$ (horizontal side) and $\sin \theta$ (vertical). The small change $\Delta \theta$ produc...

The difference quotient assumes that stupid limit too

Let me try to remember a cool proof that avoided all this nonsense

Slereah

$$\sin'(x) = \frac{\sin(x + dx) - sin(x)}{dx} = \frac{(\sin(x) \cos(dx) + \cos(x) \sin(dx)) - sin(x)}{dx} = \frac{\sin(x) (\cos(dx) - 1) + \cos(x) \sin(dx)}{dx}$$

u must evaluate the limits

0celo7

Epsilon delta proof please.

Slereah

4

Q: Nice $\epsilon$-$\delta$ proof that the derivative of $\sin(x)$ is $\cos(x)$?

Looked around a bit and all I see are proofs using the limit definition of a derivative. This is not for an assignment, I could just use the limit definition if I wanted to, but I was wondering how you could go about proving this using the epsilon-delta definition of a derivative ($\forall \epsilon

real-analysis derivatives

0celo7

Too complicated, I'm not an analyst

bolbteppa

4:51 AM

Yeah you pretty much need that limit

I can give you a different method if you're going to be very rigorous and define arc sin using integrals

If you define $\arcsin(x) = \int_0^x \frac{1}{\sqrt{1-t^2}}dt$ as your fundamental trigonometric definition you can get everything from this

Note you have to start from inverses because if you want to define $\sin(\theta)$ you first need to define $\theta$

I expected a bit of Southern hospitality and moonshine, oh well, good luck

Yashas Samaga

Why isn't mathjax enabled for the chat room messages?

$$test$$

David Z

6:01 AM

@YashasSamaga you need an addon for it. It hasn't been programmed into the chat system. See the link at the top right for some resources.

John Rennie

@ACuriousMind Saxon? The Saxon from the early eighties? I saw them at the first ever Monsters of Rock festival in 1980!

Secret

6:22 AM

Roughly ever since that message said he said I must be a genuinely insane person, he becomes a neutrino in h bar, only rarely interacted and reside here.

Of course, what other h barers said about him may also be true, cause he might be too busy on data mining

It will soon be my fate once I start working...

John Duffield

9:44 AM

@peterh : I don't know the details. But I do know that it's an imperfect world peter.

Wood

I have a question, but since I don't want to get downvoted, I'm gonna try to ask it here: about the holographic principle, isn't it trivial to translate a theory in N dimensions to N-1 by just using a bijection from R^N to R^N-1?

peterh

@Wood Google for that, explain why do you think it (using the google refs). Show that yes, it is a realistic thing in your knowledge level. Format the things well. Downs aren't very bad things here on my experience, the closes are the bad in my opinion.

@Wood Btw, downvoted and closed question will be automatically deleted in some days or weeks, and after deletion you get the reps back.

@Wood If your question is closed and you don't understand the reason, ask it is comments. If you don't get answer, answer it on the meta (politely). Your meta question will be probably downvoted, but you will probably get an answer (if you are lucky, your question will be even reopened). After that, think a lot, google again, and re-ask.

@Bass I think simulation games should be developed where SR (or even GR) and QM "features" are visible and can be easily experienced. For example, a first-person-shooter where $c$ is only around $40 \frac{km}{h}$. It would help a lot to develop the required intuition.

ACuriousMind

10:38 AM

@JohnRennie Yes, that Saxon :)

@Wood Those bijections are discontinuous and therefore don't yield a proper theory after you apply them.

@Slereah Is that why you've not been around so much to ramble about AQFT or CTCs? ;)

"Nothing but puppies.png" sounds like the most obvious virus disguise ever :D

Slereah

11:01 AM

@ACuriousMind Pretty much

Did you know @ACuriousMind

It is very hard to make AQFT work on a non-globally hyperbolic spacetime

But a few spacetimes can do it

Maybe I should reread that paper, now that I know roughly what AQFT is

arxiv.org/pdf/1504.00586.pdf

ACuriousMind

I really don't know much about QFT on non-Minkowski to begin with

Slereah

I think it's basically the same thing?

Except that you have to replace light cones with chronological futures

and the algebra changes to fit the new hamiltonian

user5444075

Hello everyone!!There is this thing bothering me..
When an led is connected in a circuit it needs a resistor so that excess current does not pad through it,but when resistor is placed after the LED how does it still work; it has high amount of current flowing through it as the resistor opposes the current after it flows through the LED.
Help appreciated!!

Slereah

AQFT is a bit tough on curved spacetimes because you lose a fair amount of axioms

Poincaré infariance

Uniqueness of the vacuum

etc

ACuriousMind

Yeah, that non-uniqueness of the vacuum is rather annoying

@user5444075 The current in a series circuit is everywhere the same

Slereah

11:06 AM

Then again vacua aren't unique in interacting theories, anyway

ACuriousMind

@Slereah Hush, we don't like to talk about that ;)

But I see your point, it's not really surprising

user5444075

The thing is I want to know what happens at a very microscopic level and no one seems to explain at that level

ACuriousMind

@user5444075 What about the usual flow analogy doesn't satisfy you? If I have a valve in a water circuit, it doesn't matter where I put it, it still regulates the flow in the entire circuit

Slereah

Wait, the Borsher algebra isn't a $C^*$ algebra?

I thought that was the point

Apparently you have to do more wizardry on it

user5444075

How do the elctrons know that there is a resistance ahead before actually flowing through it??

ACuriousMind

11:09 AM

Explaining circuits "at a microscopic level" is a) rather tricky because there are several different models of conduction and b) rarely enlightening

6

A: How do electrons "know" to share their voltage between two resistors?

Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effec...

John Rennie

@ACuriousMind I last saw them in 2008 when they supported Motorhead. Biff Byford was looking a bit old even then, though for that matter Lemmy was no spring chicken. Good show though. Do they still play "747"?

Slereah

I wonder if the F algebra business of AQFT with CTCs reproduces the "classical" result

Where time evolution isn't unitary

user5444075

Ill check in that and be back again

Slereah

I really should learn about wavefront sets and microcausality

Bass

@user5444075 Think of it as a water pipe with some pressure coming from the left, and a very narrow passage to the right (a "resistor"). How does the water coming from the left "know" that there's a narrow location?

ACuriousMind

11:13 AM

@JohnRennie I think they did (I don't actually know many of their songs, I listened to them because they were the opening act, so to speak). Biff ate the set list halfway through and let the fans shout which songs they wanted :D

John Rennie

There are a few front men who are genuinely nice blokes and Biff Byford is one. Back in the eighties when I was a teen metaller he would always come to the stage door after the show, sign autographs and chat to us.

Actually Lemmy was basically a nice bloke though he could be a little intimidating.

I might cue up Wheels of Steel now, just for old times sake :-)

ACuriousMind

Ah, they played Wheels of Steel!

Itachí Uchiha

Is curl operator applicable in two dimension?

ACuriousMind

@ItachíUchiha That...depends on what you mean by that

The natural notion of a curl of a 2D vector field $(v_x,v_y)$ is $\partial_x v_y - \partial_y v_x$.

Which you should think of as a vector of that length standing perpendicular on the 2D plane.

user5444075

11:40 AM

@Bass what if the water is flowing for the first time and there is a fan in between the small pipe and the left side and the fan cannot tollerate much water? The fan will be destroyed as soon as high amount of water touches it (fan=led)

ACuriousMind

@user5444075 Both the rules for water and for electric circuits are in equilibrium (as the answers to the question I linked up there say). When you "start up" the circuits (no matter whether water or electric), there's a short period where they don't obey the simple rules we're used to until they've settled down.

user5444075

Im so confused!! How do they come to know either way, even after they've settled down??

ACuriousMind

Did you read the answer I linked (by dmckee)?

user5444075

Yes

ACuriousMind

Then why are you still asking that question? They don't "know", it's a dynamical system that finds its equilibrum in the state where the current throughout a series circuit is everywhere the same

dmckee explains it quite well, I thought: Whenever that's not the case, the fields/forces are such that this state is attained after a short period of time. There is no "knowing" involved, this is simply the stable state of the system.

user5444075

11:52 AM

But, till it stabilizes won't the LED get damaged at first place?

ACuriousMind

Why would it?

The main "damage" components take from current is by heating up, and heating up takes time, too

Itachí Uchiha

@ACuriousMind,How do I find the magnetic field of dipole of magnetic moment at a point that is making 60 degrees with the dipole axis?Can I use this :wikimedia.org/api/rest_v1/media/math/render/svg/… and wikimedia.org/api/rest_v1/media/math/render/svg/…

Bass

@user5444075 Imagine a crowd of people entering a big concert at the same time. The resistor is the security that cannot handle everyone at the same time, so people have to queue. How do those in the back know that there's a resistor down their way? They know it because there are people standing in front of them, people who are waiting too.

ACuriousMind

@ItachíUchiha I'm not sure why you linked the images of the formulae, but sure, to compute the B-field of a dipole you use the formulae for the B-field of a dipole. (I'm not really sure what you're unsure about)

Bass

Now the only difference to the wire with the LED is that the wire is always filled with electrons. So "everyone has to queue". There's no moment when the queue is empty.

user5444075

11:58 AM

So if a suddenly attached a new resistor (2) in the closed circuit will the current be the same everywhere in the circuit @ACuriousMind

*i

Itachí Uchiha

I am new to vectors.Since the point has two dimensional,How would I progress?(like finding $A$ separately along x and y axis.,then curling them and do vector addition?)
@ACuriousMind

ACuriousMind

@user5444075 After a short period of time, yes (if you attached it in series).

@ItachíUchiha I'd say you "progress" just by plugging your values into those formulae. If your issue is that you are not comfortable with vectors and thus can't "just plug in", you should maybe first get used to them before trying to apply them in such situations. I've got to go and can't walk you through this, sorry

Itachí Uchiha

Thanks for helping. :)

user5444075

@Bass So if the circuit is open and I connect everything will the electrons at the end of the queue know that they need to go slow

Anyways thanks for the Help teally appreciate it @ACuriousMind

*really

@Bass thanks for helping but really cant get this

*I

David Z

Hey, so I might as well drop a notice in here: I'm tweeting from the @realscientists Twitter account this week and I will be talking up PSE, so we might see a bit of a bump in new members. I'm not sure if it will be significant, but let's be aware.

2

Bass

12:30 PM

@user5444075 They "know" it because they are repelled by the electrons in front of them. Like the people waiting to get inside the concert hall. They can't enter as fast as they want because there's people in front of them.

user5444075

12:45 PM

Ok perhaps now I fully understand all this?!

Anyways thanks for helping me. You were very cooperative!@Bass

☺

Bass

No problem :)

0celo7

1:16 PM

@ACuriousMind can you explain the $\Bbb R\to S^1$ thing again please with the exponentiation

The morphism $g:\Bbb R\to S^1$ drops down to a linear map $g_*:\Bbb R\to\Bbb R,t\mapsto at$ of Lie algebras

But do we have $g\circ\exp=\exp\circ g_*$?

peterh

@ACuriousMind What I understand: you are gaming, and you continously try to close & destruct newbie questions. What I don't understand: why are you doing this. For that, I have only hyphotheses. | My best hyphothese is that you want to get the site into a MO-like direction. It would be a good idea also for me, but after I've tried to gain your support for a solution, you went in ignore on a trumped reason.

@ACuriousMind I tolerate your behavior, because you are extreme highly competent in physics and produce a lot of very HQ content, but I simply can't understand what makes enjoyable to harm others while you are totally closed for any cooperative talk.

0celo7

I really don't see how to do it explicitly...

@ACuriousMind You said to do it explicitly, but how do you do that?

Every $g(t)$ is rotation, but how do you get that they fit together nicely

ACuriousMind

@0celo7 Yes. "The Lie algebra of the image is the image of the Lie algebra".

I'm afraid (not actually afraid, mind you...) I have to go to a party now, I might be away the rest of the day

peterh

@ACuriousMind If you would make this openly, you would probably get my support.

@JohnRennie Why is the culture of the physicists so different from the mathematicians? Do you think it is somehow because they need costly projects for experiments, while mathematicians need only pen & paper?

David Z

1:39 PM

@peterh take note that when you make accusations like that, you really need to back them up and make sure you are not going beyond the facts. To do otherwise is running up against our "be nice" rule.

peterh

@DavidZ I don't make accusations, if I would, I would do this on the meta. I want to understand and find a way to a cooperative talk, while I greatly honor his work on the site to make wonderful very HQ answers for the whole internet.

David Z

@peterh I don't care to argue terminology, so let me put it this way: statements like "What I understand: you are gaming, and you continously try to close & destruct newbie questions", whatever you call them, are borderline inappropriate.

peterh

@DavidZ Yes, I can back it up, but I won't make any accusations. I want to understand why he does this, if he would have a clear statement, a clear algorithm, clearly communicated motives, probably I would have at least something to cooperate with or to reject it. And it is much more probable that I would support it. But I don't have anything, and he doesn't seem to want to talk about his motives.

John Rennie

@peterh I guess it's nice to know you are willing to tolerate valued members of this site who have contributed more to it than you ever will.

peterh

@JohnRennie Exactly. Thanks.

0celo7

1:49 PM

fukn rekt by @JohnRennie

god damn

John Duffield

Anybody want to talk physics? For example, about the difference between curved space and curved spacetime? The distinction is crucial.

Bernard Meurer

@JohnRennie Holy crap

What a burn

@JohnRennie What is the secret to women?

0celo7

@JohnDuffield imo, there's no difference

Difference/distinction

John Duffield

@0celo7 : imo, there is.

David Z

@peterh Again, statements like the one I quoted are borderline inappropriate. We tend to be a little lenient if it's obvious that you're sticking to the facts, e.g. by quoting and linking, but in general, it's probably best not to say such things.

Bernard Meurer

1:52 PM

Great discussion

0celo7

If that's your opinion

John Rennie

@BernardMeurer Be nice to them and have a friendly smile.

Bernard Meurer

@JohnRennie I was hoping for more immolation of grapes in that answer

dang it

peterh

@JohnRennie Any idea, why the culture of the mathematicians differ so from the culture of the physicists? The only difference I can see is that the physicists require costly experiments. But if it is so, what is the causality chain?

0celo7

@JohnRennie worked for me

John Duffield

1:53 PM

@0celo7 : not just mine.

peterh

@DavidZ Yes, it could be done, but I won't do it, because I consider his work much more important as the harm he causes.

0celo7

@JohnDuffield multiple people can have a wrong opinion

peterh

@DavidZ What I only do: desperate tries to a cooperative talk.

David Z

@peterh OK. In that case, just drop it.

John Duffield

@0celo7 : yep!

John Rennie

1:54 PM

@peterh frankly you're making yourself persona non grata around here. Please don't ping me again.

4

Secret

Mathematician concern about rigor, while physicist can tolerate some orders of magnitude error as long the current scale the model is applied those errors are negligible

Physicists also need to constantly reference to the exprimental result and other intuition in order to ensure things make sense

To them, maths is mostly a tool in building the model, details don't matter

also physicists have a tendency to approximate nonideal things as a power series of ideal things

David Z

@peterh It's possible you're being blocked, anyway.

0celo7

@JohnRennie what did he do?

Wow what happened

peterh

@JohnRennie I am sorry for that, but honestly what I've experienced until now were my unanswered questions.

John Duffield

Ping!

Secret

1:59 PM

Persona non grata

In diplomacy, the term persona non grata (Latin, plural: personae non gratae), literally meaning "person not appreciated", refers to a foreign person whose entering or remaining in a particular country is prohibited by that country's government. It is the most serious form of censure which one country can apply to foreign diplomats, who are otherwise protected by diplomatic immunity from arrest and other normal kinds of prosecution. == Diplomacy == Under Article 9 of the Vienna Convention on Diplomatic Relations, a receiving State may "at any time and without having to explain its decision" declare...

John Duffield

Anybody want to talk some physics?

peterh

@DavidZ Possible, but...

@Secret Thanks, I know the term, but if I find it unfair, I think it says more from the community as from me.

Secret

@peterh No the reason I posted this is because it reminds me of something...

Secret

9 mins ago, by John Rennie

@peterh frankly you're making yourself persona non grata around here. Please don't ping me again.

36 mins ago, by ACuriousMind

I'm afraid (not actually afraid, mind you...) I have to go to a party now, I might be away the rest of the day

Aug 3 at 2:05, by Slereah

I sincerely believe he is a crazy man

Jul 31 at 17:48, by vzn

@ACuriousMind honestly actually not really following DSs original point apparently linking human heredity/ evolution/ DNA & art or science or the point being debated/ overall topic/ assertion etc, there seems to be some "talking past one another". but there is possibly some connection in some areas eg music in particular. anyway this all seems to be triggered by a rather provocative/ argumentative/ sacred cow-goring n00b...

I don't know what it really means, but this is like the 4th case this give me this undescribable emotion in the same month

Perhaps I am just being too paranoid about my future plans in my study and work

peterh

@DavidZ I see already a significant reason to drop the case.

Secret

2:07 PM

My brain is a tangled mess, sometimes just one word is enough to remind a whole lot of things that simply don't make sense in the eyes of anyone else (even me in some case)

Bernard Meurer

@Secret Nothing of what I think makes sense really

@JohnRennie Only you can save me

@JohnRennie I need to check if you'll get a joke. Are you around?

John Rennie

@BernardMeurer joke away :-)

Secret

@BernardMeurer This might be one reason why you so far have no records of ignoring me so far. You, perhaps like my supervisor and my friends, somehow knew what I am thinking

Bernard Meurer

@JohnRennie So I've started writing my own Kernel/OS with it's own bootloader and all

I've called it Daedalus, and this week I finished it's logo

John Rennie

@BernardMeurer Cool! :-)

Bernard Meurer

2:09 PM

Here it is:

John Rennie

The suspense ...

Bernard Meurer

Do you get it?

John Rennie

You're dissing Windows! BURN HIM!

Bernard Meurer

@Secret I can't say I know what you're thinking, but I most definitely know you are thinking, which by all my standards is more than enough

@JohnRennie Lol

I'm very of this ~~photoshop~~ GIMP

John Rennie

At it's core Windows is a very good operating system. The design of device drivers is better than any other OS I know. The trouble is they take the great kernel and wrap it in so much crap.

Bernard Meurer

2:16 PM

I think MINIX has nicer driver design than Windows

I'm not a fan of Linux's, a bad driver will fuck you up

Secret

Johnrennie also somewhat knew what I am thinking (or at least in the sense that he knows that I am thinking or something and (I don't know))

One example being way back when I ask the h bar about the time travel model (which is the first time I mentioned about it), he is the only one who tries to ask sense about it, and that brought up a whole lot of spacetime topology discussion with Slereah and co.

If there's one thing, compared to the Theoretical h bar Trio, he seemed to be a lot more tolerant to my weirdness in my messy mind and can comprehend it somewhat

Back in my real life, there are always two group of people: One is a group who somehow understood me, another will keep asking me what with me trying to explain what I have in mind

Perhaps I am lucky my supervisor is type 1, as that helps him to comprehend my lthesis and thus errors are easily squashed

peterh

@BernardMeurer There is a middle-strong unsaid direction in linux to a microkernel direction. It is probably a compromise between the uC volks and Linus.

Secret

however I do have a long track record that my cosupervisor (being the manager in a part time job, or just any second in command of any project I participate in) is always type 2

and that is a main source of frustration

It gets interesting when it is found that for some cosupervisors, they thought they always want to seek my approval while I never desired any of that

John Rennie

@BernardMeurer I have to admit I don't know MINIX at all.

Secret

so yeah, human communication is a frustrating business that cannot be described or predicted by formulae

John Rennie

2:24 PM

I think there is a limit to have far you should go in making the kernel resistant to driver bugs. At the end of the day the correct approach is not to have bugs in device drivers.

Secret

I don't know how many times I need to explain to type 2 people what I think, but at the same time, they think I just don't understand them.

So, basically both sides are utterly confused on what the other is thinking

and no amount of explanation can get it clear. PErhaps there's simply something that cannot be expressed in words that are unique to our worldviews

John Rennie

The early builds of Windows NT went quite a long way down the microkernel + client/server design route, but they couldn't get the performance up to scratch. By the time NT3.0 launched they'd taken a lot of that stuff out, and every release since has got closer and closer to the standard monolithic kernel design.

Danu

0

Q: Is virtual particle the same thing as force particle?

Is virtual particle the same thing as force particle? Which of the above categories do photon, graviton and gluon fall into? Virtual particle, force particle, or both?

quantum-field-theory forces interactions virtual-particles carrier-particles

I foresee problems!

Secret

If there's one day this communication issue between me and type 2 people can be resolved, I will be happy because we can finally understood what each other are thinking

Danu: I think not all virtual particles are force carriers

Danu

I think virtual particles are not a thing---but that has been discussed ad nauseam in this chat room, already.

Secret

2:29 PM

indeed

John Rennie

> Maybe physicists would complain less that philosophy is useless if it wasn’t useless.

Ooh!

backreaction.blogspot.co.uk/2016/08/…

Secret

There's really only one thing that is useless, and that is...
(The Suspense...)

seriousness aside, here's a joke for that
http://www.theuselessweb.com/

(It does not matter what it means, or what it is, or whether it has existed, as long it is uselessly useless)

However, philosophy is not a useless thing, nor is the complaint of it

vzn

2:47 PM

@Secret cant tell why he said that not sure what he was talking about, he calls stuff "insane" in vernacular sometimes. this? (you were talking about noncommutative qm probabilities...)

Aug 3 at 2:04, by Slereah

Is it, you insane man

John Duffield

3:06 PM

@JohnRennie : there she goes, pontificating again. And she doesn't even know the difference between curved space and curved spacetime, and between virtual particles and vacuum fluctuations.

peterh

@Danu In physics, things what don't seem to exist, but are coming very clearly from the formulas, are with time considered existing.

Itachí Uchiha

3:52 PM

How do springs work?

peterh

@ItachíUchiha Too broad, unclear, opinion-based :-)

John Duffield

All: I've asked another question. It's how does gamma-gamma pair production really work? I intend to give a 500-point bonus.

Itachí Uchiha

Does it mean that we donot know how spring really works?

peterh

@JohnDuffield Wow. :-) It seems the site could gulp your first nice initiative, although it was hard for them. But I am not sure it will happen again.

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