01:00 - 17:0017:00 - 00:00

5:00 PM
P.S. I have not plug in the details of statement 1 yet, which include the aforementioned discussion of the algebraic structure a few hours ago
I felt that need to be generalised to a set of any number of elements otherwise it is too specific

@2physics Try this

@JohnRennie you know all these tricks and methods? or you also look them up when you need?

@2physics For me I tend to google whenever I need a mathjax command to type something

@2physics I routinely use only a small subset of all the stuff the formatting can do. The more complicated stuff I Google when I need it.

Tex SE is also a good place for mathjax commands

5:10 PM
@JohnRennie okay thank you .. and you too @secret

Hai
@JohnRennie I've got a question for ya

presented the question as a gift :D

Why do I see a lot of books titled like "Vectorial mechanics for engineers". Why would a book on maths be 'for engineers' I feel like this is akin of "Quantum Physics for babies"

I don't know. I'm not an engineer and I don't read books targetted at engineers.
But engineers tend to be rather focussed in their use of maths, so I'm not surprised they would want maths books targetted at them.

5:15 PM
Engineer here

@0celo7 you're not a typical engineer :-)

I don't know what this shit is, I read grad texts in math and grad studies in math books
GTM, GSM
Standard math books really

@0celo7 You're a mathematician, you're just getting an engineer degree to make Bob happy

@BernardMeurer are you an engineer too?

No, it's because I love GDP.

5:18 PM
@2physics I'm a child
And a freshmen for CEng

You're as old as I am

@BernardMeurer I've never seen a child confessing to his/her being child
@0celo7 how old is he?

@0celo7 Yeah, but you're smarter than me

18

@2physics I'm 18

5:20 PM
What is there to confess? It's being a grown-up that you should be ashamed of! :P

@2physics I'm honest

@0celo7 how old are you

@2physics He just said we have the same age :p
@ACuriousMind Old hag :p

@JohnRennie I got so angry at Make that I'm making my own build script with Python, fuck that

5:22 PM
I would stick with make. The key to getting make to work reliably is to keep it simple.

@ACuriousMind any truth is confessable.
or maybe confessionary.. don't know the word:D

To my knowledge, "confess" in English carries a connotation of the thing being confessed being shameful.

@ACuriousMind oh really? didn't know.. so I must confess that I'm surprised.
:D

@JohnRennie I can't use it, and I don't want to learn it now, I'll make a nicer makefile once my kernel can run C code with the stdlib

@ACuriousMind maybe "declare" is better to use then.
or announce? I don't know.. or maybe accept

5:28 PM
@ACuriousMind ehhhhh

@0celo7 If I understood correctly from MSE, a broken geodesic is like a broken line graph version of a geodesic on some manifold, do the cusp joining any two segments solve the geodesic equations?
2

Let $\gamma$ be an infinite broken geodesic in the hyperbolic plane, that is a curve formed by consecutive geodesic segments. Assume also that each of these segments is longer than a certain positive constant $C$. For any two consecutive geodesic segments $[p,q]$ and $[q,r]$, we can consider the ...

@Secret I figured it out, the proof is trivial

Is it important to have the corners of each broken geodesic to solve the geodesic equation, or just their neightbourhood is sufficient?
@0celo7 The above is a completely different question

5:45 PM
The corners are not geodesics

I see, makes sense

You mean the point where they meet?

yes

It's a geodesic from each side, but not a geodesic at that point.

I guess its because they are not differentiable there?

5:46 PM
Yes
One can show that a geodesic must be smooth.

ok

Hmm. I don't remember the proof of that. I think you use a variational technique to show that a critical point of the energy functional is a geodesic, and there's a way of getting smoothness too.
Oh, no
A geodesic solves an ODE so smoothness follows from ODE theory.

6:05 PM
A video from the RSS feed question
0

Levitating a superconductor on a Mobius strip In the video the superconductor levitates upside down, how is it possible when the gravity and acting downward and by definition superconductor repels the magnetic field.

I need to give titles to my PDF collection
I need to find a book in this mess
not working very well

That reminds me some time later I need to ask Slereah about klein bottle spacetimes

6:22 PM
If I have an open set $U\subset M\times M$, can I fit $B(\delta)\times B(\delta)$ inside of it, where $B(\delta)$ is an open geodesic $\delta$-ball?
Probably.
The manifold topology agrees with the Riemannian topology, and geodesic balls thus form a basis.
So using the definition of product topology, the result follows.
Yeah, that works.

1 hour later…
7:34 PM
IMHO the answers to this question demonstrate everything that's wrong with physics stack exchange, in a microcosm.
7

We live in an expanding universe - so I'm told. But how can that be possible? Everything imaginable is attracted by a bigger thing. So, why can't gravitation stop the expansion of the universe? I know the "Big Bang" theory, but is it possible that the expansion of the universe is caused by the at...

Gravity will not make space fall down, or up, no matter how many people vote for it.

lol
-1

That too.

you used that wrong Einstein quote again
I downvote every time just for that

Einstein wasn't wrong. And the moot point is that science is not a democracy. I see correct answers from people whom I know to be experts, and incorrect "popscience" answers from people who aren't. The latter gets more upvotes, and the former votes with his feet. The expert poster becomes the ex poster.
Because the 18-year-old who thinks Einstein was wrong downvotes every time.

8:01 PM
Hi.I'm new here.Can someone help me this question please ?physics.stackexchange.com/questions/276432/…

@JohnDuffield Don't you ever get tired?

9:00 PM
@JohnDuffield no, it's really just you whom I feel compelled to downvote constantly

@BernardMeurer : no. I just get more determined. Here's why:
"I say that because there’s a standing joke in our house, that I’m the only one who can change a light bulb. But somehow it isn’t funny. If you selected a hundred people at random and tested their technical and scientific knowledge, I think the average score would be lower than that of a comparable group from fifty years ago. Yes, we’re more specialist these days, and some things are more difficult to understand.
But it seems there’s more people around who just don’t understand the basics, who have only the vaguest concept of how things work. They wouldn’t know where to start if their car broke down. It’s like there’s a low-rise, low-brow tide that doesn’t feel healthy, that slowly, insidiously, is getting worse."

Could Einstein change a lightbulb?

@0celo7 Bit difficult with those paws, isn't it? :P

@0celo7 : yes he could. He answered questions too. You haven't answered a question in two months.

@JohnDuffield I don't know/care about physics, so why would I answer questions here?
@ACuriousMind Hah

9:14 PM
@0celo7 : because that's the sort of web site it is.
OK, now if you'll excuse me, I have work to do.

@0celo7 This chat hasn't seen a cat pic in ages!

@ACuriousMind Because I haven't lived at home for a year
rewriting my $\ell^p$ proof
@ACuriousMind did you have to do this in your analysis course?
it's a little long

I didn't hear the word "Banach space" till I took functional analysis

My course is titled "analysis in $\Bbb R^n$ and Banach spaces"

Well, mine wasn't, what else can I say? :P

9:24 PM
you can help!
or at least read this when I'm done
@ACuriousMind When an inequality $<$ turns into $\le$, is that called a "sharpening"

You could call it that, yes.
Usually I'd use that term for replacing $a < b$ by $a < c$, where $c \leq b$.

This is the correct proof, finally :P
@ACuriousMind Does it pass muster

@0celo7 It looks fine

Banach is a funny name

@Bernard hi, saw your assembly code awhile back, wondering what are you working on?

9:37 PM
@BernardMeurer How so?

@vzn I'm writing my own Kernel

It makes for horrible puns: What is yellow and complete?
A banana space.

but thankfully I'm onto C now, way more high level

@ACuriousMind Can you produce a proof, off the top of your head, that all norms on a finite dim space are equivalent?

@ACuriousMind lol

9:39 PM
@BernardMeurer wow very ambitious. saw this today & it reminded me of your hackerspace, some fun stuff/ prjs/ equipment/ angles etc sparkfun.com/news

@0celo7 I think showing that they are all equivalent to the maximum norm is easy, and being equivalent is transitive so you're done.

@vzn I just needed asm to do Real mode -> Protected mode -> Paging -> Long mode

@ACuriousMind Hmm, I'm wondering why books don't do that then.

@vzn AH, yes, sparkfun, they really are awesome

My textbook for analysis does a brute force proof that any two are equivalent

9:40 PM
@DanielSank By way of apology re my previous octopus link: arstechnica.co.uk/science/2016/08/…
It looks pretty cuddly

I wonder if I can run Linux on my Linux hat...

@BernardMeurer thought if you were into hackerspaces you might have heard of em. liked the network chess kit & the jet turbine off of ebay etc sparkfun.com/news/2155 sparkfun.com/news/1890 ... same guy does battlebots =D

@vzn They have so much cool stuff

@EmilioPisanty somewhat similar to this one saw recently theverge.com/2016/8/16/12499646/…

@0celo7 Okay: Fix an isomorphism to $\mathbb{R}^n$. Then for any $x = (x_1,\dots,x_n)$, we can take the maximum $M$ of $||x||$ on the unit ball w.r.t. the maximum norm, and this from this we get $||x|| < (M+1)||x||_\text{max}$. We can also take the minimum $m$, and get $m/2 ||x||_\text{max} < ||x||$.

9:45 PM
you're assuming I know this isomorphism
when did you start assuming I know linear algebra

In fact, this should work for any norm, as long as you already know the unit ball w.r.t. the norm is compact.

Yeah, that might be tricky.
I guess show that it's closed and bounded?
It's clearly bounded by the ball of radius 2

@0celo7 Well, it's easy for the maximum norm, it's a cube, i.e. $[-1,1]^n$.

closed...I'd have to ponder that.

@BernardMeurer have you ever ordered anything from em? am probably gonna go to this in few wks, went last yr, it was big fun avc.sparkfun.com/?_ga=1.179764519.2014959095.1472144344

9:47 PM
@ACuriousMind Does that give compactness already?

@0celo7 Finite product of compact sets is compect, no?

@vzn Yep, I have some components by them, my review is 100% positive, great company, great products

The hard part of Tychonoff's theorem is doing that for arbitrary products, but for finite products, it's easy

@ACuriousMind Uh, sure.
Wait
We don't need that the unit ball in the first norm is compact?
hmm. guess we don't.
@ACuriousMind I think Tynchonoff is cheating. Did you have that in your analysis course?

@0celo7 No, I only need to be able to take the maximum/minimum over the unit ball w.r.t. the maximum (or Euclidean, or whatever you want to use) norm.
@0celo7 Which is why I said it's easy to show for finite products.
I didn't know Tychonoff, but we don't need its full strength here

9:49 PM
Ok, so it's cheating
But let's examine your proof anyway...
@ACuriousMind Is the max norm on $\Bbb R$ equivalent to the standard one
you have to prove $[0,1]$ is compact if you wanna use Tynchonoff
is the max norm topology even a product topology?

@0celo7 Explain to me how the max norm is even different from the standard norm on $\mathbb{R}$.
@0celo7 It's the absolute value, silly!

I know
But still, is the max norm topology the product of max norm topologies on $\Bbb R$?
it is because of what we're proving, but that's cheating.

@0celo7 I was about to say that that follows from what we're proving, why is that relevant?

If you want to use Tynchonoff you have to know that $[0,1]$ is compact

If you are worried that we can't know that the cube is compact in the max norm topology, then take the Euclidean norm.

9:54 PM
But you want that $[0,1]^n$ is compact in the max norm topology
But Tynchonoff talks about product topologies
@ACuriousMind how does that help?

@0celo7 My idea doesn't rely on using the max norm. You can use whatever norm you want.

I'm confused

I'm confused what you are confused by

If you don't like my "cube is compact because it's $[-1,1]^n$", then use the Euclidean norm and use the standard ball, which you should know to be compact.

9:57 PM
women, for another

Or if you don't want to use Tychonoff, then use Heine-Borel to show the cube is compact

@ACuriousMind No, we haven't proven Heine-Borel. And I don't know why the ball is closed.
(I do, but I don't know)
(We're talking about a poor analysis child, not me)

Oh, I'm not doing this silly routine. If you willfully don't want to use any other theorems, then of course you can't show the statement any other way than by brute force.

I don't see how this is silly
How do you show that the ball is compact without Heine-Borel in the Euclidean case?
Genuine question.

@0celo7 It's silly because you asked me "is there any other way to prove this", and when I give an answer, you add arbitrary restrictions to what is allowed in the proof
I find that both silly and annoying.

10:01 PM
Can you prove Bolzano-Weierstrass directly for the ball?

@0celo7 I don't care!

@ACuriousMind Calm down
You said I should "know it is compact"
Is that simply via Heine-Borel or did you mean something else?

@0celo7 I didn't think about how you know this. Euclidean Heine-Borel, finite Tychonoff, it doesn't matter. But you know this. If you want to pretend you don't for some strange reason, then fine. But that's not my problem.

@ACuriousMind Ok, and I'm curious how you use Tynchonoff to show that the Euclidean ball is compact.
Closed subset of compact set is compact?

Yes

10:04 PM
Ok, that's all I wanted
Sorry for the offense.
(you still have to show closedness)
(I'll work that out)
@ACuriousMind Let $S=\{v\mid ||v||=1\}$. Let $v_n\to v$ be a sequence in $S$. Then $|||v_n||-||v|||\le ||v_n-v||\to 0$, so $\lim||v_n||=||v||$. But $||v_n||=1$ for all $n$. Done.

10:22 PM
@yuggib what's up

1 hour later…
11:46 PM
5

I can offer up something similar to this, which is an isomorphism between something called the Tsirelson bound and the spacetime metric. This is not exactly the emergence of spacetime from quantum mechanics, but it does illustrate how spacetime could be seen as quantum mechanics in diguise. Supp...

^this answer, and probably some other answer by the same user as well, could use some closer scrutiny

why?
Poor Straumann, why did you have to fall apart
you deserved better

There is a consistent pattern of not actually answering the question posed, and using a lot of formulae and terminology while actually misusing the terminology. The answers are well-written and sound good, but if you look at them in detail at least some of them just fall apart

@DanielSank this one could probably use a good samaritan like yourself
-2

I've problem choosing between two classes. One is about Banach algebra and spectral theory; the other is PDE class. I'm a theoretical physicist. Which one should I choose. Thanks; any help will be appreciated.

For instance, in this case, I can't actually tell what the argument is supposed to be

sort of like Timaeus?

11:48 PM
The only argument I can extract from it is that the Tsirelson bound has the same sign pattern like the Minkowski metric, hence the two must be "deeply related"

seems like a perfectly valid physics argument to me @ACuriousMind

@0celo7 Ha. Ha. Ha.

in my QM lecture we went Stern-Gerlach -> light polarization -> Hilbert space + Born rule
leaps of logic are basically what physics is

01:00 - 17:0017:00 - 00:00