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12:49 AM
@Cows good
 
@0celouvskyopoulo7 i wrote a proof
 
yay!
So far the scariest thing was Union of sets and and intersection of sets
not bad so far
@0celouvskyopoulo7 I am on page 11 now. Looking at the uniqueness of measures now.
@0celouvskyopoulo7 what are you up to?
 
Learning the Choi-Wang theorem.
 
sounds hardcore. I will stick with page 11 of this text for now :D
 
1:05 AM
that's a good theorem
you only need to check that measures agree on the algebra of sets that generates the sigma algebra
 
1:30 AM
Hello
 
 
2 hours later…
vzn
3:19 AM
@GPhys very sorry to hear that but maybe a chance to chat/ type while resting/ recuperating? :) physics.meta.stackexchange.com/questions/7783/…
 
 
1 hour later…
4:39 AM
100 pages
100 pages and he doesn't prove what I need
3
I wand to end it
 
4:51 AM
aha
I have a hint
Hmm
I like Hebey's argument.
I think I can generalize it
I need to write a book.
@SirCumference ::laughs maniacally:*
I will pay you to help me
 
Uh oh...
 
AHA
The poincare inequality proof goes through for the $W^{1,2}_c$ space
I just need Rellich's theorem on $\partial$-manifolds
Guess what I'm doing tomorrow?
 
Finding it?
 
Proving it
It should be easy once I have all of it in front of me
 
5:21 AM
2 messages moved to Trash
@JaimeGallego Done
 
Hey guys, small question. When calculating electric dipole moments, is it valid to place the reference point on a charge. I.e. consider q1 and q2 separated by r, the dipole moment is given by q1(r1-r)+q2(r2-r), is it ok to put r such that it is located at q1 (or q2)?
 
A multipole expansion can be calculated about any point, in the same way that angular momentum can be calculated about any point. The point you choose for the calculation is generally selected for its physical relevance.
 
I see, thanks
 
 
1 hour later…
user228700
6:40 AM
@JohnR: Morning :-)
 
Morning :-)
Indeed it's a good morning as I've just finished that big chunk of code I was working on.
 
Although there is considerable testing yet to be done.
 
user228700
@JohnRennie Nice! :-)
 
@dmckee In my later years at Unilever I used to do preliminary interviews for applicants. I never asked them why they wanted to work for Unilever :-)
 
6:43 AM
@JohnRennie Why not?
 
@paracetamol because it's a stupid question
 
._.
 
@paracetamol the answer is (if the candidate was speaking truthfully): I need to feed my family and pay my mortgage
 
Say @JohnR , the other day...didn't you say there was no such thing as a "stupid question"?
Apr 18 at 11:11, by John Rennie
@Fawad I disagree, though I probably wouldn't use the words good and bad. A question is useful if you will learn something from the answer. If you're just going to write that answer in your homework book then it wasn't a useful question.
;D
 
@paracetamol that is not the same as saying there are no stupid questions.
 
6:50 AM
...true...
But close! :D
 
In fact I was explicitly disagreeing with the claim that there no bad questions.
 
Oh.
 
Fawad said:
Apr 18 at 11:10, by Fawad
@paracresol sorry but no question is bad
 
Yeah...
 
And I said: I disagree
 
6:51 AM
Yep...
 
Anyway I need to work now for a few minutes. Back in a bit ...
 
Oi! How the heck did something like this happen without me! :(
17 hours ago, by Physics Guy
Better than being a christ (or whatever)
@JohnRennie o/
17 hours ago, by Jim
@PhysicsGuy Please avoid conversations that focus on bashing religious beliefs. This is Physics.SE, not Offensive.SE
:(
 
 
1 hour later…
8:10 AM
Hey guys
I have a problem with xrd
Can anyone help
2
Q: Powder diffraction - spikes on diffractogram

Gennaro ArguzziWhen a powder sample of a crystal is used to known its crystalline structure, on the diffractogram there are spikes corresponding to the constructive interference, at specific angles, by different crystallography planes. Why each spike is associated with a single family of planes? If the particl...

 
9:01 AM
Reeb says the branching real line has two non-isomorphic differential structures
End my life
 
user228700
9:15 AM
Apr 13 at 7:08, by John Rennie
Given what I've read recently I think I would be a bit concerned if my daughter attempted the trip on her own ...
 
user228700
@JohnR: Eh, do u really think I should wait till I find someone?
 
@Kaumudi.H Your father is the best person to judge.
 
user228700
Right, but he only said "Maybe you shouldn't go on your own" :-/
 
All I know is what I see in the UK papers, and I doubt that gives a complete picture.
I'd guess that maybe means I would be worried about you but I wouldn't stop you
 
user228700
That's what I thought as well!
 
user228700
9:18 AM
Eh. I wish he'd let me go. I'm going to push the matter once more when he's around.
 
"A differential structure of class $C^r$ on a manifold $M$ is called regular if for every $C^r$ function defined on the neighbourhood of $x \in M$, there exists a $C^r$ function $f'$ defined on $M$ such that $f$ and $f'$ coincide on a neighbourhood of $x$."
One of the differential structure of the splitting line is not regular
The horror
Reeb says a function is "of rank 1" (de rang 1)
What does it mean
it's no matrix
"For $M$ a manifold with a regular differential structure of class $C^r$, if a $C^r$ function $f$ is of rank $1$ at a point $x$ of $M$, it is also of rank $1$ at any point $y$ not separated from $x$"
"In mathematics, the rank of a differentiable map f : M → N between differentiable manifolds at a point p ∈ M is the rank of the derivative of f at p."
Ah, there we go
 
10:12 AM
@0celouvskyopoulo7 Yes, by teaching as much of it as possible in the finite-dimensional setting.
 
10:53 AM
-2
A: Gravitational waves as dark energy?

Giovanni Maria TeofilattoGravitational radiation is simple general orbital of electron of dark mattere 313 in general garvitaional galaxies orbital 3559. Is the relativity before the Big-Bang in quantum atom 263.

@ACuriousMind ^ please :-P
 
@AccidentalFourierTransform Hmm, I clicked the delete link and it was immediately deleted. It now says "Deleted by community". I'm not sure what happened there.
 
what happened here is that an AI was faster than the other one
 
[Trigger happy] I want to see a crank post downvoted every second for 15 seconds
(this is impossible, however)
 
it used to be possible but now He is suspended so...
 
Heyoooo
Do you think he will come back after his suspension
he's pretty old, he might die
 
10:59 AM
of course he will
from time to time he wanders over here
we have our very own chat ghost
 
(To time travel researchers in h bar, I have news for you in a few seconds. Please wait for a few moments ago)
 
he observes us from the shadows, unable to interact with the alive
he's not dead, yet he's not alive
 
His blood boils whenever we discuss of GR
 
@JohnRennie I spam-deleted the user which also deletes all of their posts.
 
@ACuriousMind Ah. That's a shame - I was wondering if I had acquired a new superpower :-)
 
11:04 AM
Hm
 
@AccidentalFourierTransform No AIs here :)
 
trying to think of a single Killing vector with a flow that doesn't generate hypersurfaces
Take the torus
Take a single curve that is a space filling curve
The tangent vector to that curve will be a vector field over $M$, but will it be 1) smooth 2) a Killing vector field
I think it wouldn't be smooth, for a start
Possibly not even continuous
Wait
On the Clifford torus, there is a space filling curve that's just a single line of the same slope over the fundamental polygon
The vector field would be perfectly well defined
If we take the Lorentz torus and make the slope $< 45°$ that's a Killing vector field
So Geroch's argument doesn't apply on it, although in that case, we still have $ds^2 = - g_{tt} dt^2 + g_{xx} dx^2$
So not the best counterexample
Although of course the problem is that this spacetime also has Killing fields that don't have space filling integral curves
It slightly bothers me that such a common theorem of GR is not necessarily true
 
Five answers to this question and none correctly explains why $\mathrm{SU}(2)$ appears in quantum mechanics. Sigh.
 
11:20 AM
something something complex Clifford algebra
 
@ACuriousMind I think the group names are beautifully formatted though
 
8
Q: Functional difference of Benedict's solution and Fehling's solution

busukxuanChemically, Benedict's solution and Fehling's solution are very similar, with copper sulfate as a source of copper (II) ions, sodium carbonate and sodium hydroxide respectively as alkalis, and sodium citrate and sodium potassium tartarate respectively as chelators. They are also both used to test...

 
@ACuriousMind I tried. Is my answer correct?
 
11:38 AM
IIRC Hawking has an imprisonned timelike curve spacetime that is causal but I don't think it has any time symmetry
 
@AccidentalFourierTransform Yes
 
I was already writing my own, though, spreading the gospel of projective representations seems to be my shtick here ;)
 
6 answers
if the question gets a couple of upvotes, it will probably end up in the hot list
 
It would be a better HNQ than many of the other PSE questions that land there, imo :P
 
11:42 AM
@AccidentalFourierTransform @ACuriousMind I need physics halp
 
It's right there on the diagram, above the rectangle
 
The oscillator in the figure is composed of an ideal spring of constant $k=20N/m$, by a viscous dampener that applies a force given by $F_x = -5.0 v_x$, and by a mass $m$ that slides over the plane with no friction.
a) It is intended that the movement is (translation warning) oscillating and sub-critically damped. What conditions must $m$ obey?
b) Consider that $m=0.5 Kg$. What will be the oscillation's "period"
c) If the oscillator is pushed from it's equilibrium position, and then released, after how long will the amplitude reduce to half?
How the hell do I even do this?
 
11:46 AM
In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force, F, proportional to the displacement, x: F → = − k x → {\displaystyle {\vec {F}}=-k{\vec {x}}\,} where k is a positive constant. If F is the only force acting on the system, the system is called a simp...
the condition is $\zeta<1$
 
The hell is $\zeta$
 
the damping ratio
its defined in the wiki page
actually, all you need is to compute $\omega_0$ and $\zeta$, where these parameters are defined in the wiki page
once you have them, you just use
$$x(t) = A \mathrm{e}^{-\zeta \omega_0 t} \ \sin \left( \sqrt{1-\zeta^2} \ \omega_0 t + \varphi \right)$$
 
The heck is that?
 
the solution of the equations of motion
 
Oh jesus
 
11:50 AM
ok, first: calculate $\zeta$ from the given data
 
@BernardoMeurer Um...did you even look at damped harmonic motion before deciding you have no idea how to solve this question?
 
@ACuriousMind I don't even know what that is
I have an exam in 6 hours
I know nothing
all of the material is in Comic Sans, I can't read it
 
I can't quite fathom you being asked to solve this exercise without having been taught damped harmonic motion.
 
My only notes are, literally, drawings of spinning potatoes
 
11:52 AM
don't let your oscillators get damp
 
Apparently that was once and exercise
 
remember to cover them before leaving
 
Okay, shit
I have an analysis class
I'lll be right back
sorry
 
12:21 PM
@ACuriousMind that's my reaction to all of his school work
 
12:51 PM
~~~Beginning transmission~~~
 
Yes, I am
 
The arxiv appears to be slightly more up to date despite it is dated 2013 while the actual article is submitted on 2017
 
people do update
 
No, I mean the (seemly) more updated version is on 2013 while the less update version is submitted to the journal on 2017
(or perhaps I was wrong, cause in the journal the diagram is actually nicer)
But anyway...
So, out of all the CTC spacetimes, is this seemed more possible than the others such as godel, tipler, etc.?
(even though the WEC is still violated)
The singularities at the kink do look worrying though. What would happen to that particular null curve that hit those singularities at the top and bottom of the inner wall of the bubble
 
Oh no. I had a proof I mind for something at 1am and now I lost it
 
1:06 PM
Another more practical oriented question:
We knew that (the cauchy horizons of) CTC spacetimes tend to impose extra constraints in a cauchy surface at some time in the past.
That means, some events in the cauchy surface will have correlations that otherwise not possible if the CTC region is absent. But then, even if we do found a correlation, is it even possible to tell apart on whether they are due to the information within the cauchy horizon of the CTC, or just they happened to have those correlations and does not have any "future" cause?
 
The big problem of CTCs is mostly the divergence of quantum fields
 
chat team, I've got a project for you
The top comment on the MSE proposal definition page reads as follows:
> This was initially proposed by edanm on meta.SE: meta.stackexchange.com/questions/5534/proposal-for-mathematics-se – Dan Dumitru Jun 3 '10 at 23:04
However, MSE/q/5534 leads to this question, with a different title:
6
Q: Does the Stack Overflow software have a name?

Stephen JenningsDoes the software that runs the Stack Overflow sites have a name? I've always thought that, just as Slashdot's software is named "Slash," Stack Overflow's software should be named "Stack" or "The Stack."

Any ideas on what's going on there?
 
74
Q: Brief outage planned for Wed, May 3, 2017 at 0:00 UTC, 8pm US/Eastern (like a fire drill for computers)

Tom LimoncelliMicroVersion: Planned service degradation: All Stack Overflow/Stack Exchange sites read-only for 20 minutes on Wed, May 3, 2017 shortly after midnight UTC (8PM US/Eastern). If you blink, you'll miss it. Short version: There will be a service degradation for up to 20 minutes shortly after midnig...

 
@Yashas boring
=P
up to 20 min
 
@EmilioPisanty That comment likely refers to the old meta.stackexchange for the 1.0 SE sites. The current meta.SE was not around back then.
I have no idea what happened to the old meta.SE posts, though
 
1:22 PM
@ACuriousMind You mean, an SE-1.0-specific meta that was separate from meta.stackoverflow?
 
Hello Everybody!
 
@Yashas the top answer to that question has gathered a ridiculous amount of upvotes
 
the mother meta used to be at MSO and split in two in 2014
 
@AccidentalFourierTransform It's yours, isn't it? :P
 
I am going to resume the Matrices/LA course and just for the inspiration for subject
 
1:22 PM
@EmilioPisanty Yes. For evidence of its existence, see the second paragraph here
 
And there is a comment from the Review Queue xD
 
@ACuriousMind ah, that's mighty interesting
 
I seek few example problems you guys must have encountered in your experience as physicist which are counterintuitve to solve thorugh matrices at first glance but are well tackled with the help of MAtrices/LA?
LA = Linear Algebra
 
This does not provide an answer to the question. If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. Just kidding :-P — rand al'thor 14 hours ago
@AccidentalFourierTransform :D
 
1:25 PM
@ACuriousMind that presumably explains this, then
 
@EmilioPisanty Yeah
 
It's damn hard to read on the wayback machine, though
all the formatting's been stripped bare
 
Hmm...
> Closed causal curves are pretty bad in field theories. There are many configurations in which a field, when confronted with a closed causal curve, will just circle round and round, getting blue shifted with each cycle, and just diverging to hell. This isn't necessarily too terrible, because this may only happen for some configurations of the field.
Is there no known mechanism to make the blueshift bounded, like how when we get into relativity, velocity will be bounded above by the speed of light thus acceleration will only asymptote the velocity towards that limit?
I mean thoeretically, if such mechanism exists, then the field can be blueshifted forever and still be bounded in energy momentum density
 
@ACuriousMind done, thanks a bunch.
 
@EmilioPisanty Heh, are you attempting to become the SE chronicler? ;)
 
So, if we can design a CTC metric such that the blueshift is bounded in all regions despite increasing forever, then we might be able to get around the quantum field divergence problem
 
@ACuriousMind I just got tired of having the same conversation with dewy-eyed idealists over and over again
nothing wrong with idealism
but I would rather have them dry the dew off their eyes and understand the history, instead of projecting their wishes on it
 
I think having all this history in a single place and be able to point to it is a very good idea, and you've invested a lot of effort to find all the relevant original posts
 
1:43 PM
@Yashas You are too late. THAT is what caused me to stirr the discussion with slereah about time travel today
and in fact, that work is already done since 2013
 
@ACuriousMind yeah, it was some work but mostly it's the fact that if e.g. you did it, you wouldn't even know what to look for ;-)
 
or too early?
 
@Secret You function faster than a bot lol
 
Well, you can say it is too early if any one of us just happened to be inside such bubble while typing in this chat (modulo someone else get fried by unruh radiation at the walls)
---
About the notion of time, I actually have a question that is inspired by the following idea explosion:
\begin{align}
& \textrm{Last night dream}\to\textrm{fixed point infinite sums of the form } 2^{\sum^{\infty}=\sum^{\infty}}\\
& \xrightarrow{\textrm{Ted talking about varying $\gamma$ in $\int_\gamma\omega$ which is a functional of $\gamma$}}\xrightarrow{\textrm{infinite sets}}\\
& \textrm{Discrete countable variation idea on optimising}\sum_{\lambda}g(\lambda)\\
& \xrightarrow{\textrm{Google search}}\textrm{Discrete Lagrangian mechanics}
\xrightarrow{\textrm{Bounded sequences}}\\
& \textrm{Sequence variations $\delta \lambda$ have extra degrees of freedom in the form of density of the poin
And the question is as follows:
How do we experimentally know that time is linear. I.e. when you look at an acceleraitng truck, rather than thinking it is moving faster (covering more distance per unit time), it is equally valid to say that time flows faster while the moving distance are the same?
Both scenarios will look identical to us
 
You can just take an accelerated frame of reference in which the truck is at rest
 
1:51 PM
That's true but then no notion of time (linear or not) can be deduced if you rely on the state of the truck to be your clock in that frame
(Of course, one can argue the observer can always measure proper time, and that one take that to be linear for convenience)
I guess I need to think more carefully how will the laws of physics be changed if time is nonlinear
 
@Slereah how does one say de boglie
 
2:07 PM
de broy
 
Debrey
Except it's the french e
Which has no english equivalent
 
is there any straightforward relation between Killing vectors and the Laplacian?
given $K$ Killing, what is $\Delta K$?
 
@AccidentalFourierTransform Is it $-\mathrm{Ric}(K,K)$?
 
Do you mean divergence
 
The divergence of a Killing field is zero
 
2:16 PM
Oh delta
 
@0celouvskyopoulo7 I dont know :-P
perhaps
 
Actually that doesn't make sense
 
Well
 
I wrote a scalar but you need a vector
 
What I advise is
 
2:17 PM
Maybe $-\mathrm{Ric}(K,\cdot)$
 
Take the Killing relation
Apply $\nabla^\mu$ to it
 
Yeah you're gonna get a curvature term
 
yep, that's correct
cool!
 
@AccidentalFourierTransform Note $\nabla^2_{X,Y}K=-R(K,X)Y$
So take the trace of that
@AccidentalFourierTransform what I said?
 
yes, its $\frac12 \mathrm{Ric}(K,\cdot)$
or with a negative sign
 
2:19 PM
1/2?
 
I did what S. said and you get a $\frac12$
I think
$0=\nabla_{(\mu}K_{\nu)}$
apply $\nabla_\mu$:
$0=\Delta K_\nu+\nabla^\mu\nabla_\nu K_\mu$
and commute the derivatives
 
I dunno. $\Delta K=\sum \nabla^2_{e_i,e_i}K=-\sum R(K,e_i)e_i=-\mathrm{Ric}(K,\cdot)$.
God why use $K$
 
$0=2\Delta K_\nu+RK$
hmm
 
Uh, what?
note that $\nabla^aK_a=0$
 
I...
up to
2 mins ago, by AccidentalFourierTransform
$0=\Delta K_\nu+\nabla^\mu\nabla_\nu K_\mu$
we agree?
 
2:23 PM
I believe that equation.
Then apply the Ricci identity and note that div K =0
 
oh fuck
yes
thanks
 
old (fake) news
@yuggib greetings
@yuggib quick q: In a reflexive Banach space, is a strongly continuous function weakly lower semicontinuous?
 
2:38 PM
@0celouvskyopoulo7 Yo
If I launch a projectile with angle $\theta$
 
And I want to to reach $x=12m$ and $y=6m$
 
are you starting at 0,0
 
How do I figure out my launch speed/$v_0$
Yes
 
Do you know the angle?
You clearly can't do it with every angle
 
2:39 PM
Yes
60
This is the figure
and they say the launch angle is 60 degrees
 
What do you think you should do?
They just want you to reach it? No other stipulation?
@yuggib It might not be true but that's not a problem for what I need, actually
It would be interesting to know when it is true though
I just need it for a function derived from a norm
 
@0celouvskyopoulo7 greetings
so the function is from $X$ to $X$, $X$ reflexive Banach?
or from $X$ to $\mathbb{R}$ or $\mathbb{C}$?
 
2:55 PM
@0celouvskyopoulo7 Sorry, had a call
@0celouvskyopoulo7 I tried writing the equations of motion for the projectile
$x(t)=v_0 t \cos\theta$
$y(t) = v_0 t\sin\theta - \frac{gt^2}{2}$
so I know that I need to reach $x=12$ and $y=6$
But I don't know the time so I can't solve those for $v_0$
@0celouvskyopoulo7 Ah, do I do a system? Since I have two equations and two variables?
@Yashas You too since you know stuff
 
$x = x\tan \theta - \frac{gx^2}{2u^2 \cos^2\theta}$
$u$ is the initial velocity
 
@Yashas How did you get to that?
 
substitute t as $\frac{x}{v_0 \cos \theta}$ in $y = v_0 t \sin \theta - \frac{gt^2}{2}$
I knew that formula byheart though
dono why I don't forget them
used it over 100 times :P
 

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