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12:27 AM
0
Q: Does one observe light reach infinity in finite time in AdS spacetime?

user76284I have a question about the following passage from this article: Moschidis imagined standing in the middle of AdS space-time, which would be like standing inside a giant ball whose edge or boundary lies at infinity. If you sent a light signal from there, it would travel out and reach the boun...

 
 
3 hours later…
3:08 AM
@MohamedObeidallah When we say that electrons are fundamental particles (in the Standard Model of particle physics) that just means that they aren't composed of simpler particles. In contrast, a proton isn't fundamental because it's composed of quarks.
@MohamedObeidallah Saying that a particle is fundamental does not imply that it can't be created, destroyed, or transformed. However, any reaction that creates, destroys, or transforms particles must obey various conservation laws. Here's an example of a transformation from Wikipedia, which shows what happens when a nucleus captures an electron.
"The leading-order Feynman diagrams for electron capture decay. An electron interacts with an up quark in the nucleus via a W boson to create a down quark and electron neutrino. Two diagrams comprise the leading (second) order, though as a virtual particle, the type (and charge) of the W-boson is indistinguishable."
So in this weak interaction, an up quark is transformed into a down quark, and an electron is transformed into an electron neutrino. Or we could say that an up quark & electron are destroyed and a down quark & neutrino are created.
Note that you can't just have an up quark turn into a down quark, or an electron into a neutrino. Those reactions would break conservation of electric charge. But the combination of the two reactions is permitted.
 
 
4 hours later…
6:46 AM
@ACuriousMind Went to bed and missed your response, didn't realise it was different from regular potential, thanks a lot for your help again. Hope you don't mind the ping while you're not here.
 
 
2 hours later…
8:39 AM
Hey! Is anyone here? I'd like to discuss if someone here knows of how to possibly shift signals without circular shift in matlab. I guess a requirement for fft is that it assumes that it's been given one period of the signal in it's vector? and shifting longer than the time vector with ifft(exp(iomegatau)*fft(signal(t)) wraps around the signal. However, I'd like to not have it wrap around
 
@DakkVader I'm not sure how many MatLab users we have lurking here. Most physicists I know use Mathematica.
 
Gah!
I guess the problem isn't unique to matlab however
Did i describe the problem well enough btw? Did you "get" the issue? (just asking, wrote a non-coherent question about this in mathematics chat and they didn't get me)
 
You mean make the fft not assume a periodic signal with the period equal to the length of the array?
 
Yeah more or less. I have a time-vector and then a number of different delays i need to execute, but some of the delays are longer than the extent of the time vector, thus a "new" pusle comes in from the other side, which i don't want.
Maybe I could pad a longer signal/time-vector, shift that one, and then cut out the center window ?=
 
The FFT is inherently periodic.
Just use it and remove the frequency associated with the time of the array.
I guess you could squeeze the signal in towards the middle of the array, though that either reduces your frequency resolution or requires a larger array.
 
9:21 AM
@DakkVader There's a whole stack dedicated to signal processing:
 
 
1 hour later…
10:44 AM
2020 Particle Data Group is online! pdg.lbl.gov
 
@Slereah i don't understand a single thing there :P
Oh, it's like a database of calculations (done every year) about particles?
seems cool, i'll plug all the data into some AI algorithm and I can be ant-man
 
 
3 hours later…
2:13 PM
given you a light cone at a point of a manifold, how do you know if a light cone is a future cone or past cone?
 
2:45 PM
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Q: Test to determine if a coincident point in a pair of rotated hexagonal lattices is closest to the origin or not?

uhohA pair of hexagonal lattices with one scaled by the square root of a rational number $r = \sqrt{\frac{m}{n}}$ and then rotated will produce a variety of different coincident points. For the first lattice let $$x, y = i+\frac{1}{2}j, \ \frac{\sqrt{3}}{2}j$$ and for the second $$x, y = r\lef...

anybody here like linear algebra questions?
 
3:12 PM
@Charlie no worries, you're welcome
@Slereah what are you gonna do with it?
 
I am getting crazy with the QFT book of Itzykson & Zuber. In Eq. 5-173 they have a bunch of terms which are non-zero when the 4-momentum q satisfies $q=pn - (p1+p2)/2$. The momentum $p1,p2,pn$ are time-like. They say that the two terms do not contribute outside a region bounded by the two hyperboloid sheets, that they draw in Fig. 5-8.
They say that these hyperboloid sheets correspond to two different masses $M_+$ and $M_-$ but they don't provide the definition of these masses. Moreover, I have a counter example in 2D, where there are no hyperboloid sheets. For example, $p_n = (1,0), p_1 = (4,3), p_2 = (2,1)$. In this case, $q^2 = 0$ is null. what the hell?
 
I don't have I&Z lying around, so I'm afraid I can't really tell what's going on there
 
Also, they say that the hyperboloid sheets are centered at $\pm 1/2(p_1+p_2)$ which has no sense since $p_1$ and $p_2$ are vectors... are they talking about the modulus?
Look, I am uploading a picture
This is their Fig. 5-8 instead
These seven lines under Eq.5-173 are driving me crazy
 
Note that the y-axis there has $q_0$, so I think they're saying the centers of the hyperboloids (the tips of the triangles there) are $(0,(p_1 + p_2)/2)$ and $(0,- (p_1 + p_2)/2)$.
Ah, your problem is that $p_1 + p_2$ a vector
 
yes, I think they mean the tip, but what does $(p_1 + p_2)/2$ even mean.
probably they mean the modulus?
But this doesn't work in the following example $p_n = (1,0),p_1 = p_2 = (4,3)$
the resulting q is null, therefore in Fig.5-8 you would have a light-cone centered in 0
 
3:28 PM
What I think they mean is that the hyperboloids are really higher-dimensional hyperboloids when you draw them in the full $(|q|, q_0,q_1,q_2,q_3)$ space. In that space the center would lie at the vectors they write, but I agree it's sloppy/confusingly worded
Their $\delta$ functions do not vanish when $q$ has a chance to fulfill $q = \pm ((p_1 - p_2) / 2 + p_n)$
 
I see what you are saying...
Why the center would be the vector they write?
 
So I think the hyperboloids are the hyperboloids for the equation you get when you square that, i.e. $q^2 = ...$.
It's easy to see that when $p_n = 0$, then this is just the triangles they draw.
 
@ACuriousMind yes, I was taking the square of q
so, is my example in contradiction with them, or not?
 
And when $p_n \neq 0$, then you get additional term that correspond to a mass (because the mass hyperboloid of a massless particle is the triangle, so the deformation can be interpreted as adding mass)
I don't quite understand what your example is supposed to do - in your case, $(p_1 + p_2)/2 = (4,3)$, so they're not centered at zero
 
$p_n - (p_1+p_2)/2 = (-3,-3)$ which is the value of the vector $q$
 
3:39 PM
@ACuriousMind I'm building a fort with all my copies of the PDG
@CaptainBohemian There's no difference, fundamentally
Past or future is a choice you make
Of course in more practical cases, there are methods of determining it
ie pick the direction that increases entropy
IIRC that's the York time?
 
@Slereah nice
 
As I get at best one per year, it's a slow process
Apparently they decided to keep sending paper copies, though
 
@ACuriousMind Is my example wrong or what?
 
Sorry, I'm thinking
 
ah ok sorry
 
3:46 PM
@Slereah if you always keep MTW on one side of your house the weight will make the house gradually tilt to this side. Put a copy of the PDG on the other side and it will restore the balance.
 
The PDG isn't that big
 
@Slereah is it possible that a spacetime is not time orientable so that a light cone of it is the mix of both future and past?
 
Barely 900 pages
@CaptainBohemian It is possible that it is not time orientable, although it's always locally time orientable
 
@CaptainBohemian wouldn't that be true for any spacetime that included closed timelike curves? The Godel metric for example.
 
@ACuriousMind Take also $p_n = {1, -1/2 }, p_1 = p_2 = (4,3)$. This configuration generates two rotated hyperbolic, with the principal axis on $|q|$
 
4:01 PM
@apt45 I think I know what's going on: They wrote $q - (p_1 + p_2)/2 = p_n$ and squared that. This is the equation for a hyperboloid with mass $p_n^2$, shifted ("centered") by $(p_1 + p_2) / 2$.
 
@JohnRennie I heard that kind of metric makes time travel possible, but I haven't thought of it as a way of being time-unorientable.
 
@CaptainBohemian Any CTC means your past light cone includes trajectories that come from your future
 
(In more detail: $q^2 = m^2$ is the usual equation for a mass hyperboloid. Doing $x\mapsto x + a$ is a shift operation that must also shift the submanifold described by this equation by $a$, so $(q - a)^2 = m^2$ is the equation for the same hyperboloid shifted by $a$
But crucially, in the shifted hyperboloid, the "mass" $m^2$ is not the mass of $q$! A "hyperboloid of mass $m$ centered at $a$" does not consist of vectors with mass $m$.
So your example shows nothing - sure, $q^2 = 0$, but the "mass" they're talking about is not $q^2$, so the shape described by the equation is still a hyperboloid with non-zero "mass".
They explain it very confusingly and I think it's poor terminology, but I'd say there's nothing wrong there.
 
I was pretty sure there was no mistake in the text - I was giving counter-examples to make you understand what I was understanding :)
let me think about it
thank you very much
 
@Slereah Hello
Urgent!
 
4:09 PM
I thought this book was good; however, the terminology is very bad and things are explained very bad. I prefer by far Weinberg's book
 
@JohnRennie sir... Do you know, Crystal momentum?
 
@JohnRennie CTCs do not mean lack of time orientability
 
@Slereah Hello
Urgent help needed... plz
 
I am of no help
 
@abhas_RewCie I know what it is but not any details
You can always ask ...
 
4:11 PM
@JohnRennie sir can you solve all those 5 questions plz
XD
@Slereah plz
:'-(
 
nobody bleeding, nothing on fire, no coolant leaking, cannot be urgent
 
@apt45 Different people look for different things. Weinberg is excellent if you're looking for a comprehensive and comparatively rigorous treatment of QFT that explains most of its steps in exhausting detail (at least at first). Books like I&Z are more "casual", and many people find their intuitions and explanations nicer than the forest of indices Weinberg's equations develop into.
 
@abhas_RewCie I guess that's meant to be a 1D crystal lattice, so the electron wavefunctions will be Bloch waves.
The crystal momentum of a Bloch wave is just the wave vector times $\hbar$.
 
@JohnRennie I don't know anything about it, not even Bloch waves... Only thing, I know what is Baloch Sphere usually used to represent Qubits...
 
Sounds like a load of balochs to me :-)
 
4:14 PM
@JohnRennie Only solution will work (I'm not asking for myself, I'm asking for her - twitter.com/sayani_pandit/status/1268523743044763648 )
 
@abhas_RewCie Do not ping random people to answer your questions, especially not when they're just poor photos of exercises. You've been told this before.
 
@ACuriousMind to be fair my profile says I'm happy to be pinged.
 
@JohnRennie Slereah's doesn't.
 
True ...
 
@JohnRennie Okay, I'll come to you next time to learn more about Baloch waves, currently, a rough solution will work (specially (d)), you may leave other bits, in case you know the solution...
 
4:18 PM
@Slereah so CTC can still be time orientable?
 
@CaptainBohemian Not all of them, but overall yes
Also not all non-time orientable spacetimes have CTCs, If I remember correctly
The elliptic AdS space has no CTCs
 
Hey guys, quick question, practically speaking how do electromagnet strengths compare with rare earth magnets. I dont often see permanent magnets using a unit thats eaasy to compare with electromagnets, or they dont provide strength info per unit mass
 
@JohnRennie are you still solving sir....? I'm afraid, if I asked such lengthy question....
 
@abhas_RewCie I don't want to do it I'm afraid.
 
@JohnRennie Okay, I think, I'm about to complete it, so, don't do it... It's almost done. I'm checking it.
 
4:26 PM
Cool :-)
 
:-)
@JohnRennie Done...
Good night sir... :)
Good Night Slereah :)
Bye
 
@Skyler Isn't an electromagnet's strength generally only limited by how much current you can run through it before it melts?
 
4:40 PM
@ACuriousMind well that's a pretty certain limit, still!
Although I guess technically you can still use it after it melted
as long as it still has the correct shape
 
@ACuriousMind well technically youd hit the curie limit before that most likely
or was it called curie temp
curie temp
but how do you take the magnetic ffield produced by something like a solenoid and compare that with the magnetic forces characteristic of magnet. Like there should be some kind mass related term for a magnet shouldnt there, but you just see people throw out "this neodynmium magnet is ~1T"
 
yeah, I get what you want but I have no idea either
 
Also you can cool an electromagnet, if you're worried about it melting :p
 
4:59 PM
@ACuriousMind I understand what you said. I am still confused by the following thing. If I take $q=p_n -(p_1+p_2)/2$ and square it, I get $q^2 = p_n^2+(p_1+p_2)^2/4-p_n\cdot(p_1+p_2)$. If I draw this contour, I get the orange lines. Instead, If I plot $(q+(p_1+p_2)/2)^2=p_n^2$ I get the blue line. What am I misunderstanding here?
I think I am missing something trivial. The two equations should be consistent with each other, somehow, right?
 
What the heck is this?
 
@apt45 See the point where they intersect? That's the "real" value of $q$.
 
I am seeing this in every review queue.
Was the "Steward" badge burninated?
 
consider that you initially have an equation that determines $q$ uniquely - it's not an equation with a whole range of solutions, there's *one specific $q$" that fulfills it
 
@ACuriousMind yes, that's the only consistent point. But I would have said that the delta-function vanishes outside the orange region
 
5:12 PM
When you square the equation, you're forgetting information about $q$
With different ways of squaring it, we're forgetting different information, so you get two different graphs
 
yes, I understand this. But then, why I should I trust that their quantity C(q) vanishes in the region outside the blue line but still below the orange line?
for values of q_0<0
 
I&Z are interested in the blue graph because when you do this for the lowest $p_n^2$ of all the $p_n$, then (for fixed $p_1,p_2$) you know that all the other hyperboloids for the other $p_i$ have larger "mass"
Therefore their blue shapes lie "inside" the hyperboloid for the lowest $p_n$, and therefore no possible $q$ lies outside of it.
Hence $C(q)$ vanishes outside of this hyperboloid for the lowest $p_n$.
 
yes, I understand their logic. I am trying to understand why what I was doing is not correct. Which mistake I would have made by choosing the orange line?
 
none
 
I would have given a weaker condition, I guess
 
5:18 PM
but you wouldn't have been able to make a nice statement about where the lines for all the other $p_n$ lie, so it wouldn't have told you anything interesting about $C(q)$
 
I see. thank you very much for helping me clarify this
 
@ACuriousMind Mind explaining this? :-)
 
@FakeMod I would have said something already if I had any idea what's going on :P
Post on mother meta if it persists, not much else you can do
 
@ACuriousMind Maybe it's because of the new update going on in the review queues. Did you see the new GUI.
 
I don't use the review queues - they're not that useful for mods
 
5:22 PM
@ACuriousMind Does anybody ever do that? :P I always thought that Meta was meant for scrounging up already asked questions since every bug that you face would have been discussed about in Meta before.
@ACuriousMind So you guys have your "VIP review queues"? ;-)
 
@FakeMod Someone has to be the first to post a given bug there, no?
 
@FakeMod Sort of, but I more meant that e.g. the close review or LQ review queue is not something we're supposed to be doing (since our votes are binding)
 
@ACuriousMind Yeah, sure, but not me :-) At least, not now.
@ACuriousMind Yeah, binding votes need to be handled with special care.
Is there any way to see how many posts did you VTC or "Recommend Deletion" were really deleted? In other words, the accuracy of your reviewing. SEDE might help, but I don't know how...
 
i don't think so
 
5:29 PM
@ACuriousMind Finally finished Torment
(The second one)
 
ugh, don't remind me of that one
 
You're the one to complain
I spent 150 bucks on it :V
 
and do you feel you got your money's worth? :P
 
Eh
It's not awful
But a bit of a letdown
 
yeah, that's how I recall it, too
 
5:44 PM
Also I'm not sure it's such a good idea to drop you right in a weird fantasy world with weird words and a long history
It makes for tedious reading to catch up
 
 
2 hours later…
7:36 PM
2 hours ago, by FakeMod
What the heck is this?
 
8:00 PM
 
8:59 PM
@FakeMod Aw what'd I miss
The mods are too fast
 
 
3 hours later…
11:40 PM
@Slereah Magnets are cool! :) Especially electromagnets with superconductive windings.
in Python on Stack Overflow Chat, Dec 23 '19 at 16:29, by PM 2Ring
@Dair Here's a cute integer thing I noticed a few years ago while messing around with square roots and Pell"s equation. Obviously, 18 - 13 = 5. Not so obviously, 18^2 - 13×5^2 = -1. That is, 18/5 is a good approximation for sqrt(13). I suppose I should do a search to see if there are other numbers with that pattern...
There are other number pairs apart from (18, 13) with that property.
7²-2×5²=-1, 7²-3×4²=1, 9²-5×4²=1
 

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