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12:55 AM
In group cohomology, the second group is in one-to-one correspondence to central extensions... Now, in order to derive the Heisenberg (or Poisson) algebra from the Galilean algebra, one needs to first centrally extend the later. How neatly does that fit with the requirement of non-trivial second cohomology group when defining a symplectic form!
@enumaris WTF !?!
''I am certain that you can accelerate. [...] Until then, such theories are mere religions, with no merit.'' 0.0
 
1:58 AM
Last night dream, I was forgot except reading a blog by a guy called David Czevosky. The blog's layout is white and grey and for every page, just before the article begin there's always a grey box which reads "Rarely, there are things which were worked out in detail by science fiction writers. Thus it is my job to explore them"

Various topics were explored such as a maglev similar to the one from Black Panther and wormholes, with detailed equations and workings on predicting the speculative physics and engineering aspects. On the maglev article, he concluded that it is impossible even in s
 
2:12 AM
@G.Bergeron en.wikipedia.org/wiki/… seems like the ability to set up a symplectic structure implies a general Heisenberg algebra?
 
@bolbteppa Yes, it does (at least on ''flat'' symplectic spaces). I thought it was cute that in order to obtain the Heisenberg algebra, you need a central extension, as is the case to have a non-trivial second cohomology group that contains the symplectic form.
 
I'm fascinated by how this thinking leads to spinors
 
2:28 AM
@bolbteppa You mean the $SO(3)$ - > $SU(2)$ thingy when considering projective representations?
 
Yeah
 
2:40 AM
@bolbteppa Yeah the same reason is behind why we want the projective representations (amplitudes are not the observables) of both $SO(3)$ and the Galilean group. However, for $SO(3)$ it just happens to be the case that a representation of $SU(2)$ is a projective representation of $SO(3)$ (and hence the spinors), but for the rest of the Galilean group, the issue is more subtle as central extensions is required to cover all the cases.
I'm more excited by this connection between dynamics and those central extensions.
 
Hmm, in the representation theory section of that page it explains it holds for $\mathbb{R}^n$ en.wikipedia.org/wiki/Heisenberg_group#Representation_theory because of the non-trivial center, not just $SO(3)$?
Wait that's obvious, yeah I am not too sure about the rest of the Galilean group and how things could mess up
 
@bolbteppa Not sure what you mean... I am not saying we look at projective representations because of $SO(3)$ in both case (that's because the invariant quantities are squared absolute amplitudes which implies projective reps on those amplitudes), but rather that central extensions of $\mathbb{R}^{2n}$ or the Galilean Lie algebra leads to other different non-trivial projective representations.
 
Sys.arch.labs: There is evidence that the Marystar anomaly in math chat started to stablise. Reporting results back to skull and skillpatrol in a few minutes
 
3:19 AM
@bolbteppa, @ACuriousMind Perhaps my confusion is better outlined in the thread I made (linked below). If you had some spare time over the weekend for just a comment I would really appreciate it, cheers.
https://physics.stackexchange.com/questions/410724/clarification-on-transformation-laws-as-commutators
 
 
5 hours later…
8:05 AM
@Blue little hint for ya: (in this case) when I put something in actual quotation marks, it really is a quote... Also, to be clear, I'm normally really happy to have conversations about policy like the one you wanted yesterday. Only, because context that I hope you're now a bit more aware of, yesterday wasn't a great time...
 
Anonymous
@Mithrandir24601 I understand. And I guess things are fine now and we have been able to convey the point to user1271772. I wasn't really speaking about any specific comment yesterday, but rather in a more generic sense. Considering this issue has caused a hell lot of controversy on Physics SE, it's better to take some precautions (like making a clear meta post), while the site is still in beta phase
 
Anonymous
The aim should be: to be able to objectively classify all comments into distinct categories so that in the future there are no controversies about which comments should be deleted and which shouldn't be
 
@Blue ... I hope so. Their issue was with the specific comment that they believe I removed, unlike your issue being more general. I really should be clear: I'm not saying that I think we should just go around removing lots of comments. I also think that part of the issue is a difference in what we mean here. I really don't want to spend any time going through removing semi-useful comments anyway :P
There's a large difference between what has to be removed and what has to be kept. All I'm trying to say is that we should remove the stuff that needs to be removed. Or rather, I'm saying that what hasn't been removed, doesn't need to be removed
(there could very well be comments made that I haven't seen that do need to be removed, but I hope you get the gist of what I'm trying to say)
 
Anonymous
8:25 AM
@Mithrandir24601 Right. That's my point too: Don't remove something unless that must be removed. As for what must be removed, some clear categories are spam and rude. Something like a "Thank you", doesn't fall into either of those two categories and you don't need to remove such comments.
 
Anonymous
The particular comment user1271772 seems to be complaining about, was a rude comment. So it clearly deserved deletion.
 
Anonymous
There may be some exceptional cases when a certain user posts a lot of simply "Thank you" or "+1" comments without much additional material. In that case you should interfere and post a comment telling them to avoid it.
 
@Blue There are caveats with this though - if an answer gets flooded with such comments, then it does get in the way and I would remove them. I'm also not going to complain if such comments get removed, but on the whole, yeah, pretty much
 
Anonymous
@Mithrandir24601 Yes, that's an exceptional case i.e. if there's a flood of such comments. But till now I haven't noticed anything like that. And I don't think we need to ever worry about those situations, since our site's activity is reasonably low.
 
Anonymous
For now, let us not overthink what could happen. Rather let's focus on what happens often.
 
8:37 AM
@Blue Nor have I. There was one case (that I know of) where a 'thank you' comment was deleted, as it was at the end of a chain of comments, which also got deleted. I think this was more to reduce the number of comments (which later ended up getting moved to chat) than anything though, although the question in this case is now deleted, so I wouldn't worry overmuch about this specific one
 
Anonymous
That's fine. But in such situations, if the comment discussion has already gotten quite long and you know that you'd have to shift it to chat anyway, there's no need to delete a single harmless "thank you" message. Just wait for a few days (around 4-5 should be good enough for a discussion to conclude) and then move it to chat. That is don't tamper with messages unless absolutely necessary.
 
@Blue Yes, so serious question: do you genuinely feel that (or have any examples of when) we're deleting too many comments as 'no longer needed' on QC? If so, I'm quite happy to go about having a discussion on chat (although I'd leave it a few days so we don't start mixing conversations about what appear to be the same issues but aren't again) or meta (feel free to make a post there). If not, then everything seems to be good and we're saying different words that actually mean the same thing
 
Anonymous
@Mithrandir24601 I personally don't so but I felt it was worth a discussion since a notable user began complaining about it. I wasn't aware of what type of comments they were speaking of, though, which is why I was speaking generically.
 
Anonymous
If it was a comment by a random new user, I wouldn't have worried that much.
 
@Blue Oh, on moving to chat - you made 5 points yesterday and they were good, so a meta post on those at some point wouldn't go amiss, to help clarify things. Although it's worth mentioning that we're 3 new mods, so we're still undergoing the hard learning process of what's best to do when and how and when best to deal with flags etc.
 
Anonymous
8:48 AM
@Mithrandir24601 "we're 3 new mods, so we're still undergoing the hard learning process of what's best to do when and how"...which is exactly why I'm trying to frame some sincere feedback! :) And yes, I'll try to address those 5 points along with your points when I make the meta post
 
@Blue This is a hard 'how to best moderate' issue... I hope that people trust the mods enough to realise that me not considering the possibility I'm over-moderating has a long chain of implications about what the person making such a statement has done.
On the other hand, without being aware of that context (which, ironically enough, would have been the case if I didn't leave a certain comment) and seeing that I'm not willing to answer things like 'who did what mod action' it could be very easy to come to the conclusion that I was actually over-moderating
 
9:05 AM
@Blue And I like sincere feedback :) it's a steep learning curve for us all. To be honest, yesterday was quite stressful and I would now go about dealing with such a situation in a different way, by being more upfront about what I meant by "not constructive at best" instead of essentially saying that the mods are entitled to remove unneeded comments
(which we are, but generally I don't go around looking for mod actions to perform and only do stuff like that when I either actually need to or am prompted by a different source, such as flags, which is something I was assuming people would just know and so, figure out all the implications by themselves)
 
Anonymous
@Mithrandir24601 Only the people who have been observing the site closely (for example me) would have that trust. If someone who has made considerable contributions to the site and arguably has been observing the site closely, too, makes a complaint like that, it's fair enough to consider the possibility of over-moderation. However, speaking of myself, I do not think you were over-moderating.
 
Anonymous
@Mithrandir24601 Yes, that would indeed be better :)
 
Anonymous
If you ever feel stressed while answering to questions about your mod-action the best thing to do is to mention upfront that you'd like to continue the discussion at some other point of time. It's not like you have to answer to people immediately, or even on the same day. You should ideally take your time and ideally be answering when you are not under stress.
 
@Blue This is the sad thing - before someone gets a diamond, they've got a chunk of the community going 'we trust you to do this right', then as soon as the diamond arrives, the story can change so easily for reasons that I really cannot fathom
 
Anonymous
It's just an online site. Don't let this affect you in real life!
 
Anonymous
9:12 AM
@Mithrandir24601 Happens with all superheroes too ;)
 
Anonymous
Anyhow, take rest now. We can talk later. I gotta get back to some work. :)
 
@Blue This is another thing about being a mod - I'm so invested in the site now it's really hard for that not to happen :(
@Blue I'll keep this in mind... It's a very valid point that I really should have taken into account... :P
@Blue Fair - bye! (and thanks for having this conversation)
 
I think most of the pass chatmates that now became mods are doing fine and I don't see their personality changed much after getting the diamond
Acuriousmind personality already before mod means he is a pretty strict mod but again I don't see any evidence of him crossing lines
Most of the old mods are also fine although Shog9 can be very scary sometimes
 
@Secret That's good :) I hope everyone else realises that the person before getting the diamond is the same person afterwards
@Secret Shog's great! Although yeah, they tend to deal with people who are out of line, so if you see them in a chat room they're normally not in, it's because they have to deal with stuff :/
 
yeah often when he is in this chat room, we knew someone have made a big trouble
 
9:23 AM
Yeah, on the other hand, if e.g. Shog appeared in the Frying Pan, it's possible they might just want to talk about food or cooking (Shog appears to have an interest in cooking/baking, going by site participation)
 
10:01 AM
Could anyone give me directions on how to go about computing the trace relation $\text{tr}(M_{ab}M_{cd})$ where $M_{ab}$ are the Lorentz generators in an arbitrary representation?
Could I compute it in a specific representation (say the fundamental rep) and then use the fact that trace is basis independent to lift the result to an arbitrary rep?
 
@SigmaAlpha That the trace is basis-independent does not imply anything about the value of that expression in other representations
 
Yeah I thought that might be the case..
 
For the adjoint representation (i.e. the representation of the Lorentz algebra on itself) that there is the Killing form.
Ah: The Lorentz algebra is simple, and hence that expression is proportional to the Killing form in every representation.
So all you need to do is evaluate the expression for a single pair of matrices to get the proportionality factor
Hm, that might only be true for complex representations. For real representations you probably just need to do the computation
 
Jeez my supervisor really threw me in the deep end this week lol.
What do you mean just do the computation? It's in an arb rep so im not sure how to do that
He said that $\text{tr}(M_{ab}M_{cd})=c(\eta_{ac}\eta_{bd}-\eta_{ad}\eta_{bc})$ but I need to figure out the constant c now
I also need to compute the same thing but for a product of three lorentz generators
So I figure its best If I just learn how to do find the identity from the product of two and the somehow lift that method to three
 
 
1 hour later…
11:23 AM
Hi
 
11:35 AM
Hey guys! How would you descibe that oscillations as they are shown on paper do not exist (I mean they exist but...):

I am trying to explain how WiFi and LTE works to a kid by explaining how electomagnetic waves work, but... you know, its hard to explain that electromagnetic waves don't exist in the real world. I mean they exist, but you cannot see anything going up and down, because a wave, as its shown on paper, just graphically displays the change of the state of the magnetic field (I hope I have explained this correctly)! So the change of the magnetic field is real, but there is no osc
What would you describe it (because "doesn't" exist) isn't the right word for it....
 
its intangible?
 
Anonymous
@watchme "invisible" should be okay
 
Anonymous
You can compare it to light
 
Anonymous
And perhaps sound, too
 
Anonymous
But do point out the differences
 
11:52 AM
@Blue well, it's not only invisible, but it really doesn't exist. You will never see an oscillation as it's displayd in the above picture. It just doesn't exist. No waves are send out by someone screaming. But air which "goes together and apart" (otherwise you eardrum wouldn't vibrate). I don't know how I should explain this to you
 
Anonymous
@watchme Yes, but that "displacement of air particles from mean position" graphed on paper, is what those waves (which in your words go up and down), represent, for sound waves.
 
Yes, exactly ! But these waves don't really exist, no information is conveyd by using those waves, you know? And I cannot find a word for it
 
Anonymous
@watchme "no information is conveyed by using those waves"...depends on what you mean by "information".
 
Anonymous
The waves totally exist, at least as far as sound waves as concerned. Just that they are longitudinal waves rather than transverse waves.
 
There are no waves which go from the cell tower to the smartphone to convey digital information (as shown in the picture above). In a magnetic field, other devices can "feel" the "tug" and get that the cell tower wants to transmit a "0". :P
It doesn't really work that way, but just to convey the idea of what I mean
Everyone imagines that from this cell tower, these "waves" somehow are send away, but the aren't, no wave is send a away. The wave which many people imagine is just the "back and forth" of the magnetic field displayed graphically.
 
Anonymous
12:00 PM
12
Q: How do EM waves propagate?

opethDamnationI have read about how electromagnetic waves propagate and what I surmise is that when charged particles such as electrons accelerate they produce time-varying electric fields. These electric fields produce magnetic fields and the process goes on. Are the EM fields really moving? My textbook say...

 
@watchme I'm not sure what you mean by that. all waves are arguably "just the back and forth" of some field.
Sometimes that field directly relates to a substance you can touch, like the height of water in a lake. Sometimes it doesn't, like the electromagnetic field.
 
@ACuriousMind Indeed! But they are not real. They're just drawn on paper to see whats going on over a period of time.
 
wiggle wiggle baby
 
@watchme That you cannot see the magnetic field does not make it not real.
 
Anonymous
That ^
 
Anonymous
I still didn't understand what you mean by "real" @watchme. Do you mean "visible"?
 
@ACuriousMind Indeed! But waves as we see them on paper are not real. They're just drawn on paper to see how a state changes over a period of time.
 
The waves are real though
 
Anonymous
@watchme That's not true
 
@watchme You must have a very different idea of what "real" means. The values of the electric and magnetic field are perfectly real (in that we can in principle measure them at every point) and they oscillate precisely in the wave-like manner we draw
Sure, we can see the wave we draw on paper and not the electromagnetic one, but that's just because we simply can't directly see the values of the electric and magnetic fields.
 
12:05 PM
What conveys the digital information is the state which changes. But this itself is not a wave, right? The wave is just here to display the change of a state over a period of time, right?
 
Anonymous
Whose "state" are you talking about?
 
I don't understand what you are saying at all, I'm afraid.
 
the values of the field are actually changing in a wave-like fashion @watchme
 
Yup, the electric and magnetic fields are real and they change in a wave-like fashion, thats clear to me.

And that's what is real. When we draw that process of how a magnetic field for instance changes over time, we get a wave. But the wave itself is not send out by a tower cell "somehow", but a magnetic field which goes "back and forth" is.

Many people imagine that waves are send out. But they aren't^^ Its just a magnetic field changing in a wave-like fashion^^
 
I think you are overestimating how literal people take the depiction of waves.
 
12:10 PM
But it's exactly the ssame type of wave as a wave in water
It's just that this time the medium is not something we can sense
 
Anonymous
@watchme Again, it depends on what you mean by "send out"
 
Anonymous
42 secs ago, by Danu
But it's exactly the ssame type of wave as a wave in water
 
Anonymous
This is the main point ^
 
Anonymous
@watchme Can you first explain how waves in water are caused?
 
Just to spice things up. Schrodinger "wave" equation :p
 
12:16 PM
When someone for instance drops a stone in the water, it presses the water down (and to the site), which then goes up again. But as it goes up, the water around it is down again, which goes up again with water around it down, which wants to get up again.

Sorry for my english, but thats the best I can say^^
 
Anonymous
@watchme That looks okay. So it's basically the change in the field which is propagating, be it water waves or sound waves or electromagnetic waves.
 
My picture above. Does everyone of you think that data is conveyed like that?
That is what I mean by "the wave get send out".
No waves are send out^^ but the magnetic field changes
 
Anonymous
@watchme What's the definition of "wave"?
 
A wave represents the change of a state over time
 
Anonymous
12:24 PM
@watchme Quoting Wikipedia: "In physics, a wave is a disturbance that transfers energy through matter or space, with little or no associated mass transport"
 
Anonymous
In your case, the "disturbance" is definitely being "sent out"
 
well, you guys make me curious: How does an oscillation at one point manage to cause nearby points of (be it water, fields, the ground, air etc. depending on what type of wave we are talking about) to oscillate in sync and thus allow the oscillation to sort of spread out to many points at once in a wide area?
Or perhaps more simply: Why would oscillations propagating at all instead of staying put at one spot?
 
I don't have a deep answer here but I think
1) energy has to go somewhere
2) time

:P
 
Anonymous
The oscillation is not always in sync. If I start vibrating the source (say for sound waves) faster, the regions further away from it would not be in sync with the source. The change takes time to propagate.
 
yeah, but I guess the question is why an oscillation at some point A will be able to influence what happens to the oscillation at some point B infintesimally nearby
Is there some kind of interaction going on in general which allows disturbances to propagate from one point to another infintesimally nearby?
 
Anonymous
12:33 PM
Take an analogy: If we both are standing in the middle of a long queue with me behind you, and I push you forward, even the person in front of you gets pushed forward.
 
Look guys, I don't know how I should explain this.

When I want to convey digital information to my cousin, I would do someting like this:

press against his body for '1' and pull for '0'.

Now, this can be displayed in a wave form. Buuuuuuuuut its not a wave.
 
Blue: Ah right
 
@Blue I think he is got with physics and wanted to prank me
:(
 
Anonymous
@watchme Embedding digital signals into analogue signals is a complicated process. It would take quite some time to explain/understand
 
@Secret How the neighbouring points influence each other depends on the type of wave. E.g. a wave on a string spreads because the molecules the string is made of are coupled to each other.
 
12:35 PM
@Blue its easy^^

But the example above with my cousin: I imagine the same with a magnetic field. so "tug" for 0 and "pull" for 1.
 
Anonymous
@watchme "its easy" Can you explain it here?
 
So tug and pull are not a wave. but they can be displayed as a wave.
 
@ACuriousMind hmm... so for mediumless waves such as gravitational and electromagnetic waves, it is the spacetime curvature and the electromagnetic field, respectively, at different points become coupled in some way to be able to support the propagation?
 
@Blue Well you can convey digital information with anything you want as long as there are two states (pull/press, dark/bright, high voltage/low voltage)
 
Does anyone know how to calculate the edge states' band structure from the bulk Hamiltonian?
 
12:40 PM
I must have missunderstood your question as you're studying computer sience^^
And when you say its difficul I hink you mean another thing ^^
 
Anonymous
@lılostafa You are probably in the wrong place. We don't have solid-state people here :P
 
@Blue But HEP-theory folks are supposed to know all of the physics
I think it's a (very) straightforward calculation...but I have zero experience doing stuff like this. I need an expansion to this answer:
2
A: How to calculate Edge states of Topological insulators

physshypIt is quite straigthforward. First you write your hamiltonian in second quantization formalism in crsytal momentum basis. Then you make a fourier transform to fermionic field operators such that, now they are in lattice site basis. Now your hamiltonian is in form of $$\mathcal{H}=\sum_{i,j,\math...

 
Anonymous
@watchme I was trying to point at the fact how digital 0's and digital 1's are actually converted into high and low signals. Also, there's a transition threshold for which voltages are representative of 0 and which are representative of 1. Anyhow, perhaps you don't need that detail at the moment. Your exact question is: How is a digital 1 and a digital 0 "carried by" an electromagnetic signal? Hint: They have frequencies and amplitudes. Maybe ask this as a whole question on the main site.
 
I read this explanation
And I thought that the tug meant 0^^
 
@ACuriousMind any help would be tremendously appreciated :)
 
1:00 PM
@Blue

Imagine that digital data gets conveyed over a copper-cable (say Twister-Pair).
Do simplify all that lets say that a voltage of 5 is '1' and a voltage of 0 is '0'. (Amplitude Modulation).

When a computer wants to send 64Bit/s the voltage changes 64 times at most (with the manchester technique more often of course).

Know, this signals can be displayed as a wave (nothing is perfectly squared). BUT the wave doesn't get send over the wire, if you drew a wave along the wire as if a wave would get carried over the wire - that wouldn't be true, as there can only be one voltage at a time o
But it doesn't matter as I think I won't need it that much in my life ^^ You don't have to answer anymore, you already invested enough time in writing answers to me ^^
 
 
3 hours later…
3:39 PM
So this book says that $H_1(\mathbb R^n) = K$
The hell is $K$
 
vzn
@Secret that is very close to the Tenev+Horstemeyer formulation but it is unaccepted/ unacceptable to (current/ conventional) mainstream physics pov & ACM (et al) has said so himself in here, pushing back strongly on the idea in past. its the fluid dynamics interpretation of "spacetime fabric" etc. and notice how close it is to QM+GR unification crux/ pivot.
 
Q. What is the most general potential V (x) for which the equation of motion for x(t) is linear?
My Answer. $V(x) = \alpha x(t)$
is this correct?
or could x go to the second degree?
 
vzn
3:55 PM
@Mithrandir24601 thx for your expressed concerns on so-called "overmoderation". the concept is not widely recognized on SE. mods have high power compared to nonmods. their decisions are rarely reversed, voluntary or otherwise. deleting comments is a rather small )( issue conmpared with other major mod areas such as eg, "hotbutton/ triggering" topic around here lately, suspensions. if you can keep an open mind about what is overmoderation & take into acct others pov, think youre on the right track! :)
 
@vzn Oh yeah, deleting comments is the most minor mod-power I've got... That's one of the things that's really confusing - people seem willing to trust me with the ability to delete posts, suspend people, yet also feel that deleting a few comments that needed to be deleted is 'overmoderation'... Anyway, I'd best stop there before I say something I probably shouldn't :P
 
 
1 hour later…
5:35 PM
Oh great, the two things congress really understands
science and computers
 
More funding though! :)
 
how about they get rid of the ridiculous universities fee instead
They'll get more physicists that way
you'd better take that money before people find out quantum computers aren't magic
 
vzn
6:02 PM
@Slereah it worked out great with the SSC decades ago... not
@Slereah which university fee
 
the ridiculous fees to attend US universities
 
vzn
@Slereah lol its usually called "tuition" but yeah its skyrocketed last few decades. US universities seem almost to be a cartel system now, one of many now on long list in US (health care, banking, cable, telecom etc). bernie sanders was talking about free college edu... instead we get immigration ban + the Wall (so far). o_O
 
vzn
6:21 PM
@JohnRennie are you still willing to pay hossenfielder fee for a Physics AMA? (presuming she is still doing the "consulting"...) could work to arrange it, think it could be very engaging etc. anyway looks like she might have declined posting my congratulatory/ fan comment posted yesterday on her blog. sigh, ouch :(
 
If you invite Hossenfielder here at least invite Motl too
It will be fun
 
Anonymous
Aug 12 '16 at 8:58, by ACuriousMind
@DavidZ Just need to get the 150$ together and keep Lubos away :P
 
Damn you!
 
Anonymous
I don't think we need to do much do get Motl here... :P
 
Don't we
I've never seen him here
from the log he has never been here
 
Anonymous
6:30 PM
Maybe just start attacking his theories here, and link it in his blog
 
although I do recall him saying something about how @0celo7 was a fine young man
 
Anonymous
@Slereah Lol. They're both somewhat similar :P
 
I don't think ocelot has a good opinion of string theory tho
 
Anonymous
Ah, sad news. They'd really pair up well
 
Anonymous
Didn't Motl say somewhere that QM may be wrong, but String theory certainly isn't? I don't remember where though
 
6:33 PM
but string theory is based on QM!
For all its worldsheets and vertex operators it's still just a theory of operators on a Hilbert space
2
Q: Which specific smooth structure are we using in general relativity?

Victor Vahidi MottiIn this lecture by Fredric Schuller it is said that in the case of a non compact four dimensional manifold there is a non countable infinity of differentiable or smooth manifolds that are NOT diffeomorphic. Differentiable structures definition and classification - Lec 07 - Frederic Schuller My...

 
Anonymous
Yeah, that didn't make much sense to me. But then, I know nothing about strings...
 
I should try to actually work out some exotic spacetime really
Work out the actual measurable differences
Unfortunately the only one with coordinate charts you can represent is $S^7$
Not the worst since at least it's a valid spacetime
Though not the best either
Different smooth structures are pretty radical changes
 
7:00 PM
Jesus
I'm reading a 1956 paper on algebraic topology and it's already impossible to understand
The field was only 20 years old and it's all nonsense jargon
 
7:20 PM
Hey chat
 
Anonymous
Holla
 
How's it going @Blue
 
Anonymous
@Albas Not bad :P
 
Anonymous
Just stuck up in a certain paper
 
7:41 PM
Ah I see
 
7:57 PM
LaTeX is so much better when you have emoji.
 
vzn
@Slereah actually chat invited LuMo a long time ago! maybe someone who is friends with him can tip him on the shoulder, assuming he has any :P
 
yes but Motl hates you
and he is correct
 
vzn
@Slereah lol strong words, he does seem to hate some ppl & has expressed hateful sentiments to me repeatedly in comments, but am far from unique in that regard & doubt he would regard it as something as significant as "hatred"... hes like a semiprofessional troll & not all his words are genuine, some are something like performance art...
 
>This message has been edited 5 times
plz stop
 
Anonymous
Lol
 
vzn
8:05 PM
@Slereah last chat 297d ago
in Discussion on answer by Ron Maimon: Is it possible that QM is just GR?, Apr 7 '17 at 23:55, by Luboš Motl
Dear @Ron, as a high school teenager, I was thrilled by that sentence by Einstein about explaining quantum phenomena as something... that I would interpret as solitons. I think we were affected by the same sentence here. ;-) I found some "new solution" - a new type of solitons - as a result. What he wanted to solve was just the quantization of photons' energy and the electric charge. That's of course much more modest than what an actual competitor of quantum mechanics, as developed in the 1920s, would have to do. The character of the "main goal" changed as people understood QM more properly...
 
talk about some old timey chats
 
vzn
"skeletons in the closet" so to speak o_O
> This account is temporarily suspended network-wide. The suspension period ends on Mar 18 '92 at 16:28. physics.stackexchange.com/users/4864/ron-maimon
 
Can't wait for 2092
 
vzn
lol! cyber thought crimes worse than hitler? surely after the singularity & all will be entirely digitized by then :P
 
Anonymous
@vzn Wait, did the suspension period get increased again?
 
vzn
8:12 PM
@Blue apparently so! seems very strange. BT was just remarking on it in the other room.
 
Hm
I'm guessing the best way to construct a spacetime out of $S^7$ is to use the Hopf fibration until I get $S^1$
And then define the timelike direction along that
o no quaternions are involved
 
vzn
8:30 PM
@Blue his position seems to be very strongly that QM cannot possibly be incomplete, as in his recent review of becker motls.blogspot.com/2018/06/… because... copenhagen? shut up and calculate? [insert misc ad hominem here]? :P
 
 
1 hour later…
9:45 PM
@Mithrandir24601 The main problem with SE "comments" is that...they're called comments when they're really not intended for the "put your miscellaneous and tangential thoughts on the matter here" that many other sites use them for. So very often users get upset about their deletion because their expectations of what comments are for does not match with what they actually are for here.
@Secret Yes. The values of the fields at different points are coupled through the Einstein equations and the Maxwell equations, respectively, but there's no more "fundamental" explanation there, because these already are equations for fundamental forces.
@lılostafa I'm afraid I don't know a thing about topological insulators
 
@ACuriousMind Yep... I don't really have an issue with semi-useful stuff being left around or anything (then again, we're hardly inundated with comments) but why people think they should be designed to be permanent I don't know. I mean, we need something that's designed to be temporary, so if we made comments permanent (which is what answers or edits are usually for) we'd need something else to fulfil the purpose currently fulfilled by comments...
(although to be fair, so far, when I've gone '[this] is what I mean by "no longer needed"', people do seem to go 'actually, yeah, fair enough')
 
10:02 PM
@vzn following up on the whole anti-physics theme, now accusing string theorists of group think backreaction.blogspot.com/2018/06/…
2
 
If ma = -dV(x)/dx what could be most general V(x) for which x(t) would be linear?
$V(x) = \alpha x$ ?
or could it be $(\alpha x+\beta)$
 
@MohammadAreebSiddiqui Well, if $x(t)$ is linear, then what power law does $a(t)$, which is what you have in that equation, follow?
 
its differentiating twice so it would go the second degree?
actually third* ?
 
No, $a(t)$ is the second derivative of $x(t)$, not the second anti-derivative.
 
if $x(t)$ is linear then it's a free particle
 
10:08 PM
Note if $x(t)$ is linear then $x(t) = at + b$ while your $V$ depends on $x$ not $t$, $V = V(x)$, so you are looking at a differential equation $mx'' = - \frac{dV}{dx}$ which can give potentially insane solutions unless $dV/dx = 0$.
 
So $dV/dx = 0$
So it's a constant potential
 
Yup
Without the mathematese, this isn't surprising: A non-constant potential implies a force acting on the particle, but the content of Newton's laws is precisely that the things that are moving in straight lines are those no net force acts on.
@bolbteppa "potentially insane solutions" aka "motion"? :P
 
Im not getting one thing if x(t) is linear, couldn't it be in the form in which x(t) depends on itself?
 
what
 
@MohammadAreebSiddiqui I'm afraid I don't understand the question. $x(t)$ doesn't depend on "itself", it just depends on time.
 
10:15 PM
In classical mechanics, Bertrand's theorem states that among central force potentials with bound orbits, there are only two types of central force potentials with the property that all bound orbits are also closed orbits: (1) an inverse-square central force such as the gravitational or electrostatic potential V ( r ) = − k r , {\displaystyle V(\mathbf {r} )={\frac {-k}{r}},} and (2) the radial...
 
Does anyone here have a solid understanding of Dirac's bra-ket notation and tensor products using this notation?
 
why does x(t) has to be linear for dV/dx to be zero?
 
@nbro Yes.
 
In particular, it would be useful if someone has some understanding of quantum computing, to answer my question.
 
@MohammadAreebSiddiqui Is that not what we just explained?
 
10:16 PM
@ACuriousMind Suppose we have the state $|0000\rangle$, then it's equal to $|0\rangle \otimes |0\rangle \otimes |0\rangle \otimes |0\rangle$, right?
 
by definition, yes
 
In fact, the latter is the definition of the former
 
Now, can the state $|0000\rangle$ also be written as $|00\rangle \otimes |00\rangle$?
 
sure
 
10:18 PM
Why, in general, can we do that?
 
mathematically it's all equivalent
There's no deep reason, it's just a notation
 
I know we can do that, but I would like to have a solid understanding
I know the definition of the tensor and I know how to apply it
But if you explain it to me I may be able to understand it fully
 
One notation may be more semantically useful than the other
 
@ACuriousMind OMGGG, i just got it
 
Like if we want to break up qubits into semantic groupings
 
10:19 PM
lol x'' = 0 hence dV/dx = 0
thanks people! im stupid
 
@nbro Well, you have $\lvert 00\rangle = \lvert 0\rangle \otimes \lvert 0 \rangle$ by definition. So you can just apply it to $ ( \lvert 0\rangle \otimes \lvert 0\rangle ) \otimes ( \lvert 0\rangle \otimes \lvert 0\rangle )$ to get $\lvert 00\rangle \otimes \lvert 00\rangle$. I'm not sure what more there is to explain
 
@ACuriousMind I am not sure of the properties of the tensor product. That's it.
But, now, suppose we represent $|0000\rangle$ as $|00\rangle \otimes |00\rangle$.
 
It's a product
It's associative
 
@nbro If you can be a bit more specific what you're unsure about I'm happy to explain it
 
@Slereah Yeah, it's a good perspective to remember it
@ACuriousMind I didn't finish... I just wanted a confirmation, to start with
Now, suppose that we want to apply one quantum gate only to the first $|00\rangle$
The quantum gate should be a 4x4 matrix, correct?
 
10:22 PM
yes
Generally you can write this operator as $A \times I$
With $I$ the identity operator
 
Now, the composite state, after having applied the partial quantum gate, should be $\left( A|00\rangle\right) \otimes |00\rangle$, right?
 
@nbro Yes, it's going to be the gate $\otimes$-times the identity on the second $\lvert 00\rangle$. Doing the tensor product explicitly on matrices is called the Kronecker product
 
Well, I'd say you can represent the quantum gate as a $4\times 4$ matrix.
 
@nbro Yup
 
@ACuriousMind You mean that you can create a 16x16 quantum gate where A (assuming it's the quantum gate we want to apply to the first $|00\rangle$) is only applied to the first $|00\rangle$ and the second $|00\rangle$ is left unchanged?
 
10:27 PM
@nbro Why would it be 16x16, not 4x4?
Oh, right, each quibt has 2 states, right
Yes, that's what I'm saying.
You take the 4x4 gate $A$ and you form the 16x16 matrix $A\otimes \mathbf{1}_2$ via the Kronecker product, where $\mathbf{1}_2$ is the identity on two qubit
 
@ACuriousMind I thought you meant that you could create a quantum gate which acts on the whole composite system $|0000\rangle$, so that it changes the first $|00\rangle$ of the composite system and leaves unchanged the second part
 
I'm not quite sure what you mean by "create". The actual gate just acts on the two qubits involved. But when the "full" state of your QC is four qubits, you need to represent that gate as an operator on the full space of all four qubits, and the way you do it is taking the ordinary representation $A$ of the gate and forming $A\otimes\mathbf{1}_2$.
 
@ACuriousMind Ok, but, in general, I can "safely" apply the 4x4 gate on the "partial state" and then just tensor product the result with the remaining part and it would be equivalent to multiplying by this 16x16 matrix, right?
 
@nbro Yup
 
@ACuriousMind Yes, I suppose this is the most mathematically accurate way of representing the operation...
Now, I like to know how the tensor product behaves w.r.t. to summations
In particular, I would like to know if the following section (which I have been changing and improving) uses a correction notation:
In computational complexity theory and quantum computing, Simon's problem is a computational problem conceived to showcase the efficiency increase a quantum algorithm could have over a classic one. Although the problem itself is of little practical value, it is interesting because it provides an exponential speedup over any classical algorithm (in a black box model). The problem deals with the model of decision tree complexity or query complexity and was conceived by Daniel Simon in 1994. Simon exhibited a quantum algorithm, usually called Simon's algorithm, that solves the problem exponentially...
Specifically, after "Observe that, for every", there's a mathematical expression
In that case, we are summing over $x_1$, but $|f(x_1)\rangle$ is outside the summations, i.e. we are tensoring product it, as $\dots \otimes |f(x_1)\rangle$, but $x_1$, in that place, is a constant, so I don't think that notation is correct
My question is: can we pull the $|f(x_1)\rangle$ into the summation? Which property of the tensor product would allow us to do it?
 
10:44 PM
@nbro Well, your problem with that notation is not really about the tensor product, but that you're generally not allowed to reuse a summation variable outside of a sum
 
I actually introduced, in that article, the explicit $\otimes$ notation (for clarity).
 
To speak in programming terms: The scope of definition of a summation variable is within the sum, and if a sort of parser was to parse something like $(\sum_x x) \cdot x$, it would complain at the third $x$ with something like "variable x unknown".
 
Previously, it was not clear if $|f(x_1)\rangle$ was or not inside the summation, as there was not $\otimes$ symbol there
 
@ACuriousMind Well, unless there was another $x$ defined outside the summation. Which is something I've seen (and done) but it is a recipe for confusion.
 
@nbro Well, but a function of $x$ being outside the summation makes no sense when $x$ is the summation variable, regardless of whether there's a $\otimes$ or a normal product or whatever
@DavidZ Any compiler that does not issue warnings for shadowing variables is evil :P
 
10:48 PM
lol, but true
 
Variable collision
 
@ACuriousMind Yes, but then let's go back to our previous discussions...
Suppose we have the state $|0000\rangle = |00\rangle \otimes |00\rangle$. Suppose we apply a quantum gate to the first $|00\rangle$, say it's a Hadamard (I suppose you are all familiar with it), then the result of that application to the first $|00\rangle$ is $\left( \frac{1}{ 2^{\frac{n}{2}} } \sum_{x\in \{0, 1\}^n} \left| x \right \rangle \right)$
 
My eyes
 
I find that notation very hard to parse, but yes
 
Ops, it should be, as I first wrote, $\left( \frac{1}{2} \sum_{x\in \{0, 1\}^2} \left| x \right \rangle \right)$
 
10:55 PM
And why did you just introduce an "n" - in this case $n = 2$.
 
@ACuriousMind Sorry, my mistake. It should be $n=2$
Now, the composite state should be $\left( \frac{1}{2} \sum_{x\in \{0, 1\}^2} \left| x \right \rangle \right) \otimes |00\rangle$, right?
 
yes
 
So, the property that I didn't use in the article and that I should have used is that the tensor product can be distributed into the summation. Right?
So, the final composite system can also be written as $\frac{1}{2} \sum_{x\in \{0, 1\}^2} (\left| x \right \rangle \otimes |00\rangle)$
 
Please put brackets!
 
The property that you're using is just how products of operators act on products of wavefunctions
 
10:58 PM
E.g. $\frac12 \sum_x \left(\lvert x\rangle \otimes \lvert 00\rangle \right)$
 
but yes tensor product is bilinear
 
@Slereah I still don't understand that perspective of quantum computation. I am not really a physicist, as you can easily guess
@ACuriousMind Yes, I wish all people thought like you!!
Anyway, guys, thanks a lot for your help!!
 
np :)
 
A "thanks" with some breakbeats, if you like them
 
@nbro I currently finished my a levels and I have a passion for quantum phy and computer science both, i see you are a man on quantum computing, what degree are you pursuing exactly? can you guide me on how should I pursue this career? I'm mainly interested in research on how to build quantum computers(their working, architecture etc), not creating algorithms for them
 
11:17 PM
@MohammadAreebSiddiqui First of all, I would like to say that I had only one course in quantum computation, and (of course?) we covered the theoretical part/perspective (i.e. algorithms, etc.). Quantum computation is still a very embryonic discipline. I don't want to discourage you, but, honestly, I think that what you really need to be one of the chosen ones to build or attempt to build those quantum computers is "luck"
Apart from that, I don't think I can really give you some piece of advice. I am more into algorithms, software, etc.
I can be completely wrong (I hope), but I don't think we are going anywhere with quantum computers.
But, again, don't take me too literally. If you believe in quantum computing, go for it!!
 
So I basically should have my college degree in Computer Science to make sure I have a safe path in case some albert schrodinger proves quantum computing is impossible :P
 
@dmckee whaddayaknow, OP accepted that one physics.stackexchange.com/questions/385587/…
 
@MohammadAreebSiddiqui Right now, theoretically, quantum computing is an interesting field, if you like mathematics and, in particular, linear algebra
If you don't even like mathematics, then, definitely, do not study QC
 
11:45 PM
I love mathematics even though I suck really bad it
 

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