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3:42 AM
why has chat been so dead lately?
 
user116211
@user333355557777: what happened to you o.O
 
user116211
Meanwhile I came across this:
 
user116211
Is it some coincidence or do you know each other @user333355557777?
 
user116211
Hmm.
 
user218912
3:46 AM
oh that's me
 
user218912
made a new acc.
 
user116211
I'm confused.
 
SOCK PUPPET ALERT
 
user218912
no
 
user218912
ask the mods, none of my posts are upvoted by my accounts.
 
user116211
3:47 AM
Were you unable to access your earlier account?
 
user116211
@3750 ok but why?
 
user218912
no I just made a new one because it was dead with 2 solid state physics answers.
 
user116211
So?
 
^
 
user218912
what
 
user218912
3:49 AM
I abandoned the old account.
 
user218912
who cares
 
user116211
2 mins ago, by MAFIA36790
Were you unable to access your earlier account?
 
user218912
nah I just didn't like it anymore.
 
...what??
 
user116211
You can merge the two.
 
user218912
3:50 AM
I got tired of the colour of 3075.
 
user116211
@3750 okay, contact the SE team to delete that account.
 
user218912
I don't like that shade of blue anymore.
 
user116211
Or tell them to merge the two.
 
user218912
or I can leave it...
 
user218912
it takes days to delete and account that has answers.
 
user218912
3:51 AM
you have to e-mail them specifically
 
user116211
Nevertheless I love your new avatar.
 
user116211
Hmm.
 
user218912
thanks.
 
user116211
Well, unless you use the two for your personal gains, it seems okay ;\
 
user116211
But I'm really not getting any proper reason behind this.
 
user218912
3:52 AM
nah why would I
 
user218912
if you don't believe me you can ask mods to confirm.
 
user116211
well, that's okay ;\
 
user218912
xD
 
user116211
But still I would assert to either merge the accounts or delete the first one you are disliking.
 
user116211
No matter how much time it takes.
 
user218912
3:54 AM
fine dude
 
user116211
It would make you stay in the safe borders ;)
 
user218912
ugh I have to relog. brb
 
user218912
@MAFIA36790 done lol
 
user116211
good.
 
what is done?
deleted?
 
user218912
4:03 AM
scheduled for deletion.
 
that's crazy
 
user116211
not crazy, but weirdo ;\
 
user218912
why lol
 
user116211
WTH!
 
user116211
John Nash completed his phd in 2 years!!
 
4:11 AM
Makes you wonder what losers like @ChrisWhite take so long for.
 
user218912
that's not nice.
 
user116211
Well, his thesis was just 26 pages long with only 2 citations.
 
user116211
I've not heard of such a short thesis so far ;/
 
5:48 AM
1 message moved to Trash
 
user116211
20
Q: If the British government did not follow through with the brexit what would happen?

user1605665The referendum has called for a British exit from Europe but to exit the british parliament needs to pass on the request to the EU. If the british parliament did not pass on the request to exit from the EU what would / could happen? Brexit results http://www.bbc.com/news/uk-politics-36616028

 
user116211
ooh.
 
vzn
@Jim how about a nonlocality demonstration/ measurement using classical physics? have been musing on this a long time; its ambitious but within reach of a talented undergraduate, contact me here for details chat.stackexchange.com/rooms/9446/theory-salon
 
@MAFIA36790 the referendum isn't binding. When we elect the government we give them the right to decide on our behalf so what we subsequently say in a referendum isn't binding.
However if the government ignored the referendum it would cause an uproar. There would inevitably be an immediate election.
What would happen in that election is an interesting question. If the government had any balls they'd refuse to exit and hold an election. So the election would in effect be a second referendum on the issue.
 
user116211
It's a common sense; after all Britain is not run by a Fascist dictatorship.
 
user116211
5:55 AM
I hope they remain associated with EU like Norway.
 
@MAFIA36790 the issue is the freedom of movement that the EU requires i.e. immigration. That was the reason most people voted to leave. If the UK won't accept freedom of movement then we're going to be locked out of most of the EU.
 
user116211
@JohnRennie don't say most; they won by a narrow margin ;\
 
I meant that of the people who voted leave, most of them voted leave because of the freedom of movement that the EU requires.
 
vzn
@JohnRennie am not really questioning most aspects of big bang, only some. note that (afaik—not an expert) it didnt really have strong experimental evidence until "relatively recently". some hypotheses are expressed as math eqns. the math helps make it more precise but not necessarily more "real"...
 
@MAFIA36790 And to be fair I can see their point. The UK has the second biggest economy in the EU and it's fourth or fifth (I forget which) in the world. So it's a magnet for people from the poorer countries in the EU.
Those people are overwhelmingly hard working people who want only to do the best for their families, and they make a net positive contribution i.e. they contribute (far) more to the economy than they take out.
 
user116211
6:02 AM
If people were more aware what they were doing before they voted for exit ;(
 
But the result is that the UK infrastructure is going to need a lot of investment, and that takes a lot of money at a time when the public finances are really tight.
 
vzn
@JohnRennie similar arguments made in US. various parallels. the economic impact/ cost/ benefit of immigration is very difficult to measure, not many studies have been done and they probably vary widely in methodology/ findings, its very abstract to calculate.
 
@vzn Agreed. It's a hugely complicated issue.
 
user116211
Meanwhile, I forgot to post it here ;)
 
user116211
user image
2
 
vzn
6:04 AM
@JohnRennie eg there are cases in economics where the economists might argue its better if you lose your job ("for a better one" or whatever) but nobody can really accept/ "swallow" that personally...
my opinion, brexit is probably highly tied to spiralling wealth inequality, and wrt immigration, the public picking the wrong bogeyman...
 
However there are clear problems that immigration is causing.

For example the school system is under strain because too many people want its services i.e. there are too many children. The solution is to build more schools, but that takes a huge up front investment that the country can't afford.

Another solution is to remove several million people from the population, and blocking freedom of movement would indeed do that.

In the short term it would solve a problem, though of course in the long term it will make us Brits worse off.
@vzn but I suspect you're correct that this is what has influenced many people.
@0celo7 The science fiction of the fifties and sixties was stuffed with references to unified field theories. Though it's hard to believe, Einstein's search for a unified field theory was well known back in the day. Any SF reading student would have heard of unified field theories.
 
vzn
@JohnRennie the electoral map you posted awhile back was quite remarkable & telling & deserves much further analysis. where else do you get such high geographic cohesion/ consistency? now superimpose a map of wages or wealth inequality by region. think the results will surprise some but not others....
 
@vzn It isn't that simple. My area (Cheshire) is pretty wealthy but it voted leave (by about the same margin as the country as a whole).
 
vzn
@JohnRennie maybe its not as "wealthy" as you might think... think there is much room for further analysis in this area...
 
 
1 hour later…
7:23 AM
Hello
@JohnRennie it is quite amusing that the seminal paper on wormholes was written to help out a science fiction writer
 
@Slereah there's a lot of Physics SE members who are also members of the SciFi SE. I suspect many of us got into physics by reading SF.
 
user116211
Why are integrals always defined on compact set?
 
user116211
convergence of an integral over compact set...
 
user116211
@JohnRennie Totally true.
 
user116211
Got something:
 
user116211
7:32 AM
10
A: Why is the Riemann integral only defined on compact sets?

user12014If you use the definition without tagged partitions, the reason the interval needs to be compact is that you need the function to obtain suprema and infima on every subinterval on a partition. For example $f(x) = 1/x$ is continuous on $(0,1)$ (so it should be integrable), but it never attains a s...

 
@JohnRennie true
I wonder if there are statistics between science studies and the popularity of SF
Was there a big bump in people going for a science major towards the end of the 19th century
or beginning of 20th
though to be fair, there might be more general phenomenon at work
Maybe both were triggered by the industrial revolution
For most of history, technology did not change all that much for the average man, so maybe there was not a lot of reason to hope for a more technological advanced tomorrow or studying science
I've read a bunch of "science fiction" from before the 19th century, the future never changes all that much technology-wise
People mostly hoped for social changes
 
7:47 AM
@MAFIA36790 Take, e.g., $(0, \infty)$ and the function $f(x) = x$ on it. How can the integral make sense? The area under the curve is clearly not finite.
You can integrate over whatever domain you want if your function is compactly supported however, which is the class of functions most people restrict to.
 
@JohnRennie : at a time when the public finances are really tight. The public finances would be tight if we stopped giving away £10bn a year to the EU, or £12bn a year to foreign countries, or £15bn a year to Scotland.
 
Farage and the evidence
 
8:06 AM
@EmilioPisanty they have other answers. In any case I'm not too motivated to be accommodating to someone who, when told directly not to edit the content out of their answer, edits the content out of their answer.
 
@Slereah : no, a vote for democracy. Some people are somewhat attached to it. As opposed to having their lives ruled by a lobster-supper gilded elite who don't pay tax and who can't be voted out of office, even when ordinary people are desperate and youth unemployment hits 50%.
 
8:28 AM
@Slereah If you consider these operators $\bar{u}_L\gamma^\mu u_L\bar{u}_L\gamma^\mu u_L$ and $\bar{u}_R\gamma^\mu u_R\bar{u}_R\gamma^\mu u_R$ where $u$ is the u-quark, can you have interference terms between them when you compute the cross section of the process u u > u u ?
 
Errr I think so?
The two chiral components of quarks are couples by the mass term
 
you can consider a vanishing mass
 
Although for the up quark it's not uncommon to do $m \approx 0$
If the mass is 0 then the chiral components are decouples
You can treat them as two free particles
Well free of each other, at least
btw in the limit $m = 0$ you gain some SU(2) symmetry for the strong interaction
 
ok, I though the same thing. I am trying to do the calculations of a paper and they put inside an interference term
 
With the doublet $(u_L, u_R)$
Not too fond of meson physics but I worked for like 4 months on the SU(n) effective field theory for strong interaction so some things stick :p
 
8:52 AM
@Slereah ;)
 
 
1 hour later…
10:20 AM
@Slereah do you use feyncalc?
 
@DavidZ Yeah, that's fair enough. I hadn't seen the re-do.
 
 
1 hour later…
11:50 AM
I do not
All analytic, baby
 
1
Q: Homework and Exercise Tag

Harsh Sharma Why is there a complete homework and exercise tag, when all the question asked under that category are termed as off-topic. This tab accounts as the second largest tab but still the moderators down vote and also mark the topics under this tag even after providing a complete attempt done by the p...

 
My work on mesons was computing cross sections of meson scattering
It was some boring stuff
Also trying to compute SU(3) Clebsch-Gordan coefficient is not easy
 
I need to buy a toaster.
 
Why not just use a pan
 
user116211
why not just use a stove
 
12:03 PM
@MAFIA36790 don't have one?
 
user116211
@0celo7 Models of Philips are good, BTW.
 
user116211
and of course cheap.
 
@JohnRennie Found another way of integrating on nonorientable manifolds while browsing MO. If $M$ is nonorientable, it has an orientable double cover $M^c$. So find the Riemannian volume form $\nu$ of $M^c$, and then $\int_Mf=\frac{1}{2}\int_{M^c}f\nu$.
Although I'm not clear on how the metric of $M$ gets lifted to $M^c$...
 
Isn't the metric the same, just the atlas changing
 
Proof?
 
12:15 PM
And then if two points are ~ the same point in the original manifold the metric has the same value
Gut feeling mostly
 
I don't even know where to find that
 
my belly
 
Most Riemannian geometry books are unconcerned with such minute details
 
I vaguely recall people solving field equations on non-orientable manifolds using this method
mb check something like that
 
wonder what time it is in Japan
 
Jim
12:21 PM
@ChrisWhite I'm listening intently
@vzn nonlocality? I've not heard of such a thing existing. Are you referring to the Bell's inequality experiment? Because that doesn't show nonlocality in the least bit
 
@JohnRennie Hmm, I've never been into SF.
At least not SF for SF's sake.
 
Jim
In the words of the great Isaac Asimov, "Individual science fiction stories may seem as trivial as ever to the blinder critics and philosophers of today - but the core of science fiction, its essence has become crucial to our salvation if we are to be saved at all."
 
tfw you hear a sample and you know it was on another song but you have no clue where
Must have been in a Pendulum song
Hmm, maybe not,
 
12:46 PM
preliminary results of my analysis of the homework policy data
in case anyone wants to start thinking about it
 
I say kill them all
 
Jim
@DavidZ I may need some context. Do you have a link to an accompanying meta post or something?
 
@Slereah flag worthy?
that sounds like a threat to me
 
user215373
hi
 
wew only 86 degrees today
I can not die of heat stroke yay
 
user215373
1:03 PM
@Slereah hi
 
Jim
:30699189 play nice
 
I just want a consistent application of the rules
@Slereah did not even get suspended
 
Jim
@0celo7 you play nice too
 
boohoo
 
Jim
don't make me separate you two
 
1:17 PM
@Jim I am always nice.
 
Jim
good. Because so help me, if there's any more fighting I will turn this chat room around and then nobody gets to go to physicsland.
 
@Jim the data come from this big list we put together, and then ratings of each of the questions linked there on 9 attributes (listed on the horizontal axis). I'll make a separate meta post about it when I have done a proper analysis.
 
Hey,

I need to start learning Mathematics. I don't know programming as such and I need to learn Mathematics for the purposes of the research project(s).
How should I go about it systematically. I don't want to spend time learning/going through manuals etc. but I do want to go about it slightly systematically rather than having to reply on youtube videos in random orde
r
 
The point to start thinking about is that the factors which we supposedly are basing our new policy on - tediousness of the calculation and lack of effort - are almost entirely uncorrelated with what questions people think should be on and off topic.
And that one of the factors which is most correlated, namely "interestingness", is completely subjective and thus an absolutely terrible thing to base a closing policy on
 
Jim
@DavidZ so the numbers are correlation factors? 1 being perfectly correlated with being on-topic and -1 being perfectly correlated with off-topic?
 
1:31 PM
Ah, I knew I forgot to mention something. Yes, it's a correlation matrix, between the 9 attribute ratings and the score each answer to the meta post received. (Score is supposed to be a proxy for on-topicness)
To emphasize, this is a very preliminary analysis.
 
Jim
it certainly seems so
 
I have some better analyses in the works, but no pretty pictures from those yet.
 
Jim
am I reading this wrong or does it suggest that people consider "is this correct" more on-topic than "shows effort"
 
It does seem that way, yes
Not by much, though
 
Jim
both of which are more on-topic than conceptual questions
I'm starting to dislike this analysis
 
1:37 PM
Basically, for a factor which seems to strongly influence people's judgment of on-topicness, the corresponding column will show red, blue, red (if they think it's on topic) or blue, red, blue (if they think it's off topic)
I admit the analysis is not great, but I do think it shows that there's something to dislike about the underlying data, and not just the method of analyzing it
 
Jim
@DavidZ indeed, that is what I meant
 
ah, gotcha
 
@Jim Well I'm notorious for hating physics so that's ok.
 
Jim
@DavidZ where did the data come from, exactly?
 
The score data came from the answers to the meta question
 
Jim
1:40 PM
@0celo7 just as long as you're also okay with answering to all those that wanted to go to physicsland
 
The ratings on the 9 attributes came from me. I also have a bunch of ratings submitted by others which I'll incorporate into the analysis next
 
Jim
link? I may have missed it while I was away
 
which link?
 
Jim
the meta question
 
I posted it above
19
Q: Replacing the homework policy 1: what existing questions should be on/off topic?

David Z TL;DR: post examples of current questions which are edge cases for the new policy, look through the list, and vote answers UP if you think they should be ON TOPIC vote answers DOWN if you think they should be OFF TOPIC When we last left the ongoing examination of our homework po...

 
Jim
1:42 PM
right thanks
I'm going to go over this and see if there's a systematic cause that may be skewing the data
 
and I think this is the link to the rating form. Anyone who hasn't already gone through it is still welcome to do so. MOAR DATA nomnomnom
anyway I'm out for now
 
@JunaidAftab If I were you I would learn it in a random order from YouTube videos.
Maths is a huge subject and I think it's silly to try and learn maths. Learn the bits you need for your work as you go along.
 
Jim
2:00 PM
interesting side note: MATHS is an acronym for Mathematical Anti-Telharsic Harfatum Septomin
 
For a small subset of meanings of the word interesting :-)
 
Jim
@JohnRennie that's an interesting addendum
 
@Jim that's what my girlfriend tells me
 
Jim
@JohnRennie <---- Winner!
 
user116211
@DavidZ: You said HW tag would be removed. Can you please elaborate on it?
 
Jim
2:37 PM
@DavidZ having taken a look, it seems that some of the upvoted answers to the meta question that are contentious and most likely should be downvoted are usually also upsold by the answer poster and/or the comments. It may be that people read what others think before judging the question for themselves and that biases their vote. One might try doing this without allowing for personal thoughts in the answers or comments. Just read the linked question, vote how you think, and go.
Furthermore, it seems many questions slip through the cracks (this is not just in the example posts, but also from personal experience) by being more interesting. Which begs the question of if interest should really be a saving grace (possibly a separate debate for meta to handle)
 
@0celo7 The tangent bundle of $TTM$ is the direct sum of the horizontal and the vertical subbundle (horizontal/vertical relative to the projection $TM\to M$). Show that these two subbundles are isomorphic and you're done.
@kevinTahN. I'm afraid I don't see what I get out of this offer.
 
@ACuriousMind vat
 
@0celo7 Ehhh...the tangent bundle of $TM$.
One $T$ too many there. I'm saying that $TTM$ is the direct sum of two isomorphic bundles.
 
@0celo7 Pullback metric? Since $M \to N$ is a covering, that induces an isomorphism at the level of tangent spaces. Push two vectors in $T_pM$ forward, inner product in $T_{f(p)}N$, pull that back by that isomorphism.
 
2:54 PM
@balarka the arc length formula is derived from the hypotenuse of a triangle right?
 
I don't know what that means. You mean arclength of a curve? Sure, infinitisimally, it's Pythagoras's formula.
 
@ACuriousMind I know but why
@BalarkaSen Ah yes
 
Blatant advertising
0
A: Standing waves due to two counter-propagating travelling waves of different amplitude

Emilio PisantyIn general, a mixed wave of the form $$ f(x,t)=\cos(x-t)+r\cos(x+t) $$ will not have nodes, at least in the sense of points $x_0$ for which $f(x_0,t)\equiv 0$ for all times $t$. In general, there's relatively little to say beyond what the picture will convey: As you turn $r$ up from 0 to 1, ...

 
@0celo7 It's the direct sum of the pullback bundle $\pi^\ast(TM)$ for $\pi:TM\to M$ and the vertical subbundle that consists of the tangents to the fibers of $TTM\to TM$. That they have the same rank is obvious, that they are isomorphic might be a bit harder to show rigorously and I have no interest in doing that.
 
Just for the joy of that $r=0.6$ one.
 
3:06 PM
@ACuriousMind How do you know that?
 
What's the name for the region around an astronomical body where its own gravity is dominant. It's something like the Fred's zone or possibly Fred's radius.
 
@ACuriousMind Did you mean this partial integration? (concerning this hint of yours)
 
@JohnRennie ding! It's the Hill sphere.
 
[quantum mechanics self study part 1, entanglement summary 1.X]
(Questions and discussions will follow shortly (before I move on to chapter 8). This is just a convenient checklist that others can refer to to check my understanding so far for these 12 days of self study, when analysing my questions in case it does not make sense without me realising)

Book name: Leonard Susskind quantum mechanics theoretical minumum
Current location: Page 235

Summary of stuff learn from it:
1. Qubit systems
a. Experimental outline of the stern galach experiment (if apparatus is aligned spatially to the dire
 
@0celo7 Argh, I mistyped again, it's the fibers of $TM\to M$. My "intuition" is that $TM\cong U\times\mathbb{R}^n$ locally, where $U\subset M$ and $U\cong\mathbb{R}^n$. Now $TTM$ is locally $TU\oplus T\mathbb{R}^n$ and $TU\cong T\mathbb{R}^n$. So locally one might just switch the roles of $U$ and $\mathbb{R}^n$, thereby exchanging the horizontal and vertical subbundles - if this switch glues, then the two bundles are isomorphic.
 
3:21 PM
@ACuriousMind I have literally no clue what you just typed
 
@Bass yes
 
@ACuriousMind So if $\frac{\delta}{\delta x(t_1)}$ acts on $x(t)\ddot x(t)$, I need to take the "product rule", and then switch $\frac{\delta}{\delta x(t_1)}$ with $\frac{d^2}{dt^2}$, right?
 
@0celo7 Well, as I said I'm not making this rigorous. So, if you will, I don't actually "know" that the splitting is true except that the exercise clearly implies it and these two bundles are the only "natural" ones to choose here.
 
@ACuriousMind why is $TTM$ locally $TU\oplus T\Bbb R^n$?
@ACuriousMind I'm not looking for rigor, I'm looking for intuition
 
Hi all
 
3:26 PM
Q1. Referring to susskind's book, What is the difference between the wavefunction described by a complete set of commuting observables vs wavefunction in a composite systems?

For example a wavefunction $\psi (x,y,z)$ that describe the state of a moving particle in space has 3 set of (continuous) indices x,y,z

whereas the wavefunction $\psi (a,b)$ in the two spin system is indexed by two set of indices a and b (which take values up or down each) corresponds to Alice's and Bob's subsystem

since both $\psi$ are functions of a set of indices, is it correct to say that the moving particle is
 
@0celo7 If I have a manifold that's $M\times N$, then the tangent bundle is spanned by the subbundles $TM,TN$, i.e. it's $\pi^\ast_M(TM)\oplus \pi^\ast_N(TN)$ (...I think that's true, at least). It now occurs to me I omitted the pullbacks from the factors there.
@Bass Ah, my intention was to first switch the variation and the derivative, and then integrate, so that you get rid two dots on the factor that you don't vary
But I guess you can also just switch them now
Oh
I guess we might not need to do this after all - you asked "what to do" with the derivatives acting on the deltas, right?
I realized I never answered that: $\int f(x)\partial_x\delta(x) = \partial_x f(0)$. The derivative of the delta just shifts the derivative onto the test function when evaluated.
 
@ACuriousMind A quick QM question, if you have two observables that commute but one has degenerate eigenvalues then can we still state that they have a complete set of common eigenvectors?
 
Doing that you might also get the answer without partial integration?
 
@ACuriousMind Well, my question after all is, how can I derive the equality Nakahara uses.
 
@JohnDoe Why would the completeness change because of degenerate eigenvalues?
@Bass Yeah, my new and improved answer is: Don't use that "chain rule", the variation and the derivative commute. Then use $\int f \delta' = f'(0)$ for all derivatives acting on deltas that you might get.
 
3:32 PM
@Acuriousmind typo thus repost again:
Q1. Referring to susskind's book, What is the difference between the wavefunction described by a complete set of commuting observables vs wavefunction in a composite systems?

For example a wavefunction ψ(x,y,z)
that describe the state of a moving particle in space has 3 set of (continuous) indices x,y,z where the corresponding observables commute with each other.

whereas the wavefunction ψ(a,b) in the two spin system is indexed by two set of indices a and b (which take values up or down each) corresponds to Alice's and Bob's subsystem. The observables
 
Quiz: what's more explosive, a cubic centimetre of atomic hydrogen at STP with all the electrons removed, or the same sample but with all the electrons replaced with antiprotons?
 
@ACuriousMind OK! Gonna try again.
 
@EmilioPisanty The "replaced with antiprotons" variant appears to be obviously correct, so it's probably the other option ;)
 
@ACuriousMind I have two proof which both assume non degeneracy. The first theorem is "If two Hermitian operators, $A$ and $B$ commute and if $A$ has no degenerate eigenstate, then each eigenvector of $A$ is also an eigenvector of $B$." the second is "If two observables are compatible, their corresponding operators posses a set of common eigenstates. " Both assume it non degeneracy in the proofs, I thought maybe the result does not hold unless we have this assumption.
 
@ACuriousMind Yeah, it's really strange. You'd think that any macroscopic sample annihilating would be at the top of any catastrophe list, but the electrostatic repulsion is even stronger than that.
 
3:37 PM
@JohnDoe Ah. Well, the first statement indeed fails for degenerate $A$ - consider what happens if $A$ is the identity operator. The second holds regardless of degeneracy: Commuting observables have common eigenbases (but because of the failure of the first statement, not every eigenbasis of one of them is a common one).
 
@ACuriousMind So?
 
@ACuriousMind Shouldn't it be $\int f(x)\partial_x\delta(x)=-\partial_x f(x)$?
 
Ohhhh
 
(a sign because of partial integration)
 
vzn
@Jim its a long story, very central to QM, will take a lot of time/ commitment to pull off, but am willing to collaborate, looking to play theorist counterpart to someone more "hands on/ engr". have been studying it all close to ~2decades at this point. last thoughts on subj here, plz contact me in other room if any )( interest vzn1.wordpress.com/2015/11/03/…
 
3:40 PM
@Secret I'm not exactly sure what you're asking, but I think it's that $L^2(\mathbb{R}^3) = L^2(\mathbb{R}\otimes L^2(\mathbb{R}) \otimes L^2(\mathbb{R})$. Less formally, a wavefunction in three dimensions may be written as $\psi(x,y,z) = \sum_{n,l,k} \psi_n(x)\psi_l(y)\psi_k(z)$ where $\psi_i$ is a basis for the wavefunctions in one dimensions.
@Bass Oops. Yes, you're right
 
@ACuriousMind The second one doesn't state that it is a complete set of common eigenbasis is this implied?
 
@JohnDoe Well, I don't know if the author of that statement wanted to imply it, but it's true.
 
@0celo7 I think the reason so many of us were into SF as kids is that it opened so many wonderful possibilities, but made them plausible. It was like magic, but logical magic that made sense. Actually I still think that a lot of physics is like logical magic that makes sense.
 
@ACuriousMind Did you previously mean consider what happens if $B$ is the identity operator and $A$ is degenerate?
 
@JohnDoe No. I meant what happens if $A$ is the identity operator (which is maximally degenerate, all eigenvalues are 1!)
 
3:48 PM
Apart from QFT of course, which is magic but is neither logical nor makes sense :-)
@Qmechanic: closing this seems a little unkind:
1
Q: How large does a gravity well need to be to move particles from a planetary body?

Dupontrocks11I almost considered asking this question on WorldBuilding, however I wanted the brute mathematics on the subject, so please excuse the theoretical nature of this question. I understand the basic nature of gravity wells, in that they are analogous to literal wells in the fabric of space. Theoreti...

yes, the OP mentions an artificial gravity source, but the question doesn't depend on the source being artificial.
 
@JohnRennie Ah, that sounds very much like Clarke's third law: Any sufficiently advanced technology will be indistinguishable from magic.
 
@ACuriousMind Then every vector would be an eigenstate of $A$.
 
@ACuriousMind as a 12 year old boy in the 70s, books about magic was what the girlies read. We real men read books about rayguns and spaceships. Of course the wonderful inventions in our SF books were really just magic as well, but we had the excuse that it was SCIENCE!
 
Hmm, I see. From what you said, that a wavefunction in three dimension can be considered as a composite system of 3 1 dimensional wavefunctions (because the summation formula you wrote (and also the tensor product equation involving $L^2$ suggest the basis of the 3D system can be constructed by tensor product of the basis in each 1D system)

Clarification of the question:
Susskind introduced commutating set of observables by cosidering an orthogonal basis $\lvert a,b,c,\dots\rangle$ where the a,b,c are eigenvalues of mutually commuting observables A,B,C... Then any state in such system can
 
@ACuriousMind I see the proof would fail.
Unless $B$ was also the identity operator...
 
3:56 PM
@MAFIA36790 where did I say that?
 
@DavidZ in this reply where you say That being said, our homework policy is changing soon. Perhaps at that point, we will get rid of the tag.?
 
@Jim perhaps that would have been better, but it was an intentional decision to allow comments in the first place. I wanted to get a reflection of the consensus, not just a vote. If people had their view changed by discussion, I figured the scores should reflect those updated views.
@JohnRennie key word perhaps
 
@Secret No, it can not be done in general. Consider using the commuting operators $A$ and $A^2$.
 
@DavidZ indeed, but my guess is that's what Mafia was thinking of.
 
@JohnRennie : Noted. 'Artificial gravity' is one of those words which should be used with caution (meaning: define precisely what it is when used), not to taint Phys.SE's record of not harbouring crack-pot physics.
 
4:01 PM
I have to say I think it will be a cold attosecond in the Big Bang before the homework tag disappears.
 
@ACuriousMind What is the condition that will determine whether a basis formed by N commutating observables can be constructed from tensor product the basis for the subsystems that corresponds to each observable? Is no two observables being functions of each other (like the $A$, $A^2$ example above) a sufficient condition?
 
i like crack-pot physics
 
@BalarkaSen ikr
 
"if you go through a black hole on a spacezooter, will you end up in an interdimensional time zone?"
2
 
@Secret There is no "subsystem that corresponds to each observable". It just happens that if $\{\lvert n_1,\dots,n_k\rangle \mid n_1\in I_1,\dots n_k\in I_k\}$ is a basis for the space, then that space is abstractly isomorphic to the tensor product of the spaces $H_i$ spanned by vectors $\lvert n_i\rangle$ labeled by $n_i\in I_i$.
 
4:07 PM
@ACuriousMind This is the statement that confused me which is written after the proof of the second theorem I stated previously "What happens if an operator, say A, has degenerate eigenvalues? The specification of one eigenvalue does not uniquely determine the state of the system. Among the degenerate eigenstates of $A$ only a subset of them are also eigenstates of $B$. Thus, the set of states that are joint eigenstates of both A and B is not complete." Does this make sense given compatibility?
 
user116211
@DavidZ I may be misinterpreting, but you said some rid of in your latest answer.
 
@JohnDoe Before I go on a rant how that quote appears to be complete nonsense, what is your definition of "complete"?
 
@MAFIA36790 see above
 
user116211
@DavidZ ah! got it.
 
@ACuriousMind The set is complete if it spans the space.
 
4:10 PM
@JohnRennie the Fred Flintstone
 
@Qmechanic better now?
 
user116211
@BalarkaSen noted.
 
@ACuriousMind How does this isomorphism breaks down (as implied in "No, it can not be done in general. Consider using the commuting operators A and A2") for the $A$ and $A^2$ example?
 
@JohnDoe Then the quote is false. The set of states that are joint eigenstates of commuting operators is complete, even if the operators are degenerate. In each $n$-dimensional degenerate eigenspace of $A$ you will find $n$ linearly independent vectors that are eigenvectors of $B$.
Where is that quote from?
@Secret The hypothesis is false. There is no set of the form I gave that forms a basis.
 
@ACuriousMind Maybe I'm missing something that the author is implying. The book.
 
vzn
4:19 PM
@BalarkaSen why? not sure what you mean
 
@JohnDoe I hope you're missing something, since otherwise you need to find a new book :P
 
@mods, is this a protect candidate?
46
Q: Why is the Moon considered the major cause of tides, even though it is weaker than the Sun?

Moctava FarzánYou have likely read in books that tides are mainly caused by the Moon. When the Moon is high in the sky, it pulls the water on the Earth upward and a high-tide happens. There is some similar effect causing low-tides. They also say that the Sun does the same as well, but has smaller effect compa...

 
@ACuriousMind After what I wrote previously it then states:
To resolve the degeneracy, we can then
introduce a third operator $C$ which commutes with both $A$ and $B$;
joint eigenstates of $A$, $B$, and $C$ that is complete. If the degeneracy persists, we may introduce a fourth operator $D$...
 
...I think that book misuses the word "complete". In particular, the phrasing "only a subset of them are also eigenstates of $B$. Thus, the set of states that are joint eigenstates is not complete" is just nonsensical - the "Thus" part doesn't follow at all - subsets of sets that span the whole space can still span the whole space.
@EmilioPisanty ...do you have any idea which of these answers is so good it needs a 500 rep bounty? oO
I find that bounty suspicious.
 
@ACuriousMind Okay thanks for help.
 
4:30 PM
@Acuriousmind ok I see
Q2. I am computing expectation values of the following partially entangled state of the two spin system:
$$\lvert K \rangle=\frac{1}{\sqrt{3}}\lvert ud \rangle-\sqrt{\frac{2}{3}}\lvert du \rangle$$
The spin operators for Alice and Bob are $\sigma$ and $\tau$ respectively
I have computed the following:
$$\langle \sigma_x\tau_x\rangle=\langle \sigma_y\tau_y\rangle=-\sqrt{\frac{8}{9}}$$
$$\langle \sigma_z\tau_z\rangle=-1$$
Experimentally I know how to interpret -1, it means that the spins are perfectly anticorrelated in the z direction (i.e. whenever Alice measures +!, b
 
@Secret Your interpretation for $\langle \sigma_z\tau_z\rangle$ is wrong. These are expectation values, and they do exactly what expectation values do: They tell you what happens on average. Nowhere in the expectation values is it encoded that the spins are perfectly anticorrelated, and nowhere is any information about how often you will get which result.
 
@EmilioPisanty candidates for protection are those which have gotten a couple of deletion-worthy answers (including non-answers) from users who haven't earned 10 rep here, and seem likely to get more
That doesn't seem to be the case here
 
user54412
@Jim The easy home experiment uses something like dry ice to provide the temperature gradient, but this obviously doesn't last long. Instead, use a more mundane compressor to maintain the gradient.
 
user54412
Fill a large glass container with your favorite alcohol, and enjoy.
 
@JohnRennie : Yes.
 
user54412
4:35 PM
(Your favorite alcohol, by the way, is methanol. The others are too heavy and so they fall quickly.)
 
@ChrisWhite I don't think my favourite alcohol is methanol...
 
@ACuriousMind you can get blind drunk on it
 
user54412
@Jim If you get it sensitive enough, you'll actually want a static electric field to clear the tracks, otherwise you won't see the individual ones after a few seconds. Then you can add a magnetic field to induce Larmor motion, and start doing things like measuring charge-to-mass ratios.
 
user54412
@ACuriousMind You're not redneck enough I guess.
 
But you cannot get on average of -1 for $\langle \sigma_z \tau_z\rangle$ if there exists at least one trial which give a a pair of spins that are not opposite to each other in the z direction.

Using the definition of expectation value in terms of probabilities, to get -1 will mean no matter how many replicates of the system are being measured the z component of the spin must always be opposite in the two subsystems?
 
user54412
4:43 PM
@Jim You'll also want a pump to actively bring the alcohol to the top, and good lighting is critical. I've seen this done with a ~ 50x30x20 cm box, and you get several particles per second.
 
Using what you said that -1 is somehting that happens on average, what are the possible types of outcomes that Alice and Bob can get from this entangled state when they measure it assuming they have many replicates of it?

Because I don't know what else can pop up for the z component of spin that can give rise to -1 besides the spins being opposites always?
 
user54412
@DavidZ I see what you're doing, but for the final analysis I suggest reversing the color of the downvotes to make it easier to see at a glance.
 
@Secret Yes, for this specific system, you are correct because $\sigma_z\tau_z$ only takes values $\{-1,1\}$, so in this case you can get probabilities from the expectation values because there are only two possible outcomes. However, computing the expectation values for that instead of directly computing the probabilities for the two outcomes strikes me as rather useless.
 
I see, in that case I should try to calculate some probabilities later to check another part of my understanding on entangled states
 
@EmilioPisanty : I'm usually reluctant to protect a question with a bounty, but if the community think so, I could do it.
Ah, now I see @DavidZ has already answered you.
 
4:55 PM
Sorry
My alcohols are either ethanol, menthol or xylitol
Glycerol is fine I guess
 
@ACuriousMind well, good point. Anyways if you did I will possibly get something out of it, but eh. . . I will be a bit more persuasive if I were younger(I'm 26 and have been around) so eh. . . well if you do it who knows you might learn something about yourself, since it is statistically improbable, that you will learn some new physics skill . . . even though one never knows. Thanks for responding though,. . . if you change your mind you can message me.
 
@Secret @JohnRennie I am in dire need of a chemist
 
what's the question?
 
@Secret How explosive is a 2:1 mixture of methanol and nitric acid
At room temperature
 
@DavidZ @Qmechanic Yeah, I was leaning that way but on the fence. Several pretty low quality answers but also few that'd be excluded by protection.
 
5:08 PM
Maybe 3:1
 
Jim
@ChrisWhite cool. Will have to make that a suggestion to the physics club
 
@0celo7
https://en.wikipedia.org/wiki/Methyl_nitrate
Perhaps less explosive than in wikipedia since methanol and nitric acid react 1:1 thus you will have 2 moles of methanol left behind that might dampen the detonation velocity of 6300 a bit (gas expansion caused by the ignition of the excess methanol not considered yet)
 
It shouldn't explode upon mixing o.O
I need to make a beaker of it lol
 
Well I am estimating theoreticaly how explosive it will be. Methyl nitrate is rather shock sensitive though, thus hadle your mix with care
 
I don't think it should be methyl nitrate...
It's called Nital
And I need to run a current through it too
And drop various things in it
And not die
 
Jim
[citation needed]
 
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