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vzn
12:31 AM
@ChrisWhite dont be fear the bohmian mechanics! anyway you must have ordered something related eh? ("accidentally"? wink wink) anyway, how does one determine/ identify "bohmians with an axe to grind"? and presumably without actually readiing the book? :P
quite to the contrary it seems at times its all the "shut up and calculate crowd" with the so-called "axe to grind"... :|
was a bit shocked to find this at 6v, missed it back when it was posted! you antibohmian extremists are gonna have to work a little harder! "the nail that sticks up gets hammered down!" as the japanese say... o_O
6
Q: In the oil droplet experiments that suggest de Broglie’s pilot wave theory might be accurate, what does the fluid surface correspond to?

user90664As a particle travels to a screen, it is traveling through 3-dimensional space. In the oil droplet experiment, there are only two dimensions of any importance—the droplet merely moves along the surface of a wavey fluid. It seems like a 3-dimensional wavey superfluid would result in far more compl...

ps was surfing movie previews/ trailers yesterday & noticed that yudkowski is interviewed in this one... the singularity is near... it was definitely one of the more grandiose/ psychedelic/ scattered previews on level with a religious movie... maybe will watch it sometime for cheap kicks... o_O
 
12:48 AM
@Slereah I don't need a tux either, but I wear one when there is a need to project really ridiculous levels of swank.
Or, in all seriousness, you use unreasonable precision in a computer code because it is easy and it's one less thing to have to worry about.
On top of which lathes reliably cut things round to better than a part in a thousand, even if keeping them that way through subsequent handling is tough.
I know a machinist who cut precision inside threads on a 1 mm diameter source capsule for a colleague of mine. He always seems slightly disappointed with the projects I brought him.
Until I wanted to weld two sub-millimeter thick pieces of 316L steal.
After some research he asked if the project budget would extend to buying an electron beam welder, and after I said "No" he recommended an outfit in southern California to outsource the job.
 
1:21 AM
Woohoo! Just got gravitational raytracing working
(Schwarzschild black hole at radius 1, but I can have time dependent metrics or whatever I want in it)
 
@NeuroFuzzy Snazzy. Have a pat on the back and a cup of your favorite beverage, but don't expect a raise.
 
1:40 AM
@dmckee Ah... so still -$8/hr.
 
 
3 hours later…
4:50 AM
https://www.facebook.com/BusinessInsider.Video/videos/719673064796775/

The way these puzzles works with so few moves relies on some kind of symmetry in these figures

Is there a name to these kinds of symmetry?
As far I can observe, it seemed that the initial figure and the solved figure (which is not present until the figure becomes the solved figure) are related by a combination of rotation and reflection
 
5:08 AM
Yet another black hole render! Woo!
close-up of the above image
 
so this is not rotating black hole?
 
5:34 AM
@Secret Nope
 
Slightly more than the usual number of reallt crappy question lately, but much, much worse they are getting more than the usual ratio of really crappy answers.
Yikes!
 
6:07 AM
Grrr this is annoying
cause right now a genre savvy enough student can ask, "Why not $\frac{1}{3}?$, why not <insert any number other than half>"?
Trying to rule out genre savvy cases like these is the main reason I spent on average 3x amount of time to do a physics or maths problem
 
 
3 hours later…
9:25 AM
@TanMath I redid all the computation a few times, and realized that I had been wrong since the beginning :c since the length at start is of course the diameter 2R and nor the radius R. Now I find a somewhat coherent result.
 
 
2 hours later…
11:21 AM
hey hey
 
11:46 AM
I hate papers that are on two columns except equations
 
 
1 hour later…
1:14 PM
@Slereah I've heard that one before :D.
 
1:29 PM
@JohnDuffield cc @theNamesCross well, I would mind marking with "Edit:" a little :-P The preferred way to add content into a post (or remove from it) is to update the post so that it looks like you'd written it that way all along - there's no need to indicate in the post itself which parts are new and which parts existed originally. That's why the revision history exists, so that people who care about when parts of the post were introduced can see it.
2
But otherwise, I endorse that statement.
 
1:41 PM
@I'mmostlyjustanidiot It's a really awkward format
I guess it has its use for old timey papers when they had to be in print, to save space
But it's pretty ugly
 
 
1 hour later…
3:03 PM
"The Wightman framework also does not cover gauge theories. "
Wightman is pretty shit
 
3:33 PM
Weinberg actually mentions how we get rid of boundary conditions in the path integral
Good old Weinberg
You earned that nobel prize, my friend
 
3:55 PM
How does one show that the Lie algebra of GL(V) is End(V)?
 
4:06 PM
what is End(V)
Oh, endomorphisms
 
@Slereah Yes.
The exponential map is not surjective, can't use that.
 
5:02 PM
hey @dmckee
 
@0celo7 Use the definition in terms of left-invariant vector fields, I think.
 
@Danu Of the Lie algebra?
 
Mhm
But I'm not sure, honestly.
Yeah, that's how Leeb does it in my notes.
If you wanna be more quick and dirty:
 
I don't see it.
 
Note that $GL(n)$ can be identified with an open subset of $\mathbb{R}^{n\times n}$ and hence its tangent space (at the identity) is just $\mathbb{R}^{n\times n}$.
which is exactly $\operatorname{Mat}(n\times n)\cong \operatorname{End}(n)$
(this way you don't actually prove that the Lie bracket is the same one as the natural one on $\operatorname{End}(n)$)
 
5:15 PM
That's how Jost does it. I'm not sure why GL(n) is an open subset, I guess.
 
but it makes pretty clear how it works
@0celo7 It's the complement of a zero of some continuous function (a polynomial, even)...
Can you see which function?
 
Aha!
The determinant.
 
woop-die-doo
 
>die
 
dee?
 
5:17 PM
You've clearly been spending too much time in Germany.
 
Meh :P
I'm Dutch, remember?
 
No, I didn't know that.
 
wat
Why are you always trying to make fun of me being high and stuff, then?
 
That's behind me.
 
5:33 PM
@Danu How does one see that su(n) matrices are traceless without referring to the exponential map in some form?
 
$[T_a, T_b] = f_{abc} T_c \rightarrow Tr(T_a T_b) - Tr(T_b T_a) = f_{abc} Tr(T_c) = 0$
Course that doesn't help for $SU(2)$ since it's the only SU group with $f = 0$
 
That just says a certain linear combo of traces vanishes.
 
Well if $f = 0$ then all basis components are traceless
 
5:49 PM
Seriously? U(1) has f=0 but the basis is certainly not traceless.
 
U(1) isn't SU(1) :p
Also for U(1), $[T_1,T_1] = 0$
 
@0celo7 Take the derivative of a certain defining condition...
 
So this doesn't apply
 
@Danu Which one? Can I write M=I+tX and then take the derivative of det M=1 using the formula det(I+X)=something +Tr(X)
 
Not quite sure how to prove it for SU(2) outside of like
Exhaustively
Just checking the Pauli matrices
 
5:53 PM
You never used any property of SU(n) in your "proof."
 
The commutator totally is a property of SU(N)
 
So it's general. And in general g is not traceless.
 
I don't understand why you don't look these things up in Lee, @0celo7
 
All Lie groups have that property.
@Danu Can't find it in there.
@Slereah The bracket of left invariant vectors is left invariant. The commutation relations follow from this.
This holds for all Lie groups/algebras.
 
Example 8.47, and use a trivial modification.
You need to learn how to search better.
 
5:55 PM
@0celo7 Well then I guess it is true for all Lie groups with $f \neq 0$ :p
 
@Slereah Your methods of proof are terrible :P
 
I'm a physicist dammit
I'm no dirty mathematician
 
@Danu Whoops.
Was about 30 pages off.
@Slereah Also wrong.
All you're telling me is that some linear combination of traces vanish.
 
Am I
 
This doesn't mean that every trace has to vanish.
 
5:59 PM
I'm telling you $Tr(T_a) = 0$
 
No you wrote f^abc Tr(T_c)=0
Sum over c implied.
 
Oh right
nvm then
 
@Danu Lee doesn't do the special case, which is what I need.
And tracelessness for so(n) is trivial.
So Lee doesn't help.
 
6:37 PM
I told you, a trivial modification ;) O(n) and U(n) are almost exactly the same thing
@0celo7 Also this statement doesn't make much sense to me.
 
Which statement? I'm on mobile.
 
About requiring det=1
 
@Danu I need the su(n) tracelessness condition.
 
I hope you see it's the same thing around the identity
 
The fact that they are also skew Hermitian is trivial.
@Danu I don't see what you mean by that.
 
6:40 PM
Easier way to see tracelessness: It's a subalgebra of...
Anyways, I try not to make it a habit to tell you how to do your exercises.
 
Subalgebra of what?
 
As has been mentioned before, maybe you should create a separate room for that stuff.
 
Is det=1 not the definition of the S in SU?
I don't see why you don't understand why I'm considering it.
 
7:00 PM
@Danu If you were going to say "subalgebra of sl(n)", well, I'm trying to figure that one out, too.
 
7:26 PM
Is this a thing: $\sum_{n=0}^N a^n (N!)/(n! (N-n)!)$?
 
That's the binomial formula, no
?
Oh, not quite
 
@Slereah Yeah. Not quite.
 
Well, it is for $(a+1)^n$, I suppose
 
@Slereah ?
 
The formula is $(a+b)^n = \sum a^{k} b^{n-k} \binom {k} {n}$
So it works for $b = 1$
 
7:31 PM
Oh. Duh. Thanks.
You mean $(a + 1)^N$, of course.
 
Replace variables to fit the situation, yes
 
@Slereah I just meant that the $n$ is either the sum index, or not :-)
Thanks for the help!
 
I may not know all the math proofs but I do know the tricks :p
I recall my math year
Couldn't follow most of the classes
But when the professor did some spherical integrals, I was the only one that could :p
 
Nice.
 
TOO PRACTICAL FOR YOU, MATH PEOPLE?
We wouldn't want any math that can solve something!
 
7:34 PM
@Slereah heh
 
Anybody familiar with Bell's inequality?
 
@FrankScience Sure.
Very important.
 
I posted a question here, however I didn't describe everything good and I haven't received any satisfied answer.
 
I don't think I understand the question.
A quantum state has associated with it a probability distribution.
That distribution itself evolves by deterministic laws.
Is that the heart of your question?
 
Then?
 
7:48 PM
@FrankScience Then what?
 
There's nothing evolving. For other things, approximately yes.
 
@FrankScience I have no idea what you mean by that.
You seem to be suggesting in that post that the choice of measurement basis is itself a hidden variable. Is that right?
 
@FrankScience Well, that information isn't hidden. It's a part of the parametrization of the problem.
 
Sorry, I didn't understand what you mean by the choice of measurement basis.
You mean, experimental apparatus?
 
7:54 PM
Quoting from your post:
" there are specific choice of a,b,a′,b′a,b,a′,b′ such that correlations between them don't satisfy Bell's inequality, hence the measurements cannot be random variables of a probability space. "
 
These choices of $a,b,a',b'$ are given.
 
Unfortunately, you don't say what a, b, etc. mean in the post. Those symbols are not defined.
 
It's given in the wiki page.
 
@FrankScience Definitely a good idea to put all relevant information into the post itself.
 
And they are four constants valued in $S^2$.
 
7:57 PM
Especially notation/variable definitions.
Please note: if a question post is not self-contained, lots of would-be answerers will simply lose interest and move on.
You can't seriously expect people to read your question, read the Wikipedia article, put all the pieces together, figure out what you're asking, and write an answer.
It's the job of the person writing the question to make it easy for everyone else to understand that question and provide an answer.
This is probably a huge part of why you haven't gotten an answer.
I mean, a satisfactory answer.
 
Yes, I know that, so I realize that I need to ask for a discussion on formulating. I have difficulty organizing such a post since QM is what I am quite unfamiliar with.
 
@FrankScience Ok.
I see, do you have education in quantum mechanics?
You seemed puzzled by the term "measurement basis". Do you know what that means?
 
user54412
(I feel like this conversation is about to go on for a while, and at the end, no one will walk away satisfied...)
 
@ChrisWhite That likely depends on @FrankScience's willingness to pick up an intro quantum book and study the first hundred pages.
 
No. I started to read Tannoudji & Diu & Laloë this semester.
 
8:03 PM
I will be satisfied if I can convince him/her to do that.
 
Up until things like fundamental postulates of quantum mechanics.
 
@FrankScience Ok, in that case I definitely think you should focus on learning some basic quantum before tackling your Bell question.
@FrankScience Measurement is the most poorly explained (and understood) part of quantum mechanics, and Bell violations are a complexity on top of that.
 
user54412
I like to be more optimistic. I think a purely probability-based description can be given, and might even be better, because it avoids all the nonsense people get into over "what is a measurement?"
2
 
@ChrisWhite 100% agree.
 
On the book I'm reading, it's just related to an observable.
namely, a Hermitien operator on the Hilbert space.
 
user54412
8:06 PM
Then again, I've just spent some time going through the wikipedia pages and references, and I realized I can't point to any source that isn't itself confused on the topic.
 
Doesn't change the fact that measurement is poorly taught because most folks have a cotton ball in the part of their brain devoted to it.
4
@ChrisWhite Correct.
@FrankScience Yeah, that's true, but you're asking about whether or not the choice of observable is a hidden variable, right?
Why should it be?
 
No, I'm not referring to the choice of observable.
 
@FrankScience Then what are a, b, etc?
 
user54412
I thought the nuance of the question was this: Why does it seem that the CHSH inequality should hold for all random variables, but it can be violated by QM, which is a theory of random variables?
 
@ChrisWhite Oy. I don't know if I have the mental fortitude to explain this right now. Perhaps you were right about nobody coming away satisfied.
 
8:08 PM
@ChrisWhite Thanks for the clarification. Yes, roughly that.
 
@FrankScience The inequality does not hold for all random variables.
 
user54412
Of course, since Bell/CHSH all reference local hidden variable theories, any statement of them that doesn't distinguish locality seems incomplete, which is what I think Henning Makholm is saying.
 
@ChrisWhite Don't know who Henning Makholm is, but that is a good point.
 
user54412
^ one of the answerers on math.SE
 
^ Ah
Hey, @ChrisWhite, is this a thing: $[N!/(N-m)!] \gamma^m$?
$\gamma \ll 1$.
$N \gg 1$.
Seems like something an astroperson would know.
 
8:12 PM
hey alL!
 
@TanMath Hello.
 
user54412
@DanielSank I don't think I've ever seen a factorial in astro :p
 
@ChrisWhite Awww. I though y'all were statistics superstars.
 
user54412
::checks first ever publication:: -- hmm, seems I did include some pedantic combinatorics for statistical analysis
 
> . <
^ emoticon fail.
HTML, Y U DO ME LIKE THAT?
 
user54412
8:21 PM
It almost looks like the binomial distribution for getting $m$ out of $N$ with probability $\gamma$, except we ignored $(1-\gamma)^{N-m}$ (because $\gamma \ll 1$?), and there's an extra $m!$ (because reasons?)
 
I've edited the post to include $a,b,a',b'$. I hope it's clearer.
 
user54412
Then again, what doesn't almost look like the binomial distribution?
 
@ChrisWhite Real data?
 
user54412
@FrankScience I find the Markov chain answer funny. I'm pretty sure anyone who studies Markov chains answers everything in terms of them.
 
user54412
@dmckee lol
 
8:27 PM
@ChrisWhite What you comment on Henning Makholm's answer is sound to me, but one of my goal of posting this question is to clarify these concepts. I'm still unclear after reading Le Bellac and wiki pages.
 
@NeuroFuzzy cool! Looks like Saturn was going to be swallowed or something!
 
CHSH inequality does apply to any random variables.
 
@TanMath Yep! Well Saturn is far enough away to be treated as a celestial sphere, so you can imagine this as a pea sized black hole, and you're a tiny camera about eight pea radii from it :D
 
OK, I see.
It should not have been measurement, but physical quantity.
or measurement results.
 
 
1 hour later…
vzn
9:48 PM
@FrankScience hi, how much time do you want to dedicate to this? have studied bells thm in some depth. keep in mind its over ~½ century old now & massive amts have been written on it and there is some new thinking. there are many different ways to approach it. plz drop by here for more discussion. (hint: maybe "nobody can be satisfied" right now because the state of existing knowledge is not fully satisfactory.)
bells thm & related ideology seems to have a quality of partial knowledge as hinted in the blind men/ elephant story... not unlike QM itself...
@ChrisWhite can you elaborate on this pov? eg is it contained somewhere in the literature that can be pointed to?
 
Verrrry nice :D
I got another! :D
 
10:06 PM
@Danu Awesome!!
I'd love to see longer-term data on that. It might just be my eyes, but there seems to be a slight downward trend over time. Or perhaps a periodic chance in the trend.
 
@vzn The point is that I'm focusing on mathematical formulation and argument, not on whether it's physically accessible, so maybe nothing is not fully satisfactory.
 
10:20 PM
@HDE226868 Whoa today is my lucky day
3rd eclipsing binary now
and had several very clear transiting planet patterns too
 
@Danu Wow, that's something. I'll see if I can get anything cool in my curves.
 
@ChrisWhite Contraction of Levi-Civita symbols?
 
I am officially a member for one year!
Ooh.. A planet hunters discussion!
 

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