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12:01 AM
@DanielSank If your interested, I'll try to do a small write-up later when I have time, now I should really leave this place...
 
@G.Bergeron ok, I'd be interested to read.
You can write it as a self-answered post. Those are usually quite useful.
 
Nice that you hang out here nowadays @GBergeron---you seem to have interesting stuff to say :)
 
@Danu <3
 
It's been a long time since there was a steady stream of interesting physics talk in the h bar :P
 
@Danu oh?
 
12:10 AM
Okay, let me rephrase that [done]
 
@Danu Well, you betrayed us and turned into a mathematician! :P
 
@ACuriousMind It's not like I generated a lot of interesting physics talk.
 
I actually calculated a thing today.
Pretty happy about that.
It'll be in my theory repo in a bit.
 
Actually, ACM, I had a question. In Huybrechts' complex geometry book he says that the $N=(2,2)$ SUSY algebra is given, in mathematics language, in terms of the following operators
 
12:14 AM
@Kaumudi.H!
Your name and appearance have changed a bit...
 
@heather: :-P What confusion has this change brought?
 
You have become a flying toaster-cat.
 
A flying NAND-gate cat :-D
 
hahahaha
How is your quest to reduce the world suck?
 
$\partial,\bar\partial$ and their adjoints, $L,\Lambda,H$ where $L$ is the Lefschetz operator (wedging with the Kaehler form) and $\Lambda$ is its formal adjoint; they give rise to an $\mathfrak{sl}(2,\Bbb C)$-representation on the exterior algebra, which is completed by the "counting operator" $H$ (the Cartan element, proportional to the identity).
What the heck are $L,\Lambda$ in physics? Earlier on, he says that $\partial,\bar\partial$ are the $G$-fields (fermionic analog of the EM tensor, IIRC)
 
12:17 AM
Uh, what?
 
Which part?
I have some idea---perhaps something like this $J$ current that you get in $N=2$? idk
 
The whole thing, I don't understand how that is supposed to be "the SUSY algebra" at all
 
You know the SUSY algebra in physics notation, right
like $\{G,\bar G\}=0$ and shit
 
Sure, bunch of $Q$s ;)
 
dafuq is SUSY algebra?
 
12:18 AM
and $[T,G]\propto G$
 
::Googles::
 
Supersymmetry algebra
@ACuriousMind So I only know these infinite dimensional ones (where the infinitely many tgenerators are e.g. modes of $T$)
But Huybrechts says one should not be thinking about those
This one is the one I know
That's the superconformal one
 
Yeah, that's a very specific superalgebra and not at all what one means by a usual susy algebra
 
Okay
So I've briefly read, a while back, about those SUSY algebras featuring $Q$ and $H$
I'm pretty sure that $G$ should be $Q$
Since it's a 1-st order differential operator (it will switch commutativity rules for forms, because of the $\Bbb Z_2$-grading)
 
The usual SUSY algebra consists of the Poincaré group in the bosonic piece and $N$ spinors $Q^i$ with anticommutation relation $\{ Q^i_\alpha,\bar{Q}^j_\beta\} = \gamma_{\alpha\beta}^\mu P_\mu \delta^{ij}$.
 
12:24 AM
The $L$ wedges with a 2-form so it should be bosonic
@ACuriousMind Okay, that's easy enough.
I remember that :P
Now, if I have $N=(2,2)$ there should probably be another field/thing
 
It also has central charges $Z^{ij}$ with $\{Q^i_\alpha, Q^j_\beta\} = \epsilon_{\alpha\beta} Z^{ij}$.
 
Probably some kind of analog of a supercurrent in that $N=2$ SCFT algebra I linked earlier
something bosonic (but not commuting with everything like the $Z$'s)
 
But I have no idea how this algebra is supposed to match the operators $\partial,\bar\partial, L,\Lambda$ you listed
Neither of them appears to be a spinor
 
$\partial,\bar\partial$ are, of course
like I said, they raise your degree by one, so you'll get different commutation rule after applying them
Spinors are supposed to be odd-degree forms, right?
 
Hmmm, in the susy algebra, "spinor" quite literally means Dirac spinor.
 
12:28 AM
Sure, but right now I'm definitely not talking about Spin-structures 'n' stuff so that can't be it for now
 
Which is what confuses me
 
So actually
It really uses Kaehler structure
because else this whole $L$ thing makes no sense
Does this complicate matters on the physics side?
 
It uses Kähler structure for what?
 
For this $L$ and $\Lambda$
 
wait a moment
 
12:30 AM
The entire algebra of (super)commutators of all these operators is based on the Kaehler identities
Oh, by the way
Nope, never mind.
 
Okay, so I think he's using SUSY not for the super-Poincaré algebra I thought of, but more generally for any algebra that has "bosonic" (mapping even forms to even forms) and "fermionic" operators
 
Yes... BUt I was hoping this algebra is at least the same algebra as one of the ones in physics.
 
Witten does something similar in "Supersymmetry and Morse theory", the "supercharges" are $\mathrm{d} + \mathrm{d}^\dagger$ and $\mathrm{i}(\mathrm{d}-\mathrm{d}^\dagger)$
 
Yeah
So those are $\bar\partial$ and $\partial$ of course
So indeed those should be the supercharges.
Which was really the only option since they're the only first order differential operators
 
@Danu Aren't those related to the creation/annihilation operators?
 
12:34 AM
Danu: Comes here and complains about lack of physics talk. Immediately starts math talk.
 
@ACuriousMind Yes, that's what it seems
 
@DanielSank I didn't complain, and I'm trying to connect to physics
Of course, that doesn't take away that I'd be glad to see a lot of pure math talk here, too ;D
@G.Bergeron Sure, they're raising/lowering the spin IIRC.
So... no takers for an interpretation of $L$? Wedge product with the Kaehler form, anyone?
@QMechanic, perhaps?
 
Does it have something to do with the Laplacian?
The susy algebra made out of $\partial,\bar\partial$ should have the Laplacian as its "Hamiltonian", i.e. $P_0$.
 
@ACuriousMind I'm trying to interpret the adjoints... But yeah, that should probably be true. So let's see how that'd fit in.
So $\Delta=P_0=\{\partial,\partial^*\}$
Oh, yeah that makes a lot of sense. I guess I must've been off saying that $\bar\partial$ should be $\bar Q$, before? Is $\bar Q$ the adjoint of $Q$?
Yeah, it's something like $Q^\dagger \gamma^0$ right? Sounds like adjoint to me :D
 
12:52 AM
In mathematical physics, the 2D N = 2 superconformal algebra is an infinite-dimensional Lie superalgebra, related to supersymmetry, that occurs in string theory and conformal field theory. It has important applications in mirror symmetry. It was introduced by M. Ademollo, L. Brink, and A. D'Adda et al. (1976) as a gauge algebra of the U(1) fermionic string. == Definition == There are two slightly different ways to describe the N = 2 superconformal algebra, called the N = 2 Ramond algebra and the N = 2 Neveu–Schwarz algebra, which are isomorphic (see below) but differ in the choice of standard basis...
 
No, it's not the superconformal one. We went through this already ;)
33 mins ago, by Danu
But Huybrechts says one should not be thinking about those
 
-_-
 
It's a finitely generated one
 
The SCFT algebras always have infinite generators
@ACuriousMind Sounds like exactly what we need
 
12:54 AM
section 2.2
 
Is it possible he's defining it in the envelopping algebra?
 
Mew
quick up the greenhouse gas emissions!
 
Hmm, so $L$, $\Lambda$ are... what, in physics terminology? The $T^i$'s from that paper
 
It's the super-$\mathfrak{so}$ algebra in 3D - the $L,H,\Lambda$ are $\mathrm{SL}(2,\mathbb{C})$ as you say, and the $\partial,\bar\partial$ are the supercharges.
 
12:57 AM
Boosts?
 
@Danu generators of rotation, I think.
 
Right, same difference
We should be doing stuff in + signature anyways
 
Yeah, this is the non-relativistic/Euclidean SUSY algebra in 3 dimensions.
 
Huybrechts clearly didn't know/care what a special case he was using :P
 
I'm also confused why he says $N = (2,2)$.
 
12:59 AM
Yeah...
 
There is no $N = (n_1,n_2)$ in dimensions that don't have Weyl spinors, only a single $N$, and I also think this is just the $N=1$ algebra
The spinor is just $(\partial,\bar\partial)$
 
Oh well
A valiant attempt by Huybrechts. 6/10
 
@Danu In the article it says that the zero mode operators form a 5D Lie superlagebra
 
1:27 AM
Ah no forget it, there's the adjoints...
 
@Danu hmmmm, questionable.
 
Hm
If I have a metric of the form $(dt + dx) (dt - k(x) dx)$
is $u = t + x, v = t - v(x) dx$ an adequate coordinate transform
or something similar
 
What does "adequate" mean?
@DanielSank Huh? He rather explicitly asked what the mathematical operators correspond to in the standard way the SUSY algebra appears in physics.
 
that it will give me a simple metric form to solve equations
 
@Slereah I think $v = t - K(x)$ where $K'(x) = k(x)$ gives you just $\mathrm{d}u\mathrm{d}v$ as the metric, is that want you want?
 
1:43 AM
that would be nice, yes
Let's try it out
Well I'm guessing it won't be quite $dvdu$ since it's not Minkowski space
But something of that form would be nice
 
Well, it's immediate: The differential is linear, so $\mathrm{d}u = \mathrm{d}t + \mathrm{d}x$ and since $\mathrm{d}f = f'\mathrm{d}x$ for $f$ a function of x, you have $\mathrm{d}v = \mathrm{d}t - K' \mathrm{d}x$, right?
hmmm
 
@ACuriousMind If it's strings, it ain't physics seems to be a common attitude ;)
 
@Danu supersymmetry is totally not exclusive to string theory
 
Mew
anyone know about climate science here?
 
That's more the realm of Earth Science
 
Mew
1:57 AM
@ACuriousMind are sun spots the realm of earth science or physics?
 
Probably depends on whether your question is about the sun spots themselves or about how they influence climate on earth. The former is probably physics/astronomy, the latter earth science, although climate science/meterology is of course somewhat interdisciplinary.
 
Mew
@ACuriousMind I'm interested in the sunspots themselves
Is there a relationship between the number of sun spots and the heat output of the sun?
Sun spots are cooler areas of the sun
but is this associated with a higher radiation output overall from the sun?
is it even known if a such a relationship exists or is it just speculation?
 
2:11 AM
FFS can't I just tease you guys in peace?
 
@DanielSank No, this means war. Step up your game.
 
@ACuriousMind but I am le tired...
 
@ACuriousMind Seems odd that it would look exactly like flat space
Lemme get my notebook
$u = t + x, v = t - K(x) x \to u + v = 2t + x (1 - K(x)), u - v = x (1+K(x))$
Hm
 
hello
 
Inverting that transform isn't trivial
 
2:27 AM
@heather Yo.
 
Although
I guess I don't need to do that transform
I can just see if $dudv$ transforms to said metric with the transform I have
 
@BernardMeurer, I think I have to reinstall it, because I had to leave/shut my computer in the middle of the installation =/
hmm...there's a problem somewhere in my code, but I don't have an error to guide me...
@DanielSank, been working on any cool projects lately?
 
@ACuriousMind Now that... :P
 
Let's see... $g_{tt} = g_{uv} \partial_t u \partial_t v + g_{vu} \partial_t u \partial_t v = 1$
So far so good
 
@heather Yeah, I think I mentioned this circuit for demodulating RF signals?
 
2:41 AM
$g_{xt} = g_{uv} \partial_x u \partial_t v + g_{vu} \partial_x v \partial_t u = \frac{1- k(x)}{2} $
 
Today I calculated something I needed to know for the design.
Calculation is iq_demodulated_noise.tex here.
Probably won't make much sense unless you've worked with signal processing stuff.
::shrugs::
 
$g_{xx} = g_{uv} \partial_x u \partial_x v + g_{vu} \partial_x v \partial_x u = k(x)$ or something
Hm
That is odd
It's supposed to be the Krasnikov metric
Not really like Minkowski space
Although maybe it's flat in 2D
Let's check
fuck do I still have Mathematica installed
 
@DanielSank, no, doesn't really make much sense =)
but i think you mentioned that circuit. it sounds really cool!
 
Then again
There's nothing preventing me from doing the exact same metric transform in 4D
Argh
Hm
Thinking about it
I'm guessing that a static Krasnikov tube is identical to Minkowski space
There's no "shortening of distances", it's just what the distance always was
Fuck
I don't want to have to compute a dynamic one
 
2:57 AM
So what happens if a bounty closes with no accepted answer or highly voted one? It really just disappears?
 
Yep
All goes away
Like dust in the wind
 
On the other hand
What prevents me from doing the same coordinate transform if $k$ is a function of $t$ as well
Let's find out
 
3:33 AM
@ACuriousMind You are many?
 
 
2 hours later…
5:13 AM
Really I think that anytime you try to make a metric, the first step should be checking the Riemann tensor just in case you made Minkowski space by accident
 
5:52 AM
@DanielSank It's going okay. It could go better, but OK is OK, for now :-P
Lots of other people doing a lot of world-suck reducing over here, BTW:
These people are the heroes of now:
Oh, hey @JohnRennie: Morning! :-)
 
@Kaumudi.H Morning :-)
@Kaumudi.H while I'm not a special Vlogbrothers fan I do think making the world suck less is a good thing, and I think that's pretty much what we do here. I think the physics stack exchange does a generally excellent job of making the world suck less :-)
 
I couldn't agree more :-D When I said "other people", I meant " people other than me". I'm just sitting here, studying.
 
Every time we smile we make the world suck less. You don't have to be curing cancer or achieving world peace to make a difference :-)
 
6:07 AM
I know, I know :-) Still.
 
user246160
@JohnRennie Good morning. I need some help in Newtonian mechanics. Work done by forces is said to be frame-dependent. So if I am sitting in an accelerating frame of reference (suppose any vehicle) , all the the forces acting on the vehicle seem to cause 0 displacement. So does that imply in the accelerating frame work done by all external forces on the vehicle is 0 ?
 
Hey, @JohnRennie: Patrick Rothfuss just joined the livestream!
 
Wow! Have you read The Name of the Wind?
@TheStackExchange Work is a frame dependent quantity because kinetic energy is a frame dependent quantity. There are quite a few questions about this on the site already. I could probably dig out a couple if they would be of any interest.
 
@JohnRennie No. Unfortunately, I have not read any of his books. Definitely going to, in the summer! It's gonna be one hell of a summer :-P
 
user246160
@JohnRennie I am actually looking through a few questions on the site. But I wanted to ensure whether the work done by all external forces is actually 0 if I am sitting in an accelerating frame...
 
6:12 AM
@TheStackExchange With accelerating frames you have to decide what you want to include. For example right now I'm stationary in an accelerating frame (at 1G) and I'm pretty certain that no work is being done on me.
Or if some mad scientist puts me in a centrifuge then again I'd be in an accelerating frame but no work would be done on me.
 
user246160
@JohnRennie Okay. That seems like a nice analogy :). But suppose in a physics problem we are told to calculate work done on an accelerating body by all the external forces acting on the body, then we calculate it from the frame of the body itself or from the frame of ground ? (if frame of reference is not mentioned)
 
@TheStackExchange The question will almost certainly mean the work done as measured by the observer at rest in the lab.
 
user246160
@JohnRennie Oh now that makes it clear. BTW I wonder which "mad scientist" would do that :P Hahahaha XD
 
user246160
The Centrifuge Brain Project is a 2011 German short mockumentary fantasy film written and directed by Till Nowak. The film incorporates computer-generated imagery to create seven real-seeming fictional amusement park rides used in a faux documentary film about the construction of physics-defying rides intended for use in research efforts to improve human cognitive function. Nowak was inspired to create the project when visiting an amusement park in 2008. == Background == === Art installation === Creating the sequences for the seven rides took three months, spread out through 2008 and 2011. After...
 
user246160
Okay, someone made a movie on that concept ^
 
user246160
6:17 AM
:P
 
@JohnRennie: Are u watching it for Patrick?
 
@Kaumudi.H No. I'm busy writinng an answer about time dilation in galaxies :-)
 
OK :-) Have fun!
 
Actually I'm not that keen on his books. The trouble is that his main character, Kvothe, is a bit of an idiot. He always gets riled up and does the wrong thing. I found myself shouting at the book don't do that!
 
[Games] Caption: Perspective games do result in extremely alien looking motion when you look at it from an isometric perspective.

For example, beating this level requires first going to bird's eye view so that the white cube can move down the small alcove at the end of the slab. Once in side the slab, because there's an unobstructed viewpoint from the exit green cube to the location of the white cube, after looking at that perspective the cube is teleported to the exit and level beaten
 
6:31 AM
@JohnRennie Haha :-D
I have a quick question about submarines. Apparently, some of them have been known to travel down to around 900 feet.
 
Caption: And for this one, orientation matters whether the white cube can move pass the red patches
 
user246160
@Kaumudi.H "QUICK QUESTION" XD
 
I've googled about how submarines are able to counter this huge pressure but all I've gotten is s'thing about ballast tanks.
@TheStackExchange Oh, shut up :-P
 
user246160
@Kaumudi.H Go on buddy. Looks interesting :-D
 
user246160
6:34 AM
:P
 
In contrast, there aren't many things in physics depends on absolute orientation, but there are a lot that depends on relative orientations. for example, in collision theory of chemical reactions rates, molecules has to be aligned in a certain way relative to each other in order to react. Another example is an unentangled spin, which it's probability of being projected somewhere depends on how it is oriented relative to the measuring device.
 
Does anybody know anything about this? I haven't gotten very far :-/
 
Mew
@Kaumudi.H some can go 1600ft
 
user246160
@Kaumudi.H The shape of submarines largely helps to overcome the huge pressure...even large aquatic animals can go that deep only because of their shape
 
user246160
Shape > Material (importance) with which it is built
 
Mew
6:38 AM
@Kaumudi.H strong hulls and cylindrical shapes
 
user246160
If you blow up a balloon and hold it underwater in your sink or the bathtub. You'll see that the balloon won't change shape, because it's round. But if you were to put something hollow that's shaped with flat sides (like a box made of tin foil) underwater, it squishes in on itself pretty easily.
 
> "The hull of a standard ship is the metal outside that keeps the water out. Most submarines have two hulls, one inside the other, to help them survive. The outer hull is waterproof, while the inner one (called the pressure hull) is much stronger and resistant to immense water pressure. The strongest submarines have hulls made from tough steel or titanium."
 
Mew
very good
 
user246160
@Kaumudi.H Obviously the hull is strong ! But that is not the most important factor
 
Mew
of course it is
the most important factor
 
6:40 AM
@TheStackExchange It also boils down to the material and what's inside as well. If ur balloon was filled with some type of liquid, it would indeed get squished.
 
user246160
@Kaumudi.H "If ur balloon was filled with some type of liquid, it would indeed get squished." Naaaaaaa! I would argue that even a normal steel hulled submarine could go very deep if designed well
 
user246160
If your submarines was a cuboid
 
user246160
Then it would'nt go down even half the depth
 
Of course the shape is also key!
 
Mew
Why waste any time on what's MOST important
both are importnat
 
6:43 AM
But I don't think it's very productive to argue about which factor is more important.
 
Mew
@Kaumudi.H i just said that
 
@Mew :-P Exactly.
 
Mew
lol
 
user246160
I was trying to emphasize the shape is as important as the material :).Done.
 
@Mew Yeah, I didn't see that :-P
Wokay, thanks :-)
 
user246160
6:46 AM
@Kaumudi.H You might find this interesting scientificamerican.com/article/how-do-deep-diving-sea-cr
 
@JohnRennie Am I still saying something stupid on that: physics.stackexchange.com/questions/297333/…
 
What's going on in here?
 
meh
 
@G.Bergeron It wasn't my downvote, but I think your answer is a bit wrong headed because I think we'd need to take care to define exact what we mean by a barycentre in GR. As David says it's not where the spacetime curvature is zero.
 
@JohnRennie No I know, I mean the new version that has nothing to do with the old
 
6:50 AM
I consdiered answering the question myself then decided I'd have to think long and hard about exactly what I meant. In the end it seemed too much effort!
 
@DanielSank Dyou never get notified about my pings?
 
That was the stupid thing
 
@Kaumudi.H ? Of course I get pings.
 
this is confusing
what causes a torque on a gyroscope?
i mean i know that gravity does that
 
... a torque?
 
6:54 AM
but how does it do that?
ok a spinning gyroscope
 
Well gravity doesn't necessarily do that
 
the what does it?
 
Only if you arrange the gyroscope so that it induces a torque
 
watch from 9:00 to 9:30
then u will get a better idea of what im asking
 
In that case it's because the gyroscope is falling from an unstable upright equilibrium position. The moment this equilibrium is broken, then gravity induces a torque on the gyroscope centre of mass
The same a brick falling on its side has a torque induced by gravity
 
6:57 AM
well...what i get is
one side of the gyroscope is lower
so gravity on that side is more
so there is more downward force
 
No variation of the gravitational force is minuscule at that scale on earth
 
then what exactly is causing the torque?
 
It's because the 'point of attach' is not the centre of mass it's the bottom par that sits on a table
gravity
Forget the gyroscope
An upright pen falling down has a torque on it's centre of mass induced by gravity
 
You must first understand that before understanding the gyroscope which has another nice effect
 

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