The h Bar

ACuriousMind

00:00

@Sofia I've read that article. What about it? Interaction-free measurements are weird, but nothing new. It's just interesting that thev're managed to implement it.

NeuroFuzzy

Isn't this question a duplicate of some question about why electrons stay inside matter and don't jump off? I think you might hold the accepted answer on the one I'm thinking of @Sofia
http://physics.stackexchange.com/questions/170253/what-stops-charges-from-jumping-across-a-capacitor

ACuriousMind

It's all just a very elaborate Elitzur-Vaidman bomb test.

0celo7

abstrusegoose.com/455

ACuriousMind

I love that webcomic.

Sofia

@ACuriousMind my dear one, I am not as quick as you. Let me see what's going, and I won't do this by night. And I need the basic article of that Hatim Salih fellow. Don't you have time?

ACuriousMind

00:05

@Sofia I have all the time in the world. Why do you think I don't have time?

0celo7

abstrusegoose.com/463

These are pretty good.

Sofia

@ACuriousMind I do think that yes you have time, it was a rhetorical question. And since you have time I'd like to see what's going. There is Zeno effect there and all sort of things, but bottom line, I'll look what's they obtained, in fact.

@NeuroFuzzy Before I see that Hatim Salih paper I know of no duplicate.

NeuroFuzzy

Oh I must be thinking of this post... physics.stackexchange.com/questions/95889/…

0celo7

abstrusegoose.com/474

It's like string theory textbooks.

ACuriousMind

@0celo7 Heh, probably

Sofia

00:15

@KyleKanos I asked the user to find that paper of Hatim Salih. I see from the indications in what the user provides, that the issue smells of faster-than-light communication. So, please, I wait for the paper, on which all the commentators rely. Please, I want to see what's going.

ACuriousMind

@Sofia If we need to read a paper to find out whether a question has merit, then it's not a good question.

0celo7

@ACuriousMind That's debateable.

You need a book to really appreciate some questions.

ACuriousMind

@0celo7 You might need to read a paper or book to understand a question, but it should not appear non-sensical to you without further reading.

0celo7

@ACuriousMind I'm sure string theory would appear nonsensical to Maxwell back in the day.

ACuriousMind

@0celo7 Yeah. Asking about ST on a 19th century physics.SE would also not be a good question ;)

0celo7

00:27

I think this is how the field equations were first "derived".

ACuriousMind

@0celo7 Proofs.

Stan Shunpike

@ACuriousMind Are C* algebras important to learn for QFT?

Sofia

@ACuriousMind ahhh! It's you, or it's I who will read? I got the paper already. But, I have no wish to deal with it now. I don't expect that smbd. discovered a way to do FTL. I just want to see what is so much noise around this article, what that Salih could have said that we didn't know yet, s.t. people so talk of it? I guess that the user got mislead by all sort of sayings around it. So, let's make light.

0celo7

bolbteppa

@StanShunpike remember I told you everything in QM comes back to the notion of 'expected value': https://i.sstatic.net/iTIPS.png Pretend f in that picture is velocity and f(j) represents the j'th measurement of the velocity. Another way to say it is that that calculation represents the 'mean value' of the velocity, or the average velocity we expect. Well, isn't it sheerly absolutely fantasmogastically amazing that one can formulate the notion of 'mean value' using eigenvalues!!!???
http://books.google.ie/books?id=J9ui6KwC4mMC&lpg=PP1&pg=PA10#v=onepage&q&f=false

0celo7

00:36

I think you did some consulting work...

ACuriousMind

@StanShunpike "Important" is pretty vague, eh? No, they're not needed for anything, but they're a valid, nice, clean formulation, in my opinion.

bolbteppa

That right there is a demonstration of why axiomatic QM approaches are useless, simply none of them give you that intuition

ACuriousMind

@0celo7 Who? Me? ::looks around innocently::

bolbteppa

They actively shield you from such intuition

ACuriousMind

@bolbteppa As they should.

0celo7

00:38

@ACuriousMind It's frustrating that people think physics solves philosophical problems.

bolbteppa

Exactly, that is the death of science

0celo7

It's almost embarrassing to say that we simply don't know how some stuff makes sense.

ACuriousMind

@Sofia I'm not sure what you want from me, but I read two of the links in that question and I am not wasting anymore of my time on this.

0celo7

People don't give a shit that we can derive the energy levels of H from $\mathrm{SO}(3)\times\mathrm{SO}(3)$, they want to know what happens to that damn cat.

ACuriousMind

@0celo7 Do...we know how making sense makes sense?

Word games, all of it.

Sofia

00:40

@ACuriousMind did I say that I want smth. from you? I didn't.

ACuriousMind

@Sofia You addressed me (@), and there was a question mark in there, so I guess you asked me something or wanted something from me, no?

0celo7

@ACuriousMind If we put the cat in the box and attach a nuke to the detector, will the nuke only go off if we open the box?

ACuriousMind

@0celo7 Not if we are in the world with the angry monkey.

0celo7

@ACuriousMind While that was a serious question, I appreciate the comic.

bolbteppa

@StanShunpike I took that picture from here books.google.ie/…

Sofia

00:45

@ACuriousMind I want nothing from you. Good night, it's late, I am going to sleep.

bolbteppa

I thought Schrodinger's point was to show the ludicrousness of thinking the cat was half alive half dead, such an interpretation is only possible if you think the wave function is a physical notion, rather it just represents the probability for something happening for which we have no information on?

ACuriousMind

@0celo7 The nuke doesn't care.

0celo7

@ACuriousMind But isn't the nuke in a superposition of boom and not boom?

ACuriousMind

@0celo7 But the boom state isn't isolated - it spreads rapidly, interacting with the enviroment, inducing either collapse or decoherence, and so forces an outcome

bolbteppa

yeah that's not a closed system

0celo7

00:49

I see.

What about this: Can the cat be an observer to his own death?

ACuriousMind

Or you can take the statistical view that bolbteppa took. My explanation is what you would call "ontic", I think, while his is "epistemic".

@0celo7 I don't know what that would mean. But I guess the answer ultimately boils down to the fact that we have no idea how consciousness fits into all this anywhere.

bolbteppa

My only question is - did he mean Sean Carroll? lol

0celo7

@ACuriousMind Attach a bell to the poison vial. The cat hears the bell, which signals that he will die. So has he collapsed the wave function without anyone having to open the box?

bolbteppa

@ACuriousMind what's the difference between ontic and epistemic?

ACuriousMind

@0celo7 That's a meaningless question - if the box is soundproof, we don't know what happened in the box either way - how would you distinguish a "non-collapsed wavefunction" from "there's a chance X the wavefunction is collapsed into dead cat and a chance Y it's collapsed into alive cat"?

0celo7

00:56

@ACuriousMind The cat can rot if left dead long enough. If he collapses the wave function and dies, and then rots, (lost train of thought)

ACuriousMind

@bolbteppa Ontic standpoints treat the quantum state (or the wavefunction) as something "real" you can talk about (I said "the state is spreading") that represents a probability in the world. Epistemic standpoints treat the quantum state as an encoding of our limited knowledge, a probability for the observer.

bolbteppa

Wait you think the wave function is real?

ACuriousMind

@bolbteppa I don't know what that would mean.

But my manner of phrasing "the state is spreading" is usually characteristic of an ontic standpoint.

I honestly don't care, I see no empirical difference between any of these interpretations.

bolbteppa

lol http://motls.blogspot.ie/2015/01/tom-siegfrieds-delusions-about-reality.html
"The words are derived from Greek roots – because the philosophers incorrectly assume that they may hide their stupidity in front of everyone by using words that sound Greek to others – but they may be translated in a simple way. The wave function is "ontic" if it "objectively exists" (the root means "to be") while the wave function is "epistemic" if it describes what we "know" (the root "episteme" means "knowledge" or "understanding")."

0celo7

:'D

bolbteppa

01:03

So we may have picked "epistemic" because "ontic" was totally wrong. "Ontic" claimed that the wave function describes just another classical (probably deterministic) theory – that the wave function is a kind of a classical wave – which is surely not the case. Fine.... But the second option, "epistemic", is also wrong if you study what these philosophers actually imagine under "epistemic" in detail.

ACuriousMind

@0celo7 Essentially a more elaborate version of xkcd.com/435

0celo7

@ACuriousMind I knew what that was going to be before I clicked it.

You've infected me.

ACuriousMind

@0celo7 Good :D

I usually don't like Motl's polemics on the blog, but his disdain for quantum interpreations I can fully share.

0celo7

@ACuriousMind Too bad all laypeople care about are the interpretations.

ACuriousMind

@0celo7 Yeah. It can make me furious at times.

Hmmm... AFuriousMind?

Stan Shunpike

01:10

@ACuriousMind Are you talking about Sofia?

bolbteppa

"It's a lot of amazing and mathematically accurate work that is going to lead to many followups but one may still have the feeling that the real "paradigm shifts" are mostly taking place elsewhere. So while Witten writes fundamental papers in the early stage - that are later followed by lots of other researchers – I would still say that there has almost always been a "paradigm shift" done by someone else before Witten wrote his important paper." Fighting words?

ACuriousMind

@StanShunpike Huh? I responded to 0celo7's comment about laypeople

Stan Shunpike

Lol bad joke I guess. She just always seems in an uproar about something or other.

ACuriousMind

Oh :D

0celo7

@StanShunpike Word.

ACuriousMind

01:12

@bolbteppa Dem's fighting words for sure.

0celo7

Witten sounds like a girl.

That's right, I said it.

ACuriousMind

I have no clue how accurate they are, though^^

@0celo7 ...you couldn't find an insult more cliched?

0celo7

No cliché. Cold, hard truth.

I'm so dumb. If I have a list of two adjectives, do I need two commas?

Stan Shunpike

@0celo7 in vocal timbre you mean?

0celo7

@StanShunpike Engrish.

@ACuriousMind If $\Lambda^k T^*M$ is the $k$-th exterior power of the cotangent bundle, what is $\Lambda^0 T^*M$?

(Besides the 0th power whatever.)

Is there some space $A$ such that $A\otimes V=V$?

ACuriousMind

01:21

@0celo7 Yes

And it is exactly that space

0celo7

@ACuriousMind Based skull. Proof?

What is the space called?

ACuriousMind

@0celo7 It's just the field you are taking the tensor product over

I.e. $\mathbb{R}$ or $\mathbb{C}$.

0celo7

I'm not completely sure why that doesn't do anything.

ACuriousMind

You can give the isomorphism as $\mathbb{F}\otimes_\mathbb{F} V \to V, x\otimes v \mapsto x\cdot v$.

For a basis $v_i$ , the basis of the tensor with the field is just $1\otimes v_i$

0celo7

@ACuriousMind The identity operator?

@ACuriousMind What does $1\otimes v_i$ mean?

ACuriousMind

01:26

@0celo7 Well, $1\in \mathbb{F}$ and $v_i\in V$, so$1\otimes v_i \in \mathbb{F}\otimes V$.

...unless you are using some weird definition of the tensor product

0celo7

@ACuriousMind Can you give an explicit example please, say for $\mathbb{F}=\mathbb{R}$?

ACuriousMind

@0celo7 I'm not sure how explicit this is, but do you know that $\mathbb{R}^n\otimes\mathbb{R}^m = \mathbb{R}^{n\cdot m}$?

0celo7

@ACuriousMind Yes...

ACuriousMind

Set $n=1$. This shows that tensoring with the base field does nothing, since $n\cdot m = m$, then.

0celo7

Ah.

ACuriousMind

01:30

And the $1$ that I talked about up there is just the canonical basis element of the field, considered as a vector space over itself.

bolbteppa

@0celo7 the vector space $\mathbb{R} \otimes \mathbb{R}^3$ is just the set of tuples $(\lambda,\vec{v})$ where $\vec{v}$ is an arrow in space, we can think of this tuple as intuitively representing how much we've scaled the arrow by when we write $\lambda \vec{v}$, and we formalize this intuition by showing the map $F : \mathbb{R} \otimes V \rightarrow V | (\lambda,\vec{v}) \mapsto F(\lambda,\vec{v}) = \lambda \cdot \vec{v}$ preserves addition and scalar multiplication (i.e. is an isomorphism)

If you replace mathematics with physics in that picture I think that's me haha (up to the chem stuff, then I'd modify it just to make sense!)

Oh man abstrusegoose.com/strips/angry_string_theorist.png

0celo7

I don't get what they mean by "isn't defined nonperturbatively"

bolbteppa

@StanShunpike there should come a moment with Kolmogorov when this abstrusegoose.com/strips/beautiful_theorem.png happens, I wont blame you

@0celo7 I think this superstringtheory.com/blackh/blackh4a.html is what they mean, not sure but it looks awesome (it's short)

I don't actually see how a beta function says anything about conformal field theory

ACuriousMind

01:51

@bolbteppa Conformal field theories have vanishing beta functions - that's why you often hear that conformal invariance implies "scale invariance", but not vice versa.

0celo7

@bolbteppa Beta function in the RG sense, not Euler.

ACuriousMind

@0celo7 I think they mean that there's no non-perturbative string scattering - the "sum over worldsheet topologies" is a perturbation series, in essence.

And, in contrast to the QFT case, that sum is often taken as defining the string scattering

0celo7

@ACuriousMind In contrast?

bolbteppa

My understanding is that you can take any measurable operator $P$ that has a coupling constant $g$, assume it's all a function of some variable $a$, $P(a,g(a))$, then $\dfrac{d}{da}P(a,g(a)) = 0$ gives you the beta function, right?

ACuriousMind

@0celo7 You can define in/out spaces without doing perturbation, and QFT scattering amplitudes are simply the inner product of the in state with the out state

Also, you may be able to exactly solve the path integral without expanding anything

While I've not seen that kind of formulation for string theory (doesn't mean it doesn't exist, it probably does)

0celo7

01:57

@ACuriousMind So there's no way to exactly sum the topologies?

Oh you don't know.

@ACuriousMind Aren't string amplitudes the same thing? String field theory is just crazier QFT.

bolbteppa

I suppose $a$ is supposed to represent the scale, so a conformal field theory obviously produces operators depending on a coupling constant that do not depend on scale, which is why their beta function vanishes. But some other theories may possibly exist that do not depend on scales but are not conformal on principle right? It's not just some weird counterexample is it?

ACuriousMind

@0celo7 Well, I've not seen anyone define the string-S-matrix not-perturbatively. There's no exact expression that would be expanded into the sum over different worldsheep topologies (like there is in QFT where an exact expression is expanded into the series of Feynman diagrams), the sum is taken as the definition.

bolbteppa

ACuriousMind

@bolbteppa Yeah, the $a$ there (often $\mu$ or $\Lambda$) is the energy scale at which you are looking that the theory, as far as I understand it.

Oh god, have I now infected two of you? :D

NeuroFuzzy

@bolbteppa apparently Witten is coming to my university on the eighteenth

0celo7

02:02

@NeuroFuzzy Tell him he sounds like a woman.

NeuroFuzzy

@bolbteppa ...w/ all the reservations filled already. I think I have to try to sneak in.

Hahahahaha oh yeah totally

0celo7

@ACuriousMind Infect me with some knowledge of how to do this god damn direct sum.

ACuriousMind

@0celo7 Weren't you struggling with tensor products moments ago? Which direct sum?

bolbteppa

Interesting, so a massless Klein-Gordon equation is part of a conformal field theory, but when you add a mass it becomes scale dependent, and since mass = energy, and you can show the equation gives a $(m\lambda)^2$ term, we see the constant is literally part of the energy scale!

0celo7

@ACuriousMind That thing you agreed looked horrible.

$$\Lambda^kT^*_\mathbb{C}M=\bigoplus_{j=0}^k\Lambda^{j,k-j}M$$

bolbteppa

02:04

@NeuroFuzzy cool, he referred to one of my lecturers as "my old friend" after a talk he gave to us haha

0celo7

I verified $k=1$.

ACuriousMind

@0celo7 I don't know what a $\Lambda$ with two indices is supposed to mean

0celo7

@ACuriousMind I got you. ::types::

Let $T^*_\mathbb{C}M$ be the complexified cotangent bundle. I have already verified that $T^*_\mathbb{C}M=T^*M\oplus \bar T^*M$, where $T^*M, \bar T^*M$ is the holomorphic (resp. antiholomorphic) cotangent bundle. We also have $\Lambda^{p,q}M:=\Lambda^p T^*M\otimes \Lambda^q \bar T^*M$.

bolbteppa

yeah!

0celo7

@ACuriousMind For $k=1$ we have $$\Lambda^1T^*_CM=(\Lambda^0 T^*M\otimes \Lambda^1 \bar T^*M)\oplus(\Lambda^1 T^*M\otimes\Lambda^0 \bar T^*M)=\Lambda^1 \bar T^*M\oplus \Lambda^1 T^*M$$

Which is a true statement I think.

I was thinking induction, but I'm not sure how to proceed.

Kyle Kanos

02:12

@alarge I had a phone interview; this one will be on-site. I expect programming & math challenges (e.g., given some function f(x) with undefined value at say x=0 (e.g., f(x)=(cos(x)+1)/x), write an efficient function that will compute this without the DivByZero error). But even that I'm not sure about.

bolbteppa

$\Lambda^2 T_{\mathbb{C}}^*M = \Lambda^2 (T^*M \oplus \bar{T}^*M) = \Lambda^{0,2}M \oplus \Lambda^{1,1}M \oplus \Lambda^{2,0}M = [\Lambda^0 T^*(M) \otimes \Lambda^2 \bar{T}^*(M)] \oplus[\Lambda^1 T^*(M) \otimes \Lamda^1 \bar{T}^*(M)] \oplus [\Lambda^2 T^*(M) \otimes \Lambda^0 \bar{T}^*(M)]$ such that any differential form in this space is of the form $f(z,\bar{z})d\bar{z}_id \bar{z}_j + g(z,\bar{z})dz_i \wedge d \bar{z}_j + h(z,\bar{z}) (dz_i \wedge dz_j)$

0celo7

@bolbteppa You mind formatting that?

ACuriousMind

@0celo7 Uh. I'm not sure how to proceed either :)

bolbteppa

Ahh sorry

Stan Shunpike

I'm glad I ordered that book on exterior algebras @bolbteppa mentioned because this is too complexified for me at the moment.

0celo7

02:19

@bolbteppa What's up with the third expression?

@ACuriousMind How the hell does one pronounce Nijenhuis?

@bolbteppa How does it follow from the second?

ACuriousMind

@0celo7 I have no idea. @Danu probably knows.

Stan Shunpike

ney-in-house

0celo7

House? How does huis become house?

I've been saying Neeshenwee

:D

Stan Shunpike

LOL

bolbteppa

So
(1) $\Lambda^2 T_{\mathbb{C}}^*M = \Lambda^2 (T^*M \oplus \bar{T}^*M) $
right? But then
(2) $\Lambda^2 (T^*M \oplus \bar{T}^*M) = \Lambda^{0,2}M \oplus \Lambda^{1,1}M \oplus \Lambda^{2,0}M$
okay? But this can be written as
(3) $\Lambda^{0,2}M \oplus \Lambda^{1,1}M \oplus \Lambda^{2,0}M = [\Lambda^0 T^*(M) \otimes \Lambda^2 \bar{T}^*(M)] \oplus[\Lambda^1 T^*(M) \otimes \Lambda^1 \bar{T}^*(M)] \oplus [\Lambda^2 T^*(M) \otimes \Lambda^0 \bar{T}^*(M)]$
Any differential form in this space is of the form $f(z,\bar{z})d\bar{z}_id \bar{z}_j + g(z,\bar{z})dz_i \wedge d \bar{z}_j + h(z,\bar{z}) (d…

0celo7

02:23

@Sean You come to join the teach 0celo7 complex geometry session?

ACuriousMind

@0celo7 forvo.com/word/nijenhuis

Stan Shunpike

That's the page I found lol

0celo7

@bolbteppa I don't get that (2).

@ACuriousMind That is such BS.

Stan Shunpike

Is that circle cross a direct sum?

0celo7

Now I have two things to talk about in my Nobel speech. Something about strangled chickens and Neeshenwee.

bolbteppa

02:25

@0celo7 you defined (2) above chat.stackexchange.com/transcript/message/20540287#20540287 (the huge circled plus post of yours)

I chose the case of k = 2 and worked it out explicitly

Sean

@0celo7 oh gosh, you really give me more credit than I deserve. Also, are those spherical chickens that you will talk about in your Nobel speech?

0celo7

@bolbteppa OH, you motivated (3) and used the definition to get to (2), which is my equation?

Sean

Haha

0celo7

@bolbteppa Noice.

Sean

http://physics.stackexchange.com/a/170263/60201
So I can't believe I'm saying this, but I think Floris actually got an answer wrong...

0celo7

02:27

@bolbteppa I don't see how step (1) was necessary though.

bolbteppa

Yeah and I even motivated the whole thing as just being a fancy way of defining the space of all differential forms
$f(z,\bar{z})d\bar{z}_i \wedge d \bar{z}_j + g(z,\bar{z})dz_i \wedge d \bar{z}_j + h(z,\bar{z}) (dz_i \wedge dz_j)$
where you have to take wedge products, and sum over all permutations (I never remember how to write those sums), yeah (1) was just an unnecessary reminder step but it helps when you see how crazy (3) looks

I'm not sure about the $dz_i \wedge d\bar{z}_j$ term tbh

alarge

@KyleKanos Not programming as in OOP, language specific stuff etc., though?

Kyle Kanos

@alarge I suspect, if anything, they'd ask simple stuff about inheritance & such

0celo7

@bolbteppa Well from what I understand, $\operatorname{span}\{dz^i\}=T^*M$ and $\operatorname{span}\{d\bar z^i\}=\bar T^*M$, right?

bolbteppa

yeah

0celo7

02:33

So the $dz\wedge d\bar z$ term is just a holomorphic 1-form space tensor producted with the antiholomorphic one, i.e. $\Lambda^1T^*M\otimes\Lambda^1\bar T^*M$.

alarge

@KyleKanos Does Fortran do OOP? For example, dynamic polymorphism (or have you been reading up on C++ or something else)

0celo7

@bolbteppa What's wrong with it?

Kyle Kanos

@alarge F03 & F08 have introduced aspects of OOP

e.g., fortranwiki.org/fortran/show/Object-oriented+programming

bolbteppa

@0celo7 yeah if you write $d\bar{z}_i \wedge d \bar{z}_j = d \bar{z}_i \otimes d \bar{z}_j - d \bar{z}_j \otimes d \bar{z}_i$ then everything is a sum of $\otimes$'s, I just don't think it's useful to think of wedge products with those $\otimes$ because I don't see the geometry, but I guess it explains the factorial stuff mathoverflow.net/a/54353/38721

Stan Shunpike

@ACuriousMind what is a probability density matrix? I have heard of PDFs but never a matrix.

Kyle Kanos

02:45

Which basically says that dynamic polymorphism doesn't appear in the standard to date (and I'm not sure it's in the plan since it's actually slow)

ACuriousMind

@StanShunpike Uh, they probably mean the density matrix.

bolbteppa

@StanShunpike again that's explained in the (not my!) quantum mechanics expected value picture I posted earlier i.sstatic.net/iTIPS.png better than anywhere else

ACuriousMind

It's the way to do quantum statistical mechanics.

0celo7

@bolbteppa Perhaps the $\Lambda^k$ spaces are defined with the wedge on them already, so the tensor product of two of them is understood as the wedge.

Stan Shunpike

@bolbteppa I'm not gonna lie, I have been procrastinging learning that because it looks so horrible, but I just now understood in under 3 seconds.

bolbteppa

02:48

From that 4 line calculation you see we can either choose wave functions on which we can base our QM dynamics on, or we can choose Hilbert space analogues of those wave functions (using the l2 L2 isomorphism) to base the dynamics on, or we can choose the density matrices in that calculation to base our dynamics on (and get the Von Neumann equation), all should be equivalent, however the density matrix allows for closed systems with all the off-diagonal entries

Stan Shunpike

Actually, I don't think I could have prior to today

bolbteppa

That took me like 3 years to realize

Stan Shunpike

I just got Ballentine and now what u said makes sense

bolbteppa

@0celo7 that is standard notation for a space of k-forms

0celo7

@ACuriousMind Finished Abstruse Goose from like 400 to present.

bolbteppa

02:49

But the tensor product thing is different, you've defined a tensor product on spaces of forms

Stan Shunpike

Abstruse Goose....what is that?

ACuriousMind

@0celo7 Heh, still ~400 to go, then ;)

bolbteppa

@StanShunpike now go back and read my eigenvalues post! :D

It's so unbelievably unifying it's crazy

Stan Shunpike

On it brb

0celo7

@bolbteppa So is your argument wrong?

bolbteppa

02:52

Here we are again, I had no idea why Landau decided to define expected values via eigenvalues, I literally had NO idea until earlier today, Landau was years ahead of me AGAIN, even something as simple as how he defined expected value has such deep significance

alarge

@KyleKanos Well yes, you do incur some (small) costs as you'll have to be chasing pointers to the heap and probably doing some cache misses in the process, but I think it would be quite difficult to design basically anything in the spirit of OOP without.

bolbteppa

I don't even fully understand defining mean values via eigenvalues tbh but I know it can be done

@0celo7 no nothing I said is wrong

0celo7

@bolbteppa So $dz\wedge d\bar z\in\Lambda^1\otimes\bar\Lambda^1$?

Kyle Kanos

@alarge Eh, as long as you're Objects have Methods applied to them, it's OOP in my books. I 'm not convinced that development time is shortened using dynamic vs static polymorphism, but then again, I've not really used dynamic and only static in a few cases.

Stan Shunpike

@bolbteppa That's crazy. Expectation values are like the coolest thing ever.

Kyle Kanos

02:56

Though, reading some more, it does appear that some run-time polymorphism can be hacked with F95+

Stan Shunpike

@bolbteppa I've never heard it summarized quite that way with one picture. That's a really neat picture

bolbteppa

@0celo7 I just had an issue with the notation, I'm pretty sure it should be $dz \otimes d \bar{z}$ and the reason is that you don't want it to matter whether you write $dzd\bar{z}$ or $d\bar{z}dz$, if you wrote it with a $\wedge$ then the order would matter

0celo7

@bolbteppa Well a form has to be in a graded algebra, so we need the wedge.

bolbteppa

@StanShunpike did you see how Landau introduces it? See how he doesn't explain why, and quite honestly it's fucking crazy that he defined it that way, I could not understand why, but YEARS later you see he was more right than you were, you had to raise your game lol, there are TEN books like this! Why read any others?

Stan Shunpike

@bolbteppa That's amazing. My book Landau just shipped. I can't wait to get it. You won't believe it, but I got my copy for $10

0celo7

02:59

I have to know what all this fuss is about.

alarge

@KyleKanos Well, to take a very rudimentary example, suppose I have a game, and in it I have a purely virtual class Enemy, and the concrete classes SmallEnemy and BigEnemy that inherit from it (and have different implementations of the methods). Now I want to throw small and big enemies at you at random, so they are in a list that contains Enemies. I want to loop over the list and call some method (action) on each Enemy. How would you do this in Fortran?

ACuriousMind

@0celo7 The tensor algebra is also graded. I believe $\mathrm{d}z\wedge\mathrm{d}\bar z$ is not really good notation, because $\mathrm{d}z$ and $\mathrm{d}\bar z$ live in different spaces (the $\Lambda$ and $\bar\Lambda$, respectively). Writing $\mathrm{d}z\otimes \mathrm{d}\bar z$ or omitting the sign between them altogether is better.

bolbteppa

@StanShunpike could you figure out the whole writing mean vaues of functions using eigenvalues thing and explain it to me in a way I'll understand? I think it only holds for linear functions/operators, so you can't use eigenvalues for mean values except in a special case, but still I don't fully see the out-of-thin-air intuition for eigenvalues as mean values when I think of $\dfrac{x_1 + x_2 + ... + x_n}{n} = x_1p_1 + x_2 p_2 + ...$ as what a mean value is?

0celo7

@ACuriousMind So how do we make the connection with the wedge then if we don't use it?

ACuriousMind

(you can't commute them anyway, because $\mathrm{d}\bar z\otimes \mathrm{d}z$ would be in $\bar\Lambda\otimes\Lambda$ instead of $\Lambda\otimes\bar\Lambda$, so you can't define the wedge by dividing the usual ideal out of the tensor algebra, anyway)

bolbteppa

03:03

@0celo7 that doesn't affect the grading and @ACurious mind is right

Kyle Kanos

@alarge Simple: store only the enemy type and call the differing functions based on the current enemy type: if(enemyType == BigEnemyType) then; call BigAttack; elseif(enemyType == SmallEnemyType) then; call SmallAttack; endif

0celo7

@ACuriousMind I thought forms are in the exterior algebra and now you're telling me general tensors are in there?

bolbteppa

Well you can commute them in the sense that when you operate on everything and end up in $\mathbb{R}$ it wont matter whether you started from $dzd\bar{z}$ or $d\bar{z}dz$, we are just trying to formalize that baby example anyway, @0celo7 the exterior algebra is built out of the tensor algebra as a bunch of sums of $\otimes$, e.g. here mathoverflow.net/a/54353/38721

Kyle Kanos

(obviously all contained within a do-loop, maybe a do-while if you've made a linked-list)

alarge

@KyleKanos The problem here is that if you later decide to add an enemy type or an action type, you'll hate your life.

0celo7

03:06

@bolbteppa I know how the exterior algebra is built. It only contains fully antisymmetric covariant tensors.

But it if we just write $\mathrm{d}z\otimes\mathrm{d}\bar z$, we lose that antisymmetry.

ACuriousMind

@0celo7 Well, it gets a bit confusing in this complex case. You essentially have the completely antisymmetric holomorphic forms and the antisymmetric antiholomorphic forms, and then arbitrary $\otimes$ combinations of these, I think. The holomorphic and antiholomorphic pieces each live in an usual exterior algebra.

bolbteppa

@StanShunpike which Landau did you get? Volume 3 will be extremely hard if you haven't read 1 & 2 and you will absolutely not get much out of it without having built the intuition in the earlier volumes anyway, and they are hard enough, I call Gel'fand's CoV as Landau book 0 because book 1 is hard enough to require that lol these are grad books in the US in many places so be warned ;)

0celo7

@ACuriousMind So do the $dz$ and $d\bar z$ not anticommute when wedged?

Or is the wedge not even defined?

ACuriousMind

@0celo7 You can't wedge them.

You only get wedges out of spaces that are n-fold tensor products of an underlying space, but the $z$ and the $\bar z$ are considered to live in different spaces, as far asI see it.

0celo7

@ACuriousMind Then why do I see them wedged in every geometry and string theory book out there?

ACuriousMind

03:10

But I don't know complex geometry as such, so I could well be wrong.

Kyle Kanos

@alarge Yeah, that's not doing anything to convince me. Basically any modifications made at later times are going to be "tough" to finish through, I don't care what programming paradigm you follow.

ACuriousMind

@0celo7 Because they're sloppy, probably :D

0celo7

I think they both live in the overarching complexified tangent bundle.

Kyle Kanos

The hydro code I use is a hybrid (uses a lot of OOP paradigm in a largely procedural context)

And it's a pain to fix mistakes

Whether it's the procedural parts or the OOP aspects

bolbteppa

I think it's okay to wedge them in general but you have defined something that requires more care, I think you are using something more advanced that has it's own theory with it tbh

ACuriousMind

03:12

@0celo7 Ah, yes, but are you considering the tensor/wedge over $\mathbb{C}$ or over $\mathbb{R}$. I think the issue here is that the fields the products and dimensions are considered over are always just implicitly assumed, and not written out.

0celo7

@ACuriousMind ::squints, reads, yawns:: I can't keep up with you vampires. I have to be somewhere in 8 hours, and I need sleep.

bolbteppa

Oh maybe that's it, maybe your thing is over $\mathbb{C}$ and the $dz\wedge d\bar{z}$ is over $\mathbb{R}$?

ACuriousMind

You should probably really find someone who knows complex geometry, I think I'm butchering this :P

alarge

@KyleKanos Well the point of OOP is really that if properly done, it is easy to add functionality. In this example, nothing in the execution code would change if a new enemy type were added. You would not even need to compile those parts of the code again (which can be an issue if compile times are long).

0celo7

@ACuriousMind There are no string theorists who frequent the chat.

bolbteppa

03:14

If you start from real numbers and complexify, like you do in string or conformal field theory, then $dz \wedge d \bar{z}$ is okay, but if you are working with complex manifolds from the get-go then it's $dz \otimes d \bar{z}$, at least I think

0celo7

Well what about the difference between the hermitian metric and the Kahler form?

The only difference is an overall factor and the wedge.

Kyle Kanos

@alarge Unless you can provide an example where I can add code, not recompile it and expect it to work as if I did recompile it, I do not believe you on that aspect.

0celo7

@ACuriousMind @bolbteppa How do you reconcile that the Kahler form is defined with a $dz^i\wedge d\bar z^{\bar j}$ in literally every reference?

ACuriousMind

@0celo7 I don't reconcile it, it shows I really don't know complex geometry ;)

Kyle Kanos

(though, if you use a scripting language (a la Python) to get around that statement, I'll take it as a win for myself).

Stan Shunpike

03:21

@bolbteppa Like, one thing that confuses me is that....consider the average role of a die is 1/6 per side right. So the expected value is 3.5. And (1+2+3+4+5+6)/6 = 3.5. But in that case, we can just pull out the 1/6 from the expected value and we suddenly get the term on the LHS of the equation you stated. But in this case, the probabilities aren't the same, so we can't factor them out. Does that make any sense or am I just rambling nonsense?

0celo7

@ACuriousMind And there has to be some type of antisymmetry going on because of the Kahler equation $J=i\partial\bar\partial K=-i\bar\partial\partial K$.

I have to sleep for real.

This will be continued.

bolbteppa

@0celo7 I think there is some merit to what I said tbh, this en.wikipedia.org/wiki/… seems to sync with what I said, the only associate a wedge product after all the constructions as being the imaginary part, thus it gives equivalent information but still comes from the symmetric tensor product stuff!?

Stan Shunpike

@bolbteppa Oh, I just saw your Landau comment. Darn! well, I am considering getting the others as well. I suppose I can't go wrong with Landau so probably worth the investment. lol grad books don't scare me :p

alarge

@KyleKanos Well, you can for example look at the pimpl idiom to see a simple example of how by moving things around you can avoid compiling a lot of things.

Also, perhaps more relevant (to the general discussion about OOP, but also to compilation), is the open-closed principle.

bolbteppa

@StanShunpike in the case of rolling a die, the probability of each side is 1/6, so the expected value of the die is 1*(1/6) + 2*(1/6) = 3.5 = (1/6)(1 + 2 + 3 + ... + 6). For my example I should have written x_1(1/n) + x_2(1/n) + ... = (x_1 + x_2 + ...)/n my point was that I intuitvely think of averages as something like (x_1 + x_2 + ...)/n and the formal definition is just
x_1(1/n) + x_2(1/n) + ... = x_1 p_1 + x_2 p_2 + ... where in general p doesn't have to be 1/n, but it's a good thinking aid (for me) to remember this stuff, but eigenvalues!?

@StanShunpike you should get the other 2, those books probably have more aha moments than the QM one does

Again, the amazon review for the first one used the word poetry lol, and it's tiny

Stan Shunpike

03:33

@bolteppa lol I live for ah ha moments

and I'll take your word for it. I'm putting Landau at the head of my list. Budgeting book buying. Grrrrrrr

lol

Eigenvalues are just numbers though, right? So it's not like we are trying to average an operator...

Kyle Kanos

@alarge AFAIK, you still need to recompile the .c/.cpp file (though not the .h file)

bolbteppa

@StanShunpike look at the last equations on this page cobalt.chem.ucalgary.ca/ziegler/chm373/rudiment/quanmath/… I can understand how starting from <A> and ending up with $\sum |c_n|^2 a_n$ shows that the concept of mean value leads to using eigenvalues and that this calculation explains why we can interpret eigenvalues as measurable quantities. But Landau reverses this calculation and says the end result is the "usual" definition of mean value,

so there should be an out-of-thin-air motivation for defining mean values with this definition and a conceptual picture with it

Kyle Kanos

And it's effectively a 1D finite difference code used at the bank. That can be done in like 250 lines (albeit fairly compact). It's not a million+ sloc project here, maybe 1-2k lines. Debugging, compiling, editing are trivial in that case

Maybe OOP has a benefit for the larger projects we see (e.g., VGs), but for something as trivially small and procedural as a 1D finite difference code, I just don't see the point.

Stan Shunpike

04:13

@bolbteppa Okay, let's consider position space. In classical mechanics, we can talk about all the points as a vector space $\Bbb{R}^3$. But in quantum mechanics, we want to think about it differently. We still treat each point as member of a set. But we want to find the probability for realizing a particular point in the set. This should immediately call to mind random variables right?

Isn't that like what we are doing here?

I mean it's not exactly the same since we can have a superposition of states, but maybe that relates to mean values somehow...

NeuroFuzzy

So apparently the seminal paper on monte carlo methods has a very symmetrical name: "Metropolis, N.; Rosenbluth, A.W.; Rosenbluth, M.N.; Teller, A.H.; Teller, E" "(The story goes that the authors searched in vain for another physicist named Metropolis, so that the paper could have the symmetrical author list Metropolis, Metropolis, Rosenbluth,Rosenbluth, Teller and Teller!)"

Stan Shunpike

LOL that's redonkulous

wtf

People will do anything for symmetry

NeuroFuzzy

It still has "Rosenbluth, M.N" and "Metropolis, N."

which is nice

hahah

Stan Shunpike

lmao

@NeuroFuzzy why do we say extend condolences?

Like a relative of mine died and someone said I extend my condolences. Its a very odd phrase

Its not like a hand or something where I can stick out out and extend it.

I would think one sends or gives condolences. Wait sry I'm on the wrong SE. I'll pop over to the English guys and see what they think

Boom. Posted it.

Alright peace out

NeuroFuzzy

04:35

Hahaha

peace "out"?? It's gone??? And what exploded?!

i hope you weren't hurt in the explosion. You have my condolences.

Stan Shunpike

apparently all logic (exploded)

NeuroFuzzy

Please return them with a full tank.

2

David Z

05:05

1

Q: History of energy

Can anyone suggest me some book/article explaining how and why the quantity named 'Energy' made its way into Physics? I have gone through "Lectures on Physics" (Vol. 1) by R.P. Feynman and have been convinced that at first scientists were in search of a quantity which remains constant w.r.t. any...

That one might plausibly be off topic... thoughts?

Stan Shunpike

06:16

@DavidZ I don't understand what he is asking. Is he confused about the nature of energy or is he asking about the historical origins of basic laws involving energy?

One reason I am confused is he doesn't pick a tense and uses "we" vaguley. I am not sure who the we is. Does he mean we as in modern day phycists? We as in ancient physicists and therefore mankind as a whole?

Like does he mean energy as in $L=T-V$?

I guess I can ask

David Z

07:25

Yeah, I don't know anything more about these questions than you do

David Z

07:55

0

Q: High pitched noise as hot object hits water?

https://www.youtube.com/watch?v=9qSEfcIfYbw Just wondering what this phenomenon was called, and its cause? Would have googled, but I don't even know what it's called.

thermodynamics water

I think that's off topic but I'd rather not close it unilaterally

alarge

10:44

@KyleKanos Obviously you need to compile something when you add code, but the point here is that you need not compile the core logic. You might be using a precompiled, even purchased, library that you can't have the source for (for example the one that processes all the enemies), yet you want to be adding functionality (by adding new enemy types).

2k LoC is a very small program and in scientific computing odds are it's basically throwaway: Built for a single purpose. There's often very little programming involved in these coding projects. I doubt a bank is hiring you to write simple PDE solvers as these are textbook material and are likely to have been implemented decades ago. Even if they were, they'd probably have a 1m+ LoC library that you'd have to interface with to get all the data for your computation.

Danu

10:56

I recently had a small applied physics question

I was trying to cut something quite hard and dense (very much like plant sap), and doing it by pretty much pushing a knife into it without any saw-like motion

As I pushed it in, it suddenly snapped through when I was almost exactly halfway through, breaking through the bottom half in one go

Is it a coincidence? Or is there some physics to why it'd do that halfway through?

alarge

@Danu Well I suppose the whatever you were cutting was getting thicker until the midpoint, so you were increasing the force on the knife all the time. After the thickest point, though, this constant force (as I imagine you wouldn't have had the time to react) would cut through the rest.

Kind of reminded me of this TEDx talk.

2

Danu

11:22

That makes some sense

Danu

11:33

@alarge I hated how the guy kept on saying 'fundamental physics' haha

alarge

11:45

@Danu Haha, I can see your frustration.

Danu

fundamental $\neq$ basic

alarge

@Danu Well to be fair if you define the word physics as statics, as one often seems to do (not saying it's right, but that this is what physics means to a lot of people, especially if you want to distinguish between say physical and chemical effects), I suppose there's not much of a difference.

infinitesimal

in common usage they are synonyms

fundamental: of the basis of a subject

basic: forming a basis

Danu

now look up "basic physics" and "fundamental physics"

:P

infinitesimal

12:06

perhaps "elementary physics" is more to your liking

Danu

of course, since it actually means "basic physics" as well ;)

infinitesimal

but "basic" may not be simple

elementary: dealing wit the simplest facts of a subject

Danu

I don't really care about the meaning without "physics" appended

infinitesimal

sub in "physics" for "subject" in the above "common usage" definitions

@alarge Thanks for sharing :-)

Danu

12:49

Happy 3.1415 everybody! ;D

2

Sofia

13:48

@ACuriousMind Why this cruelty? You said that you have time, I asked time to read the article of Salih. I don't have even the possibility to vote for reopening.

Sofia

14:04

@KyleKanos Am I, a researcher in fundaments of QM, the last of the stupids from your point of view? I asked so much to leave this question, because it is answerable. I explained that it is about some apparent claims of FTL communication, therefore its absolutely on-topic, and I will handle it. I have no credibility in your eyes? About the part with the GR I have no idea, but there is a part of QM. Nobody noticed it? Why should have been this question closed?

Kyle Kanos

@Sofia the question doesn't make any sense. It is literally asking if QM & GR describe different things, the answer is an uninteresting yes.

GR describes the geometry of space-time, QM describes the nanoscale interactions of particles (ignoring gravity)

infinitesimal

sounds like a good opportunity for IBL

some lessons are best learned that way

Kyle Kanos

IBL?

infinitesimal

the moore method

Kyle Kanos

Moore method?

infinitesimal

14:16

inquiry-based learning

Kyle Kanos

Isn't that how Plato taught?

infinitesimal

sort of

Kyle Kanos

@Danu Pi is still wrong

I suppose this was done before Manish & Qmech became mods, but I just came across a post closed by three mods

ACuriousMind

@Sofia Votes are to be cast immediately, and based on the present state on the question. I don't see what my (or others') voting to close has to do with you (or anyone else) reading an article.

@KyleKanos Do you mean the Socratic method?

Kyle Kanos

@ACuriousMind Almost immediately after pressing enter, I started thinking Maybe it was Socrates & not Plato....eh, whatever

But that is what I was thinking

ACuriousMind

14:29

@KyleKanos Well, all we know about Socrates is what Plato wrote about him, so you're not really that wrong :)

(Yeeees, there's also a bit from Xenophon about him, before someone asks :P )

Kyle Kanos

@ACuriousMind No idea who that guy is, but reading Wiki now....

infinitesimal

14:54

@ACuriousMind where?

ACuriousMind

@infinitesimal Where? You mean, what the work from Xenophon about Socrates is called?

infinitesimal

@ACuriousMind yes

ACuriousMind

The most interesting thing is probably that Xenophon has his own version of the apologia, and that his Socrates is more arrogant and altogether less "noble" than Plato portrays him.

@infinitesimal Everything else I'd have to cite from Wikipedia's list. My Greek lessons are a few years in the past now

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