The h Bar

user54412

17:01

@JohnRennie So the well-to-do ones? ;) Have you seen @dmckee's recent post? It links to this chart, which shows I didn't quite pick the right major.

0celo7

@ACuriousMind get ready to be disappointed

what does "infinite dimensional representation" even mean

ACuriousMind

@0celo7 It's a representation on an infinite-dimensional representation space, duh.

John Rennie

@ChrisWhite I'm a bit surprised at that, though I guess physics is a very broad discipline - not every physicist gets to work as a quant.

Slereah

instead of the boring old finite dimensional ones of SO(3,1)

0celo7

@ACuriousMind fine

Slereah

17:03

The rep $(p,q)$ is... $p+q$ dimensional?

I forget

wait no

That's too low

ACuriousMind

@Slereah $2(p+q+1)$ since the spin-$p$ rep of SU(2) is $2p+1$-dimensional.

Slereah

Ah yes

0celo7

@ACuriousMind So you're not going to answer seriously?

Slereah

6 D for vector fields

user54412

@JohnRennie Very true. I keep hearing about how "there's always finance," and certainly all my friends who aren't doing PhDs are quants and traders, but that just can't apply to everyone, right? There are surely far more physics undergrads than people working in finance?

Slereah

17:05

Which is indeed the dimension of the Lorentz group on vectors

Wait

That would mean that the scalar rep is 2D?

Isn't it 0D

ACuriousMind

@0celo7 Hmmm?

@Slereah Well, 0 is an exception.

Slereah

aight

ACuriousMind

@Slereah Vector fields are $(\frac{1}{2},\frac{1}{2})$.

Slereah

Oh right

ACuriousMind

So the dimension is indeed 4D, as it should be for 4-vectors.

0celo7

17:07

@ACuriousMind "infinite dimensional representation space" does not help and you know it

John Rennie

@ChrisWhite When I graduated (first degree not PhD) I got offered a job by Marconi Space and Defence Systems working on microwave tech. That's probably representative of lots of jobs for physics graduates.

Slereah

Oh that's the dimension of the object, not the transformation

ACuriousMind

@Slereah The 6D (1,1) is the traceless symmetric 2-tensor.

0celo7

Explain like I'm a German high school student.

Even that might be too much maturity.

ACuriousMind

@0celo7 I don't know what more you want to know. You know what a representation is, right?

Slereah

17:08

Hm, are spinors 3D?

It's two complex numbers

What's the restriction that causes them to be 3D

0celo7

@ACuriousMind No clue. That's what terrifying me. I'm forgetting everything.

ManishEarth

o/ @ChrisWhite

Slereah

Do they have to be normalized

ACuriousMind

@Slereah Hm, no there are 0s again

ManishEarth

How's the grad school life

Slereah

17:09

Oh

WHEN WILL YOUR LIES CEASE

ACuriousMind

Spinors are $(\frac{1}{2},0)\oplus(\frac{1}{2},0)$,

Slereah

Well full spinors, yes

user54412

@ManishEarth phdcomics.com/comics/archive.php?comicid=47

Slereah

But not left and right spinors

ManishEarth

@Slereah define "3D" :)

They're the weird kind of 3d that quantum 2 state systems are

@ChrisWhite :D

Slereah

17:10

3D as in defined by 3 basis vectors :p

ManishEarth

yeah

user54412

@ManishEarth It's job season. So it's all about giving talks and writing research statements.

ManishEarth

Nice

user54412

Also having results to talk and write about.

ManishEarth

I, on the other hand, probably will not stick in this field :/ Still enjoy it, but I've realized I'll have more fun programming

ACuriousMind

17:11

@Slereah: My formula was false. The correct one is $(2p+1)(2q+1) = 4pq + 2(p+q) + 1$

This one should really be correct.

Slereah

WHY MUST YOU LIE TO ME

Hm, that's still 3D for spinors

ACuriousMind

Hm, no, it's also false :/

Slereah

Same question!

ACuriousMind

@Slereah wat

Plug in $p=1/2,q=0$. That's 2.

Slereah

Oh yes

Hm

Does two complex numbers count as 2D?

ACuriousMind

17:13

wat

ManishEarth

4 mins ago, by ManishEarth

@Slereah define "3D" :)

Slereah

Well a spinor is just $\mathbb{C}^2$ no?

ManishEarth

The term 2D is meaningless

ACuriousMind

My formula is false for $p=q=1$, though

ManishEarth

until you specify what you want it for

are you looking for independent variables? what?

ACuriousMind

17:14

@Slereah Ehhh, wouldn't phrase it like that but yeah

Slereah

I'm looking for what that formula corresponds to :p

Well how would you phrase it!

0celo7

@ACuriousMind A representation is a homomorphism from the group into GL(V) where V is a vector space, right?

If yes, then yes, I know what that is.

ACuriousMind

@Slereah The dimension of the representation space as a complex vector space

Slereah

I mean books like to overcomplicate it by bringing in Clifford algebras and spin groups

ACuriousMind

@0celo7 Indeed.

0celo7

17:15

And yes, I know what a homomorphism is.

Slereah

But in the end all of those still just act on a section of a $\mathbb{C}^2$ bundle

ACuriousMind

@Slereah Don't get me started on physicists and group theory again

user54412

@ManishEarth I think I called that years ago.

ManishEarth

@ChrisWhite A lot of people did

Slereah

@ACuriousMind : Well how would you describe it :p

0celo7

17:17

@ACuriousMind So what is an infinite dimensional rep?

ManishEarth

@ChrisWhite The only thing stopping me from doing programming was that I wasn't sure how I'd enjoy working on stuff other people tell me to, rather than personal projects. But that's the case with physics too

A programming internship straightened me out

ACuriousMind

@Slereah A Weyl spinor takes values in $\mathbb{C}^2$ with the $(\frac{1}{2},0)$ or $(0,\frac{1}{2})$ representation of the Lorentz group acting on it.

Slereah

Well yes, that is what I am saying!

ACuriousMind

@0celo7 The case where $V$ is infinite-dimensional. I really don't know what else you wanna hear

@Slereah You said "A spinor is just $\mathbb{C}^2$". :P

0celo7

@ACuriousMind Why is V infinite dimensional for the Lorentz group?

Don't say because it's non-compact

ACuriousMind

17:19

@0celo7 Uh, no, that's not how it works.

$V$ can be finite-dimensional for representations of the Lorentz group. Just not for unitary representations.

0celo7

;_; what's a unitary representation

I literally know nothing...

Slereah

It belongs to the unitarian church

ACuriousMind

@0celo7 A representation where the image is not just in GL(V), but V is a Hilbert space and the representation is into U(V)..

Slereah

Is it not the case for the finite dimensional ones?

0celo7

Ahh. So the physically useful Lorentz group representations are unitary matrices on an infinite-dimensional Hilbert space?

Slereah

17:23

$R^4$ is a Hilbert space and $SO(3,1)$ is unitary, no?

ACuriousMind

4

A: Lorentz Algebra Representation and QFT

The confusion here arises because we are not fully analogous to non-relativistic QM here. Given a (quantum or classical) field $\phi$, we usually specify whether it is a "scalar", "spinor", "tensor", whatever field. This refers to a finite-dimensional representation $\rho_\text{fin}$ of the Lore...

@Slereah No.

user54412

@ManishEarth So the question is where will you be programming?

Slereah

what part

ACuriousMind

$\mathbb{R}^4$ is only a Hilbert space with the standard Euclidean product, but $\mathrm{SO}(1,3)$ does not preserve that product, so it is not unitary.

ManishEarth

@ChrisWhite I have a pre placement offer from Microsoft India. Base case. Google recently contacted me. Not sure if I want to go to Google, but I'll leave that option open. Going to apply to Apple (compilers) and Mozilla (browsers) too.

also maybe some startups

user54412

17:25

@ManishEarth All in India?

0celo7

Can I please have a group theory and QFT book at the level of German middle schoolers that doesn't suck?

Please?

ManishEarth

@ChrisWhite No. Only Microsoft is in India

Slereah

Oh I see

Gennaro Marco Devincenzis

why do we ignore spin in the solution to the hydrogen atom?

ManishEarth

@ChrisWhite I'm a US citizen so no visa troubles

Slereah

17:25

@GennaroMarcoDevincenzis Do we?

Gennaro Marco Devincenzis

@Slereah I mean the elementary solution, the one in most textbooks on non relativistic quantum mechanics.

ACuriousMind

@GennaroMarcoDevincenzis Because it decouples. Adding spin gives just two (or more) copies of the spinless solution.

Slereah

pretty much yeah

Unless you count the magnetic field of the nucleus

user54412

@ManishEarth Apple makes compilers? Is this why they're silently getting rid of gcc from their OS? Some sinister plan to subvert all software that runs on their machines? :p

ACuriousMind

@0celo7 Only when I'm finished writing it ;) I didn't learn all those things from a single book or lecture, I pieced it together over about two years.

ManishEarth

17:28

@ChrisWhite They never had GCC on their OS

@ChrisWhite

administrators-Mac-mini:~ administrator$ gcc --version
Configured with: --prefix=/Applications/Xcode.app/Contents/Developer/usr --with-gxx-include-dir=/Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.11.sdk/usr/include/c++/4.2.1
Apple LLVM version 7.0.0 (clang-700.0.72)
Target: x86_64-apple-darwin14.5.0
Thread model: posix

:)

Apple is heavily invested in clang. Mac gcc is just an alias to clang.

hwlau

@ManishEarth Are mozilla really hire any people?

ManishEarth

@hwlau careers.mozilla.org/en-US/

0celo7

@ACuriousMind If you ever write a book, write it with me in mind. Physics books are inaccessible for me. I don't know why.

ManishEarth

@hwlau If I could make open source stuff my job I'd be really happy

Esp if it's servo or something

user54412

@ManishEarth I know. Recent OSX machines come with gcc as an alias to clang. Which we found out when people in our group started reporting that our code doesn't compile with gcc on their machines, even though we know it compiles on gcc.

ManishEarth

17:31

@ChrisWhite "recent"? I thought this has been going on for ages

Though I don't use macs, I wouldn't know

user54412

@ManishEarth I'm old. The years fly by.

ACuriousMind

@0celo7 It will be a pure definition-theorem-proof style mathy book. What else did you expect from me? ;)

hwlau

@ManishEarth It seems there are lots of opening, how can they hire that much poepel

user54412

@ManishEarth And you want to be a programmer? :P

ManishEarth

@hwlau Why not?

hwlau

17:32

@ManishEarth sure it is

ManishEarth

@ChrisWhite linux ftw

ACuriousMind

@ChrisWhite lol, what? Instead of telling you the program is not installed, it executes are different program?

hwlau

@ManishEarth where is the income from?

ManishEarth

@hwlau Yahoo, actually

used to be Google.

user54412

@ManishEarth ::caresses macbook:: ::affectionate purring::

ManishEarth

17:33

whoever the default search engine is. Also some revenue from other search engines

@ChrisWhite I have had nothing but bad experiences developing on macs

You need sudo to run a debugger

0celo7

@ACuriousMind That would be nice. But you'll probably write it at a level that will be way beyond me

ManishEarth

sudo.

for debugger.

why

Also sometimes command line tools need you to re-accept a license. Also needs sudo.

user54412

@ACuriousMind Yeah pretty evil. But when was the last time you were legitimately surprised by evil in the world?

hwlau

seem bad security model

Slereah

@ACuriousMind : Is the CSR similar to like

A wavefunction?

Vector with a continuous index

And is it a particular Hilbert space

ACuriousMind

17:38

@Slereah No, no, no, I don't think so. They're just representations which appear because the little group of massless particles is this weird $\mathrm{ISO}(2)$-thingy or whatever, and not just an $\mathrm{SO}(3)$ so that we get spin.

"Continuous spin" probably because they just aren't representations of $\mathrm{SU}(2)$.

Slereah

Hm

But what object is it :O

I still don't know

On which object does the representation act

ACuriousMind

@Slereah On what "objects" do the other infinite-dimensional unitary representations act?

Slereah

yes

On what elements of what vector space, if you will

ACuriousMind

No, I mean, what is the answer to your question for the spinor case, for instance?

Slereah

Well $\mathbb{C}^2$ I guess

0celo7

17:43

If they're infinite dimensional why can I write down the transformation rules for the fields explicitly

ACuriousMind

@Slereah The infinite-dimensional unitary representation the states the spinor field creates transform in is certainly not $\mathbb{C}^2$.

Slereah

I mean in the finite case

0celo7

Like a vector transforms with a 4x4 matrix

Slereah

I don't know what it is in the infinite dimensional case!

that is what I ask

0celo7

That's not infinite dimensional, is it?

ACuriousMind

17:44

@0celo7 I thought I discussed that by posting this answer

0celo7

Never seen that axiom before.

ACuriousMind

@Slereah Then your question doesn't make sense, I'm afraid. Contrary to the other cases, the CSR does not have an associated finite-dimensional representation of the Lorentz group.

Because you can classify all those finite-dimensional reps and they're just the spin $(p,q)$-spin ones.

@0celo7 Yeah, most people are pretty bad at distinguishing between the non-unitary representation the fields take value in and the unitary representations the states live in and the fields are operators on.

dmckee

@ChrisWhite I still remember the first time I went gcc -v and got a clang answer. Everything stopped for half an hour while I webbed p one side and down the other. BUt I write reqasonable portable code, so I was able to adapt fairly easily.

Until I went gdb. That change really bit into by work flow.

Slereah

@ACuriousMind : It still has a representation with some $GL(V)$, though, right?

I am just wondering what V is in this case

ACuriousMind

@Slereah An infinite-dimensional separable Hilbert space. They're all isomorphic as Hilbert spaces, so nailing down any specific space is rather useless.

Slereah

17:49

Hm

ACuriousMind

You may think of the $V$ (except in the CSR case) as being the one-particle space inside the Fock space of the field with the corresponding finite-dimensional representation.

Slereah

What would a component of a Hilbert space vector correspond to, if that makes sense?

Like how it would be an eigenstate of some operator in QM

ACuriousMind

@Slereah Vectors in Hilbert spaces only have "components" after you choose a basis.

Slereah

Is there a basis that suggest itself naturally in this case, though

ACuriousMind

Ehhh...in the abstract case, no. In the Fock space case, well...no, because the momentum states aren't actual states in the Hilbert space :D

Slereah

17:53

(RIGGED HILBERT SPACE BABY)

But I mean for the CSR :p

ACuriousMind

@Slereah No idea.

Slereah

dang it

It remains a bit abstract :p

And what would be the EOM for a CSR too D:

So many questions

ACuriousMind

@Slereah No e.o.m. because it doesn't belong to any field, I think.

The CSR is really, totally and completely unphysical from a QFT standpoint.

Since you can't fulfill that Wightman axiom about an associated finite-dimensional representation.

...what sparked your interest on those, anyway?

Slereah

Not physical doesn't mean it hasn't been studied :p

I was looking up Wigner stuff because I was looking at tachyonic representations

Because there really isn't much for tachyons

0celo7

what do

don't know what to learn

Slereah

18:02

learn about birds

0celo7

I meant math/physics wise

I'm not ready for Arnold

Slereah

bird physics

0celo7

I'm not ready for, well, any of the books I have

Slereah

landau's book on QM

0celo7

maybe

I just want to know what I'm doing wrong

Slereah

18:05

It's genetics

Engineers can't learn physics

0celo7

I guess...

well, what do I do now? my faculty advisor pulled strings to get me into analysis next semester also I'm taking probability 1

I can probably change my classes

0

Q: preservation of $d\Omega^2$ maintains spherical symmetry?

Minkowski metric is found to be $$ds^2=-dt^2+dr^2+r^2d\Omega^2$$ where $d\Omega^2$ is the metric on a unit two-sphere. Why is it said that once we maintain the form of $d\Omega^2$, we maintain spherical symmetry where everything else can be multiplied by other coefficeints?

general-relativity

What?

Slereah

I dunno

0celo7

if you multiply by $xyz/l^3$ you lose spherical symmetry, right?

$l$ is for dimensional reasons

0

Q: Covariant derivative acting on $\delta R$

In this question I am seeking a nice methodology. Maybe the experts on the site can help me on this. I am trying to find the GHY term for a higher derivative theory of gravity. Partly, I have figured how to find the boundary term, it is long calculation and somehow irrelevant to this question so ...

general-relativity tensor-calculus modified-gravity

indices...they are the tools of the devil

0celo7

18:40

@ACuriousMind On the bottom of page 224, Arnold says we identify the action of $I$ with mutliplication by $\mathrm{i}$. But the condition $I^2=-E$ says only that the eigenvalues of $I$ are $\mathrm{i}$ or $-\mathrm{i}$ and if I remember some complex geometry, this is where we say half are $\mathrm{i}$ and the other half are $-\mathrm{i}$. Is this a typo?

@ACuriousMind Also how is $\mathrm{GL}(n,\mathbb{C})$ the group preserving the complex structure? Why $n$ and not $2n$ (and I don't see why $\mathrm{GL}$ is the right group).

Slereah

GL is always the right group

0celo7

then answer my question

ACuriousMind

@0celo7 No, it is not a typo. The idea is that since $I^2 = -E$, we can define a multiplication of a vector $v$ by a complex number $a+b\mathrm{i}$ by $(a+b\mathrm{i})v = av + bI(v)$. Hence $I$ is "multiplication by $\mathrm{i}$".

0celo7

Another thing I can't remember

Great.

ACuriousMind

@0celo7 There is a basis in which $I = \begin{matrix} 0 & -E_n \\ E_n & 0\end{matrix}$. But the identity matrix is preserved by all invertible matrices, so $I$ is preserved by $\mathrm{GL}(n,\mathbb{R})\times\mathrm{GL}(n,\mathbb{R}) = \mathrm{GL}(n,\mathbb{C})$.

0celo7

18:52

@ACuriousMind Ah, well that makes sense.

@ACuriousMind I've never seen it done that way.

ACuriousMind

@0celo7 You can see it another way. Taking my above form for $I$, the $2n$-dimensional space $V$ splits naturally into two $n$-dimensional spaces $V_1,V_2$, and you can think of it as $V = V_1\oplus\mathrm{i} V_2$. Call $V_1$ the "real" part and $V_2$ the imaginary part. Then multiplication by $I$ throws real elements to imaginary elements without change of sign, but imaginary elements to real elements with a change of sign. This is precisely what multiplication by $\mathrm{i}$ does.

0celo7

19:16

@ACuriousMind Well I know it when you put it like that...

Huy

19:27

a student did some measurements of the intensity of some proton beam. however, it isn't very bundled, i.e. the protons have slightly different energies. the goal of the measurement was to compute the absolute difference of the "penetration depth" (?) of the protons. I was asked to help clear some things up but I don't know anything about this kind of physics anymore. what formula will be used here? @0celo7 do you by any chance know?

0celo7

@Huy my knowledge is that of a German middle schooler

I have no idea

Huy

@ACuriousMind: Do you know by any chance?

0celo7

"anymore"

when the hell did you learn about proton penetration depth

Huy

idk, maybe at some point I did

I don't remember 90% of the stuff that isn't maths that I learned at some point

@0celo7: what are you doing with complex structures?

0celo7

nothing, apparently

Huy

19:29

I somewhat recall proving $G(4,2) \cong S^2 \cup S^2$ using them

oriented Grassmannian

that was pretty surprising

ACuriousMind

@Huy Well, penetration depth probably refers to the $\lambda$ in the Lambert law $I(x) = I_0\mathrm{e}^{-x/\lambda}$.

Where $x$ is how far into the absorber we are, $I(x)$ is the intensity measure there and $I_0$ is the initial intensity

Huy

@ACuriousMind: the student sent me their data and asked me to look at it so I can get an overview of it. I have like 300 entries of "current in magnet", "intensity" and then for some reason their product. should I be able to find $\lambda$ from this?

0celo7

@ACuriousMind when did you learn about proton penetration depth

ACuriousMind

@Huy Uh, for that I'd need to know the experimental setup. What "magnet" is that? What are we targeting with the proton beam and where do we measure the intensity?

Huy

I wish I knew. I'll ask. It was done at the PSI, if that helps at all

(but I doubt it)

ACuriousMind

19:34

@0celo7 I did have some physics courses that weren't pure theory and I also had those terrible labs.

0celo7

smh

ACuriousMind

@Huy You have to ask. Without a description of the experimental setup, data is pretty worthless

@0celo7 ?

Huy

@ACuriousMind: anything else I should ask?

0celo7

@ACuriousMind nothing

I'm just...tired. Dunno why. Math and physics are exhausting me, this is a new experience.

ACuriousMind

@Huy Yeah, what exactly we are trying to compute. "absolute difference in penetration depth" sounds as if we measure the beam at different energies and look how far it penetrates into a fixed target for varying intensities, so that we get a curve telling us how penetration depth varies with beam energy.

Huy

19:38

ok

"those horrible labs" aka the main reason I didn't want to study physics ever

ACuriousMind

@Huy The sense of relief after we had completed the last one was one of the best feelings ever, though :)

0celo7

labs are the best way to increase GDP

Huy

I imagine it is similar to what I will feel after I completed the last pedagogy course

Terry Bollinger

"All Clifford algebras can be translated into power-of-two, diagonally limited binary connection matrices." Why bother? Objects like spinors become a simple hierarchy of recursive, almost hanging-mobile-like, major-minor diagonal transformations on the multi-scale quadrants of the resulting overall matrix. The various unnamed square roots of 1 (and -1) start to take on a much more definitive geometric flavor, also.

Loong

@ACuriousMind "penetration depth" = CSDA range?

ACuriousMind

19:51

@Loong I Yes, I think it depends on the field what you call it. It's also "attenuation length" and "mean free path" in some cases.

(I had to look up CSDA, never heard that before)

Loong

@ACuriousMind yes, and CSDA refers to the used model.

The ICRU report 49 includes the stopping powers and ranges for protons.

ACuriousMind

@Loong Have we stumbled upon your field of expertise here?

Loong

@ACuriousMind just a grazing shot

originally, I am a radiochemist who did much radiation protection and radiology and ended up in building nuclear power plants

Slereah

0celo7

20:07

@Loong Is that what you do now?

Loong

No, I quit that about a year ago.

0celo7

Are you German? I can't imagine a German building nuke plants.

Loong

@0celo7 Yes I am German. And yes, our new-build projects were not in Germany.

of course, we had various small projects at the existing plants in Germany

0celo7

I need to practice German but no one wants to speak German with me :(

@Loong Is there a German language chat?

Loong

deutschsprachiger Raum

General discussion for german.stackexchange.com. You may speak...

but it is not really busy

vzn

20:36

@0celo7 controversial STEM clockmaker ahmed mohamed meets obama etc / Tmachine blog

0celo7

> personally was hoping Mohamed would better document more somewhere how he built his clock so that naysayers could fade out some, but that seems not to have happened

Well he didn't build it, so there was nothing to document...

Garbage.

vzn

@0celo7 afaik/ afaict he has not described how it was "built". there is a lot of internet speculation that he just took apart/ disassembled a working clock.

0celo7

Speculation? People found the model of clock he used. He had a printed circuit board for crying out loud. Not a bread board. Commercial printed circuit board.

vzn

@0celo7 think the attacks are unfair afaict. agreed its not really "invented" but he seems to have made some kind of modifications...? he does not seem to have said/ asserted anything outright false/ fabricated. think internet trolls are "over the top" on this one.

0celo7

Internet trolls? He's the troll.

vzn

20:49

@0celo7 a lot of ppl seem to be "projecting" on this. doubt he had any nefarious intentions.

0celo7

Then he's an idiot. Either way not worthy of the praise.

vzn

@0celo7 alas, a 14-yr old polarizing figure :|

0celo7

What does his age have to do with anything?

Children can be evil and/or stupid.

vzn

@0celo7 lol (ouch!), seems pretty relevant to intrepreting the whole situation. guess youve never raised kids eh? :P

0celo7

You're telling me "he's only 14" is an excuse for not knowing "his" clock looked like a bomb? But he's supposed to be some kind of genius? Give me a break.

You're also ignoring the fact that he showed the clock to a teacher who told him not to show it to anyone else.

So he also doesn't follow teacher's advice.

Yeah, real great kid.

Now leave it be.

vzn

20:56

ok, dude, think there is reason to find all this a tricky situation to say the least, but the excoriation of the kid seems out of line. dodging your (numerous) straw men. havent heard anyone calling him a genius.

Hans de Vries

@TerryBollinger Hi Terry, I missed your message. See you next time!

vzn

sometimes events take on a symbolic/ amplified significance far out of line with their "surface appearance". esp in hot-button political areas.

Terry Bollinger

@HansdeVries Hi Hans! Again, that was some interesting work you did a while back in terms of giving different perspectives on standard items.

vzn

0ce, fyi, more on the facts of the case (wikipedia)

Terry Bollinger

Also, Hans, this is likely more along your fellow-computer-science lines: A lot of this seemingly esoteric stuff about Cliffords and Grassmans and Hypercomplexes oh my really does seem to have some remarkably simple binary interpretations. I did not expect that. I just wanted a cleaner understanding of why spin matrices look so asymmetric. They really are not; it's just bad representation.

Must go now. @0celo7, shame on you, give it a rest. No kid should be arrested like that, and if you sincerely think he deserved that, you really need to control your imagination.

0celo7

21:09

@TerryBollinger Give it a rest? I'm not the one who brought it up today.

0celo7

21:56

Slereah

heheh

0celo7

user685252

Terse

0celo7

22:23

@ACuriousMind physics.stackexchange.com/questions/213676/…

Why don't they form a vector space? :)

Slereah

@0celo7 : Have you ever seen the shortest math paper of all times

$user image$

0celo7

lel

why .-.

who would ever build such a device

bolbteppa

@0celo7 you sound jealous and willing to delve into insane conspiracy theories to justify inadequacy, wtf?

0celo7

What?

bolbteppa

Why would a kid with a nasa t-shirt play with the circuitry of a clock? Shocking, must be a conspiracy

0celo7

22:36

For the love of fuck, why can't people drop this?

skull petrol

f*ck

It's like a meme

bolbteppa

You're the one who is saying he didn't build a clock, by this logic chair-makers don't really make chairs since they didn't make the wood, I'm just responding to this lunacy

0celo7

Lunacy? Block me then.

I really don't give a shit anymore.

bolbteppa

Now that you're called on all this you don't, but 5 minutes ago you had him sussed out

It's not a big deal, relax

0celo7

5 minutes? Like two hours ago.

skull petrol

22:40

Internet time :-)

bolbteppa

Good point

0celo7

I'm tried of people giving this kid more than 15 minutes of fame. Drop it.

bolbteppa

haha

skull petrol

Internet time :-)

bolbteppa

It's not about the kid though, it's about fighting against bigotry

skull petrol

22:40

Asynchronous

0celo7

Wow. You actually said that. Just stop talking.

bolbteppa

I would expect that to be difficult for someone who suspects his motives

0celo7

What part about drop it do you not get?

skull petrol

The "drop" part has no meaning without gravity :P

0celo7

Don't talk to me.

skull petrol

22:46

You know where the ignore button is pal.

0celo7

Yup, done.

I don't care, he's blocked.

skull petrol

That song is a ear worm.

"I don't care"

For psycho chicks :P

0celo7

@ChrisWhite @skullpetrol for every unit that block A above moves, block B moves...half as far?

(above = picture up a bit)

skull petrol

I don't see any "picture up a bit"?

ACuriousMind

@0celo7 That does not work for anyone but you. It instead sends everyone a prompt to login.

0celo7

22:56

ah

well

Oh my picture changed!

here we go

it's actually double

I think

skull petrol

Thanks @ACuriousMind I was wondering what that was.

0celo7

yes, B moves twice as fast/far as A

does anyone agree?

deutschsprachiger Raum

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