The h Bar

Balarka Sen

17:00

@0ßelö7 don't you like horror games

Phase

what does one do to get both diagonal

Balarka Sen

He's probably trying to simultaneously diagonalize A and B

shai horowitz

what does one do to get one diagonal

0ßelö7

@BalarkaSen I played Metro 2033 and it's absolutely horrifying. I really can't into the horror genre as a whole.

I'm so terrified of jump scares

shai horowitz

how do you feel about cabin in the woods

best horror movie

0ßelö7

17:01

never heard of it

I actively avoid that stuff

Emilio Pisanty

@ACuriousMind @dmckee @DavidZ @Qmechanic, maybe get the ball rolling asking the mods at CS and TCS whether this one is migratable to their sites?

0

Q: Solve Travelling salesman problem with light machine

I got a crazy idea how to solve the famous Travelling salesman problem using light machine, and I want your input if such machine can be constructed. It starts by shooting a white(full spectrum) light. Each node(city) will be a device that will absorb part of the light and redirect it to all the...

visible-light

Balarka Sen

bleh I wouldn't call Cabin the best horror movie

@0ßelö7 play one without jumpy scaries then

0ßelö7

@BalarkaSen Such as?

Balarka Sen

surival horror like Last of Us

v good game

shai horowitz

hows 1408? as an actual scary (horror) movie its my favorite

Balarka Sen

17:05

Oh I have heard of that one

Never seen it though

I might try it

Phase

I'm never gonna get an answer :c

0ßelö7

@BalarkaSen isn't that PS4 game?

Phase

I asked on the Linear Algebra room 12 hours ago and it's still barren

0ßelö7

@Phase I have no idea what the doubt is

Phase

what do you mean

Balarka Sen

17:06

@0ßelö7 oyeahprobably

Phase

I'm asking for what the formal method of doing it is

Emilio Pisanty

@Phase what's the question?

Phase

Diagonalising two hermitians A and B that commute, where one [or both] are degenerate

Emilio Pisanty

what about it?

0ßelö7

is this not in Shankar

Phase

17:07

I'm asking what the formal method of doing it is

Emilio Pisanty

diagonalize A, then look at the action of B on each of the eigenspaces of A (which is closed because [A,B]=0) and diagonalize B on each individual eigenspace of A, and you're done

shai horowitz

cant you just define C=AB=BA then diagonalize C using normal methods

Emilio Pisanty

@shaihorowitz that's a terrible method

shai horowitz

im just gonna go with your right, do your method

Phase

How much freedom is there to choose the degenerate parts of the eigenbasis?

if both are degenerate

Emilio Pisanty

17:09

@Phase in the initial part (diagonalize A) you can choose any eigenbasis you want

after that, when you're diagonalizing B, it is crucial that you don't try to diagonalize B directly, but only its action on the A eigenspaces

at which stage you can choose any eigenbasis you want (within the preexisting A eigenspace)

Phase

So

If im not mistaken

which i probably am

If both are finite, like say 2x2 or 3x3

ACuriousMind

@EmilioPisanty Can do. Do you have a reason to believe it's not on-topic at CS though? (We're not really supposed to bother mods unless whether or not the question is in-scope is unclear. If we possibly wouldn't migrate it for quality reasons, just don't migrate)

Emilio Pisanty

@Phase that's too small to be useful

Phase

You could take the eigenvectors in the basis that A is diagonal in, and assemble them into the identity matrix and use B?

Oh

Well I just mean finite matrices

Emilio Pisanty

yeah, but 2x2 won't fit interestingly different degenerate eigenspaces

Phase

17:11

because I'm not sure how decomposition works in infinite dimensional matrices

yeah good point

wait im a bit confused still

Emilio Pisanty

Say, take $A=\begin{pmatrix}0&1&0&0\\ 1& 0&0&0 \\0&0&0&1 \\ 0&0&1&0\end{pmatrix}$

Phase

What do you mean by look at its action, could you just write down a general form for that?

Oh thanks

Emilio Pisanty

and, say

wait

$B=\begin{pmatrix}0&0&1&0\\ 0& 0&0&1 \\1&0&0&0 \\ 0&1&0&0\end{pmatrix}$

god, I hope those commute

Phase

^ physics in a nutshell?

Emilio Pisanty

excellent, they commute

so

first you diagonalize A

and you find two eigenspaces, one spanned by $e_1=(1,1,0,0)$ and $e_2=(0,0,1,1)$ with eigenvalue $+1$, and one spanned by $e_3=(1,-1,0,0)$ and $e_4=(0,0,1,-1)$ with eigenvalue $-1$.

Then you find the action of $B$ on those eigenspaces

so $Be_1=e_2$ and $Be_2=e_1$, and ditto with $Be_3=e_4$ and $Be_4=e_3$,

Phase

17:18

so

Emilio Pisanty

note that $Be_1$ does not involve either $e_3$ or $e_4$, which is crucial

Phase

A's trace would be 1, 1, -1, -1 and so would B's, because it's just switching 1 and 1

and switching -1 and -1

in their diagonalised form i mean

Emilio Pisanty

@Phase that's true but not specifically useful here

Phase

o ok

Emilio Pisanty

So, in the A eigenbasis we just found, the matrix representation of B is

Phase

17:20

if $Be_n = e_m$, does that mean that it's nth eigenvalue is the value of A's mth?

Emilio Pisanty

@Phase no

that's not an eigenvalue relation because $n\neq m$

$[B]_{A\text{ eigenbasis}}=\begin{pmatrix}0&1&0&0\\ 1& 0&0&0 \\0&0&0&1 \\ 0&0&1&0\end{pmatrix}$

^ that just encodes this chat.stackexchange.com/transcript/message/40356181#40356181

Phase

O

Emilio Pisanty

hmmm, wait

Phase

if only one is degenerate

dmckee

:40354732 That message was causing strange layout artifacts on my machine, so I have deleted it for the nonce. If this was unique to me I can restore it, but it seemed to be a problem.

Phase

17:22

Why couldn't you just use the matrix of eigenvectors of A, sandwiched with B in the middle

and then choose the eigenbasis for the degenerate part

^given A is non-degen, and B is degen

Emilio Pisanty

@dmckee can you reply to the message just below? what message was this?

Student404Mus

Hi people

Emilio Pisanty

@Phase the non-degenerate case is a strict special case of the degenerate case and it does not require additional attention once you're doing the general situation

dmckee

@Secret Reply to the post below the one I killed. as requested by @EmilioPisanty.

Student404Mus

I have a question

dmckee

17:24

The artifact was a huge vertical blank space that didn't allow any further posts to be displayed.

Emilio Pisanty

more specifically, if a given eigenvalue of A is nondegenerate then you'll get a 1x1 block in the B matrix over that eigenbasis, which you then proceed to diagonalize (by doing nothing)

@dmckee ¯\ _(ツ)_/¯ no idea

@Student404Mus don't ask about asking, just ask

(we should choose that as the h bar motto)

4

Student404Mus

Hhhhhhm... What are the 1 St concepts I should learn in QFT

dmckee

@Student404Mus What is "1 St" in that comment?

0ßelö7

First

@Student404Mus QFT is a lie.

Student404Mus

😊

0ßelö7

17:26

@Slereah can confirm

Student404Mus

What?

0ßelö7

@ACuriousMind might argue but will eventually agree

ACuriousMind

@dmckee I think it's supposed to mean "first"?

Secret

@dmckee O yes, that message of Slereah's somehow infinitely stretch the screen when the latex is rendered, but it display correctly on the main

Student404Mus

I mean fist

First

Emilio Pisanty

17:28

@Phase that's incomprehensible to me

Phase

So like

take the matrix of eigenvectors of A, like you do when you just diagonalise A

and call it V

Student404Mus

I never toke any knowledge In relativistic QM

dmckee

You need to know (0a) enough classical mechanics to know what a Lagrangian is and how to use it; (0b) enough electricity and magnetism to be able to write the energy in an electromagnetic wave; (1) some quantum mechanics; (2) the ladder-operator solution to the harmonic oscillator in quantum mechanics; (3) probably a few other things, but you can pick them up as you need them.

Phase

why cant you use $V^{-1}BV$, being aware that it's degenerate with some values and choosing those afterwardS?

Emilio Pisanty

@Phase you can

but

it will have a block structure

0ßelö7

17:30

@dmckee Lie theory,

Phase

wait

Emilio Pisanty

like my last matrix above

which is exactly that

Phase

I just went back and looked at your B in A eigenbasis again

Emilio Pisanty

and

dmckee

@0ßelö7 Oohhh, yes. That would be good.

0ßelö7

17:30

You need to know that beforehand because no physicist is able to explain it.

And no mathematician cares about the Lorentz group

So it's quite the conundrum

Emilio Pisanty

if you neglect that block structure

then you'll be left with an eigenbasis of B that's not an eigenbasis of A

Phase

Why isn't B diagonal in it's eigenbasis if it shares eigenvectors with A?

Emilio Pisanty

so, for instance

@Phase [A,B]=0 guarantees that they share an eigenbasis, not that they share every eigenbasis

so, say, take $\tilde B =[B]_{A\text{ eigenbasis}}=V^{-1}BV=\begin{pmatrix}0&1&0&0\\ 1& 0&0&0 \\0&0&0&1 \\ 0&0&1&0\end{pmatrix}$

Phase

But how come B doesn't even have a single element in it's diagonal? Is that just because every eigenvalue is at least twofold degenerate?

Emilio Pisanty

@Phase because you haven't diagonalized it yet

"have elements in its diagonal" is a meaningless criterion

Phase

17:32

._. I'm not cut out for this

Emilio Pisanty

it's either diagonal (or block diagonal) or not

Here's the thing

Semiclassical

there is at least one fact you can infer from the diagonal elements being zero: the trace of the matrix is zero, so the sum of the eigenvalues is zero.

Emilio Pisanty

take $\tilde B =[B]_{A\text{ eigenbasis}}=V^{-1}BV=\begin{pmatrix}0&1&0&0\\ 1& 0&0&0 \\0&0&0&1 \\ 0&0&1&0\end{pmatrix}$ as above and say that you want to diagonalize it. You might say "hey, $(1,1,0,0)$ looks like a handy eigenvector. Or you might say "hey $(1,1,1,1)$ looks like a handy eigenvector".

do you see the difference between the two?

Semiclassical

and similarly the determinant is 1, so the product of the eigenvalues is 1. that sets some pretty strong constraints on it. that'd allow eigenvalues like 1,1,-1,-1; but, it'd also allow 2,-2, 1/2, -1/2. (that doesn't happen here, but the facts i've quoted aren't enough to prevent that)

Emilio Pisanty

keeping in mind that in that basis $\tilde A =[A]_{A\text{ eigenbasis}}=V^{-1}AV=\begin{pmatrix}1&0&0&0\\ 0& 1&0&0 \\0&0&-1&0 \\ 0&0&0&-1\end{pmatrix}$

$(1,1,0,0)$ stays inside the upper 2x2 block and it is therefore guaranteed to be an eigenvector of $A$. $(1,1,1,1)$ breaks out of that block and it is therefore not an eigenvector of $A$.

this is why it is crucial that you diagonalize $B$, on the $A$ eigenbasis, on a block-by-block (i.e. eigenspace by eigenspace) basis

Semiclassical

17:37

@EmilioPisanty that's in the Math chat room description

so it's not a bad idea (though it gets ignored all the time)

Phase

i see

so because in a twofold degenerate eigenspace, any two orthogonal vectors in it are eigenvectors right? So you can break it into two 2x2 matrices each representing the eigenspace?

*an eigenspace

Emilio Pisanty

@Phase break what?

Phase

Break the 4x4 matrix into the two 2x2 matrices

Emilio Pisanty

but yes, any vector in an eigenspace is an eigenvector

Phase

and ignore the 2 2x2 matrices with zeros in all the elements

yeah but I mean when it comes to choosing what the eigenvectors are

Emilio Pisanty

17:40

@Phase No, that's not trivial, and you strictly require [A,B]=0 for that to be possible

Semiclassical

ignore, misread you

Phase

?

Oh ok

im confused

Emilio what is your comment in response to?

Emilio Pisanty

@Student404Mus you want to take this abstrusegoose.com/272 from step 3 onwards

@Phase the fact that the representation of B in the A eigenbasis is block-diagonal

Phase

Sorry im not familiar with the term block-diagonal

Emilio Pisanty

i.e. that it can be broken down into the direct sum of two 2x2 blocks

Phase

17:42

oh right

Well

Semiclassical

@EmilioPisanty oh man, that's great

Phase

huh

ok thanks emilio

Emilio Pisanty

@Phase $\begin{pmatrix}P& 0 &0 & \cdots\\ 0& Q & 0 & \cdots \\ 0& 0&R&\cdots \\ \vdots & \vdots & \vdots & \end{pmatrix}$ where $P,Q,R$ are submatrices and the zeros are rectangular matrices of zeroes

Semiclassical

for a physics grad student it's more like "well, i've got half the books on my table. but do i really want to do the other half?"

ACuriousMind

@EmilioPisanty I still love the "Witten to English Dictonary"

3

Emilio Pisanty

17:43

@ACuriousMind yup

Phase

wait how would you expand that

Emilio Pisanty

for some reason it always takes me ten minutes to find that one every time I want to cite it

Semiclassical

reminds me, there's a paper of his I still need to read

Phase

Is A the whole matrix, and then B is the whole matrix minus A's column and row and so on

Emilio Pisanty

@Phase exactly like the B representation on A in the original example

Semiclassical

17:46

I'm interested in Lefschetz thimble stuff

and Witten has some good stuff on that in one of his papers

Alex K Chen

Kip Thorne's brilliant interview after winning the Physics Nobel of 2017 !

Emilio Pisanty

@AlexKChen ah. well played.

Alex K Chen

Yeah, isn't it very interesting ?

dmckee

@EmilioPisanty Hey. I forgot to list Fourier analysis in the prerequisites. I guess that should be (-1a).

0ßelö7

@dmckee not to be too pratronizing but what do you even need? Definition and formal manipulations?

vzn

18:00

@Semiclassical ... solitons ... =D

Alex K Chen

Hi @vzn

vzn

@AlexKChen hey

Alex K Chen

Is the book by Klepneer and it's exercises good for learning basic Newtonian mechanics ?

Anonymous

@dmckee I have a C question: When I use scanf("%3d %3d %3d",&a,&b,&c); here why does b get the value 4 and not 456? Using %3d %3d %3d should extract non-whitespace consecutive (3) characters and place them in the variables a, b and c respectively, isn't it ? Any ideas?

Anonymous

18:38

Ok, I got it now. "three consecutive non-whitespace characters"

Anonymous

"3" is the maximum field width here

David Z

18:55

Thoughts on whether this question is clear enough to be reopened?

0

Q: Detecting light path

Is it possible to build a machine that will be able to detect a path chosen by a ray of light? My idea is to created stations along the path of the light, each station is a device that absorb part of the light and redirect it to all the other stations. My goal is to detect when the light have ...

visible-light computational-physics optimization

@Blue yeah, when you provide a field width, that's only the maximum number of characters it will read. It still stops on spaces etc.

I didn't think using field widths with scanf was very common but I'm not sure

Icemybread

@AlexKChen flagged

Anonymous

@DavidZ Yup, got it!

Anonymous

@DavidZ I found it in a book. I too have never seen real code using field widths for scanf before.

David Z

@Icemybread don't say you're flagging things. If you're going to flag a message, just flag it, there's no need to announce it.

@Blue yeah it sounds like the kind of thing that would appear in a book and nowhere else

0ßelö7

@BalarkaSen Is Hirzebruch's book on algebraic geometry good? I was just handed it

Balarka Sen

19:09

i heard of it a few times

too heavy 4 me

You mean topological methods in algebr geom right?

0ßelö7

19:21

@BalarkaSen Yeah

My advisor says that Hirzebruch explains what the mysterious A-roof genus is

@Semiclassical but things change based on us observing them...

dmckee

@0ßelö7 Well, what you really need is the understanding that you can form arbitrary(*) functions by summing over the basis functions, and that there is a methodical approach to finding the coefficients of the sum for any desired result.

(*) For 'physicist' values of arbitrary, of course.

0ßelö7

@dmckee you know me too well

@BalarkaSen I'm about to give Talk 1 a run and see how long it takes me. Do you want to Skype

Balarka Sen

19:40

@0ßelö7 i don't skype. use a rubberduck

0ßelö7

"Here, B is any so-called Bott manifold i.e. a simply connected
8-dimensional spin manifold"

Wonderful

Balarka Sen

wott manifold

0ßelö7

@BalarkaSen What's the bordism group of $n$-dimensional spin manifolds

I'm in algebraic topology hell, pls halp

Balarka Sen

Don't think there's a simple description

0ßelö7

what the hell does any of this mean

Balarka Sen

19:50

Oh, right there's a Pontryagin Thom theorem for cobordism groups with additional structure

0ßelö7

How would I even go about finding out what this stuff is

Balarka Sen

shrug

0ßelö7

Are you being unhelpful on purpose?

Balarka Sen

nopeee

0ßelö7

So you don't know what it is either?

Balarka Sen

19:52

I don't know what the general form of Pontryagin-Thom is, no, but I have definitely seen it flying around

Maybe check out tom Dieck

0ßelö7

@BalarkaSen dam son

I'm in the library right now

this book looks lit

Balarka Sen

tom Dieck?

0ßelö7

yeah

Balarka Sen

I liked that book

There's a chapter on Thom spectra iirc so that might be useful

0ßelö7

thx

Icemybread

19:54

@EmilioPisanty 7: me irl

Balarka Sen

np

0ßelö7

I have 3 weeks to learn that stuff

and write it into the talk

and somehow not seem like an idiot

...

Balarka Sen

Teach me after you learn it

0ßelö7

@BalarkaSen kk. Really I just need enough to understand the statement of the theorem lol

I'm trying to find the state of the art on obstruction theory for positive scalar curvature metrics

it's a homotopy theoretic mess

Balarka Sen

unsurprisingly

dem homotopy theorists everywhere

0ßelö7

19:57

@BalarkaSen My advisor pulled that stunt again

Gave me a paper that he definitely does not understand

gave me a false sense of hope

Slereah

Who you calling a homotopy theorist

Wrichik Basu

20:29

@ACuriousMind or @DavidZ or @dmckee or @Qmechanic would you mind paying a bit of attention to this question? Under the answer by user171097, there are quite a lot of comments that are more like quarrels and will not be useful to future users. Please delete them as necessary. Requesting here as it's not possible to flag every comment.

ACuriousMind

@WrichikBasu Just flag one of the comments in such cases, we'll deal with it

Wrichik Basu

-2

Q: Why centripetal force does not pull the object towards the center?

I know that the force is used to change the direction of the velocity but my question is while it has 3 options either to pull the object towards the center or change the direction of velocity or both. It would have been more appropriate if it did both.

newtonian-mechanics rotational-dynamics centripetal-force

dmckee

@WrichikBasu The usual thing is to flag one comment with a custom reason and explain that there is a mess.

Wrichik Basu

@dmckee should I do it now, or you'll look to it?

I'll keep that in mind from next time.

dmckee

We're looking at it now.

Wrichik Basu

20:31

@dmckee ok. :-)

lılostafa

:\

nasil

how can i find the function of a string that is between two points

Anonymous

@nasil What?

Anonymous

What string?

nasil

you take a rope

Anonymous

20:42

@nasil Ok, then?

Anonymous

What function are you looking for ?

nasil

take arbitary two points on that line

then pin those points on some arbitrary points in 2d space

Anonymous

So, you're looking for standing waves on a string.

Anonymous

Plenty of information about that stuff on the net

Anonymous

Just look around a bit

nasil

20:45

i am looking for the right word

and that is not what i am looking for

Anonymous

hyperphysics.phy-astr.gsu.edu/hbase/Waves/string.html

Anonymous

Start from here

nasil

lets make up two cordinates

y and x

x horizontal

and is f(x)

every x y pair represents a point on the rope

David Z

@dmckee cc @WrichikBasu FWIW I usually look at the full comment thread even if there's a standard flag on just one comment.

nasil

@Blue did you understand the question?

Anonymous

20:47

Standing wave

In physics, a standing wave – also known as a stationary wave – is a wave which at each point in its medium has a constant amplitude. The amplitude of the wave's oscillations may vary at different points in space, but are constant in time. The locations at which the amplitude is minimum are called nodes, and the locations where the amplitude is maximum are called antinodes. Standing waves were first noticed by Michael Faraday in 1831. Faraday observed standing waves on the surface of a liquid in a vibrating container. Franz Melde coined the term "standing wave" (German: stehende Welle or Stehwelle...

nasil

length of the rope is given

its not a standing wave

it is just a stationary rope

David Z

That sounds like the catenary problem

nasil

thanks a bunch

Anonymous

Oh. That should be hyperbolic.

Anonymous

So you were not looking for waves

nasil

20:49

nope

Anonymous

Catenary

In physics and geometry, a catenary (US: , UK: ) is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends. The catenary curve has a U-like shape, superficially similar in appearance to a parabola, but it is not a parabola. The curve appears in the design of certain types of arches and as a cross section of the catenoid—the shape assumed by a soap film bounded by two parallel circular rings. The catenary is also called the alysoid, chainette, or, particularly in the materials sciences, funicular. Mathematically, the catenary curve is the...

Anonymous

"Mathematically, the catenary curve is the graph of the hyperbolic cosine function. "

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