The h Bar

Danu

12:00 AM

I want to convert to $T_z\mathbb C$-notation

What is the appropriate expression for $X(z)$ in this notation?

NeuroFuzzy

Oh yep I didn't read that far back :D

Danu

I guess I can piece it together from the "real coordinates" expression

$\partial_z=\partial_x+i\partial_y$ and take it from there

So something like

$-y\partial_x=-(z-\bar z)/2 (\partial_z+\partial_{\bar z})/2$

So we get

NeuroFuzzy

Although I really don't understand... don't you have to make a choice of what isomorphism to use? Why isn't it $iz \partial_z$? Again, you're probably way ahead of me. I'm just typing out words.

Danu

$X(z)=-\frac{z-\bar z}{4}(\partial_z+\partial_{\bar z})+\frac{z+\bar z}{4}(\partial_z-\partial_{\bar z})$

@NeuroFuzzy This is the start of my thought process too (when I first asked the question). I think I have the solution now

...and understand why the notes only give the "real" expression and not the complex one

because it is kind of non-obvious

Rewriting: $$X(z)=\frac{\bar z}{2}\partial_z+\frac{z}{2}\partial_{\bar z}$$

much neater

:)

Actually, preferable over the "real" expression because of the nice symmetry. I'll edit this into the notes :)

NeuroFuzzy

@Danu I don't understand at all... if I have something in the tangent space of R (isomorphic to R) then, if the vector field is X(x)=x^2 component-wise, shouldn't the vector be x^2 \partial_x?

bolbteppa

12:09 AM

Maybe what it means by saying that $\mathbb{C} \cong T_{z} \mathbb{C}$ is that the space $\mathbb{C}$ (generated by the 'basis vectors' $1 = 1 + oi$?) & is isomorphic to the space $T_{z} \mathbb{C}$ generated by the basis vector $\partial_{z}$ in the sense that $T_{z} \mathbb{C} = f(z,\bar{z}) \partial_{z} + 0 \partial_{\bar{z}}$ so instead of writing $X(z) = iz = iz \cdot 1 $ you should write it as $f(z,\bar{z}) = X(z) \partial_{z} + 0 \cdot \partial_{\bar{z}}$?

Danu

Guys, my solution above is probably correct, no? It looks too good not to be true :P

@bolbteppa I don't think so, because straight-up translation of the "Real" expression gives my version

...which does not agree

lol that typo

@bolbteppa it's \cong by the way

for isomorphisms

bolbteppa

Well, on general grounds what you seem to be doing is taking a holomorphic vector field and randomly expressing it in terms of holomorphic and non-holomorphic parts for no reason... The meaning of $\mathbb{C} \cong T_{z} \mathbb{C}$ is to say that $\mathbb{C}$ is isomorphic to the tangent space w.r.t. the $z$ variable on $\mathbb{C}$, but we know the tangent space on $\mathbb{C}$ is in general generated by the $\{\partial_{z},\partial_{\bar{z}} \}$ basis.

Danu

So your proposal is what exactly? $X(z)=$?

I'd also like to understand what's wrong with the expression I got

$$X(z)=-y\partial_x+x\partial_y=\frac{\bar z}{2}\partial_z+\frac{z}{2}\partial_{\bar z}$$

NeuroFuzzy

$$X(z)=-y\partial_x+x\partial_y$$ is given to you?

with X(z)=iz$ in its non- tangent space form?

Danu

12:24 AM

Yes, that part is for sure

(and it makes sense, as per DanielSank's initial explanation)

because $iz=-y+ix$

bolbteppa

I am focusing on the fact you have written $\mathbb{C} \cong T_{z} \mathbb{C}$, I'm pretty sure this comes from the fact that in general you would have $T_{z,\bar{z}} \mathbb{C} \cong \mathbb{C}^2$, but because $X(z,\bar{z}) = X(z)$ you have $X(z,\bar{z}) = iz \partial_{z} + 0 \cdot \partial_{\bar{z}}$ i.e. $X(z) = iz \partial_{z}$

Danu

The main issue here is that I do not know anything about complex manifolds

@bolbteppa Okay, so you're saying; Simply put the $X(z)$ before the $\partial_z$ and it's done.

I'm willing to accept that; but how do you reconcile it with the expression in real coordinates?

(because it's not the same, right?!)

bolbteppa

It's basically the same as saying that $X(x,y) = (-y,x)$ is just $X(x,y) = -y \partial_{x} + x\partial_y$

Danu

Sure, like I said I'm willing to accept it

bolbteppa

Only using $X(z,\bar{z}) = (iz,0)$

Danu

12:28 AM

But how do you reconcile the two?

NeuroFuzzy

Maybe it's just the choice of isomorphism?

Danu

$$-y\partial_x+x\partial_y\neq iz\partial_z$$

bolbteppa

What notes are you reading?

Danu

Notes on my course at LMU

I'm editing them

When I finish, I'm moving on to the notes on Riemannian geometry

and by then, there will be new courses with new notes ^^

Anyways, time for me to sign off for now

Hope you guys have some kind of explanation!

bolbteppa

Okay I think I see where you went wrong

Danu

12:36 AM

ah

drumroll

bolbteppa

Wait, where did this $-y \partial_x + x \partial_y$ come from?

Danu

It's da truth!

That is actually in my notes; and one can "derive" it from $iz=-y+ix=-y\hat x+x\hat y$

In the same way, it is tempting (and possibly correct) to say $iz=iz\hat z$

...but how to reconciiiiile

bolbteppa

But what about $iz \partial_z = i(x + iy)[(1/2)(\partial_x - i \partial_y)] = ...$?

Danu

that's exactly the problem

that I don't see how to resolve

Also did you mix up a sign there?

bolbteppa

No? math.stackexchange.com/questions/970540/…

Danu

12:41 AM

Oh, of course, my bad

I think I have some sign errors then

aaaand they resolve everything... maybe?

Meh, I still don't see how this works

@bolbteppa This yields the redundant terms $ix\partial_x+iy\partial_y$ right

...and the factor 1/2 is scary

really gotta go sleep now

I'll pray for waking up with a solution posted here ;D

gnight

bolbteppa

Well, just quickly

My thinking is they probably mean that the vector space $\mathbb{C}$ over the scalar field $\mathbb{C}$ generated by the basis vector $1$ is isomorphic to the vector space $T_z \mathbb{C}$ over $\mathbb{C}$ generated by the basis vector $\partial_z$ and that's it

Danu

I still don't see how this jives with the other expression

I'll ask a question about it on Mathematics

...unless @ACuriousMind can resolve it, of course ;)

or someone else...

Sean

Question: What's the general policy for approving edits to closed questions?

bolbteppa

Ohhh

ACuriousMind

@Sean I don't think we have one :P

Sean

12:54 AM

Ok

because I was looking at a suggested edit for a question that had been closed as homework

My guess is the question might have been written by a non-native English speaker, and so the edit did improve the readability somewhat

but didn't really do anything that would have caused the question to be reopened

So i wasn't sure whether it was worth approving the edit, and the guy getting +2 rep for editing a terrible question, that would likely stay terrible

ACuriousMind

Well...I guess I would reject it if it doesn't address the reason for closure

Sean

sometimes i feel like i'm a tougher reviewer than others haha

obe

@FenderLesPaul @ACuriousMind Hey can you help me out with some classical mechanics?

Look at this from arnold ch1.

Can you explain this intuitively?

so parallel transport on the affine space defines a vector space $R^4$ and if that is mapped to $R^1$ it defines time?

ACuriousMind

1:16 AM

@obe That linear map is essentially just a projection - it singles out one direction in the $\mathbb{R}^4$ as time.

bolbteppa

@Danu this is driving me crazy, wtf does it all mean

0celo7

3:21 AM

@FenderLesPaul

skill patrol

3:40 AM

Don't be such a "bandwagon" fan @0celo7

A double overtime lose is nothing to be ashamed of pal.

Triple H - The Game Lyrics

►

Stan Shunpike

4:03 AM

forbes.com/sites/nataliesportelli/2015/09/10/…

Apparently even Nobel Laureates must take the SAT

@ACuriousMind Would this happen in Germany? lol Just curious how the US compares on this sort of thing.

0celo7

4:31 AM

@skillpatrol I'm not ashamed

just sad

@StanShunpike I don't see the issue

everyone has to take it

Huy

7:01 AM

@0celo7: Is that a good or a bad thing?

Danu

8:58 AM

@StanShunpike Her getting the Nobel Prize was so ridiculous :P

Slereah

9:24 AM

Hello

Danu

Hi

skill patrol

9:58 AM

hi

Danu

11:15 AM

@bolbteppa

0

Q: How do I express this vector field, given in terms of an element of elements of $\mathbb{C}$, in the standard notation for tangent spaces?

In a set of lecture notes (not available online) that I'm currently working through, one is given the following set-up: Consider rotational vector fields on the plane $\mathbb{R}^{2}\cong\mathbb{C}$ with coordinates $(x,y)\cong z=x+\imath y$. Let $X$ be the vector field given by $...

differential-geometry

ACuriousMind

@StanShunpike Well...it couldn't for the simple reason our "end-of-school tests" aren't really standardized

Danu

They're not? Wow!

ACuriousMind

They became more standardized in recent years by something called Zentralabitur, but it's still quite far from everybody taking the same exam regardless of school.

@Danu Closest I can find is this

Danu

@ACuriousMind Oh well

I guess it was kind of an unnecessary rant anyways ^^

ACuriousMind

Well, I find the existence of global objects kinda important :D

Also, many people are confused about this, as the recent "prove this is a tensor"-question shows.

Slereah

11:27 AM

Well a symmetrized tensor is a linear combination of tensors

Hence a tensor :p

Danu

@ACuriousMind I know!

Especially since the abstract object is, in some sense, a lot simpler

The transformation rule is kinda wtf when first encountered

I hope someone answers my question about that vector field

Secret

11:57 AM

Slereah

whaaat

Secret

something curious when trying to mark out points when I am evaluating a function iteratively based on the defintion of this lecture notes

It seems fixed points somehow act like stationary points in calculus

Danu

lmao

You have a unique way of thinking :P

ACuriousMind

I have no idea what that picture is supposed to convey

Slereah

Why is the Google in chinese

Secret

12:00 PM

(Because my comp is in chinese)

because if your value deviates slightly from that of the fixed, point, then after iteratively applying the functions, it slips alway or towards said fixed points

This could be seen for the example $y=x^2$ and $y^{(k)}=(x^2)^k$

If x=1, it stays 1 as k increases
simialrly if x=0

If 0<x<1, then after iterating, the values tends to zero as k increases

If x>1 the value tends to infinity at a rate of x^2

Rigor

are you asian?

Secret

My birth country is HK

but I am australian

Rigor

sorta like Terry Tao :-)

Slereah

Is he the Tao wormhole guy

Or is that a different Tao guy

Secret

(cont.) So by staring at those values, it seems if you plot the results against k, it seemed to form some kind of weird surface which it slips towards infinity for x>1 and slips towards zero if 0<x<1, and only stays put at the fixed points x=0,1,-1

Danu

12:03 PM

Tao as in famous mathematician

Rigor

^

Danu

PDE's & number theory

also childhood genius

Secret

(cont.) This is why it gives me a feeling it resemble stationary points in calculus

But I need to check more to see if my hypothesis holds

Rigor

he was doing algebra in grade 2

ACuriousMind

@Secret Fixed point have not much to do with stationary points - for one, a fixed point has no condition on the derivative - e.g. $f(x) = x$ has all points as fixed points.

Slereah

12:06 PM

Neeerd

Hm, what was I doing in grade 2

Grade 2 is like, 9 years old?

Rigor

7 years old

Slereah

Hm

Don't remember too much

That was when I got my first console

Game boy

With the zelda game

Rigor

zelda has become the game of war girl

:P

Secret

@ACuriousMind I know, what I mean is that it behaves in an analogous way, at least for y=sin and y=x^2 applied iteratively

If we have some kind of mathematical tool that is analogous to a derivative, but it measures how much the value of an iterative function changes with k, then by the above observation you would get zero if x is a fixed point (because it stays at that value no matter how many iterations you apply on that value), but will be some nonzero value for other x values

For the case of y=x^2, it seems this change will be (something that tends to zero as k increases) for 0<x<1 an…

Danu

Iteration is not continuous

Secret

12:15 PM

But then how to describe those change in values of y with number of iterations I saw from that experiment above?

They seemed to change in some kind of systematic way at least for y=sin x and y=x^2

Danu

Why do you insist on drawing an analogy with derivatives

It's not a very complex phenomenon, right

You can just understand it directly

Don't let this discourage you from experimenting though

I guess searching for connections like this is a good thing

Rigor

he who is taught the best is told the least :P

Secret

I am not sure if derivatives is the right ananlogy

Ok
What I am observed is the following
for the example $y=x^2$
if 0<x<1, repeated application of y will bring the value close to zero which "slows down" the higher the no. of iterations
But for x>1 and x<-1, repeated application of y brings it towards +infinity and -infinity respectively
for x=0,1,-1, the values remained unchanged no matter how many times you apply y onto it

Putting the above results together, it seems there's a set of x that "slips away" from x=1 and x=-1 at increasing rate with the no. of iteration k, and then there's a…

Danu

Obviously

But the pattern is very simple

It's obvious what's going on

I don't understand what you are searching behind it

What could be an interesting question is investigating what conditions on a function $f:\mathbb{R}\to\mathbb{R}$ guarantee that, in the limit of applying it $k\to\infty$ times, there are only finitely many outcomes.

ACuriousMind

12:34 PM

Am I getting more grumpy or are the questions getting worse?

Rigor

Perhaps the two are related?

ACuriousMind

Why would me getting more grumpy lead to worse questions?

Rigor

vice versa

ACuriousMind

Ah

Well, that is possible

Secret

1:01 PM

Example 1 $y=x^2$
0<x<1 converges towards fixed point x=0
x>1 diverges to infinity

-ve values first mapped to +ve values, and then follow the convergence trends above

Comments: "Looks like some kind of surface"

Example 2 $y=\sin(x)$

$x\in [-1,1]$ converges towards fixed point x=0

Any x beyond this interval is first mapped into this interval ,then exhibit this convergence pattern

Danu

Again, I don't think this is as profound an observation as you seem to hope.

ACuriousMind

I don't understand what you are on about

Stuff converges to attractive fixed points. Is this somehow surprising?

Danu

^

Rigor

^^

Secret

Example 3 $y=\cos(x)$

$x\in[-1,1]$ converges towards 0.73908513321

Any x beyond first got mapped into this interval then converges to that

Given I have little background on dynamical systems, it will seemed kinda surprising to me

(Still need to try out $y=e^x$ because I think it will tell me it is nothing really special)

But right now, the 3 experiments I have tried so far seemed to suggest things roll towards or away from fixed points, as if they are lying on some kind of surface where the height is the number of iterations

ACuriousMind

1:15 PM

Attractors have an attractive range within which everything moves towards them - there are not many possibilities when something can do - it can move towards an attractor, or it can move away from it. If it wouldn't move away from it but stay in a convex, bounded region, then the fixed point theorem would mean a presence of another fixed point in that region, towards which it then should move

@Secret There simply aren't any other things the starting points can do - either they move toward an attractor/fixed point, or they have to always move away from them.

There also might be occurences of cycles, where the iteration does not converge or diverge at all, but these should be rare.

Secret

But is there anything that governs how fast or slow something move towards or away from the attractor/repellers?

Because in the sin(x) example, as they inch closer to the attractor (the fixed point x=0), they seemed to slow down the closer they are to the attractors, as if by analogy, that area is "quite flat" thus the points don't move much

is there something that describe this "flatness" I have in mind?

ACuriousMind

I think that'd be the discrete version of the Lyapunov exponent.

Danu

@Qmechanic Thanks for your response on my question ;)

ACuriousMind

Or something related to that - there is a whole theory of attractors

Danu

@Secret Sine and cosine are the same experiment

yuggib

1:24 PM

@ACuriousMind It's a while now...good questions can be counted, in my opinion, on one hand per month

Danu

@yuggib That's too strict.

The questions by the hard-core of users on this site probably already net those 5

and other people ask good questions too, on occasion.

yuggib

@Danu In my opinion the good questions are quite rare

Qmechanic

@Danu : You're welcome. Btw, there still seems to be a factor half missing in your $\partial_z=\partial_x-\imath\partial_y$.

Danu

@Qmechanic Yes. Hah.

yuggib

and there is always the long-lasting problem of almost no research-level questions

Danu

1:27 PM

@yuggib What about the one I recently asked? :D

yuggib

I agree with @ACuriousMind and his grumpiness

Danu

What is the link to the homework FAQ?

ACuriousMind

57

Q: How do I ask homework questions on Physics Stack Exchange?

What is the policy on asking homework questions on Physics Stack Exchange? What kinds of questions are considered homework questions? Are homework questions allowed? What should I include in a homework question? Why don't you provide a complete answer to homework questions?

discussion faq homework

Danu

Thanks.

ACuriousMind

@yuggib Not sure if someone agreeing with it makes me more or less grumpy ;)

yuggib

1:29 PM

@Danu The maths one? Is not on physics :P

@ACuriousMind at least makes your grumpiness feel less alone :D

Secret

@ACuriousMind Hmm, I should have a look at that concept you mentioned, I think it will be very soon because it might be covered in some form in those numerical analysis lecture notes I need to read for my honours project

@Danu I know that $\sin(x)=\cos(\pi/2-x)$
However as shown from my excel calculations, the convergence behavior is very different

The sin case converges smoothly towards the attractor x=0
While the cos case it converges with a asymptotically decreasing oscillation towards the fixed point x=0.73908513321...

Danu

22

Q: How does one correctly interpret the behavior of the heat capacity of a charged black hole?

Consider the Reissner-Nordström metric $$ds^2=-f(r)dt^2+f^{-1}(r)dr^2-r^2d\Omega^2\hspace{2cm} f(r)=1-\frac{2M}{r}+\frac{M^2q^2}{r^2}$$ where I defined the charge-to-mass ratio $q:=Q/M$, which satisfies $|q|\leq 1$ to avoid naked singularities. As is well known, there are two horizons (or only a...

general-relativity black-holes hawking-radiation qft-in-curved-spacetime black-hole-thermodynamics

^@yuggib

yuggib

Ah ok, that one

Danu

@yuggib Ah, you were thinking about my vector fields question. That one was just sigh confused, and not very interesting at all, nor about physics, I agree ;)

yuggib

@Danu Yes I was thinking about that...the other one is almost two months old, so not supporting much your thesis

ACuriousMind

1:32 PM

@Danu In this case, not really, because e.g. $f(x) = x$ has infinitely many fixed points, but $f(x) = x + 1$ has none.

yuggib

(even if it is one of the interesting ones)

Danu

@yuggib But I'm only one of the users in this chatroom. Taken together, I think we all contribute something like maybe 5 a month?

@ACuriousMind But that's not periodic

Obviously, if all $x\in\mathbb R$ are attracted to the same point by repeatedly iterating a periodic function, then the same holds for the shifted periodic function.

yuggib

@Danu maybe...not sure (we can ask the "PSE statisticians")

ACuriousMind

@Danu Hmm...I've asked 6 questions in 15 months, and one of them is a self Q&A.

Danu

@ACuriousMind Self-answer still counts!

ACuriousMind

1:37 PM

I don't see us getting to 5/month :P

Danu

I've asked 20 in 29 months

Not all were good, I'll grant you that.

ACuriousMind

Why is @0celo7 so dark?

Danu

(most weren't)

He's sad they lost the football game?

yuggib

He should be used to that...SEC is tough

0celo7

ACuriousMind

1:38 PM

But now he looks like @TheDarkSide, this is confusing :P

0celo7

THAT WAS NUTS

Oh my god that was so American

Danu

Yeah @0celo7 go on, change your gravatar pliz

0celo7

"tribalism" does not even begin to cover it

I think I have tinnitus

Rigor

do you still have your voice?

0celo7

Haven't tried speaking

Rigor

1:40 PM

:'(

0celo7

Seriously, my ears hurt

@Danu to what

it was time for a change anyway

Rigor

wear earplugs next time

Danu

To something nuclear stuff whatever related

A picture of a power plant? Their shapes are neat

0celo7

dude I want to work in fusion, not fission

Rigor

Rocky Marciano Jr. reacts to Floyd Mayweather tying father's 49-0 record

Danu

1:41 PM

@0celo7 Then I know just the right picture

0celo7

that's not square

Danu

also that stuff is like the shittiest physics ever, just sayin' :D

0celo7

ok no fusion batteries for you

that leaves more for me

Danu

Magnetohydrodynamics, not even once

0celo7

yeah donovan told me I need to become an expert on that

HDE 226868

1:44 PM

Quick question, regarding this comment:

Danu

I guess that, as an engineer, you won't be doing much physics anyways ;)

0celo7

Danu

Building these tokamaks would be pretty fun, I guess :D

HDE 226868

This makes no sense. No organism, or any other device, can use thermal energy as an energy source. It is thermodynamically impossible to convert heat to other forms of energy, because it would reduce total entropy. — Roland 46 mins ago

My response is that the system in question is not a closed system; energy is being put into it. Is that correct?

Danu

@0celo7 Also decent

0celo7

1:45 PM

looks suspiciously like that twistor thing

Danu

googling "JET tokamak" gives you lots of nice pics

0celo7

I think a twistor formulation of nuclear fusion is not so far fetched

Danu

lel

goforit.jpg

@HDE226868 Yeah

His comment is total nonsense

Do you even cold-blooded animals, bro?

Secret

http://i.stack.imgur.com/c2TXF.png

This is what I had in mind, Acuriousmind suggest a discrete version of the Lyapunov exponent might be what measures the "flatness" of this surface

0celo7

ugh now my eye is broken

losing football games is physically detrimental

HDE 226868

1:46 PM

@Danu Thanks.

@Danu Cool picture.

Danu

@Secret I think everybody already got your analogy

Nice pictures :D

Nice pictures everywhere

except in my line of work :D

Secret

@Danu is this a helpful analogy for me to have when studying numerical methods and dynamical systems?

0celo7

@Danu what does that even mean

Danu

@Secret As I said before, I don't regard it as particularly useful because I don't find it surprising. But if you find it helpful, why not?

@0celo7 A close friend of mine wrote his BSc. thesis on it

he worked at a Dutch institute for fusion

He told me it's hell

mostly computer programming

Oh god

Do you think i am a school going kid? I know it pretty well that giving the solutions as it is for a "HOMEWORK" can be bad for a student. I totally agree with you and the site policies. As per my case your assumption is totally wrong. Surely the questions that i ask on this site may seem very basic to you but for me its not. I have just "STARTED" to explore this fascinating world of physics and just wanted to know more and more about it and you just cannot tag my effort as a "HOMEWORK". Hope i have made myself clear. Hoping to get some positive responses from you guys to help me clear my doubt — Tajammul Saleem 1 min ago

0celo7

@Danu That's how I felt when I had a question closed.

Danu

1:52 PM

@0celo7 Yeah, I saw you stirring up some drama haha

0celo7

I think I even once told Jamal that "high schoolers need help with string theory" when I thought he was a grad student

Danu

over here

4

Q: How to calculate $3\otimes 3$ and $3\otimes 3\otimes 3$ in $SU(3)$?

EDIT: I have boiled my question down to How many independent components does a rank three totally symmetric tensor have in $n$ dimensions? A derivation would be nice too. OP: I know that I can represent $3\otimes 3$ by a rank two tensor. Then the symmetric part is $3(3+1)/2=6$ and the antisymm...

homework-and-exercises group-theory representation-theory

It's funny, because Jamal himself was also a high schooler (he's the same age as you)

0celo7

(why do you think I mentioned that?)

Danu

Because it's already notable that he thought you were a grad student

May have been worth mentioning without further context

0celo7

what

dude did a bunch of comments get deleted on that

Danu

1:54 PM

...because it's kind of cool to be mistaken for a grad student as a high shcooler

3

@0celo7 Yes

I flagged the back-and-forth as obsolete

0celo7

can you see them?

Danu

No, but I remember them.

Once I get 10k I will

0celo7

well I wrote half of them

Danu

I know

and you got mad

0celo7

I wrote a curse tablet for the people who closed it

ok where did that string theory question go

did they delete it

Danu

1:56 PM

Hey, that conversation ended up taking a turn for the best!

Neither i am taking it personally. I know its really difficult for you guys to respond to each and every question and i am really thankful that you guys try your best to solve our problems. I will surely give this question another go and see if it works.! — Tajammul Saleem 1 min ago

0celo7

@Danu see, that's an instance where I, as someone who has self studied a considerable amount, has a lot of sympathy

it was made clear in class that the fragments behave that way

but I don't recall the book emphasizing that

Danu

@0celo7 Well, try to give him a better hint then

(without explicitly solving it... how much more can you do?!)

0celo7

well I disagree with the policy of not explicitly solving, so

Danu

...so?

Your answer will be deleted if you do it anyways :P