The h Bar

5:00 PM

@0celo7 Sigh

@0celo7 While that is strictly true, please consider the tone of your statements.

Also, while it's not an "excuse", it is quite natural that less experienced users will need more effort and guidance to ask good questions.

@ItachíUchiha Do you understand what divergence is?

0celo7

ooo, I do

Bernard Meurer

@0celo7 You need to work on getting your message across without being an asshole, man

0celo7

it measures the variation of the volume element along integral curves

DanielSank

@BernardMeurer You need a comma after "asshole".

0celo7

5:03 PM

@BernardMeurer I don't see anything wrong with what I said.

Danu

@DanielSank What if someone is just really into assholes? :P

Bernard Meurer

@0celo7 That's exactly the issue

DanielSank

@Danu I'm playing the probabilities here, man.

^

Danu

^^

DanielSank

\(^^)/

Danu

5:04 PM

Wish I knew how to do smileys

Bernard Meurer

~(^^)~

0celo7

@Danu Google.

Danu

d(-_-)b ?

Good enough

DanielSank

@BernardMeurer Dancing samba?

@Danu Listening to music?

Danu

@DanielSank Yea

Itachí Uchiha

5:05 PM

@DanielSank No.I am expecting a simple analogy as an answer.I have tried reading about that but I didnot get it except divergence=flux/volume

Bernard Meurer

@DanielSank That's me dancing Talking Heads

0celo7

(=^ェ^=)

DanielSank

@BernardMeurer That's why the Poles like you.

@ItachíUchiha You know what a vector field is?

Vector field

In vector calculus, a vector field is an assignment of a vector to each point in a subset of space. A vector field in the plane (for instance), can be visualised as: a collection of arrows with a given magnitude and direction, each attached to a point in the plane. Vector fields are often used to model, for example, the speed and direction of a moving fluid throughout space, or the strength and direction of some force, such as the magnetic or gravitational force, as it changes from point to point. The elements of differential and integral calculus extend naturally to vector fields. When a vector...

^ That's a good illustration of a 2D vector field.

0celo7

@BernardMeurer I'll accept criticism, but the "criticism" in this chat consists of calling me an asshole, which is pretty useless.

DanielSank

@0celo7 No, it's not. It should draw your attention to your message and prompt you to think about whether or not it used effective tone.

Bernard Meurer

5:07 PM

@0celo7 It's not about calling you an asshole it's trying to show you when you're being an asshole

DanielSank

And @BernardMeurer, you may want to avoid name-calling. That's less than effective as well.

0celo7

Being a high schooler isn't an excuse for asking a bad question. What's wrong with that?

Itachí Uchiha

@0celo7 What you have said is right. But I stated my age here because I can only understand very simple explanation not because I want an excuse to ask bad question.IF that question is bad,go downvote it.I donot have any problem.

DanielSank

@0celo7 It doesn't help anyone. That's what's wrong.

Bernard Meurer

@0celo7 The way you said it comes off as harsh and unnecessarily so

@DanielSank I'm his friend, if I wasn't I wouldn't do that :)

DanielSank

5:08 PM

@BernardMeurer The real problem is that it's completely unhelpful.

Danu

To me, the problem is mostly that that is the first thing you said

DanielSank

@Danu And that.

Danu

It's not really that rude as a message in itself

0celo7

What, I should have said hi first?

Bernard Meurer

Oh boy here comes the mods

DanielSank

5:09 PM

@0celo7 you do have a rather solid history of making unhelpful posts in the chat.

Danu

Yeah, perhaps

@BernardMeurer No flags

Bernard Meurer

Oh, in that case, hey @HDE226868!

Danu

lol

0celo7

@DanielSank Of course I think they are all helpful.

Danu

asshole ;)

HDE 226868

5:09 PM

Yeah, I'm just here to talk physics.

0celo7

So once again, bold claims.

HDE 226868

Although my timing clearly could have been better.

@BernardMeurer Hi there!

Secret

@ItachíUchiha check if my answer is clear enough for you in the PSE page

DanielSank

@0celo7 No, you should have either 1) Offered constructive criticism on how to actually improve the question, or 2) Helped the user understand the issue at hand.

Bernard Meurer

@0celo7 Here's the deal, instead of saying what you said, you could've shown him how to make a better question

0celo7

5:11 PM

I'm not interested in doing that.

Bernard Meurer

Then don't do empty criticism

user218912

@DanielSank do I need to taylor expand anything?

0celo7

It wasn't empty.

I was pointing out there are no special rules for high schoolers

DanielSank

@IceLord No.

Bernard Meurer

Okay, then it was filled with nulls

DanielSank

5:12 PM

@IceLord, let me write down the two equations we have, again...

Bernard Meurer

@0celo7 Now you're appealing to formalism to defend your point while knowing you were wrong

Danu

Also, you're not accepting the criticism right now :P

Itachí Uchiha

@0celo7 I WAS NOT DEMANDING RELAXATION OF RULES FOR ME.I stated that I am a high school student because I need a simple explanation.

Bernard Meurer

See, this is what I'm talking about

user218912

@DanielSank I have them down.

Bernard Meurer

5:12 PM

@ItachíUchiha Chill out

DanielSank

1) $\exp([A,B]/2) \exp(A+B) = \exp(A) \exp(B)$

2) $\exp(-[A,B]/2) \exp(A+B) = \exp(B) \exp(A)$

0celo7

@Danu Because I disagree that my criticism was empty.

Perhaps it would have been better if I didn't say anything, but I don't see anything strictly wrong with what I said.

DanielSank

@IceLord Using nothing but basic, high school algebra, please now solve for $\exp(A)\exp(B)$ in terms of $\exp(B) \exp(A)$.

@ItachíUchiha We know. It's ok.

@IceLord, let me try writing this in a simpler way:

user218912

@DanielSank but we're dealing with operators.

user218912

is it the same algebra?

DanielSank

5:17 PM

1) $\exp(X/2) Y = Z$

2) $\exp(-X/2) Y = W$

Danu

@IceLord As long as you're not trying to commute anything, many things work quite analogously

DanielSank

Solve for $Z$ in terms of $W$.

@IceLord Well, note that $\exp([A,B]/2)$ is a scalar in our particular case.

Danu

In particular, you can still left- or right-multiply equations by factors

DanielSank

^ And that.

Danu

@DanielSank Damnit Daniel :P It's still an operator

0celo7

5:19 PM

DanielSank

@Danu It's also a scalar. What I said is true.

0celo7

@IceLord For reference. Taken from Hall's Lie groups book.

Danu

Depends on what you mean by a scalar: I usually take it to mean "an element of the field (field being a mathematical field)"

user218912

@0celo7 thanks.

Danu

In that sense, it is not true (i.e. it amounts again to saying it's a number)

0celo7

5:20 PM

Just say it's central and be done with it.

Danu

Right

DanielSank

@0celo7 I have never heard that term before @Danu used it the other day.

So I strongly doubt using it with IceLord would be effective.

user218912

@DanielSank $$\frac{e^{2A + 2B}}{e^Ae^B} = e^Be^A$$ ?

0celo7

@Danu Did you cover the uniform topology on $\Bbb R^\omega$ in your topology course?

DanielSank

I try to balance what I think is useful communication against rigor.

Danu

5:22 PM

Hmm, well when accompanying it with a definition it's really no problem

0celo7

@DanielSank Hmm.

DanielSank

@IceLord I don't know what that means, sorry.

Danu

@0celo7 We never dwelled on artificial things much.

DanielSank

Dividing by operators?

user218912

idk

Danu

5:23 PM

I don't think there is any point to considering things like ordinals for someone interested in geometry & topology

user218912

let me rewrite it

Danu

Anyways, I gotta run

DanielSank

@IceLord go back to here:

0celo7

Cheers.

DanielSank

5 mins ago, by DanielSank

1) $\exp(X/2) Y = Z$

and please solve for Z in terms of W.

Secret

5:23 PM

I think he means $e^{-A}(e^{2A+2B})e^{-B}=e^{B}e^{A}$

user218912

@DanielSank okay in that case $Z = \frac{Y^2}{W}$?

0celo7

@DanielSank Can you explain the notation please?

What's $W$?

DanielSank

A variable.

Some algebraic object. Never mind the details.

0celo7

But what is it in terms of $X,Y$?

user218912

@Secret yes

0celo7

5:25 PM

@Secret Is that true?

user218912

but how are we able to do this without using BCH?

DanielSank

@0celo7 You'r job is to solve for Z in terms of W. What part of that do you not understand?

0celo7

@DanielSank I don't know what $W$ is!

DanielSank

@IceLord Multiply both sides of the second equation by $\exp(X)$.

user218912

@0celo7 you don't have to know?

0celo7

5:26 PM

You're telling me to solve $y=f(x)$ in terms of $k$, I don't know what that means.

DanielSank

@0celo7 Scroll up.

user218912

@DanielSank which equation?

Secret

@0celo7 Using the algebraic properties of exponential matrices, the inverse of $e^{A}$ is $e^{-A}$ and thus matrix mutiplication cancel out one of the $e^A$,$e^B$ leaving $e^Ae^B=e^Be^A$

0celo7

$W=e^{-X/2}Y$?

DanielSank

@0celo7 Solve for Z in terms of W.

$\exp(X/2) Y = Z$

$\exp(-X/2) Y = W$.

user218912

5:28 PM

@DanielSank I did it.

user218912

now what?

user218912

I replace the exponentials?

DanielSank

@IceLord Well what did you get?

0celo7

@IceLord what is the result

user218912

final result or before that?

0celo7

5:29 PM

I got $Z=e^XW$.

DanielSank

@IceLord it's a bit tiresome to keep having to prompt you to show your steps. Could you just please write down what you got for $Z$ in terms of $W$?

@0celo7 Very good, but I'm not trying to help you.

You already know how to solve the problem in question.

0celo7

^^ yay

user218912

$exp(X)Z = exp(X)\frac{Y^2}{W}$ ?

0celo7

@DanielSank I do? I don't know what it is tbh

user218912

wait no

user218912

5:31 PM

i'm retarded

user218912

yes I got what 0celo7 got

user218912

I divided them

DanielSank

Very good.

user218912

thanks

user218912

^_^

DanielSank

5:33 PM

Now, can you go from this example and plug in the A's and B's?

0celo7

@IceLord use \exp

user218912

alright

0celo7

Ok, for $f:\Bbb R\to\Bbb R^\omega, t\mapsto (t,2t,3t,\dotsc)$ I think I have $$\bar d(f(t_0),f(t))=\sup_{i\in\Bbb N}\{\max\{i|t_0-t|,1\}\}$$

so...what now

user218912

$$e^Ae^B = e^{[A, B]}e^Be^A$$? @DanielSank

DanielSank

Where'd you get that minus sign?

0celo7

5:36 PM

the ether

user218912

fixed sorry I confused myself

DanielSank

@IceLord Ok, so now what are $A$ and $B$?

user218912

I can take it from here @DanielSank

user218912

I know what to do now

DanielSank

@IceLord Very good.

Now let's go over a few things, do you mind?

0celo7

5:38 PM

Ok, I think this function is discontinuous.

user218912

sure

0celo7

Interesting.

DanielSank

Ok, item #1:

We had two equations:

0celo7

@IceLord Let's prove BCH.

DanielSank

$\exp([A,B]/2) \exp(A+B) = \exp(A) \exp(B)$

$\exp(-[A, B]/2) \exp(A+B) = \exp(B) \exp(A)$.

Since $[A, B]$ is proportional to the identity, we can treat it like a plain number and forget about operators. So that's why you can multiply both sides of the second equation by $\exp([A,B])$ without thinking too hard.

user218912

5:40 PM

@DanielSank because $[a, a^\dagger] = 1$?

DanielSank

On the other hand, you can always multiply both sides of an equation by an operator as long as you multiply both sides on the left or both sides on the right.

@IceLord Yes.

Item #2: there's another way to do this problem.

$\exp(z a^\dagger)|0 \rangle = \sum_{n=0}^\infty z^n (a^\dagger)^n / (n!) |0\rangle)$

user218912

I saw a way to do it using the displacement operator.

user218912

but it didn't apply to my case.

DanielSank

$=\sum_{n=0}^\infty (z^n / \sqrt{n!}) |n \rangle$

Similarly, $\langle 0 | \exp(z^* a) = \sum_{n=0}^\infty \langle n | ((z^*)^n / \sqrt{n!})$

and then you just do the sandwich.

user218912

yep

DanielSank

5:45 PM

Done.

user218912

that looks like a difficult sandwich

DanielSank

No screwing around with the BCH formula.

user218912

i wouldn't know how to do it

user218912

well i didn't try

DanielSank

@IceLord Are you kidding? $\langle m | n \rangle = \delta_{mn}$

user218912

5:45 PM

oh alright

DanielSank

Dude, that's a trivial sandwich.

Don't be afraid of stuff. Just do it.

user218912

alright I got it.

user218912

thanks @DanielSank

DanielSank

Ok, item #3:

I suspect it may be a very good idea for you to put off QFT for a while. Focus on QM and other physics/math courses. Get a good foundation.

Then take QFT when you're in a position to absorb it well.

user218912

I'm absorbing it well so far though.

user218912

5:47 PM

I understood everything we learned.

DanielSank

Otherwise, you might be using a lot of your own time in an inefficient way.

@IceLord Can you solve the problems?

user218912

yes

DanielSank

"Understanding" is meaningless without doing calculations.

user218912

with some help

DanielSank

Ok, best of luck then.

user218912

5:47 PM

I can understand the derivations

user218912

after class I always repeat the derivations

user218912

from my notes

DanielSank

@IceLord That's not the same thing as doing calculations on your own.

0celo7

scratches head

DanielSank

But anyway, good luck.

user218912

5:48 PM

I do them without looking

user218912

thanks

0celo7

How does BCH work like that

DanielSank

I have to go make a chocolate cake now.

0celo7

I thought the matrices had to be "small"

Bernard Meurer

@IceLord I think you might be approaching learning this stuff with the wrong angle

user218912

5:48 PM

@DanielSank I learned a lot in this process and I can move on now, I love you.

user218912

@BernardMeurer what do you mean?

0celo7

ew.

DanielSank

@IceLord I'm married.

Bernard Meurer

@IceLord Mind if I explain anecdotally?

user218912

@DanielSank xD

user218912

5:49 PM

@BernardMeurer sure

DanielSank

@0celo7 Why "ew"?

user218912

@BernardMeurer @DanielSank just sayin there are people in my qft class who haven't taken QM

user218912

mathematicians

user218912

so I'm good

0celo7

@DanielSank PDA.

Bernard Meurer

5:50 PM

@IceLord I really really REALLY wanted to figure out how drivers (CS stuff) worked

And I didn't want to wait for college to do it so I started right away

so I learned basic C well

then I learned Systems architecture

then I learned about Linux in depth

DanielSank

@IceLord That reasoning does not make sense to me, for whatever it's worth.

Bernard Meurer

then I did the how LFS guide

then I started reading about drivers

and now, a year and something later I have an okay understanding of how they work

But I only started looking into drivers a couple months ago

user218912

@DanielSank if they can do it, then with a lot of time and effort I can too if I try my best.

DanielSank

@IceLord very well

Bernard Meurer

I feel like you're trying to learn QFT jumping through the basis that it requires

and that'll just be frustrating man

user218912

5:53 PM

in my defence

user218912

I did once learn all of QM but didn't absorb all of it

DanielSank

@IceLord You don't need to defend yourself. We're trying to help you.

user218912

so I know what things are

user218912

but I just am not used to them

0celo7

you did not learn all of QM

user218912

5:53 PM

or experienced with them

DanielSank

@IceLord Yes, this is what I'm worried will happen to you with respect to QFT.

0celo7

What a ridiculous thing to say

user218912

@0celo7 yes I did I read all of shankar

0celo7

That is not "all of QM"

user218912

I mean all of shankar

user218912

5:54 PM

you know what I mean xD

DanielSank

@IceLord You think that means you understood and could use all of the material in Shankar?

0celo7

Shankar is a giant book

DanielSank

@IceLord No, I did not know what you meant. You seem to have a very unrealistic concept of what it means to understand something.

user218912

@DanielSank true

user218912

guys don't put me down, I'm going to take QFT no matter what and try my best.

user218912

5:55 PM

I even enrolled in the course officially

DanielSank

@IceLord We're not putting you down. We're trying to help you.

user218912

yes and I really appreciate that

user218912

I'll need all the help I can get

user218912

to get a good mark in the course

Bernard Meurer

@IceLord Just trying to show there's a better way to achieve your goal

DanielSank

5:56 PM

Thinking that people advising you to do something different from what you're doing is "putting you down", is a terrible attitude.

Bernard Meurer

@IceLord Your mark doesn't matter

What you actually learn matters

DanielSank

@BernardMeurer I disagree with that, but whatever.

user218912

@DanielSank well I can't change it so there's no point of telling me to do something else.

user218912

it will just make me feel regret

user218912

that's what I mean

0celo7

5:58 PM

@ACuriousMind I'm confused about BCH. The general lore is that it holds for matrices whose matrix norm is sufficiently small, and that's what's proven in Wiki's source for it.

Why do physicists use it without that consideration?

DanielSank

@IceLord Fine, but frankly given that you really need to revise QM, I am now uninterested in helping you with more advanced topics. Perhaps you can convince other users here to feel differently, but I doubt it. The point is this: if you try to do something and you need help, you should be responsible about asking for that help.

0celo7

And why doesn't the article mention that?

DanielSank

If the problem is, fundamentally, that you're taking a course that's beyond your level, that's your mistake and other users won't want to spend time fixing it for you.

@0celo7 Because it's true without that consideration.

user218912

@DanielSank I know I'm taking responsibility to learn what I'm missing.

0celo7

@DanielSank I can't find a reference for that.

user218912

5:59 PM

@DanielSank the good thing is that I am familiar with QM so I can easily find out what I'm missing and learn it.

DanielSank

@IceLord Well, sort of, except that @ACuriousMind and I both spent a lot of time getting you through what was in the end a basic algebra issue. I'm not trying to put you down. I'm trying to point out that you might want to direct your resources more efficiently.

Good luck with your studies. I'm happy to help you with QM issues in the future.

chocolate cake time.

user218912

thanks.

Secret

Ok I think I understood now. You literally impose the constraints onto the system's trajectory to force it into a constrained trajectory. I think I now understood D'Alembert now To be continued tmr, must go to sleep, it has been a long day...

0celo7

@DanielSank According to Hall, the BCH formula is not true in general.

The mess of commutators need not converge.

However, it is true in the case when $[X,Y]$ is central, and he has a full proof of it.

0celo7

6:24 PM

@BernardMeurer been learning any good calculus?

Have classes started for you?

Bernard Meurer

@0celo7 They start monday

and they use Apostol as bibliography :/

Fuckign hated Apostol

0celo7

bibliography?

Bernard Meurer

the book they use is Apostol

user116211

@BernardMeurer Apostol is a good book.

Bernard Meurer

I hated it

0celo7

6:25 PM

why?

user116211

@BernardMeurer O.o

Bernard Meurer

It was too hard for me

user116211

@BernardMeurer Man, if you want some mild treatise, you can use the Berkeley lectures on Calculus.

user116211

-2

Q: What is the cause of "spooky action at a distance"?

Introduction: If I want to accelerate a box, I need to push it by physically contacting and touching the box. Gravity on the other hand manages to accelerate a box or any thing that has a certain distance called r from it without touching it. If I give a electron under the influence of magenti...

user116211

I don't know what OP is up to. 6 edits within a min.

0celo7

6:27 PM

@BernardMeurer Maybe reading Shankar is not a good idea then.

user116211

As Lanczos writes, "The mysterious force of universal attraction is interpreted as a purely geometrical phenomenon - a consequence of Riemannian structure of the space-time world." — MAFIA36790 3 mins ago

0celo7

Riemannian?

Hardly.

Bernard Meurer

@0celo7 I'll go through Linear algebra and Calc I first

0celo7

Riemannian geometry is completely different from GR.

user116211

@BernardMeurer You can go through Paul's notes available online; they are quite intuitive.

Bernard Meurer

6:28 PM

@MAFIA36790 Who's Paul?

0celo7

google it

Bernard Meurer

0celo7

Sigh.

user116211

@0celo7 His words; I'm in his Variational Mechanics book; way better than Goldstein.

0celo7

@MAFIA36790 Sigh.

I'm starting to not trust physicists when it comes to physics even.

user116211

6:30 PM

@0celo7 Why?

user116211

@0celo7 hmm.

0celo7

@MAFIA36790 Because it's wrong.

user116211

@0celo7 He was a mathematician too.

0celo7

Spacetieme is not Riemannian, that's what makes it so interesting and mathematical GR so hard.

user116211

@0celo7 No idea.

0celo7

6:30 PM

@MAFIA36790 I don't trust mathematicians when it comes to physics either.

user116211

hmmm.

0celo7

The statement is wrong, spacetime is not a Riemannian manifold.

Some basic theorems carry over, yes.

Bernard Meurer

tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

user116211

@0celo7 Why did he even write that at the very first place if it is wrong?

0celo7

But there are two issues:

user116211

6:31 PM

@BernardMeurer yes.

0celo7

The Ricci tensor is pseudo-hyperbolic, not pseudo-parabolic/elliptic.

Bernard Meurer

Sweet, thanks

0celo7

There is no distance function on a Lorentzian manifold

The mathematics of hyperbolic equations is completely different from elliptic ones.

user116211

@BernardMeurer For the next level, you can go to Spivak; a dry but good book. Apostol has quite a good exercise though.

user116211

@0celo7 ohh.

0celo7

6:33 PM

@MAFIA36790 That's the analytic perspective.

There are also topological considerations.

Not all manifolds are even Lorentzian.

So the usefulness of Lorentzian geometry for probing topology is not great.

You already need to know a lot about your manifold to conclude it even has a Lorentzian metric!

user116211

@0celo7 sure.

0celo7

@MAFIA36790 Besse has an explanation of this, let me copy it.

user116211

What I'm not getting why he did even write that if it is wrong to say so.

0celo7

> It would seem that Riemannian and Lorentzian geometry have much in common: canonical connections, geodesics, curvature tensor, etc... But in fact this common part is only a common disposition at the onset: one soon enters different realms. For exampie, looking at the Einstein condition, the Riemannian geometer will proceed as in the present book: existence, uniqueness, moduli... [cont.]

> On the other hand the physicist starts from a spacelike hypersurface and propagates the induced Riemannian metric to get a Lorentzian metric on space-time using Einstein's equation. The Riemannian will meet elliptic PDE's and the Lorentzian hyperbolic ones. Moreover, the latter will work essentially on non compact objects and meet singularities.

> The interaction between the two fields is best illustrated by the follow- ing picture from C.N. Yang in CYan]:

user116211

@0celo7 This sums the whole point. Now, I'm getting he might be flexible in his writing then.

0celo7

6:37 PM

@MAFIA36790 Hah. If he had said "pseudo-Riemannian" I would have had no issue.

@MAFIA36790 I have a rather lengthy, advanced book on global theorems in Lorentzian geometry. It's an incredibly difficult endeavor to try to salvage global results from Riemannian geometry.

The positivity condition on a Riemannian metric is incredibly powerful.

user116211

Would be looking further for it in the future. Thanks @0celo.

user116211

Now, I will read the book with lots of salt ;)

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