The h Bar

0celo7

12:12 AM

@BalarkaSen No clue.

I have no clue what your idea is.

My solution works.

So either tell me or don't, but I won't waste any more time thinking about it.

NeuroFuzzy

12:55 AM

@bolbteppa Can you share a bit more on this? I'm somewhat unhappy with the treatment of electrodynamics in Griffiths. (not mathematical enough/doesn't go anywhere else)

bolbteppa

2:15 AM

@NeuroFuzzy In Griffiths you spent your time studying the electric field, then the magnetic field, then a bit unifying them, only at the end doing relativity, but logically you can do it all in general in reverse unifying the microscopic theory in the most mathematical way possible with special and general relativity amazon.com/Classical-Theory-Fields-Fourth-Theoretical/dp/… themindofatheorist-theblog.blogspot.ie/2010/09/…

before even touching the macroscopic theory amazon.com/Electrodynamics-Continuous-Media-Second-Theoretical/… doing it all from first principles, really only needing to know elementary classical mechanics amazon.com/Mechanics-Third-Course-Theoretical-Physics/dp/…

0celo7

@bolbteppa just marry Landau already

skill patrol

(0:

0celo7

what is that abomination

looks like a mutated clown.

skill patrol

:(

0celo7

@bolbteppa My girlfriend has agreed to read Zee after she takes vector calculus. Should she memorize the Riemann tensor

user218912

2:28 AM

what program is she doing?

0celo7

music

skill patrol

srsly?

user218912

why does she need vector calculus? 0.o

user218912

random elective?

0celo7

@3750 yes

user218912

2:37 AM

I bought this awesome new blue ink gel pen.

0celo7

Going old school with Tha Carter III.

Get on my level, you'll ... need a ladder that's forever. - Lil Wayne.

How can one man be so based

bolbteppa

@0celo7 She should read Landau's derivation of the Riemann tensor using Stokes theorem ;)

0celo7

@bolbteppa one does not derive a definition

bolbteppa

One does

0celo7

the Riemann tensor is a definition, it cannot be derived

unless you're talking about Lovelock or something

i.e. motivate it via physics

bolbteppa

2:42 AM

That's ridiculous

0celo7

ok, how does he derive it

what is the starting point

it's the obstruction to local flatness in a sense that can be made precise

is that what you mean?

bolbteppa

Integrating the partial derivative $\partial_l$ of a vector field $A_j$ whose covariant derivative $A_{j;l}$ is zero $\int \frac{\partial A_j}{\partial x^l} dx^l$ around a closed loop then using Stokes then anti-symmetrizing then expressing in terms of Christoffels and you get $R^i_{jkl} = \partial_k \Gamma^i_{jl} - \partial_l \Gamma^i_{jk} + \Gamma^e_{jl} \Gamma^i_{ek} - \Gamma^e_{jk} \Gamma^i_{el}$ ;)

0celo7

How is that a "derivation"

bolbteppa

If you actually work that out you can remember wtf those indices are about, e.g. the $i$

0celo7

what did you derive

I think that's a variant of "local obstruction to flatness"

bolbteppa

2:54 AM

I derived the total change in a vector due to parallel displacement around a closed loop, i.e. the total change $\Delta A_j$ in $A_j$ around a closed loop $\int \Delta A_j = \int \frac{\partial A_j}{\partial x^l} dx^l$, if it's covariant derivative is zero (parallel displacement) yet something still changes, it means two things have to change to counterbalance one another, since the covariant derivative derives not only the field but the basis also,

I derived what happens to the basis by starting from the field

0celo7

I thought that was only an approximation

I'm 99% sure it's in Zee

You need "infinitesimally small loop" shenanigans

bolbteppa

I think Zee did it using only derivatives which is what most books do which is why he used infinitesimal loops

0celo7

let's get out Zee and check

bolbteppa

Alright, one minute

0celo7

ahh, wonderful smell

page 546

bolbteppa

2:58 AM

Page 341 he does something ridiculous with commutators

0celo7

I've never been convinced by that computation

It's never been clear to me what the commutator has to do with curvature

a priori, that is

bolbteppa

Me neither, but I know it does, and Tao says something genuinely insane about thinking about non-abelian groups being symmetries of curved spaces even in general, and I think there's a lot to that

Even finite groups as graphs and graphs as Riemann surfaces makes it makes sense (I think)

Lot of words in that derivation around 546, but equation 10 is what you should get

I have no idea why he's Taylor expanding either :o

0celo7

@bolbteppa Ok, the totally rigorous version of what Landau shows is: Parallel transport is path independent in some neighborhood $U$ if and only if the Riemann tensor vanishes on $U$.

@bolbteppa One can also show that an open set $U$ is isometric to some open set of Euclidean space if and only if the Riemann tensor vanishes on $U$.

But I would not say these are "derivations" of the curvature tensor...

bolbteppa

Look at what Zee says on page 546, amazing: "In theoretical physics, when faced with a fairly involved calculation, it is always a good habit to anticipate the answer" I should have just quoted that earlier

0celo7

@bolbteppa I agree with you on that point!

Although @ACuriousMind dislikes Zee, so it would not have done much good.

bolbteppa

3:07 AM

@0celo7 That is a derivation of the Riemann tensor, maybe not what Zee is doing but what Landau does is, it's black and white a implies b, I derived the expression for how the basis changes in response to the change in a vector field, then I define it after knowing wtf it does, this is the whole point of the concept and what it does even explaining what the indices are, any Wald-style exposition of it is just making it chart-independent which this is anyway,

Why not define Riemann tensors in a homology book? Why in one specific section of a diff geom or GR book? Because you need it for certain things that arise naturally and are worth putting the label of a definition onto

0celo7

@bolbteppa How are you even defining the Riemann tensor...

What does "Riemann tensor" mean to @bolbteppa

bolbteppa

Let me see if this posts in a nice way or an ugly way

$\vec{A}_{;j} = \partial_j (A_i \vec{e}^i) = A_{i,j}\vec{e}^i + A_i \vec{e}^i_{,j} = A_{i,j} \vec{e}^i + A_i \vec{e}^i_{,j} \vec{e}_k \cdot \vec{e}^k = A_{i,j} \vec{e}^i + A_i (\vec{e}^i_{,j} \cdot \vec{e}_k) \vec{e}^k = A_{i,j} \vec{e}^i - A_i (\vec{e}^i \cdot \vec{e}_{k,j}) \vec{e}^k = A_{i,j} \vec{e}^i - A_k (\vec{e}^k \cdot \vec{e}_{i,j}) \vec{e}^i $

$ \ \ \ \ \ = A_{i,j} \vec{e}^i - A_k \Gamma^k_{ij} \vec{e}^i = (A_{i,j} - \Gamma^k_{ij} A_k) \vec{e}_i = A_{i;j} \vec{e}_i$ (Using $(\vec{e}_i \cdot \vec{e}^k)_{,j} = (\delta_i^k)_{,j} = 0 \rightarrow \vec{e}_{i,j} \cdot \vec{e}^k = - \…

0celo7

fuck dude

bolbteppa

Are you happy with that? That is the covariant derivative of a covariant vector field

0celo7

there are some typos

what is $\vec e^i_{,j}$

should it not be $;j$

bolbteppa

3:12 AM

Where? $\vec{e}^i_j = \frac{\partial }{\partial x^j} \vec{e}^i$, no it shouldn't be $;j$

0celo7

then no, it's wrong

bolbteppa

Nope

0celo7

how does $;j$ turn into $\partial_j$

in the first equality

bolbteppa

It doesn't, a covariant derivative is just a derivative of a vector field where not only the vector field, but also the basis, are position-dependent

Then connection coefficients arise by expressing the derivative of that basis vector as a linear combination of the original basis

I'm sorry all that differential geometry you learned obscured all this ;)

haha

Go read old Russian differential geometry books dude

haha

They are insanely good

Borisenko Vector Analysis page 185 onwards

So with that background:

$\Delta A_j = \oint_{\partial S} dA_j = \oint_{\partial S} \frac{\partial A_j}{\partial x^l} dx^l = \int_S \frac{\partial^2 A_j}{\partial x^k \partial x^l} dx^k \wedge dx^l = \int_S A_{j,lk} dx^k \wedge dx^l = \int_S \frac{1}{2} ( A_{j,lk} - A_{j,kl}) dx^k \wedge dx^l$

$ \ \ \ \ \ \ \ = \int_S \frac{1}{2} ( \partial_k A_{j,l} - \partial_l A_{j,k}) dx^k \wedge dx^l = \int_S \frac{1}{2} [ \partial_k (\Gamma^i_{jl}A_i) - \partial_l (\Gamma^i_{jk}A_i)] dx^k \wedge dx^l$ (Using $A_{j;l} = 0 = A_{j,l} - \Gamma_{jl}^iA_i$)

0celo7

Oh yeah, that result is exact (I think). But if you want to pull the Riemann tensor outside of that integral (like in Zee) you get an approximation.

bolbteppa

3:18 AM

If you ignore the basis when you take derivatives, then you have to actually define a covariant derivative as something new to take account of this children's notion

So it seems scary

0celo7

@bolbteppa I know what the Christoffel symbols are.

But it's not correct that $A_{;j}=\partial_j A$.

user116211

Hey, @yuggib, do you know anything about the book Methods of Mathematical Physics by Hilbert and Courant?

bolbteppa

@0celo7 I didn't say it was

0celo7

@bolbteppa That's what your first equality is!

$\vec{A}_{;j} = \partial_j (A_i \vec{e}^i)$!

bolbteppa

For a vector field it is true that $\vec{A}_{;j} = \partial_j (A_i \vec{e}^i)$ because I am taking derivatives of both the basis and the components, it is not true that $A_{i;j} = \partial_j A_i$ which is deriving just the components of $\vec{A}$, since I am neglecting deriving the basis as you do on a manifold

0celo7

3:23 AM

What the heck is $\partial_j(A_ie^i)$

bolbteppa

If $\vec{A} = A_i(u_1,u_2,u_3) \vec{e}^i(u_1,u_2,u_3)$ and you try to calculate $\frac{\partial }{\partial u^j} (A_i(u_1,u_2,u_3) \vec{e}^i(u_1,u_2,u_3))$ what happens?

0celo7

I do not know what the derivative of a vector means on a manifold.

*The partial derivative

bolbteppa

What about in the plane?

Or in space rather

$\mathbb{R}^3$

0celo7

There is a canonical parallelism there which allows us to represent the Levi-Civita connection with $\partial_i$.

bolbteppa

Just take any vector field in the plane and put it into polar coordinates in a polar coordinate basis, as you'd find in chapter 1 of Kleppner's Mechanics, where they calculate this 3 ways (and don't even use the words covarant derivative)

haha

0celo7

3:28 AM

Namely, in Cartesian coordinates one can define $\nabla_YX=\mathrm d(X^i)(Y)\partial_i$.

bolbteppa

Just be a child and calculate that derivative, don't use any fancy words

Treat it as a calculus problem

0celo7

I don't know how to calculate it!

bolbteppa

Do the product rule

0celo7

What is $\partial_j \vec e$?

bolbteppa

It's just a vector

It's a vector you can then re-express as a linear combination of basis elements

0celo7

3:30 AM

I think you should read a geometry book for mathematicians. What you're asking me to do does not make sense.

I'm not a physicist, I'm afraid I can't read your mind

bolbteppa

What are the coefficients multiplying the elements of the basis? Those things you have said multiple times you know - connection coefficients

I am asking you a children's problem, any high school kid knowing multivariable calculus can do this man, you can too

0celo7

Connection coefficients involve a connection, you're not asking me to do anything with a connection.

bolbteppa

Pretend you didn't know any modern math and just invent it as you go along

I am explaining to you why people even thought to define connection coefficients in the first place

0celo7

That's how it works.

bolbteppa

The rest of the stuff is just a game to define what I am saying so that you can do this on a manifold where you use deficient coordinates and have to piece things together, still just the same baby idea

You just posted something that did exactly what I did in a very awkward non-intuitive way

0celo7

3:34 AM

Except mine is correct.

I'm done with this conversation.

I can recommend some books.

But right now you're just babbling.

bolbteppa

No you just don't understand how simple this stuff is and prefer to pretend it's complicated

You gave me the covariant derivative of a contravariant vector field btw

It's amazing you think I'm babbling when I asked you to compute the derivative $\frac{\partial }{\partial r}$ of a vector field like $\vec{A} = (2x + y)\vec{e}_x + (3x^2 + y^3)\vec{e}_y$ after converting this to polar coordinates and changing the basis to a polar coordinate basis

@0celo7 Here is an explicit calculation books.google.com.vn/…

Just a few lines to get that scary expression for $R_{ijkl}$ in terms of the metric, but oh well ;)

0celo7

3:50 AM

I don't care.

bolbteppa

@0celo7 If you want to understand any of this try and compute $\frac{\partial }{\partial r} (x^2 \vec{e}_x + y^2 \vec{e}_y)$ using nothing but elementary calculus i.e. taking derivatives and no more, no manifolds concepts, no isometries, by the end you'll understand what a covariant derivative is, use the link to compare answers if you get stuck, this is not a physics problem it's a math problem, good luck

0celo7

You're nuts, sorry.

bolbteppa

haha ok come back to this in a year or 3 when it's (I'm?) less nuts

yuggib

6:14 AM

@MAFIA36790 is it also in english? I have a german version but I don't read german T__T anyways, it surely is a bit outdated nowadays; more of an historical piece

user116211

6:32 AM

@yuggib amazon.in/Methods-Mathematical-Physics-Vol-1/dp/0471504475

user116211

The German edition was banned by the Nazis... after Courant went to US, he re-published it again collaborating with K.O.Friedrichs due to a license issued by the US Govt. to re-print it by InterScience publications, part of the Wiley.

user116211

It contains additional material unlike the German counterpart.

user116211

The whole Vol 2 is solely dedicated to Differential Equations, PDEs.

user116211

@yuggib Thanks for the feedback.

skill patrol

7:09 AM

Hi @ChrisWhite are you back on the right side of the pond yet?

;-)

Slereah

8:23 AM

Hey advertisers, maybe stop making ads hinging on your product being "fresh"

Any product is fresh if frozen

I could have myself a nice frozen turnip and it would be just as fresh

skill patrol

Subway isn't frozen.

Slereah

"Fresh" is a word that has several semantic values, yes, thanks for pointing that out

Will Smith was also Fresh in the prince of Bel Air but I did not mean that either

skill patrol

np ;)

The Frozen Prince of Bel Air?

Slereah

he was not frozen either

arxiv.org/pdf/0809.2576.pdf

Damn it

Another of my idea stolen!

Stolen 5 years before I even had it

How can I compete

arxiv.org/abs/0807.0289

user116211

8:41 AM

Andre Gsponer

user116211

Independent Scientific Research Institute?

user116211

What does that mean?

Slereah

Hey

That other paper is written by Colombeau himself

also lol he has a wanadoo email address

user116211

9:08 AM

@Slereah oh man.

Noah P

9:21 AM

Hi, anyone here know a little about cosmology in particular?

Slereah

Well ask your question and find out

Noah P

For a given galaxy, the force exerted on it is F=-(GMm)/r^2

However, if we consider a repulsive, anti-gravity force, proportional to r, then we get:

F=-(GMm)/r^2 + (Amc^2r)/3

Where A is the cosmological constant

Would the integration of that give the potential energy of the galaxy?

Slereah

Why do you have the cosmological constant in a classial argument?

Noah P

General relativity isn't required for the friedmann equations

Its an assumption, not a statement

Slereah

9:38 AM

1

A: Particle Data Group book

From the last email of the particle data group : "The 2016 edition will be published in summer. PDG Books containing Summary Tables and review articles as well as Booklets will be mailed in fall. Starting this year, the Data Listings will only be published online."

Why do I have to define what "fall" is

user116211

@Slereah Well, this should be known to everyone, I guess; this would be the first guy who doesn't know fall ;/

Noah P

0

Q: Integration of Force to obtain Potential Energy

Given that for a force $F$, $F=- \frac{GMm}{r^2} + \frac{Am c^2 r}{3}$, where A and M are constant, and the potential energy $V$, where $V= - \int Fdr$, would I be correct in saying that $V=- \frac{A c^2 m r^2}{6} + \frac{GMm}{r}$. I did try the physics SE for this, but no joy. Thanks!

integration

Noah P

10:00 AM

No one home..?

Noah P

10:33 AM

Hellooo?

Emilio Pisanty

@lucas fine, but you need to be brief

lucas

@EmilioPisanty Sure.

Emilio Pisanty

... so?

lucas

This was my question:

Hello sir! I have a question:

If I say I don’t accept quantum mechanics and relativity, what will you do?

a) You ridicule me?

b) You get angry?

c) You want to kill me?

d) You don’t care me and say with yourself *”leave him to die in idiotism”*?

e) None of options

Thank you very much!

Emilio Pisanty

@lucas Honestly? I think that that's below the tone I'd want the AMA to be at.

If the AMA questions were all like that, I'd leave before five minutes had passed

If 30% of the AMA questions were like that, I'd still leave

Whether it should be moderated out is another question

lucas

10:46 AM

But why?

Emilio Pisanty

Because I don't find the question interesting, nor would I be interested in the speaker's reply

Asking a multiple-choice question in an AMA is terrible form to begin with

In any case

lucas

Most people ridiculed and offend me at past days because of my opinion. And I wanted to know why. for example see the first answer to my question and its insults to me. physics.stackexchange.com/questions/267302/…

We are in a scientific place and I think we mustn't insult the others because of their opinions. Am I right?

Emilio Pisanty

@lucas The AMA format is specifically for questions where you want the interviewee's opinion specifically about something

Is there some specific reason why yuggib's answer to your question matters more than anyone else's?

lucas

No he/she hadn't to answer. But at least I think I had right to ask my question.

Emilio Pisanty

@lucas Look. I did not delete your question, and I explicitly asked you to keep this brief and concise.

lucas

10:53 AM

OK, I don't want to waste your time. Thank you very much because of your attention and time! Sorry again for that day! Have a good time!

Emilio Pisanty

I think your question was a terrible fit for the format. In an unmoderated AMA, you should have been allowed to post it, with the explicit possibility that the speaker simply ignores your question. In a moderated AMA, that question should have been allowed as a proposal but not posed to the speaker.

The AMA format has several different traditions and there's a bunch of associated conventions

lucas

You are right. I am not familiar with official stuff so much.

Emilio Pisanty

Your question breaches some of them, so there's no use starting a drama because it got deleted.

There was a specific debate as to what format the AMA events here should take

8

Q: What format should we use for Physics Stack Exchange Ask Me Anything events?

There's been talk in the hbar of starting an "Ask me Anything" (AMA) series in which featured guests would answer questions from other users. These questions would be about physics, the guest's career, or anything else. What format would work well for a Physics Stack Exchange AMA?

discussion chat-events

The clear winner was a moderated AMA

In any case, before you kick up more drama about the name, I would seriously advise you to look up how the AMA format works in other sites. This is mainly in reddit (reddit.com/r/iama), but also on multiple blogs (e.g. Scott Aaronson's).

John Rennie

@NoahP I suspect that your statementGeneral relativity isn't required for the friedmann equations has left us all a bit stunned.

lucas

@EmilioPisanty Thanks a lot for teaching me. You always are kind to me. Thank you very much. One thing else, I saw the video in your profile. You are so handsome:-)

Noah P

10:59 AM

@JohnRennie The friedmann equation is the same whether you derive it from GR or from newtonian gravity

I'm going with newtonian

Emilio Pisanty

@lucas No worries. Have a good day, and keep it easy on the drama.

lucas

@EmilioPisanty Thank you again and again!

John Rennie

@NoahP you go where you want, but don't be surprised if no-one wants to go there with you.

Noah P

@JohnRennie Ok, understandable. But, to quote 'An Introduction to Modern Cosmology' by Andrew Liddle, 2014 Edition: The Friedmann equation turns out to be the same whether derived from GR or newtonian gravity. However, the derivation is not completely rigorous, and GR is required to fully patch it up.

Hence, my question with the combination of classical physics and GR

@JohnRennie Understandable now?

David Z

@EmilioPisanty cc @lucas one other thing to keep in mind: regardless of whether the AMA itself is moderated or not, it is still taking place in a moderated chat room. I'm not saying that particular question was inappropriate for the chat room, but the point is that even though we consider it an "ask me anything"-type event, it doesn't mean literally anything is acceptable. We don't suspend the normal rules and standards for chat content.

@lucas by the way: nobody flagged that answer. In particular, you didn't flag that answer, which you should have done if you think you're being insulted.

0celo7

11:51 AM

@lucas dude

0celo7

12:11 PM

@yuggib Is writing $D_\boldsymbol{\alpha} f$ for a multiderivative OK outside of the context of PDE?

lucas

@DavidZ Thank you very much because of your attention! You are right, I didn't flag. I wouldn’t like to use flag because I think here is a scientific place and there is no need to secretly (hidden) works. There is no fight or war here and all of us should be logical. So, I never use flags because maybe someone thinks that I have used flags for reporting low quality posts. If I think that I should say something, I say my opinion by comments and I see no reason for hiding my opinion.

In short, I need a zero number of flags in my profile page. Maybe someday I lose that zero number, but I think today isn’t that day.

@0celo7 dude :-)

0celo7

@skillpatrol to be fair, that number would be half if I just abbreviated to wtf

what the fuck star dude

We all know who it was

Davey Jones.

skill patrol

Please consider using the abbreviation :-)

0celo7

@skillpatrol Why?

skill patrol

Manners.

0celo7

12:21 PM

My philosophy:

Offended by the "N word"

►

skill patrol

They can get away without saying it on CNN, right?

If they did say it there would be no more CNN.

Physics Guy

What role does cobordism play in topological quantum field theories ?

skill patrol

wtf = 450/827 FYI @0celo7

user116211

Hey @JD.

user116211

Nice to see you ;}}

John Rennie

12:36 PM

@NoahP There is an apparently convincing "derivation" of the Friedmann equation using Newtonian gravity, however it is based on tinkering with infinite sums to achieve the required result. It is not even close to mathematically rigorous.

That shouldn't stop you having a play with it if you want, but I fear you'll find little enthusiasm for it here.

Physics Guy

How can you derive the Friedmann-Equations from newtonian Gravity ? In this model The expansion of the universe depends of the energy momentum tensor.

*on

John Duffield

@MAFIA36790 : Hi Mafia. I was just reading a few things. Having a little diversionary break from some writing I'm doing.

skill patrol

Another book?

John Duffield

12:51 PM

Yep, I've said before, my working title is The Physics Detective.

skill patrol

Cool.

Physics Guy

What is it about ?

skill patrol

Who is the target audience?

0celo7

It should be a slam piece of PSE

John Duffield

@PhysicsGuy it's about fundamental physics, and how to break the impasse.

@skillpatrol : as I speak the target audience is physicists. But that might change to the general public.

0celo7

12:55 PM

@JohnDuffield Are you qualified to work on a physics book?

2

John Duffield

@0celo7 : I have no physics qualifications, but I know a lot of physics. IMHO that qualifies me. It's not as if I'll be making things up.

0celo7

I think your downvoters on this site would disagree...but ok.

skill patrol

Did you think about what said about swearing @0celo7

0celo7

@skillpatrol Wasn't paying attention.

Physics Guy

@John On what level is that book ?

John Duffield

12:58 PM

@0celo7 : not everybody thinks the downvoters on this site are the arbiters of truth. Like I said, I've just been reading a few things.

0celo7

@JohnDuffield Reading insane ramblings, aha.

You keep on keeping on.

skill patrol

Do you swear around your professors @@0celo7

0celo7

@skillpatrol Around one, yes.

skill patrol

How about your dad

0celo7

Yes.

skill patrol

1:00 PM

Mom

0celo7

Yes, but less

skill patrol

Ok, why not all your profs then?

0celo7

I don't think they would appreciate it.

John Duffield

@PhysicsGuy : it isn't at some level like a textbook is. My aim is to make it conversational and explanatory and intuitive, with good evidence and references.

skill patrol

Why only one

John Duffield

1:02 PM

OK, I got go. Bye.

skill patrol

Do you think the profs in here appreciate it @0celo7

Physics Guy

@John Aha, good luck, then.

skill patrol

Bye

0celo7

@skillpatrol Beats me, I don't know them.

skill patrol

So that makes it ok

0celo7

1:03 PM

My thought exactly.

Glad you agree.

skill patrol

You don't respect strangers

0celo7

I actively disrespect them, actually.

David Z

@lucas well, if you choose not to use flags, that's your right - we can't make you. But flags are the way to point out situations that may need to be dealt with by moderators. If you don't flag problematic content, mods don't see it, can't deal with it, and in the end it hurts the site. Plus, it's a bit hypocritical to complain in chat or the comments about something that you don't flag.

0celo7

flags are for snitches

snitches get stitches >:)

skill patrol

The world is full of snitches

Like it or not

Jim

1:06 PM

the world is also full of stitches

Balarka Sen

@0celo7 So, what's up with the problem?

Jim

correlation implies causation, therefore all snitches must get stitches

Balarka Sen

@PhysicsGuy What do you mean by role? A TQFT is literally a cobordism-invariant, that's how they are defined. Or is that a physics question?

Jim

@0celo7 don't think of it like you're being a snitch. Think of it like playing minesweeper. You're flagging the mines so nobody explodes them and loses the game

2

Balarka Sen

Physical motivation, etc?

David Z

1:10 PM

@Jim ooh, good analogy

0celo7

@BalarkaSen I've solved it, but not by your method.

And I'm not going to try to figure out what you did.

So you can tell me or not.

Balarka Sen

Just to confirm, you have proved that any compact hypersurface in $\Bbb R^n$ has trivial normal bundle?

John Rennie

@NoahP See this discussion:

Mar 23 at 18:25, by John Rennie

BTW any thoughts on:

Balarka Sen

Without using J-B, that is.

0celo7

I proved the NX-X thing by hand, no sphere bundle.

Balarka Sen

1:12 PM

@0celo7 Oh, ok. Tell me your proof.

0celo7

I told you yesterday.

You said it was too complicated.

Balarka Sen

You said something vague, I couldn't make sense of it. Please retell.

Jim

@DavidZ I'm not playing several games of minesweeper right now.... What would make you think that?

Balarka Sen

I didn't say it was complicated, just that the idea can be simplified. But if you made a proof out of it, that's great.

David Z

@Jim no reason :-P

Emilio Pisanty

1:18 PM

brianbuccola.com/…

> How to cite in LaTeX without the citation appearing in the bibliography

↑ Prize for most bizarre thing to want to do in LaTeX, maybe?

Jim

@EmilioPisanty I'd like to see the other nominees first

Emilio Pisanty

@Jim Admittedly there's probably plenty on that list.

Noah P

@JohnRennie physics.stackexchange.com/questions/267854/…

John Rennie

@PhysicsGuy see hyperphysics.phy-astr.gsu.edu/hbase/astro/expuni.html but note this is not a rigorous derivation.

yuggib

@0celo7 in general in functional analysis is used ; and you can even drop the boldface

also you can use $\partial$ instead of $D$; often $D$ means $-i\partial$

Slereah

1:30 PM

So who's the next AMA fellow

0celo7

@yuggib I just need to write "all derivatives up to and including order $i$ of $h$ are bounded by $C_i$"

So can I write that as $\partial_\alpha h\le C_i,|\alpha|\le i$?

Jim

@DavidZ I think you're right that a date-based cutoff is inappropriate. Which is why I thought to choose a material based cutoff, like having to involve modern physics. If that effectively amount to the same thing, then perhaps we should brainstorm a different standard

David Z

Well, I guess I should clarify that I don't think a material-based cutoff is any better than a date-based cutoff. IMO

0celo7

@BalarkaSen No, I don't have a proof; don't care enough for that. Given some time I could certainly iron out the details. But it would get ugly because I'd likely have to deal with charts and stuff.

Balarka Sen

Oh well.

That's fine by me.

0celo7

1:38 PM

So you can either tell me what you did or not.

Balarka Sen

I won't :)

lucas

@DavidZ First: I think there must be a way to contact moderators that not to be secret. Second: I prefer to be known as a hypocritical rather than a hidden reporter in a scientific place! If the owner (any owner not only high rep users) of the post was seeing my flag, I certainly was using flags. But in the current conditions I prefer to not use.

David Z

There's the seed of a criterion in my mind which is based on whether a question could be answered without considering what actually happened in history. It'll take some thinking to turn it into an actual proposal though.

@lucas There are already ways to contact moderators that are not secret - we have this chat room. I'm always pingable here, and probably most of the other mods as well.

0celo7

@BalarkaSen Ok, I guess your proof is wrong.

Balarka Sen

It's not, not really.

0celo7

1:40 PM

Then you have no reason to not tell me.

How are you even supposed to join the math chat?

Balarka Sen

If you don't care for the problem, then why care for the proof?

0celo7

Where is the damn button on MSE

Jim

@DavidZ Does this seed come with any examples of what you mean?

0celo7

@BalarkaSen I don't care to spend another three hours on it.

Balarka Sen

@0celo7 Well, then, don't. I said that was fine.

lucas

1:41 PM

@DavidZ But you called me hypocritical because I had said my claim in chatroom or by comments!

David Z

@lucas However, there are not public official ways to bring problematic content to the moderators attention. (And there are very good reasons for that.) I mean, yes, you can ping one of us about it here, but that's unofficial, and your notice doesn't go on the record. We much prefer to have complaints put on the record as flags.

0celo7

@BalarkaSen But I gain nothing from that unless you tell me your idea.

I tried, and I give up.

David Z

@lucas I said it's hypocritical to complain that you were being insulted and that the moderators didn't deal with it when you didn't report that content to the moderators in the first place.

Balarka Sen

You probably won't gain much if I tell you how to do it either. But I am tired of this conversation: take a Riemannian metric on NX, look at the unit S^0 bundle. If NX - X is disconnected, so is the unit bundle inside. But that's a two fold covering space, so is like X x S^0. Pick the copy of X x {1} inside. That's a section of NX.

So NX cannot be but trivial.

0celo7

@BalarkaSen You're supposing I know that the existence of a section of a line bundle implies triviality.

I do not know this.

David Z

1:45 PM

If it's the private nature of flagging that bothers you, I recommend posting a comment when you cast a flag. You can publicly share your objections to the content using the comment, and you can get something done about it using the flag.

Balarka Sen

Oh well.

lucas

@DavidZ I think there is no need to report to a moderator. Moderators should check the site. I have a question, if no close votes cast, then moderators won't close any question? I have seen plenty of questions that have been closed by a single person, a moderator.

0celo7

You don't need a Riemannian metric if you use GP's definition of $NX$, either.

@BalarkaSen Well, I do know it, but I was trying to solve it using methods from GP.

David Z

@lucas In that case you are wrong. There is a need to report things to moderators. We do check the site, but we only see a small portion of what is posted. Most of what we deal with is brought to our attention via flags.

0celo7

And it appears the only way to solve it that way is to explicitly construct the diffeomorphism.

Balarka Sen

1:46 PM

Since X sits inside R^n, it has a Riemannian metric, which gives NX a Riemannian metric, yes.

0celo7

@BalarkaSen Once again, GP does not contain "Riemannian metric".

So please don't give me a problem I am unable to solve.

lucas

@DavidZ No one won't check all of my comments to know that why I have used flags. They just will look to the flags number!

Balarka Sen

OK, if you do not want challenging problems, you should have mentioned that when you asked for them.

0celo7

Why would I want challenging problems?

All they do is waste time.

Being able to solve challenging problems has nothing to do with understanding material.

David Z

@lucas I doubt that anyone looks at that flag number ever, and even if they do, it shouldn't matter

Balarka Sen

1:50 PM

I'm abandoning this conversation.

0celo7

@BalarkaSen Do you talk like that in real life?

I've just never seen anyone with your mannerisms, forgive me.

Maybe that's my problem with @Danu. I've just never met anyone who behaves like him and I'm at a total loss at how to respond.

@DavidZ Flag number?

lucas

@DavidZ It is not matter that who sees the flag number or how many people will see that. Until I believe that secret flagging is not a reasonable and good thing, I won't use it even if no one sees my profile.

David Z

@0celo7 on your profile it shows how many helpful flags you've cast

0celo7

@DavidZ Oh, right.

David Z

@lucas well, again, you're not required to use flags. Just keep in mind that complaining about bad content, when you haven't flagged it, will not get you much sympathy, and will probably not get anything done about that content.

lucas

1:56 PM

@DavidZ I see. But I hope moderators kindness. And even if they do nothing I won't claim.

Emilio Pisanty

@DavidZ cc @lucas, yeah, that, pretty much.

lucas

@EmilioPisanty What is pretty much?!!

Emilio Pisanty

@lucas It means I agree with David

btw @DavidZ

this one

@lucas ←this one