@ACuriousMind Is the derivative normal exponential map off a totally geodesic hypersurface an isomorphism of the orthogonal subspaces of the tangent vector of some geodesic

I estimate the chance of you having any idea what I'm talking about at roughly 1%

Oh, assuming no focal point along the geodesic.

@0celo7 Good guess :P

@ACuriousMind Understandable, this is beyond Milnor. I think your caring about Riem geo ends about half-way through that book.

@Obliv I think you actually can now.

What's your motivation for using Word though?

@DanielSank I can put Excel tables in Word easily.

@0celo7 Indeed.

Is that a good thing?

Images are easy too. Also our lab reports have to be TNR 12pt double spaced.

@0celo7 TNR?

Times new roman

I see.

@DanielSank last lab had like 200 data points from excel. So yes.

What do you mean "from excel"?

I recorded the data in excel during the lab

Does the program running the experimental apparatus automatically put the data into Excel?

@0celo7 Ah, so it was user error.

No, but I used Excel to compute all the stuff

Or as they say, PEBKAC

@DanielSank sigh.

@DanielSank uhhh

Problem Exists Between Keyboard And Chair :)

Positron Emission Boys Kill All Children?

Oh.

Anyway, Excel/Word are fine.

You can dump your excel file as csv though.

I type my upper level homework in TeX.

But I'm not going to fuss over EM lab reports, word is fine for that.

tablesgenerator.com/latex_tables

The biggest gripe I have with Word is that it's not human readable text so it's terrible in version control.

@DanielSank I once did a high school lab report in TeX

It's also disappointingly annoying to write things modularly.

@0celo7 Didn't like it?

Got a 100, teacher said it was amazing

Took too much time!

@0celo7 Surprised to hear that. Were you using a crap editor?

Beats me. I use the first thing that popped up when I googled TeX

It works

But writing the report took too long.

So now you're using Word.

Interesting.

Yes. Making tables, putting images, etc.

Images are hard in TeX?

I had to insert pictures and diagrams.

You want my magic macro that makes it easy?

@DanielSank Yeah. I have to look up the syntax every time, it's impossible to remember.

@DanielSank I don't know how to use macros.

alright gimme a sec...

@0celo7 -_-

% Figures. Example usage:
% \quickfig{\columnwidth}{my_image}{This is the caption}{fig:my_fig}
\DeclareRobustCommand{\quickfig}[4]{
\begin{figure}
\begin{centering}
\includegraphics[width=#1]{#2}
\par\end{centering}
\caption{#3}
\label{#4}
\end{figure}
}

Put this in your preamble and then to use it do what it says on the second comment line.

The arguments are width, image file, caption, label.

For width, you can use \columnwidth.

@0celo7 I'm almost done with a public journal post about using LaTeX effectively. Will post here when it's done.

@DanielSank My senior thesis will be a lengthy manifesto on geometry...no pictures allowed.

But the TNR requirement for physics labs makes Word pretty much mandatory.

What is the trouble with using the TNR font in TeX?

@ACuriousMind Can one do that?

Of course you can change the font!

Huh.

Computers are weird.

@0celo7 wtf?

@0celo7 You are generally aware that computers exist to make our lives easier, yes?

@DanielSank I struggle with them too much for me to believe that.

I see.

A common problem.

Simply seek help as you do for other topics.

By begging?

No.

Beware: computer people typically expect appeals for help to come with explicit description of what the asker already tried to do.

In a sense, the computer whiz community is the ideal Stack Exchange user base. Not an accident, of course.

Of course, some hacker communities are too discouraging of noobs, which is bad.

I once tried to change my TeX font.

It looked absolutely horrible to do and I gave up.

I have a question for the mods: are we allowed to make critical statements of published materials? For example, can I say "Griffith's quantum textbook is not good and I wouldn't recommend it to be used by any student"?

@DavidZ @dmckee

Follow up question: if the answer to the above is "yes", can we make such statements regarding materials published by users of this site?

@DanielSank Do you mean answers or textbooks/notes/JD's book?

@DanielSank Where do you want to make such statements? In chat, in comments, in answers? All of the above?

@ACuriousMind Chat.

And I don't want to. I'm just asking how the rules work.

Because I think there's a difference between leaving a comment/chat message on a post recommending Griffith and writing an entire answer denouncing it.

@0celo7 Interesting question. I was thinking about published materials that are not posts on the main site. That said, I would extend the question to include posts on the main site.

@ACuriousMind I'm only talking about chat.

@ACuriousMind That does bring up a new interesting question though: if I think a particular book degrades the quality of physics education, can I say that in chat or in a post on main?

These are all interesting questions.

@DanielSank I personally would think that such statements about textbooks are fine, and maybe sometimes necessary

For now, let's stick to comments made on chat which criticize any published works.

I think there are several examples of me and @0celo7 decrying the quality of string theory textbooks already :P

@ACuriousMind Interesting. So IYHO we're still being nice if the criticism is limited to the work itself?

@ACuriousMind Weinberg too.

@ACuriousMind Doesn't mean it wasn't against the rules.

@DanielSank True, just saying there's precedence

@DanielSank Yes, and this would extend to any other published material. I don't see any policy which "I think X is wrong/misleading/misguided because Y" would violate

@ACuriousMind Indeed.

So I can say "ACuriousMind's paper on XYZ is wrong because ABC. Don't read it or you'll pick up all kinds of misinformation"?

I really dislike ACM's BS thesis. Garbage book.

@DanielSank I see no issue with that, though if you did that with something I actually wrote that would of course start a lengthy discussion ;)

@ACuriousMind Well, I meant what I said. Your paper on XYZ is pretty bad.

Whenever JD does that to me someone gets banned.

Maybe it's not allowed.

I mean, you completely forgot the results of Deed and Treadit.

@DanielSank Ah, but you have misunderstood me! It's not about "XYZ", it's about WXY! I never claim to do Z.

@ACuriousMind It's strongly suggested that Z is just a special case of the other claims in the paper (which I think are wrong anyway), so you should have been more clear in the conclusions section. You're misleading hapless readers!

::Puts on crown::

Ah, but had I been more clear the paper would have looked too obvious!

Can't have that, can we?

::stumbles due to knife in the heart::

You've won the day, @ACuriousMind. But I'll be back.

I'll await the rematch ;)

::wraps self in cape and disappears::

@ACuriousMind does a compact set contain its sup and inf? Surely yes because you can take a minimizing/maximizing sequence that has to converge.

Hmm, it doesn't have to converge.

Maybe it needs to be monotone

@DanielSank Do you know how I can do this? I have to use word because my home computer's keyboard controller broke. Unless I want to do these derivations with an on-screen keyboard I'm forced to use word. (Not physically able to install new software on these computers)

@ACuriousMind You there? I'm not convinced by the Dyson series.

I'm here. What are you not convinced of?

@Obliv Do I know how you can do what?

@ACuriousMind So the idea is to find a first order approximation to $U(t,t_0)$, then plug it back in and generate higher order terms. Am I right?

Yes

How do I know I'm not losing information this way?

You don't, a priori. You just take the Dyson series at the end, plug it into the evolution equation and find it solves it

And since it solves the full equation, not just an approximation, you didn't lose information

@ACuriousMind So you need a Picard-Lindyhop for operator ODEs?

@0celo7 Uh...if you want to know that that solution is unique, I guess you do

@ACuriousMind Does such a thing even exist?

I don't know

I mean, what the hell does $\partial_t U(t,t_0)$ even mean?

@ACuriousMind I thought you liked the mathy bits of this stuff

The theory of operators that depend on more than one parameter tends to get ugly quickly

For one-parameter $U(t)$, you have wonderful stuff like Stone's theorem. For $U(t,t_0)$, I don't know of any "nice" results

@ACuriousMind What

it's still one parameter

$U(t,t_0)=U(t-t_0)$

@0celo7 That's only if your Hamiltonian isn't time dependent, in which case there's no need to do the Dyson series, the evolution operator then is just $U(t,t_0) = \exp(-\mathrm{i}H(t-t_0))$.

@ACuriousMind Do we know that $\exp (-\mathrm iH(t-t_0))$ converges?

What does that expression even mean?

But if you are in a situation where the Dyson series is necessary, then it's not so simple, and $U(t,t_0)$ is a true function of two parameters

@0celo7 Yes, by functional calculus (the way of defining functions of operators by applying the function to the eigenvalues we already talked about very recently)

I don't remember that

what was the conclusion

You don't remember complaining that you only had a "physicist's proof" of what the commutator $[P,f(X)]$ is?

I complained to my E&M TA who is also in my class

He was confused about what a "physicist's proof is"

If you want to refresh your memory, the conversation starts here:

Aha, yes, I remember.

@ACuriousMind So that operator converges even if $H$ is unbounded?

Yes

@ACuriousMind what conditions do you need on $H$

Self-adjoint.

Do you know the proof?

In what topology are we considering $\exp\mathrm -i H$ anyway

@0celo7 Yes, like the rest of this, it's in Reed & Simon, as I said.

@ACuriousMind ayy Reed & Simon is PhD level analysis

@0celo7 I don't think we need a topology (the space of unbounded operators might not carry a nice one) since we aren't really using the power series definition of $\exp(-\mathrm{i}Ht)$ when we do functional calculus

So we don't need a notion of convergence for that series. It's been a while since I've read about this, there might well be a topology here.

Hi, everybody.

@DanielSank Hi doctor nick!

@ACuriousMind What are we using??

@ACuriousMind My probability book says "recall from algebra that $\sum r^k=(1-r)^{-1}$

How is that algebra???

15 YEARS LATER: ON THE PHYSICS OF HIGH-RISE BUILDING COLLAPSES / Jones, Korol, Szamboti, Walter, europhysics news

@0celo7 have you made sense of Sard's proof

@bolbteppa sure.

What is the jist of it

Induction

I'm busy right now

search the chat for a summary of the proof I gave Balarka.

How long ago

looking.

Think I understand most of step 1, but it's a bit wobbly

What book are you reading?

Milnor

what part is confusing

step 1

I don't have milnor right now

what part of step 1

gp is basically the same

I know

I'm not at home

I have Cheeger & Ebin with me and do Carmo

So you're analyzing the Jacobian right, when is $\mathrm{rk}(df) < p$ in $f : R^n \rightarrow R^p$, well it can happen when all partial derivatives $\partial_i f/\partial x_j$ are zero, or when only some of them are

what is the rank of the identity matrix

oh wait, is $n>p$?

or $p<n$

or does Milnor not make any assumption

No assumption

It seems like you can already predict we'll use Fubini from this information

let me get out my notes on the proof

I did understand all the details at one point.

wtf

where does he talk about rank

in step 1?

@DanielSank what?

Mississippi ratified the 13th amendment in 1995.

@DanielSank You're 21 years late on that TIL, bub

If you analyzed the case where not all $\partial f_i / \partial x_j$ were zero, say $\partial f_1 / \partial x_j$ then we expect the critical values to be determined by the terms $(f_2,\dots,f_p)$, so the critical points live in $R^{p-1}$ implying Fubini would lead to measure zero

@0celo7 That's just silly though.

Milnor says rank in the statement of the theorem

@bolbteppa No, not all derivatives have to be zero

There are a few examples of late ratifications of crazy stuff I think

that's why $C-C_1$ isn't empty

but step 1 is showing that it has measure zero

the idea of the proof is basically this:

you have your critical set $C$

now, this set contains the parts where $df_x$ is singular, but also where ALL partials vanish

and also the one where all SECOND partials vanish, along with all first ones

so that's how you construct $C\supset C_1\supset C_2\supset\cdots$

Now, the goal is to reach a $C_k$ that you can prove has measure zero explicitly.

Hold up

Why even think to consider second derivatives

@bolbteppa Because you want $C_k$.

The proof is backwards, honestly.

there is a good reason

Read step 3 first.

No, there is a cool reason, I figured it out yesterday

Why did he think of that insane decomposition

He didn't, the proof is not due to Milnor.

There is no reason to think of 2nd derivatives given the statement of the theorem

But you can constructively get them and that decomposition naturally

@bolbteppa I think Hirsch's proof is better.

Don't remember Lee's proof.

I'm not doing your homework.

Lee is basically Milnor too, is Hirsch different

How would I remember?

I was playing BF4

I used to be good at physics :(

I really don't remember

probably.

yeah?

So again, we are analyzing the Jacobian, we know it is zero either when all derivatives are zero, or when only some are zero. First take the case when some derivatives are non-zero, assume $\partial f_1/\partial x_j$ is non-zero for simplicity, then we expect the critical values to be determined by the $(f_2(x_1,\dots,x_n),\dots,f_p(x_1,\dots,x_n))$ terms, not $f_1(x_1,\dots,x_n)$, but then we would be analyzing the measure of $R^{p-1}$ in $R^p$ and could apply Fubini's theorem,

so lets formalize this intuition. Maybe we could find a change of variables for $f$ so that the first coefficient $f_1(x_1,\dots,x_n)$ became trivial,

@IceLord Probably.

So this seems to be why gp/Milnor even use that stupid $h$ function in their proof

You want to make sure $(x_1,\dots,x_n) \mapsto h^{-1}(x_1,\dots,x_n) = (f_1^{-1}(x_1,\dots,x_n),\dots )$ so that $f \circ h^{-1} (x_1,\dots,x_n) = (f(f^{-1}(x_1,\dots,x_n)),\dots ) = (x_1,\dots )$

@IceLord Exercise: if $\partial M=0$ for $M$ a manifold, then there is a manifold $N$ such that $\partial N=M$.

It's related to cobordism, and topological QFTs are functors on the category of cobordisms.

you didn't "forget it"

you didn't learn it

So how do we make sure some $h^{-1}$ exists so that the above will hold? We want to construct some $h$ motivated by the above, obviously $h(x_1,\dots,x_n) = (f_1(x_1,\dots,x_n),x_2,\dots,x_n)$ where putting $f_1$ in here now makes complete sense, the rest doesn't matter so long as the jacobian is non-zero I think