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22:07
@Sanya btw just added screenshots for the 'worthy' questions that are currently deleted
it took rather longer than I'd hoped, though.
@EmilioPisanty thank you, that is helpful :) sorry to dump that work on you
@Sanya no, that was me trying to be helpful and then realizing halfway through just how long the chores would take
btw, in terms of average questions/week, that's available here
@EmilioPisanty :D
PSE gets 84 questions/day on average
23 questions a week is something like 4%
your other 64 questions are closer to 11%
it is worth a debate whether we should not consider allowing well-formulated, thought-through, well-written homework questions showing work put in on the site then ... although that might of course quickly make the numbers explode
22:15
In the last 90 days we closed 29.4% of questions, 35,6% of those with the homework close reason. (<10k (?) link) Taking your 1/4 "with effort" for granted, we get that 2.6% of all questions were closed as "homework" when they showed effort.
@ACuriousMind how'd you get to that tab?
I can see it but I don't know how to get there
@EmilioPisanty Go to the 10k "moderator tools", go to the close tab, click on "question close stats" at the bottom.
@ACuriousMind gotcha
@ACuriousMind we need to say that honestly less than 10% of that 1/4 could remain open in their current state - but thanks for the numbers :)
@Sanya you're welcome to start that debate
I'm not sure people will be very happy
but then I've been surprised before
22:20
I actually don't think it will be popular either - I have to admit though that my personal dream would be that the homework question gets replaced by low effort
@ACuriousMind I think the more striking number is thatof those h/w-closed questions, only 0.67% got reopened
meaning we would not allow those "look xy up for me pleeeaaaaase" questions anymore, but high-quality homework instead
btw @ACuriousMind in this page physics.stackexchange.com/… do you see a bunch of Google Plus icons?
@EmilioPisanty Well, there isn't too much reopening going on with the rest, either...
@EmilioPisanty lol, yes. Pretty sure that should be an icon for an arrow or something like that
@ACuriousMind Yes. It's a bit on the low side, but we don't reopen much
(for those keeping score, 4% of Too Broad closures were reopened, and that's the highest there is.)
0
Q: Google Plus signs on the 10k tools migration stats

E.P.This is how the 10k tools currently looks like on Physics, more specifically the migration tab. Are those Google Plus icons intentional? 'cause I smell a bug =).

lolz
22:32
Hi, everybody.
@DanielSank, hello. Thanks for the link!
Sure. Too bad there's nothing on Hamiltonian mechanics.
I found a couple of textbooks online (goldstein, landau/lifshitz) that I think with youtube and google will be able to get me through =) my one debate is whether or not to get an actual copy of either.
@ACuriousMind Do you maybe understand the definition of formal linear combinations given in the question? I don't see how he gets $F = \sum_{i=1}^{m}a_ix_i$?
yo y'all, look what I found: http://cdn.sstatic.net/Sites/physics/img/sprites.png has
user image
3
22:40
@Alex That equality is basically the definition of a formal linear combination of $x_i$.
@EmilioPisanty Heh, someone wanted to save space and now the migration page is taking its image from the wrong point in that png?
... so apparently all the arrows and whatnot are just directly cut out from a single image?
Hello

I have a short question related to electromagnetivity.
@ACuriousMind Oded says that the "sprite" thing is wrong - there should be an arrow where the g+ is
What is $j\omega \epsilon$ in this formula: imgur.com/a/Ve6PX
@heather Goldstein is the standard grad level book.
22:41
but that's some messed up html black magic right there
It's ok, I guess.
@heather as much as I hate myself for saying this, but my experience is that the books you can not find as pdf cost enough to make buying the ones you actually have as pdfs seem pretty unattractive
@DanielSank I'll second that
@Sanya wait, what?
@EmilioPisanty is there a good book on Hamiltonian mechanics?
22:42
@DanielSank ah
so I learned my analytical mechanics from Lanczos
actually, yeah, scratch the hesitation, Lanczos is awesome
@Sanya, I'm sorry, but I'm confused as to your meaning.
@ACuriousMind Yeah I know that is the usual definition but I don't see how gets from defining the formal linear combinations as mappings from a set $S$ $\to \mathbb{R}$ and then ends up with that definition. Do you see how he gets that?
@EmilioPisanty, Lanczos? okay, googling...
@trilolil $\omega$ should be the frequency, $\varepsilon$ the permittivity, $j$ might be a current, but I'm not sure without having looked closer at the context
"the variational principles of mechanics"?
Dover therefore dirt cheap
oh, yep, okay
cheap = happy
I think I was always confused by the way physics books fail to distinguish between variables and functions.
this is the source: it is on the first page, you won't have to read a lot. That is what I thought as well but was not sure. I am just not sure whether j=current...
http://whites.sdsmt.edu/classes/ee382/notes/382Lecture32.pdf
22:44
wow =) it looks good
@Alex You define a map from those functions to the formal linear combinations in the usual sense by $f\mapsto \sum_{s\in S} f(s) s$, and dropping all the terms with $f(s) = 0$. This is a bijection.
I also had Gantmacher for that class, it's slightly older but it's still plenty good
@heather this must be the german habit of too long sentences. What I meant is - in my experience, I need to spend enough money on the books that I cannot find as pdf anywhere already. Therefore, I feel that buying the books that I do have as pdfs is a luxury I can't really allow myself. I hope that was better :|
@EmilioPisanty yeah, Dover is great.
22:46
@Sanya, oh, that makes sense. because books that you can't find as pdfs cost so much, you shouldn't buy books you already have, because that's even more money?
but that one might be harder to get hold of
oh, geesh
my edition is in Spanish and published by Editorial URSS
$500 (::gulps::)
As an aspiring book writer, I'm somewhat annoyed that people so often download illegal copies without a second thought.
22:47
@heather yup, that's what I meant
@heather that's unlikely to be representative
but it's out of print so it'll be harder to find
@Sanya you sure?
@DanielSank did you ever contact a publishing company? Springer offered us a laughable amount for an estimated 250-300 pages book - so not like the author is going to suffer. Not that I want to defend it - the ebooks I have are usually ebooks I am allowed to have.
EUR 50 at the publisher
@trilolil no, I actually think his $j$ is the imaginary unit
see his second slide, down
22:49
€18 in Spanish =P
@Sanya I think so as well. But are you sure about $\omega$ ? This would imply omega is in Hz or so....
if the current should be $j Q \omega$, then $j$ shouldn't have units
yeah, omega seems to be used quite uniquely as frequency only in my experience
@trilolil yes
$\omega$ is the angular frequency
@trilolil which would mean his current is in units of C/s which makes sense
and $j^2=-1$
22:51
@Sanya uh huh. And how many times have you mailed the authors a check for what you think the book is worth?
If that is an engineering textbook, then $j=-i$, where $i=\sqrt{-1}$.
@EmilioPisanty What makes you think that if I may? Indeed \omega usually refers to some angular frequency, but maybe it might not be the case here.
@DanielSank never, of course. I'm just saying that it isn't really the author who suffers big time from that - on the other hand, I won't say that stealing is good, of course
Agreed
@trilolil that would be a breach in convention of such a magnitude that the blame for the misunderstanding is on them
22:54
@EmilioPisanty It is just in a very strange form IMO. This would mean that the higher the frequency of the current the lower the electric frequency.
@ACuriousMind I'm studying independently hence I lose some of the usual methods. So is the general result: If $\{ f : f: s \to \mathbb{R}~~~ \text{where} f(s) \neq 0 ~~\text{ for finitely many } s \}$ then we can define a bijective mapping $\Phi$ on that set $$\Phi(f) = \sum_{s \in S}f(s)s$$?
@trilolil Why are you surprised?
You've got $j\omega \epsilon \vec E = \nabla \times \vec H$.
@EmilioPisanty I would intuitively expect the opposite.
that's the Faraday-Lenz law right there
@trilolil if you change the frequency then the mode also changes and $\nabla\times\vec H$ also changes
@EmilioPisanty what do you mean by "mode"?
22:56
7
Q: What is a mode?

CasimirAdmittedly, this seems like a very simple question. The word mode pops up in every field of physics, yet I can't remember ever having read what I felt was a precise and sensible definition. After having searched fruitlessly on this site as well, I feel that even though it seems like a trivial qu...

the spatial dependence of the force fields
mode = pattern of a wave.
(in physics)
@trilolil yes
roughly
@EmilioPisanty force?
@DanielSank force fields = E and B
@EmilioPisanty oh, I suppose. We have modes of a string too...
23:01
@DanielSank oh, sure
@Alex yes
@ACuriousMind :) Okay thanks. How is that standard result? Do you maybe know where I can read up on this?
@Alex I'm not sure what's left to say, and I don't know many books, sorry
2
Q: An attempt at a homework-closure statistic and a bit of input concerning the ongoing policy debate

SanyaSummary I made a week-long list of all the questions closed due to the homework flag and tried to gauge whether they were showing any effort at all. This idea came up during a chat session when we were talking about the new physics Q&A site proposal and our homework policy. The hope is to give a...

@ACuriousMind Okay no prob, I'm just interested to see at what level this result is introduced.
23:09
to everybody/anyone:
Is there a good trick to be able to know when people are writing the nable operaton whether they mean gradient, curl or gradient?
I always confuse them all as they all seem to use the same operator.
@trilolil Sure: $\nabla X$ is a gradient, $\nabla \times X$ is a curl and $\nabla \cdot X$ is a divergence.
@ACuriousMind what is the difference between gradient and divergence?
they both seem to be a multiplication.
OOh nono, gardient: differentiation and divergence is just a dot multiplication. Right?
@trilolil In one case, $X$ is a scalar, in the other, a vector. I wrote $\cdot$ purposefully there - if you think of $\nabla$ as $(\partial_x,\partial_y,\partial_z)$, those should make sense to you
@ACuriousMind mhm
@ACuriousMind $\Nabla X$ might mean $\partial_i X_i$ or $\partial_j X_i \vec{e_i} \otimes \vec{e_j}$ depending on the author, wouldn't it?
23:13
@ACuriousMind thx :)
actually, never mind, if $X$ is a scalar ... sorry
@ACuriousMind But how is the addition of $\Phi(f) = \sum f(s)s$ defined if $S$ is just a set and not a vector space?
23:33
@EmilioPisanty cool!
@Alex Well...the $\sum_{s\in S}a_s s$ is really just a notation for a tuple of numbers $(a_s)_{s\in S}$ indexed by elements of the set $S$. If you want to be really precise, you should probably just start with the functional definition of the free vector space and define writing "$s$" for the function that's 1 at s and 0 everywhere else. Then clearly you can write every function as $f = \sum f(s) s$.
@ACuriousMind Is $\sum$ in $f = \sum f(s) s$ a sum in the usual sense? or a tuple?
@Alex Yes, it's pointwise addition of functions if we defined $s$ to be the function 1 at s and 0 elsewhere.
23:49
@ACuriousMind As you go through the sum $\sum_{s \in S}f(s)s(s)$ would you get $s(s) = 1$ for every $s \in S$?
and would this then effectively give $f(s) = f(s_1) + ...+f(s_n)$?

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