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12:53 AM
@bolbteppa that’s not quite right. You can turn on link sharing and allow anyone to view the file by sharing the “view only” link. To edit a file one needs an account, even through the link sharing. For someone (not the owner) to access the file through her or his directory (which implies an account), the file must be shared. It is a step back from Overleaf_v1, especially as one can only share the file with a single collaborator unless you have a pro account (not i expensive).
1 hour later…
2:21 AM
What ever happened to 0celo?
1 hour later…
3:45 AM
That book has an article exactly on the topic I've been writing about
I should read more experimental books!
just realized i've been mispronouncing lie groups up till now
textbooks ought to have the pronunciation in the definition
Did they...
lie to you
4 hours later…
7:53 AM
@ACuriousMind I mean that seems like a way to make math feel arbitrary and confusing
not that i claim more knowledge about the subject than von neumann
but that quote always seemed weird to me
1 hour later…
9:05 AM
@JohnRennie In that video the narrator says that nobody knows where those dark theme came from in Waters’ mind. What are your speculations about the origin of dark thoughts in his mind?
@Knight I don't know Roger Waters so I can only guess.
I don't think he had a happy childhood, and I suspect he's a bit prone to feelings of angst.
And being in a famous rock band is enormously stressful. Lots of rock stars have had breakdowns under the pressure. In fact many have committed suicide.
A: Cause of weightlessness

HelenI AM Making a Good Salary from Home $6580-$7065/week , which is amazing, under a year ago I was jobless in a horrible economy. I thank God every day I was blessed with these instructions and now it's my duty to pay it forward and share it with Everyone. go to home media tech tab for more detail r...

Is this what we call spam? The first spam I have seen on this site.
Yes. Flag as spam if you have enough rep. If there enough spam flags it will automatically delete the post.
9:26 AM
@JohnRennie Wow my spam has deleted the answer, I never thought I hold so much of the power.
@Knight :-)
@JohnRennie Was his childhood something similar to Jake Lamotta?
One of the key ideas of the SE system is that as your rep increases you can start helping to moderate the site by doing things like flagging spam posts.
Yes. I think it would take me just one and only one millennium to get reputation equal to yours (the current reputation of yours) :)
:-) My reputation just builds by itself these days. I hardly answer questions any more because I'm busy all the time in the chat, but old answers keep attracting upvotes.
9:32 AM
have you ever thought of doing some philanthropy? I mean like donating some of your reputations to people like me?
I suppose I could start awarding random bounties to transfer rep.
I think you could write my name in your will of reputations of SE
9 mins ago, by Knight
@JohnRennie Was his childhood something similar to Jake Lamotta?
I don't anything about Jake Lamotta's childhood.
He was born in a locality where killing, murdering and what not was a common thing. When he became 6 years old his father used to exhibit him in front of all locality by making to fight vigorously with other kids (bigger than him).
Waters came from a reasonably wealthy family so he didn't suffer from poverty. But his father was killed in the second world war so he grew up with no father.
He was sent to boarding school from a young age and I think he didn't like school.
And I'd guess, though I'm not sure, that he didn't get on with his mother.
So, completely unlike Jake Lamotta!
9:43 AM
So having such a good life (Waters was married also) what caused him such dark thoughts. Was he same as philosophers? You will find that most of the philosophers were wealthy therefore they never did any work.
Waters has been married and divorced multiple times. I suspect he's a hard person to live with :-)
It may be that his voice is so heavy and eardrums may need to be operated time to time for his partners.
No, I suspect he's just a hard person to live with.
You're always right, reality may be occasionally wrong.
I must make it clear I don't know Roger Waters. I have never met him (though I have seen him on stage) so I have no personal knowledge of what he's like.
But the other big name from Pink Floyd is David Gilmour, and in contrast to Waters David Gilmour seems a relaxed and likeable chap.
9:51 AM
If you ever get time then please just listen this song
@Knight yes, it's a nice song.
@JohnRennie What did you feel by the song?
@Knight it's a nice song. It isn't really my sort of music.
@JohnRennie Oh! Do you like screamo?
10:03 AM
with full strength
I'm not sure how to best characterise the music I like. I like lots of different styles of music but I have a taste for music that is a bit weird. Some of the music I like is loud, but some is soft.
For example I like Black Sabbath and I also like Tangerine Dream, and they are about as different as you can get.
I have to listen to them, I have never heard about them,
There are more bands than you could listen to in a lifetime, so I wouldn't worry about having to listen to everything other people mention.
Just listen to what you like.
10:09 AM
@JohnRennie But I hope that eventually you will like Poets of the Fall.
10:20 AM
@JohanLiebert Do you know @AbhasKumarSinha at any other site?
@Knight no I never asked him about that. Though it is possible that might have a fb account.
@JohanLiebert Would you please search for him?
@Knight OK! But I don't know how he looks like. So I would search for his city related info.
@JohanLiebert Which city is he in?
@Knight Bandhabahal
@Knight I searched though I wasn't able to pin point his account
11:07 AM
The mystical experience of the ghosting emotion:
whenever I detected an instance of (usually Facebook) ghosting, the following experience is felt:
1. There is a loud tinnitus that floods the background
2. Time stops in its track, visual field locked in the current zoom
3. Vision turned pitch blood red as an intense feeling of freefall more than 3 g is felt
4. The redness fades back to normal, along with the ceasing of the freefalling, with an aftertaste of intense dread, now felt extremely tangible black cubic soot, locked in a rocky cavern
11:21 AM
@Mithrandir24601 Note sure if you recall my previous Big-O notation query. You stated the following: $$\partial_\theta\left(\mathcal O(\theta^2) + \mathcal O(\theta^3)\right) = \partial_\theta\mathcal O(\theta^2) + \partial_\theta\mathcal O(\theta^3) = \mathcal O(\theta) + \mathcal O(\theta^2)$$Am I correct in stating that the derivative of these big-O terms can only be proved to exist at $\theta = 0$, hence we assume the above holds only for small $\theta$?
A: Cause of weightlessness

Manoj KumarThis question is answered really well by the answers above. Of course, as explained in all answers the statement quoted from your textbook is incorrect. The phenomenon of weightlessness could be explained in two different ways: The object orbiting the planet is in free fall (as described in oth...

@JohanLiebert the question has been protected now so that should stop any more spam answers.
@JohnRennie yes I looked at the edit history and it looks like the community user has protected it (due to the fact that two spams were deleted today?)
@JohanLiebert yes. I didn't realise there was a mechanism to automatically protect questions.
11:54 AM
@JohnRennie That feature is rather handy, especially on SO, where popular questions often attract a bunch of low quality or copycat answers (eg answers that duplicate code in existing answers, but with no explanatory text). In extreme cases we clean them up, but you can chew up a lot of delete votes rather quickly that way.
Feel free to flag questions for moderators to look at, if worried that they need protecting
The spam ones tend to get caught quickly and auto-protected, but drive-by VLQ answers tend to accumulate
My theory is that new answerers use such questions for practice, figuring that they can't do too much harm if they're wrong, and they might even score an upvote or two if they're lucky... or if they have school friends reading their posts...
12:26 PM
@JohnDoe I mean, technically, $\mathcal O$ means that it's upper bounded in the limit of $\theta \rightarrow \infty$ (and yeah, I could and will below argue that they're abusing this notation in that paper, but practically, I think it's understandable), but that statement is still right $\forall \theta$. However, that equation given in the paper is only true for small theta
So, what they did in equation 25 is technically wrong (but understandably correct, if that makes sense?)
They gave that $$\langle S_m \rangle = 1-(\theta^2/2)\mathcal{Q}(G) + \mathcal{O}(\theta^3)$$
Can anybody confirm that absolute state of rest in space is impossible.
That, s what Einstein has said to us!
But they also gave that $$\langle S_M\rangle = \sum_{m=0}^\infty \frac{(i\theta)^m}{m!}\text{tr} \left(\left(\sum_{i=0}^n X_i\right)^m G\right)$$
So, unless all terms with $m>0$ cancel, saying that this is $= \mathcal O(\theta^3)$ is actually technically false
So, this only applies in the limit of small $\theta$
And using $\mathcal O$ is using this notation in a regime where it just doesn't apply
So the equations they give only apply in the limit of small $\theta$
@YuvrajSingh... I'd argue that the statement should be more along the lines of saying that 'absolute rest' as a concept is wrong. It's not 'possible' or 'impossible', rather the concept itself is just not a thing. I picking very fine hairs here though. Like, to say it's impossible to me is implying that it can be defined, whereas I'd argue that being at 'absolute rest' isn't something properly definable. Might just be my odd way of putting things though
12:48 PM
@Mithrandir24601 You state " that statement is still right $\forall \theta$"...but how would you show that $\partial_{\theta}\mathcal{O}$ exists for all $\theta$?
It's really just an asymptotic upper bound statement, this is not enough to conclude differentiability for all $\theta$
@JohnDoe Doesn't need to be differentiable $\forall\,\theta$ though, does it? it's just saying that it's differentiable $\forall\,\theta > \theta_0$ for some $\theta_0$
But what they appear to mean in this paper (this is physics remember!) is that $\mathcal O(\theta^m) = x\theta^m$ for some matrix (or number or something) $x$ that's independent of $\theta$ and this is why I'm saying that it's technically an abuse of notation
@Mithrandir24601 Yes the abuse started with the approximation of the higher order Taylor series terms using the Big-O notation to begin with...but as you say it is physics...
1:48 PM
Does anybody understand the "Killing form"? I'm trying to show that the $(2,0)$ Killing tensor is invariant under the $\text{Ad}$ homomorphism: $K(\text{Ad}_A(X),\text{Ad}_A(Y))=K(X,Y),$ with $X,Y\in \mathfrak{g},\hspace{1mm}A\in G,$ and $K(X,Y)=-\text{Tr}(\text{ad}_X\text{ad}_Y)$. I currently have no idea how to interpret that trace.
Trying to peer into it, I'd write $\text{Tr}(\text{ad}_X\text{ad}_Y) = \text{Tr}([X,[Y,\cdot]]),$ which just heightens my feeling that this needs an argument to make any sense. How am I to interpret that trace??
Find the magnetic field at point P on the axis of a tightly wound solenoid (helical coil) consisting of n turns per unit length wrapped around a cylindrical tube of radius $a$ and carrying current $I$ (Figure 25). Express your answer in terms of $\theta_1$ and $\theta_2$ (it's easiest that way). Consider the turns to be essentially circular, and use the result of example 6. What is the field on the axis of infinite solenoid (infinite in both directions) ?
My doubt is how $\theta_1$ gonna become 0 and $\theta_2$ $\pi$ as the length of solenoid gonna go to infinity. Please help me in seeing that.
@Mithrandir24601 OK then how do we measure age then?
@YuvrajSingh... Number of oscillations of a caesium-133 atom that you're stationary relative to
@YuvrajSingh... No, I'm not angry with you. And I wasn't angry with you yesterday either, as I said before. I was upset & disappointed with you, but I forgave you after you apologized, deleted the plagiarised material, and promised not to do it again.
@Mithrandir24601 actually I do not know about that, I heard about it that we use wavelength of caseium,!
@PM2Ring ah, thanks sir, I was in stressed from yesterday! That I might angry you.
2:04 PM
@PM2Ring Do you understand the "Killing form"?
@YuvrajSingh... We use the frequency of a caesium oscillation, not the wavelength. I suppose we actually use the period of that oscillation as the basis for our definition of the second.
@dsm Not well enough to answer your question properly.
Alright no worries. I have no clue how to interpret what is meant by that trace...
@PM2Ring make me correct that mean excitation of electron from ground to higher orbit is that what it mean?
@Knight it becomes infinite in both directions, not just one. Hence the ends are infinitely forward from the origin (theta = 0) and infinitely backwards (theta = pi)
There's a nice description here that gives the main details without getting too technical: NIST-F1 Cesium Fountain Atomic Clock
2:09 PM
If you make it infinite in only one direction, then you’re at one end of an infinite solenoid rather than at the center of one
@PM2Ring it is not opening sir!
It opens for me just fine, for reference
One more question it is quite basic one what make ice to change it s phase, is it heat or is it the temperature difference. Question can be little bit stupid when one can say heat depend upon temperature difference but still if we look at the graph of ice Vs temperature graph is initially straight and then become parallel to axis then again go straight.
@YuvrajSingh... Well heat is what actually drives it to change phase. That's what the latent heat of fusion measures. The temperature determines when the added heat goes into changing phases instead of changing temperature, but it's the actual heat that melts it.
@YuvrajSingh... The caesium clock uses transitions between the two hyperfine ground states of caesium-133 atoms. It's a bit hard to explain exactly what that means, but roughly speaking, instead of there being a single ground state for the valence electron of caesium there are 2 states caused by the interaction between the electron's magnetic dipole & the electric field created by the nucleus & the other electrons. See en.wikipedia.org/wiki/Hyperfine_structure
2:21 PM
Having said the above, I’m having a hard time getting the trig to work out properly
@j Mac @PM2Ring Thanks.
@YuvrajSingh... That's a shame. It's a USA government site, I don't think it would be blocked in India. Wikipedia has an article, but it's not as good, and it doesn't have the nice diagrams that the NIST site has. en.wikipedia.org/wiki/Atomic_fountain
@Semiclassical Sir if you don’t mind much can you please make a sketch of it?
Can’t. I’m on mobile
Okay. Let’s say that distance from forward center to point P is $d$.
2:25 PM
@Semiclassical what was that book you recommended for me again? Can’t find it in the transcript :p
I’ll suggest $z$ instead, to avoid conflating with differential d
@JakeRose shankar
Okay. Let’s call it $z$
It’s a standard grad QM book. Big thick and red
Seemingly we should have $\tan \theta=a/z$
$\sin \theta_1 = \frac{a}{a^2 + z^2}$
2:27 PM
But that goes haywire for small $z$
@YuvrajSingh... I remember "discovering" this when I was a child, playing with icecream in a cardboard cup. I discovered that if I quickly stirred the icecream it would start to melt but it also made the cup feel much colder: it was pulling heat out of my hand so it could do the phase transition.
@Semiclassical Means?
$\tan \theta_2 = a/l $
l is the length of the solenoid.
no. $\tan(\theta_2)=a/(l/2)$
@PM2Ring then how you find the reason?
We want it to be infinite in both directions, so half is in front of the origin and half is behind
So $\tan(\theta_1)=a/(-l/2)$
2:31 PM
But I think $\theta_2$ is the angle that the farther end makes with the point P, ?
Yes, but the front end is $+l/2$ away from the origin and the back end of $-l/2$ away
Otherwise you’d be stretching it in one direction, not both
@Semiclassical any particular sections? Just got it from the mathematical library so all is good
@YuvrajSingh... Well, I didn't understand it when I was 9 years old, but it made sense a few years later when I learned about latent heat.
Don’t remember off the top of my head
It’s a big book tho
Sir, I thought that I should calculate for the normal case and then take the limit as $l \to ~ \infty$
2:34 PM
So probably look up the sections on spin 1/2
Yes, and that’s what I’m doing.
But I didn’t get when you wrote $l/2$ in the present case of the picture
The normal case has both ends equidistant from the origin. For it to have length $l$, that means one end is at $z=+l/2$ and the other is at $z=-l/2$.
@PM2Ring it feel especial to learn something new at early age, I M not sure about how Australian schools are. But Indian school they teach with no logic just mug up all the thing irrespective what actually they mean I must say our high school teacher pattern is pathetic.
If you take $z=l$ and $z=0$ as the ends of your solenoid, then your observation point is at one end of the solenoid. That’s an interesting problem in itself, but it isn’t what the question is asking
@Semiclassical Yes I agree. But we agreed to call the distance from the center of forward end to P as $z$?
2:38 PM
The reverse process is responsible for generating heat inside the Earth. The inner core of the Earth is solid, mostly iron. It's surrounded by the liquid outer core. As liquid iron crystallises onto the inner core latent heat is released, which heats up the outer core and slows down the crystallisation process.
Ah. I’ve been using $z$ as the coordinate along the axis, not as a particular position
@PM2Ring wow. Does this hest have significant effect!
@Semiclassical My mistake was that I was considering this a real problem.
@YuvrajSingh... Bad teachers seem pretty universal. I had some great teachers and some terrible ones in high school. An anecdote I like to share is about my grade 11 "advanced" math teacher. The teacher didn't even understand rounding. We were expected to round 0.45 up to 1 because 0.45 rounds up to 0.5 and that rounds up to 1... Bad teachers exist everywhere.
2:41 PM
@Semiclassical What question is saying ? I understood it that one end is at $z=0$ and the other at $z=l$
. The question tells you to take the solenoid to be infinite in both directions
@YuvrajSingh... I don't know what it's like in Australian high schools these days, I went to school several decades ago. Back then, we did have to memorize some facts, but there was always a strong emphasis on understanding the principles and logic. In exams, you might lose a mark or 2 if you got the memorized facts wrong, but if you demonstrated that you understood the principles then you'd generally get good marks.
That means the origin is at the middle, with one end in the positive direction and the other in the negative direction
@j Mac yes that, s problem I recently passed high school and I have gone through this experiences.
@Semiclassical okay.
2:43 PM
“ What is the field on the axis of infinite solenoid (infinite in both directions) ?”
For specificity
@PM2Ring that, s very much good.
@PM2Ring To add on to this, and what I was saying about some teachers being bad, my experience was quite a mixed bag. I had several teachers (thankfully precalc/calc and science teachers mostly) who put heavy emphasis on understanding the concepts and applying that understanding. I had other teachers who heavily emphasized just knowing the procedure and applying it blindly. It all depended on the teacher, not the system in general.
@YuvrajSingh... Definitely! IIRC, it's more important than the heat released by the radioactive decay of uranium, thorium, and potassium (which are the main sources of radioactive heat inside the Earth). David Hammen has written some great answers that mention this.
All right. So, how $\tan \theta_2 = \frac{a}{l/2} $ please explain
From my thinking $\tan \theta_2 = \frac{a}{l+z}$
@Semiclassical Are you here?
How to capture my ideas and thoughts at the moment?
Like, let's say I'm outside and I think of something brilliant or just a thought/idea I want to have in a database. How do I do that?
I don't have the time to stop and type on my little phone.
2:52 PM
@JMac Understood. I was lucky to have some pretty good maths & science teachers. But I'm pretty sure that the syllabus in my state had a strong emphasis on teaching concepts and not just rote memorization of facts. For a couple of years, my science teacher was actually one of the co-authors of our science textbook, which was pretty cool. He was the principal author of the astronomy & chemistry sections.
@NovaliumCompany Doesn't your phone let you record audio?
I’ve given my reasoning already. One end of the solenoid is $+l/2$ from the origin, so $\tan\theta_2=a/(+l/2)$
@PM2Ring have you done a science project during school days?
@NovaliumCompany Carry around something to take notes on. I was watching an interview with a writer last night where he talked about that exact problem. Sometimes it's hard or impossible to actually leave a note at the time. I don't know if there's a good solution besides just thinking of some way to record ideas. Taking notes with Siri (or whatever android has) is one possible workaround.
The endpoints can’t be functions of $z$ if $z$ is to denote the coordinate along the axis
@PM2Ring on what topic it was
3:01 PM
@Semiclassical Please elaborate.
@JohnRennie rennie sir, what was your experience. Of high school teaching.!
I’ve elaborated multiple times. If what I’ve said so far isn’t enough, then I don’t see how further repetition is going to help
@JohnRennie how is UK high school teaching.?
@YuvrajSingh... you mean what was the teaching like when I was at school? If so, I was lucky because the teachers at the school I went to were very good.
3:03 PM
@JohnRennie Sir please help
1 hour ago, by Knight
Find the magnetic field at point P on the axis of a tightly wound solenoid (helical coil) consisting of n turns per unit length wrapped around a cylindrical tube of radius $a$ and carrying current $I$ (Figure 25). Express your answer in terms of $\theta_1$ and $\theta_2$ (it's easiest that way). Consider the turns to be essentially circular, and use the result of example 6. What is the field on the axis of infinite solenoid (infinite in both directions) ?
@JohnRennie very good! On what basis?
1 hour ago, by Knight
user image
@YuvrajSingh... they knew their subject well, and they were good at encouraging the pupils.
1 hour ago, by Knight
My doubt is how $\theta_1$ gonna become 0 and $\theta_2$ $\pi$ as the length of solenoid gonna go to infinity. Please help me in seeing that.
@YuvrajSingh... Not really. But I did science stuff at home. I started playing around with electronics when I was around 9 or 10. A year or so later I had a small chemistry lab in the basement. At school, in the science labs, we had access to much more dangerous chemicals than they let kids handle these days.
I was on good terms with the science master, and had permission to use the chemistry lab for my own experiments during the lunch hour, as long as I didn't do anything stupid. I was mostly sensible, but I did get banned for a week one time for making a large "volcano" with ammonium dichromate, which was rather messy :)
3:06 PM
@Knight the field on the axis of an infinite solenoid is the usual $B = \mu_0 n I$, where $n$ is the number of turns per unit lenth.
I'm too lazy to search for my phone or take it out of my pocket to record my idea/thought. Not only that, but I need it as text, so I can read it out later easily. (and not listen to my jibberish)
Yes. My doubt is how $\theta_2$ gonna become $\pi$ as length goes to infinity
@JohnRennie wow your teacher allowed you to use lab as if you want, when I was in high school we are not allowed to enter in lab without teacher permission.
@JohnRennie what kind of dangerous material are you referring for.
I think I have a solution for my problem. BOOM business idea.
@PM2Ring I haven, t ask you what kind of course you had done during University days.
3:14 PM
@YuvrajSingh... I started doing science & computer stuff, but I didn't last very long, for various reasons (which I don't wish to discuss). I dropped out in less than 6 months. :(
@PM2Ring oh!
nvm my idea sucks
apparently amazon echo loop exists
they've integrated f*cking Alexa into a ring...
you literally have the whole internet and many apps on your finger
you can record, take notes, take calls, google...
3:50 PM
user image
4:19 PM
Please describe what kind of mice you want to catch otherwise your question would be closed.
4:34 PM
Yes, it's too broad needs more focus. Or primarily opinion based.
5:23 PM
The new CEO has just made his first post on the network:
Q: A Note To Meta on my 2020 Kickoff Blog Post

PchandrasekarHello members of the Meta community - I wanted to share the blog post I wrote to kick off 2020 and reflect on my first 90 days at the company. It’s intended for a wide audience, the tens of millions of people we serve, but also because I know you all want to hear directly from me. My style is to ...

I am not going to read all that stuff in that blog post.
It's only about Stack Overflow anyway.
do users who haven't been here in months still get pinged if you reply to their messages?
No, at some point this doesn't work anymore
aw man
5:49 PM
@JohanLiebert If possible make a Facebook account with name Johan Liebert Stackexchange and send a friend request it to all Abhas Kumar Sinha in the list (with the name of the city which you told me). We should bring him back.
I can do it but there are more chances of you to find him because you are his paisano
@Knight actually today i talked to him a bit.
@Knight we are friends now on fb
@JohanLiebert Wow! What happened then?
@Knight he came here for some time in general chat and shared his fb profile (which is there till now).
@JohanLiebert Please give me the link here. Why he is not coming to the h bar?
@Knight you know all that( what happened yesterday ). There is no point in repeating it.
5:59 PM
2 days ago, by ACuriousMind
@AbhasKumarSinha You may come back tomorrow, provided you can abstain from advocating Nazi ideology then.
I guess he couldn't stop advocating for nazi ideology
Actually he was unaware of what he actually did and so he got punished for that.
well he'll come back if he wants to
But he shouldn’t leave like this, I cannot understand what hurts him?
not sure how you can be unaware of advocating Nazi ideology
that's a pretty extreme ideology
6:03 PM
Or wanting someone who advocates nazi ideology to come back
@Knight He made a very poor reference to Nazi ideology. I don't think it was actually trying to promote it (IIRC it was a bad joke); but it wasn't the right thing to say at the time. I think he might be overreacting to the ban too though. From my understanding, it was a slap on the risk to get him to think about saying things like that; but he seems to have taken it more seriously.
If they’re willing and able to come back, they will
@Slereah @SirCumference he wasn't spreading any ideology. He just made a sarcastic comment on it (and intended it to be a joke and nothing serious).
@JMac @Slereah @SirCumference I have read his chat in the other room and the main problem with him is that he still doesn’t know what was his mistake, that’s immensely terrible but if you take his age into consideration you will find that he didn’t mean what his words meant.
@JohanLiebert Just send this chat transcript to him on Facebook and I’m sure he will come back for giving clarifications (I know him)
in General chat, 5 hours ago, by Abhas Kumar Sinha
@tpg2114 thanks sir, I agree to the point that the joke was tasteless and terrible.
6:09 PM
He just said it for formality
He should come back here and learn from a real German I.e. ACM about what actually went wrong with his reference of that book to me.
@Knight I think that might be part of the problem. He "didn't know what his words meant"; but he said them, and they had pretty significant meaning.
@Knight I don't think he will come back. And by the way its his decision why should I interfere?
i think it's a little early to tell
it's only been a day or two
Yes it’s early but I should convince him to come back before eeveything fades away
@JMac So what you propose?
give him a day or two to cool off?
6:15 PM
@Knight why do you think there is a necessity for him to come back? I mean he can always make a new account if he wants to ask a question.
@JohanLiebert I’m not feeling good without him.
@Knight What the heck am I supposed to propose here? He said something he shouldn't have, got a 1 day time-out, and then decided to not come back. I propose that if that's what he decided, it's his choice to make.
@JMac I solemnly reject your proposal
@Knight You reject that it's his own choice to make...?
@Knight try to accustom yourself to this. You should nor try to force people against their will.
6:20 PM
@JMac “ I propose that if that’s what he decided, it’s his choice to make” I reject this proposal of yours
@Knight On what grounds?
@JMac In all human sorrows nothing gives comfort but love and faith.
On this ground
@Knight So you think it is not his choice, so we should force him back into the chat?
@Knight How is that a reason to reject my proposal? As far as I can tell, it's a non sequitur.
@JMac That’s a very nice reason for rejecting your proposal. You can see it clearly.
Aha! ACM is here.
6:27 PM
So anyway, anyone learn anything cool recently?
@Knight I don't this is good to continue this discussion.
@Knight No, it's not a reason at all. It's a non sequitur. It doesn't follow from what I said. How is "In all human sorrows nothing gives comfort but love and faith" the grounds to reject that we should respect his choice? For one thing, I don't even necessarily agree that the quote is correct, let alone that it's relevant.
@Slereah I'm offering a 500 point bounty for the tiger's answer.
@ACuriousMind I mean really though, the tiger has it figured out. The cat had an X Y problem.
I think some sites call that sort of answer a "frame challenge"
6:29 PM
@ACuriousMind Yeah X Y problem might not be the exact right wording
But the OP has not done any proper research on this topic. So its off topic.
@JohanLiebert Depends on where you ask sadly. Some SE sites might eat that up.
@JMac lucky cat though!
Lack of research has never been conclusively settled as an off-topic reason on most sites - certainly not on ours! People occasionally vote to close because of "no prior effort" as a custom close reason, but we've never really gotten a consensus on that one as far as I know
@ACuriousMind I feel like it becomes more of an issue if the answer can easily be found
6:37 PM
I just finished watching 1917
Everything was great but I expect a better ending.
@NovaliumCompany I thought the edit was fitting. But that is all I will say since it is a movie that just came out
@AaronStevens Yeah, that's one of the reasons consensus on this is hard because people have wildly differing standards for what "easily found" means
@AaronStevens I saw the question titled "W = Fd does not explain anything!" and just knew I was in for a great misunderstanding.
@JMac Yeah I am trying to think up a better title now
Sure, if I paste the title of your question into google and get the answer, it's clear you didn't do any research. But does that mean the same question with the same amount of prior research gets on-topic if OP just chooses a bad title (and is generally bad at expressing themselves)?
6:44 PM
@JMac I have come across questions like these recently. "Why is this term not defined this way instead?" It could be a good question if the OP makes the case that the definition has a certain goal in mind, but another definition would achieve that goal in a better way. But questions like that never go like that
@ACuriousMind Rawr >:3
@ACuriousMind On a somewhat related note, what is the current view on questions that are mostly likely only to be useful for the person asking and not useful to a broader audience? Based on the very limited old posts I have seen, it looks like that used to be a reason to close a question. I think more recently it was mentioned in the homework close banner (the one before the current one). Now it seems like that is not as much of a reason anymore, or maybe I am just missing something?
@AaronStevens Probably because if a different definition would achieve a goal in a better way, that definition likely already has a word associated with it and is being used in that context. Especially with classical mechanics questions, where most topics have been studied extensively for centuries.
@JMac Right. And that happens in this question, which is why I discuss impulse in my answer :)
@AaronStevens It used to be a standard close reason as "too localized", but it was abandoned years ago. We use "questions should be useful to a broader audience" as one reason to try and motivate people to make the HW questions more about concepts and less about how to solve a specific problem, but it's not seen as a close reason on its own anymore
6:49 PM
@AaronStevens Personally, I often like answering questions like that. If I can show the person that they actually were kinda onto something; just maybe not exactly what they expected, I have hopes that it will encourage them to keep thinking about the concepts and maybe feel some pride that they were thinking about it at all.
@JMac True, that is a good point. I think it also is something a lot of students get hung up on. They think they need to understand "why is this called work?", when instead they need to take the definition as a given and learn how to apply it. I suppose it is a difference of trying to arrive at a definition vs. starting at a definition.
Often learning how to apply it shows how the definition is useful, which then shows why it was given a special definition in the first place
@ACuriousMind My respect for you is increasing day by day. Even today your maturity (by not saying a single in that discussion) has won my heart.
7:41 PM
"I was traveling exactly the same speed as the other guy. There's no way we could have collided!" — WillO 16 mins ago
@AaronStevens I think it's even worse than that. It's like, "the guy in front of me was driving and spilling oil the whole time. If we were going the same speed, how the hell did I wind up being affected by the oil skid left behind"
@JMac Haha right, exactly.
I wish I had the ingenuity to come up with questions like that
Ones where you can kind of see where it came from, yet where the question ends up is very odd.
2 hours later…
In the context of statistical mechanics, does the classical limit start to disappear within low or high temperature ranges?
@dsm $\mathrm{ad}_X$ for fixed $X$ is a linear map from the algebra to itself (if you choose a basis you can represent it as a matrix), and $\mathrm{ad}_x \mathrm{ad}_Y$ is just the composition of two of these (matrix multiplication!). Linear maps from vector spaces to themselves (square matrices) have traces.
@ACuriousMind sent you an email XD
I'll see it tomorrow, then :)
sounds good :D
Hope I didn't send it to the wrong person lol
9:26 PM
A colleague of mine gets about a wrong email a day because he's got a name in common with about five other employees, so...it happens xD (my name is unique as far as I know, though)
Would be awkward for this email since I said like "Greetings from Enumaris" on it
9:49 PM
@psa you start to approach the classical mechanics at high temperatures
not sure what you mean by "disappear"
poor wording, I didn't know how else to ask it
thanks though, that's what I had figured
10:49 PM
You know what I still don't get?
Viewing differential operators as unit vectors in a vector space
they don't gotta be unit
11:08 PM
why is "vibrational" kinetic energy distinct (i.e. we count it as another degree of freedom) from "translational" kinetic energy?
there's two quadratic degrees of freedom in $\frac{1}{2}kx^2 + \frac{1}{2}mv^2$, but haven't we already accounted for $\frac{1}{2}mv^2$ by just calling it a translational degree of freedom?
energy and degree of freedom are not the same things
sure, but I'm talking about the equipartition theorem here
@enumaris even just vectors then
are we just distinguishing CM motion from internal motion?
motion of center of mass vs individual motion of particles
11:14 PM
@psa vibrations, and translations are two distinct things. i.e. they have no dependence on each other. Thus they each get an additional bit of energy by equipartition theory
cool okay thanks
which could be rotational or vibrational w.r.t. CoM
@JakeRose They form a vector space so you can consider them as vectors
right, thanks both of you
Mhmm. Is there any more intuitive explanation
Or should I bend my intuition to the axioms of vector spaces
Do they technically form a hilbert space?
11:17 PM
an inner product is not defined naturally on all manifolds
Such as?
well you generally have to define a metric to have an inner product
giving rise to Riemannian or Semi-Riemannian manifolds
but you need not define a metric...in which case the inner product is not defined
A generic vector space need not have an inner product defined in it.
Ah. But you can define a Hilbert space from them, just not generally?
Are you trying to deal with infinite-dimensional manifolds or something?
Why the focus on Hilbert spaces?
I'm not dealing with anything. Just generally interested.
11:22 PM
@enumaris huh, i thought i heard that we can always find an inner product on a real/complex vector space
nvm apparently that needs the axiom of choice
@JakeRose You can probably intuit, using Euclidean geometry, that for any vector, there is an associated derivative operator that "goes along with it". So e.g. $\partial _{\vec{v}} = \sum_i \vec{v_i}\cdot \partial _{x_i}$. Then you can just choose the $\partial_{\vec{v}}$ as "vectors" and $\partial _{x_i}$ are your basis vectors
A: Is there a vector space that cannot be an inner product space?

Christian BlatterI'm assuming the ground field is ${\mathbb R}$ or ${\mathbb C}$, because otherwise it's not clear what an "inner product space" is. Now any vector space $X$ over ${\mathbb R}$ or ${\mathbb C}$ has a so-called Hamel basis. This is a family $(e_\iota)_{\iota\in I}$ of vectors $e_\iota\in X$ such t...

@SirCumference There's a difference between "a vector space that cannot be an inner product space" versus "are all vector spaces inner product spaces?"
well yep, a (real/complex) vector space isn't an inner product space until we choose an inner product
11:29 PM
crud misread the conversation
np :)
11:41 PM
if I'm trying to approximate $\sum_{j=1}^{\infty} (2j+1)e^{-j(j+1)\beta\epsilon}$ with $\int_{0}^{\infty} (2x+1)e^{-x(x+1)\beta\epsilon} dx$, I should be taking very large $\beta$, right?
because if so I'm confused on the physics
11:58 PM
@psa if you're approximating it with a Riemann sum then you to evaluate it at some point, and have some width $\delta x$.
So it looks like $\int f(x) \approx\sum\limits_n (2x_n +1)e^{-x_n(x_n + 1)} \delta x$

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