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3:44 AM
This morning's review posts have been a flagging goldmine. I've already flagged 6!
@Secret FAKE NEWS!
 
 
1 hour later…
5:00 AM
@GodotMisogi waddaya mean "flagging goldmine"?
 
@ZeroTheHero Lots of homework questions and non-answers, so I can raise lots of flags for badges
 
oh dear!
:D
 
(Coincidentally, I'm currently reading your answer to this question)
 
@GodotMisogi oh dear :D
 
A more interesting question would be the physical consequences of different representations. I recall that locality breaks in the momentum representation
 
5:09 AM
I seem to recall that some nuclear potentials are sensitive to momenta and yes that causes problems, although locality is the least of your worries in this case.
oh! I should add that the gravitational potential is also linear..
 
@ZeroTheHero Possible typo? You've written $V(x)\psi(p) \to ik\hbar\frac{\mathrm d}{\mathrm dp}\psi(p)$. Shouldn't it be $V(x)\psi(x)$?
 
well when you use the momentum representation your wavefunctions are p-dependent...
I mean: $\hat x$ acts on functions of $p$ by derivative.
 
Oh, I've been reading it as (position space) → (momentum space)
So, in general, $V(x)\psi(x) \mapsto V(i\hbar \frac{\mathrm d}{\mathrm dp})\psi(p)$
 
The more accurate statement would be that
$\langle p\vert V(x)\vert\psi\rangle= V(i\hbar d/dp)\langle p\vert\psi\rangle$
After all one determines if you are in the $x$- or $p-$ representation by closing with either $\langle x\vert$ or $\langle p\vert$. The ket $\vert\psi\rangle$ is just a vector and doesn't care about the basis.
anyways I have to prep for the morning classes.
TTFN
 
I understand. I think I misunderstood the presentation as eigenfunction representations on both sides
 
 
4 hours later…
Anonymous
9:39 AM
Motl's on our site!
 
Anonymous
2
A: Optimal strategy to a quantum state game

Luboš MotlWe simply translate the binary result of a qubit measurement to our guess whether it's the first state or the second, calculate the probability of success for every possible measurement of the qubit, and then more find the maximum of a function of two variables (on the two-sphere). First, someth...

 
Anonymous
Could we get any more publicity than this? :P
 
Praise be
 
0
Q: How can "torus shape" planet be "non-mainstream" physics?

GravitonRefer to this question: How would the rotational and orbital mechanics differ on a torus-shaped planet as compared to a spherical planet? This is asking about torus-shaped planet rotational and orbital mechanics. To me this is a question can be answered completely using Newtonian physics, and i...

 
Anonymous
@Slereah We've both Duffield and Motl now. Our site's going places. :)
 
9:44 AM
Time to sink that ship
 
10:25 AM
23
Q: sed on OSX insert at a certain line

CRThazeSo I've been using 'sed' on linux for a while, but have had a bit of difficulty trying to use it on OSX since 'POSIX sed' and 'GNU sed' have so many little differences. Currently I'm struggling with how to insert a line of text after a certain line number. (in this case, line 4) On linux I woul...

ffs stop saying sed. It is useless in macs
Whoever came up POSIX should be thrown into a blackhole
 
@johndecker Hi, what's the question?
 
I want to learn about solving PDEs on curved surfaces.
For example, the 2d wave equation on a sphere: tap the north pole, a wave spreads with decreasing amplitude to the equator, then the amplitude should increase again as the wave continues past the equator toward the south pole. This is different from solving the 3d wave equation and then narrowing the solution to a spherical surface.
I believe there’s a rule “replace derivatives with covariant derivatives” in the PDF. Is that it?
Where do I study this? Textbooks? Online lecture notes? Examples?
(Culix suggested I post here, as it's supposed to be a helpful chat room.)
 
@johndecker that's a bit outside my comfort zone, but yes using the covariant derivative sounds a plausible approach.
The area of maths involved is differential geometry
In fact it's the same maths as used in general relativity, and for the same reason - solving equations on a curved manifold.
 
10:41 AM
@JohnRennie, exactly. I've studied introductory differential geometry and GR, but the textbooks I have only solve trajectories (geodesics), not PDEs (except Einstein's equations themselves which solve for the geometry, not classical fields IN that geometry).
 
I think it's a general principle that any differential equation on a Euclidean space can be applied to a curved manifold just be replacing derivatives by covariant derivatives. The results can get a bit messy though.
The Christoffel symbols for a 2-sphere are pretty simple though.
 
11:21 AM
It's amazing that most of the issues Lisi failed to understand/understand-properly, e.g. the $SU(5) \times SU(5)$, $F_4 \times G_2$ subgroups of $E_8$ or $E_8$ and it's real reps motls.blogspot.com/2007/11/exceptionally-simple-theory-of.html , are in GSW's string book, but Lisi kept bashing strings so clearly never checked the usual reference for this stuff
 
11:33 AM
Ideally we should put the works of the blogosphere physicists behind us and get on with physics that is actually fun and interesting.
2
 
0
Q: Why can't I suggest "Chemistry Stack Exchange" as another site when flagging as off-topic?

Tausif HossainI've recently noticed that when I flag something as off-topic by suggesting it to another site(in order to get better answers for that question) I only see the options of Physics Meta Stack Exchange and Mathematics Stack Exchange (shown below in the picture), like no Chemistry Stack Exchange or B...

 
12:13 PM
@johndecker Try Friedlander maybe
 
user351417
12:54 PM
user image
2
 
user351417
@Blue Does that look strange or what?
 
1:29 PM
that noob
 
1:48 PM
When we solve definite integrals, in what (I forgot the word) do we get the area? In meters squared or in cm square...?
mathsisfun.com/calculus/integration-definite.html Here they just get a number for the area, but is it $m^2$, or maybe $cm^2$?
Maybe it's $banana^2$ xD
 
well it depends on what units you use to measure your $x$ and $y$
if $x$ is meters and $y=f(x)$ is meters too then $f(x)\mathrm{d}x$ is $m^2$
 
@user2723984 Thx.
 
user351417
2:04 PM
@NovaliumCompany Indeed, if your $x$ and $y$ axes both have units '$\rm bananas$', you will actually find the integral as '$\rm bananas^2$'.
 
2:21 PM
 
rob
2:41 PM
-1
Q: If the jerk is 2, then acceleration is 2t, velocity is $t^2$ and distance is $t^{3}/3$?

Kim Are my graphs and equations right? I just began learning about these..

I case a "homework-like" close vote there, but I'm having second thoughts: it looks more like "playing around" that I'd prefer we encourage. Any thoughts from others?
I think the photograph threw me off, and I would have parsed it differently if it'd been typeset.
 
user351417
3:34 PM
physics.stackexchange.com/questions/457220 Yikes! There were 5 answers, including upvoted ones there!
 
Anonymous
@Chair Oops! SE is broken. ;)
 
rob
@Chair User removed for violating system restrictions. The removal script deletes negatively-scored content.
 
user351417
@rob That's a pity, because the answer which was highest scored at the time I last looked was quite different from all the answers in that and the duplicate.
 
user351417
Is a merge or something possible?
 
3:55 PM
0
Q: Notification to the user whose comment is deleted

harshit54I strongly believe that a user, whose comment is deleted by a moderator should be notified about it. Many moderators ask the user to remove their comments however some just remove them without informing the user at all.

 
4:38 PM
fun times
 
5:38 PM
@Chair We do not merge users that were created in bad faith (and it's not possible anyway after deletion).
 
user351417
@ACuriousMind Nono, I was talking about the questions, not the users. I've already found that mods don't merge accounts anymore, even for active users (chat) :P
 
user351417
My point was that the deleted question had answers which could (IMO) be appropriate under any of the surviving 'high school whatnot rant' questions.
 
@Chair Oh, well. Question merges are rather messy because they confuse the histories, so they tend to only happen in exceptional cases. Really, if anyone is attached to their answer there I'd much rather they just re-post it as an answer to the dupe target.
 
user351417
@ACuriousMind Will the people with <10k rep will be able to find that page?
 
user351417
I think it'd show up in their rep histories as non-public events, so I can imagine them having links...
 
5:49 PM
@Chair Admittedly not but we could certainly provide them with the text of their answer nevertheless if they ask
 
user351417
Yeah, I guess they'd decide to flag a random post and ask for a screenshot of their answer or something. Thanks for clarifying!
 
Anonymous
6:22 PM
The lack of editing culture on PSE (all over SE actually) sort of bothers me. This has received two answers and yet no one edited to improve the typesetting in the question. I don't know if there's any solution to this i.e. encouraging users to edit more and to use proper typesetting.
 
7:34 PM
I am being told that one cannot focus infrared radiation emitted by an object at ambient temperature, using gold or aluminum dishes in an answer to my question i post on SE
 
@Blue Edits that correct the typesetting are too often rejected on PSE.
 
Supposedly because of the 2nd law of td. But i cannot see why. What happens to the radiation any object emits (as well as absorbs) even at ambient temperatures, when it hits a gold dish while the object is at the focal point? Shouldn't that direct all radiation coming from the object and hitting the dish into one direction which then could be focused by a second dish?
 
Anonymous
@Loong That is even more reason to be worried.
 
Yes
 
Anonymous
Anyway, given the current state of PSE, I don't think it can be salvaged anymore. Seems like a lost cause now.
 
7:38 PM
lost cause D:
 
it is what it is
 
Anonymous
On QCSE too...if I leave the site for a couple of weeks the front page becomes a complete mess. :/
 
Anonymous
Fortunately it's still a small site and I can manage editing everything myself‌​.
 
@Blue maybe we should clone you a few times
 
-2
A: 2016 Moderator Election Q&A - Question Collection

LoongEffective communication in the physical sciences, technology, and education relies on extensive internationally standardized terminology, nomenclature, quantities, units, symbols, and typography. How seriously do you take conformity with these requirements with regard to this site? If applicable...

You can see the votes.
 
Anonymous
7:43 PM
Sigh...
 
@Loong I think you and Blue are ralking about different kinds of "correcting the typesetting" :P
 
maybe
 
I'll bet that the edits you have in mind are those pertaining to correct typesetting of units or otherwise conforming to standards comparatively few people know or care about, and so the edits will get rejected because the average reviewer will think they are a stylistic choice rather than conforming to a standard
But I would be highly surprised if an edit that corrected something like the typesetting of the enumeration in the example Blue linked would get rejected on our site.
@Blue I'd argue that you should perhaps view this a bit mor compassionately with the non-editors: Being able to edit other users' posts is a rather unusual feature of SE compared to other sites. It's somewhat hard to train oneself to keep in mind that one can edit badly formatted posts one didn't author if one otherwise spends a lot of time on sites where that isn't possible..
And I'll say that SE treads a really weird line between "posts have owners who get points for their quality" and "posts are collaborative resources everyone should try to improve"
If gamification is the driving element of engagement on SE, then it is no wonder there are so few edits: You only get at most 2 measly points for them, and especially if you've been around for a while you'll get none
 
Anonymous
8:00 PM
True. We need to somehow lure the folks into editing with rep points... :P
 
Anonymous
Can't think of a good model for that though, at the moment.
 
@Blue Not with the current model. I do not want people to be able to gain close votes through editing.
 
Anonymous
Yup.
 
Who owns SE?
 
Anonymous
@pZombie Stack Exchange Inc.
 
Anonymous
8:04 PM
Jeff Atwood and Joel Spolsky were the founders. Jeff left a couple of years back. He still has some shares I think.
 
SE generates income?
 
Anonymous
Of course. It's a for-profit company. But most of the income comes from Stack Overflow.
 
@pZombie sure: from ads, the jobs platform, its enterprise versions, teams,...
 
Do skilled moderators get paid a share?
 
No, moderators are unpaid volunteers
 
8:08 PM
that's stupid
 
you might get a t-shirt
 
maybe there is your answer why there are not enough skilled reviewers which care to edit
 
and a hat
 
a man needs to eat
 
Anonymous
Hmmm...actually hiring professional editors wouldn't be a bad idea (once they start generating sufficient profit). :P
 
8:09 PM
sure, but I don't go to SE looking for food :P
(unless I'm on Seasoned Advice, I guess)
 
Anonymous
It would take away some of the democracy from the community though. Not that it matters.
 
@Blue Ugh, no. You'd have to freeze posts they edit otherwise some users will just overwrite their edit because they disagree with it and you paid for effectively nothing
 
Anonymous
@ACuriousMind Is overwriting good edits that common? :/
 
No idea, but it certainly would happen, especially since some users would be opposed to the very principle fo having paid editors
But, more importantly, paying someone for their contribution to SE changes the whole deal: How is gamification via virtual internet points supposed to work if there's a way to earn real money?
 
yet same users would not be opposed to SE getting paid for some services it offers
 
Anonymous
8:14 PM
I'm speaking of professional editors as in company employees. It should definitely be possible to lock their reputation at 101, in case they're being paid by the company.
 
Anonymous
For PSE, just one full-time editor would suffice. For Math SE, maybe 3-4.
 
Anonymous
Just like there are jobs for CMs, they could have jobs for editors.
 
If you think SE could possibly employ full-time editors on a per-site basis you have a much more optimistic view of their revenue than I do :P
 
Anonymous
Anyway, none of this happening until they can come up with a better profit model.
 
Anonymous
@ACuriousMind Yeah, that's the only problem.
 
8:17 PM
And I'm not sure this is really desirable. Having paid editors contribute to the site changes the collaborative atmosphere. I believe there's good reasons Wikipedia doesn't employ paid editors, either.
 
Anonymous
That's true too, yes...
 
the whole world comes to SE seeking valuable answers to problems. It is quite unique and efficient
It would even justify global government funding
 
@pZombie But everthing's at the same temperature, and you aren't adding any energy into the system (or removing it), so no arrangement of mirrors or lenses will change the temperature of any component. Remember, all ray paths are reversable, so for any fancy arrangement of lenses & mirrors that's suppused to heat up object B using the IR from object A works equally well to heat A using the IR from B.
But it can be hard to fully understand this principle. Take a look at this relevant XKCD forum thread, which contains just about every argument on both sides of the fence. forums.xkcd.com/viewtopic.php?f=60&t=113444
 
-1
Q: Am I allowed to ask questions about sound on main?

LunaI am fundamentally curious and want to make sure before I am able to ask my next question

 
8:36 PM
@Blue It sure is on SO, the prime offenders being the OP themselves.
FWIW, there are badges for editing. If a question is good enough to be worth answering, but it could be improved by an edit, it deserves to be edited. OTOH, one shouldn't waste time &energy editing rubbish that's going to be closed anyway.
 
@PM2Ring Thanks, your answer concerning the reversibility of the mirror/lenses/dishes arrangement actually seems to make sense. Much better than most i get as in the 2nd law of tm forbids it
 
@pZombie No worries.
 
@PM2Ring however, i am not sure if this applies here for all mirror etc arrangements. The material which gets hit by the concentrated infrared radiation, emits radiation itself radially. What if in the final stage i focused all IRR to hit it straight. This does not seem to be reversible
@PM2Ring my first dish would have the large object in its focal point, resulting in the IRR reflecting in a straight line towards the second dish, which would focus it. Then i would use a some setup which would shoot the IRR as focused as possible in a straight focused beam towards the target.
@PM2Ring while surely some of the radially emitted IRR from the target would make their way back in reverse, it should be negligible
 
@pZombie It doesn't help. That's discussed in the XKCD thread. And people who don't trust the theory have tried all sorts of clever arrangements of mirrors, lenses, chambers, baffles, etc. None of it works, if you stick to the conditions of an isolated system, with no external temperature reservoirs.
 
@PM2Ring my system is NOT isolated though. It is a large cave with the temperature inside the cave being at about 15°C and kept at 15°C through the surrounding environment
@PM2Ring all isolated systems will eventually succumb to thermal equilibrium. My system is NOT isolated
 
8:50 PM
@pZombie It is isolated. The cave is isolated from stuff with a temperature that isn't 15°. So it's already in thermal equilibrium, and will stay that way.
 
@PM2Ring Let me ask differently then. Every object inside this cave emits and absorbs IRR at all times, being at a thermal equilibrium at 15°C. If now i hit a given object with more IRR radiation than it usually gets from its environment, it will heat up, will it not? It won't matter where this IRR came from or what its source was, would it?
@PM2Ring likewise, if i was able to shield all incoming IRR towards the object, such that it cannot absorb any, yet i allowed the object to emit IRR (by some magical setup) it would cool down, wouldn't it?
if you agree on both of those statements, then all i would have to do is to program a simulation with objects inside a cave emitting and absorbing radiation, along with arrangements of mirrors and lenses you say would not permit such a thing.
 
9:06 PM
The keyword there is "magical". As I said earlier, if radiation can travel from A to B, it can go in the reverse direction. However, we are only considering a system composed of passive components. If we used active components, it's a different story. But that also requires breaking the isolation of the system so that we can dump unwanted entropy somewhere that won't affect the cave.
Anna V gives the example of using solar panels to power the LHC, thereby having a solar powered system that creates temperatures higher than the Sun's photosphere. But that only works because the solar panels can dump entropy.
 
@PM2Ring That is of course a correct statement. If radiation can travel from A to B then it can also travel from B to A. But the radiation traveling from A to B emitted by the large object is much greater than the radiation going from B to A with a proper arrangement of dishes, even if the objects on both ends are the same
 
Please read that XKCD link! There's no point in me repeating those arguments here.
@ACuriousMind Agreed. If editors deserve to be paid then good answer writers also deserve to be paid. ;)
 
@PM2Ring This is my last reply to you as you have not addressed my 3 dishes setup but kept only pointing me to some link. I drew some less than artistic picture of the dishes arrangement here imgur.com/a/HxaohLa . You are telling me that in such an arrangement, the left object would get hit by the same amount of radiation coming from the right object as the right object from the left object? If yes, then i guess i am wrong, but i cannot see why
 
Anonymous
9:25 PM
@pZombie It might be better to tone down a bit, given that you're the one asking for help. Remember that folks here are simply volunteers and no-one is obliged to answer.
 
where is the difference between asking for help and wanting to discuss an idea?
 
Most people come to Stack Exchange to read answers, and questions are just a means to create and locate those answers. But it's an intrinsic feature of the Stack Exchange model that the vast majority of questions are written by the least competent members. They aren't experts in the knowledge pertaining to their question, and they aren't familiar with the SE standards & culture.
 
Anonymous
@pZombie There's not much difference. If you're wanting to discuss something that's still asking others to invest their time in your discussion. Again, they're not obliged to participate in your discussion at all. So saying things like "you have not addressed my 3 dishes setup but kept only pointing me to some link" is not acceptable.
 
And on sites like Physics & Mathematics there's the additional problem that most newbies aren't familiar with LaTeX / MathJax. But I don't know how to encourage competent members to edit questions.
 
@Blue i am not asking others do anything by force. I am investing my time as well
 
9:29 PM
Maybe give all editors the 2 point edit bonus. True, that may encourage excessive trivial edits, but IMHO that would be better than the current shambles.
 
Anonymous
@pZombie I understand. But what you said there appears slightly hostile. That's why I requested you to tone down.
 
@pZombie Once again, that won't help. There's no such thing as a one-way mirror. Sure, you can make a partially silvered mirror that kind of functions as a one-way mirror, as used in interview rooms, but that only works because the interview room is bright and the observer room is dim.
 
@PM2Ring did you look at the drawing?
i never said that there is something like a 1 way mirror
i admitted that radiation would also go from B to A ... just that it would be less
 
@PM2Ring And yet they are an integral part of the whole setup - because most "experts" are terrible at anticipating the questions newcomers to a topic ask
 
@pZombie Yes, I did. And such schemes, and variations on it, are analysed in the XKCD thread.
 
9:37 PM
@PM2Ring Rather than reading a whole thread which might or might not contain some useful information, i would rather you tell me why what i drew would not permit to assume that less radiation from B would hit A than from A to B
you as in anyone else
of course, if it bothers you too much, you are free to not discuss it
 
hmmm
 
@ACuriousMind Indeed! It can be difficult for experts to remember how the newbies think, which is why the SE system works so well. Recently in the SO Python room, one 20k+ regular jokingly suggested that writing questions should be a privilege restricted to those over 1000 points. :) But of course that would never work. Another option is for all questions to go through review before they become publicly visible, but that would require a vast army of competent reviewers.
@pZombie That thread does contain a lot of useful info, that's why I posted it. It also contains some "fluff", and various incorrect arguments, but most of the incorrect stuff is successively argued against. It's extremely unlikely for anyone to come up with an argument, right or wrong, on this topic, that isn't covered by that thread & the links it contains.
 
All right, Imma try to make a serious ad for the chat. Any ideas?
 
Anonymous
@SirCumference I already told you mine. :P
 
Anonymous
Jan 25 at 19:11, by Blue
@SirCumference Insert "asynchronous communication protocol" somewhere, maybe? ;)
 
Anonymous
9:49 PM
I'm serious.
 
Not sure most people are gonna understand "asynchronous communication protocol" :P
 
Anonymous
They don't need to understand it. People are attracted to fancy words which they don't understand. :P
 
I wonder whether ads even reach most people. I haven't seen an ad in ages and I'm not quite sure why, though it might be related to the adblockers :P
 
Anonymous
When they drop in, we can explain. ;)
 
@ACuriousMind Yeah, same...
I suspected it was my ad block but turning it off seems to do nothing
 
Anonymous
9:51 PM
I see the ads when I log out (sometimes).
 
@pZombie This kind of arrangement is defeated by conservation of entendue, which is discussed in that XKCD thread.
 
@Blue Ah, yes. Surely logging out is the point in time when we want to alert people to chat :P
 
Anonymous
Lol
 
Anonymous
Something is wrong with the ads for sure.
 
@PM2Ring I see. This will take some time for me to digest
 
Anonymous
9:55 PM
IIRC there was something to do with user reputation. Low rep users are shown ads more often.
 
@Blue Then you should see many ads on physics.SE :P
 
Anonymous
I'm asking in the Tavern...
 
@ACuriousMind [status-bydesign] for sure
In serious if none of us are seeing ads we could bring it up on meta
 
Well, they are getting some clicks, so someone is seeing them :P
(the low totals are because it took a while for ads to reach the 6 vote threshhold to be shown at all)
 
@ACuriousMind Perhaps only the people who visit the Meta page?
 
Anonymous
10:02 PM
in Tavern on the Meta on Meta Stack Exchange Chat, 21 secs ago, by Mithrandir
The new Community Ads posts just went up, so not many have reached the votes needed to show up yet.
 
@Blue We do have ads above the threshold, so that's not the issue here.
 
@Blue But at least on Physics there are quite a few that reached the threshold. I'm not seeing any tho
 
Anonymous
in Tavern on the Meta on Meta Stack Exchange Chat, 1 min ago, by Blue
@Mithrandir Hmm, but on one of the sites I frequent at least 3-4 ads have crossed the threshold already.
 
I didn't know chat ads even existed until I saw you guys discussing them a few days ago. For that matter, I didn't know Stack Exchange had chat rooms for my first few months as a member, and then I didn't bother investigating them for several months more. Eventually, curiosity got the better of me, and I checked out a few of the SO rooms. These days, I mostly hang out in the SO Python room, and I've been one of that room's owners for a couple of years.
 
I try to get onto the meta chat and
Oh SE
 
10:06 PM
The meta.SE chat is on a different server. If you've not been there before, you might not have a chat profile there. For most actions it should be created automatically, but...
 
Anonymous
@SirCumference Huh, that's strange.
 
Anonymous
Do you have an account there?
 
@Blue Not the first time. Often happens when I try to get onto Meta's chat
 
Anonymous
Try in Incognito mode?
 
@Blue Yep, I'm logged in. Yet trying to get into the chat from your link yields
 
10:08 PM
You need to be logged in to your meta.SE account in order for meta.SE chat to log you in. Try visiting a meta.SE post, see if you're logged in there, and if you are, try again
 
SE and its bugs
@ACuriousMind I am. If e.g. I click on the "logged in" link from that pic I just posted, I get this
 
Anonymous
Even if you're not logged in, it shouldn't be saying "Oops! Something Bad Happened!".
 
And naturally I click the "click here" at the bottom of that page. Which brings me back to
 
Anonymous
Oh, I remember facing that issue once.
 
@Blue That happens when I try to get in from the meta site itself
 
Anonymous
10:10 PM
If you don't have any posts there, just delete your Meta SE account and re-create it. It's a bug (I faced that too).
 
When I search for you in the meta.SE chat users, you don't show up.
 
Ah, wait, it's working now
 
Anonymous
The chat account was somehow not created or got deleted.
 
Welp never mind. I was able to load into the chat selection without that error, but once again
 
Anonymous
Oh, you have some posts there. Can't re-create right away.
 
Anonymous
10:13 PM
@SirCumference Try logging out of your main account once (if you haven't already).
 
@Blue Ok, just did that
And we're back to the cat meme
 
Anonymous
Lol
 
And same log in issue when I visit from your link
 
Anonymous
5
Q: I have a 148 rep in meta, but can't login to the tavern to chat

JoeI can see my rep at the top of the page, so I'm logged in, but the bottom of the page says I have to be logged in to chat. There's no message entry box. What am I doing wrong? This is my first time trying to chat (in a long time?). Clicking on "logged in" just refreshes the page. I looked at a ...

 
I'm on Safari tho, I don't have HTTPS Everywhere
Tried turning off my ad blockers, no luck
 
10:17 PM
Maybe you're underage and the tavern bouncer simply isn't letting you in :P
 
Anonymous
@SirCumference Time for a bug report!
 
@ACuriousMind Damn, he's found me out
@Blue Eh I'll get to it soon enough
I don't use the meta chat that often tho
 
Anonymous
5
Q: Can't log in to chat

g .I've tried several times to get chat working over the last couple months from several stackexchange sites, followed all of the instructions and it doesn't work. The help pages reports that all the tests are ok. I've followed the instructions, logged out, logged back in, cleared cookies and cache...

 
Anonymous
This was a Safari issue too. ^
 
Hello
 
10:22 PM
$\mathfrak{L}$
 
Anonymous
BTW I think superpinging creates a chat account. Maybe we could request the Meta SE mods to superping you.
 
@Blue Worth a try :)
Well actually I oughta be working right now
 
Anonymous
Cya! Journeyman Geek and Tinkeringbell don't seem to be around. I'll ask them when they come by.
 
Let $G$ be an abelian group of order $pq$, where $gcd(p,q) = 1$. Suppose $G$ contains elements $a,b$ of order $p$ and $q$ respectively. How can I show $G$ is cyclic?
Not sure if I should go for a proof by construction or by contradiction
Doesn't this group only contain the identity element?
 
@Lozansky Look at the subgroup generated by $ab$.
 
10:30 PM
oO
 
@Lozansky If it contained only the identity, its order would be 1, not $pq$ :P
 
@ACuriousMind I guess this subgroup generates elements $b^k$ if $k|p$ and $a^k$ if $k|q$. Should I note something else?
 
@Lozansky What is the order of this subgroup?
 
$\max\{p,q\}$?
 
nope
 
10:42 PM
Oh $pq$
 
Yep :)
 
Ah I guess this does it
 
Note that, in general, if $a$ has order $p$ and $b$ has order $q$, then $ab$ has order $pq/\mathrm{gcd}(p,q)$.
 
Makes sense
 
@Blue Welp they tested, no luck
Weirder is that I got pinged with Pies, which was my username on chat 3 years ago
 
10:55 PM
It's still appropriate for your profile picture, though :P
 
Anonymous
@SirCumference This is your chat profile. It seems to belong to a deleted account.
 
Anonymous
Probably you changed your login method since then or added an alternate login.
 
@Blue Yeah, didn't think that would cause a problem
 
Hey guys, someone posted this in a non-physics guys group (as myself) asking how much energy could be produced with that kind of setup. The general consensus seems to be multiplying: height * water quantity (litres, which is equal to its mass, right?) * gravity
Nevermind the text, it's basically asking "How much electricity could be produced"
My question is: isn't the pipe diameter relevant in this?
 
Anonymous
@SirCumference I asked in the Teachers' Lounge. You need to write up a meta post or send a mail to the CMs. I think destroying your chat profile should solve the problem (which only the CMs and Meta mods can do). ;)
 
11:02 PM
@Blue I'll get to it sooner or later. Thanks :)
 
All problems can be solved by destroying users.
 
Anonymous
Alternate logins are really buggy.
 
@ACuriousMind ~Chat mod ACM, 2019
 
I'm thinking the difference in pressure between the generator's input and output is relevant. Is it not?
 
@RodrigoSilva What's the "bomba de agua"?
And what's the rectangle on the right it's connected to?
 
11:06 PM
@ACuriousMind Water pump, sorry.
 
@ACuriousMind Reservoir of 3000 liters
 
Anonymous
Sounds like "water bomb" to me. :P
 
The "upper line" is basically a pipe with a generater inbetween
 
Solar panel?:P
 
@SirCumference The rectangle, not the cylinder! :P
 
11:06 PM
@ACuriousMind Fly swatter?
 
Looks like it, the author didn't specify. The upper line is a water pipe with a generater in-between, where the water "comes down" and electricity is generated.
 
Must be a hell of a fly if its swatter is larger than a 3k litres reservoir!
 
Can we close the author's question as low-effort?
 
The below line is another water pipe supposed to "push" water upwards again
 
Drawing is pretty crude lol
 
11:08 PM
@SirCumference I have to agree
I just got curious at how "easily" they are calculating the generated energy
 
If you're just letting the water fall down and pump it up again to the same height then you're not going to generate any energy
It will cost you the exact same amount of energy to bring it back up as you gained from it falling down.
 
@ACuriousMind That makes sense. But how do you actually calculate the generated power? Is it as simple as I mentioned?
Basically: height * mass * gravity?
By generated power I mean, the power generated during half the process (when the water comes down).
 
Yes, if you want the energy you can potentially generate from the water falling down then that is correct
 
Let's ignore the fact that no actual energy is gained from the whole process.
 
Of course, in reality you're going to have losses due to friction, inefficient generators, etc.
 
11:11 PM
Is this one of Elon Musk's latest inventions?
2
 
The roast is real haha
I'll screenshot this and send it back to my guys.
Unleash internet hell
@ACuriousMind So the difference in water pressure between the input and the output of the generator has nothing to do with it?
And consequentially, the pipe diameter?
 
The pipe diameter only affects how quickly you're going to generate the energy, i.e. it will determine the power generated.
But if you just want the energy you can get from going from a full upper reservoir to an empty one, it is irrelevant
 
But it does affect the whole "circuit" pressure, right?
So it would only be relevant if, for instance, the generator were to have a higher efficiency for a specific pressure value.
 
That's correct
 
11:18 PM
Alright, thank you :)
What about the pipe angle?
And by adjusting the pressure, can it affect the amount of "wasted energy"?
 
Probably the angle will also affect the pressure, although we're hitting the limits of my hydrodynamics "knowledge" here :P
@RodrigoSilva Whether the pressure has any effect on the efficiency really depends on the specifics of the generator
 
@ACuriousMind Right, makes sense.
I'm thinking maybe I could apply F=ma to see how much the angle affects the stream
 
I found a better way i think to pose my question. Should i delete the old one which already got a reply and create a new one?
 
@pZombie If your question was answered, then you really should only delete it if you think it didn't make any sense. (And you can't delete it if the answer was upvoted, anyway)
 
@RodrigoSilva FWIW, actual large-scale pumped hydro energy storage systems have efficiencies approaching 70%. Some of the loss is due to evaporation, but it's not considered worthwhile covering the reservoirs because the extra construction & maintenance costs outweigh the small energy saving.
 
11:33 PM
Basically, it goes like this. Imagine two lightbulbs switched on. They both emit radiation into all directions. Both lightbulbs get hit by the same amount of radiation/em waves from each other logically. Now if we put one lightbulb inside a torchlight, and point it towards the other, it SHOULD hit the other bulb with more radiation than the other way around. Now apply this to ANY object which emits radiation (as well as absorbs) even at ambient temperature.
The mirror inside a torchlight is aligned such that it focuses the light towards one direction while the other bulb spreads its rays all around. The same SHOULD happen with an object just at room temperature. One could focus its infrared emissions. I cannot see how this would not work.
 
@PM2Ring I think the author had just wanted to play around with a generator, this is not meant for a large-scale production :P.

Evaporation, due to temperature, icnreased by pressure? So it's basically *engineering* the sweet spot for the entire system configuration?
Anyway, this was a good break from what I actually need to do hehe. Thanks for the clarification guys :)
 
I ll take a few days i guess. I am not even sure why i was concerned about the object i target would send back the same amount of radiation to the source. This would actually be rather beneficial if it did
 
@RodrigoSilva Understood. I just happened to remember that efficiency figure for large systems. I assume small systems are worse due to larger relative friction losses, both in the pipes & in the generator turbine. Pumped hydro reservoirs lose a fair bit of water, especially on sunny days, or when there's strong winds. OTOH, the upper reservoir can pick up some extra "free" energy through rainfall.
 
@PM2Ring I guess the rainfall could make up for the losses, if only we weren't in Portugal :P
 
How about a reservoir just below the clouds?
 
11:47 PM
Now that you mention lightbulbs, have you guys ever heard that riddle about the lightbulb and the 100 prisioners?
No googling
 
@RodrigoSilva :) We have some pumped hydro here in Australia, which is a rather dry continent. But of course our hydro schemes tend to be placed in fairly wet locations.
 
A stack of reservoirs below the clouds kept in place via a spacelift or hydrogen allowing to get back the energy of the falling rain to a greater extend than usual
 
Eg, there's some pumped hydro capability in our Snowy Mountains Scheme.
The Snowy Mountains scheme or Snowy scheme is a hydroelectricity and irrigation complex in south-east Australia. The Scheme consists of sixteen major dams; seven power stations; one pumping station; and 225 kilometres (140 mi) of tunnels, pipelines and aqueducts that were constructed between 1949 and 1974. The Scheme was completed under the supervision of Chief Engineer, Sir William Hudson and is the largest engineering project undertaken in Australia.The water of the Snowy River and some of its tributaries, much of which formerly flowed southeast onto the river flats of East Gippsland, and into...
 
how much potential energy is there from the falling raindrops daily on average i wonder
 
@PM2Ring Apparently there are 49 hydroelectric plants in Portugal. I was not expecting it, what the...
@pZombie COUNT ALL THE DROPS
 
11:55 PM
@pZombie Interesting concept, but I would't like to live underneath one. :) It'd block the sunshine, and would be scary if it fell on you.
 
Australia not running 100% on regenerative energy would be unexcusable if one considers the low population density, high sun exposure and topology allowing for all other kinds of regenerative energy sources
 
@PM2Ring Eh, once we've built a Dyson sphere you won't see the sun anyway
 
I like the idea of simply popping a long, well insulated pipe several hundred meters into the ocean and as a result getting cold water inside the pipe. Benefiting from the temperature differences of the water inside the pipe vs the hotter water on top of the ocean
an endless supply of cold water from the bottom of the ocea
 
@pZombie The low population density also means big losses are incurred with long-distance electricity transmission.
 
Can we just have wireless charging everywhere already? Come on guys.
 

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