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12:04 AM
@ACuriousMind Yeah it's an interesting interview.
Well, luck does play a role, for sure. But the pragmatic won't count on it when hiring others. Past records are the most tangible testimonials when gauging how useful a person will be for your company/organization.
2 hours later…
2:20 AM
Original Faddeev-Popov article is interesting too sciencedirect.com/science/article/pii/0370269367900676
3:16 AM
re simons theres an old expr wrt the elite/ "1%" referring to economic inequality/ mobility... pulling up the ladder behind them ... just read lengthy/ remarkable articulate/ erudite article on that... the 9.9% is the new american aristocracy theatlantic.com/magazine/archive/2018/06/… but at least simons is involved in a lot of philanthropic activity much of it science/ grant related. also driving force behind great quanta magazine...
other remarkable (astronomy) news this reminds me of benard convection one of the remarkable aspects/ phenomena of emergence... Weird hexagon on Saturn is way bigger than scientists thought nbcnews.com/mach/amp/ncna906541 en.wikipedia.org/wiki/Rayleigh%E2%80%93B%C3%A9nard_convection
2 hours later…
5:48 AM
@JohnRennie I had updated from Windows 8 to Windows 10. Then I got my laptop formatted so idiotic Windows 8 is back. How do I get windows 10 back?
You can download the Win10 install files from Microsoft. Let me dig out the web address ...
6:15 AM
Hey @JohnRennie I am thinking of buying a laptop today. I just wanted to ask you if 725 USD is a good deal for an i7-8550U processor, 256GB SATA SSD, 8GB RAM, non-backlit keyboard, 15.6 inch FHD non-IPS display. Actually it is this model laptopmedia.com/review/…
@JasperLoy I'm not up to date with the current laptop market, so I'm not the best person to advise you.
@JohnRennie I see. Thank you!
I invariably buy laptops on eBay that are two to three years old because they are dramatically cheaper. But you do need to know what you're doing to go down that route.
Yeah, I don't trust ebay.
But Dell Inspirons are well respected. It certainly won't be a bad laptop.
6:25 AM
Why is entropy for the system $$\frac{Q_{rev}}{T}$$ but for the surroundings $$\frac{Q_{actual}}{T}$$
@JasperLoy getting a new laptop gives you security because it comes from a trusted supplier and you get a warranty. That's the safe course. I always buy from eBay but then I'm a long time computer nerd and I'm happy to repair the laptop myself if anything goes wrong with it.
@AvnishKabaj who said
@Abcd Bruce h mahan
@AvnishKabaj please send pic (to provide more context)
Not at my house right now
6:34 AM
@JohnRennie, hi John, can we continue the discussion on the door (handle vs. hinge) problem?I think I cannot pull (nor push ) the door, only an impulse (like a kick) can work
@user157860 I'm working for about half an hour. Back later.
How is programming applied physics?
Both are completely unrelated
6:45 AM
@AvnishKabaj Which edition do you have? Is it coloured or black and white? Also, is it a good book for preparation?
@user157860 any experimental physicist (I was an experimental physicist for 15 years) will tell you to immediately try the simplest experiment you can. Even if it ends up failing it will tell you how to improve the experiment. So get a luggage balance, or similar, and try the experiment. See what happens.
15 mins spent trying the experiment is worth any number of hours of discussion about how hypothetically it might be done.
@JohnRennie, I meant we cannot consider a force pulling/pushing the door but only an impulse, because a force pushing in one direction and progressing with the door will soon not be perpendicular any more, shall I draw a sketch if it is not clear?
If you're holding the end of the balance move your hand to keep the balance normal to the door.
7:29 AM
@JohnRennie, one should advance and rotate the force at the same time without losing contact or changing the applied force, that can't be scientific, can it?, isn't it better an impulse?
7:54 AM
@user157860 you're overthinking this. Try with a balance first and see what happens. Keeping the balance normal and at exactly the same constant force will be hard and there will be some experimental error as a result. But you can estimate the experimental errors and see how much they affect your final result.
@JohnRennie, I posted my first answer: physics.stackexchange.com/questions/173987/…, can you tell me if it is correct?
what do you make of the accepted answer, it's wrong, but do you get what it's trying to say=
8:23 AM
magic burning a piece of wood at a distance
maybe able to be made via some kind of counterfactual communication setup with a high power laser such that the destructive interference ends only at the surface of the wood
This has the advantage of a lasting action that cannot be blocked because nothing passes the middle
2. Quantum computing provide a computational memory impossible with classical computers as the whole atom of universe is used, at a finite region
this means two finite numbers can be infinite relative to each other if extra constraint such as the size of the universe is imposed
1 hour later…
9:53 AM
Soooooo something itneresting i noticed with these two cubes on my desk
their 1cmx1cmx1cm cubes of machined titanium
and I've been rubbing them together (you can see the maching marks showing they were cut in a perticular way
however now I've managed to get them to 'stick' together
and I'm wondering if it's just friction o_O
Someone just linked me to this
which seems like the same thing is happening
1 hour later…
10:55 AM
@djsmiley2k If the surfaces of your cubes are very flat, then it's possible that it's wringing.
the more important question is why have you got two cubes with such a flat surface? :P
11:24 AM
because they were cool :D
@djsmiley2k Jo blocks are available in most machine workshops. And they're designed to have completely flat (and by extension nearly frictionless) surfaces. The sticking of Jo blocks, I guess, is a pretty well known phenomenon
So the reason certainly isn't friction :P
a distinct lack of friction?
@djsmiley2k Sure
I just found it cool
11:34 AM
@djsmiley2k It's not only cool but also useful! Jo blocks are normally stacked to create desired reference lengths
Also, metals are easier to flatten than other materials
Over and above, titanium has a very low value of $\alpha$. So its length doesn't change with temperature (practically). So those blocks form good references
what happened to your avatar? @Blue
@user2646 Changed!
Well, sorta inspired from the Watchmen
11:41 AM
Watched it last week. It was a pretty cool movie. Don't know how I managed to miss watching it earlier. :)
the other one was more sinister loooking
That one was a dark Deadpool logo :P
This current avatar stands for apathy
Q: Why did Dr. Manhattan choose the hydrogen atom for the symbol on his forehead?

Ankit SharmaIn the alternate versions of Watchmen, when Dr. Manhattan's is forced to choose a symbol, he made one on his forehead representing a hydrogen atom. Why did he choose this symbol? How does it relate to him?

11:48 AM
I highly recommend watching the movie btw
thanks, i'll put it on my list
12:16 PM
Sup everybud. We know that $a = \frac{dv}{dt}$. If $v(t) = t^2$ for example, then at $t = 3$ the velocity will be $v = 9m/s$ and the instantageous acceleration at that point in time will be $a = 6m/s^2$ since the derivative of $\frac{dv}{dt}$ will be equal to $2t$. Is my basic understanding of derivatives correct?
@NovaliumCompany Yes, but your usage of units could be a bit more consistent ;)
@ACuriousMind Sorry :D?
how would you correct that @NovaliumCompany?
(usage of units)
@NovaliumCompany What unit is $t$ measured in?
12:21 PM
Oh, I have forgotten the seconds in $t = 3s$?
Oh xD My mistake sorry
Well, that's one issue. The other is that if you plug $t=3\mathrm{s}$ into your equation for $v(t)$, you get $v = 9\mathrm{s}^2$, not $v=9\mathrm{m}/\mathrm{s}$.
Well then the velocity has units of $s^2$
Well, that's weird. :D
12:23 PM
2 mins ago, by user 2646
how would you correct that @NovaliumCompany?
@user2646 You are asking how I would make the units of the velocity not $s^2$ but $m/s$?
Pff, I have no idea :D
This week’s colloquium talk: history of density functional theory. Not bad
@NovaliumCompany Well, since this is just an example pulled from thin air, the simplest way is to introduce a constant in front of the $t^2$ such that their product has the correct units.
whatever happened to the Langlands project lectures?
@Semiclassical It's Sundrum! I've even asked a question about his and Randall's model. Why are you ugghhhhing? :P
So $v(t) = ct^2$ where $c = \frac{d}{t^3}$. Pretty sure this doesn't make sense but.
Because “fifth dimension” just sounds silly and speculative to me
@NovaliumCompany How about $c = 1 [m/s^3]$?
12:31 PM
Maybe that’s too much of a knee jerk reaction
@Lozansky Isn't this the same as what I wrote?
12:46 PM
@NovaliumCompany Not really. I am saying $c$ is a constant equal to $1$, measured in units of $m/s^3$. You are saying $c$ is a function inversely proportional to the cube of $t$ (the independent time-variable). Also, you have introduced a mysterious $d$ (which I know is meant to be a constant with value $1$ measured in $[m]$, but this is not clear from the expression).
@NovaliumCompany $v(t) = t^2$ is terrible form, and it should never be used if at any point downstream from that formula you're going to put things in with units.
$v(t) = ct^2$ is slightly better, though I would strongly discourage you from a variable name (like $c$) that is associated with a velocity.
$c=d/t^3$ is wrong, though.
(a) you haven't defined $d$, but more importantly (b) you've already used $t$ for a variable, which would mean that $c=d/t^3$ is not constant.
if you just want to indicate dimensionality, you use square bracket notation
i.e. $[c] = [d/t^3]$
but preferably $[c] = [L/T^3]$.
Got it, so the best thing would be to do what @Lozansky said? $c = 1 [m/s^3]$
@NovaliumCompany that's an incorrect usage of the square bracket notation.
You just say $c=1\: \rm m/s^3$.
@EmilioPisanty It's how engineers use it
@Lozansky If you're writing for an engineering publication and there is a solid engineering-specific style guide to back it up, then it's ugly notation but acceptable in that context.
12:54 PM
Ok what is the right notation and overall what's the right thing to do with the case of $v(t) = t^2$ to get the right units? Maybe just not use it at all?
Absent either of those conditions, the notation falls under SI Brochure or NIST Guide To The SI style-guide territory, and both make it very clear that that usage is incorrect.
@NovaliumCompany personally, I would use $$v(t) = (1\:\mathrm{m/s}) \times \left(\frac{t}{1\:\mathrm{s}} \right)^2$$
or just $v(t) = kt^2$ with $k=1\:\rm m/s^3$.
@EmilioPisanty two rep limits in a row. Don't you just love the HNQ? :-)
I prefer the last one :D
depending on the context and which aspects require more clarity and emphasis.
As I said above, though, I would above using $c$ for anything that isn't a velocity.
i.e. if $[c]\neq[L/T]$ then $c$ is probably the wrong symbol to use.
@EmilioPisanty I'll just use $b$ for example?
12:59 PM
@NovaliumCompany that works.
@JohnRennie I do.
@EmilioPisanty :-)
@EmilioPisanty We also write stuff like $\dim(k) = LT^{-3}$, is that correct at least? :P
Q: Square bracket notation for dimensions and units: usage and conventions

Emilio PisantyOne of the most useful tools in dimensional analysis is the use of square brackets around some physical quantity $q$ to denote its dimension as $$[q].$$ However, the precise meaning of this symbol varies from source to source; there are a few possible interpretations and few strict guidelines. ...

you're probably referring to usage 4 in that answer's list
Yup, cool
it's one possible correct usage
the only thing that matters is that you use your notation consistently, and that you provide appropriate definitions for any non-standard notation.
everything else is just style
1:04 PM
So in $v(t) = ct^2$ where $c = 1m/s^3$. Where $c$ is just a constant equal to 1 with units of ($m/s^3$). Is that correct?
... which can also be 'wrong' if you're writing in a venue that enforces a strict style guide, of course.
but, as documented in that answer, the official style guides differ on that issue.
@NovaliumCompany no. $c$ is a constant equal to $1 \: \rm m/s^3$.
"equal to 1 with units of X" is language that's just going to cause you more confusion down the line.
@EmilioPisanty Ok, for now I'll just accept it although I don't really understand the $1 m/s^3$ part?
@NovaliumCompany what's there to not understand?
you're already OK with writing $c=1 \: \rm m/s^3$, right?
just replace the $=$ with "equals".
Yes, but what is this number 1, having units of m/s^3?
@NovaliumCompany No. It is a physical quantity, period.
It happens to have the numerical value $1$ when expressed in meters per second cubed.
1:10 PM
Hmm, so I shouldn't try to make sense of the $1 m/s^3$ part? It's just a physical quantity with those units?
You can say "the numerical value of $c$ in m/s$^3$ is 1", though
it's also smart to choose notation which isn't liable to be misinterpreted
$b=1 \: \rm m/s^3$ is a physical quantity of what?
and $v(t)=ct^2$ definitely lends itself to misinterpretation given how frequent $c$ is used to label speed
"$c=1$ in units of m/s^3"
So the constant $b$ is just a physical quantity of nothing really, created just to get the units of $v(t) = bt^2$ right, is that it?
1:15 PM
note that $v'(t)=a(t)=2bt$ and $a'(t)=2b$
so one way to understand $b$ is that it's (half) the rate at which the acceleration is changing
Ok, I'm gonna stop with $b$ here, becuase it's getting even more consfusing.
so if we choose to express our positions/times in units of meters and seconds respectively, then acceleration is expressed in units of m/s^2 and so b would be in units of m/s^3
Do I have to use such constant every time I use derivatives in physics?
It really has nothing to do with derivatives per se.
The point is that physical expressions need consistent units.
And we create these constant to get the right units?
1:20 PM
Well, suppose I have an object in free fall. I'll have v(t)=-gt. Would you say that I "create" g=9.8 m/s^2?
Nor would I
I mean, g=9.8 m/s^2 is a human creation insofar as the definitions of meters and seconds are human creations
but g itself exists regardless of those definitions
@NovaliumCompany you have to use such a constant every time you use expressions that would otherwise have different units on either side of an equation.
By contrast, $v\propto t$ or $v\propto t^2$ is fine
@NovaliumCompany well, it looks like it's "of nothing" in this context because that's where your $v(t) \propto t^2$ came from in the first place
1:23 PM
Can you give me one more simple example like $v(t) = -gt$ so I can get my head around it.
In the interest of getting this out of the mechanics context, consider the case of some system in thermodynamics e.g. some volume of gas
if you were doing this in a real-world physical situation then you would have specific details of the forces and kinematics that produced your $v(t) \propto t^2$ motion, and the constant of proportionality would then have a clear physical meaning in terms of the details of those forces and kinematics.
If I want to increase the internal energy of said gas by $\Delta U$, I'll need to increase the temperature by $\Delta T$
it looks "unanchored" in this situation because your entire calculation is unanchored.
or reduce the volume?
1:26 PM
@NovaliumCompany Sure, I should say "at fixed volume" (and fixed particle number for that matter)
to the extent that the change in internal energy is small, I expect the change in internal energy to be proportional to the change in temperature
i.e. $\Delta U\propto \Delta T$
However, temperature is measured in units of Kelvin and energy is measured in units of Joules
A very very stupid question, does proportional mean equal? Or it means as the change in internal energy goes up or down, then change in temp. goes up or down respectivly?
$A\propto B$ (or $A\sim B$---same idea) says that if I change $B$ by a certain factor, then $A$ changes by the same factor
In a real-world application, you might have e.g. the requirement that the particle's acceleration increases linearly from zero to $a_0$ between $t=0$ and $t=t_0$. That would then mean that the acceleration reads $a(t) = a_0 \ t/t_0$, and you integrate that to get $v(t) = \frac 12 \frac{a_0}{t_0} t^2$. Then has the form $v(t) = bt^2$ and now the constant of proportionality $b=a_0/2t_0$ has a clear physical interpretation in terms of your initial description of the situation.
A factor is a number basically?
a pure number, yes. no units
1:29 PM
e.g. if I have an object moving at a fixed speed, then waiting twice as long means that the object moves twice as far
So time and distance in this case are proportional?
If you don't provide that context, then $b$ won't have an 'explanation' because there's no context to explain it in.
No. Distance and time are proportional, when the velocity is fixed
Oh, yep, sorry, my mistake.
1:30 PM
also, distance relative to the position at time t=0
more generally, fixed velocity means that the change in position is proportional to the change in time
and the (constant) velocity expressed the relation there: how many units of position change per unit of time
Got it. Meaning, if the distance increases with +2 for example, the time would also increase by 2?
Suppose I'm moving at 60 miles per hour. Then in one hour I"ll move 60 miles and in two hours I"ll move 120 miles
@Semiclassical btw I suspect you'll get a chuckle from this charge distribution.
Yep, got the idea. I've also heard of inversely proportional which I suppose is if one increases twice for example, the other will decrease twice?
If by "increase twice" you mean "doubles", then yes
Proportionality is not a story about additive change. It's about multiplicative change
1:36 PM
Yep, got it now :D Thanks
so an inverse proportionality relation would mean that doubling one quantity will halve the other
in the case of interest, you have $v(t)\sim t^2$
which would mean that doubling $t^2$ would double $v(t)$
or, in terms of $t$: doubling $t$ would increase $v(t)$ by a factor of 4
Pff, I don't know but I feel so stupid, my mind isn't working properly, I can't concentrate ;\
However, that's not enough to actually tell you what $v(t)$ is. You could have, at t=1 second, that $v(t)=1$ meter/second. In that case, $v(2t)=4 m/s---the velocity has increasd by a factor of 4
but you could just as well have that v(t)=343 m/s when t=1 second i.e. the speed of sound (if I'm remembering that right) in which case v(2t) = 1372 m/s
Ok, I'm gonna take a small break because my mind is overheating and about to explode, I don't know why. I'll think about these things and come back later, thanks so much for the help!
Exercise for myself: Suppose $|\psi\rangle = |\uparrow\rangle^{m}|\downarrow \rangle^{ n}$ (in the sense of m- and n-fold tensor product). Compute $S^2|\psi\rangle$ where $\vec{S}=\sum_{k=1}^{m+n} \vec{S}_k$
1:45 PM
sounds like it's gonna be some Clebsch Gordan bullshit
to be clear, $|\uparrow\rangle = |\frac12\,\frac12\rangle$
so there's not a lot of clebsch-gordan BS possible there given how simple the actions of $S_z,S_\pm$ on $|\frac12\,{\pm\frac12}\rangle$ are
Still gotta write it out tho
1:59 PM
@Semiclassical that's why they invented the notation $|{↑}⟩^{\otimes m}$.
yeah, i'm just being lazy
2:28 PM
What would be the best place for discussing about programming ?
depending on the programming, here might be a good place if only b/c of how many of the physicists here have transitioned to programming (quite a few machine learning people)
To improve your code^ (code review)
Everyone knows that ^
so I finally figured out what I was remembering from Penrose. he was referring to a construction of Majorana, one which was later picked up by Schwinger and used to formulate his boson representation for spin-s states
which I have seen before but forgot about. So that wraps it up nicely
2:53 PM
@Semiclassical, hi guys, anyone willing to answer a simple query on math?
@user157860 Hi. Just ask your question, without asking if anyone is willing to answer you. And then, wait patiently. It is absolutely useless to ask whether anyone is willing to answer your simple query on math, because they don't know what your "simple query" is, and whether it is "simple" in the first place (by their standards).
$a = \frac{dv}{dt} = \lim_{\Delta t\to 0}\frac{v(t + \Delta t) - v(t)}{\Delta t} = \frac{d^2x}{dt}$. Is this right?
@Blue,thanks Here:en.wikipedia.org/wiki/Harmonic_series_(mathematics) ...wiki says that the harmonic series is divergent, but haven't we all studied that the sum (or integral) tends to $e=2.4828...$
(I wasn't sure math was allowed here)
@NovaliumCompany yes.
@user157860 "haven't we all studied that..." No, I haven't studied anything like that. The proof of divergence is clearly given on that page
3:04 PM
also, $e=2.71828...$
@user157860 It is allowed, but you need to install the bookmarklets from here: math.ucla.edu/~robjohn/math/mathjax.html
one does have $e=1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots$
but not $e=1+1/2+1/3+1/4+...$
Yup, I think @user157860 is confusing the second for the first ^
it's funny how many series amount to "take reciprocals of some infinite subset of the positive integers and try to sum them"
BTW, we need something like this for our room: sopython.com/salad!
3:08 PM
if that subset is the factorials, then you get e. if it's the subset of squares, you get pi^2/6
That's the Stack Overflow Python chat's website ^. I wonder who is paying for hosting the site though
if it's the subset of cubes, no one knows :)
@Semiclassical I think it's to be expected, isn't it? Taking reciprocals of a growing series of integers is the easiest way to produce a sequence that goes to zero, which is necessary for the series to converge
oh, sure.
3:10 PM
@Fawad Eh?
@Semiclassical @Blue, please check this proof 1 + sum of inverse even numbers=1 + sum of uneven numbers 1/3+1/5.... which is surely less than previous , then total less than 3. What is wrong?
it's a natural enough. but when one starts out with series one just sees a bunch of seemingly arbitrary sums
How to proceeed?
@user157860 you're trying to manipulate a divergent series as though it's a convergent one. that won't work
1+1/3+1/5+1/7+... is divergent, just like 1+1/2+1/3+... is
if you want to do something like 1+1/3!+1/5!+... < infty, then such manipulations will make sense
@Semiclassical, 1/2 +1/4 tends to 1, right?
3:11 PM
@Fawad Not feeling like attempting to prove combinatorial relations now. Ask on the main site maybe :P
@user157860 1/2+1/4+1/8+... +1/2^n+... does. 1/2+1/4+1/6 +...+1/2n+... does not
the latter is again a divergent series.
@Semiclassical, I see, thanks
in Mathematics, 53 mins ago, by Manish Kundu
How do I find : S=∑k^2 * (n-k)C(n-r) from k=1 to k=r
3:14 PM
@Fawad Please do not ping random users in an attempt to get a question answered. If someone wants to answer your question, they will do so on their own.
@Semiclassical Isn't "a bunch of seemingly arbitrary symbols" a good description of any math field one is only starting out with? :P
except geometry. then it's a bunch of seemingly arbitrary pictures
3:19 PM
What did fawad post that got him banned?
@Abcd What does it matter to you?
@Semiclassical Depends on the geometry - I know plenty of geometers who only rarely draw pictures ;)
(plus a picture is itself a symbol, just visual rather than textual)
Or only draw..."pictures". I will never forget the algebraic geometry lessons with a fuzzy bunch of lines on the board and the TA going "This is the generic point" :P
Speaking of which.....does anybody still do research in Euclidean geometry? Or is it just "competitive math" for high schoolers these days
@ACuriousMind Curiosity
3:23 PM
@Blue I don't think there are many unanswered questions in pure Euclidean geometry
I think the main things you see nowadays in Euclidean geometry are finding other ways to prove what's known.
Or games of "how weak can I make these axioms and still show stuff?"...
actually, one realm in Euclidean geometry where you probably still do have active work is polyhedra in higher dimensions
But technically that's not Euclidean geometry anymore then, more like sub-Euclidean geometry :P
which does connect to group theory inasmuch as one wants to describe the symmetry groups of said polyhedra
2D or 3D Euclidean geometry seems pretty well exhausted, tho
3:43 PM
So there's this interesting question over at World Building: worldbuilding.stackexchange.com/questions/3474/…
I could make the question a little more formal by asking something like: how would gravitational potential behave in a non-Euclidean space?
@TannerSwett That's just General Relativity :P
You could easily define a metric space that's "Euclidean space with a portal added".
@ACuriousMind Oh, perfect. :D Sadly, I don't know general relativity.
So, it's pretty easy to define gravitational potential in a non-local way. Something like: gravitational potential is the sum, over all massive particles in the universe, of the mass of the particle divided by the square of the distance to that particle.
Can I characterize gravitational potential locally, though?
@TannerSwett Damn, the theorists would make good money if they went into the movie-making/story-writing industry :P
If I simply look at some region of space near a massive object, what does the gravitational potential look like?
If I knew the gravitational potential at the boundary of that region, are there some differential equations I could solve to find the gravitational potential everywhere within the region?
@TannerSwett the problem is that gravitational potential defined in that way is observer dependent.
3:49 PM
@TannerSwett Unfortunately, there's no such thing as a well-defined "gravitational potential" in general relativity.
Let's set aside general relativity, then, and assume I'm just talking about Newton's gravity.
Perhaps the Laplacian of gravitational potential is zero in regions which don't contain any matter?
Aha, this is Poisson's equation for gravity, right? en.wikipedia.org/wiki/Poisson%27s_equation#Newtonian_gravity
The Laplacian of gravitational potential at a point is proportional to the density of matter at that point.
that's true
@Abcd your curiosity is your curiosity none of his curious mind
4:29 PM
@Fawad What did you say?
Apparently we're discussing this on meta now, so let's not additionally discuss it here.
Woah I dont believe this.
Apparently they're blue now and the ask question button is still red... There you go! — Catija ♦ 6 mins ago
@ACuriousMind btw, let me know if you update that one question re: the trivial representation
Q: Bad behaviour from mod

FawadRecently I was struggling with a problem so I asked for help from two people. 1st from Blue . I talked to him before and he is nice dude ready to help. Then I saw semiclassical who is also cool and very active on math chat. He helped me so many times (just some proof) before also in math chat so ...

4:46 PM
Just when I thought only AntiFa vs far right is a clusterf8888
No, a clusterf888888 is too mild, we are dealing with Condensationf888 here
5:18 PM
@PhysicsMeta oooohhhh, thanks for digging this up
I'd forgotten about it
@ACuriousMind (to be clear: what's not acceptable is telling other users to shut up, not the mere usage of the acronym)
I would suggest making the assertion more precise. Telling other users to shut up is rude but, depending on context, may be acceptable if done politely. Using profanity while doing so is what takes it over the line.
please do not star any of that.
Anyhow, the situation looks pretty irreparable now, especially since the user in context has some sort of personal vendetta which has lasted for more than a year. (The last comment by the OP makes it clear)
why john rennie delete his answer
5:24 PM
@EmilioPisanty I guess you're right. I added "in a rude manner" as a qualifier. Which should not be taken to mean it's always acceptable to tell others to shut up if you're doing it politely.
@EmilioPisanty Maybe we should add a "Don't post anything you don't want to see on the starboard" to the room description :P
That ^
Hmm, yeah. There might be someone flagging these
5:30 PM
@Semiclassical I'm still thinking about whether it's salvagable at all.
@ACuriousMind it isn't.
I don't know what you're talking about, but it's almost certainly not salvageable.
@EmilioPisanty I was about two sentences into a rebuttal before I read that second message :P
holy snail I just discovered I have the rights to check the upvotes and downvotes to questions.... WOW. but it's been months I could do that
that's freaking amazing. unfortunately I am interested to get the statistics of the website, so I need like 25 k points lol
@user54826 I think the feature is intentionally a bit hidden because it's somewhat expensive on the database
5:35 PM
@user54826 You can get almost all the same data and more using Stack Exchange Data Explorer.
There were some user scripts to view upvotes/downvotes before reaching the rep threshold
Maybe some exist on the Stack Apps
Haven't checked in a while tho
@Blue I've got one installed as a Chrome extension.
man so that question physics.stackexchange.com/questions/258837/… has 2 upvotes and 2 downvotes. WOW geezus im so freaking happy i feel like a super powered man
5:37 PM
Q: "View Vote totals" without 1000 rep

Rob W Screenshot About The vote counts are a great tool to determine whether an answer is disputed or not. Unfortunately, not many of us have enough time to join all Stack Exchange websites and get 1000 reputation. This script unlocks the "View Vote counts" feature for those who are not logge...

@user54826 they're... less useful than you think
> LaTeX Warning: A float is stuck (cannot be placed); try class option [floatfix]
@ACuriousMind is this a situation where it's OK to tell RevTeX to shut up?
@Fawad you use an i-phone?
@EmilioPisanty Umm, is it compiling fine otherwise? That issue seems to be very common while using revtex
how do you like my new username
@EmilioPisanty rise up!
5:42 PM
not changed?
@Abcd no ipad
I'm occupying this chat, you can't make me leave except by extraordinary measures!
@Fawad Okay.
don't tase me bro
strange my username is changed on my profile page but not here
5:43 PM
@EmilioPisanty The float is...stuck? Does that mean it didn't find any position where it fit?
@coniferous_smellerULPBG-W8ZgjR Chat profiles update with a delay.
I see @ACuriousMind thanks
@Blue oh, it compiles fine. The figures are right where they need to be. Toggling floatfix doesn't do anything to the float placement or make the warning go away.
...unless a mod is feeling nice and presses the "refresh profile" button, which I did.
Oh, floatfix is pretty useless :P There are other ways to explicitly shut off those warnings tho
@Mithrandir huh
5:44 PM
I don't remember the exact command, but lemme check
@Blue oooohhh, are there? if you have links handy, I'm interested.
Q: How much electrons can absorb a single photon?

ArturIn the one hand, energy can be emitted\absorbed only by a discrete portions. In the other hand - according to Coulomb law the higher distance - the lower force, because count of force lines per area decreases with distance. This fact always confused me: the problem is that it's difficult(or imp...

anyone feeling brave?
A: RevTex4-1 warning: "Repair the float package"

egregThe revtex4-1 class defines several similar messages, for instance 351 \@ifx{\eqnarray\eqnarray@LaTeX}{% 352 \class@info{Repairing broken LaTeX eqnarray}% 353 \let\eqnarray\eqnarray@fleqn@fixed 354 \newlength\eqncolsep 355 \setlength\eqncolsep\z@ 356 \let\eqnarray@LaTeX\relax 357 \let\eqna...

Iirc the silence package works well
If you have to block specific warnings
@EmilioPisanty I mean the silence won't help you fix the root of the issue i.e. the "float" placement issue, but it will just suppress those warnings. I don't know if that's what you're looking for
5:55 PM
Gandalf? Is that you?
@Blue yeah, I'll take it.
Like I said, I just want revtex to shut up and leave that space for useful warnings instead.
not nice to tell off revtex bro, what did revtex ever do to you? It's trying it's best...
@EmilioPisanty I can understand the trauma with revtex :P The worst part: It moves footnotes to bibliography unless you force it not to!
The journal guys who maintain revtex sure have some .... issues
> "Gandalf," the old man repeated, as if recalling from old memory a long disused word. "Yes, that was the name. I was Gandalf."
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