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rob
8:01 AM
So for an old ascii fogey, o/ (:-p)
 
/o \o how do you type emoji?
 
rob
@Secret I don't
 
user228700
@rob Yes!
 
user228700
It looks like this on my device:
 
user228700
 
rob
8:21 AM
DONE
 
user116211
Is it some new version of XKCD?
 
user116211
user image
3
 
user116211
@JohnRennie This is awesome ;))
 
user116211
 
$$\textrm{TASTY}\cap\textrm{CHEAP}\cap\textrm{HEALTHY}=\emptyset$$
actually at leats to me, it is not empty: The initersection is salad
 
user116211
8:57 AM
Some Olives, a bit pepper, lettuce, cabbage leaves, cucumbers, tomatoes and carrots.
 
user116211
Yeh, I do like this combo.
 
Vegetable risotto would be my candidate. Take whatever veg you have lying around, chop it up, add fresh chilis, garlic and vegetable stock and make a risotto.
 
user116211
I love a bit burnt dried chilli; it adds an all new flavour to the salad.
 
user116211
Hey @BalarkaSen.
 
9:26 AM
Could anyone link some mole concept notes and problems? Thanks.
 
@SwapnilDas that's an awfully vague request. What sort of thing did you have in mind?
 
Challenging problems and conceptual notes :-)
 
About what a mole is i.e. 6.023 x 10^23 particles?
 
9:47 AM
I've got a mole problem, could you help me? @JohnRennie
 
OK, what's the problem?
 
If the atomic weight of C-12 was 24 u, then find the value of the duplicate avogadro constant.
Is it 2x N_A?
 
Well N_A is defined as the number of atoms in 12g of Carbon-12. The question isn't clear, but I assume it's asking how many atoms there would be in 12g of Carbon-24.
 
Yes, perhaps.
Thoughts?
 
My thoughts are that it's a stupid question because it isn't clear what it is asking.
 
10:07 AM
@MAFIA36790 nice
cool find
 
Since the molar mass of $^1$H is 1gmol-1, we can perhaps assume the molar mass of protons and neutrons are 1gmol-1, thus the isotopic mass of $^{24}$C should be roughly 24gmol-1
then we should be able to obain the no. of moles of 12g $^{24}$C that way
 
@JohnRennie wait, what?
where are you going to get this magical carbon 24?
you can get ahold of some carbon 22
for something like a few milliseconds
 
@EmilioPisanty It's a stupid question. Trying to look too deeply into it is pointless.
 
Carbon (C) has 15 known isotopes, from 8C to 22C, of which 12C and 13C are stable. The longest-lived radioisotope is 14C, with a half-life of 5,700 years. This is also the only carbon radioisotope found in nature—trace quantities are formed cosmogenically by the reaction 14N + 1n → 14C + 1H. The most stable artificial radioisotope is 11C, which has a half-life of 20.334 minutes. All other radioisotopes have half-lives under 20 seconds, most less than 200 milliseconds. The least stable isotope is 8C, with a half-life of 2.0 x 10−21 s. Averaging over natural abundances, the relative atomic mass for...
@JohnRennie I dunno
you could ask "If I did manage to put together 12g of carbon 24, what would happen?"
to which the answer is probably "a small nuclear explosion"
@JohnRennie that's the thing
how large of an explosion?
 
$^{24}$C is probably not really that unicorn sounding, thus can still be reasoned with. But yes it wil be extremely radioactive, probably more so than Ts
 
10:16 AM
@Secret how is it not unicorn-sounding?
it's got three times as many neutrons as it should
 
Well no laws of physics say we cannot make such isotope, except it will almost immediately fell apart based on the isotopic table trends
probably first via beta decaying into some higher elements by converting excessive neutrons into protons
A unicorn concept is a different thing, it is forbidden by the laws of physics thus you cannot realise it even in principle or in theory
The isotope table below shows isotopes of the chemical elements, including all with half-life of at least one day. They are arranged with increasing atomic numbers from left to right and increasing neutron numbers from top to bottom. Cell colour denotes the half-life of each isotope; if a border is present, its colour indicates the half-life of the most stable nuclear isomer. In graphical browsers, each isotope also has a tool tip indicating its half-life. == Isotope table == == See also == Island of stability == References == == External links == Interactive Chart of Nuclides (Broo...
Ok nvm, it seems I knew almost nothing about nuclear chemistry
 
10:31 AM
Hey @JohnRennie
on that track
so carbon 24 isn't on any list
but I'm getting some contradictions on how carbon 22 decays
Wikipedia puts it at beta decay into nitrogen 22
but Mathematica's data puts it mostly at BetaDelayedNeutronEmission into nitrogen 21
with <1% going to nitrogen 22 through BetaDelayedTwoNeutronEmission
 
There may be some amusement factor in learning about improbable isotopes of carbon, but the original question was just supposed to check students understanding of what Avagadro's number is. And the question is so poor as to be meaningless.
 
@JohnRennie oh, yeah, forget about the original question
hmmmm
 
That carbon-22 exists at all surprises me. Ten extra neutrons and it still exists long enough to have a measurable lifetime?
 
or it looks like the Mathematica curated data might not be that accurate
@JohnRennie yup. and in the millisecond regime.
then again it's a weak decay, which can be a fair bit slower
but still
anyways, what do you make of these branching ratios?
oh, never mind
> Note on percentage normalization:
> ec β+ 100%
> β+ p 23%
> means that all decays are ec β+, of which 23% is β+ p
 
user116211
10:49 AM
An entire book on self-force:
 
user116211
WoW!
 
user116211
Those two chapters of Feynman in Vol I and vol II were really some sort of badass.
 
A bit of a vague question, but that's also because I can't make a clear picture in my head. Say I have a diagram with two spins, just some arrows pointing up and down with some dotted line or something in between. Does anyone have a good idea on how to depict a vibrational environment in such a diagram? As in, to depict that the entire system is shaking, in a way. Some finite temperature regime.
I've been looking for diagrams in papers where they include some photonic environment pictorially but I can't seem to stumble upon one
 
10:59 AM
@user129412 what's your hamiltonian exactly?
that description is a little vague
 
@EmilioPisanty I'm writing it up now, its a bit involved
 
@user129412 do the spins couple individually, or collectively?
 
Individually, in principle. But this might get very confusing in a picture.
 
@user129412 ...but in practice collectively?
If it's collectively, I'd just put a circle around the two spins and then wriggly-line-couple that circle to some box
If it's individually, I'd put the box at the bottom with (fainter) wriggly lines to each of the spins
 
No, not collectively in practice. But as a first approximation in the picture.
@EmilioPisanty it would be something along the lines of $$H = -\frac{\omega_1}{2}\sigma_{z,j} -\frac{\omega_2}{2}\sigma_{z,j} + J_{12} (\sigma^{+,1}\sigma^{-,2} + \sigma^{+,2}\sigma^{-,1}) + \sum_{j=1,2} \sum_l(a^\dagger^j_l a^j_l + 1/2)$$
 
11:10 AM
@user129412 something went wrong with that tex
 
Okay, that didn't work out just yet.
Still updating it.
 
$$H = -\frac{\omega_1}{2}\sigma_{z,j} + -\frac{\omega_2}{2}\sigma_{z,j} + J_{12} \left(\sigma^{+,1}\sigma^{-,2} + \sigma^{+,2}\sigma^{-,1}\right) + \sum_{j=1,2} \sum_l({a^\dagger}^j_l a^j_l + 1/2) + \sum_{j=1,2} \sum_l \chi^j_l \sigma_{z,} ({a^\dagger}_l^j + a_l^j) $$
yeah, there we go
 
Right, with a redundant plus sign and some terrible notation.
 
ok, hang on
why the double index on the $a$s?
or do you mean
$$ \sum_{j=1,2} \sum_l \chi^j_l \sigma_{z,j} ({a^\dagger}_l^j + a_l^j) $$
 
Because each of the spins has its own phononic bath
 
11:13 AM
on that last term
 
Yes, indeed
That was a typo, to forget the j.
 
no worries
 
But indeed, I'm trying to make a diagram of this situation
 
then I would make two separate boxes under the arrows
and match them with wriggly lines to each spin
 
Hm. Okay, I have to say I don't completely see that exactly. The box would indicate the environment, in some way?
Sort of connecting them to a mass via a spring?
 
11:15 AM
don't think of it as a spring
make it gray
 
I see, I see. Good idea.
 
and make the wriggly line tightly-wound half-circles
 
Sounds like a plan.
 
anyways, I'm off to lunch
 
Great, I'll probably give this a try then. Thanks for the idea
 
user116211
11:23 AM
And here is the Frenkel theory: dx.doi.org/10.1016/S0031-9163(64)91989-4
 
hello everyone
 
user116211
hello
 
user228700
Hi, g'morning :-)
 
user228700
12:22 PM
Anybody familiar with parabolas and circles here?
 
user228700
I've a quick question:
 
user228700
in Mathematics, 46 mins ago, by Kaumudi
I've a quick question (again :-P). I'm given the equation of a parabola and been asked to find the equation of the circle having center at the focus of this parabola and it's also given that the circle touches the parabola.
 
user228700
in Mathematics, 46 mins ago, by Kaumudi
I've attempted it and got the answer but I want to know what the perfect reasoning is, behind my steps. I'm afraid mine may not be correct, even tho I got the correct answer.
 
user228700
Anyone? What would u take the radius to be?
 
user228700
12:45 PM
Anyone? Pls? _/_
 
btw, if I may
 
user116211
Did you draw a figure?
 
Qmechanic's current Illuminator progress is 485/500 (!)
and I also notice that there are 34 eligible answer-and-edit combos from him/her sitting at score zero
so...
... just saying.
 
user116211
Qmech edits like monster.
 
user116211
The numbers are simply mammoth.
 
12:57 PM
@Kaumudi I fear I might be misunderstanding the question, because the circle centered at the focus touching the parabola is obviously just the circle through the vertex, isn't it?
 
Hello hooligans
 
@EmilioPisanty I think no one ever doubted he'd get to that badge as the first
 
@ACuriousMind oh, indeed
s/he's at well over twice the next contestant
@PhysicsGuy source?
 
user228700
@ACuriousMind OK, how is that obvious? :-P
 
user228700
1:12 PM
BTW. This is my fav. thing on the internet. It cheers me up every time I feel down. Dunno if it'll work for u, but check it out if u feel like it. It's absolutely <3
 
user228700
 
Hey guys, have a EM conceptual question:
Consider a cylindrical piece of magnetic material with a constant radial magnetisation pointing out from the centre of the magnet
 
@Kaumudi Because, pretty much by definition, the vertex is the point on the parabola closest to the focus?
 
then clearly at the curved edge and also at the axis, the div(M) there is nonzero
 
So if you want a circle centered at the focus that touches the parabola, it has to touch it at the vertex.
 
1:24 PM
however I recall back in griffith's book mentioned that if the system has cylindrical symmetry, then div(M) can be guarenteed to vanish. How to recoincile this, is M being cylindrically symmetric not a sufficient condition for vanishing div(M)?
 
user228700
@ACuriousMind What dyou mean it has to touch it?
 
user228700
Ur reasoning makes complete sense, and yes, that is the correct answer but I'm unable to understand how it's so obvious that it has to touch the vertex.
 
user228700
Yes, it's the point closest to the focus.
 
user228700
But...so what?
 
user116211
Why? I mean take a circle which shares a common chord equidistant from the vertex and touches the parabola and having its center at the focus; it doesn't pass through the vertex of the parabola.
 
1:26 PM
@Kaumudi 1. The vertex is the point on the parabola closest to the focus. 2. If circle is to touch the parabola, it in particular may not intersect it. 3. If the circle has a radius smaller than the distance between focus and vertex, it doesn't touch the parabola at all (or 1. would be false). 4. If the cricle has a radius larger than the distance, the vertex lies inside it, and so it intersects the circle at two points left and right of the vertex.
So the only option that's left is that the radius must be exactly the distance between the focus and the vertex.
 
user228700
@ACuriousMind Yes, OK, I understand. Thanks so much :-)
 
Can I have proofs for these things?
 
user228700
@ACuriousMind It sounds like we're slowly increasing the radius of the circle from 0, yeah? And then it reaches the vertex and once it surpasses that, it'll start intersecting some other points and no longer just touching one, hm...
 
user228700
@MAFIA36790 In this case, the circle wouldn't be touching the parabola, it would be intersecting it, as ACM has pointed out.
 
user228700
@0celo7 Why?
 
1:34 PM
In fact, if we build a torus out of magnetic material of this form, then we can end up with a system with torodial symmetry, but div(M) not vanish at the surface of it?
 
user228700
@ACuriousMind: Oh, nvm. Read ur answer again.
 
user116211
1:46 PM
-6
Q: Why does the viscous force not depend on the normal reaction because of the mass of fluid above them?

AashishWhy does the viscous force not depend on the normal reaction because of the mass of fluid above them?

 
user116211
Check v1 of the post.
 
2:04 PM
Posted a new EM question. hopefully it is not a stupid question...
 
 
2 hours later…
3:42 PM
@DanielSank I think that quote is usually attributed to Wheeler
 
user228700
How come this place is so quiet? O.o
 
I am writing a letter
 
@Kaumudi I have just eaten a huge lunch and have fallen into torpor.
 
user116211
3:59 PM
@0celo7 To some author? For books? To some prof?
 
Birthday card
 
user116211
WTh ;/
 
4:21 PM
@Kaumudi We're planning an attack.
@MAFIA36790 o/
 
user228700
@JohnRennie Ah, I see :-)
 
user228700
@0celo7 I write letters too, for birthdays!
 
user228700
@DanielSank Why am I not in on it? -__-
 
user116211
@DanielSank \o
 
@Kaumudi it's the only civilized thing to do
 
user228700
4:27 PM
@0celo7 Some people buy pens/pencil boxes -__-
 
user228700
In fact, u're like the 3rd person Ik who writes letters.
 
@Kaumudi hmmmm, you can be, if you know the password.
@Kaumudi Letters are great!
 
user228700
OK, who knows the password here? Except the cunning Sir Daniel Sank!
 
user228700
@DanielSank Yes, they are! :-D
 
user228700
Bed is calling. Goodnight! :-)
 
4:37 PM
@JohnRennie So, this explanation is more similar to the effective mass idea that ACM was talking about right? Does this new quasiparticle have mass?
 
@Obliv yes, polaritons have a non-zero mass.
 
@Kaumudi huh?
 
@JohnRennie (In this particular use of "polariton")
 
Wtf is a polar iron
 
@johnR When the polaritons leave the medium (back to the incident medium), what causes the de-coupling (if thats the right word)
 
user228700
4:40 PM
@0celo7 Huh?
 
user228700
Omg omg omg:
 
@DanS I think it's an exciton-polariton but I'm not sure.
 
user228700
K bye! (:-P)
 
> The Nature survey revealed that most researchers only actually spend around 38 percent of their time doing research - the rest is a mix of admin duties and teaching responsibilities, which are usually essential to getting a job at a university. And we wonder why there are no modern Einsteins or Feynmans.
> As depressing as the situation is, the good news is that more and more people are now talking about the issue, and trying to figure out how to change things for the better.
> Many researchers are now bypassing peer-reviewed journals altogether and instead choosing to publish their work on pre-print sites to be poured over by their colleagues, and others are setting up initiatives that help support and nurture young scientists, as well as make them aware of the realities of the career before they graduate.
> Hopefully these changes will lead to more young scientists spending time where we need them - in the lab. Because it's not just an issue for researchers, but anyone who wants to world to progress thanks to accurate and well thought-out science... just like the prediction of the Higgs boson back in 1964.

"We've got to reward people who do something differently," Bruce Alberts, a biochemist at the University of California, San Francisco, told Nature, adding that the current pressure on researchers encourages "mediocre science".
Why am I seeing the EXACT same pattern of things happening across all disciplines?
 
4:43 PM
@Obliv the polariton is an interaction between the electric field of the light and the electrons in the medium. When the light leaves the medium that interaction ceases. The polariton is a quasiparticle like a phonon. It isn't the case that light changes into another particle when it enters the medium then changes back as it leaves.
 
It's as if since somewhere around 2012. the world is universally burning out inside and outside, with stress and underpaid efforts skyrocketing everywhere
Take sustainable technologies: Farming, energy storage etc. becoming more intensive, with the returns not increasing much
Take work: Working hours become extremely long, but the salary just keep dropping
Take politics: Post truth politics flooding everywhere, and everyone is arguing nonstop
Take academic: The peer review system with its impact factor and publish or persih is a mess, and there are growing number of non reproducible results
etc. etc. etc.
 
@johnR but.. isn't light neutrally charged?
 
See? It's all one and the same pattern. Something is going wrong with the world since around 2012 but what is it?
 
@secret illuminati
 
@Obliv Yes it is, but a light wave is associated with an oscillating electric field (and an oscillating magnetic field). Electrons feel an oscillating force in response to this field so they interact with the light.
 
4:48 PM
@Obliv why are you doing such advanced physics?
 
@johnR Hmm, but I thought that description of light was specific to classical systems. The idea of polaritons seems to be quantum.
@0celo7 Just trying to poke into some deeper physical descriptions of refraction/reflection so I can reference some stuff in my lab report
 
user116211
@Secret Darkseid is coming.
 
@Obliv Have you looked for info on polaritons? The Wikipedia article is quite good if rather terse.
 
@johnR Yeah I looked at the wiki page. It does seem pretty empty, though. Would light still slow down if in a dense medium of neutral particles with no free electrons?
 
5:04 PM
@Obliv To a first approximation no it wouldn't. I'm being cautious because there may still be a weak interaction. For example light does interact with neutrons, though far less strongly than it interacts with electrons.
 
5:16 PM
Has anyone here heard of the ARCC program at Arecibo?
 
@ACuriousMind how does one show that the 2j+1 blocks are irreducible?
 
@0celo7 What 2j+1 blocks?
 
@ACuriousMind angular momentum in QM
 
You'll have to be more specific.
 
The claim on the top of page 197 in Salurai
 
5:32 PM
@EmilioPisanty Source for what ?
 
@johnR have you or anyone any idea whether light necessarily polarizes a medium by some non-zero amount, when passing through it?
 
@PhysicsGuy For your anecdote about Hilbert
@0celo7 Eh, that's just standard representation theory of SU(2). Every representation is decomposable and the irreducible representations are exactly those of dimensions 2j+1.
 
I guess @Obliv is German
@ACuriousMind but why?
 
@0celo7 E.g. by the argument I outline here
 
So much Deja vu
I swear to god we've had this exact conversation at least two times before
 
5:41 PM
That is entirely possible
 
I'm sorry, I don't know how to remember things
How do you do it?
 
@0celo7 das ist nicht korrekt
 
5 messages moved to Trash
 
dumpster'd
 
Please refrain from trying (and failing) to write German insults :P
4
 
5:47 PM
<|8O(
 
@ACuriousMind Ahem. My insults were fine.
 
Alright so now I have good ideas about the theory section of my report but no references. Can I use the h-bar and wikipedia as a reference?
 
@HDE226868 Why duplicate comment but no duplicate close vote here?
 
@ACuriousMind I was distracted by a mod issue on Worldbuilding after posting the comment. Close vote added.
 
Holy fuck some people got 5 and 15 on the QM test
 
5:53 PM
@0celo7 did u get a 100
 
99
 
proof?
 
@0celo7 obviously changed 94 into 99. Not convinced, proof needs more steps.
 
He changed that :P
I can show you the grade book
 
6:00 PM
how would you have access to that.. o.e
 
Blackboard?
 
Oh, we don't use that. We don't get to see our grades online for the most part (mostly because professors are too lazy to update them)
 
6:24 PM
@ACuriousMind do you recognize the book in the background of my picture?
 
@0celo7 Judging from the typeface, it's Milnor
 
7:04 PM
@Obliv Yes, all media have a non-zero polarisibility, though the polarisibility may be very small. So all media are polarised by light to some extent. This equivalent to saying that no media have a refractive index of exactly one.
 
7:35 PM
@ACuriousMind yes!
 
7:48 PM
@ACuriousMind do they have ranch dressing in cabbage land?
 
@PhysicsGuy that Hilbert anecdote
 
@0celo7 We know of it, but it's not very popular
 
@ACuriousMind I just saw some guy putting it on pizza
 
@0celo7 Did you just call Germany cabbage land?
nowhere near as many cabbages as there are potatoes in my experience
 
8:02 PM
now I'm hungry.
 
@EmilioPisanty cabbage is kraut. I think it was understood instantly.
 
@0celo7 I mean, yes
 
And cabbage is very popular, at least where I came from. Potatoes too, ofc.
 
but it's the potatoes that will drive you out of your mind first
 
maybe you
I love potatoes
 
8:07 PM
@0celo7 nothing wrong with a few taters now n then
but
 
in any case, potato land would be Ireland
 
potatoes every day?
no thanks
In other news, though
it seems that this poorly researched question just became my top answer ever, in something like 18 hours
It beat even Can a car steer on a frictionless surface? and Have I discovered how to calculate the proton's mass using only integers?
 
Wow only 72 votes?
 
@0celo7 yeah
 
ACM once got like 150 with one sentence.
I think, "muons experience time dilation"
 
8:10 PM
@0celo7 not quite
115
A: Can radioactivity be slowed through time dilation?

ACuriousMindYes. The classic example is that this is the only reason muons produced by cosmic radiation high up in the atmosphere live long enough to reach the ground.

↑ That is actually the entire answer in that onebox
 
I mean...that's basically what I said.
 
@0celo7 but
look at the timeline
70 votes over the first two days
 
8:41 PM
@EmilioPisanty, well, it is a good answer
 
@heather ACM's or mine?
 
@EmilioPisanty, I was talking about yours when I wrote that, but both are good
Most votes I've ever gotten on an answer is 27
27
A: What on Earth is this drawing of Venus?

heatherAccording to the caption for that picture on the same Wikipedia article, it is Francesco Fontana’s drawing of the supposed satellite(s) of Venus. Woodcuts from Fontana’s work (1646). The fringes of light around Venus are produced by optical effects.$^1$ Fontana lived from around 1580 to ar...

 
Are there any truly optically isotropic materials?
 
yes
the vacuum
 
@heather No, it's an OK answer to a pretty bad question, but the votes are completely out of proportion.
 
8:44 PM
@0celo7 vacuum is not a material
it is the absence of material.
NICE TRY
 
I knew it was HNQ so I was reluctant to answer because of that, but in the end I did feel that if a ton of traffic was going to go there we did on balance need a better answer than was already present
 
@EmilioPisanty (::shrugs::) I thought it was pretty good. =)
 
@heather It's literally three google searches with the <5th result thrown in
 
@EmilioPisanty, perhaps, but it is a nice compilation of information that is well written, and includes a couple puns.
 
@heather the puns are all accidental
it was 2am at the time
 
8:47 PM
@EmilioPisanty, they still make it pretty funny - for instance, the "a bit of digging" one was especially good.
You wrote that answer at 2am?! I'm impressed! My "2am answer" would probably be flagged as low quality =P
 
@ACuriousMind I think I remember a picture where in some german food store there was ranch dressing labeled as "American Sauce" or something lol
nevermind it was thousand island dressing imgur.com/a58fVda
 
@Obliv, =P that is hilarious
 

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