Graphite ain't no molecule

You might as well say it's $C_{10^{26}}$

$\mathcal{F}[e^{-x}H(x)](\omega) = \dfrac{1}{1+i\omega}$ or $\dfrac{1}{1+i\omega} +\pi \delta(\omega+1)$?

H is what, heaviside?

I mean, is it composed by 6 athoms positioned like a hexagon, right?

@Slereah Yeah!

looks like it's just the first one

[Random] Partial deconvolution by linearisation:

$$\int_0^\infty e^{-x(i\omega + 1)} dx = [\frac{e^{-x(i\omega + 1)}}{(i\omega + 1)}]^\infty_0 = 0 - \frac{1}{(i\omega + 1)}$$

up to various signs and constants that I'm too lazy to write

Given some n dimensional data. For n-1 axes, pick the most dominant ordering which will shuffle the data points so that it looks roughly linear. Then, the final result will illuminate the trend in the nth axis

I mean, O2 means that each molecule has 2 atoms of O and these molecules are connected by London forces.

Graphite is composed by graphenes and each one has 6 atoms of C

@Slereah But given the transform pair $H(x) \mapsto \dfrac{1}{i\omega}+\pi \delta(\omega)$ and the shift theorem, shouldn't you get the second one? :O

@Curio C_6 is unstable, so it needs to form a hexagonal lattice

Well you have $\mathcal{F}[fg] = \mathcal{F}[f] * \mathcal{F}[g]$

But on the other hand

$e^{-x}$ isn't a function you can Fourier transform

So I'm not sure this apply

what is the shift theorem?

@Secret why then do we write it as C and not as C6?

@Slereah Yeah that's true. $e^{i \Omega t}f(t) \mapsto F(\omega - \Omega)$

Ah right

because there are many carbon allotopes

That only applies for real $\Omega$

Neeever mind

aight

@Secret what about C60?

actually... what's the formula unit of diamond and graphite?

C

yeah, I think people distinguish between them via crystal structure, since the smallest repeating unit is just carbon. The unit cell of each carbon allotrope differs, however

The unit cell of diamond is a tetrahedron so its formula would be C5

Why then do they sometimes put the number of atoms per molecule (C60 for example)?

this is because C60 is molecular, not a covalent network solid

so its repeating unit is the molecule itself

So if the solid is covalent, do we take away any number?

nope, for example, quartz is SiO2

:/

since these bends come together to form the tetrahedral network

so the bends are the smallest repeating unit

But diamond is a composed by a lot of tetrahedrons too

but they are all carbon, thus the repeating unit is even smaller (formula unit ignores crystal structure if it is a pure element, unless it is molecular)

OK

Thanks!

Actually, the reason why SiO2 has a formula unit of such is because its unit cell has 2 oxygens in it and one Si

whereas for C network solids, the unit cell is just the carbon atom since any translation will generate the crystal structure

I think that is the correct way to understand it

so I was contradicting myself earlier: Formula unit does depend on the unit cell

It's just for carbon covalent network solids, to generate the whole structure, you only need a carbon atom and translation, thus a carbon tetrahedral is redundant as the unit cell is smaller

It's interesting how the starboard functions at times. On some occasions, it alerts you to conversations you really wish you had been there for. On others, it indicates conversations you are so very very glad to have missed.

Starboard has two main functions: To ridicule, or to praise something

Ok thanks

(pinned star drew attention to important admin stuff)

@Semiclassical And now you have to wonder how the star on your message will be seen ;)

::Note to self not to star the above message else a star chain reaction will occur::

Hi guys, a total noob here. I was looking at this question (physics.stackexchange.com/questions/233899/…) and it was helpful. But my question is, which of these (if any) types of scattering explain why a blue chair looks blue? Because I am under the impression that non-fluorescent objects are visible to us because of scattering, just not sure which type of scattering. Any ideas?

Anyone do any good april fools pranks?

@SirCumference I've been pretending to be an elderly GR crackpot for the past ten years

(april fool!)

You got me

I've been creating some real-time machine learning tools in python and might share those today depending on how I feel. In the mean time, I am doing some android application development live on youtube youtube.com/watch?v=DaIXFQIOoOY

Does Not Exist = acepsessl

guys, any ideas about my question?

ok, I'm live again

let's see how far i get hehe

made some changes

Isn't this image wrong? goo.gl/images/SLoFnd

It's the opposite of this one goo.gl/images/FmYGZb

Communicating vessels are bugged

@nitsua60 thanks for your comment. My understanding that it is scattering comes from the text of this question (physics.stackexchange.com/questions/87094/…), which says - "The wavelength that is not absorbed reaches our eyes and we perceive it as "color". This "color" phenomenon is often described as light of a certain wavelength being reflected. I think it is more appropriate to call it scattering". Please correct me if I am wrong

Why is word on mac so buggy

@user1993 Ah, okay, so it's using "scattering" in the broadest sense--"interactions between light and matter"--rather than the more colloquial sense.

@nitsua60 so my question is that are any of the 3 types of scatterings (Rayleigh, Raman, Compton) responsible for a blue chair appearing blue, or is it just plain reflection? I am sorry if the question is too dumb

Petition to change "April fool's day" to "april prank's day" has been issued recently by a couple radical meme groups

(that was itself an april fools joke, nobody really petitioned it)

get prank'd son

Flagged as offensive.

ur face is flagged as offensive

yo mama's so fat when she walked by the tv we missed two goals

no u

rekt

Guys, I am trying to understand how we see normal things like chairs, and how we see fluorescent objects. I was reading this paper (pubs.acs.org/doi/pdfplus/10.1021/ed081p705) and it says that the main difference is that -

You don't know me fool @BalarkaSen

@Cows I think I misread that. What sort of learning tools does a real time machine need?

Is it possible to have a continuous group that's not a Lie group?

@Slereah Yes

I imagine so, but I can't think of one. If it weren't possible, the standard definition of a Lie group wouldn't require differentiability.

I think Lie requires $C^2$ (that's still an interesting result, a group that'S $C^2$ is already Lie, i.e. smooth)

@ACuriousMind C^0 implies analytic for locally Euclidean groups.

if he did it at all

You know I wonder how Galileo described Galilean invariance

you probably have to interpret somewhat to get galilean invariance

@ACuriousMind FC5 is good

the story is basically an acid trip

@0celo7 That was a good description of FC3's story as well :P

@ACuriousMind this game has more freaky acid trips

a third of the map is literally covered in a hallucinogen

and another third too, really

the ending is Not Good