The h Bar

Anonymous

7:00 AM

@BernardoMeurer Meninists wouldn't like that statement.

Bernardo Meurer

@Blue What's a meninist?

Anonymous

@BernardoMeurer Opposite of feminist.

Anonymous

You're old for sure

Bernardo Meurer

Ah

Why not?

Anonymous

To keep both groups happy you need to superpose "men" and "women" in that statement to give equal status to both. One shouldn't come before another. lol :'D

Anonymous

7:02 AM

Use your CS skills to do that. :P

Anonymous

Meh...that was a bad joke. I should just get back to studying Electronics.

Bernardo Meurer

I have placed a GIF into a pdf before

For my lab report on designing a CPU with VHDL

Balarka Sen

Bernardo Meurer

It's cover picture was the death star shooting it's laser

Balarka Sen

This is one of the most absurd funny things I have seen in a while

Bernardo Meurer

7:03 AM

and then in the end there was a gif of it exploding

John Rennie

@BalarkaSen Crash Bandicoot?

Balarka Sen

yup, the cancer of the memeworld

user228700

@BalarkaSen Yes, please!

Bernardo Meurer

@JohnRennie What's the word for an email you send out to someone after the first one which they have not responded to

It's not reminder

Secret

(This is not spam) VVVVVVVVVVVVVVVVVVVVV

google it

Balarka Sen

7:14 AM

@Kaumudi Right now?

John Rennie

@BernardoMeurer follow-up?

Bernardo Meurer

@JohnRennie Exactly!

Thank you

The National Task Force against homosexuality in Uganda

►

Here's a video to pay you back for your services

Balarka Sen

loooool

Secret

This, however is spam:
$Capitalism^{Capitalism}Capitalism^{Capitalism}Capitalism^{Capitalism}Capitalism^{Capitalism}Capitalism^{Capitalism}Communism$

user228700

@BalarkaSen Whenever you can.

Secret

7:16 AM

The above is open to interpretations

urbandictionary.com/define.php?term=metaspam

Balarka Sen

@Kaumudi.H Ok, let me give a teaser trailer

You know what a graph is?

user228700

@BalarkaSen Lol.

Secret

user228700

@BalarkaSen I think so. I watched those videos and got an idea, yes.

Balarka Sen

@Kaumudi So you have three kind of objects coming from a graph. (1) Vertices (2) Edges (3) Faces

user228700

7:19 AM

Right.

Balarka Sen

Vertices are just the dots, edges are the lines joining the dots and faces are areas bounded by the edges (think of a triangle, though it may not necessarily be a triangle)

user228700

Yes, right.

Balarka Sen

For a given graph denote the number of vertices, edges and faces to be $v$, $e$ and $f$ respectively.

user228700

Hmm.

Secret

ϖ--------------ϖ
|
|
|
ϖ--------------ϖ

A graph with 4 vertices and 3 edges

Balarka Sen

7:21 AM

And zero faces.

Secret

indeed

user228700

Right.

Balarka Sen

@Kaumudi.H A small note. Notice that your graph needs to be "planar" for faces to be well-defined.

This, is not planar. Edges intersect each other. So the idea of a "face" is not well-defined.

GPhys

@BalarkaSen are you good with tensor notation? :3

user228700

@BalarkaSen Planar=None of the edges intersect? That's, hmm, OK.

Balarka Sen

7:22 AM

Planar means it can be drawn on a plane without intersections :)

Secret

Is it meaningful to define some notion of hyperfaces for graphs, or just vetices, edges and faces are sufficient?

Balarka Sen

You can draw the graph on the 3D space without intersections say

Just slightly lift up some edges towards you from the page

user228700

OK...

Balarka Sen

But that's a digression

Consider the number $\chi = v - e + f$

user228700

Ah, that's the Euler characteristic, isn't it?

Balarka Sen

7:24 AM

@Kaumudi.H Draw your favorite graph and compute $\chi$ for me.

Yep, it is. Do you know the story?

user228700

Well, only a little. In the video, she went on about how it is -2 for some contrived graph and I didn't understand that at all.

Balarka Sen

If you draw a planar connect graph, then $\chi = 1$ always ever

Try computing some examples!

Start with the dumb example, square.

user228700

Ah. I don't think I caught that specification.

Balarka Sen

Then square with a diagonal

Alex K Chen

Does it have any connection with number of holes in 3d objects ?

user228700

7:27 AM

@BalarkaSen Thanks, I was about to ask exactly how the square has two faces :-P

Alex K Chen

I read it somewhere that chi is somehow connected to twice the number of holes in a solid body.

Balarka Sen

@Kaumudi.H Heh. The thing is I was thinking of the compact case; you can consider the exterior of the square to be another face.

It doesn't matter.

user228700

The exterior, erm, OK.

Anonymous

I had seen a physicsy proof of that stuff once (using electrical charges and induction)...forgot it though

Balarka Sen

@Kaumudi.H That idea will be important, but later. You should compute some examples just of the fun of it

Alex K Chen

7:29 AM

Yeah there's a proof on the book Mathematical Mehcnaic by Mark levi

Balarka Sen

It's one of the most interesting observations in mathematics, gateway to topology

user228700

@BalarkaSen It does seem remarkable that this general statement would apply to all cases under the given condition.

Balarka Sen

@Kaumudi.H Let me explain in a few words why it should be true.

user228700

Wokay.

Balarka Sen

Here's a thing. It's better to think about polyhedron than planar graphs at this point. Eg, tetrahedrons, octahedrons, etc

You can similarly easily define $\chi$ for polyhedrons, right?

user228700

7:32 AM

Yep.

Balarka Sen

Can you compute the $\chi$ of a tetrahedron for me?

user228700

Oh, dang, sorry about that.

user228700

Right, that would be 2, no?

Balarka Sen

Yup.

A cube?

user228700

2 again.

Balarka Sen

7:35 AM

So, fact: Any convex polyhedron has $\chi = 2$

user228700

Right...

Balarka Sen

The reason for this is this following buzzphrase: "Any two convex polyhedrons are topologically equivalent"

What this means is nothing fancy

Take a cube, inflate it by pumping air inside. What do you get after full inflation? A sphere, right?

user228700

Ah, right, you can change one to the other without cutting etc., right?

Balarka Sen

Right, right, exactly.

Alex K Chen

@BalarkaSen But your "pumping" proof works for concave polyhedral without holes too, right ?

Balarka Sen

7:37 AM

Any convex polyhedron is "homeomorphic" to the 2 dimensional sphere.

@AlexKChen Nah.

Think about a polyhedron with one vertex lying on the center of a face.

Alex K Chen

Okay, what happens when you pump up a "cashew" ? It looks like a ball, right ?

Anonymous

Balarka Sen

@AlexKChen Not if the cashew has two sides of it touching each other.

Anonymous

Ah, found it :)

Balarka Sen

It'll inflate up to wedge of two spheres. It's a little complicated.

@Kaumud.H So the point is $\chi$ is a topological invariant.

It doesn't care about the shape of the polyhedron, only the topology of it.

user228700

7:39 AM

Right.

user228700

According to the host, so are Betti numbers, yes? The number of holes...but that bit went completely over my head.

Balarka Sen

The $v$, $e$ and $f$ are Betti numbers indeed.

Let me explain the holes business real quick.

Take the following picture

user228700

Wait, they are Betti numbers?! Wtf.

Balarka Sen

@Kaumudi.H Mhm :) It's really very simple

user228700

@BalarkaSen OK.

Balarka Sen

7:41 AM

Ok, once you have that picture, can you compute $v$, $e$ and $f$ for me? It's a little convoluted because of so many vertices and edges but you can do it

user228700

OK, hang on...

user228700

Is v=30 even remotely correct?

Balarka Sen

I think $v$ is $24$.

Like, there are 6 vertices on the top

6 on the bottom

6 on the two sides

user228700

Ah, I counted the middle one twice.

Balarka Sen

so 6x4 = 24, right?

user228700

7:44 AM

Sorry.

Balarka Sen

Yup

It's okay

I can only count upto 10

user228700

Right, so e now. Hang on...

user228700

@BalarkaSen Lol.

user228700

Is it 48? ::Winces::

Balarka Sen

Ah, yes.

user228700

7:47 AM

Oh, thank God.

Balarka Sen

What's $f$? :)

user228700

On that...

user228700

I'mma say 30 but I can almost guarantee that it must be wrong; I dunno how to count faces in this context.

Balarka Sen

Alright, think about each of those colored trapeziums

user228700

Hmm...

Balarka Sen

7:50 AM

On the top front band there are 6 of them, right?

Bottom front, another 6

user228700

Yeah.

Balarka Sen

Similar on the other side we can't see

user228700

Right.

Bernardo Meurer

Tim and Eric - The Universe + Extanded Scenes (Complete)

►

Balarka Sen

So a total of 4*6 = 24 faces

That makes $\chi = 24 + 48 - 24 = 0$

user228700

7:51 AM

I seem to have counted another 6 at the bottom :-/

Balarka Sen

heh

It's ok

user228700

@BalarkaSen Right.

Balarka Sen

Hard to see these things without interactive objects

@Kaumudi.H The disparity is easy to explain. If you inflate this polyhedron, you don't get a sphere.

You get a donut. Or, as a mathematician calls it, a torus.

user228700

Ah, I see.

Balarka Sen

The torus and the sphere are NOT topologically equivalent

So $\chi$ are immediately different

user228700

7:53 AM

I geddit.

Balarka Sen

Indeed, if you come up with polyedrons which inflate to the double torus, which looks like this:

Then you'd get $\chi = -2$

I won't give an example though, that'd be messy to compute :)

But yeah, so that means $\chi$ actually is intimately related to the number of "donut-holes" on the surface after inflating the polyhedron up.

user228700

Ah, see, I couldn't compute $\chi$ for that object if you couldn't "deflate" to a polyhedron.

Balarka Sen

Right, not with our definition so far.

What is remarkable is that it doesn't matter what polyhedron you deflate back to (Eg both cube and tetrahedron are deflated spheres, but their $\chi$ is the same) :)

user228700

@BalarkaSen Uhh, OK...

skullpatrol

hi, peeps :-)

user228700

7:56 AM

@BalarkaSen Right, of course, yes :-)

user228700

@skullpatrol Hey, man.

Balarka Sen

@Kaumudi.H You could consider surfaces of higher and higher genus

"genus" means "number of donut-holes"

the above surface has genus 3

Torus has genus 1, the double torus above has genus 2

user228700

Donut holes, nice, OK...

Balarka Sen

Fact: For a polyhedron, $\chi = 2 - 2g$ if the polyhedron, after inflation, takes the shape of a genus $g$ surface.

(Agrees with our observation; if $g = 0$ (it's a sphere! There are no donut-holes), $\chi = 2$, if $g = 1$, $\chi = 0$)

user228700

Ah, wow, nice.

Alex K Chen

7:59 AM

@BalarkaSen LOL did you every seen a donut cashew ? I speak for normal holeless cashews, the proof holds, right ?

skullpatrol

what happens to the Klein donut?

Balarka Sen

@AlexKChen If you merely have a polyhedron with a small concavity, then it works, sure. If the opposite side touches with a vertex, it stops working, was my point.

I wasn't really giving a proof anyway, this is an informal explanation sort of

@skullpatrol Nonorientable surfaces are too badass to fit within the margin of this chat

@Kaumudi.H Well, so there you go. That's the end of the teaser trailer.

skullpatrol

Please continue

user228700

@BalarkaSen ::Claps:: Thanks so much :-) It did give me a better sense and helped me to understand some concepts I couldn't grok when the host rushed through them.

Balarka Sen

These are beautiful swaths of mathematics that can be explained through pictures, I love it

You can actually find a connection with electric circuits and whatnot as @Blue was saying I think

user228700

8:04 AM

:-) Clearly. It is quite remarkable.

user228700

@BalarkaSen Yes, I will absolutely, 100% look it up.

Balarka Sen

After all electrical circuits are graphs; basically homology can be explained using Kirchhoff's laws

There's a John Baez post about this

But that's a story for another day. Ask for more later :P

user228700

@BalarkaSen :-P Will do.

Balarka Sen

I haven't slept all night so I'm going to get a nap now lol

seeya

skullpatrol

Cya pal

user228700

8:06 AM

@BalarkaSen Wtf.

user228700

Have a good one :-) Bye!

GPhys

thinks he solved the question

Anonymous

8:25 AM

@JohnRennie Do free electrons in a lattice have only one energy eigenvalue? I think so...but just wanted to ensure... The expression comes to be $E=\frac{\hbar^2k^2}{2m}$ I think

Anonymous

"A particle constrained to a finite interval has quantized energy. A "free particle", that can move any where in space, has continuous energy. Mathematically, that is because the eigenvalues on a finite interval (where you can use a Fourier series) are discrete while the eigenvalues on an infinite interval (where you can use a Fourier integral) are continuous.

Reference https://www.physicsforums.com/threads/energy-of-free-particle-not-quantized.597934/"

Anonymous

Okay...I think I got something:

GPhys

8:51 AM

@0ßelö7 https://i.imgur.com/L4s0G1i.png - I think (?) I solved it. The idea is the vector $V$ is related between the $e_{(\mu)}$ (signature $(p,n-p)$) and $f_{(\nu)}$ (signature $(q,n-q)$) bases by a matrix equation, and so if I let $V^{q+1}=\dotsb =V^{n}=0$ in the $f_{(\nu)}$ basis and assume $q>p$ then there is nonzero $V^1,\dotsc ,V^q$ in $f_{(\nu)}$ basis such that $V^{1}=\dotsb =V^{p}=0$ in the $e_{(\mu)}$ basis. (there are $q>p$ variables for $p$ equations).

Now if I take $V$ as above and consider $g_{\alpha\beta}V^\alpha V^\beta$ which is the same in either basis, then for the nonez…

I think the idea of the solution is fine (it's just linear algebra), but is that notation fine? I wasn't sure a better way to write it than constantly noting what basis I was writing the vectors in

John Rennie

9:14 AM

@Blue free electrons in a lattice is an oxymoron. Assuming you mean Bloch states then each state has a well defined energy.

skullpatrol

electrons free to roam around in the lattice?

John Rennie

9:29 AM

Ha! Finally I got to an easy book ID question first! Usually one of the full time SFFSE geeks beats me to it :-)

skullpatrol

nice work

+1

You're saying that the responsibility for avoiding miscommunication lies entirely with the listener, not the speaker, which explains why you haven't been able to convince anyone to help you down from that wall.

Anonymous

@JohnRennie A lattice for which $V_0=0$. :P Anyhow...forget the term lattice. Free electrons aren't quantized. Right?

John Rennie

@Blue Correct.

The free particle energy eigenstates have a continuous range of energies.

Quantised energy levels are generally the result of a confining potential.

Anonymous

9:46 AM

I think that may be a reason for why we take $\alpha=k$ here: physics.stackexchange.com/questions/356582/…?

Anonymous

0

Q: Are some steps wrong in this derivation (Kronig Penney Model)?

Here, Wikipedia has the derivation for the equation: $$\cos(ka)=\cos(\alpha a) + P \frac{\sin(\alpha a)}{\alpha a}$$ I didn't understand one of the steps: $\beta^2-\alpha^2$ should be equal to $\frac{2m(V_o-2|E|)}{\hbar^2}$ and not $\frac{2m(V_o)}{\hbar^2}$. I don't understand how they m...

Anonymous

Instead of $\alpha a = 2n\pi \pm k a$

Anonymous

Not very sure though

ACuriousMind

What are you trying to do? :P

Anonymous

@ACuriousMind I'm trying to copy Adam's comment here. It's not working

Anonymous

9:49 AM

<https://physics.stackexchange.com/questions/356582/are-some-steps-wrong-in-this‌-derivation-kronig-penney-model?#comment798150_356582>

ACuriousMind

You have to click on the comment's timestamp, then copy the link.

Anonymous

physics.stackexchange.com/questions/356582/…‌‌-derivation-kronig-penney-model#comment798150_356582

Anonymous

0

Q: Are some steps wrong in this derivation (Kronig Penney Model)?

Here, Wikipedia has the derivation for the equation: $$\cos(ka)=\cos(\alpha a) + P \frac{\sin(\alpha a)}{\alpha a}$$ I didn't understand one of the steps: $\beta^2-\alpha^2$ should be equal to $\frac{2m(V_o-2|E|)}{\hbar^2}$ and not $\frac{2m(V_o)}{\hbar^2}$. I don't understand how they m...

quantum-mechanics homework-and-exercises condensed-matter semiconductor-physics

Anonymous

Damn...again

Anonymous

I did click on the timestamp and paste the link

Mr. Xcoder

9:50 AM

@Blue try this:

https://physics.stackexchange.com/questions/356582/are-some-steps-wrong-in-this‌-derivation-kronig-penney-model#comment798150_356582

.

Anonymous

Okay...pasting that:

ACuriousMind

For the first question : $E$ is finite, even in the limit $=V_0\to\infty$, so it can be neglected. For the second question : we are interested in the lowest energy state, which is the one with $n=0$. — Adam 15 mins ago

Anonymous

physics.stackexchange.com/questions/356582/…‌‌-derivation-kronig-penney-model#comment798150_356582

Mr. Xcoder

For the first question : $E$ is finite, even in the limit $=V_0\to\infty$, so it can be neglected. For the second question : we are interested in the lowest energy state, which is the one with $n=0$. — Adam 15 mins ago

Look it works ^

skullpatrol

^

ACuriousMind

9:51 AM

@Blue Are you posting from mobile or from a desktop?

Anonymous

See what happened when I pasted what Mr Xcoder wrote ^ :P

Anonymous

@ACuriousMind Laptop

Anonymous

Something is wrong with Chrome

Mr. Xcoder

@Blue Oh... Try adding https:// in front

John Rennie

For the first question : $E$ is finite, even in the limit $=V_0\to\infty$, so it can be neglected. For the second question : we are interested in the lowest energy state, which is the one with $n=0$. — Adam 17 mins ago

ACuriousMind

9:52 AM

@Mr.Xcoder There is a https in front in what he posted

John Rennie

It works here ...

Mr. Xcoder

@ACuriousMind :o I'm blind

Anonymous

@Mr.Xcoder I did.

Anonymous

:P

ACuriousMind

@Mr.Xcoder The chat just automatically removes that when it linkifies the, well, link. You can't see it.

Mr. Xcoder

9:53 AM

@ACuriousMind I tried looking at the history of the message to see its source, but @Blue didn't edit it, unfortunately

And that message now appears 100 times....

skullpatrol

lol

Mr. Xcoder

Is it just me or all the chat icons are misaligned?

skullpatrol

just you

Mr. Xcoder

lol

Anonymous

I think there was some problem as the comment was an edited one.

Anonymous

10:02 AM

Sandbox

Where you can play with chat features (except flagging) and ch...

3

Anonymous

Now it seems to work fine

Secret

10:28 AM

Allow me to introduce:

existentialcomics.com

Jasper

@Mr.Xcoder LOL I thought you added that UFO inside.

Mr. Xcoder

@Jasper Hehe

Jasper

@0ßelö7 Learning French would be a good investment. If you do math in grad school in the US, usually they require reading knowledge of French, German, or Russian.

Anonymous

10:49 AM

This looks nice: webshop.elsevier.com/languageservices/translationservices. It's paid though.

GPhys

@0ßelö7 morning :3

lılostafa

11:32 AM

Where is it that people have electrical sockets in bathrooms? I’d never have noticed this as none of my bathrooms have electrical outlets! — Tim 2 hours ago

@Tim - so you plug your washer, hair dryer etc. into... what? I bet my ass that this is gonna be another American "ban Kinder Surprise because it's too dangerous" thing. — Davor 2 hours ago

Anonymous

@lılostafa I believe most Eastern countries don't have sockets in bathrooms as they are wet bathrooms unlike the Western countries which have dry bathrooms (or rather the dry and wet zones are separate).

Anonymous

In our house we have a room just outside of the actual bathroom where we plug in washer, hair dryer, etc. That's commonly called the dressing room.

ACuriousMind

@Blue If you read the comment thread you'll see that Tim lives in the UK and they apparently don't have sockets in bathrooms.

Anonymous

@ACuriousMind Ah. Interesting. I didn't know that about UK. My impression was that mostly the Asian countries have wet bathrooms. Sockets are not installed to prevent electric shocks and stuff

GPhys

this GR homework is giving me math major PTSD from when I took some math master's courses

ACuriousMind

11:45 AM

In any case, I purged the comments because they didn't have anything to do with the question itself :P

0ßelö7

@GPhys ?

GPhys

@0ßelö7 I think I solved one of them (above)

Anonymous

@ACuriousMind lol....you spoilsport mod :P

0ßelö7

Morning. My phone does this thing where it logs me on when I'm afk

GPhys

and I think I just figured out why/how to solve one, but it doesn't seem to be the way the hint is suggesting

I summarized my solution for the sylvester one above :3

lılostafa

11:49 AM

@Blue Shouldn't have posted here....the question was hot and receiveing a lot of cross-SE views, and we probably could have a nice extensive discussion on this in the comments

GPhys

I don't like my notation on it though. I guess even now I wrote it in a way that implies the eigenvalues of the metric are actually +-1, but they don't need to be

I'm not sure what the best way to write e.g. $V^1V_1$ is in a way that includes the metric, but not the full summation

$g_{11}V^1V^1$ I guess?

lılostafa

@JohnRennie at least we don't go to school anymore. this is what I say to myself these days, when I feel stuck with many problems.

ACuriousMind

@lılostafa I regularly check the highly commented posts, anyway, so you'd only have delayed the inevitable

Anonymous

I'm a bit confused....what would the area under the curve of F-D stand for?

ACuriousMind

12:00 PM

@Blue Why do you think it stands for anything?

0ßelö7

Usually the integral of a distribution means something @ACuriousMind

Anonymous

@ACuriousMind I was thinking of an analogy with M-B distribution where $N=\int f(v)dv$ ($N$ is the number of molecules in a gas)

Anonymous

I thought the integral might actually stand for something

0ßelö7

@GPhys I guess. I'd have to see the full proof to know what you mean for sure

GPhys

@0ßelö7 did you see it above, or want more?

ACuriousMind

12:05 PM

@Blue Well, that depends on how you normalize it. "The" Maxwell-Boltzmann distribution is usually your $f$ normalized by $N$, so that $f(v)/N$ is the probability density that any given particle has velocity $v$.

Anonymous

@ACuriousMind Umm, without normalization the $f(v)$ just means the number of gas molecules with a certain velocity, right? Similarly, $f(E)$ should mean the number of electrons with a certain energy $E$. If I normalize it by (suppose) total number of electrons, then the area under the curve should give me the total number of electrons, right? Otherwise it doesn't mean anything. I think I get it...

ACuriousMind

@Blue It doesn't mean the "number of electrons" in your picture, though. The graph axis is clearly labeled "probability of occupation", not "average occupation number".

(which makes sense since fermions can actually occupy a state more than once due to Pauli)

Anonymous

@ACuriousMind Well, so what does probability of occupation mean ? I suppose it is just means (average occupation number of a certain energy E by electrons)/(total number of electrons) ?

Anonymous

Maybe they even took the spin factor of 2 into account

ACuriousMind

@Blue That's one way to say it, but since you won't find more than one electron in any state to begin with, that's kinda a misleading way to think about it. Just think of it as the probability that if you look at any state with energy $E$, then that function gives the probability there's an electron in it.

The Fermi-Dirac distribution is for one state per energy $E$ - if you have more than one state with the same energy, you need to know the degeneracy $g(E)$, also known as "density of states".

That is, to answer your original question, the F-D distribution is just a probability distribution and hence the area under it will always be 1, if you deal with the units properly.

To get the total number of electrons you need the additional information $g(E)$.

Anonymous

12:18 PM

@ACuriousMind Aha...that makes it clear! I was confused about where they are adjusting for the spin (i.e. the factor of 2). The density of states takes opposite spin of electrons in the $E-k$ space into account. Gotcha.

Anonymous

Thanks :D

ACuriousMind

@DavidZ Did you ever figure out what the issue here was?

0ßelö7

@GPhys I did not see it and can't find it

I won't be at a computer until this afternoon though

GPhys

@0ßelö7 chat.stackexchange.com/transcript/message/39931408#39931408

Anonymous

@ACuriousMind One more question. I'm not sure which units to use so that the area under the F-D distribution comes out to be $1$. Neither $J$ not $eV$ seems to work. (In the FD graph I posted above) Is that graph wrong or I am missing something?

Anonymous

12:23 PM

With eV as unit the area is $5$

ACuriousMind

@Blue You need to recognize that the probability density is not unitless

But it may well be that the graph is wrong :P

Yeah, it's wrong, and you're right with the 5 :P

Anonymous

@ACuriousMind Umm, what's the unit for probability density?

ACuriousMind

Someone drew this not thinking about it having to be a normalized density

Anonymous