The h Bar

Why is distribution of holes in valence band given by $\rho(E)=g_{v}(E)[1-f_{F}E]$. Basically I can't understand why distribution function of holes is $1-f_{F}(E)$ (where that $f$ is Fermi-Dirac distribution function)

Semiclassical

math/physics

well, what's the distribution of electrons in the valence band?

vzn

@Semiclassical double major? so do you have BA/BS in both? ps (trying to remember) have you heard of our chat speaker sessions?

Semiclassical

BS

and, no

@BalarkaSen math.umn.edu/events/multiscale-theory-and-computation

and semiclassical analysis is in the realm of multiscale theory, so there you are

Balarka Sen

aha

Anonymous

4:18 PM

I suppose that is given by the F-D function itself

Anonymous

(for valence band electrons)

vzn

@Semiclassical ok, we are always looking for new spkrs, plz consider it, esteemed group now, popular event, consider you qualified. are you working on a Msc physics? info re sessions physics.meta.stackexchange.com/questions/7783/…

Anonymous

But as some of the valence band electrons move to conduction band...

Semiclassical

right. so $\rho_e(E)=g_v(E)f_F(E)$ and evidently $\rho_h(E)=g_v(E)(1-f_F(E))=g_v(E)-\rho_e(E)$

@vzn physics phd student. but I'm not really interested at this point.

So evidently $\rho_e(E)+\rho_h(E)=g_v(E)$.

0ßelö7

"evidently"

Semiclassical

4:21 PM

in the sense of "this is equivalent to what you're trying to show"

vzn

@Semiclassical bummer ok understand plz let me know any objections or if youd be willing to reconsider in future, conditions etc, realize time investment is can be low/ informal.

Semiclassical

So I suppose the interpretation is that every state in the DOS is filled by either an electron or by a hole.

Anonymous

@Semiclassical Aha...got it now!

Anonymous

@Semiclassical Exactly!

Anonymous

I think that's a reasonable assumption? Or no?

Semiclassical

4:22 PM

I think so

Anonymous

32 secs ago, by Semiclassical

So I suppose the interpretation is that every state in the DOS is filled by either an electron or by a hole.

Anonymous

Okay

Semiclassical

indeed that seems to me the point of having a hole distribution

I wonder if that's the only explanation, though. Consider the following

vzn

@Semiclassical not sure what you mean by that...?

Anonymous

I think some quantum states may remain unoccupied by both electrons or holes...I'm not sure though

Semiclassical

4:24 PM

$$1-f_F(E)=1-\frac{1}{e^{\beta(\epsilon-\mu)}+1}=\frac{e^{\beta(\epsilon-\mu)}}{‌e^{\beta(\epsilon-\mu)}+1}=\frac{1}{e^{\beta(\mu-\epsilon)}+1}$$

Jasper

It must have taken you a long time to type that.

0ßelö7

get rekt by chat

Semiclassical

bah, almost right

There we go

so for the holes it's as though the energy $\epsilon$ and the chemical potential $\mu$ changed sign

Anonymous

@Semiclassical What's the physical interpretation of that fact? (If any)

Semiclassical

@Blue I don't think so. If memory serves, a hole is just a way of talking about unoccupied electron states.

Balarka Sen

4:28 PM

I should learn some more Riemann surface theory

Semiclassical

@Blue I think there's an interpretation but I can't remember what it is. It does seem consistent with holes being unoccupied electron states, though.

Anonymous

@Semiclassical Okay. Thanks a lot for the help :)

Anonymous

I think that is a good question for the main site

Anonymous

(Provided it gets answered :P)

Semiclassical

@Blue see here for some discussion of the chemical potential of a hole: physics.stackexchange.com/questions/7470/chemical-potential

0ßelö7

4:30 PM

@BalarkaSen but it's so algebraic

learn GMT instead

Semiclassical

In particular: "for fermions, both positive and negative $\mu$ is OK. Also, it is easy to see that if both particles and antiparticles exist, $\mu$ of the antiparticle has to be minus $\mu$ of the particle because only the difference $N_{particles}−N_{antiparticles}$ is conserved; this is true both for bosons and fermions. So if the potential for electrons is positive, the potential for positrons or holes (which play the very same role) has to be negative, and vice versa."

Balarka Sen

@0ßelö7 its just pictures so far

forster looks full of analysis

Anonymous

@Semiclassical That does make sense. Hmm.

Semiclassical

I think the $\rho_e+\rho_h=g_v$ explanation is basically a 'Dirac sea' interpretation (all states filled up to the Fermi level, and holes are just empty electron states)

whereas the $\mu_h=-\mu_e$ is more to what you'd see in particle physics re: vacuum

0ßelö7

@BalarkaSen if you think that's full of analysis then you don't want to read federer

vzn

4:34 PM

@JohnRennie interesting yet theres no consistent/ uniform policy on the topic, its quite easy to find articles on the subj at that site. theyve also covered mass extinction science. my impression is their science coverage is "not biased, not bad, sometimes excellent". cf an entire section etc dailymail.co.uk/sciencetech/article-4888574/… dailymail.co.uk/sciencetech/article-4888574/…

0ßelö7

@BalarkaSen The No True Analyst Theorem says your book isn't analysis unless it looks like i.gyazo.com/d4cfb391327940f3751efe977efea652.png

(for the record I have no idea what is going on there)

Balarka Sen

i am still in chapter 1

2 looks like shitload of analysis

0ßelö7

looks like sheaves to me

oh there are some norm estimates too

Balarka Sen

those are what scared me

Semiclassical

@JohnRennie Eh, for the sake of our crazy old loons we'd better hope people keep buying it

vzn

4:41 PM

rats copy paste fail, corrected dailiymail climate change section link, extremely thorough/ in-depth, maybe even in line with or exceeding NYT etc dailymail.co.uk/news/climate_change_global_warming/index.html

0ßelö7

oh my god it's 12:40

@BalarkaSen this is what happens when you do analysis

you watch Rick and Morty while pretending to compute a limit and forget breakfast and lunch

Balarka Sen

#rekt

0ßelö7

0

Q: Deriving $E=mc^2$ with calculus

I am trying to derive $E=mc^2$ by using the work energy theorem. I start with the definition of work: $$ \Delta E = \int_{0}^{x_f} \frac{dp}{dt}{dx} = \int_{0}^{v_f} \frac{dp}{dv}\frac{dv}{dt}vdt = \int_{0}^{v_f}\frac{dp}{dv}vdv $$ by parts this gives $$ \Delta E = \Big[pv\Big]^{v_f}_0 - \int_...

homework-and-exercises special-relativity

@Semiclassical ???

Does this work?

Semiclassical

twitch

hrm

Jasper

I never understood the meaning of E=mc2.

I guess I need to study more physics, which is what Einstein told Nash at Princeton.

Semiclassical

4:51 PM

Here's my quick gloss on it.

One of the bits of special relativity which people like to talk about is length contraction / time dilation

0ßelö7

@BalarkaSen Oh my goooooooood now people are defining this thing as having the uniform limit property

goodbye cruel world

Jasper

@0ßelö7 Same time here, but convert PM to AM, LOL.

Balarka Sen

AM to the PM, PM to the AM funk

@0ßelö7 Which property?

Semiclassical

@0ßelö7 my response to that would be: youtube.com/watch?v=5FjWe31S_0g

0ßelö7

@BalarkaSen $P_tf\rightrightarrows f$

Balarka Sen

4:53 PM

so the theorem you are proving is not really a theorem?

just a definition?

or what

0ßelö7

@BalarkaSen I...maybe?

There's no loss of generality because it is true, but the proof I know is really roundabout.

Balarka Sen

D4NNY - Goodbye (Official Music Video)

►

0ßelö7

There should be an easier proof

Semiclassical

@0ßelö7 yeah, it works

The first two terms of his last line can be written as $$\gamma m_0 v_f^2 +\dfrac{m_0 c^2}{\gamma}=\gamma m_0(v_f^2+c^2/\gamma^2)=\gamma m_0 c^2$$ since $1/\gamma^{-2}=1-v_f^2/c^2$.

So his final result is just $\Delta E=\gamma m_0 c^2-m_0 c^2$. If we interpret the first term as the total energy at the final velocity, then at rest the energy must be $m_0c^2$ as expected.

(Note: By "it works" I mean "the equations work out.")

0ßelö7

@BalarkaSen Hmm. I think this book is just written backwards. It's actually clear from the construction of this particular $P_t$ that the convergence is uniform, but not directly from the definition. Thankfully, I understand the construction.

So I give up. I am defeated. And I'm fucking hungry. Off to get some pizza.

(Costco is amazing)

Balarka Sen

5:04 PM

@0ßelö7 So now the song I linked is 100% applicable.

Semiclassical

As is mine :)

0ßelö7

@BalarkaSen But in other news I somehow managed to construc the Dirichlet heat kernel completely elementarily. Wonder where the mistake in my proof is

@BalarkaSen One point that I need to inspect more carefully: Consider a manifold $\bar M$ with boundary, and let $D$ be the double. Is the map $f:D\to D$ which takes a point to its "opposite", a smooth map?

I think so because far away from the joint it's the identity and near the joint it's just a reflection.

Balarka Sen

Reflection along the boundary, you mean?

For sure.

0ßelö7

@BalarkaSen Right. Hmm.

This proof is really obvious when you think about it. I must have made an error.

@BalarkaSen Do you know what a heat kernel is?

Or what the method of reflections is?

Balarka Sen

nop

but maybe i'll tell you to teach that to me some other day

0ßelö7

5:08 PM

yeah not now

Semiclassical

I feel like I should know what te method of reflections is

0ßelö7

@BalarkaSen heat kernel is just an integral kernel that gives solutions for the heat eqn

@Semiclassical you solve a PDE without caring about the boundary conditions, then add a reflected solution across the boundary to cancel out bad boundary behavior\

you can get Dirichlet or Neumann conditions that way

Semiclassical

ahh. so pretty similar to the method of images (image charges)

0ßelö7

oh, that's what I meant :P

I just added the heat kernel so its reflection across the boundary in the double and got a really sensible result for the heat kernel with Dirichlet data

Semiclassical

I forget what exactly the free heat kernel looks like. I want to say it's $G(x,t)=e^{-x^2/2t}$ and therefore $f(x,t)=G(x,t) \star f(x,0)$

0ßelö7

5:11 PM

$(4\pi t)^{-n/2}e^{-x^2/4t}$

Semiclassical

but I say that mostly because I saw somethign similar yesterday and I'm not sure I'm remembering right

what's $n$ here?

dimensionality?

0ßelö7

yeah

Semiclassical

kk. but that's just normalization, really

0ßelö7

no, the $t^{-n/2}$ behavior is absolutely completely crucial

Semiclassical

hmm

Isn't $\int_{\mathbb{R}^n} G(x,t)\,d^n x=1$?

0ßelö7

5:14 PM

estimating that pole is why things like the Atiyah-Singer index theorem work :P

@Semiclassical yeah

Semiclassical

then I'll defend my comment re: normalization. You start with $G(x,t)\sim e^{-\|x\|^2/4t}$ and normalize to have unit mass.

0ßelö7

I objected to the "really" part there

Semiclassical

to "really" or to "just" ? :P

0ßelö7

It might be normalization but that normalization factor blowing up spawned a huge amount of math

@Semiclassical er, just

Semiclassical

heh

yeah, that's fair.

humble origins, but deep implications

0ßelö7

5:18 PM

yep

Semiclassical

of course, this is the free kernel

so not what you're doing

0ßelö7

I'm working with kernels on Riemannian manifolds

Semiclassical

hm

0ßelö7

which are basically that times a power series in $t$ with some other stuff

Semiclassical

interesting

0ßelö7

5:20 PM

Yep, one writes down the naive kernel and derives corrections

Those eventually converge to the real deal

right now I'm worrying about exactly what $\lim_{t\downarrow 0} h=\delta$ is supposed to mean

Abcd

What is the purpose of the radial component of force that we apply to make a body rotate? I understood that the tangential component provides the necessary centripetal force but I can't find the function of radial component anywhere.

vzn

hmmm, mathematicians vs heat eqns? try this one :) michaelnielsen.org/polymath1/… polymathprojects.org/2013/08/09/…

Abcd

I meant "TORQUE" not centripetal force.

alarge

5:45 PM

@0ßelö7 So like heat kernel expansions then?

0ßelö7

@alarge yes

But there's a few expansion formulas

I'm thinking of the Minakshisundaram-Pleijel one where it's that modified Gaussian times a series

Cows

6:50 PM

Hey guys. . . please straighten me out . . physics.stackexchange.com/questions/358875/…

@0ßelö7 ^

@BalarkaSen^

I need some feedback

0ßelö7

7:22 PM

@Cows I know nothing about instantons.

Cows

@0ßelö7 oh ok no worries.

It's been a while, . . how are you . .

nice photo btw

green is a nice color

0ßelö7

@Cows not terrible, but I've got more questions than answers right now

you?

Balarka Sen

@0ßelö7 doesn't have the best avatar for the century.

vzn

@Cows did someone say soliton? =D

Anonymous

@BalarkaSen It seems you do ;)

Balarka Sen

7:34 PM

Mine has potential. I'll give it a decent to strong 9.

Anonymous

I was expecting something more weird from you though

Anonymous

You disappoint me

Balarka Sen

I am not sure if you have the background to understand the avatar.

Have you actually heard vaporwave?

Anonymous

N a h

Anonymous

"Vaporwave is a microgenre of electronic music and an Internet meme that emerged in the early 2010s"

Balarka Sen

7:36 PM

Then you clearly don't understand the avatar.

Anonymous

Most normal people wouldn't. :'D

Balarka Sen

Yes, normies are normies.

Anonymous

I am malware though

Anonymous

I can give it a try

Balarka Sen

Try Macintosh Plus's breakthrough album

0ßelö7

7:38 PM

@Blue this

Anonymous

Floral Shoppe

Floral Shoppe (Japanese: フローラルの専門店, Hepburn: Furōraru no Senmon-ten) is the seventh studio album by the American electronic musician Vektroid (under the one-time alias Macintosh Plus), released on December 9, 2011 by the independent record label Beer on the Rug. It was one of the first releases of the vaporwave microgenre to gain popular recognition on the Internet. Since then, Floral Shoppe has been considered by many to be the essential defining album of the vaporwave genre, and has been recognized by some as a classic. == Background and composition == Vektroid's alias for Floral Shoppe...

Anonymous

It's not even on youtube "Booting" (ブート Būto) "Tar Baby" by Sade (1985)

Balarka Sen

The full album is on youtube, what are you talking about

Anonymous

Macintosh Plus - ＦＬＯＲＡＬＳＨＯＰＰＥ [Full Album]

►

Anonymous

Finally

Anonymous

7:40 PM

I dug it from the depths of the mariana trench

Balarka Sen

@0ßelö7 tfw you recommend cancer to someone

and they unironically take the reco

Semiclassical

@BalarkaSen ...huh

0ßelö7

@BalarkaSen that's me every time I tell someone to read federer

coffee + pizza = interesting

Anonymous

Sounds better at 2x

Anonymous

Not as cancerous as I expected

Anonymous

7:43 PM

@0ßelö7 pizza dipped in coffee sounds like a good idea

Balarka Sen

@Blue Inhale the aestheticc

Anonymous

0ßelö7

@BalarkaSen my reaction to that is NSFW

gotta clean up, jeez

@BalarkaSen I just felt a strong desire to learn about handlebodies

any tips?

I have Kosinski but it's too hard

Balarka Sen

Nope, none

0ßelö7

do you know about them?

Balarka Sen

7:50 PM

Hardly. I don't know how to use handle decompositions in any reasonable way

Heegaard is as far as I go

0ßelö7

wtf is heegaard

Balarka Sen

heegaard decomposition

breaking up 3-manifold into two handlebodies

0ßelö7

@BalarkaSen idk, but any topology that requires me to be in a certain dimension scares me

@BalarkaSen I finally figured out that conformal transformation

I had the wrong picture the whole time $z+1/z$ maps the exterior of the unit disk to the plane but with a branch cut along the $x$-axis. For some reason I thought it was mapping the interior of the disk to the UHP.

So it actually maps a larger disk exterior to the exterior of an ellipse. When the disk has radius $1$, the ellipse degenerates to a line and there's the branch cut.

0ßelö7

8:44 PM

@ValterMoretti Hey, did you ever see math.stackexchange.com/questions/2422664/… ?

peterh

@Blue Wow, it looks a little bit post-modern...

Anonymous

@peterh I intend to try it out

0ßelö7

@Blue nasty!

Semiclassical

@Blue on one hand, that just looks weird to me. on the other hand, I pretty much only ever drink milk when I'm having pizza

peterh

Does anybody know, what is happened to MAFIA36790? His account is deleted. :-(

0ßelö7

8:54 PM

@Semiclassical milk is like the one thing you should never drink with pizza :/

Semiclassical

pfff

0ßelö7

what do they do to you people in Minnesota

Semiclassical

dairy with dairy

Anonymous

I once tried curd with chicken. Tasted good

Semiclassical

Milk is Good Stuff.

Anonymous

8:56 PM

Someone told me I'd die if I eat mint with coke.

0ßelö7

what is it with you Indians and coke, my god

do you and @BalarkaSen just snort all day?

ACuriousMind

@peterh Generally, an account is deleted if a) the account is a sockpuppet b) the users themselves request deletion c) their only contribution is spam.

Semiclassical

@Blue the usual urban legend is coke + mentos

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