The h Bar

What kind of analysis is that?

That's ring theory.

user218912

I know the proof for 0x = 0 but why is there a negative?

Ask an algebraist.

@IceLord Adapt the usual proof of $-a=(-1)\cdot a$ in a unital ring.

user218912

12:31 AM

we're not doing rings but alright i'll follow that proof.

user218912

@0celo7 to get everyone up to speed

What speed?

user218912

ikr

@IceLord You just have to prove that $0+(-1)\cdot 0=0$.

It's stupid.

Just factor a $0$.

$0+(-1)\cdot 0=(1+(-1))\cdot 0=0\cdot 0$.

You said you know that $0\cdot 0=0$.

user218912

yep

12:36 AM

Did you prove that $0$ is unique though?

user218912

yes

user218912

I have it in my notes

Weird analysis class.

I need someone good with degenerate gases

Anyone?

I'm a degenerate, does that count?

12:37 AM

Well you are good with QM...

user218912

@0celo7 we're doing basic properties of $\mathbb{R}^n$

Don't think it qualifies tho

user218912

@SirCumference I'm taking a graduate condensed matter course right now so ask me in december and I'll help you.

user218912

but you need it now right?

@IceLord D:

Yep

Say $E_{F}≫kT$, so we have a completely degenerate gas

user218912

12:40 AM

@0celo7 reminds me if I do well in all my graduate courses and also in analysis and algebra I'll know so much by the end of sem1. Like literally 50x where I am right now in knowledge and experience.

user218912

I'll try my best.

It's a basic question, don't worry

Let's say we have a neutron star, filled with tons of completely degenerate neutron fermi gas

Ignore my redundancy

As the neutron star gets more massive and compresses, why will the Fermi energy increase?

user218912

we learned about fermi gas last class but not enough to help you.

user218912

he just defined what it is

Oh :/

Thanks tho

@0celo7 You said you weren't interested in stars, right?

That count neutron stars?

12:43 AM

I used to know something about them

But I purged my mind of all physics.

user218912

@0celo7 and... you're taking qm right now.

What?

Ridiculous.

@SirCumference Because the density of particles rises while the available states do not change. And clearly, if you have to fit more particles into the same states (starting from the lowest energy state), the energy of the highest occupied state will end up higher

German outta nowhere

@ACuriousMind Oh wow, somehow that didn't hit me...

I feel a bit dumb

Well, here's the next question I have

in a partially degenerate gas, will all the fermions still occupy the lowest energy levels up to the Fermi energy?

user218912

12:45 AM

@SirCumference btw why are you asking these questions? xP

user218912

are you in a course?

he's a nerd

The only difference between a partially degenerate gas and fully degenerate gas that I seem to see is that there's a different Fermi energy

@IceLord As 0celo7 says, I just like learning

Though he put it in different terms

user218912

@SirCumference but... are you really learning?

@IceLord Hopefully :P

Neutron stars and white dwarfs are hella interesting

12:47 AM

@IceLord Oh no you don't

user218912

Ikr

Huh?

user218912

I shouldn't be talking xD

user218912

@SirCumference nothing just has to do with me enrolling in a bunch of classes which are apparently above my level. I do feel like I'm absorbing everything though.

Also, one more question

12:48 AM

@SirCumference I don't know what a "partially degenerate gas" is supposed to be

user218912

we'll see in december.

@ACuriousMind Where $E_F \sim 10 kT$

@ACuriousMind $X\times Y$ is compact iff $X$ and $Y$ are compact?

Oh, Tynchonoff.

I'm stupid.

user218912

@0celo7 wanna bet Lee?

@SirCumference I don't know what that's supposed to mean. The Fermi energy is defined for a zero temperature Fermi gas, what's that $T$ in there?

12:51 AM

@ACuriousMind My prof wrote $\overline{pq}\vee\overline{qr}$ for "$\overline{pq}$ joined with $\overline{qr}$." Does that have to do with the wedge sum?

@IceLord What are we betting Lee on?

@0celo7 Assuming those are two intersecting lines, that is a wedge sum with the distinguished point the intersection point

@ACuriousMind Ok, good.

user218912

@0celo7 I'll get above 75% in my QFT course and condensed matter course(with the generous help of ACM and DS) and above 70% on the final exams which I will have to do all on my own in class.

user218912

If I fail I'll buy you Lee.

Oh no no, no generous help.

Delete your account, then yes.

I'm not betting on how much help you can extract from them.

user218912

12:54 AM

@0celo7 but the final I'll do all on my own.

Back

@ACuriousMind $kT$ is the thermal energy of the particles

user218912

@0celo7 if I get above 70% on it it proves I actually learned stuff.

If I make this bet, then I'm betting you fail. I don't want you to fail.

So it's a shitty bet.

user218912

@0celo7 ignore what I said before, how about we bet on the final, is that fair?

user218912

I just want to prove I'm actually learning

user218912

12:56 AM

I'm trying really hard, seriously.

Then just do well, why do you have to make a bet?

@IceLord Then stop asking questions that are answered in chapter 1 of Shankar.

user218912

@0celo7 :(

Did you do that integral from earlier?

user218912

yes

user218912

which?

user218912

12:57 AM

my integral or yours?

$\int_0^L\sin(n\pi x/L)\sin(m\pi x/L)\,\mathrm dx$.

user218912

yes

What is it?

user218912

$\delta_{nm}$

@ACuriousMind Oh, you should check out the discussion from earlier about a divergent integral.

I'm pretty sure it's physics math gone rampant.

@IceLord Close.

user218912

1:01 AM

$L \delta_{nm}$

user218912

if you include the term in front of it.

user218912

@0celo7 no actually I'm right I'm seeing it right in front of me in this pdf on orthogonality of sine and cosine.

Close.

user218912

:|

Missing a factor.

user218912

1:04 AM

the 2?

Either check Shankar or work it out for yourself.

Did you work it out yourself?

user218912

no I found it on wikipedia and in a pdf on fourier series integrals.

user218912

brb I have a "floor meeting" with my don. 5 mins

don?

@ACuriousMind Can I get you to write a solution for this "easy" exercise?

Please?

Uh, how about no?

1:10 AM

Why not?

Why should I?

In any case, I'm about to go to bed. G'night.

Night.

Because it would help me greatly.

I can't get the Russian anthem out of my head...

Commie.

Every day a bunch of kids outside my school start chanting it loudly

It's what wakes me up

1:21 AM

Jesus.

What kind of school are you going to?

I know, I have no idea who they are or why they do it

Northeast region, private school

Was hard as hell to get into

That's all I'm elaborating on it

@0celo7 But I don't even think they go to the school

Just looked it up, I was wrong

It's the USSR's national anthem

Commies.

I say run them over.

user218912

@0celo7 I procrastinated my analysis homework working on qft and cm.

user218912

and it's due tomorrow

user218912

:O

1:29 AM

Cool.

user218912

not cool.

user218912

I'm screwed, gonna have to do an allnighter

user218912

there are like 30 things to prove

user218912

and no i'm not asking for help

Link

user218912

1:30 AM

this is basic stuff

@IceLord Preaching to the choir

user218912

@0celo7 it's on a sheet.

user218912

@SirCumference huh?

@IceLord I'm taking a statistics class for the hell of it. The guy wants a huge paper due tomorrow

I've got 18 pages in so far, single spaced

user218912

@SirCumference gl

1:32 AM

Complete with graphs and everything

What the hell?

user218912

what?

18 page paper?

So far at least

Still writing

user218912

weird statistics class you're taking.

1:34 AM

He gives us an admittance list

user218912

the one i'm taking next year is just proving a bunch of shit using analysis and probability theory.

For the university

user218912

wait are you in university?

"Define an ideal college student. Analyze a graph for each variable, compare them to each other, and describe the trends. Explain how each relationship can help us find our ideal college student."

user218912

ideal college student is @0celo7

1:35 AM

what

I can't prove anything in topology :(

user218912

you're in 2nd year chill

that was unrelated

I'm failing at topology right now though

user218912

the class you're taking?

no

user218912

like <50%?

user218912

1:36 AM

oh

no, transformation groups are hard

I hope we cover some advanced metric space stuff in topology

user218912

what do you mean by topology?

user218912

the class?

topology 1

user218912

so you're taking it.

1:37 AM

ffs just part 1 is 10 pages

yes

user218912

o.k.

That's an idea of what I've written for part 1

Still gotta finish part 2 and 3

user218912

if you put that all together it's like 5 pages of text.

Nope, 8 pages

That's for the first part tho

user218912

1:39 AM

anyway I have work to do bye.

There's 3 we have to do by tomorrow

@IceLord did you learn about the compact open topology in that topology course you took

user218912

@0celo7 what topology course I took?

user218912

the one I went for 1 week back in like 2015?

Yes.

user218912

1:43 AM

no we only learned the definition of a topology and topological space and also some review stuff like proving de morgan laws.

user218912

I didn't continue because I would get home at like 11pm

Q: If there is no gravity on the moon why is the american flag waving?

I think I've found the best question on the site

-18

If there is no gravity on the moon, how could this flag be flapping in the wind? (see link) http://www.stumbleupon.com/su/2wD6eg/hea-www.harvard.edu/~fine/images/desktops/Armstrong.jpg

user218912

best... question... ever.

This guy's clearly proven the moon landings were fake

Dude

wtf

Look at this

user218912

there are worse questions like "werewolves have sided"

user218912

1:47 AM

if I remember from long ago.

Q: Do we need Maxwell's Equations since they fail to account for an experimental fact at least in one occasion?

Eh, there's this

-25

This question is an outgrowth of What is the difference between electric potential, potential difference (PD), voltage and electromotive force (EMF)? , where @sb1 mentioned Faraday's law. However, Faraday's law as part of Maxwell's equations cannot account for the voltage measured between the rim...

electromagnetism maxwell-equations

user218912

Mar 11 at 20:13, by 0celo7

-1

Q: Werewolves Have Sided

What would happen if the werewolves never existed would we exist? or would we cease to be on earth ? i have a theory that if the moon shines one the humans then it will degenerate the blood which would cause it to shift to the werewolf so if the werewolves didnt exist we wouldnt either what to do...

collision

user218912

what am I doing I have work to do

Same...

@dmckee Oh, you're here

Howdy

You know about degenerate gases by chance?

Ok, I have shown that the compact open topology is actually a topology.

dmckee

1:53 AM

Only in a vague sort of way. I've more or less forgotten all the mathematical detail.

Deer god, this problem is going to murder me.

So difficult :o

@0celo7 Why do you pray to deer gods?

A google search actually returned this

Because they are cute and delicious

@0celo7 You don't seem qualified to own a pet...

Cute and delicious aren't things that should be in the same sentence

Why not?

Pigs, calves, rabbits are all in that category.

Lambs.

Are you a vegan as well as a commie?

A: Why does adding kinematical conditions between the coordinates prevent the Riemannian line element to "preserve the Euclidean structure"?

2:11 AM

@0celo7 I'm Jewish

Most of those are forbidden :P

user116211

I guessed it! Qmech has answered:

user116211

0

Lanczos is essentially saying that if one imposes constraints on a $3N$-dimensional Euclidean space (which is an affine space with an Euclidean metric of Euclidean signature), one gets (under certain regularity assumptions) an embedded $n$-dimensional submanifold, with the generalized coordinates...

user116211

Now, I need to wrap my mind on this....

good god that's a long post

user116211

@0celo7 Hmm; it was a great confusion indeed. John and I were discussing it yesterday but didn't come to a fruitful conclusion...

user116211

2:16 AM

Qmech says the term Euclidean has different meaning in Physics and Mathematics.

A: Clarifying what metric counts as flat space

@MAFIA36790 Okay. So what does it mean in physics?

user116211

@SirCumference I'm still reading...

user116211

@SirCumference He told to see this post:

user116211

9

1) OP is asking about the use of the word flat metric. It means a pseudo-Riemannian metric (of arbitrary signature) whose corresponding Levi-Civita Riemann curvature tensor vanishes. 2) However, the word Euclidean space may potentially cause confusion among mathematicians and physicists. For a m...

What's going on in here?

2:27 AM

I have a 9 part functional analysis problem I have to solve.

I've done part 1, no clue how to start part 2.

k

user116211

I know that the curvature tensor vanishes when the metric is flat.

Anyone know how to compute the mean speed of a particle in the ideal gas without manipulating probability distributions?

@MAFIA36790 Good.

That's nontrivial, so good.

What's your proof?

user116211

@DanielSank Dude, you are the expert at stat mech here.

user116211

2:31 AM

@0celo7 Wait... I need to finish reading the answers of Qmech; he linked to many of his posts.

@MAFIA36790 It's funny that I would have that reputation, but in any case I'm asking if anyone knows a particular way to recover the result.

@MAFIA36790 You need a nontrivial result about isometries and the Riemann tensor. See Kobayashi-Nomizu or my blog for a proof.

user116211

@0celo7 oh, sure.

I miss middle school...back when everything was about differentiating between opinions and facts

Now I'm in college

I feel old

Well, hopefully it'll be uphill now that I'm taking pretty cool classes

@SirCumference You mean like $d\text{opinion}/d\text{fact}$

user116211

2:36 AM

@SirCumference Are you living on your grandpa's pension in an old Siberian hill?

@DanielSank When I only did part of the work, it was $\partial \mathrm{opinion} / \partial \mathrm{fact}$

@SirCumference hah

@MAFIA36790 Er...no?

user116211

@SirCumference Then you are not old ;P

I guess you're right.

2:40 AM

What?

@0celo7 What what?

In your butt.

Q: Higher-derivative constraints at the boundary for a scalar field in Hamilton's principle

3:48 AM

0

Hamilton's principle, when applied to classical field theory, constrains the physical evolution of a classical scalar field via the equation of motion $$\frac{\delta S}{\delta \phi}=0$$ In the derivation of the Euler-Lagrange equation (using Hamilton's principle) for a Lagrangian $\mathcal{L}...

lagrangian-formalism field-theory boundary-conditions classical-field-theory

0

Q: Partition function for continuous energy

The partition function has usually two definitions: the first is for discrete microstates with energies $E_i$. In this case it is defined as $$Z = \sum_{i} e^{-\beta E_i},$$ where $i$ ranges over all microstates. The second definition takes place on the phase space $M$ with hamiltonian $H : M\t...

statistical-mechanics

This is sounding more like science fiction

user228700

4:20 AM

Hello! Today is the LAST day that I will be tormenting this thread with doubts regarding thermo. so you can all breathe a sigh of relief :-)

user228700

Today's first doubt is this; when given a PV diagram, how to properly discern which curve corresponds to what sort of process?

user116211

4:43 AM

Man, I forgot that I downloaded the pdf of Nakahara M.: Geometry, Topology and Physics...

user116211

@JohnRennie: I asked it here yesterday...

user116211

0

Q: Why does adding kinematical conditions between the coordinates prevent the Riemannian line element to "preserve the Euclidean structure"?

I was reading Kinetic Energy and Riemannian Geometry in The Variational Principles of Mechanics by Cornelius Lanczos; here is the concerned excerpt: Let us define the line-element of a $3N$- dimensional space by the equation: $$\overline{\mathrm ds}^2 = \sum_{i\,=\,1}^Nm_i~(\mathrm dx_i^2 + \...

classical-mechanics differential-geometry

user116211

Qmech has answered it; I'm currently looking to the wiki articles for the terms I'm not still familiar with.

@KaumudiHarikumar Isobaric and isochroic are easy (horizontal and vertical paths respectively).For isothermal and for ideal gas, make sure PV is constant throughout the path (give a hyperbola in the PV diagram). Finally for adiabatic, and for ideal gas, ensure PV^{\gamma} is constant throughout (looks steeper than a hyperbola)

user228700

@Secret Yes, the problem is with identifying adiabatic curves. The rest, as u pointed out, are really easy.

user228700

4:58 AM

How to know if a given path really is adiabatic or just isothermal..?

@KaumudiHarikumar At first glance an adiabatic and an isothermal curve look pretty similar.

The only way to tell them apart is to calculate PV along them and see if it is constant.

user228700

@JohnRennie What do you mean "along them"?

Compute PV along all points of that path in question

user228700

@Secret "all points"?!

@KaumudiHarikumar enough points for you to be confident you know what the curve is

5:03 AM

Another way to check is that the pressure and volumes behave differently as you move through the adiabatic or isothermal curve

user228700

@JohnRennie What if I don't know the exact values of $P$ and $V$? What if I dunno what type if gas it is, so I dunno gamma?

The shape of those two curves depends on the equation of state. In your level, cycles are often in the form of ideal gas, thus the above relations will apply

So yes it is a bit harder to tell them apart for other types of gases

user228700

No, I mean monoatomic vs. diatomic and not real vs. ideal...

@KaumudiHarikumar Unless the person setting the exam question is a psycopathic maniac that situation isn't likely to arise :-)

If I were presented with this in real life I'd graph log(P) against log(V). The gradient will be 1 for an isothermal process or $\gamma$ for an adiabatic one.

user228700

5:08 AM

@JohnRennie See what I mean?

Do your course cover the details of gammas, or they just give the value of gamma to you for ideal gas?

Based on the picture alone i don't think there's any way to tell what process the F to G step is. What does the text of the question say?

user228700

Says "One mole of a monoatomic gas is taken along two cyclic processes E→F→G→E and E→F→H→E as shown in the following indicator diagram. The processes involved are purely isochoric, isobaric, isothermal, or adiabatic".

because the difference between diatomic and moatomic ideal gas is the gamma differs by the degrees of freedom of these gases (rotational, vibrational, translational)

user228700

And clearly, the list of nature of processes doesn't correspond to the steps of the cycle in the given order.

5:12 AM

I think the question means that all processes are one of those three types.

The adiabatic curve will look steeper because $P=\frac{constant}{V^{\gamma}}$ and $\gamma$ is $> 0$

So FG and FH are either isobaric or adiabatic. And it should be obvious which is which.

user228700

@JohnRennie *Four(Sorry, typed it wrong. Edited it.) And yep, that's what it means...

user228700

So in this case I just look at the slopes to figure it out, right? In all other cases, I do the thing with the multiplication of P and V..?

user116211

Okay, Euclidean signature is $(+,+,+,+)$ ....

5:15 AM

@KaumudiHarikumar you're never going to get a question where you'd have to study the curve in that much detail. For example in this case the answer should be obvious without needing any detailed calculation.

user228700

@Secret Yes, that I know and for this question, we can identify using this but I had been asking for all kinds of questions. I mean, even where there is no isotherm to compare our adiabat(or vice versa) to.

user228700

@JohnRennie OK. Thanks, both of u :-)

user228700

5:30 AM

@JohnRennie Are u there?

Yes, still here :-)

user228700

OK :-) Um, this Wien's displacement law..? Looking at the graph, what am I supposed to infer? So the wavelength of the radiation emitted by the black body decreases as the temperature increases...or what?

user228700

I don't exactly get it. Wien's displacement law talks strictly about black bodies, yes?

Wien's law is a special case of Planck's law.

Planck's law tells you what the spectrum of energy emitted by a black body is at any temperature.

Wien's law just tells you how the maximum of the spectrum changes with temperature.

The laws apply only to black bodies but in practice most things are approximately black bodies.

user228700

Okay...

user228700

5:40 AM

Looking at the graph though, the max. emissive power is plotted against the wavelength, yes?

user228700

Which wavelength? I don't understand that part...

A black body emits light with a continuous range of wavelengths.

user228700

Okay...

The Planck law tells you the intensity at every wavelength

user228700

Ohh, right.

user228700

5:43 AM

And at higher temperatures, the intensity reaches a max. for lower values of wavelengths, correct?

Yes. Which is obvious really. Think about heating up a piece of metal. As it gets hotter is glows red, then yellow then white hot.

user228700

Yes, OK. And exactly what does the area under the curve signify..?

The Planck law is one of those density distributions. If I recall correctly it gives the power per unit wavelength.

So the power radiated for all wavelengths between $\lambda$ and $\lambda + d\lambda$ is $B(\lambda)d\lambda$.

So if you integrate $B(\lambda)$ from $\lambda = 0$ to $\infty$ you get ...

Care to have a guess?

user228700

Ohh, right. I'm afraid I dunno what $B$ is; you see, I don't need to learn this in much detail. Just the outline of it is given in my textbook.

Well if $B(\lambda)$ is power per unit wavelength and you integrate over all wavelengths you get ... ?

user228700

5:51 AM

Total power emitted? But what does that even mean?

user116211

@JohnRennie Full Power!!!

@KaumudiHarikumar Correct! It just means the total energy per second emitted by the black body.

user228700

@JohnRennie Wait, this is a constant? For a given body, I mean.

The sun can be approximated as a black body, thus using the above one can estimate the power the sun output and received by earth.

The whole black body spectrum depends only on the the temperature, thus the power is constant for a fixed T

user228700

@Secret Okay, I understand this somewhat, I guess. It's OK. I don't need to learn this in too much detail anyway. Just needed to know the interpretation of the graph.

6:01 AM

@KaumudiHarikumar For a black body the total energy emitted per unit area of the black body depends only on temperature. This is Stefan's law.

It's the energy per unit area of the surface, so a big black body emits more energy than a small black body simply because it has a larger surface area.

In fact, integrating the black body spectrum $B(\lambda)$ over all wavelengths and the area of the black body give you stefan boltzmann law

user228700

@JohnRennie Thought Stefan's law gives the wavelength corresponding to max. intensity at a given temp..?

user228700

Oh, crap! THAT'S Wien's displacement law!

user228700

Sorry, my bad.

user228700

OK, I think I understand this now, thank you. One more...what if the curve didn't have a peak wavelength?

6:08 AM

Stefan's law and Wien's law are empirical laws. They were based on observation at a time when black body wasn't fully understood.

That won't happen for black bodies and most things act like black bodies anyway

When Planck came along and formulated his theory of black body radiation it became obvious that the earlier laws were special cases of Planck's law.

user228700

@Secret Well, what if it did? What would that imply..?

user228700

@JohnRennie Okay, I see...

@KaumudiHarikumar The Planck distribution always has a maximum.

It's mathematically impossible for it not to have a maximum

user228700

6:10 AM

God. I have a question and it doesn't -_- Why is my textbook like this?

Well the first thing is that such weird body is definitely not a black body and planck's law cannot model it. Other than that the properties of this body will depend on some other things...

@johnrennie actually, based on what we knew about quantum systems, is it possible to construct an object with this kind of emission spectrum (because all spectums I saw in spectroscopy have nothing that look like a uniform distirbution)?

user228700

Okay, this $B$? In my textbook, it's given as E_lambda, defined as the emissive power per unit wavelength. And if I integrate this over all the wavelengths from 0 to infinty, I get back the emissive power, which is the total energy radiated per unit time per unit area along the normal to the area. This is the same as $B$, right? (I was confused before; sorry I said I dunno what $B$ is.)

@KaumudiHarikumar I lose track of all the various ways you can write the power emitted. Energy per unit time per unit wavelength per unit area per unit solid angle per unit elephant.

user228700

@Secret Okay :-( Of three bodies-a glowing bulb, a tungsten filament(I dunno how this is supposed to be different from a glowing bulb :/) and the sun, none are "weird"? All are approximately black bodies?

@KaumudiHarikumar all have quite close to black body spectra

user228700

6:16 AM

@JohnRennie LOL XD OK. Was confused. Sorry :P

user228700

@JohnRennie Yeah, so apparently, one of them doesn't have a peak. Great.

@KaumudiHarikumar What? All three will have a peak in their spectrum.

user228700

See:

user228700

All three curves have a peak ...

user228700

6:20 AM

Preeetty sure that it's not a printing mistake.

The one for the Sun is off the top of the scale but it's still there

user228700

Wait, they meant off the top of the scale?!

The curve can't be discontinuous, so they must mean it goes off the top of the scale.

It's a silly graph.

user228700

Okay, awesome, I suck at this. My bad again then, sorry for completely wasting your time :P

No problem. I'm just drinking coffee and postponing doing any real work :-)

user228700

6:23 AM

@JohnRennie All this postponing you do early in the morning definitely helps me out! :-)

Johnrennie, is it possible for any emission spectrum (I am obviously not talking about black bodies here) that can have a uniform distribution or close to it, thus the lack of peaks?

@Secret I can't think of any way for that to happen.

ok