@AaronAbraham I'm reluctant to make that an answer because I think the question isn't well-posed (for the reasons I listed).

this is visually crazy (NB gif don't embed, click pics to see)

Occasionally

@obe Dark gray Jordan Horizons

@BernardMeurer loves them

They are absolutely hideous

@secret that is pretty amazing.

hi @obe

@obe Birds in the Trap Sing McKnight

@ACuriousMind likes them too

Screw you then

i see u ocelo

@MartianCactus it clearly depends on the direction that you apply it

@MartianCactus not you

I meant @obe

oh

why is foce a vector quantity?

@martian why not?

soyou ALWAYS specify the direction of force too?

yes.. by definition force is a vector quantity.

and if we multiply a scaler quantity with a vector quantity , what will the result be?

If you don't care about direction then force would mean something else (magnitude of change in speed of a mass)

but we clearly do, which is why acceleration and velocity are defined as vectors.

(magnitude of change in speed of a mass)--isnt force that only?

magnitude of change in velocity of a mass is force

velocity $\ne$ speed

but we dont include direction in the definition

$m\frac{dv}{dt} \ne m{ds}{dt}$ where s is speed

@martian yes we do

what definition are you looking at

also what is the result of $$vector*scaler$$

how come?

You should check out khan academy or google

i did watch the first video on vectors/scaler

Scalar is just a number. Vector is an object in a vector space that specifies a direction and magnitude. Multiplying a vector by a number just gives you the vector with a scalar multiple

oh

what if we multiply 2 vectors which are acing in opposite directions?

will one of them be negative?

no that would make the whole thing negative

how do i try it?

idk what that means XD

how can we even multiply directions?

@martiancactus you can't

that's a meaningless idea

you need to develop a different idea of vector multiplication

hm..

@obe huh?

@Obliv tensor product?

the dot product of two vectors yields a number (not a vector) that gives how much of each vector is in the direction of the other. Can you derive such a definition? @martian

idk

what definition are you talking about?

I'm asking you.

@obe yes

idk

actually i gtg

will talk about it next time!

@obe meh

@0celo7 what is $\text{inf} \left\{...$ followed by a sum mean? Infimum of each term? also the terms are written $\mathcal{l}(I_k)$ where $I$ is an interval what is $\mathcal{l}(I)$?

dude i hate latex holy shit

okay just look at this en.wikipedia.org/wiki/Lebesgue_measure#Definition

oh $\mathcal{l}(I)$ is just the length of the interval

Why are you doing Lebesgue measure @Obliv

@0celo7 was reading wiki article on fubini's theorem for double integrals and it mentioned something 'embarrassing' about it not being applied to $\mathbb{R}\times \mathbb{R}$ with lebesgue measure. so, I wanted to know what it was but I don't care anymore

@Obliv link?

@0celo7 en.wikipedia.org/wiki/…

@Obliv Oh, I think I knew that.

I wonder if it's a problem when one rejects AoC though

Because then there are no non-Borel measurable sets

@Obliv For Riemann integrals it doesn't matter, don't worry.

Okey dokey.

@Obliv In the typical treatment one just has to be careful with integrating over null sets

But that's fairly technical

Does that translate to holes?

@Obliv I think one can have $X\times Y$ be a null set (zero volume) in the product measure, but $X$ not be zero measure wrt. its initial measure

or even unmeasurable

but that certainly requires AoC, which is false, so you're ok

Huh, I didn't know \eqref works in MathJax.

what does it do

$x\eqref y$

Hm, let's see if it works in chat

$\tag{1}$

$$ \text{This is an equation} \tag{1}$$

$\eqref{1}$

\eqref{1}

bitch

I was already doing it

doesn't look like it works, no

Anyway, look here to see it in action on the main site

probably because the chat posts are disjoint

@Obliv measure theory can get crazy

you probably learned that a line has measure zero, right?

well, you can induce a measure on the line, which will not be complete! That means, you can find a nonmeasurable set on the line

Assuming the Devil's Axiom, of course.

So you have something that has measure zero, and there's a subset of that which doesn't have measure zero.

@ACuriousMind Did you know that the nowhere-diff'ble functions are dense in the set of continuous functions?

Makes you wonder why it's so hard to write one down

@0celo7 Yes.

Stumbled across it when I wondered whether one could restrict the path integral to "nice functions", and it turns out the differentiable functions have Wiener measure zero :P

@ACuriousMind I'm assuming Wiener measure is a measure on the Banach space of bounded functions?

Continuous functions

something like that

@ACuriousMind How would one polygonally approximate a continuous function on a compact interval?

@0celo7 Depends how abstract you want to be, but the "standard" Wiener space would be something like all finite paths starting at a fixed point. There's also a conditional Wiener measure that is on the space of paths with fixed start and end points

@0celo7 I know that you know about Stone-Weierstraß, so I don't understand the question.

@ACuriousMind Polygonal functions don't form an algebra.

I have $f:[a,b]\to\Bbb R$ and I want to get an $\epsilon$-close PL function.

(in sup norm)

Oh, polygonal. Wtf is a polygonal function?

(I read "polynomial")

@ACuriousMind PL!

piecewise linear

So, partition $[a,b]$ into $n$ parts, then I can find a sequence $(p_n)$ of PL functions.

With $|f(x)-p_n(x)|<\epsilon_n$.

But how to see that $\lim \epsilon_n=0$.

There's a clear candidate for the $p_n$.

@ACuriousMind This is weaker than SW because PL functions aren't smooth.

So the question is how to estimate $\epsilon_n$.

So, let the partition be $a=x_0<x_1<\cdots<x_n=b$.

Now $p_n$ is linear on each $[x_{i-1},x_i]$.

And we set $p_n(x_i)=f(x_i)$, right?

This determines $p_n$ on $[a,b]$.

Your "clear candidate" is just the straight lines between the $x_i$, right?

Right.

So $$|f(x)-p_n(x)|\le \sum_{i=1}^n\sup_{x\in[x_{i-1},x_i]}|f(x)-p_n(x)|$$

We can bound the total error by a sum of errors on each interval

$f$ is continuous.

$\sup_{x\in[x_{i-1},x_i]}|f(x)-p_n(x)|$ can certainly be estimated somehow

So $|f(x)-p_n(x)|$ is a continuous function on $[x_{i-1},x_i]$

it goes from $0$ back down to $0$ at the end

So it has a supremum.

Since you are on a compact space, you can use that $f$ is not only continuous, but uniformly continuous

Agreed, I was thinking about that.

Maybe I need my intervals to fit inside the magic uniformly continuous $\delta$?

Yes. To each $\epsilon$ you can find then intervals such that $f$ doesn't vary more than $2 \epsilon$ within them

Which should then give you that $\lvert f(x) - p_n(x)\rvert$ goes to $0$ for $n\to\infty$.

Hmm, but how small do I make $\epsilon$?

Well, what you have to show is that for all $\epsilon > 0$ there exists $n$ such that $\lvert f(x)-p_n(x)\rvert < \epsilon$

Yes

So fix $\epsilon$, and by uniform continuity there's a $\delta$ such that $\lvert f(x)-f(y) \rvert < \epsilon$ for all $x,y$ with $\lvert x - y \rvert< \delta$.

Agreed, but I think we run into problems shortly

Then choose $n > 2/\delta$.

$2/\delta$?

I think it depends on $a$ and $b$

Oh, yeah, I'm doing this on $[0,1]$, I guess

But we can work on $[0,1]$, I'm not committed.

So make $n$ partitions.

What now?

I got this far too

So we have that $\lvert f(x) - p_n(x) \rvert = \lvert f(x) - ((f(x_{n-1}) - f(x_n)) (x - x_{n-1}) + f(x_{n-1}))\rvert$.

Oh man, you manage to write down $p_n(x)$, I didn't know how to do that

I might have fucked it up :P

But it looks something like that :D

Ok, so assuming such a thing

Leeme think more carefully

PhD line through two points

The slope is $m_n := \frac{f(x_n) - f(x_{n-1})}{x_n - x_{n-1}}$, so we have $p_n(x) = m_n (x - x_{n-1}) + f(x_{n-1})$.

Yeah, that's the right $p_n$.

Right.

So we do triangle inequality and get $\lvert f(x) - p_n(x)\rvert \leq \lvert f(x) - f(x_{n-1})\rvert + \lvert m_n(x-x_{n-1})\rvert$.

Sure

The first thing is $<\epsilon$

The second is $<m_n\delta$?

Well, $m_n (x-x_{n-1}) = (f(x_n)-f(x_{n-1})) \cdot \frac{x -x_{n-1}}{x_n - x_{n-1}}$.

The first factor is $< \epsilon$, the second is $< 1$.

That's not good :P

Why

We have now $ \lvert f(x) - p_n(x)\rvert < 2\epsilon$, don't we?

oh, wtf

well still

$\epsilon+1<2\epsilon$ only in special cases

Huh?

oh, factor

I'm stupid

Right, so my original method worked but I had to actually write down the PL function

thanks

@ACuriousMind I wonder if one can try a lebesgue-type method. Partition the range, not the domain.

Meh, somethign to think about

@ACuriousMind Time to play FNV :3

@ACuriousMind Can you explain the Ulysses story?

Idk what that dude's problem is. Something about Bears and Bulls and couriers

@0celo7 His story is part of all four add-ons and probably is only understood through playing all of them. I'm reluctant to say something because it will probably be a spoiler.

Hmm, ok. So playing the hopeville one first was a bad idea?

Ah, that's the last DLC, chronologically. It's the only one where he appears physically, but yes, the other three would have given you more backstory.

whoops

Dunno if I should play more of NV or play F4 again

or be a decent member of society and do some math

Howdy

@ACuriousMind You good with degenerate gases?

@SirCumference Your course seems to be densely packed with degenerate material

@rob No, I'm just trying to figure out whether my understanding is correct or not

The course is going into so many different things

@SirCumference How often do we have to do this dance where I tell you to just ask your question because I'll not commit to answering a question before hearing it? :P

@ACuriousMind All I was asking is whether you're good with them or not...

@SirCumference He's not sure he wants to deal with the pressure

@rob If he's up for dealing with pressure, then degeneracy is certainly relevant

@ACuriousMind So...are you?

@rob You sure do like your puns, hm? ;)

@SirCumference No.

@ACuriousMind Oh...well can I ask you a question about it?

@SirCumference Just ask the question.

@ACuriousMind I think that buns are the best thing since sliced bread

Wait ... typo

All right, just tell me if my understanding in why density leads to degeneracy in massive objects is correct

Hehe

Is that correct?

hello

@heather At first glance I thought that was ACM saying "hell no" XD

@SirCumference I'm not sure I understand what you're saying. If by "reach their zero point energy" you mean that the atoms will be in their ground states, then I don't understand what that has to do with lowering the gravitational potential energy, and I also don't understand your argument for why the atoms cannot get closed to the center of mass

@SirCumference, lol, no

@ACuriousMind Okay, as they get closer to the center of mass their GPE lowers, right?

Yes.

If the GPE keeps lowering, will the total energy of the atoms eventually put them in their ground states?

@SirCumference This is kind of a subtle question.

I don't think the gravitational energy has anything to do with the "state" of the atoms. When we talk about the ground state, we mean that all the electrons in the atom are in the lowest possible orbitals

We get ground states in atoms because electrons are bound and the energy levels are discrete. One of the states has to be "the lowest".

So losing gravitational potential energy won't affect electrons in higher orbitals?

Inside of a star the temperature is high enough that the electrons aren't bound to the nuclei, so they're "free" in that sense.

The gravitational potential energy of the atom as a whole has, to my mind, nothing to do with whether it is excited or not, i.e. where its electrons are

The electrons are gravitationally bound to the star, but the star is so large that those states are essentially continuous.

@ACuriousMind Then take the GPE of the electron

Shouldn't that affect its energy, and as a result, its orbital?

That also hasn't got anything to do with the state the atom is in.

But in a white dwarf for example, electrons aren't even bound to the atom

They still occupy their lowest energy configurations

Ah, yes, but then you're in a situation where you should model the star as a degenerate fermion (electron) gas, not as a bunch of atoms!

There comes a point in the collapse where the atoms break apart

@ACuriousMind Then these electrons can still reach their zero point energies, right?

(meaning lowest energy configs)

They will fill up all the lowest energy states, yes (only one will be in the lowest state due to Pauli exclusion, though)

@ACuriousMind Technically two, both of opposite spins

But shouldn't they be unable to get closer to the center of mass, since doing so would require losing GPE?

Eh, depends on whether there's a magnetic field or not, and is also not really relevant

@ACuriousMind In a white dwarf there will be :P

@SirCumference Don't get too hung up on gravitational potential energy. Near the center of the star, the shell theorem tells you that the gravitational potential is more or less unitform. So the relevant parameter is the pressure from the outer layers of the star.

@SirCumference In equilibrium, any given electron will be unable to get closer the center of mass because all the energy states up to its energy will be filled up.

@ACuriousMind So they won't get closer since the GPE would have to lower, right?

I'm not sure why you keep mentioning "losing GPE", what matters is just that in the degenerate gas all the states are filled so no fermion can move unless it gets more energy from somewhere

@ACuriousMind Huh? Why can't they move?

@SirCumference With "move" I mean "go to another state"

@ACuriousMind Why would them being unable to change states cause them to release a pressure?

@SirCumference Because they therefore resist compression unless enough energy is applied to actually lift one of them into an unoccupied state of higher energy.

@MetaEd

quick comment on this one

> Cast a closevote, if the post is a question (3 000 reputation needed). Because there is no predefined close reason for plagiarism, use the close reason “Off-Topic > Other” and include a respectful comment (see above).

You can compute the degeneracy pressure from the energy density, but I don't remember the details, for a formula you'll have to consult a textbook

@ACuriousMind All right, I'm lost. Why would they resist compression because they are occupying low energy states?

@ACuriousMind This is a thought-provoking interpretation: a degenerate gas as a band-gap insulator against mechanical motion.

keep in mind that <3k users can flag questions for closure

@SirCumference Here's a totally orthogonal way to think about it that might be useful.

The electrons are rattling around in the gas with some temperature and some typical velocity distribution.

Associated with each electron's velocity is a de Broglie wavelength, $\lambda = h/p$.

Also associated with the gas density and temperature is some mean free path between collisions for each electron.

As the density of the gas increases, the mean free path gets shorter and shorter.

Eventually the mean free path becomes comparable to the electron's wavelength.

@SirCumference Well, because they can't compress. They cannot move inward because there are no free states of their energy, so any force acting on them has to supply the energy needed to lift them into the next free state in order to achieve anything.

That's where you get stuck: the wavelength represents a minimum uncertainty in the location of the electron. So it's continuously being repelled by its neighbors.

This transition from intermittent collisions to wavefunction overlap is the transition from a normal gas to a degenerate gas.

@ACuriousMind Sigh...why would them moving inwards require there to be free energy states?

@ACuriousMind Do you know what the "section property" of a product $\sigma$-algebra is?

@SirCumference How could they "move" if there's no state for them to be in?

@ACuriousMind You basically reversed the question I asked.

That's called the Socratic method my boy

The Socratic method is kind of cool, actually. Figuring it out yourself but with guidance via the questioning.

@SirCumference Well, I'm not really understanding what the problem is. Let's take a toy model: I have a two boxes with certain energy levels in them, let's called them $\lvert n \rangle_i$ where $i$ denotes the box and $n$ the $n$-th energy level in the box. Suppose I have 1 fermion in $\lvert 1\rangle_2$ and one fermion in $\lvert 1\rangle_1$.

@heather It can easily be condescending though

Now, if I wish to move the fermion in the second box to the first box, then to be in the first box, it needs to occupy one of the $\lvert n\rangle_1$.

But the only state at or below its energy, $\lvert 1 \rangle_1$, is already occupied.

@0celo7, true. I guess how well it goes is dependent on the questioner.

So it must at least occupy $\vert 2 \rangle_1$ to actually be in the first box

So to move it there, I must at least supply the energy difference between the first and second state.

@ACuriousMind So how does this relate to degenerate stars?

The same thing happens in a degenerate gas, just that we have densities of states instead of discrete boxes.

But you may think of the star infinitesimally being made up of such boxes

@ACuriousMind What do you mean by "densities" here

wtf is transfinite induction

@heather do you want to do some analysis?

@0celo7, yeah, sure!

@SirCumference The way one formally gets the pressure is that one computes how many available states there are per energy and unit volume, then determines the total energy from that and the particle density. Just like ordinary internal thermal energy of a gas, this energy then exerts a pressure as per $\mathrm{d} E = P\mathrm{d}V$.

@heather Were we discussing $\sup(0,1)$?

@ACuriousMind I like how this functional analysis book is very careful to not use AC :)

hello

this is a very weak attempt of mine to try and not waste even more time on the SE network:P

@0celo7, indeed