The h Bar

0celo7

18:04

@yuggib My algebra book proves the first statement.

yuggib

@0celo7 the converse should not be so difficult

but I never thought about it

0celo7

@yuggib Oh I misread. It proves induction given the WOP.

So really it proves the second statement.

yuggib

ok

induction $\Rightarrow$ WOP I don't know if it's easy

0celo7

@yuggib Yeah, remember that theorem I cited about induction? The proof is pretty much "use WOP and this other lemma."

Converse...no clue.

yuggib

set theory goes definitely from WOP to induction

0celo7

18:11

I wouldn't be able to come up with $\Rightarrow$ on my own, much less $\Leftarrow$.

Er.

Other way around.

privetDruzia

Hello @0celo7

0celo7

@privetDruzia Hello.

privetDruzia

Hello @Danu, @ACuriousMind

I have a short question

which is quite easy

if u don't mind?

Danu

Just ask away---I won't guarantee that I'll help you though!

0celo7

Just ask it, someone will answer if they can/want.

privetDruzia

18:19

well my question is in this comment

Please provide a reference (link) for your graphs. I think they're about inter-nuclear energy in a molecule. — Gert 3 hours ago

wait I can't post a link, the title of the post is: "Why is the potential energy minimal when the repulsion and attraction force between molecules is 0?"

the 3 comments below the omly answer

Danu

Have you seen this formula before? $F=-\nabla V$?

privetDruzia

never

I know what it means

0celo7

never? you haven't taken a calculus based physics class?

privetDruzia

but I can't understand that this is correct. Applying the nabla operator on a scalar field of potential energy points

Danu

@privetDruzia Any idea why it's true?

privetDruzia

18:24

returns a vector, which appearantly in my case might be the intermolecular force

Danu

@privetDruzia oh

privetDruzia

but yes thatts the problem, I can't see why this is true

Danu

Think physically

0celo7

that's the worst way to think

Danu

If you are pushing an object and it takes you energy (don't think about friction for now)

You are essentially putting energy into it

0celo7

18:26

such intuition

Danu

What kind of energy? Well, that depends on the type of thing that's causing it to cost you energy.

As an example, think about lifting an object on Earth.

It takes some force: But why?

FenderLesPaul

do you even lift bro?

Danu

You're increasing its potential energy

privetDruzia

mhm

Danu

And the amount of force it takes is exactly the rate at which you're increasing its (potential) energy

(here I'm always considering "infinitely slow processes" so that the kinetic energy plays no role)

In equations, that's just $F=\frac{\mathrm d V}{\mathrm d h}$ where $h$ is the height

In this one-dimensional problem, this is of course equal to the gradient.

privetDruzia

18:28

OK but, so if you are lifting it very high, the potential is high and the force is high

Danu

No

privetDruzia

so no derivative/gradient so far

Danu

Be careful

If you're lifting it here or on Mount Everest, the force should be the same (ignoring variation of the gravitational potential, for now, since mountains are small compared to the radius of Earth)

Don't you agree?

privetDruzia

yes

I do

Danu

Okay, so the amount of force doesn't really depend on the height

privetDruzia

18:31

so the faster you want to lift something the more force

doesn't seem as a derivative yet I think

Danu

Let me think for a little bit on how to put this more clearly.

0celo7

@Danu $F=mg$. I don't see where it depends on "how fast"

Danu

@0celo7 I know :D I messed up because I wanted to ignore kinetic energy, but then didn't.

privetDruzia

:3

0celo7

@Danu Let me take a shot?

privetDruzia

18:32

plz

yes

Danu

In any case, we've established that adding an arbitrary constant to the potential does not affect the force.

privetDruzia

while he s thinking

Danu

That's already a good start, since that's something derivatives also do.

0celo7

@privetDruzia Are you willing to accept that the total energy (kinetic + potential) is constant in time?

Danu

Yeah, I guess you can do it that way ^

privetDruzia

18:34

if its constant the derivative is 0

...hmm ok why not let s try

Danu

Of course you can just think about particular examples

0celo7

Yes. Are you also willing to accept (1) $F=ma$ and (2) $E(x,v)=\frac{1}{2}mv^2+V(x)$?

Danu

e.g. a free particle which is not affected by any potnetial has constant velocity (i.e. $F=0$)

0celo7

The latter being the constant energy.

DanielSank

@BernardMeurer Hello again.

0celo7

18:35

@privetDruzia Then $dE/dt=mva+V'(x)v=0$.

This implies $V'(x)=-ma=-F$

For more than one dimension it works similarly.

privetDruzia

If a force acting on an object is a function of position only, it is said to be a conservative force, and it can be represented by a potential energy function which for a one-dimensional case satisfies the derivative condition

?

0celo7

Yes.

Danu

@privetDruzia This is basically what Ocelot just showed you

$V(x)$ is an assumption

privetDruzia

mhm

but writing it as: $F = \grad{U}$ or $dU/dt = F$

what s the difference

0celo7

Uh, $dU/dt$ is not $F$

$dU/dt=\nabla U\cdot v$

chain rule

privetDruzia

18:39

times V?

0celo7

Dot product with $v=dx/dt$.

privetDruzia

what is v? The potential energy? Isn;'t that U?

0celo7

Small v is velocity.

Potential energy is capital U.

privetDruzia

ow ok, but does velocity in my case really matter?

i.ie intermolecular forces

0celo7

Intermolecular forces are radial, right? So I think they're conservative.

privetDruzia

18:42

ok so v doens't matter

0celo7

It does matter in the derivation.

privetDruzia

(if they are indeed radial)

0celo7

Where are you concerned about them mattering?

@Danu I'm assuming it's kosher to take diagrams from a textbook to put in a PSE post (specifically Lee Smooth) if I cite them?

privetDruzia

yes but

what s the difference between writing $-dU/dx = F(x)$

Bernard Meurer

I'm back, hey @DanielSank

privetDruzia

18:45

so in one dimension

?

Danu

It's with a minus

privetDruzia

typo

Danu

but the difference is non-existent in one dimension

0celo7

# of dimensions

Danu

In arbitrarily many dimensions, $F(x)-\nabla U$ is correct

privetDruzia

18:47

sO IN one dimension I can write both?

ow OK

Danu

yeah,

privetDruzia

so conclusion: there is no difference?

0celo7

The difference is basic calculus

privetDruzia

ok I ll keep it siimple

and not write the nabla operator

0celo7

Actually

privetDruzia

18:48

just $-dU/dx = F(x)$

0celo7

The difference is really the difference between abstract tensor notation and component notation

privetDruzia

so the derivative of the position with respect to the position = a force vector(which varies dependent on x)

0celo7

@DanielSank Do you have access to Wald's General Relativity (1984)?

privetDruzia

One really last question

imgur.com/46rcbV7

on the scheme you can see that in the beginning

Ep is very high/maximal, but the derivative of that curve, i.e. the one below,

isn't approaching the value 0 at all for that point

Danu

@KyleKanos Wanna play (correspondence) chess sometime on chess.com? ;)

privetDruzia

18:54

@Danu, an idea for that last question please?

I ll leave you playing chess in peace after that

Danu

@privetDruzia $E_p$ does not reach a maximum there

It "blows up"

privetDruzia

"blows up" = goes to infinity?

Danu

Yes

privetDruzia

and the derivative of infinity

Danu

This reflects the fact that you cannot force these particles to occupy the same point in space

@privetDruzia A derivative if a number does not exist.

privetDruzia

18:55

which is a max is not zero...

Danu

But the derivative of a function that steeply approaches infinity does, perhaps ;)

privetDruzia

ow ok

Danu

@privetDruzia No, $\infty$ is not a maximum

You cannot work with $\infty$ like with other numbers

0celo7

@privetDruzia what does "ow" mean

Danu

Oh

privetDruzia

18:56

ok but if you would calculate the limit... that shoudl approach zero

Danu

No

It keeps on getting steeper and steeper as it approaches $r=0$

0celo7

@Danu What about the diagrams?

Danu

Probably technically not.

Probably not a big deal though.

privetDruzia

ok so, infinity is a special case, I ll just have to live with the fact that for some reason the derivative (the lower curve) of the Ep which approaches infinity is not equal to 0

Danu

You could ask Lee personally

@privetDruzia This should not be something hard to accept for you. This should be very clear and natural

If you're not really at ease with it, you should try to work on that.

(not just "sigh, accept it")

privetDruzia

18:59

yes I am working very hard on it

every day

actually

just learned last week what a gradient was

6 months ago I didn t even know what an integral was...

*almost every day

can we try once more, one clear answer...
Why isn t it equal to zero?

( I reread your answers above

Danu

It's a function that goes steeper and steeper as it approaches $r=0$

but it never "caps off" to a maximum

It just keeps on growing, in unbounded fashion

It never reaches a maximum

I understand that infinity might not be an easy concept to deal with at first

It always keeps on getting larger and large but you can always go an $\epsilon>0$ closer to $r=0$ and achieve even higher values

privetDruzia

I understood, as the potential energy is extremely high, you cannot put those two particles at the same spot (meaning F=0)_

Danu

In this particular model, the potential energy keeps on growing in unbounded fashion

privetDruzia

ok

that s clear so far

0celo7

@Danu Will he know what PSE is?

privetDruzia

19:03

how can you now relate that to the intermolecular force?

Danu

@0celo7 He's very active on Mathematics. Didn't you know?

0celo7

@Danu Oh right :P

Danu

@privetDruzia This particular model just says: As you try to put the molecules closer and closer, they repulse each other stronger and stronger (in unbounded fashion, again).

privetDruzia

ok

Danu

It will cost you infinite force (which does not exist, so this cannot happen) to put them at exactly the same spot

Because in the limit $r\to 0$ the $E_p(r)$ function is infinitely steep

FenderLesPaul

19:05

the infinity force

privetDruzia

E_p(r) is inifinitely steep so F-curve is infinitely steep?

Due to the potential energy you have those forces which occur

DanielSank

@0celo7 Nah.

0celo7

@DanielSank Ok, Lee Introduction to Smooth Manifolds (2013)?

FenderLesPaul

Beast of Burden has to be the best Rolling Stones song

Danu

@privetDruzia No, $E_p$ is infinitely steep so $F$ is infinitely large. Remember: $F=-\nabla V$

If $E_p$ being infinitely steep would imply $F$ being infinitely steep we would have something like $\nabla F\sim -\nabla V$

but that's not true

Note that in all of this, I'm using the word "steepness" as a placeholder for "value of its derivative"

DanielSank

19:15

@0celo7 Nah.

privetDruzia

That clarified what I needed!

DanielSank

Why?

privetDruzia

@Danu, that clarified what I needed

0celo7

@DanielSank You don't have Springer access o.o

privetDruzia

So this statement would be true when trying to explain this: "the force is a result of Ep"?

DanielSank

19:18

@0celo7 Oh do I have access. Yeah. I don't have 'em in the house.

0celo7

@DanielSank Ok, I wanted you to take a look at some diagrams in it. I'll just post them here.

Danu

Locally Euclidean hur dur

Transition functions hur dur

0celo7

@Danu ?

Danu

@0celo7 Just looking at your pictures

0celo7

@Danu "hur dur"?

I'm trying to explain this stuff in very basic terms.

DanielSank

19:25

@0celo7 Yeah those are great.

0celo7

^ See, @Danu.

Danu

@0celo7 Just felt like saying that :)

It was not a judgement

DanielSank

When we discussed this topic yesterday, by criticism was mostly that I think it's better to discuss tensors in the component/coordinate free picture first.

Danu

@DanielSank I thought you were mister "I dun' need no manifolds except $\Bbb R^n$"? ;)

0celo7

@DanielSank Aight, I'll shoot Dr. Lee an email...aw fuck what if it goes to spam :(

DanielSank

19:26

@Danu Lol wut?

@0celo7 :(

0celo7

@DanielSank I bet the IT department is closed today, too.

dmckee

@0celo7 Write it from your university account.

Do explain who you are and why you are writing.

0celo7

@dmckee That's the one I'm talking about.

dmckee

Do use complete sentence and no smileys.

Do be polite and respectful.

0celo7

@DanielSank's work email sent my email straight to spam.

DanielSank

19:27

@dmckee Hahaha, no, see... his IT department slouched and now his emails go to spam.

dmckee

Don't include anything about nigerian banks or penis enlargement.

2

DanielSank

o_O

0celo7

@dmckee I've emailed plenty of profs before, including t' Hooft. I know how it works.

Danu

@dmckee But what if I'm just a poor lonesome Nigerian viagra salesman?

0celo7

@dmckee Lee sounds Asian, he might be interested in the latter.

Danu

19:28

@0celo7 't Hooft

dmckee

@0celo7 OK. Then forget I said anything. Except the bit that got stared, of course.

Danu

@0celo7 He's not Asian

0celo7

@Danu Seriously?

I know I checked that when writing the email.

@Danu How do you know?

dmckee

@0celo7 Er ... Could be like, I don't know, related to General Lee...

user54412

@FenderLesPaul better than Can't Always Get What You Want?

0celo7

19:29

@dmckee Perhaps.

Was General Lee Asian?

dmckee

^ Probably not.

Danu

Feb 15 '15 at 21:16, by Danu

I am unconvinced that it works that easily on manifolds, @DanielSank

Took me a while to dig that up

dmckee

At the time he'd have had trouble getting into West Point, doncha know?

Danu

@0celo7 Yup.

It's short for Het Hooft

DanielSank

@Danu eh? I stand by that.

Danu

19:31

which is derived from Het Hoofd (the head)

0celo7

@dmckee Can I send your professional email a test? I need to see if the thing with @DanielSank was a one-off thing.

Danu

@DanielSank Sure you do :)

0celo7

@Danu I know.

DanielSank

@0celo7 It will work now because I responded to your mail.

dmckee

@0celo7 Sure. I assume you've found it already.

0celo7

19:31

I should have known that, I'm sorry @Danu

@dmckee No, but I can find it.

DanielSank

It might actually work elsewhere too now because of that.

0celo7

It's insane that a .edu account would go to spam.

dmckee

@0celo7 Universities have been infested by botnets from time to time. No one is immune.

0celo7

There are Oak Ridge researchers that have utk.edu accounts.

@dmckee It's very imaginatively titled "Test"

dmckee

@0celo7 I'd never have thought of that.

@0celo7 Came through. No trouble.

0celo7

19:37

@dmckee Ok, thanks.

DanielSank

@dmckee What domain is your mail on?

0celo7

mssu.edu

dmckee

@DanielSank My school's .edu

Yeah. What he said.

DanielSank

I see.

dmckee

My rather dingy ivory (or rather brick) tower is called "Reynolds Hall", which we (chemical and physical sciences) share with math and bio so it's a lot of fun.

Alas, we have no moat. Much less sharks with laser beams.

0celo7

19:40

(wat)

dmckee

On the other hand, after a couple of years in Lego Land (repurposed FEMA trailers) we've getting new labs.

DanielSank

@dmckee Shiny.

dmckee

@DanielSank On the architect's drawings they are very Shiny. CSI-like even. But I'll settle for larger, better laid out and equipped.

DanielSank

@dmckee Having recently done a lab build-out I can offer one gem that I think will really help:

dmckee

Please.

DanielSank

19:44

Check this out

See how all the cables are overhead? There's nothing on the floor.

Danu

Wuw

DanielSank

We did this in our new lab and OH MY GOODNESS it's nice.

@Danu That's not my lab, just illustrating over-head cable trays.

Danu

Still a very nice picture.

And I agree: Vacuuming between my cables is shit!

dmckee

@DanielSank I've had occasion to work in both kinds of space. Overhead is so worth it.

DanielSank

Yeah.

If you can get the architects to install the trays it's so awesome.

dmckee

19:46

But our "cluster" is four quad-core computers in one of the mathematicians offices. Our computational nano-scale person is trying to rustle up some money for a little more horsepower or at least access to a grid.

We're actually getting overhead power fixtures for the teaching lab (instead of sticking up through the tables). I've never tried that, but a colleague tells me it is good.

Danu

@dmckee That's sad :P

0celo7

@dmckee We have that in our newer buildings.

dmckee

@Danu I think he said he was getting two more boxes this year and that would about double the capacity of it. He's a crypto guy, so he need some horsepower and memory and disk.

0celo7

Also we had them in high school.

dmckee

@0celo7 The labs I've been working/teaching in date around my fifth or sixth birthday.

0celo7

19:51

@dmckee Wow they had power fixtures back then?

Danu

They had electricity back then?

dmckee

As long as you could keep the mastodons in the treadmill...

0celo7

@Danu Lightning has been around for a while.

dmckee

There were a lot of blackouts during mating season.

4

Danu

@0celo7 Says who? ;)

@dmckee Maybe they should've gone for a thrust-based machine... Seems like animals generally don't get enough of that! Except panda's.

dmckee

19:54

The star wall is starting to make be look unhealthily obsessed.

Danu

@dmckee Hahaha!

0celo7

@Danu How are open sets of the $n$-dim half space defined?

Subspace topology inherited from $\mathbb{R}^n$?

Danu

Of course

...unless you have a different topology, obviously

0celo7

Just checking.

@Danu Hmm, what should the subject line for the email (to Lee) be?

Danu

20:15

Don't know, don't care, honestly.

FenderLesPaul

20:35

@ChrisWhite yes

possibly tied with "Can't You Hear Me Knockin'"

0celo7

Dr. Lee is on a sabbatical. I wonder how long I will be waiting for a response :/

FenderLesPaul

You won't get a response ever

just give up now

Danu

Just discard all your hard work

0celo7

:(

Without the diagrams it's all useless

Danu

Or better yet, learn how to make TikZ pictures and stop stealing pictures like a plebeian

0celo7

20:44

It'll be some Bourbaki style thing that no one can read except for German functional analysts

FenderLesPaul

is Lee still at U Wash?

0celo7

yah

Danu

@FenderLesPaul No he's at I Wash now

[apologies for that terrible... thing]

0celo7

@Danu Teach me, o wise fibration

Danu

TeX - LaTeX is your friend

I learned it too ;)

0celo7

20:46

I thought you were my friend :'(

FenderLesPaul

wise fibration sounds like a dope indie album name

thanks for the idea

0celo7

no response yet :(

privetDruzia

Ideal gas law,

An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces.

0celo7

yes

privetDruzia

why are those conditions so important for the ideal gas law?

0celo7

20:47

Do you want the derivation

Warning: it's long

privetDruzia

euhm lets start with a pragmatic explanation

that might be easier :)

0celo7

I wrote down the proof somewhere once

FenderLesPaul

not sure what kind of answer you're looking for

privetDruzia

for fun?

0celo7

I wanted to show it to my Chem teacher in high school

FenderLesPaul

20:48

the ideal gas law derivation is not valid if you assume intermolecular forces

0celo7

She doesn't know calculus so it was a waste

privetDruzia

lol

0celo7

There's also a derivation using special relativity

Not sure if it's a derivation or a "derivation"

privetDruzia

An explanation about the principles/reasons will be fine enough

FenderLesPaul

there's also a derivation using loop quantum gravity

0celo7

20:50

Link?

FenderLesPaul

@0celo7 looks like it was erased from existence

how unfortunate

privetDruzia

www.funWithQuantumLoops.com

0celo7

You basically look at the trace of the energy momentum tensor averaged over some volume

FenderLesPaul

@privetDruzia your question can't be answered without talking about at least some aspect of the derivation

privetDruzia

is it long?

Ok you know what? let's go!

(don't make it too long plz)

0celo7

20:52

you need the viral theorem

privetDruzia

i already have so many proofs to study...

0celo7

or the equiparition theorem

FenderLesPaul

@privetDruzia you can just google it

0celo7

then you need the Maxwell distribution

FenderLesPaul

google "ideal gas law kinetic theory"

privetDruzia

20:53

is this the easiest way?

FenderLesPaul

and you'll see why you need intermolecular forces to be negligible and elastic collisions

privetDruzia

ok

0celo7

you need to calculate the variance of the velocity distribution

I think Stokes theorem comes in as well

FenderLesPaul

stokes theorem always comes in

they should just rename physics to applications of stokes theorem

and taylor series

0celo7

eww there are integrals on the cotangent bundle

eww eww

FenderLesPaul

20:56

like ew

?

0celo7

yah

like...just ew

FenderLesPaul

ermagherd ew

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