The h Bar

Bernard Meurer

00:11

Phew, interview done

skull petrol

How did it go?

Bernard Meurer

Pretty damn good to be honest

The interviewer really knew the stuff, so it was very easy to discuss

skull petrol

How long was it?

Bernard Meurer

An hour and 10 minutes; standard for UPenn

skull petrol

That is standard?

Bernard Meurer

00:26

Yeah they say it will last about an hour

skull petrol

Ah, I see.

When will they let you know their decision?

Bernard Meurer

April-ish

end or march-ish

skull petrol

Good luck and I hope for all the best.

Bernard Meurer

Thanks man!

0celo7

00:54

What the heck do they talk for an hour about

Bernard Meurer

Tech in general mostly

what do you do extracurricularly that stuff

as I think @FenderLesPaul said, they just want to make sure you're not nuts

or that you're nuts in the right way at least

skull petrol

01:13

Interesting: There's roughly 4 times as many Lol's in the Math chatroom as there are in the Physics chatroom

Bernard Meurer

We actually know how to laugh

skull petrol

:D

0celo7

@Secret what do you know about Sturm-Liouville eigenvalue problems

heat equation with convection is killing me

I have no clue how anything we learned in class applies

0celo7

02:24

@dmckee Is it true that if $V_0$ in the convective heat eq. $u_t=ku_{xx}-V_0u_x$ is too large, the equation cannot be solved? (The Dirichlet problem, that is.)

@dmckee Maybe this makes sense? If you have too much convection, the heat flow gets messed up?

@dmckee Oh, wait, the first few modes will be pure exponential but eventually one gets some oscillatory stuff?

But that doesn't make sense -- the eigenvalue problem won't have a solution.

0celo7

02:44

@dmckee Ignore ^, I figured it out.

Jon Snow

Very Large Telescope

The Very Large Telescope (VLT) is a telescope operated by the European Southern Observatory on Cerro Paranal in the Atacama Desert of northern Chile. The VLT consists of four individual telescopes, each with a primary mirror 8.2 m across, which are generally used separately but can be used together to achieve very high angular resolution. The four separate optical telescopes are known as Antu, Kueyen, Melipal and Yepun, which are all words for astronomical objects in the Mapuche language. The telescopes form an array which is complemented by four movable Auxiliary Telescopes (ATs) of 1.8 m aperture...

I just found out there's a big telescope called "Very Large Telescope"
Goes to show how creative physicists are...

0celo7

@FenderLesPaul I figured it out!

Jack

03:37

Can someone please tell me youtube.com/watch?v=CZgqGTxL9cA&t=9m49s why in this simple circuit with one resistor, the voltage across the resistor is just the supplied voltage (12V)??? I thought resistor reduces the voltage. So there must be a voltage drop, given that the battery that supplies the voltage goes from positive to negative end across the circuit. So the voltage is at positive when it enters the resistor and is negative when it leaves... right?

you don't need to watch, just pause it there to see the circuit

dmckee

03:50

Recall that potential (voltage is properly called "electrical potential") like potential energy has an arbitrary zero: only differences matter. So 0 -- 12 is the same as -6 -- 6. And the difference between the two side is the voltage drop; it might be easier to think about Kirchoff's voltage rule: the total voltage changes around a circuit must be zero because each point in the circuit has to have a unique voltage.

(...but beware what large scale inductance does to the rule...)

0celo7

Oh, if anyone cares, the solution for the above PDE w/ Dirichlet BCs and initial temperature $u(x,0)=f(x)$, $0<x<L$ is $$u(x,t)=\sum_{n\ge1}A_n\exp\left[\frac{V_0}{2k}x-\left(\frac{V^2_0}{4k}+k\frac{‌n^2\pi^2}{L^2}\right)t\right]\sin\frac{n\pi x}{L}$$ $$A_n=\frac{2}{L}\int_0^Lf(x)\mathrm{e}^{-V_0x/2k}\sin\frac{n\pi x}{L}\,\mathrm{d}x$$

Interestingly, the thing has an overall exponential growth factor.

@dmckee Is there a physical reason for this?

dmckee

@JonSnow The experiment currently called DUNE (Deep Underground Neutrino Experiment) was for years called by the placeholder name LBNE (long Baseline Neutrino Experiment), but that was considered to plan and pedestrian.

@0celo7 Sure. The electrical potential is actually the zero component of a Lorentz 4-potential, so of course it isn't immune to boosts.

0celo7

@dmckee Uhhhh

What does that have to do with heat flow in a rod?

dmckee

Not a clue. You question linked to my line about inductances.

0celo7

@dmckee I didn't hit reply.

dmckee

03:56

I don't know the first thing about poorly behaved solutions to the heat equation.

0celo7

No arrow.

@dmckee It's not poorly behaved.

But there might be some interesting physics.

dmckee

Oh. I didn't have ChatJax running. But I still don't know. I haven't done those boundary value problems since grad school. Do you get that from a Green's Function approach?

0celo7

@dmckee We derived that form of the heat equation by supposing that the molecules move at some average velocity

@dmckee No, separation of variables.

It was pretty hellish, but it works out.

@dmckee But notice that one can pull out a total factor of $\exp \frac{V_0}{2k}x$

dmckee

Yeah. Well, it has most of the right features.

0celo7

Why does the heat grow exponentially as you go along the rod

The factors $\sim\mathrm{e}^{-\lambda t}$ make sense.

(heat flows out of the ends)

dmckee

04:02

I don't actually recall what the boundary conditions are, but there is a factor of $\exp -\frac{V_o}{2k}x$ in the integral for the $A_n$, and while you can't pull it out it's effect is going to hang around.

0celo7

@dmckee The BCs are Dirichlet, $u(0,t)=u(L,t)=0$.

@dmckee Maybe I should ask an engineer

Or PSE?

dmckee

Well, I have some kind to vague confidence that I could still do that math, but it would take me quite some time.

0celo7

@dmckee Do what math?

dmckee

Your best bet is to ask someone else who has been looking at it in the last few years.

0celo7

That solution is correct.

I want to know what the physics behind it is

dmckee

04:06

I believe you, but I'll understand how the terms enter better if I have done it myself recently enough to remember.

I would be asking myself "Where did that thing first show up" in search of a hint as to what it was about.

0celo7

Right...

Let's see

dmckee

The most challenging thing I've done lately myself was re-proving the shell theorem and then trying to get some juniors to understand the consequences.

0celo7

I'm stuck on a topology proof right now

dmckee

But that at least means giving them the "Drill a hole pole-to-pole through a uniform Earth and drop something in it" problem. Which is always fun.

0celo7

Distracting myself with overanalzying my PDE homework

@dmckee Ok, so we write $u(x,t)=X(x)T(t)$ as usual

Anthony

04:09

Hello!

0celo7

then the $X$ equation turns out to be $kX''-V_0X'+\lambda X=0$.

dmckee

Gotta love separation of variable. It's almost algorithmic.

Anthony

Does anyone know how to change the grand potential for a single-component homogenous fluid?

0celo7

@dmckee now, for standard heat flow, $V_0=0$ and you get the good old $X''-\lambda X=0$

which has solutions $\sin \frac{n\pi x}{L}$ w/ BCs $X(0)=X(L)=0$

Anthony

I don't really understand what is supposed to change. We have $\Psi = -pdV - SdT - Nd\mu$, what information do I have that lets me change this? (Apparently it will be only in terms of T,V,P)

0celo7

04:11

@dmckee But for the modified equation you plug in the ansatz $X=\mathrm{e}^{rx}$

this gives $kr^2-V_0r+\lambda=0$

dmckee

HEre's the thing. It's after 10 here and I have to be back at work in less than 12 hours. I'm not going to watch you solve PDEs.

0celo7

Wolfram tells me this implies $r=\frac{-V_0\pm\sqrt{V_0^2-4k\lambda}}{2k}$

:(

I'm showing you were that term comes from

DanielSank

Guys how do I deal with a Hamiltonian like this:

$(p-p_0)^2 + V(x)$

0celo7

@DanielSank ewww

DanielSank

and $V(x)$ is periodic.

0celo7

04:14

@DanielSank help me and I'll help you

DanielSank

The solutions are Bloch waves. How does the momentum offset affect things.

@0celo7 What do you need?

0celo7

@DanielSank The solution for heat in a rod w/ convection has a funny exponential growth factor

I'm wondering where it comes from

DanielSank

@0celo7 I have no idea. Is there some analysis I can look at?

0celo7

24 mins ago, by 0celo7

Oh, if anyone cares, the solution for the above PDE w/ Dirichlet BCs and initial temperature $u(x,0)=f(x)$, $0<x<L$ is $$u(x,t)=\sum_{n\ge1}A_n\exp\left[\frac{V_0}{2k}x-\left(\frac{V^2_0}{4k}+k\frac{‌n^2\pi^2}{L^2}\right)t\right]\sin\frac{n\pi x}{L}$$ $$A_n=\frac{2}{L}\int_0^Lf(x)\mathrm{e}^{-V_0x/2k}\sin\frac{n\pi x}{L}\,\mathrm{d}x$$

And posts thereafter.

The equation is $u_t=ku_{xx}-V_0u_x$

I think $V_0$ is supposed to be the average velocity of molecules in the rod, modulo constants of the material.

DanielSank

I could probably solve my own problem faster than I can help you with this. Sorry, bro.

Actually wait... this is a linear equation.

0celo7

04:19

@DanielSank Indeed it is.

DanielSank

@0celo7 so what's the question?

0celo7

@DanielSank See that factor $\exp \frac{V_0}{2k}x$?

You can pull it out of the sum.

DanielSank

sure

0celo7

Is there any physical reason why we have some exponential growth as we go down the rod?

DanielSank

What's the physics?

This is a rod heated on one end?

0celo7

04:21

Of the PDE?

@DanielSank No, Dirichlet.

Reservoir on both ends.

DanielSank

Each end is held at a certain temperature?

0celo7

Or heat sink or whatever the hell you call it

@DanielSank 0.

DanielSank

@0celo7 What does "0" mean?

0celo7

@DanielSank The unique whole number less than one.

DanielSank

@0celo7 Don't be a pain in the ass :)

0celo7

04:23

It's a fixed temp, yes.

DanielSank

@0celo7 Different temperature on each end?

0celo7

No, $u(0,t)=u(L,t)(=0)$.

Since the eq. is linear, making them a different constant temp. would just amount to adding the temp to the whole thing.

DanielSank

What's $k$?

Oh, it's in the original equation.

0celo7

Thermal diffusivity.

I don't know what that actually means.

I might take a course on statistical mechanics.

DanielSank

Hmmm, each mode is a sinusoid multiplied by an exponential.

0celo7

04:26

@DanielSank The $\exp -t$ terms make sense.

The ends of the rod are not insulated, so the heat can flow out of them.

DanielSank

Yes.

The spatial part is confusing.

Where is the left/right symmetry broken?

0celo7

Huh?

FenderLesPaul

@0celo7 nice!

Bernard Meurer

Cough

FenderLesPaul

@DanielSank will you be in UCSB around the 7th of April?

DanielSank

04:28

:27685541 Another deleted post by you.

All hail the deleted post King!

2

@FenderLesPaul Probably.

Why?

FenderLesPaul

@DanielSank just asking because that's when the visit day is

Bernard Meurer

@DanielSank He doesn't have the guts to call someone else a troll

DanielSank

@FenderLesPaul Oh. Groovy, let's meet up.

I can be the meets-all-the-internet-people King.

FenderLesPaul

@DanielSank yea!

0celo7

@FenderLesPaul Turms out the trick was to pull an $\mathrm{i}$ out of a square root

FenderLesPaul

04:29

I can meet you then I can go meet Gary Horowitz and Don Marolf

Bernard Meurer

@DanielSank You already are

Also, Groovy?

0celo7

You have to force the exponential to be complex to solve the boundary conditions, prof really should have warned us about that

FenderLesPaul

@DanielSank did you ever have interactions with Steve Giddings?

Bernard Meurer

Are we in woodstock or in an austin powers movie?

DanielSank

@BernardMeurer Yes, groovy.

@BernardMeurer We're in my braaaaaaaaaain.

Bernard Meurer

04:30

@DanielSank Not sure how to feel about that

0celo7

Jesus people, did you have to derail the chat?

Bernard Meurer

How did you like the cube btw?

DanielSank

@BernardMeurer Your existence is a simulation in my biological computer.

0celo7

I was finally talking about physics and you ruined it

DanielSank

Put that in your pipe and smoke it.

Bernard Meurer

04:31

The hell did you have man hahahaha

DanielSank

@0celo7 I thought you just said something about putting an $i$ in the exponential. Sounded like you solved your problem.

Bernard Meurer

You shouldn't ski

0celo7

@DanielSank What

Knowing how to solve the equation != understanding the physics of the solution

DanielSank

@0celo7 Well telling me that the exponential is complex is a pretty important help, dammit!

0celo7

It's not complex.

What are you talking about?

Bernard Meurer

04:32

Women are complex

DanielSank

@0celo7 Oh ffs. Didn't you just post something about putting an $i$ is a square root somewhere?

0celo7

@DanielSank Yeah, so? It falls out in the end.

DanielSank

@0celo7 I'm really confused now.

0celo7

@DanielSank If $\phi$ is a complex solution to a linear DE, the solution you really want is $A\Re (\phi)+B\Im(\phi)$

DanielSank

@0celo7 Anyway, I don't understand why the modes have that exponential dependence on $x$.

That makes no sense to me.

0celo7

04:36

@DanielSank Ok, that's the thing

The modes don't

The whole thing does!

The whole thing is multiplied by an $x$-exponential

DanielSank

@0celo7 The modes do.

Look at the equation for $A_n$.

0celo7

What

@DanielSank that's a constant

DanielSank

$n$ enumerates the modes.

The amplitude of each mode is given by the overlap integral you see in the equation for $A_n$.

0celo7

each $A_n$ is constant

DanielSank

@0celo7 Yes, but the overlap integral has the $x$ exponential in it.

I don't understand why.

FenderLesPaul

04:38

mmmm insomnia cookies

0celo7

@FenderLesPaul rich

@DanielSank Mathematically?

Or physically?

FenderLesPaul

@0celo7 I'm riding the wave of my parents' money

DanielSank

@0celo7 Both. They're the same to me.

FenderLesPaul

before I become a broke grad student

DanielSank

@FenderLesPaul I saved a ton of $$ as a grad student.

Bernard Meurer

04:39

@DanielSank That's because you had slaves

0celo7

@DanielSank Ok, mathematically, it follows from the fact that you need to use a weighted inner product on the space of functions as this is a Sturm-Liouville eigenvalue problem

DanielSank

@0celo7 That statement contains zero information.

The wave equation is Sturm-Liouville but I don't get real exponentials.

0celo7

@DanielSank the eigenfunctions have a real exponential part because $\sigma$ is a real exponential

put the $X(x)$ equation in SL form

DanielSank

@0celo7 There's no $\sigma$ in anything you posted.

0celo7

SL equation: $(p\phi')'+q\phi+\lambda\sigma\phi=0$

DanielSank

04:41

@0celo7 I don't know SL theory for real.

I'm pretty sure I can't help you.

0celo7

ok, let me teach you

DanielSank

@0celo7 Nah.

0celo7

I don't need help with the math!!!

DanielSank

I've got way too much work on my plate. Sorry, man, I just don't have time.

0celo7

It's the physics that's bothering me

DanielSank

04:42

@0celo7 Fine, well I don't understand it physicsally either.

physicsally is a real word now

0celo7

@DanielSank ok, but let me say one last thing

Bernard Meurer

Physics of sally?

0celo7

If $\{\phi_n\}$ is a set of eigenfunctions satisfying the SL equation, then we have the general Fourier series $$f\sim \sum a_n\phi_n$$ where $$a_n=\frac{\int f\phi_n\sigma}{\int\phi_n^2\sigma}$$

DanielSank

Duh.

0celo7

@DanielSank that's where the exponential comes from, $\sigma$ is an exponential for this problem

DanielSank

04:45

K, I don't get it.

0celo7

Ok, and you won't let me prove it

we seem to be at an impasse

DanielSank

@0celo7 Dude, I just don't have the bandwidth.

Explain it to the chat. Maybe it will help you figure it out.

Programmers call it "rubber ducky" debugging.

Bernard Meurer

I only do hulk debugging

0celo7

What is there to explain?

Maybe I should ask PSE

We need an actual physicist in this chat

Bernard Meurer

Goodnight folks!

DanielSank

04:49

@BernardMeurer 'night

@0celo7 I don't understand why the modes of this problem don't respect the symmetry of the problem.

0celo7

@DanielSank Oh, about $L/2$?

DanielSank

I'm sorry that's not the answer you were looking for, and I'm sorry you choose to throw around dumb-ass insults when people are trying to help you.

0celo7

Interesting, you're right, that doesn't make sense.

DanielSank

@0celo7 Yes. I already said that and you went off on some dumb rant about SL theory.

0celo7

@DanielSank You complained about the exponential.

It has to be there.

I don't know how that makes sense physically

DanielSank

04:52

@0celo7 We need literate users in this chat.

24 mins ago, by DanielSank

Where is the left/right symmetry broken?

since apparently, you cannot read.

Ok, I've restored the ridiculous insult karma of the chat. I'll stop with that now.

0celo7

@DanielSank I blame my stroke

DanielSank

@0celo7 Fair enough.

Why do you use such an assholish type of humor here, btw?

0celo7

@DanielSank What

I didn't know I was using humor, or that it was assholish.

DanielSank

7 mins ago, by 0celo7

We need an actual physicist in this chat

@0celo7 I think I understand what's going on now

You mentioned convection?

0celo7

@DanielSank Oh, stroke

I think I'm only an asshole to you

<3

DanielSank

04:58

</3

0celo7

wow

@DanielSank yes

Oh!

The particles are moving in one direction

DanielSank

@0celo7 Yes.

0celo7

There is no symmetry

DanielSank

@0celo7 I'm still not sure where the asymmetry comes from unless this "rod" is actually a bunch of gas particles and there's gravity.

Is that the case?

0celo7

@DanielSank ummm

no?

DanielSank

05:02

@0celo7 Then I don't know what's going on.

0celo7

I'm not sure how physical this actually is

@DanielSank I don't either

DanielSank

It would be easier to help you if I knew what system you're asking about.

0celo7

@DanielSank My (math) text doesn't give a specific situation.

DanielSank

@0celo7 Then why are you asking for physical motivation?

Go back to the PDE and ask yourself what physics could account for the $\partial_x$ term.

That will solve all your problems.

0celo7

@DanielSank The particles moving with velocity $V_0$

DanielSank

05:06

@0celo7 Nope. That has no direction preference.

0celo7

@DanielSank ok, according to the book

we have a heat flux $\phi$

Fourier's law says $\phi=-K_0u_x$

DanielSank

That sounds right.

0celo7

Ok, now the book wants us to modify the heat flux like this: $\phi=-K_0\partial_xu+c\rho uV$

DanielSank

What are $c$, $\rho$, $u$, and $V$?

0celo7

lol

specific heat, mass density, temperature, velocity

DanielSank

05:11

velocity of what?

Mean particle velocity?

0celo7

@DanielSank yes.

DanielSank

Well ok then that makes sense.

0celo7

@DanielSank so do you see how the symmetry is broken?

DanielSank

I don't.

Try solving the problem yourself.

Separation of variables.

0celo7

what

Is it not clear that I solved the equation?

I am the one who came up with the above solution!

For the last time, the math is clear!!

DanielSank

05:17

@0celo7 I know.

Jeesus, dude.

You manage to be kind of dickish every single time I talk to you.

0celo7

I'm sure it's a sign of affection.

@DanielSank Then why did you tell me to solve it again?

I spent a solid hour trying to solve it earlier.

I finally got it, and am now wondering about the physics.

DanielSank

@0celo7 I thought you got the solution from a book or something.

Perhaps this is incorrect.

0celo7

@DanielSank The solution is in the back.

DanielSank

I don't even understand what physical system the PDE you wrote down represents.

I can't help you because I don't know what system you're trying to understand.

Sorry, man, but you need to help me help you.

You had some equation for heat flow.

Ok, well, the heat energy at a position $x$ changes in time by the divergence of the heat flow, right?

0celo7

@DanielSank yes

DanielSank

05:22

Ok so we get something like:

0celo7

and we also add a term proportional to the velocity of [whatever]

DanielSank

$\partial_t u = -K_0 \partial_{xx}u + c \rho V(x) \partial_x u$,

right?

0celo7

@DanielSank sure

I think you have to integrate the heat flux and use conervation of energy

something like that

DanielSank

@0celo7 Ok, if $V\neq 0$, then we have broken the symmetry and the exponential factor makes some sense.

However, $V(x)$ isn't just some fixed number in a real system.

Or, well, I suppose you could have situations in which it is fixed.

Ok, well this sort of makes sense now.

If the particles are flowing through the pipe with a position independent velocity $V$, then you get the exponential part.

Let's see, does $V$ show up in that exponent?

Why lookey there, it sure does.

user54412

05:40

::reads conversation::

DanielSank

@ChrisWhite Noooooooo! Keep in mind I haven't physicsed in a week.

My brain is no work.

0celo7

@ChrisWhite your thoughts are appreciated

user54412

I'm confused by the fuss

0celo7

That's DS in a nutshell

(now he'll call me an asshole)

@ChrisWhite wait

what fuss are you talking about

user54412

05:55

your fuss

user54412

why the fuss

user54412

so much angst

0celo7

I've never understood what "angst" means

It's a German word.

But I do not have angst in the German sense.

user54412

angst: that feeling you get when you have too many consonants in scrabble, making you frustrated enough to import a foreign word and pretend it's allowed

2

0celo7

@ChrisWhite what angst do I have

@ChrisWhite what is my angst???

hmm, three ?s seems a bit much

user54412

06:00

I'm still trying to figure out what the problem is

0celo7

@ChrisWhite why does the solution of the heat equation have an exponential factor

skull petrol

perhaps, newtons law of cooling?

0celo7

what

user54412

the time exponential -- because everything is diffusing away

skull petrol

why haven't you asked on the main site?

user54412

06:03

the space exponential -- are you so sure it's there?

0celo7

@ChrisWhite YES

I solved the PDE, it must be there because of Sturm-Liouville theory and it's there in the back of the book

user54412

I mean, plug in t=0. Hopefully you get your initial conditions out.

user54412

And those don't have to be exponential, right?

0celo7

@ChrisWhite what

@skullpetrol Newton's law of cooling is a less sophisticated version of the heat equation and I'm solving a more sophisticated version of the heat equation

I'm going to bed.

I'll ask one of my engineering profs/TAs, if they can't answer it I'll post on the main site.

user54412

My point is u(x,0) doesn't look like e^x, and by extension u(x,t) doesn't for any t. Just because you have e^x doesn't mean it's the only thing there.

0celo7

06:11

@ChrisWhite hmm

user54412

Your basis functions look weird and asymmetric, but meh. They're basis functions, whatever they look like they can add to any function you want.

skull petrol

Heat equation

The heat equation is a parabolic partial differential equation that describes the distribution of heat (or variation in temperature) in a given region over time. == Statement of the equationEdit == For a function u(x,y,z,t) of three spatial variables (x,y,z) (see cartesian coordinates) and the time variable t, the heat equation is More generally in any coordinate system: where α is a positive constant, and Δ or ∇2 denotes the Laplace operator. In the physical problem of temperature variation, u(x,y,z,t) is the temperature and α is the thermal diffusivity. For the mathematical treatment it is...

user54412

umm

user54412

that's not really his problem

user54412

Convection–diffusion equation

The convection–diffusion equation is a combination of the diffusion and convection (advection) equations, and describes physical phenomena where particles, energy, or other physical quantities are transferred inside a physical system due to two processes: diffusion and convection. Depending on context, the same equation can be called the advection–diffusion equation, drift–diffusion equation, or (generic) scalar transport equation. == Equation == === General === The general equation is where c is the variable of interest (species concentration for mass transfer, temperature for heat transfer), D...

user54412

06:16

also, I don't know why the rest of the world refers to it as convection

user54412

it's clearly advection

skull petrol

Ooops, sorry :-/

GPhys

06:32

@FenderLesPaul and there goes U Chicago

FenderLesPaul

06:56

@GPhys @0celo7 got my official acceptance to UChicago!!!

:)

@0celo7 I can finally ask Bob Wald all the questions we've ever had from his book

skull petrol

Congratulations @FenderLesPaul

FenderLesPaul

@skullpetrol thank you!

I get to meet my hero

my high school self is so happy right now haha

skull petrol

Wald?

FenderLesPaul

yeah

skull petrol

Cool.

Fermi went there to, right?

FenderLesPaul

07:07

yeah

skull petrol

Good luck and I hope for all the best.

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