The h Bar

Semiclassical

04:00

something's screwy

but i'm not sure what

0celo7

ohhhhhhhhhhhhhhhhhhhhhhhhhhhh

I did the wrong initial condition

I evaluated $F=$ something

it needs to be $f=0$something

Semiclassical

F[0]=0, F'[0]=0, F''[0]=p ?

0celo7

and for $\nu=1$ they're the same

it needs to be $F'(\infty)=(\nu a)^{-1/2}$

Semiclassical

huh

i'm a little surprised by that, since in the presentation on Wikipedia: en.wikipedia.org/wiki/Stagnation_point_flow

it gets the initial condition in dimensionless variables as just F'(infinity)=1

0celo7

hmmmmmmmmmm

there's a typo in the homework then

lemme see what we did in class...

Semiclassical

04:03

my guess would be that, since F' itself is a derivative

it'll pick up a factor from the rescaling of the differentiation variable

and this hopefully cancels out the other side

it's also possible that your problem contains more parameters, but i'd start there

0celo7

yeah wiki is right

the problem set has a typo

the notes are right

Semiclassical

kk

0celo7

but the graph is still garbage

Semiclassical

yeah, something's goofy

0celo7

kek

Semiclassical

04:07

my guess is that the function provided by NDSolve doesn't like this kind of manipulation. but i frankly don't see why

0celo7

@Semiclassical the solution is to dimensionalize the NDSolve equation

Semiclassical

ugh

that's probably right

but ugh

0celo7

that's also literally awful

Semiclassical

let's try to avoid that

0celo7

mhm

Semiclassical

04:09

i mean, the simplest solution to this

0celo7

uh

why are the arrows going up

Semiclassical

would be to label the y-scale as (stealing from wikipedia) $y\sqrt{k/\nu}$

0celo7

@Semiclassical Maybe it's a chain rule issue.

Semiclassical

but this is screwy. If I plot F[y/a] from y=0 to a, I should get ths same plot as F[y] from y=0 to 1

0celo7

yeah but it's the derivative term that's mucking things up I bet

the $xf'(y)$ guy

Semiclassical

04:12

yeah

I've been trying to avoid that by writing it as F'[y/a]a, though

and it still comes out as non---wait maybe I see it

0celo7

@Semiclassical yes?

Semiclassical

okay

try doing the following plot, to have a common reference point:

GraphicsRow[{
StreamPlot[Evaluate[{x F'[y*.1]*.1,-F[y*.1]}/.sol[1.23259]],{x,0,5.1},{y,0,5.1/.1},PlotRange->{{0,5.1},{0,5.1/.1}},
StreamStyle->Red],StreamPlot[Evaluate[{x F'[y*1]*1,-F[y*1]}/.sol[1.23259]],{x,0,5.1},{y,0,5.1/1},PlotRange->{{0,5.1},{0,5.1}}]}]

sigh. it never indents how I want it to

0celo7

r/restofthefuckingowl

where are you getting this from

Semiclassical

anyways. the plot on the right is the original stream plot, and on the left it's with x rescaled by a

0celo7

mhm

Semiclassical

04:20

key thing is that, between the two plots, I'm pretty sure the streamlines which are sown on the left are actually the same as the ones on the right

the scales are different, but if you imagine overlapping them I'm pretty sure it's the same

the trouble is that, apparently, StreamPlot is mighty dumb

and as such it isn't sampling the plot for the graph on the left as the one on the right

as a consequence, it's coming out weirdly sparse

and tbh I'm not sure how to fix it

0celo7

I buy that.

So should I just turn in the nondimensional one?

Or go full insane mode and put dimensions in the NDSolve

I have good notes with all the dimensional equations

Semiclassical

what i'd do is just label the y-scale as $y\sqrt{k/\nu}$ (wiki's notation)

and tbh that seems like the right thing to do physically. sqrt(nu/k) sets the scale for y, so plotting y/sqrt(nu/k) makes sense to me

0celo7

ok sir

thank you

I'm going to have some cookies and milk in your honor

Semiclassical

lol

i'm going to drink a glass of orange juice

also, one thing you can do as a quality check of this

0celo7

orange juice at night??

better have some vodka

Semiclassical

04:26

the wikipedia page lists the asymptotic form of F for large eta

and that's something you can check with your numerical solutions

(you can also sorta eye-ball the initial slope of F' from their plot as being in the ballpark of 1.2)

0celo7

it does look linear

slope 1-ish?

Semiclassical

a bit more than one, if you look at how the plot scales

0celo7

Semiclassical

oh, that plot

yeah, slope = 1 there

0celo7

oops

Semiclassical

04:31

it'd better to match the boundary condition

lol

0celo7

lmao

Semiclassical

yeah, I'm not sure how well you should expect it to work as you push n_max larger and larger

0celo7

I don't know what to do

Semiclassical

the precision you need on p=1.23... in order to avoid divergence gets sharper and sharper

0celo7

right...

Semiclassical

04:32

to the point where the main thing you learn at large n_max is that it's really hard to avoid numerical divergence

so probably 100 is too big

0celo7

how did people do this in 1911

Semiclassical

now there is the question

0celo7

Semiclassical

hah, nice

if you do 1.24 there I bet it blow up

this reminds me of finding eigenvalues for the schrodinger equation

0celo7

Semiclassical

04:35

where any numerical deviation from the exact eigenvalue means that the wavefunction ultimately diverges

use Evaluate[] on the argument to make that plot work

0celo7

there's a finite-time blow-up or something

@Semiclassical nope

Semiclassical

huh

0celo7

Plot[Evaluate[F[n] /. sol[1.24]], {n, 0, 1}, AspectRatio -> 1]

massive error

but changing it to 1.23 works

Semiclassical

weird

oh, i get you

what's your n-max?

0celo7

100,000

Semiclassical

04:37

lol

yeah, that's not asking for trouble :P

0celo7

ok changing it to 100 fixes that

WOW the difference between 1.24 and 1.23 is insane

Semiclassical

yeah

0celo7

god damn dude

Semiclassical

not entirely surprising, though, since p=1.23259 is closer to 1.23 than 1.24

0celo7

Semiclassical

04:39

if you do 1.23 and 1.235 i bet the results won't be quite so disparate

0celo7

this is pretty cool

Semiclassical

yeah

it makes it pretty clear how sensitive this problem is

if you miss at large values, you miss by a mile

0celo7

@Semiclassical Yeah

I wonder what the long-range behavior for the flow is because of this

Semiclassical

there's probably some smart explanation for why it's so sensitive

0celo7

nonlinear ODE

Semiclassical

04:43

point

0celo7

ahhhhh

Semiclassical

some explanation in terms of the asymptotics of such solutions, though

0celo7

this is cool

Semiclassical

you can see why they call it the shooting method, lol

0celo7

Semiclassical

04:44

nice. not exactly physical, but nice :P

0celo7

the part near the bottom is the good flow

this is the most physics this room has seen in months

Semiclassical

really?

0celo7

and really it's just mathematica

@Semiclassical really what?

Semiclassical

oh, I misread that picture

...huh

0celo7

@Semiclassical the solution is accurate in a neighborhood of the origin

Semiclassical

04:46

oh, i see it now

0celo7

OH

Semiclassical

so bottom in this case means "below y=20"

0celo7

that's why the other flow is bad

Semiclassical

other flow?

0celo7

beyond like n=50, the solution is garbage

the rescaled flow

the rescaled flow is using long-range values

Semiclassical

04:47

ah

i'd question even n=30 there, tbh, since by that point the flow starts pointing towards x=0 rather than away

are you having to do some kind of matching between the short-range flow and the long-range flow?

0celo7

@Semiclassical Because I had 1.23 there

Semiclassical

kk

oh, nice

0celo7

rounding has a crazy effect here

@Semiclassical no, thank god

Semiclassical

lol, ok

0celo7

now around n=130 it gets bad again

Semiclassical

04:51

nice

one thing I conclude is that, however Hiemenz did it, he probably didn't use the shooting method

0celo7

@Semiclassical lol

Semiclassical

or if he did then his precision was probably pretty low

0celo7

he used an abacus

Semiclassical

it does make you wonder how on earth they did it

0celo7

@Semiclassical time to do fortran now...

Semiclassical

04:56

godspeed

given that Hiemanz wrote in german, i doubt i'd get much out of it

0celo7

@Semiclassical does wiki have the reference?

i can read it

Semiclassical

it does

0celo7

dissertation

Semiclassical

lol

0celo7

no way that has been scanned

Semiclassical

04:57

yeeah

the Howarth paper apparently improved on the numerics

so probably it'd give at least some sense of what the original paper did

on the other hand: AERONAUTICAL RESEARCH COUNCIL LONDON

seems like that'd be hard to track

actually: aerade.cranfield.ac.uk

"Due to problems with the underlying software, we can currently only offer part of the original AERADE service. We are taking this opportunity to redesign the site and restore the data, and hope to be able to launch it in the near future."

nuts

0celo7

it's a conspiracy

Semiclassical

yeah, their software conspired to put that paper out of reach

i do wonder why only some portion of the reports have been digitized: aerade.cranfield.ac.uk/listarcrm.php

0celo7

"The design of a sensitive yawmeter"

this is some serious applied stuff

Semiclassical

there's a lot of papers there, but it clearly isn't exhaustive

the names are occasionally interesting

the surname Hartree shows up, for instance

and seems to be the same Hartree as the Hartree-Fock equations

0celo7

@Semiclassical did he develop bombs for the military?

Semiclassical

05:10

no clue

"A solution of the laminar boundary-layer equation for retarded flow"

sounds about as theoretical as one could get in that context

0celo7

05:50

@JohnRennie gyazo.com/5760594138b58abd19fd32c929372cbc

John Rennie

That looks fantastic.What was the bread?

0celo7

@JohnRennie idk, it's the costco bread lol

@JohnRennie corriganscookday.files.wordpress.com/2012/12/…

@JohnRennie it's too bad feta costs an arm and a leg

John Rennie

Just regular bread then. It looks more like a flatbread of some kind but maybe that's just the angle of the photograph ...

0celo7

@JohnRennie it's sliced in half and toasted

it gets pretty thin after the toasting

John Rennie

Aha!

0celo7

05:59

@JohnRennie I should source some pita

and just eat pounds a day

John Rennie

I can take or leave pita bread.

0celo7

that British weather must be messing with your mind

John Rennie

Though to be fair the pita bread the supermarkets sell in the UK probably isn't best quality.

0celo7

I've had great pita at restaurants. I've never bought it in a store.

John Rennie

Naan is my no. one favourite bread, but it has to be eaten freshly made and it needs a special oven, so you can only get it at restaurants.

0celo7

06:05

@JohnRennie I've never had it

HsMjstyMstdn

Some of the best naan cooks are pretty much honorary volcano divers

John Rennie

Putting your arm into an oven at several hundred degrees always looked scary to me :-)

HsMjstyMstdn

There's an "extreme" version of it in Armenia called Lavash. Some of them who handle the bigger ovens tip in at the waist !
https://www.youtube.com/watch?v=6go1batGbnk

Whew

0celo7

seems easy.

HsMjstyMstdn

to fall in, maybe.

@JohnRennie would it be accurate for me to say that the "canonical" model of friction most physicists are taught is the Amonton-Coulomb model ?

John Rennie

06:18

I don't think friction is really taught much. We learn Amonton's law at school, but you'd only start going into detail about the mechanism if you did a specialist tribology course.

HsMjstyMstdn

But our oft-used F= meu x N is from that model, no ?

John Rennie

@HsMjstyMstdn It's not really a model - it's more of a phenomenological law. I think I posted an answer on it a while back. Let me go look ...

Fawad

@JohnRennie hi. I need some help

John Rennie

7

A: How is frictional force dependent on normal reaction?

You need to start by considering the microscopic origin of the frictional force. In most circumstances surfaces are rough so when to touch two surfaces together they actually only make contact at the highest points on the surfaces. We call these high points asperities, and in the diagram below I...

@Fawad Hi. What's the problem?

HsMjstyMstdn

@JohnRennie Aight thanks, will digest.

Fawad

06:31

why comparing energy density due to electric field ($\epsilon_o \frac{E^2}{2}$) and energy density due to magnetic field ($\frac{B^2}{2\mu_o}$) gives relation $B_o=\frac{E_o}{c}$

@JohnRennie

John Rennie

@Fawad Electrodynamics isn't my strongest area, but I think that's only true for a plane wave i.e. it's a special case. It comes out of Maxwell's equations.

That is when you obtain the plane wave solution from Maxwell's equations you find that $E/B=c$.

It's really $$\frac{E}{B} = \frac{1}{\sqrt{\epsilon_0\mu_0}}$$ but of course $c^2 = 1/\epsilon_0\mu_0$

Fawad

Ok👍

Another question. If surface is a perfect reflector then radiation pressure (and force) on durface will be 0? Radiation pressure $$p=\dfrac{U}{c}$$

:/

@JohnRennie

John Rennie

06:54

The radiation pressure is the force per unit area, and the force is the rate of change of momentum. So the question to ask is whether the momentum of the light changes when it is reflected, and if so by how much.

Sir Cumference

Quick question

You can't add two quantities of different dimensions, right?

But you can multiply them, even though multiplication is just adding several times

Fawad

@JohnRennie no. That different story.

@JohnRennie help Livesavior

Hi @Blue can you help me?

John Rennie

07:12

@Fawad that's a somewhat misleading description of how to solve the problem, because it gives the impression the energy has to be absorbed for momentum to be transfrred and that isn't the case.

The momentum of a photon is $p = h/\lambda$

The energy of the photon is $h\nu = hc/\lambda$ so the photon flux, i.e. number of photons per second per unit area, is the energy flux $U$ divided by $h\nu$.

$$ N = \frac{U}{h\nu} = \frac{U\lambda}{hc} = \frac{U}{pc} $$

And the total momentum of the photons is $Np$ giving: $$p_\text{total} = \frac{U}{c} $$

Fawad

So in solution given in my book is as if only for 1 photon meets surface?

John Rennie

If the photons hit the surface and stop the momentum goes from $U/c$ to zero, so the momentum change per second is $U/c$.

If the photons hit the surface and reflect back the momentum changes from $+U/c$ to $-U/c$ i.e. the momentum change per second is $2U/c$. So for a perfect reflector the force is twice as great.

It's no different from throwing a ball at a surface. If the ball bounces back the momentum change is twice as big as if it hits the surface and sticks.

Fawad

@JohnRennie yes. Thank you 😊

Abcd

08:34

According to second law of thermodynamics, the entropy of the universe is increasing which can be proved using Clausius inequality. What is the limit of this entropy? Can the universe bear infinite entropy?

John Rennie

@Abcd I don't think it's safe to apply the second law to the universe as a whole. Firstly it may well be infinite, and that causes all sorts of problems, and secondly since spacetime is expanding that also causes difficulties with interpretation of thermodynamic quantities e.g. energy isn't conserved.

There is a limit to the amount of entropy because the maximum entropy objects we know are black hole event horizons.

But exactly what this means for the entropy of the universe as a whole I don't know.

@Blue no, you can't invalidate flags on your own posts. I know this because I have tried it :-)

John Rennie

09:07

Right! Work done! Plan for now is to make a large coffee and stare out of the window at the rain ...

Phase

10:37

Oh I see, thanks @0celo7 that would've confused me. I gotta dash to lectures now

Emilio Pisanty

@JohnRennie Wouldn't you rather take that coffee and look out the window to a beautiful sunlit morning?

John Rennie

@EmilioPisanty Tomorrow is forecast to be a sunny day. Right now we have what the Met Office describe as a belt of intense precipitation.

Emilio Pisanty

@JohnRennie if the met office calls it intense precipitation then it can't be good

but anyways, y'all have Stockholm syndrome with your weather

John Rennie

Looking out of my window I find I agree with the Met Office's description :-)

To be fair Chester doesn't get much rain as we're in the rain shadow of the Welsh mountains when the wind is from the west, and in the rain shadow of the Peak District when the wind is from the east. I cycle into town most days and it's relatively rare that I get wet.

Emilio Pisanty

10:57

just think, you could have weather like this every day

Anonymous

@EmilioPisanty Where's that?

Emilio Pisanty

@Blue out of my office window, just now

(instead of, y'know, convincing yourselves that shit weather every day (or close to) is not that bad, and that a wet gray windy dark day can be pleasant)

John Rennie

Oh well, variety is the spice of life.
(It isn't really. Garlic is the spice of life. Variety is nice though :-)

Anonymous

I like rains and storms more than clear sunny weather

Anonymous

Fair weather is boring :P

John Rennie

11:01

Tomorrow I'll post a picture of Chester in the sunshine just to prove it does happen occasionally :-)

Anonymous

(Also, we get too much sunny weather here, all the time XD)

Anonymous

In the summers, it's enough for a sunstroke

Mithrandir24601

@JohnRennie I.e. a light drizzle :P

Emilio Pisanty

@JohnRennie that's the Stockholm talking

@Blue rains and storms is a mistaken description of rain in England

John Rennie

@EmilioPisanty Hostages develop an intense desire for garlic?

Emilio Pisanty

11:03

@JohnRennie yes, it's a well-known psychologic phenomenon

Anonymous

@EmilioPisanty Yeah, I guess. It's more of a drizzle I suppose

John Rennie

@Mithrandir24601 the rain belt is on it's way towards Bristol ... :-)

Emilio Pisanty

@Blue quite often it's not even a drizzle

it's just that the air itself is wet

Anonymous

I was talking in general about the type of weather I like

Emilio Pisanty

and also it is gray and cloudy

plus windy and cold

John Rennie

11:04

Well that's got rid of the UK then :-)

Anonymous

Afaik it rains quite a lot in London (?)

John Rennie

Just as well the European mainland is getting rid of us!

Emilio Pisanty

@Blue by volume, no

by rainy days a year, yes

Anonymous

@EmilioPisanty Ah, makes sense

Mithrandir24601

We did get a hurricane last month - I'd never before seen such good weather in October :)

John Rennie

11:05

@Blue the UK weather is dominated by westerly winds blowing in off the Atlantic, and laden with water they've picked up from the Atlantic.

Emilio Pisanty

but by volume it rains 30% more in Rome

John Rennie

The west side of the UK gets most of the rain. London is on the east side and is relatively dry.

Mithrandir24601

@Blue at home, I simply assume that it will 'rain' at some point in the day. Usually, I'm right :P

Emilio Pisanty

@JohnRennie you're in that little cranny northeast of Wales, no?

John Rennie

@EmilioPisanty Yes. In between two regions of intense rain, but our annual rainfall is quite low :-)

Emilio Pisanty

11:09

@JohnRennie still, I imagine your overcast-day count is rather high

John Rennie

@EmilioPisanty don't know to be honest. I notice the rain, because cycling in the rain is pretty miserable, but I'm not that fussed whether it's sunny or not.

Slereah

11:47

@0celo7 someone answered the Godel question

0

A: Homotopy proof of the lack of foliation of the Gödel metric

Assume there is a foliation and consider the normal bundle of a closed time-like geodesic $\gamma$, call it $\nu$. This bundle is stably trivial, because its Whitney sum with $T \gamma$ (which is trivial) is isomorphic to the restriction $T\mathbb{R}^4$ (which is also trivial) to $\gamma$. But th...

it is not a simple answer

"consider the normal bundle of a closed time-like geodesic γ"

But Godel has no closed timelike geodesic

ACuriousMind

@Slereah I don't think the argument in that answer depends on the curve being a geodesic.

Slereah

that is good

now I just have to understand most of it

Balarka Sen

Is this a topological foliation or foliation with more structure

Slereah

There's "Whitney sum" in it so that's not gonna be easy

Balarka Sen

(What is the manifold we are foliating?)

Slereah

11:53

$\mathbb R^4$

the manifoldest of manifolds

Balarka Sen

Whitney sum is just direct sum of bundles

ACuriousMind

@Slereah The Whitney sum is just adding vector bundles pointwise, it's not difficult.

Slereah

Oh

What is the "normal bundle"

Balarka Sen

@Slereah Foliating by what dimensional submanifolds?

Slereah

Spacelike hypersurface

So 3D

ACuriousMind

11:55

@Slereah Pretty much what it says - the bundle of vectors normal (i.e. perpendicular w.r.t. to the chosen metric) to the curve (or rather its tangent vectors)

Slereah

ah yes, hence the cylinder or moebius strip thing

And since the spacetime is orientable, it's a cylinder

Balarka Sen

So these spacelike hypersurfaces are compact?

Slereah

Well you see

The spacelike hypersurfaces do not exist

that's the point

I'm guessing that since we're doing $\mathbb R^4$, they are not gonna be compact if they existed, though?

I dunno

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