The h Bar

dmckee

00:32

Concerning the stared link that involves mathematicians and bedroom gymnastics. Google "observed with a short wavelength probe", and follow one of the links that leads to a Sydney Harris cartoon...

Don't worry it is safe for work.

And for some reason not on the website of the man's cartoons.

HsMjstyMstdn

01:33

Hey @dmckee, is the uncertainty principle in quantum mechanics a fundamental fact of our observations of experiments and reality that just so happens to be model-able by the "Fourier Trade-Off" or is it a sign that the mathematics underlying the "Fourier Trade-Off" have a deep, significant truth to them, in addition to being mathematically accurate and elegant, that occurs in reality ? Do we know/can we know ?

Or is it one of those philosophical questions about information and knowledge that really doesn't make sense for me to ask ?

or anyone, really

dmckee

02:20

Surely this is just the question of Plato's Cave, right? So the only way to know is for some genius to discover a mechanism by which the underlying truth can be laid bare. That is: a breakthrough.

HsMjstyMstdn

Hmm

So the question remains there for a breakthrough to tackle, is there a better and different way to represent the principle itself ?

Doesn't look like it's going to undergo a thorough revision anytime soon, though, does it...

Cows

04:19

screwed around with new camera. Did a real simple problem just to get a feel for things . I will be computing GR and QFT stuff on camera . . . once I figure out the right settings and set things up. In the mean time, I will be doing some basic, vector and tensor calculus, and differential geometry to test lighting and get comfortable in front of camera , and to gauge feed back from you donkeys

Will do some stokes theorem and other related concepts, then metrics, and connections and few other things just to test the lighting

then I will do some GR calculations . . . . . fast

and some qft calculations . . . at the speed of light

BTW This is all going to be spontaneous, as in no real prep. I might even go live and recieve questions and slap them on the fly . . . If I get stuck I'd ask you guys for help

at any rate, I am going to record another video of Stokes et all related stuff really fast at another location at my apartment with different lighting and so forth.

CooperCape

04:53

When you accidentally pull an all-nighter

Could be bad tomorrow

Today*

Might not see today

skullpatrol

Why "accidentally"?

Totally lose track of time...

dmckee

05:15

I suspect that relativity is attractive to people who want to over-throw the established order and bring clarity and common sense back to physics because they figure that they won't every have to put their claims to an actual test.

skullpatrol

Interesting suspicion.

HsMjstyMstdn

@dmckee As in special relativity ?

dmckee

05:31

@HsMjstyMstdn Well, either.

skullpatrol

They must include special relativity, since it originally got its name from being a "special" case of general relativity.

HsMjstyMstdn

I'd suspect some/most people steer clear of GR when they glimpse the math needed to go with it...

SR, as it is now, accessible and a tad "easier" than GR really does attract LOADS of people, questions and probings

Perhaps because it is the lowest "hanging fruit" on the tree of Physics that is uber-cool, seemingly about time-travel and mindbending paradoxes

skullpatrol

Even Einstein himself believed relatively was accessible to able teenagers.

HsMjstyMstdn

We underestimate teenagers, sometimes.

Maybe even most times...

skullpatrol

05:54

> Many years after his Pythagorean proof, Einstein shared this lesson with another twelve-year-old who was wrestling with mathematics. On January 3, 1943, a junior-high-school student named Barbara Lee Wilson wrote to him for advice. “Most of the girls in my room have heroes which they write fan mail to,” she began. “You + my uncle who is in the Coast Guard are my heroes.” Wilson told Einstein that she was anxious about her performance in math class:

> “I have to work longer in it than most of my friends. I worry (perhaps too much).” Four days later, Einstein sent her a reply. “Until now I never dreamed to be something like a hero,” he wrote. “But since you have given me the nomination I feel that I am one.” As for Wilson’s academic concerns? “Do not worry about your difficulties in mathematics,” he told her. “I can assure you that mine are still greater.”

David Z

@HsMjstyMstdn if you're interested in a second opinion, I would consider the HUP primarily a fundamental property about our mathematical models. That property happens to be reflected in real-world observations. If it weren't, we would have reason to abandon (or at least improve) the models.

But really, I do think that's getting into philosophical territory. The answer to your question doesn't seem to make a whole lot of difference as far as practical science.

HsMjstyMstdn

Noted. Thanks for the input

Always nice to toe the deep dark waters of philosophy every now and then, though.

David Z

Yeah, nothing wrong with that, as long as you're aware of how "deep" you're getting.

Anonymous

@HsMjstyMstdn There's a rabbit hole underneath :)

HsMjstyMstdn

@Blue There's the abyss underneath

skullpatrol

05:59

Struggling with these kinds of questions on your own does have its merits.

skullpatrol

06:35

Then again, not struggling with it leads to

1 hour ago, by dmckee

I suspect that relativity is attractive to people who want to over-throw the established order and bring clarity and common sense back to physics because they figure that they won't every have to put their claims to an actual test.

*ever :P

Tanuj

07:11

@JohnRennie Good Morning :)

John Rennie

Morning

Tanuj

I have a question.Can we move to PSS ?

Nehal Samee

08:01

@Blue...The capacitor looks like this ... ------Ⅰ Ⅰ-------. Can't provide a picture...It's not being uploaded...

Anonymous

@NehalSamee Upload here imgur.com and post link

Anonymous

Click on "New Post"

Nehal Samee

@Blue...Got it ... You can look up the picture here...Actually I was looking for it...I want to ensure my idea regarding it...

Anonymous

Okay, so what is the problem?

Tanuj

@Blue do have a look here when you're done with the other doubts .chat.stackexchange.com/transcript/message/43052313#43052313

Anonymous

08:14

Okay, I'll check when I'm free

Nehal Samee

I wanted to know whether the height of capacitor(top to bottom) in figure 1 affected it's capacitance...My idea is that the portion of capacitor plates with the dielectric reduce the electric field only by that amount...

That is , if top half is filled with dielectric , then electric filed will be half reduced and capacitance will also be half ... Is it correct .?

Anonymous

In the first one the the capacitors can be considered as two different capacitors in parallel

Anonymous

Use the formula for parallel combination of capacitors...

Anonymous

You'll get your answer

Anonymous

In the second one there are two capacitors in series

Anonymous

08:18

The capacitance increases by $K_1$ or $K_2$ in each of those capacitors

Anonymous

And if there is no dielectric in some region you can consider $K=1$

Mr. Xcoder

Hello folks

skullpatrol

Heya

Tanuj

@Blue are you done with this doubt ?

@Blue chat.stackexchange.com/transcript/message/43052313#43052313

CooperCape

09:08

Boo

ACuriousMind

09:56

Is the site very slow today for anyone else?

skullpatrol

Yup, me too.

Mithrandir24601

@ACuriousMind It's faster than usual for me... :P

ACuriousMind

I'm getting error messages for about half the pages I try to load

skullpatrol

Perhaps it's time for a reboot?

ACuriousMind

@skullpatrol I don't think I can reboot the SE servers :P

skullpatrol

10:10

What?!? You can't. I thought all mods had super-powers :P

(I meant your device)

ACuriousMind

@skullpatrol I literally just booted it.

skullpatrol

Is kicking your computer a form of electronic abuse?

;-)

Anonymous

Maybe it is related to this:

Anonymous

70

A: Systematic "This is our fault." on some links

I unfortunately don't have an awesome answer here. We don't have an exhaustive list of contributing factors yet. I woke up to a server farm on fire as well. I'll share what we know, it's all I've got. There were 2 web servers (of 9) affected. ny-web01 and ny-web04 exhausted connection pool capac...

Anonymous

I not facing any issues here though

ACuriousMind

10:18

@Blue Just found that, probably it is

skullpatrol

I thought they moved the serves from ny?

Anonymous

@skullpatrol Iirc that was their backup server

skullpatrol

Right. Apparently not all of them.

Anonymous

Live footage of the investigation and work being done to fix this: i.stack.imgur.com/jJOlp.gif — Adam Lear ♦ yesterday

3

Anonymous

lol

skullpatrol

10:31

Is that what happens when you don't keep all your servers at one location?

user55789

11:06

Hi all, I have a differential equation with a Bessel function as a solution

Rewriting it appropriately does not appear to me instantly, so any good methods or books I can look at to squeeze out the right answer?

ACuriousMind

Rewriting it appropriately for what

user55789

For completeness I want to solve the above Bessel eqn

Rewriting it appropriately to see that it's a Bessel equation

The above equation doesn't have a first order derivative so it most certainly isn't in standard form

user55789

11:34

Different example: suppose I have y'' + k^2 x^4 y = 0

Using y = u\sqrt{x} and z = \frac13 kx^3, this reduces to the standard form of a Bessel equation

One could argue that such a substitution is not trivial (terms closer to "obscure" come to mind)

But I could imagine that recipes exist for finding these substitutions systematically

CooperCape

12:23

ohai

ACuriousMind

hai-o

CooperCape

ang about

o-hai-o

oosh

Balarka Sen

13:08

Heilung | LIFA - Krigsgaldr LIVE

►

O_o

Youtube recommends this to me out of nowhere

This is 10/10

CooperCape

13:28

Can’t watch right now but looks

Well

European

Balarka Sen

yeah they're a band from Denmark

It's always the Scandinavians who do the very primitive, gritty, ritualistic, mythical folk/black/atmospheric metal like this.

I love it

xuBe

14:36

halo

i need a intellectually stimulating hobby

any suggestions

Tanuj

@JohnRennie Hi, are you free thigh now?

John Rennie

Eating lunch

Thigh?

Loong

@BalarkaSen I just had a long walk trough the snow at Sammallahdenmäki. Now I find that quite appropriate.

Tanuj

@JohnRennie stupid autocorrect ! I'm back on my laptop now . It was supposed to be "free".

Balarka Sen

@Loong Hah.

John Rennie

14:47

I'm sorry, but wearing antlers on your head is just silly.

Tanuj

lol what ?

Loong

I didn't do that part.

Balarka Sen

@JohnRennie I like the dramatic presentation.

bolbteppa

@user55789 bessel equation in your form and the substitution given here

encyclopediaofmath.org/index.php/Bessel_equation

Balarka Sen

I have never actually heard of Heilung. Looking through some of their videos it seems their stage performances are always uber-theatrical

bolbteppa

14:51

@user55789 if you reduce it to the Laplace form on that page you can solve it nicely in a way that applies to many examples phys.uconn.edu/~rozman/Courses/P2400_16S/downloads/…

Tanuj

@JohnRennie did you get that other question from a few hours back ?

John Rennie

The rolling disk?

Balarka Sen

Eg, this

Tanuj

yea

John Rennie

I realised I had misunderstood the question

Tanuj

14:53

oh i thought we did that part correct

what was the mistake we were doing ?

@JohnRennie Read your reply on PSS , I get it now (the question)

@JohnRennie Can we move to PSS and have another discussion at it ?

John Rennie

15:10

I'm currently eating lunch ...

Tanuj

@JohnRennie Yeah , enjoy it ;) .

John Rennie

Southern fried chicken and curly fries.

Tanuj

@JohnRennie Yea go ahead make me jealous now ! As if I wasn't already frustrated for not getting the questions.

user55789

15:43

@bolbteppa Hmm the problem arises because i have a +\nu^2 not a -\nu

Bessel equations of complex orders are just completely beyond me

I've been at this for 8 hours now, and I'm completely out of ideas

#hopeless

Tanuj

@JohnRennie God ! How much do you eat !

user174174

Hey

bolbteppa

@user55789 just guessing, if you set $\nu = i \mu$ it becomes a Bessel equation right, then if you reduce it to Laplace form with that substitution it becomes a 2nd order ode with first order polynomial coefficients, so you can use the Laplace method, and will derive a Bessel function integral representation, at which point you can set $\mu = - i \nu$ no?

Manolis Lyviakis

yooo

bolbteppa

Bessel's equation holds for complex parameters, to see it really clearly, if you invert your form of the Bessel equation to the standard form with the substitution in that link you will get the normal Bessel equation with a complex parameter, as wikipedia says

Bessel function

Bessel functions, first defined by the mathematician Daniel Bernoulli and then generalized by Friedrich Bessel, are the canonical solutions y(x) of Bessel's differential equation x 2 d 2 y d x 2 + ...

Manolis Lyviakis

15:55

hey bolb

bolbteppa

and that Laplace things hows you how to derive the general integral representation valid for arbitrary parameters very easily in a few lines, you will just get what is basically a Cauchy integral representation of a Bessel function

(In Landau's QM appendix you will see this is a standard method for dealing with ODE's in QM)

user55789

The above just doesn't look good

It might to you but it doesn't to me

This is not something I can do a series expansion from

Manolis Lyviakis

Particle moving with a force towards a point O.the initial velocity is against the force . at t=0 i have that the initial distance between the particle and O is $a$ .Laos $F=\frac{mγ}{r(t)^3} $ where r(t) is the OP vector. Prove that the particle goes to infinit if $u(0)^2 > \frac{γ}{r(0)^2} $ im pretty noob i wrote down the equations of newtons law. and got $ u(t)-u(0)=-γ \int \frac{1}{r(t)^3} dt $ now what i tried taking limits

user55789

Of course the last line is simple, but a generic Bessel function can't be expanded in a series

Or is it easy to divide e.g. Y_nu / K_nu?

John Rennie

@ManolisLyviakis just compute the total energy of the particle i.e. potential + kinetic

Manolis Lyviakis

16:08

Oh i should use energies?

bolbteppa

My guess is, if that's the solution they gave you, it's because they noted it's a Bessel equation, with complex parameter $\nu = i \xi$, and so wrote down the general solution in terms of two linearly independent Bessel functions $J$ and $Y$,

user55789

Eh, they didn't give me that solution

Manolis Lyviakis

I cant make an argument with just these? @JohnRennie

user55789

I have the Hamiltonian

Well not Hamiltonian but equation corresponding to H

And the above gives Bessel functions in WolframAlpha

The problem is a bound state problem

John Rennie

@ManolisLyviakis Take your zero of the potential energy to be at infinity. Then bound objects have a negative total energy and unbound ones have a positive total energy.

user55789

16:09

So I have split up the integral up to a small segment epsilon

Within epsilon from X=0 the potential is set to 0 and the bessel functions simplify

bolbteppa

So Wolfram said that wave function is the solution?

user55789

Yeah Wolfram Alpha confirms the first line above for Psi_E(X)

Manolis Lyviakis

im just learning mechanics so what do you mean by bounded unbounded.

John Rennie

@ManolisLyviakis a bound object has velocity lower than the escape velocity while an unbound object has higher than escape velocity. An unbound object can escape to infinity.

user55789

Actually the only solution I'm given is that the relative coefficients B_O / A_O give rise to a bound state energy of the form E_n = -C epsilon^{-2} \exp(-\frac{2\pi n}\xi)

Manolis Lyviakis

16:13

ohh

and how did you come up about the signs of their energies

user55789

With n = 1,2,3

etc

bolbteppa

That looks normal, the Hamiltonian leads to a Schrodinger equation with a Bessel ode as part of it, and the solutions end up involving Bessel functions, and we can see why your ode is a Bessel equation with complex parameter, so that's the general solution of a Bessel equation, there is a series for $J$, on wiki en.wikipedia.org/wiki/Bessel_function the $Y$ expansion is insane when you can do it, just trying to find the wave function you posted is all one would usually do tbh

user55789

@bolbteppa I need to find the pole of the reflection coefficient

That is what is asked

This pole will give the bound state energy

And as stated above, I have R \sim B_O / A_O, which is a massive clusterfuck of families of Bessel functions of the first and second kind

John Rennie

@ManolisLyviakis If you take the PE to be zero at infinity then because PE is lower near O the PE has to be negative. And KE is always positive because it's a square.

user55789

Even in the small epsilon\sqrt{-E} limit, there's almost no simplification possible

Manolis Lyviakis

16:16

ok ill give it a try. There is no other way right?

user55789

@bolbteppa University is closing up, I'll be back in +/- 30 if you're still willing to discuss it then

(I know I am...)

bolbteppa

Where was the problem given

user55789

It's some hand-made problem set

With hyper convoluted phrasing

Verbatim:

bolbteppa

This is a bit mental the way it stands, probably an easy way to do this

John Rennie

@ManolisLyviakis there may be other ways, but using the energy is quick and easy.

user55789

16:19

"Now we will consider g= -\frac14 - \xi^2 with \xi > 0. In order to make sense of these potentials (which appear to be unbounded from below) consider imposing a cutoff X > epsilon on the space where the wave function is defined. Impose boundary conditions psi(epsilon) = 0. Solve for the wave functions in the limit epsilon\sqrt{-E} << 1 for X \sim epsilon. Show that the boundary conditions can only be satisfied for the bound state energies: E_n = -C e^{-2} exp(-2pi n/\xi) for n = 1,2,3,...

for some constant C>0. As we have predicted we encounter a problem as we remove the cutoff epsilon --> 0. The spectrum becomes continuous and unbounded below. Notice that for the approximation to be self consistent we must have \xi << 1. Explain why this is the case."

The original Hamiltonian was: H = P^2 + g/X^2, i.e. that of a m=1/2 particle

This corresponds to H = A^\dagger A + (g+1/4) / X^2 via the usual mode expansions

Which makes the g=-1/4 the edge case at which we have bound states or not manifest

Manolis Lyviakis

@JohnRennie ill use the conservation law of energy?

@JohnRennie made a thread to see an answer in detail to understand what is happening :

John Rennie

@ManolisLyviakis If the force is $F = m\gamma/r^3$ then the potential energy is $V = -m\gamma/2r^2$. Yes?

bolbteppa

@user55789 idk, maybe those bessel functions only appear without the cutoff leading to a continuous spectrum or something, bit crazy tbh

Manolis Lyviakis

if the force is conservative

John Rennie

(because $F = dV/dr$)

Manolis Lyviakis

16:26

right?

John Rennie

Do you have any reason to think the force is non-conservative?

Manolis Lyviakis

no but i dont know how ou check it

user55789

@bolbteppa I'll go home now and then rejoin the chat - My best bet is that a division of the second and first Bessel functions lead to a simplified result in the series expansion (somehow)

But we'll see..

Brb

John Rennie

In problems like this you can generally assume the force is conservative

Manolis Lyviakis

also the force is afunction of t . isnt the derivative in respect to t of the potential energy the force. and not in respect of the $r(t)$

$-\frac{dV}{dt}=F$

Tanuj

16:31

@JohnRennie I just had dinner , I hope you're done with lunch as well ?

John Rennie

@Tanuj shall we go to the problem solving room

Tanuj

sure :)

Jannik Pitt

If a physical law is referenced as "covariant", does that mean it is invariant under some transformation? I guess "covariant" as in the transformation property of a linear functional doesn't make sense in that context?

Manolis Lyviakis

can you help me? physics.stackexchange.com/questions/388629/…

John Rennie

16:49

@ManolisLyviakis that's a homework question and shouldn't have been posted. You gave been on the site long enough to know we don't accept homework questions on the main site.

Manolis Lyviakis

different policy on math exchange

John Rennie

This isn't the math exchange

Manolis Lyviakis

yeah i meant im not on this site long enough. im mostly in math.

John Rennie

I've explained how to do this. The escape velocity is when the total energy is zero i.e. $PE + KE = 0$.

Manolis Lyviakis

oh i didnt understand that yo said that

John Rennie

16:51

$KE = \tfrac{1}{2}mv^2$ and $PE = -m\gamma/2r^2$

Manolis Lyviakis

also why is it escaping when the total energy is zero?

ill use the conservation law right? to derive the formula

my initial speed

must be the escape velocity

i think i got it

ill figure initial energy and final energy

Jannik Pitt

@ManolisLyviakis The kinetic energy has to equal the potential energy at infinity for the object to escape

0celo7

17:13

@BernardoMeurer @BalarkaSen timed link youtu.be/HI-mXMr8glQ?t=4m16s

DanielSank

17:28

@JohnRennie Windows machine humming along nicely. Thank you.

Tanuj

Guys can any help me with my physics problem ? Me and John were working on it but then he had to go attend a phonecall and now he is gone

Manolis Lyviakis

17:44

the derivative of potential energy is int respect with t or r where r i mean the position vector

$-\frac{dV}{dr}=F(t)$ or $-\frac{dV}{dt}=F(t)$

Tanuj

@dmckee can you help with a physics problem ?

DanielSank

I came to a puzzling issue that you may find amusing. I wanted to install a bit of software. That software is distributed as a 7zip archive. Unfortunately, the 7zip website doesn't use HTTPS. So I didn't install the software. Then I posted on the software's forum to the effect that they might think about releasing normal zips or gzips. Their response was that if I care, I can read the 7zip source code and compile it myself.

@Tanuj Just ask the damned question.

Cows

Just got up. recorded myself computing Christoffel symbols

vid up

DanielSank

That's a rule here: do not ask if you can ask. Just ask.

Tanuj

okay

ACuriousMind

17:46

@Tanuj Don't start pinging random people for help, please. They all can read the chat - if someone has time and wants to help you, they will. Also, just ask your question instead of asking people if you can ask them a question.

DanielSank

^ + 100

Long time no nothing, @ACuriousMind. How goes it?

Tanuj

It is a damned question , even JR didn't get it first up

loocsieulb

save ur breath people will keep asking for asking questions

no point

ACuriousMind

@DanielSank It's going well, although this weekend I've mostly been battling a cold :/

DanielSank

@ACuriousMind gross

ACuriousMind

17:48

@DanielSank Yeah, I agree :D

DanielSank

I am dealing with a really annoying problem: there's no way to securely download an installer for a program that unpack a 7zip archive.

...on Windows.

Tanuj

ACuriousMind

@DanielSank Wait, they only have a 7zip and nothing else?

Tanuj

It is to be assumed that the disc is moving rightwards and executing pure rolling

DanielSank

@ACuriousMind Correct.

It is ridiculous.

I posted on their forum and the response was kind of lame.

Tanuj

17:50

@DanielSank Winrar unpacks 7zip too

DanielSank

Ha! Ok! Winrar uses HTTPS.

Cows

@DanielSank what are you working on?

Tanuj

@DanielSank what do you mean by that ?

DanielSank

> Get WinRAR FREE with TrialPay

Lol, nope.

@Cows Unzipping a software installer.

@Tanuj I mean their website uses HTTPS so I know who I'm downloading from.

The 7zip site does not.

Tanuj

okay

look i just posted the question and no help , you see I'm forced to ping people

ACuriousMind

17:55

@Tanuj You are not entitled to people helping you here at all.

Tanuj

i know @ACuriousMind I'm just too frustrated by that fuckin question

ACuriousMind

Well, then don't complain how you're "forced" to ping people when no one wants to answer your ill-posed question.

Tanuj

@ACuriousMind take a chill pill dude , whats wrong ?

Cows

@Tanuj I'd help you but I must confess I'm not very good at multiple choice physics answers. Also I don't have any formal physics qualifications so I am always afraid to say anything in the way of advice. . . .

hang tight someone will help you

Tanuj

@Cows I'd be glad to get any sorta help

DanielSank

17:58

@Tanuj You're being childish, thinking that everyone is focused on helping you. You say "I just posted the equation", but I don't see it.

I noticed you posted a black box, but I didn't click on it because it looks like an image.

Now I'm clicking on it.

Tanuj

I'm being childish ? You didn't even read what i actually wrote , I didn't write equation , i wrote question !

DanielSank

@Tanuj Yes I did.

I just didn't read the image.

I don't actually see a question anywhere.

Link?

Transcript for