The h Bar

Rishi

00:23

> It isn’t pretentious either since it gets bigger by feeding off of matter.

Wang Yun

01:22

Hallo, everbody

good morning

Scientist Smith YT

@peterhsaysreinstateMonica I was meaning they are destroying the users by not keeping them in the loop, like they used to. Not by adding more rep.

Wang Yun

What 's the meaning of "population" here?

Does anyone know about it?

user8718165

01:51

Hello :) @Slereah , I want to ask a question.

JMac

03:19

physics.stackexchange.com/review/suggested-edits/280198 Yikes.

The Monk

06:45

I know it won't be relevant to ask this question on the main site. Hence I am posting it here ...

I'm completely burnt out. I loved doing physics. Can you guys help me ?

I'm more inclined towards the mathematical side of physics. It would be helpful if you can get me out of this.

John Rennie

@TheMonk whereabouts are you in the educational system? Degree? PhD?

The Monk

@JohnRennie Undergrad

John Rennie

Which year?

The Monk

@JohnRennie 3rd (final) year

John Rennie

@TheMonk OK so the stress is starting to build up with finals approaching?

The Monk

07:18

Of course not @JohnRennie

Sorry I had to go out for some personal business.

And we don't have finals like Cambridge or so. It is semester system here.

John Rennie

07:34

@TheMonk so is it that you've just got bored with physics?

The Monk

@JohnRennie "Bored" is not the right word to describe it.

I'm still strongly attracted to physics. But I can't give my attention to it while I try to study.

My mind goes off the track.

John Rennie

Sounds to me like you haven't been getting enough sleep.

That's exactly the symptoms I get if I have been missing sleep.

The Monk

@JohnRennie Believe me, I am getting enough sleep. I am a heavy sleeper :D

I'm clueless now. Don't have any idea what to do.

John Rennie

How many hours a night do you sleep?

The Monk

@JohnRennie 8+ XD

John Rennie

07:41

OK, that should certainly be enough.

The Monk

Yeah. I know. I'm worried a lot.

I knew it was coming evatually. I started to feel these just after I started undergrad

The weird thing is that I never find physics hard.

John Rennie

07:54

I don't know then. I've certainly felt this way in the past, and I'm sure most of us have. But it's always been associated with being tired or maybe unwell (e.g. having a cold).

The Monk

Yeah, it's okay to experience that while being ill. But my case is strange. :(

Slereah

09:10

morning

slatestarcodex.com/2019/11/18/…

You are startled. “This room is locked,” you tell him. “And how did you know what I was doing abroad? Wait a second. Are you secretly the Devil?”

“Untestable, therefore irrelevant!” says Carroll. You wonder if he has always had bright orange eyes. “But being indifferent between ‘wavefunction branches’ and ‘wavefunction branches, and then somewhere we can’t see it one branch mysteriously collapses’ is the same kind of error as being indifferent between ‘acid and base make salt’ and ‘acid and base make salt and water, and then somewhere we can’t see it a star mysteriously goes supernova’.”

My sides

Captain Bohemian

10:14

several days when I sorted out my books, I found I had a photocopied book written by Carlo Rovelli.

is there is property of a matrix which is the indication that it is diagonizable and vice versa?

Slereah

Eigenvalues?

Captain Bohemian

may be right

Slereah

Solve for $$\det(M - I \lambda) = 0$$

Diagonalizability requires $n$ linearly-independent eigenvectors

Captain Bohemian

even if some of them have the same eigenvalues

Slereah

Well I mean $I$ has the same eigenvalues

But its eigenvectors are linearly independents

Captain Bohemian

10:32

I think by I you mean the identity matrix.

Slereah

Yes

That is the standard symbol for it

Captain Bohemian

@Slereah I has n the same eigenvalues 1, but do they have linearly independent eigenvectors?

Slereah

Sure

$(1,0,0)$, $(0,1,0)$ and $(0,0,1)$

those are very linearly independent

Captain Bohemian

10:58

I think of exactly the three ones.

if n=3

then is there a property of a matrix which indicates these eigenvalues are real and vice versa?

Captain Bohemian

11:18

I know a hermitian matrix must have real eigenvalues, but nonhermitian matrices may also have real eigenvalues.

Aaron Stevens

12:47

@tpg2114 physics.stackexchange.com/questions/514415/…

Do you mean if they fix the answer?

tpg2114

@AaronStevens What do you mean? We can undelete complete solutions at a later point also -- once the danger of turning in the complete solution for a grade passes.

Unless I'm misunderstanding your question

The Monk

14:26

3

Q: Where does the magnetic field energy of a charged particle moving with a uniform velocity come from?

Consider a charged particle initially at rest with respect to an inertial frame. Let a force act on it so that it gains a velocity 'v'. It now produces a magnetic field that has some energy associated with it. My question is where does this energy come from? If it comes from the work done b...

electromagnetism

I've started a bounty. Any explanation would be nice.

😊

Aaron Stevens

14:38

@tpg2114 That doesn't make sense to me. Then others could come looking for homework solutions and get answers. And it would be confusing to users who post solutions to new questions based on the solutions they see on other questions.

tpg2114

14:52

@AaronStevens The official policy is "temporarily deleted" though. Each mod has a different approach to how long is long enough before undeleting. Usually I don't go back unless somebody pings me about it a lot later to do it.

Most of the time, the person who posted the answer decides to either edit it so it isn't a full solution, in which case it can be undeleted immediately, or just let's it drop (and hopefully doesn't post more full solutions later)

Novalium Company

sup

Aaron Stevens

@tpg2114 Interesting. Thanks for the info

tpg2114

15:11

@AaronStevens No worries. I suppose nobody ever said "temporary" had to be any timescale less than the heat death of the universe... so I tend not to worry about undeleting things unless asked after awhile.

Secret

newscientist.com/article/…

That will require string theory to be true

autoevolution.com/news/…

-> fail

user8718165

@tpg2114 hello :) Everybody seems to talk about thermodynamics these days :)

Abhas Kumar Sinha

15:33

@Secret Is there any proof that $$|\psi|^2 \stackrel{def}{=} |e^{i S/\hbar}|^2 $$ is probability?

Secret

no, it's a postulate in quantum mechanics. you need to go beyond quantum to establish this

Abhas Kumar Sinha

@Secret For $$\psi (x, t) \stackrel{def}{=} u(x) e^{i \omega t} $$, then probability is $ u^2(x) $ and independent of $e^{i \omega t}$ na?

@Secret ..? Waiting till eternity....

Secret

the phase term will drop out after squaring as it becomes real

Abhas Kumar Sinha

@Secret ..? I wanted to ask what $e^{i \omega t}$ has to do here?

bolbteppa

@AbhasKumarSinha that's not how it's defined

Abhas Kumar Sinha

15:45

@bolbteppa ..?

bolbteppa

It's just $|\psi|^2$

$\psi \neq e^{iS/\hbar}$

Abhas Kumar Sinha

@bolbteppa approximately?

bolbteppa

Only in the classical approximation

Abhas Kumar Sinha

@bolbteppa if $\psi \neq e^{i S/\hbar}$ then why it is used to define momentum operator?

bolbteppa

You can use the fact that $\psi$ basically reduces to this form in the quasiclassical case

Abhas Kumar Sinha

15:52

@bolbteppa oh okay... So, there's no proof that it works actually in the quantum state and is meant to be assumed?

bolbteppa

No, they derive the operator of an infinitesimally small displacement and discuss the case where it commutes with the Hamiltonian and then explain how it relates to homogeneity of space (which is how classical momentum is determined via Noether's theorem) and then show this quantity reduces to the classical momentum in the quasiclassical limit so it is the quantum form of momentum, it is a proof

Abhas Kumar Sinha

@bolbteppa ah okay...

@bolbteppa one last question....

21 mins ago, by Abhas Kumar Sinha

@Secret For $$\psi (x, t) \stackrel{def}{=} u(x) e^{i \omega t} $$, then probability is $ u^2(x) $ and independent of $e^{i \omega t}$ na?

bolbteppa

Yes (probability density)

Abhas Kumar Sinha

@bolbteppa then for probability does $e^{i\omega t }$ make contribution?

bolbteppa

No, they explain that it is independent of time in section 10

yuvraj singh

16:01

@PM2Ring how are you.

Abhas Kumar Sinha

@bolbteppa if it has no contribution to probability, then why keep it?

@yuvrajsingh person, I think?

yuvraj singh

Typo bro typo.

John Rennie

@AbhasKumarSinha the time dependent term is important for understanding how states interact.

It has no effect on the probablility density because the moduus squared of $e^{i\omega t}$ is just unity.

Abhas Kumar Sinha

@JohnRennie I'm confused between probability and probability density, are both $|\psi|^2$?

John Rennie

$|\psi|^2$ is the probability density.

Abhas Kumar Sinha

16:11

@JohnRennie then $|\psi(q + dq)|^2$ is probability...?

John Rennie

You get probability by integrating $|\psi|^2 dV$

Abhas Kumar Sinha

@JohnRennie $\int |\psi|^2 dV$?

John Rennie

Yes, that integral over some volume $V$ gives the probability of finding the particle in the volume $V$.

Abhas Kumar Sinha

@JohnRennie okay, then still have no idea, what $ e^{i\omega t} $ has to do here? It will straight become one.

Ezze

@AbhasKumarSinha That's the point, it does not influence the probabilities

Abhas Kumar Sinha

16:15

@Ezze then why physicist want it here? (I too want that it doesn't influence probabilities...)

John Rennie

@AbhasKumarSinha have a look at Emilio's answer here.

That shows how the time dependent term is critical for understanding the interaction between states in transitions.

Abhas Kumar Sinha

@JohnRennie If I want to find the probability only, then taking $\psi (x) = \sin k \omega x$ for some constant $k$ should also work then?

John Rennie

Yes, if your wavefunction is a plane wave, though infinite plane waves can't be normalised so they aren't physical states.

Semiclassical

Ran into a signal-sampling issue lately in an intro lab which I'm wondering how to quantify

I've got a signal $x(t)=\cos(\alpha t^2/2)$ (x-coordinate of a point on the unit circle undergoing constant angular acceleration $\alpha$)

Abhas Kumar Sinha

@JohnRennie $\psi = ae^{i S/\hbar}$ then what $e^{iS/\hbar}$ has to do here?

John Rennie

16:22

What is $S$? The action?

Abhas Kumar Sinha

@JohnRennie yes

John Rennie

That looks more like a path integral. I haven't seen the action used in a regular wavefunction.

Semiclassical

I sample that signal at a rate of $\nu$ (i.e. samples per second)

$e^{i S/\hbar}$ is the typical ansatz in WKB

Abhas Kumar Sinha

@JohnRennie uh sorry, yes it's an approximation of path integral...

Semiclassical

(well, along with some prefactor stuff)

John Rennie

16:24

I don't know anything about path integrals I'm afraid.

Abhas Kumar Sinha

@JohnRennie oh okay...

@JohnRennie Thanks :-)

Ezze

@JohnRennie To me this looks like a Borh-Sommerfeld semiclassical approximation, not neccesarily a path integral

yeah, WKB, to be exact

Abhas Kumar Sinha

@Ezze both approximation are same with their first term under consideration

WKB

Semiclassical

anyways, because the intro labs are crude, one computes samples of $v_x=\dot{x}=-\alpha \sin (\alpha t^2/2)$ by doing $\Delta x/\Delta t$

Abhas Kumar Sinha

@Semiclassical You have any idea?

Semiclassical

16:27

I mean, the point of the ansatz is that it makes it possible to do approximations. It's not going to be something you derive rigorously

The logic is basically that of geometric optics if I remember right

Abhas Kumar Sinha

@Semiclassical yes

Ezze

Landau-Lifsitz has a whole chapter on semiclassical in Vol. 3, explaining the ansatz in detail

Abhas Kumar Sinha

@Ezze ch-2

I <3 L&L.

Semiclassical

Main point is that you shouldn't expect prob ~ |e^(i S/hbar)|^2 to be anything deep. It's the Born rule plus a choice of ansatz that's appropriate for the semiclassical limit

Abhas Kumar Sinha

semiclassical

Semiclassical

16:30

:)

Abhas Kumar Sinha

:)

I don't find Griffits conceptually clear. Its childish, (except the last part)

Ezze

@Semiclassical So what is funny about your lab results?

Semiclassical

16:46

Let me say it in a simpler way first. I've got a signal $z(t)=e^{i \alpha t^2/2}$ which I sample every $T$ seconds. From that, I numerically differentiate that as $v=\Delta z/\Delta t$ to get 'samples' of $z'(t)$.

I then compute the speed at each time as $|v|$. In principle, one should get $|z'(t)|=\alpha t$.

Abhas Kumar Sinha

@Semiclassical Use this $\dfrac{d^2 z}{dt^2} = v^2 \dfrac{d^2z}{dx^2}$ to compute velocity...

Semiclassical

However, since the angular velocity keeps increasing, it will eventually exceed the sampling rate $1/T$

Abhas Kumar Sinha

@Semiclassical uh okay, I finally understood your question... You need to learn python for that and use wave library...

Semiclassical

and hence one should not expect to accurately recover $|z'(t)|=\alpha t$ for long times

And, indeed, I find that the reconstructed speed as a function of time is not $\alpha t$. What's more interesting is that it seems to be characteristically $\alpha t-\beta t^2$

I'm trying to figure out how to express this error term in terms of $T$ and $\alpha$

I guess the right approach is this: the nth sample of $z$ will be $z_n=e^{i \alpha (n T)^2/2}$, so $v_n=\frac{z_{n+1}-z_n}{T}=\frac{1}{T}(e^{i \alpha (n+1) T^2/2}-(e^{i \alpha n^2 T^2/2})$

Abhas Kumar Sinha

@Semiclassical I actually need some coffee before I check some complicated stuff..

@Semiclassical works correct for $T=0$ (with a random sample) not sure for all $T$

bolbteppa

16:54

@AbhasKumarSinha The Schrodinger equation is $i \hbar \frac{\partial}{\partial t} \psi(x,t) = \hat{H} \psi(x,t)$ and one can show $\psi(x,t) = a(x) e^{i\omega t}$ sometimes, in which cases the time-dependence does not affect probabilities, it is not something people just want it's a derived result, and $\psi \neq a e^{iS/\hbar}$

Semiclassical

@bolbteppa I mean, you can always choose to write $\psi = e^{i S/\hbar}$ (same as you can write any complex number in polar form). It's just that that's not regarded as anything fundamental.

bolbteppa

Right, but $S$ wont be the classical action

Semiclassical

(Unless you're doing Bohm's formulation of pilot wave stuff wherein $\Psi = Re^{i S}$ is taken rather more seriously. But that's definitely not the usual line, and $S$ there won't be the classical action regardless.)

Abhas Kumar Sinha

@bolbteppa For classical action $S_0$ and other action $S_1, S_2. S_3, \dots$ (non classical ones) then $\psi = \sum \limits_{n=0} e^{i S_n/\hbar}$ is correct or not?

bolbteppa

Not correct

Well

Semiclassical

16:58

not in a way that's useful

bolbteppa

I mean you're just writing corrections in exponential form and ignoring modulus factors

Abhas Kumar Sinha

@bolbteppa modulus factors?

bolbteppa

$z = r e^{i\theta}$ where $r$ is the modulus $r = |z|$

Ezze

@Semiclassical I guess there's no way to tell how many periods have passed between the first two measurement points based on the signals alone. Maybe there is a way to fit the number of periods passed between two measurements? Damned if I remember anything about cross-correlations...

Abhas Kumar Sinha

@bolbteppa okay, then $\psi = \sum a e^{iS_n/\hbar}$ now correct? (assuming $S_n$ are all the possible posibilities)

Semiclassical

17:02

after doing some algebra, I seem to get $|v_n|=\frac{2}{T}\left|\sin\left(\frac{(2n+1)T^2 \alpha}{2\pi}\right)\right|$

bolbteppa

@AbhasKumarSinha not sure, you could probably write it that way sometimes

Semiclassical

Which for small $\alpha T$ is $|v_n|\approx \frac{2}{T}\frac{(2n+1)}{2\pi}\alpha T^2=\frac{2n+1}{\pi}\alpha T$

Which looks wrong. hrm

oh

hmm

Abhas Kumar Sinha

@bolbteppa okay...

Semiclassical

That -almost- makes sense.

one has $|v_n|\approx \frac{2+1/n}{\pi}\alpha (n T)\approx \frac{2}{\pi} \alpha t$

So if that $2/\pi$ weren't there I'd be perfectly happy.

Abhas Kumar Sinha

@bolbteppa he quantity whose conservation for a closed system follows from the homogeneity of space is called momentum ... It can be force also from this definition because it is independent of coordinates..

bolbteppa

17:08

@AbhasKumarSinha you need to read these things more carefully

Abhas Kumar Sinha

@bolbteppa ok........

@bolbteppa not containing time explicitly...?

@bolbteppa oh sorry got it.

It should also be related to $\hat H$

@bolbteppa But is there any proof for it? The usual definition of momentum is different...

Novalium Company

@AbhasKumarSinha How is AI going?

Abhas Kumar Sinha

@NovaliumCompany great buddy... :)

bolbteppa

@AbhasKumarSinha you need to read these things more carefully, they even reference book 1 where they define momentum in the same way they are using it in book 3

Semiclassical

oh, found my error. just some carelessness

so it should be $|v_n|=\frac{2}{T}|\sin((n+1/2)\alpha T^2/2)|$

Abhas Kumar Sinha

17:12

@bolbteppa I remember the definitions in book one it just changes the coordinates to $q + \delta q$ and tells them the RHS should be same hence cancells so the quantity under $\frac{d}{dt}$ is constant, hence it is momentum.... (as far I can recall)

bolbteppa

Go revise it and compare

Abhas Kumar Sinha

@bolbteppa okay sir :)

Semiclassical

which for small $n$ recovers $|v_n|\approx \alpha\cdot (n+1/2)T=\alpha t$

Abhas Kumar Sinha

@bolbteppa You must have read multiple books, how do you remember exact topics and section number associated with the book without forgetting?

Semiclassical

For larger $t$ (but not so large that $\sin$ reaches its max) one instead gets $$|v_n|\approx \frac{2}{T}((n+1/2)\alpha T^2/2-(n+1/2)^3\alpha^3 T^6/8)=\alpha\cdot (n+1/2)T-(n+1/2)^3 T^5 \alpha^2/4$$

aka $|v_n|\approx \alpha t-\frac{1}{4}T^3 \alpha^2 t^3$

So that's neat.

Abhas Kumar Sinha

17:18

@Semiclassical Good luck bro... Are these coming in your exams? XD LOL LAMO... Better leave these topics to cover at last...

Semiclassical

No, this is stuff my students were seeing in their experimental results (before I went back and made them take more samples) which I wanted to be able to explain.

Abhas Kumar Sinha

@Semiclassical Then good luck for your students... I hope these long equations won't cause physicsophobia.

Semiclassical

The lessons being 1) the reason they weren't seeing constant angular acceleration initially is because their sampling rate was too low, and 2) if you measure this system long enough, you'll inevitably run into these issues because the system keeps spinning faster and faster.

Abhas Kumar Sinha

@Semiclassical What sample? Speed of spin of washing machine? or quantum mechanics?

Semiclassical

more like the former

they attached a mass via a pulley and string to the edge of a heavy disk and used that to apply a constant torque

then recorded a video to determine the (x,y) coordinates of a point on the disk

Abhas Kumar Sinha

17:22

@Semiclassical Then your students be like - Sir, our answer is both correct and incorrect until you write grades

Semiclassical

from that, they got x,y (which can be conveniently represented as $z=x+i y=R e^{i\theta(t)}$

and then determined $v_x,v_y$ by numerical differentiation

I then had them plot the magnitude of the velocity and see if they recover $|v|=R\omega(t)$

Abhas Kumar Sinha

@Semiclassical Lesson 1 1> QM should be based on vectors, complex numbers are outdated...

Semiclassical

The point being that, initially, they didn't

and the reason being that they weren't sampling enough when the disk was spinning fast

Abhas Kumar Sinha

@Semiclassical okay, that's interesting...

@Semiclassical Radius of the points from the disk - $r$, then torque/radius = force, which is mg

Semiclassical

Not quite. For the disk you have $I\alpha=r T$ where $I$ is the moment of inertia, $\alpha$ is the angular acceleration, and $T$ is the string tension. For the hanging mass, you have $ma=mg-T$

and the angular acceleration is related to the acceleration of the string as $a=r\alpha$

So you end up with $a=\frac{mr^2}{I+mr^2}g$, and in principle you can use that to deduce $I$

Abhas Kumar Sinha

17:27

@Semiclassical again, I've thinking problems without coffee... XD

@Semiclassical ah sorry, I assumed disk as ring...

Semiclassical

no problem. there's a reason I was writing $I$ instead of saying what the moment of inertia was

main point is that you need to factor in the inertia of the hanging mass

Abhas Kumar Sinha

@Semiclassical yes yes...

@Semiclassical suppose you got the answer and friction be like - Am I a joke to you?

Semiclassical

lol

yeah

tbh, that's why I mostly focused on them relating $x,y,v_x,v_y,\theta$

because I don't have a lot of confidence that the angular acceleration, even if it's constant, can really be accounted for while ignoring friction

Abhas Kumar Sinha

@Semiclassical hmm... Modern Problems require Modern Solutions

Is this true $$ \dfrac{\partial}{\partial x} \int f(x, y, z) dt = \int \dfrac{\partial }{\partial q} f(x, y, z) dt $$?

@Semiclassical

sorry, it's $x$ not $q$

Novalium Company

18:13

Does the stock price of a company have any correlation to the people's interest in that company? For example, Tesla's stock is ~360$ per share, and Facebook is ~200$ per share. Does that mean that people demand Tesla's stock much more than Facebook? What insights can we get from a company's stock price?

Loong

@NovaliumCompany no

Novalium Company

@Loong I didn't take shares amount into consideration?

Loong

exactly

You are looking for "market capitalization".

Novalium Company

@Loong thx

When we say a CEO is worth 5 billion for example, does this mean he has the amount of money right away to spend and invest, or that 5 billion is simply the large percentage he owns of his company and therefore would have to sacrifice a part of his ownership to buy let's say a yacht?

Novalium Company

18:43

Also, does high market cap of a company mean that there's high demand for that company's stock? (I'm taking into consideration not only the price of a single stock but also the amount of shares)

And therefore the price of a share doesn't reveal anything reasonable. We must include number of shares, - market cap.

DanielSank

19:27

Hi, everybody.

Novalium Company

20:52

In genetic AI algorithms, how do you create a new generation of neural nets based on the previous one? How do you make the random fluctuations on the new generation? (I'm not programming anything, just theory)

Nvm I think my question doesn't make sense

fielder

22:11

Hello everyone, is anyone here in the process of graduate school applications?

alarge

@NovaliumCompany Well what you're describing is a genetic algorithm, and I don't see why you couldn't apply that to neural networks usually: i.e. you create a bunch of neural networks by applying random noise to the previous generation in some way and then you check which ones do best and select those to generate the next generation. Normally though in neural networks you update the weights by back propagation (basically algorithmic differentiation)

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