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12:23 AM
> It isn’t pretentious either since it gets bigger by feeding off of matter.
 
1:22 AM
Hallo, everbody
good morning
 
@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.
 
What 's the meaning of "population" here?
Does anyone know about it?
 
1:51 AM
Hello :) @Slereah , I want to ask a question.
 
 
1 hour later…
 
3 hours later…
6:45 AM
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.
 
@TheMonk whereabouts are you in the educational system? Degree? PhD?
 
@JohnRennie Undergrad
 
Which year?
 
@JohnRennie 3rd (final) year
 
@TheMonk OK so the stress is starting to build up with finals approaching?
 
7:18 AM
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.
 
7:34 AM
@TheMonk so is it that you've just got bored with physics?
 
@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.
 
Sounds to me like you haven't been getting enough sleep.
That's exactly the symptoms I get if I have been missing sleep.
 
@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.
 
How many hours a night do you sleep?
 
@JohnRennie 8+ XD
 
7:41 AM
OK, that should certainly be enough.
 
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.
 
7:54 AM
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).
 
Yeah, it's okay to experience that while being ill. But my case is strange. :(
 
 
1 hour later…
9:10 AM
morning
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
 
10:14 AM
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?
 
Eigenvalues?
 
may be right
 
Solve for $$\det(M - I \lambda) = 0$$
Diagonalizability requires $n$ linearly-independent eigenvectors
 
even if some of them have the same eigenvalues
 
Well I mean $I$ has the same eigenvalues
But its eigenvectors are linearly independents
 
10:32 AM
I think by I you mean the identity matrix.
 
Yes
That is the standard symbol for it
 
@Slereah I has n the same eigenvalues 1, but do they have linearly independent eigenvectors?
 
Sure
$(1,0,0)$, $(0,1,0)$ and $(0,0,1)$
those are very linearly independent
 
10:58 AM
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?
 
11:18 AM
I know a hermitian matrix must have real eigenvalues, but nonhermitian matrices may also have real eigenvalues.
 
 
1 hour later…
12:47 PM
Do you mean if they fix the answer?
 
@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
 
 
1 hour later…
2:26 PM
3
Q: Where does the magnetic field energy of a charged particle moving with a uniform velocity come from?

UserConsider 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...

I've started a bounty. Any explanation would be nice.
😊
 
2:38 PM
@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.
 
2:52 PM
@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)
 
@tpg2114 Interesting. Thanks for the info
 
3:11 PM
@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.
 
That will require string theory to be true
-> fail
 
@tpg2114 hello :) Everybody seems to talk about thermodynamics these days :)
 
3:33 PM
@Secret Is there any proof that $$|\psi|^2 \stackrel{def}{=} |e^{i S/\hbar}|^2 $$ is probability?
 
no, it's a postulate in quantum mechanics. you need to go beyond quantum to establish this
 
@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....
 
the phase term will drop out after squaring as it becomes real
 
@Secret ..? I wanted to ask what $e^{i \omega t}$ has to do here?
 
@AbhasKumarSinha that's not how it's defined
 
3:45 PM
@bolbteppa ..?
 
It's just $|\psi|^2$
$\psi \neq e^{iS/\hbar}$
 
@bolbteppa approximately?
 
Only in the classical approximation
 
@bolbteppa if $\psi \neq e^{i S/\hbar}$ then why it is used to define momentum operator?
 
You can use the fact that $\psi$ basically reduces to this form in the quasiclassical case
 
3:52 PM
@bolbteppa oh okay... So, there's no proof that it works actually in the quantum state and is meant to be assumed?
 
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
 
@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?
 
Yes (probability density)
 
@bolbteppa then for probability does $e^{i\omega t }$ make contribution?
 
No, they explain that it is independent of time in section 10
 
4:01 PM
@PM2Ring how are you.
 
@bolbteppa if it has no contribution to probability, then why keep it?
@yuvrajsingh person, I think?
 
Typo bro typo.
 
@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.
 
@JohnRennie I'm confused between probability and probability density, are both $|\psi|^2$?
 
$|\psi|^2$ is the probability density.
 
4:11 PM
@JohnRennie then $|\psi(q + dq)|^2$ is probability...?
 
You get probability by integrating $|\psi|^2 dV$
 
@JohnRennie $\int |\psi|^2 dV$?
 
Yes, that integral over some volume $V$ gives the probability of finding the particle in the volume $V$.
 
@JohnRennie okay, then still have no idea, what $ e^{i\omega t} $ has to do here? It will straight become one.
 
@AbhasKumarSinha That's the point, it does not influence the probabilities
 
4:15 PM
@Ezze then why physicist want it here? (I too want that it doesn't influence probabilities...)
 
@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.
 
@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?
 
Yes, if your wavefunction is a plane wave, though infinite plane waves can't be normalised so they aren't physical states.
 
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 $\alpha$)
 
@JohnRennie $\psi = ae^{i S/\hbar}$ then what $e^{iS/\hbar}$ has to do here?
 
4:22 PM
What is $S$? The action?
 
@JohnRennie yes
 
That looks more like a path integral. I haven't seen the action used in a regular wavefunction.
 
I sample that signal at a rate of $\nu$ (i.e. samples per second)
$e^{i S/\hbar}$ is the typical ansatz in WKB
 
@JohnRennie uh sorry, yes it's an approximation of path integral...
 
(well, along with some prefactor stuff)
 
4:24 PM
I don't know anything about path integrals I'm afraid.
 
@JohnRennie oh okay...
@JohnRennie Thanks :-)
 
@JohnRennie To me this looks like a Borh-Sommerfeld semiclassical approximation, not neccesarily a path integral
yeah, WKB, to be exact
 
@Ezze both approximation are same with their first term under consideration
WKB
 
anyways, because the intro labs are crude, one computes samples of $v_x=\dot{x}=-\alpha \sin (\alpha t^2/2)$ by doing $\Delta x/\Delta t$
 
@Semiclassical You have any idea?
 
4:27 PM
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
 
@Semiclassical yes
 
Landau-Lifsitz has a whole chapter on semiclassical in Vol. 3, explaining the ansatz in detail
 
@Ezze ch-2
I <3 L&L.
 
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
 
semiclassical
 
4:30 PM
:)
 
:)
I don't find Griffits conceptually clear. Its childish, (except the last part)
 
@Semiclassical So what is funny about your lab results?
 
4:46 PM
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 $|z'(t)|=\alpha t$.
 
@Semiclassical Use this $\dfrac{d^2 z}{dt^2} = v^2 \dfrac{d^2z}{dx^2}$ to compute velocity...
 
However, since the angular velocity keeps increasing, it will eventually exceed the sampling rate $1/T$
 
@Semiclassical uh okay, I finally understood your question... You need to learn python for that and use wave library...
 
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 $v_n=\frac{z_{n+1}-z_n}{T}=\frac{1}{T}(e^{i \alpha (n+1) T^2/2}-(e^{i \alpha n^2 T^2/2})$
 
@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 $T$
 
4:54 PM
@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 $\psi \neq a e^{iS/\hbar}$
 
@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.
 
Right, but $S$ wont be the classical action
 
(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.)
 
@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?
 
Not correct
Well
 
4:58 PM
not in a way that's useful
 
I mean you're just writing corrections in exponential form and ignoring modulus factors
 
@bolbteppa modulus factors?
 
$z = r e^{i\theta}$ where $r$ is the modulus $r = |z|$
 
@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...
 
@bolbteppa okay, then $\psi = \sum a e^{iS_n/\hbar}$ now correct? (assuming $S_n$ are all the possible posibilities)
 
5:02 PM
after doing some algebra, I seem to get $|v_n|=\frac{2}{T}\left|\sin\left(\frac{(2n+1)T^2 \alpha}{2\pi}\right)\right|$
 
@AbhasKumarSinha not sure, you could probably write it that way sometimes
 
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
 
@bolbteppa okay...
 
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.
 
@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..
 
5:08 PM
@AbhasKumarSinha you need to read these things more carefully
 
@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...
 
@AbhasKumarSinha How is AI going?
 
@NovaliumCompany great buddy... :)
 
@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
 
oh, found my error. just some carelessness
so it should be $|v_n|=\frac{2}{T}|\sin((n+1/2)\alpha T^2/2)|$
 
5:12 PM
@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)
 
Go revise it and compare
 
@bolbteppa okay sir :)
 
which for small $n$ recovers $|v_n|\approx \alpha\cdot (n+1/2)T=\alpha t$
 
@bolbteppa You must have read multiple books, how do you remember exact topics and section number associated with the book without forgetting?
 
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.
 
5:18 PM
@Semiclassical Good luck bro... Are these coming in your exams? XD LOL LAMO... Better leave these topics to cover at last...
 
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.
 
@Semiclassical Then good luck for your students... I hope these long equations won't cause physicsophobia.
 
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.
 
@Semiclassical What sample? Speed of spin of washing machine? or quantum mechanics?
 
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
 
5:22 PM
@Semiclassical Then your students be like - Sir, our answer is both correct and incorrect until you write grades
 
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 $|v|=R\omega(t)$
 
@Semiclassical Lesson 1 1> QM should be based on vectors, complex numbers are outdated...
 
The point being that, initially, they didn't
and the reason being that they weren't sampling enough when the disk was spinning fast
 
@Semiclassical okay, that's interesting...
@Semiclassical Radius of the points from the disk - $r$, then torque/radius = force, which is mg
 
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 $I$
 
5:27 PM
@Semiclassical again, I've thinking problems without coffee... XD
@Semiclassical ah sorry, I assumed disk as ring...
 
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
 
@Semiclassical yes yes...
@Semiclassical suppose you got the answer and friction be like - Am I a joke to you?
 
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
 
@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 $q$
 
6:13 PM
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?
 
@NovaliumCompany no
 
@Loong I didn't take shares amount into consideration?
 
exactly
You are looking for "market capitalization".
 
@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?
 
6:43 PM
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.
 
7:27 PM
Hi, everybody.
 
 
1 hour later…
8:52 PM
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
 
 
1 hour later…
10:11 PM
Hello everyone, is anyone here in the process of graduate school applications?
 
@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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