« first day (3302 days earlier)      last day (1610 days later) » 

Rishi
Rishi
12:23 AM
> It isn’t pretentious either since it gets bigger by feeding off of matter.
 
Wang Yun
Wang Yun
1:22 AM
Hallo, everbody
good morning
 
Scientist Smith YT
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
Wang Yun
What 's the meaning of "population" here?
Does anyone know about it?
 
user8718165
user8718165
1:51 AM
Hello :) @Slereah , I want to ask a question.
 
 
1 hour later…
JMac
JMac
3:19 AM
 
 
3 hours later…
The Monk
The Monk
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.
 
John Rennie
John Rennie
@TheMonk whereabouts are you in the educational system? Degree? PhD?
 
The Monk
The Monk
@JohnRennie Undergrad
 
John Rennie
John Rennie
Which year?
 
The Monk
The Monk
@JohnRennie 3rd (final) year
 
John Rennie
John Rennie
@TheMonk OK so the stress is starting to build up with finals approaching?
 
The Monk
The Monk
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.
 
John Rennie
John Rennie
7:34 AM
@TheMonk so is it that you've just got bored with physics?
 
The Monk
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
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
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
John Rennie
How many hours a night do you sleep?
 
The Monk
The Monk
@JohnRennie 8+ XD
 
John Rennie
John Rennie
7:41 AM
OK, that should certainly be enough.
 
The Monk
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
John Rennie
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).
 
The Monk
The Monk
Yeah, it's okay to experience that while being ill. But my case is strange. :(
 
 
1 hour later…
Slereah
Slereah
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
 
Captain Bohemian
Captain Bohemian
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?
 
Slereah
Slereah
Eigenvalues?
 
Captain Bohemian
Captain Bohemian
may be right
 
Slereah
Slereah
Solve for $$\det(M - I \lambda) = 0$$
Diagonalizability requires $n$ linearly-independent eigenvectors
 
Captain Bohemian
Captain Bohemian
even if some of them have the same eigenvalues
 
Slereah
Slereah
Well I mean $I$ has the same eigenvalues
But its eigenvectors are linearly independents
 
Captain Bohemian
Captain Bohemian
10:32 AM
I think by I you mean the identity matrix.
 
Slereah
Slereah
Yes
That is the standard symbol for it
 
Captain Bohemian
Captain Bohemian
@Slereah I has n the same eigenvalues 1, but do they have linearly independent eigenvectors?
 
Slereah
Slereah
Sure
$(1,0,0)$, $(0,1,0)$ and $(0,0,1)$
those are very linearly independent
 
Captain Bohemian
Captain Bohemian
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?
 
Captain Bohemian
Captain Bohemian
11:18 AM
I know a hermitian matrix must have real eigenvalues, but nonhermitian matrices may also have real eigenvalues.
 
 
1 hour later…
Aaron Stevens
Aaron Stevens
12:47 PM
Do you mean if they fix the answer?
 
tpg2114
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
 
 
1 hour later…
The Monk
The Monk
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...

electromagnetism
I've started a bounty. Any explanation would be nice.
😊
 
Aaron Stevens
Aaron Stevens
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.
 
tpg2114
tpg2114
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)
 
Novalium Company
Novalium Company
sup
 
Aaron Stevens
Aaron Stevens
@tpg2114 Interesting. Thanks for the info
 
tpg2114
tpg2114
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.
 
Secret
Secret
That will require string theory to be true
-> fail
 
user8718165
user8718165
@tpg2114 hello :) Everybody seems to talk about thermodynamics these days :)
 
Abhas Kumar Sinha
Abhas Kumar Sinha
3:33 PM
@Secret Is there any proof that $$|\psi|^2 \stackrel{def}{=} |e^{i S/\hbar}|^2 $$ is probability?
 
Secret
Secret
no, it's a postulate in quantum mechanics. you need to go beyond quantum to establish this
 
Abhas Kumar Sinha
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
Secret
the phase term will drop out after squaring as it becomes real
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@Secret ..? I wanted to ask what $e^{i \omega t}$ has to do here?
 
bolbteppa
bolbteppa
@AbhasKumarSinha that's not how it's defined
 
Abhas Kumar Sinha
Abhas Kumar Sinha
3:45 PM
@bolbteppa ..?
 
bolbteppa
bolbteppa
It's just $|\psi|^2$
$\psi \neq e^{iS/\hbar}$
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa approximately?
 
bolbteppa
bolbteppa
Only in the classical approximation
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa if $\psi \neq e^{i S/\hbar}$ then why it is used to define momentum operator?
 
bolbteppa
bolbteppa
You can use the fact that $\psi$ basically reduces to this form in the quasiclassical case
 
Abhas Kumar Sinha
Abhas Kumar Sinha
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?
 
bolbteppa
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
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
bolbteppa
Yes (probability density)
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa then for probability does $e^{i\omega t }$ make contribution?
 
bolbteppa
bolbteppa
No, they explain that it is independent of time in section 10
 
yuvraj singh
yuvraj singh
4:01 PM
@PM2Ring how are you.
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa if it has no contribution to probability, then why keep it?
@yuvrajsingh person, I think?
 
yuvraj singh
yuvraj singh
Typo bro typo.
 
John Rennie
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
Abhas Kumar Sinha
@JohnRennie I'm confused between probability and probability density, are both $|\psi|^2$?
 
John Rennie
John Rennie
$|\psi|^2$ is the probability density.
 
Abhas Kumar Sinha
Abhas Kumar Sinha
4:11 PM
@JohnRennie then $|\psi(q + dq)|^2$ is probability...?
 
John Rennie
John Rennie
You get probability by integrating $|\psi|^2 dV$
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@JohnRennie $\int |\psi|^2 dV$?
 
John Rennie
John Rennie
Yes, that integral over some volume $V$ gives the probability of finding the particle in the volume $V$.
 
Abhas Kumar Sinha
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
Ezze
@AbhasKumarSinha That's the point, it does not influence the probabilities
 
Abhas Kumar Sinha
Abhas Kumar Sinha
4:15 PM
@Ezze then why physicist want it here? (I too want that it doesn't influence probabilities...)
 
John Rennie
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
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
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
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
Abhas Kumar Sinha
@JohnRennie $\psi = ae^{i S/\hbar}$ then what $e^{iS/\hbar}$ has to do here?
 
John Rennie
John Rennie
4:22 PM
What is $S$? The action?
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@JohnRennie yes
 
John Rennie
John Rennie
That looks more like a path integral. I haven't seen the action used in a regular wavefunction.
 
Semiclassical
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
Abhas Kumar Sinha
@JohnRennie uh sorry, yes it's an approximation of path integral...
 
Semiclassical
Semiclassical
(well, along with some prefactor stuff)
 
John Rennie
John Rennie
4:24 PM
I don't know anything about path integrals I'm afraid.
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@JohnRennie oh okay...
@JohnRennie Thanks :-)
 
Ezze
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
Abhas Kumar Sinha
@Ezze both approximation are same with their first term under consideration
WKB
 
Semiclassical
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
Abhas Kumar Sinha
@Semiclassical You have any idea?
 
Semiclassical
Semiclassical
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
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@Semiclassical yes
 
Ezze
Ezze
Landau-Lifsitz has a whole chapter on semiclassical in Vol. 3, explaining the ansatz in detail
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@Ezze ch-2
I <3 L&L.
 
Semiclassical
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
Abhas Kumar Sinha
semiclassical
 
Semiclassical
Semiclassical
4:30 PM
:)
 
Abhas Kumar Sinha
Abhas Kumar Sinha
:)
I don't find Griffits conceptually clear. Its childish, (except the last part)
 
Ezze
Ezze
@Semiclassical So what is funny about your lab results?
 
Semiclassical
Semiclassical
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 $|z'(t)|=\alpha t$.
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@Semiclassical Use this $\dfrac{d^2 z}{dt^2} = v^2 \dfrac{d^2z}{dx^2}$ to compute velocity...
 
Semiclassical
Semiclassical
However, since the angular velocity keeps increasing, it will eventually exceed the sampling rate $1/T$
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@Semiclassical uh okay, I finally understood your question... You need to learn python for that and use wave library...
 
Semiclassical
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
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
bolbteppa
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 $\psi \neq a e^{iS/\hbar}$
 
Semiclassical
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
bolbteppa
Right, but $S$ wont be the classical action
 
Semiclassical
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
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
bolbteppa
Not correct
Well
 
Semiclassical
Semiclassical
4:58 PM
not in a way that's useful
 
bolbteppa
bolbteppa
I mean you're just writing corrections in exponential form and ignoring modulus factors
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa modulus factors?
 
bolbteppa
bolbteppa
$z = r e^{i\theta}$ where $r$ is the modulus $r = |z|$
 
Ezze
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
Abhas Kumar Sinha
@bolbteppa okay, then $\psi = \sum a e^{iS_n/\hbar}$ now correct? (assuming $S_n$ are all the possible posibilities)
 
Semiclassical
Semiclassical
5:02 PM
after doing some algebra, I seem to get $|v_n|=\frac{2}{T}\left|\sin\left(\frac{(2n+1)T^2 \alpha}{2\pi}\right)\right|$
 
bolbteppa
bolbteppa
@AbhasKumarSinha not sure, you could probably write it that way sometimes
 
Semiclassical
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
Abhas Kumar Sinha
@bolbteppa okay...
 
Semiclassical
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
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
bolbteppa
5:08 PM
@AbhasKumarSinha you need to read these things more carefully
 
Abhas Kumar Sinha
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
Novalium Company
@AbhasKumarSinha How is AI going?
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@NovaliumCompany great buddy... :)
 
bolbteppa
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
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
Abhas Kumar Sinha
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)
 
bolbteppa
bolbteppa
Go revise it and compare
 
Abhas Kumar Sinha
Abhas Kumar Sinha
@bolbteppa okay sir :)
 
Semiclassical
Semiclassical
which for small $n$ recovers $|v_n|\approx \alpha\cdot (n+1/2)T=\alpha t$
 
Abhas Kumar Sinha
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
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
Abhas Kumar Sinha
5:18 PM
@Semiclassical Good luck bro... Are these coming in your exams? XD LOL LAMO... Better leave these topics to cover at last...
 
Semiclassical
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
Abhas Kumar Sinha
@Semiclassical Then good luck for your students... I hope these long equations won't cause physicsophobia.
 
Semiclassical
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
Abhas Kumar Sinha
@Semiclassical What sample? Speed of spin of washing machine? or quantum mechanics?
 
Semiclassical
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
Abhas Kumar Sinha
5:22 PM
@Semiclassical Then your students be like - Sir, our answer is both correct and incorrect until you write grades
 
Semiclassical
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
Abhas Kumar Sinha
@Semiclassical Lesson 1 1> QM should be based on vectors, complex numbers are outdated...
 
Semiclassical
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
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
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
Abhas Kumar Sinha
5:27 PM
@Semiclassical again, I've thinking problems without coffee... XD
@Semiclassical ah sorry, I assumed disk as ring...
 
Semiclassical
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
Abhas Kumar Sinha
@Semiclassical yes yes...
@Semiclassical suppose you got the answer and friction be like - Am I a joke to you?
 
Semiclassical
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
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
Novalium Company
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?
 
Loong
Loong
@NovaliumCompany no
 
Novalium Company
Novalium Company
@Loong I didn't take shares amount into consideration?
 
Loong
Loong
exactly
You are looking for "market capitalization".
 
Novalium Company
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
Novalium Company
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.
 
DanielSank
DanielSank
7:27 PM
Hi, everybody.
 
 
1 hour later…
Novalium Company
Novalium Company
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…
fielder
fielder
10:11 PM
Hello everyone, is anyone here in the process of graduate school applications?
 
alarge
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)
 

« first day (3302 days earlier)      last day (1610 days later) »