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12:59 AM
Howdy
 
user218912
did you finish calc?
 
I'm trying to make a mental model for rc circuits. Could someone please explain what happens with charges at 0, pi/2 pi, and 3pi/2 in a parallel cap vs series cap?
@obe I did I and II but this is not math.se, so try to keep it brief
 
user218912
what?
 
@obe do you have a calc question?
 
user218912
no.
 
1:31 AM
@heather Answered via email.
@JoeStavitsky Huh?
What do you mean "charges at 0, pi/2..."?
 
@DanielSank, as in, series vs parallel rc circuit and charges stored in cap cs sinusoidal source
 
1:48 AM
I waited in line 2 hours for this
An autograph from Neil deGrasse Tyson on my favorite book
 
@SirCumference good on him for signing autographs for 2 hours
 
@JoeStavitsky I know right
I was scared he'd leave because his hand was on fire
 
Especially ppl like that who actually work for a living, he could have easily declared that shit a waste of his time
 
@JoeStavitsky oh
So what's your question?
You have a parallel or series LC...
then what?
 
@DanielSank, thanks I got it pretty much straightened out elsewhere. I was under the mistaken impression that different circuit configuration cased different capacitor behavior
 
user218912
2:04 AM
@SirCumference astrophysics black guy?
 
@obe Yep
 
user218912
you met him in person?
 
@obe Yep
Asked him and two other astrophysics professors with him if an event horizon could ever exist, since the apparent horizon would take an infinite amount of time to reach that point
He thought I was talking about gravitational time dilation, told me that objects would move slower and slower towards an event horizon, and moved on to the next question...
Dammit!
 
user218912
because they don't actually know real physics.
 
user218912
they're science popstars.
 
2:10 AM
@obe I'd thought that too. But the book they were handing out dealt with real GR, calculus and higher mathematics
Written by them
Their lecture touched on some pretty deep stuff about cosmology
Two of them were professors at princeton
 
user218912
yes
 
I'd thought they would've been able to answer...
 
user218912
the profs did all the work.
 
user218912
@SirCumference they didn't? D:
 
@obe One actually was a serious general relativist who worked on developing some cosmology model, and even he misunderstood the same question...
Goddammit all!
 
user218912
2:12 AM
maybe you asked it bad.
 
I even said I was talking about the Oppenheimer-Snyder metric
I just asked if a black hole could exist in reality, since according to the OS metric, an apparent horizon would take an infinite time to reach the radius of an event horizon
Essentially word for word what I wrote ^
They thought I was talking about gravitational time dilation increasing to infinity as you approach a black hole...
 
user218912
idk much GR, is that an open problem?
 
rob
Offering technical answers to technical answers at a non-technical talk is always a gamble. Audience members fall into three groups:
 
@obe Not as far as I know
 
rob
(1) science fans who want pop-science explanations (2) crackpots who want to take over your talk with their bizarre pet theory (3) actual scientists who don't have an avenue to introduce themselves
 
user218912
2:16 AM
yeah
 
rob
A pop-science answer will satisfy (1), probably shut down (2), and leave (3) quietly disappointed.
 
Neil did actually read The Realm of the Nebulae, by Edwin Hubble, so he knew about the book I asked him to sign
 
rob
A technical answer will confuse (1), encourage (2), and provoke thoughtful follow-up questions from (3)
 
user218912
where does sircumference fit?
 
2:17 AM
.-.
 
rob
So basically offering complex technical answers is a high-risk, low-reward venture at a public talk.
 
;-;
Although they were talking about some advanced cosmology concepts without defining them beforehand
 
user218912
(4) fact dude who wants a technical fact
 
I assumed they would understand my question
 
rob
@obe I think that @SirCumference is a solid (3)
 
2:19 AM
@rob :)
 
user218912
yes he's an actual scientist.
 
@obe A brilliant one, mind you
 
I posted an answer on Academia yesterday (my time) which a garnered more than 100 upvotes in less than 24 hours (which seems to be a thing on that site, popular answer get huge numbers of votes even though the site doesn't seem to have a participating user base of the scale that supports it).
 
user218912
I should just leave science forever because @SirCumference is destroying all the unanswered questions.
 
user218912
there won't be any left for me or others.
 
2:20 AM
But the amusing part is that due to the timing I got circa 300 rep out of it. The tiing of the UTC day (used for the rep cap) was about as bad as it could have been.
 
@obe You can get my autograph if you want :)
 
user218912
what will happen when @SirCumference discovers the final fact about the universe?
 
rob
@obe A typical answer begets two new questions ... exponential growth is a helluva thing
 
I wouldn't care, except that I've been waiting for months for my rep on academia to exceed what I have on superuser, and I'm still a few hundred short. ::sigh::
 
@obe "the final fact"?
 
user218912
2:23 AM
yes
 
Not to sound nihilistic, I don't think I'll live that long
But anyways, meeting Neil deGrasse Tyson was definitely worth the 2 hours
 
user218912
2:39 AM
@SirCumference based off the fact that you met him today I can guess which university you're at.
 
@obe Go ahead, but you're probably wrong :3
I took the subway for a while to meet him
 
user218912
columbia?
 
user218912
I think 0celo7 guessed that before
 
user218912
it's probably right
 
user218912
xD
 
2:54 AM
I'm stupid
But also very clever at the same time
I worked out the path integral formalism in detail in high school
So now I can copy and paste my old work onto my QM homework :)
I might actually be a genius
Then the integral evaluates to
$$e^{-i\delta tV(q_j)}\int\frac{dp}{2\pi}e^{-i\delta t(p^2/2m)+ip(q_{j+1}-q_j)}=\sqrt{-\frac{im}{2\pi\delta t}}e^{[im(q_{j+1}-q_j)^2]/2\delta t-i\delta tV(q_j)}=Ce^{i\delta t\{(m/2)[(q_{j+1}-q_j)/\delta t]^2-V(q_j)\}}$$
yass
 
user218912
omg that's a hard integral
 
3:10 AM
Zee might not be so god-awful after all.
 
user218912
wtf how do you even do that integral
 
user218912
if i had that in my homework i'd be fucked
 
I did it in high school lol
If you read Zee you'd know.
 
user218912
he does it?
 
of course
Zee, Shankar and Lee contain basically all relevant math and physics.
 
user228700
3:12 AM
@0celo7 Now who's doing PhD math?
 
And HE I guess.
You, @Kaumudi
 
user218912
this isn't ph.d math @Kaumudi
 
user218912
@0celo7 you said you didn't take notes
 
@obe I lied.
 
user218912
wow
 
user218912
3:14 AM
and here I was thinking I should not take notes too
 
I didn't take notes
I thought for some stupid reason my AP stat teacher would want to see a nice intro to QM
so I wrote one
 
user218912
why would they want to see that?
 
I was young and stupid
don't ask me now
 
user116211
What sort of answer is it:
 
user116211
0
A: Observer Effect : Is it THE proof that reality is not real, but something we choose to make?

A.AlimI think the mistake lies not on whether physical reality is an illusion but on our persistent ignorance of thinking that our limited concepts are well enough to cover our perception . It's deeply mistaken since Newtonian physics since ignore our existence and measurement is a important part of p...

 
user116211
3:17 AM
?
 
user218912
@0celo7 send me them please?
 
@obe no, they are for me alone
 
user116211
0
Q: How does space between ends of a wormhole work?

VaesperI asked this on the worldbuilding stackexchange but realise it may just be a straight physics question. My understanding of wormholes consists basically of an extrapolation of the way I understand a flatlander's experience of the following illustration of an Ellis wormhole: Right now, there ar...

 
user218912
:'(
 
user116211
Too broad?
 
user116211
3:19 AM
-1
Q: Radio waves bouncing off of the firmament

Jerry KobylskiWouldn't you think that once people realize that the earth is not a globe but flat with firmament above,it would start to make more sense?

 
user116211
Earth is not a globe...
 
rob
3:49 AM
@MAFIA36790 Supposedly, in the 1960s, the president of the Flat Earth Society was shown the photo of Earth as a globe taken from the first Apollo mission to leave low-Earth orbit. The flat-Earther was dumbfounded for a minute, then began "It's easy to see how a photograph like this could fool the untrained eye ..."
 
user218912
lol
 
user116211
And they are born-PhD trained.
 
user228700
@0celo7 Sure sure.
 
user228700
@obe OK, whatever u say.
 
user218912
4:04 AM
My life sucks.
 
why?
 
user218912
I have no friends.
 
user116211
You don't need any.
 
user218912
(in real life)
 
ffs @MAFIA36790
 
user218912
4:05 AM
yes I do.
 
user116211
._.
 
user218912
I wish I knew you guys irl.
 
user218912
like the h bar was a real physics lounge.
 
rob
@obe What's local time for you?
 
user218912
4:08 AM
12:07am
 
@rob are you in Europe?
 
rob
@obe If you're depressed and alone in the middle of the night, get off the computer and go to sleep.
 
ratio of periods ought to be sqrt( Iyrx / Ixry ) from pendulum period eqn. setup yields Ix = 128 Iy, rx = 4*ry, or am i misinterpreting?
 
user218912
@rob but you guys are my resolve :(
 
woah, that looks gross. nevermind
 
user116211
4:10 AM
I meet more or less daily with my classmates; but I hardly interact with them.
 
user218912
@MAFIA36790 same
 
rob
 
user116211
Morning @JohnRennie.
 
rob
@dm__ ?
 
yes! man, that's awesome. how'd you do that?
anyways
T = 2pi*sqrt(I / mgr)
take the ratio and you get the sqrt i mentioned above
except i meant Iy = 64 Ix
getting the ratio for (C), not (B)
 
rob
4:13 AM
$I$ is moment of inertia?
 
Morning :-)
 
yeah, of the hoop through the pivot, which would be 2Mr^2
M -> 4M and r -> 4r makes that 64x as big
morning!
 
rob
@dm__ Okay ... how did you get a factor of four between periods?
 
with the r -> 4r included. so, if rx and ry denote the distances to their centers of mass, the ratio would be Ty/Tx = sqrt( Iy/Ix * rx/ry) = sqrt( 1/64 * 4) = 1/4
and above i meant Ix = 64 Iy, sorry!
 
rob
@dm__ Hmmm. I start from a different place.
For each hoop we know $M$, $R$, motion governed by $g$. Dimensional analysis says $T \propto \sqrt{R/g}$, with the same constant for both hoops.
Smaller pendulum is faster by $\sqrt{R_{small}/R_{big}}$
 
4:24 AM
so Ty/Tx = sqrt ( Ry / Rx) = sqrt(1/4) = 1/2, and i see that, it is beautifully simple, but it is killing me why i can't see what is wrong with the moments
 
rob
Your way should also work if you're doing it right
But you'll never finish the GRE if you insist on solving all the problems in detail
And as much as that seems like a stupid test-taking cop-out ...
 
Hi, everybody.
 
haha, ahh, I know, but it was very quick to setup, and now, with all the time in the world, i still can't see the error
 
rob
... when you're faced with a novel problem in the real world, you want to first come up with some ballpark estimate before you buckle down and get lost in details.
 
user116211
@DanielSank o/
 
4:26 AM
\o
@rob My freshman year physics prof said "Never do a calculation until you know the answer".
2
 
i'll definitely go with the simple for a similar problem like that. but still, the missing error will haunt me in my dreams
thanks for taking a look!
 
rob
@dm__ You'll solve it in the shower over the weekend. Move on.
 
everything is solved in the shower.
 
@DanielSank that doesn't even make sense.
 
@0celo7 Ok, well, call Steve Girvin and tell him.
 
rob
4:31 AM
19 mins ago, by dm__
T = 2pi*sqrt(I / mgr)
@dm__ If $I \sim mr^2$, this statement is also $T \sim \sqrt{r/g}$
13 mins ago, by dm__
with the r -> 4r included. so, if rx and ry denote the distances to their centers of mass, the ratio would be Ty/Tx = sqrt( Iy/Ix * rx/ry) = sqrt( 1/64 * 4) = 1/4
In your square root here, you're missing a ratio of the masses.
 
ah ha!
and that's the 4 i ned.
need*
thank you sir
but, back to the simple, shoulda canceled things to begin with
 
rob
It was interesting to scan through the GRE after you asked me a question from it the other day
Something like 3/4 of the questions I would solve today just with dimensional analysis.
 
haha yeah, i love those problems. still getting used to looking at things like that, so much quicker
 
@rob That's how I was taught to attack it.
You can usually throw out three of the five answer choices via dimensional analysis and thinking about limits when something goes to zero or infinity.
 
@Obliv Martin's answer to that question is rather misleading. The light interacts with the electrons in the medium to form an entangled system, so what is propagating is not just light but a mixture of light and electrons.
When the coupling is strong e.g. in a Bose Einstein condensate the system forms massive quasiparticles called polaritons and being massive these propagate slower than $c$. Under normal circumstances (weak coupling) the polariton description isn't appropriate but it's the same physics that causes the slowdown.
 
rob
4:51 AM
Argh, cheating students. First parts (a,b) of the exam question way wrong and have no erasures; later part (d), heavily erased, uses results from (a,b) correct to four significant figures
... to get (d) wrong, even.
 
@JohnRennie To add to this: slower than $c$ travel of light in a medium happens in classical physics too. We don't need entanglement.
 
@DanielSank yes, the effect can be calculated using classical electrodynamics. Essentially the system is a driven oscillator so the response has a phase lag. The EM reradiated by the oscillating electron density interferes with the original wave and causes the reduced speed.
 
user228700
Hihi!
 
@rob can we unpin your comment from the star board now that the flagging issue seems to have gone away?
 
rob
@JohnRennie I think that's fine
 
user228700
4:59 AM
I tried calculating the COM of a hollow hemisphere:
 
Thanks. I get a bit OCD about keeping the star board relevant :-)
 
user228700
 
user228700
I'm off by a factor of 3. The answer is supposed to be R/2 :/
 
rob
@JohnRennie How is that?
 
user228700
(Yes, I avoided the possible double integral)
 
5:01 AM
@rob :-)
NOW UNPIN IT!!!!
@Kaumudi the area of the element is something like: $$ dA = rd\theta r\sin\theta d\phi$$ where $\phi$ is the equatorial angle.
Or do I mean $\cos\theta$?
 
user228700
Nooo.
 
It's actually far simpler than it appears
 
user228700
I tried to avoid polar coordinates :-P
 
Ak OK
But I think your area is wrong.
 
user228700
Hm. I took the rectangle...
 
user228700
5:07 AM
With $dy$ breadth and circumference of circle as length.
 
Yes, but the circular hoop isn't vertical so its area isn't $dy \times 2\pi r$. You need to replace $dy$ by $rd\theta$.
 
user228700
Oh, crap. $2\pi r$!
 
You're doing this the hard way. In polar coordinates it's a lot easier ...
 
user228700
But there must be a way to do this without polar coordinates!
 
user228700
Why isn't it $dy$ times $2\pi r$?
 
user228700
5:10 AM
The strip is on the side of the hemisphere.
 
It can be done, but the height of your strip would be $\sqrt{dy^2 + dx^2}$
 
user228700
How so?!
 
It isn't a vertical hoop, it's a slanted hoop. And the higher up the hemisphere you get the more slanted it is.
 
user228700
Oh, shucks :/
 
Polar coordinates? ...
 
user228700
5:13 AM
@JohnRennie How is the height?
 
Because the hoop is slanted, as you move up the surface of the sphere a distance $dy$ you also move inwards a distance $dx$, so by Pythagoras the total distance you've moved is $ds^2 = dx^2 + dy^2$
 
user228700
Sigh. So I tried using polar coordinates for a solid hemisphere but I realized that I might have to use a triple integral! So, I stuck to the normal coordinates.
 
You've already done a double integral, and a triple integral is just as easy.
 
user228700
And it worked out OK, 'cause I took a sort of cylinder as my element.
 
In addition to $dr$ and $d\theta$ you have the equatorial angle (longitude) $d\phi$
 
user228700
5:18 AM
Yes, but I arrived at the correct answer pretty quickly using the Cartesian coordinate system.
 
user228700
Do u think there's any way for me to know when using polar coordinates would be a lot easier?
 
If you have a system with circular or spherical symmetry polar coordinates are always easier. It's just a matter of getting used to them. You'll be using them all the time at university.
 
user228700
Since all objects I will ever deal with will be 3 dimensional, a triple integral would be the most I'd have to do...but it takes some time to get used to this system...
 
user228700
Say, I also have a unit in which I'm required to do similar things to find the electric field at different points.
 
Triple integrals look scary when you write them down, but they're actually very straightforward. You did the double integral just fine.
 
user228700
5:20 AM
These polar coordinates will be useful for that too..?
 
rob
@Kaumudi Your optimism is endearing :-)
 
If your system has spherical symmetry, e.g. the field around a point charge, then yes polar coordinates are probably easiest.
 
user228700
@rob :-P
 
user228700
@JohnRennie Hm. So maybe better get used to these...
 
user228700
OK, one second.
 
5:22 AM
@Kaumudi it's a skill that will be enormously useful in years to come.
 
user228700
Hm...
 
user228700
I was just wondering if I need to try to get used to them now, since I'm running low on time.
 
user228700
But I do think it might be useful for my exam as well, maybe...
 
user228700
OK, anyway.
 
user228700
I will try to do the hollow hemisphere using polar coordinates.
 
5:25 AM
Do you want some hints, or are you going to try unaided?
 
user228700
I've asked one of my friends if he learned to work with polar coordinates when he was preparing for the sane exam. If his answer is affirmative, I will try to do the solid hemisphere using polar coordinates too...
 
the hemispherical shell and solif hemisphere calculation differ only by an extra $\int dr$
 
user228700
@JohnRennie I will try :-P Don't count on me getting there, tho. I might need ur help...
 
You have about an hour before my first task of the day drags me away.
 
user228700
OK, that's plenty of time! Starting now...
 
5:27 AM
Do you know how to use polar coordinates in 3D i.e. what the $\phi$ angle is?
 
user228700
I don't. That's why I've asked my friend if he learnt it.
 
Do you know what latitude and longitude are on maps?
 
user228700
If he did, I will too...
 
user228700
Now let me try to do that problem!
 
user228700
:-)
 
5:28 AM
Because $\theta$ is just the latitude and $\phi$ is the longitude.
 
rob
Oh thank goodness, there were some good exams at the bottom of that stack.
 
user116211
Set Theorists work in T?
 
user116211
9
A: Do set theorists work in T?

David FernandezBretonCaveat number 1: strictly speaking, no one actually works in the theory $T$, just as no one actually works in the theory $\mathsf{ZFC}$. Mathematicians work by means of carefully used natural language and not within a formal system. Formal systems are formulated as approximations that try to mode...

 
user228700
@JohnRennie: Ahh, I seem to be making some silly computing error >.<
 
It's the sort of calculation that is really easy when you know how to do it, but intimidating if you've never done it before.
 
user228700
5:39 AM
OK, let me tell u what I did.
 
user228700
The area of the strip is given by $R^2(2\pi )cos\theta d\theta$
 
user116211
\cos
 
user228700
'Cause the radius of every strip is $R\cos\theta$
 
user228700
The width is $Rd\theta$
 
user228700
5:41 AM
So the area would be $2\pi r$ times $Rd\theta$
 
user228700
Substituting for $r$ as $R\cos\theta$, I got what I got.
 
user228700
Is this correct?
 
@Kaumudi Yes. You've done an implicit $\int d\phi$ there and you've got the correct answer.
 
user228700
Huh. Implicit? Okay, we'll get back to that. So the $dm$ is given by multiplying that with $M/(2\pi R^2)$
 
I would just leave $\sigma$ in for now. It keeps the algebra simple.
 
user228700
5:45 AM
And I get $dm=M \cos \theta d\theta$
 
So $dm = \sigma 2\pi r^2 \cos\theta d\theta$
 
user228700
Actually, no, the $2\pi R^2$ term cancels so it becomes easier to integrate.
 
Ak, OK, yes I get $dM = M\cos\theta d\theta$
So we agree
And the next step is ... ?
 
user228700
Yup! And then I integrate $R\sin\theta M \cos\theta d\theta$
 
Yes
And there's a sneaky trick here ...
 
user228700
5:49 AM
So yes, I have made some silly mistake while integrating.
 
user228700
Hang on, I'll check...
 
How are you doing the integration?
 
user228700
Ah! I forgot to change the limits after substitution!
 
user228700
Oh wait :/
 
user228700
Oh, wait!
 
user228700
5:54 AM
Ahh.
 
user228700
 
user228700
0! Hurray!
 
user228700
-__- I still can't seem to figure out where I've gone wrong!
 
Your limits are wrong
 
user228700
OK. What should they be?
 
user228700
5:59 AM
..?
 
Remember that you're integrating rings. When $\theta=0$ the ring is at the bottom, and as you increase $\theta$ the ring moves up until when $\theta=\pi/2$ the ring is at the top. Yes?
 
user228700
Ah...
 
user228700
Alright. So I get $-R/4$ time $-2$ and get $R/2$ and that's all!
 
user228700
Damn, I didn't pause to think about the limits, sorry :/
 
And that's the correct answer :-)
 
user228700
6:02 AM
Yes! OK! Phew. Now this $\phi$. If my friend says that he did indeed learn that too, I will try to learn this from the internet before pestering u.
 
And for the solid hemisphere just integrate again wrt $dr$
 
user228700
What should I be searching for?
 
Let me draw you a quick picture ...
 
user228700
OK :-)
 
user228700
6:06 AM
Oh, wow.
 
$\theta$ is the angle going vertically up the sphere and $\phi$ is the angle going horizontally around the sphere.
 
user228700
Hm, OK...
 
Look at the square with the sides I've labelled $a$ and $b$. That's the square area element we need to use. Does that make sense so far?
 
user228700
Yeah, I think so...
 
So $a = r d\theta$. Yes?
 
user228700
6:11 AM
Yes.
 
Now $b$ is part of a circle with radius $r\cos\theta$. Is that OK?
 
user228700
Uhh...
 
Shall I draw another diagram?
 
user228700
U know what, sir? Nvm this for now. I'll ask u if needed later?
 
user228700
6:13 AM
^ 'Cause I dunno if needed.
 
user228700
OK, thank you! :-)
 
rob
The only thing that I would hate worse than grading this giant pile of horrible exams would be explaining tomorrow to my colleagues and students why they aren't graded.
 
user228700
@JohnRennie: I think I'm getting better at this. I didn't exactly use polar coordinates but I just correctly found the COM of a solid cone! :-D
 
user228700
@rob U're a prof?
 
user228700
How do u find the time to be a moderator?! :-o
 
6:27 AM
@Kaumudi Since a cone doesn't have spherical symmetry polar coordinates aren't that useful. Though you could use cylindrical coordinates.
In fact I'm willing to bet you used cylindrical coordinates without realising.
 
user228700
Idk if I did :-P I definitely didn't use polar coordinates.
 
user228700
@DanielSank: Hi :-)
 
Hello.
 
@JohnRennie I guess maybe putting the point of the cone at the origin? With a flat base it'd still be a bit rough though.
 
user228700
I put the vertex at the origin and took a disk as my infinitesimal volume element. Then, to find the radius, I used the angle of the cone.
 
6:41 AM
Gone a bit semipractical recently. Some basic concepts in quanutm information theory was understood and thus questions were generated. Craig wll fill in the last detail for me soon. For interest, this is the question
 
user228700
$dy$ being the height of the disk (...cylinder) and $\pi r^2$ the area of the disk.
 
@DavidZ I'd guess it was done by integrating wrt $dz$ summing the infinitesimal disk masses $dm(z)$. Those infinitesimal disk masses are obtained by integrating wrt $dr d\phi$. So it's implicitly cylindrical coordinates.
 
rob
@Kaumudi Well, I figured @dmckee seems to manage.
 
@Kaumudi it's another triple integral, but the first two integrals just give you the mass of the disk so you bypassed them. That just leaves the integral wrt $dz$.
see, triple integrals are so easy you did one without even realising :-)
 
user116211
17
Q: Is cardinality a well defined function?

DirkbossI was wondering if the cardinality of a set is a well defined function, more specifically, does it have a well defined domain and range? One would say you could assign a number to every finite set, and a cardinality for an infinite set. So the range would be clear, the set of cardinal numbers. B...

 
6:45 AM
@JohnRennie indeed ;-)
 
user116211
As always, Asaf answered it.
 
user116211
Hmm, I don't know $\mathsf{ KM}$ :(
 
user116211
I should check Jech.
 
user228700
7:11 AM
@JohnRennie Ah, OK...
 
7:36 AM
A duplicate? I'm reluctant to dupehammer it.
0
Q: What is the speed of a black hole

nrmenesesSo we know that light moves at a rate of 299,792,458 m / s. We also know that light cannot escape a black hole. Since light cannot escape a black hole, this would also mean that light speed is not the fastest speed. So my question is: What is the speed of a black hole? Is there anything out ther...

 
user116211
Hmm, I have seen related posts many times in the past.
 
rob
@JohnRennie Your suggested duplicates are useful.
 
My question was satisfactorily answered. Learnt some new things about how to do bell measurements in the lab
 
rob
One section left to grade. Dance break.
nn ts nn ts
 
user228700
@rob :-)
 
user228700
7:53 AM
Here's a high-five: 🙋 (😛)
 
user116211
0
Q: Why is a quantum state described by a complex vector in the Hilbert space?

user58226In classical theory (e.g., classical mechanics and electromagnetic theory), we introduce the complex values for the mathematical convenience, e.g., A=(a+a*)/2. There, the use of complex values (or vectors) is not necessary as we can in principle describe everything with real values (or vectors). ...

 
user116211
I'm tired of this.
 
rob
 
user116211
There are humongous number of posts on this in Phys.SE.
 
rob
@Kaumudi bad fonts :-(
 
user116211
7:54 AM
Did OP bother to check those?
 
user116211
@rob yeh, it happens when you are in computer.
 
user116211
 
rob
@MAFIA36790 It's twice as good on your machine. @Kaumudi, what's the box supposed to be?
 

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