The h Bar

user116211

11:09

1

Q: Significance of $i$ in the Schrödinger equation

There's an imaginary $i$ in the Schrödinger equation, which I guess is to define the position of the particle in a space-time involving a complex function. But what is the real physical significance of $i$ in the equation?

user116211

8

Q: Where does the $i$ come from in the Schrödinger equation?

I am currently trying to follow Leonard Susskind's "Theoretical Minimum" lecture series on quantum mechanics. (I know a bit of linear algebra and calculus, so far it seems definitely enough to follow this course, though I have no university physics education.) In general, I find these lectures f...

quantum-mechanics operators schroedinger-equation hamiltonian unitarity

user116211

@ACuriousMind: Is the later a dupe of the former? IMO, they are different. What say?

ACuriousMind

I just now voted to close as duplicate. Where do you see the difference?

user116211

@ACuriousMind Well, he asked about the significance of $i$ and the later asked where it emanated in the derivation.

ACuriousMind

And how could you understand the "significance" of a term if not by examining its origin?

user116211

11:12

But, I wasn't sure; that's why I asked it to you.

user116211

@ACuriousMind: Also, have you noticed CuriousOne's practice of writing answers in comments?

user116211

Most of the comments he leaves beneath the question are actually answer, though ;/

ACuriousMind

Uh, yes. I think he has too, but can't really help it. (I think most of these comments start as a one-liner and then he adds things without realizing that could be an answer)

user116211

@ACuriousMind I always point at that to turn these to answer, but of no avail.

user116211

I like this character though; his general comments are blunt but correct ;)

ACuriousMind

11:16

That's strange, I generally get either a reason why it's not an answer or a "Okay, I'll make that an answer" as reply.

user116211

@ACuriousMind At least not to me, I made many attempt to make them answers; there are very few to which he replies and most time, they are negative.

ACuriousMind

Well, if you think it's an answer and he refuses to make it one, you can always write the answer yourself.

user116211

@ACuriousMind Nooooooooo

user116211

One of his amazing comments I liked the most but seemed a little harsh:

user116211

If you think that you have a wiring problem CALL AN ELECTRICIAN! We can not (neither technically nor legally!) solve this for you. Anybody can guess their way around your problem, only you and the people in that room can get hurt or die from it. — CuriousOne 9 hours ago

user116211

11:21

And then his VTC reason:

user116211

I'm voting to close this question as off-topic because the community here can not and should not try to debug serious electrical wiring problems over the internet. — CuriousOne 9 hours ago

user116211

Damn true but really quite blunt ;/

user116211

@ACuriousMind: I know you are asked by many which sometimes lead to annoyance.

user116211

So, may I ask when I should ask you about something?

user116211

@ACuriousMind: You helped me earlier also at ergodicity, thermodynamics; stat mech; QM; degrees of freedom etcetera.

ACuriousMind

11:26

You see, what I get annoyed with is not people asking questions. It's asking the question specifically to me when there is nothing that indicates I'm somehow the best to address the question to.

user116211

@ACuriousMind: But I want to know at what topics specifically which you'd feel comfortable....

user116211

@ACuriousMind Yes, that's a point; but that is true also, you are next to an AI ;)

ACuriousMind

This is a chat, not my office. So unless we've talked about the specific issue before, or you think I would find the question really interesting, don't ping me with your question. Just ask it. If I want to answer a question, I will answer it even when it's not addressed to me.

user116211

@ACuriousMind That's it. Thanks! But you seemed to answer even the dumbest question of mine ;)

user116211

@ACuriousMind We didn't talk about thermo or ergodicity earlier either; I just asked you whether you'd like to answer; you did.

ACuriousMind

11:32

Yes

user116211

12:17

@JohnDuffield: o/

user116211

I'm not getting one point.....

user116211

I thought at equilibrium, the net current $$i_\textrm{drift} + i_\textrm{diffusion}= 0$$.

user116211

But Feynman says

user116211

> Under equilibrium conditions the net diffusion current must be zero.

John Duffield

@dmckee : I've never said an electron is a spinning ball of charge. I said you take a field variation and wrap it into a Möbius strip like the spinor depiction, and then you've got what looks like a standing field. Try it.

Nihal Jalaluddin P

13:06

Re questioning:we are saying particles are created by field disturbance. on the other, we say fields are excited by the presence particle. so it’s a bit chicken and egg. which comes to being first then?

Slereah

13:29

Fields.

The fundamental quantity in quantum field theory is fields.

FenderLesPaul

14:15

@0celo7 You know I can read things you say

cries

Slereah

14:35

So I asked Ellis to post the proof on SE

"Well does it make sense to you? Standing next to yourself with the arrow of time going in the opposite direction for your Doppelganger would be pretty weird … probably impossible but that depends how you view the nature of time (is it relative to surfaces, or relative to timelike lines)?

Only post if it works!"

Not quite the level of certainty I expected!

ACuriousMind

Well, at least it's an honest answer!

Slereah

True, but considering this a proof everyone refers back to, I expected slightly better :p

I mean the proof sounds good enough, but I don't have the rigorous mathematical proof for it in my head

Should give it a try I s'ppose

user116211

14:51

@Slereah: Got answers to the questions you posted at Math SE?

Slereah

No

user116211

Damn.

user116211

Sue them ;(

Slereah

For what, Stack Exchange points?

0celo7

@ACuriousMind Hmm...

But aren't all diffs just GL transformations?

ACuriousMind

14:58

No, they induce one through their Jacobian, but the diffeomorphism itself acts on the manifold, which has no notion of "GL", since it has no linear structure

Bernard Meurer

Inducing a jacobian sounds slightly pornographic

0celo7

@Slereah I'm still not convinced.

There's nothing in his construction that guarantees the curve must intersect more than once.

And there's nothing wrong with a closed curve intersecting a hypersurface once.

I'm thinking their proof is simply wrong.

Or they have to do what Freire said and pick a more restrictive class of hypersurfaces.

@Henry Yes, that's correct, but not very enlightening. It's best to use indices for such computations. — 0celo7 12 hours ago

@ACuriousMind I...I don't know what came over me...

Slereah

@0celo7 : But if it intersects only once, then it is still an odd number of time

But still can't be contracted to a point

0celo7

@Slereah Take the hypersurface that is the hemisphere of $S^2$ in $R^3$

Pick some closed curve that instersects it once

Do you agree that this is possible?

> Does the 0 indice represent time

Slereah

Hemisphere isn't boundaryless tho

0celo7

15:09

@Slereah define fucking boundaryless

Take the open disk then

That's boundaryless

Homeo, maybe diffeo to the plane, in fact.

The same construction works.

Slereah

Lemme check how HE defines boundaryless

Hm

Not quite sure he defines it there

Maybe we should do like

A GR jam

And then post the results to Ellis

To see what he has to say about it

0celo7

A...GR jam?

Slereah

We jam about GR

Brainstorming, if you will

0celo7

@Slereah I have the correct proof if $\mathcal{M}-\Sigma$ has two connected components.

But the homotopy thing is nonsense.

Slereah

Do u want me to post it to Ellis

the proof

0celo7

15:14

@Slereah They define a manifold with boundary as one that's just homeo to $\frac{1}{2}R^n$ everywhere instead of the full $R^n$.

But the open disk satisfies the original thing that every point has a nbhd homeo to $R^n$

It's a manifold w/o boundary.

Slereah

Yeah but I suspect "boundaryless" here means like

Compact or infinite

I mean you can't really foliate spacetimes with open disks and call them Cauchy surfaces

0celo7

Ok, let me ask my prof if I can rewrite his proof and have you send it to Ellis.

Slereah

Calm down my man

0celo7

Rewrite because there's comments on the PDF he sent me that are unrelated.

user116211

Does anyone know what first-order variation is?

0celo7

15:23

Yes.

Slereah

$\delta S$?

user116211

@0celo7 Can you tell me?

0celo7

@user36790 Yes.

user116211

@0celo7: Wait, let me give you the context....

0celo7

If you have a functional $F(f)$, then make the variation $F(f+\epsilon g)$

user116211

15:24

1

Q: Trying to understand the derivation of magnetic field around a current loop.

I'm reading Edward M. Purcell's Electricity and Magnetism. I'm having trouble in conceiving his derivation: The vector potential is $$\mathbf A(0,y_1,z_1)= \frac{\mu_0I}{4\pi}\int \frac{\mathrm d\mathbf l_2}{r_{12}}\;.$$ He then wrote about the variation in the denominator $r_{12...

calculus

0celo7

uh, wait, how does this continue?

user116211

@0celo7 Well let me google what functional is.

Slereah

A functional is a function from a function space to $\Bbb R$

0celo7

or $\mathbb{C}$

user116211

@Slereah vector space?

Slereah

15:25

Like $\mathcal C^0 \rightarrow \Bbb R$

For instance

The ur-example of a functional is a definite integral

0celo7

Here $F$ is the first variation

user116211

What is $O$?

Slereah

Big O notation

user116211

@Slereah What does that mean? Some series?

Slereah

$\mathcal{O}(h^2)$ means that it is a function such that $\frac{f}{h^2}$ doesn't diverge

user116211

15:32

@0celo7 Quite understood.

user116211

But, I'm not getting how to apply it at the present context :(

0celo7

It might be the wrong context for what I gave you!

user116211

@Slereah Does function space mean power set?

Slereah

N... no?

0celo7

Huh?

user116211

15:35

Got the point.

user116211

> As a special case, the power set of a set X may be identified with the set of all functions from X to {0, 1}, denoted 2^X.

Slereah

We are not talking about such functions, though.

user116211

But the point what did Purcell want to mean in the derivation?

0celo7

how do you do ö in TeX?

user116211

Those damn in Math SE are busy in answering pretty HW questions; but they have no time to answer some real questions ;(

Slereah

15:38

\"

ACuriousMind

@0celo7 \"o

user116211

@0celo7 $\ddot{o}$

Slereah

By the way writing in French in Latex is kind of a hassle

ACuriousMind

@user36790 No! \ddot is a math symbol, you do not use math symbols to emulate diacritics

Slereah

Since you have to input all accents by hand

user116211

15:39

@ACuriousMind Sorry!

ACuriousMind

@Slereah You can switch input encoding to UTF-8 and type them as you normally would

yankeefan11

Hey guys, can I link a question here for a short discussion?

0celo7

@yankeefan11 make that @ACuriousMind

yankeefan11

What do you mean

ACuriousMind

@0celo7 Stop trolling.

@yankeefan11 Ignore him, and just post your question.

yankeefan11

15:45

0

Q: Question about electrostatic boundary conditions

So I am trying to solve a problem similar to that of how Jackson(and many other books) do the corner. My geometry is shown in the picture. But I have a line charge near a corner with a slab of dielectric material ($\epsilon$). My concern is with boundary conditions. I expand my potential (i...

homework-and-exercises electrostatics

user116211

@0celo7: That's why I like you.

user116211

15 hours ago, by 0celo7

@Danu I've never embarrassed myself here.

0celo7

@Slereah What's the Hawking-Ellis notation for the tangent to $\gamma(t)$?

is it $\dot\gamma$?

or is it $\partial/\partial t$?

@ACuriousMind :(

Slereah

It seems to just be $\frac{\partial}{\partial t} f$

user116211

@Slereah: Got answer! From Ross Millikan!

Slereah

15:49

wot

user116211

Damn it was just a Taylor expansion ;/

0celo7

everything is a taylor expansion

user116211

@0celo7 it was an easier Taylor expansion.

user116211

I missed it ;/

user116211

@0celo7: But thanks, I could know new things like functional.

DeNiSkA

16:06

@NihalJalaluddinP well i am 63070000 seconds elder to you!!

user54412

16:26

8

Q: Will unattended nuclear power plants actually blow off?

Scenario: Deadly virus wiped out 80 - 90% world population and it did it pretty quickly (in 2 months) Although it is plausible that among these 10 - 20% of survivors will be people who actually know how to run nuclear power plant, it's safe to assume that they will have different tasks to solve ...

science-based reality-check apocalypse power-sources nuclear-power

user54412

I give up. Humanity is too stupid.

Bernard Meurer

@ChrisWhite?

user54412

@BernardMeurer People actually think nuclear power plants are some sort of risk.

user54412

Decent answer, but see Fukushima. While the circumstances differ (natural disaster vs. lack of manpower), part of the problem was that much of the redundancy was defeated by concurrent failures with local infrastructure (plausible in a manpower scenario as well). Also, the effects from just that one incident have been felt globally (certainly far more than "the nearest few km"!), so you might want to consider revising the impact portion of your answer. Finally, atomic bombs are certainly not pressure containers, either! :-) — type_outcast 4 hours ago

Bernard Meurer

@ChrisWhite Oh, but they are! If they blow up like that xernobil it'll make us grow new limbs and stuff

user116211

16:30

@ChrisWhite Not all are aware; many are ignorant. But I hate such questions, unresearched....

user116211

@ChrisWhite Worldbuilding!

user116211

Most of the wrong Physics come there.

user54412

I still don't get how people think anything went wrong at Fukushima. I'm exposed to more background radiation in New Jersey than anyone there received.

Bernard Meurer

Fukushima got a Tsunami earthquake combo and survived

user116211

@ChrisWhite Ha!

Bernard Meurer

16:32

I can't think of any other system that would do as well as a nuclear plant in that condition

user54412

Nuclear power: the only feasible power source that won't add greenhouse gasses to the atmosphere. Also causes fewer deaths per unit energy than anything else.

user54412

Physics gave the world a magic, essentially side-effect-free solution to its most pressing problems, and people choose to go around being afraid of it.

Bernard Meurer

@ChrisWhite But... but, nuclear waste! and it makes the workers radioactive!

Curious fact: Brazil has only one geological fault in it's entire territory. Guess what's on top of that only fault?

ACuriousMind

@ChrisWhite Well...at least in Europe, the authorities have proven to be not diligent in the disposal/storage of the waste. That doesn't exactly inspire people to believe the plants themselves are safe.

user116211

@BernardMeurer Brazil?

user54412

16:39

@BernardMeurer a nuclear power plant?

Bernard Meurer

@ChrisWhite Yep, our only nuclear plant

it's almost ironical

user116211

@BernardMeurer: Is Brazil earthquake-prone?

Bernard Meurer

@user36790 Not at all, never happened afaik. But if it ever does happen it'll be there

user116211

@BernardMeurer: You are fortunate.... Heard of Nepal Earthquake?

Bernard Meurer

@user36790 Yep, it was all over the news

user54412

16:43

It's like they were playing a game of Sim City and decided to plan ahead for the "destroy everything" phase.

user116211

@BernardMeurer Even India succumbed to it. I fell it when I was at the roof with my pup; it became damn nervous!

Bernard Meurer

@ChrisWhite It was built back in the military dictatorship days, so that theory is rather plausible :p

Update: UCD - Denied, UCSD - Accepted, UCB - ?, UPenn - ?, Cornell - ?, Waterloo - ?, NC State - ?, VA Tech - Accepted, GA Tech - Waitlisted, UIUC - Denied

At least the score is 2-2 now :p

John Duffield

17:04

@Slereah : "Standing next to yourself with the arrow of time going in the opposite direction for your Doppelganger would be pretty weird … probably impossible but that depends how you view the nature of time". Tsk. Somebody is lost in maths. The nature of time is straightforward.

@ChrisWhite : and ignoring thorium.

0celo7

17:49

@JohnDuffield lol

user116211

@0celo7: JD is always serious about his comments.

0celo7

Yup, that's why he's so funny

user116211

@0celo7 ha! JD is funny. If he could understand that.

ACuriousMind

18:09

10

Q: In a world with escape velocity greater than the speed of light, would a civilization be able to reach space?

Htrae is a planet with an abnormally strong gravity. In Htrae, the escape velocity is greater than the speed of light. Suppose a civilization like humans (call them snamuhs) evolves, with the unique ability to withstand Htrae's abnormal gravity. If the snamuhs have the ability to become as techno...

technology physics gravity

Nine answers for a question that amounts to "Can I leave a black hole?"

user116211

@0celo7: Just saw JD's layman explanation of wave-particle duality:

user116211

@Jan : it's like Chinatown: She’s my sister. Slap. She’s my daughter. Slap. She’s my sister and my daughter. The electron is a particle, but that particle isn't some billiard ball, it's a wave. This is best appreciated with the photon, see Wikipedia. Look at the first line: A photon is an elementary particle. Then see this: the energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ). And that wavelength can be tens of metres. It isn't necessarily microscopic. — John Duffield Nov 14 '15 at 18:42

user116211

How I am still alive ;/

Hari Prasad

18:47

@user36790 lol

Slereah

Hari Prasad

@Slereah whatssup in chinatown?

Slereah

Forget it.

user116211

@Slereah Best remedy.

Hari Prasad

@user36790 which board are you?

user116211

18:53

@HariPrasad .....

Hari Prasad

@user36790 what?

user116211

@HariPrasad After the exam.... not now ;P

Hari Prasad

@user36790 ok which board did you studied?

user116211

@HariPrasad cbse.

Hari Prasad

@user36790 nice

0celo7

18:57

@Slereah I don't get it either.

Slereah

Yes but that's because you have never seen a movie

Hari Prasad

@Slereah may i know which movie? is there a black hole in chinatown?

Slereah

imdb.com/title/tt0071315

user116211

chinatown.

user116211

When will people learn the art of googling ;/

Hari Prasad

18:59

@user36790 Right now!

user116211

@HariPrasad Quick learner!

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