The h Bar - 2020-10-19

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3:26 AM

Hello everyone! Good morning/afternoon/evening/night!

First of all: I hope you all are healthy.

So, I would like to know if there more than just Casmir experiment, I mean, are there more experiments which reveal quantum effects? More precisely, there are more semiclassical experiments which results negative energy tensity phenomena, like a "casimir-like" class of experiments?

5 hours later…

8:05 AM

@M.N.Raia You can observe the Casimir effect in principle for arbitrary arrangements of the "metal plates" - they don't need to be straight, they don't need to be the same metal, etc - and people have actually computed the different Casimir forces for that, but in the end it's always the same "base experiment" - you have some enclosed space and you get a force from the different vacuum energy densities inside and outside.

4 hours later…

Stupid question inc

11:42 AM

hmm guess I need to figure out myself -_-

huh?

Stupid question inc

oh I give up physics

not my thing to learn

why is physics so boring?

I mean a lot of people don't find classical mechanics thrilling, maybe study something more interesting :P

@Stupidquestioninc It is not boring -_-

Stupid question inc

The language is pushing me back a lot which makes me tired.I just want the idea stated specifically and give some examples!

The book makes physics super boring! But sometimes interesting only sometimes!

@YouKnowMe depends on your frame of reference

I find most authors don't know how to write a book nicely 😕 also they don't say why but everything is just facts facts and facts!

12:29 PM

@Stupidquestioninc start studying physics from cool books like Landau and Lifshitz

1:23 PM

I had some problems logging in to the chatroom a few minutes ago. It said that the whole chat StackExchange did not exist.

Anyone else had that?

no

What's the purpose of a Dirac distribution? Why not define a function f(x) where 1 if x = 0 and 0 if x not = 0?

Because that function doesn't behave like a Dirac function

Its integral is 0, not 1

Are you talking about Dirac Delta?

It's not 1

It's infinity at a given point

or undefined

rest points, it's 0.

it's not "infinity", it's a distribution that doesn't have specific values at points

1:29 PM

whatever... But, it's the correct Dirac Delta na?

@Slereah I can't seem to grasp the difference between a Dirac delta distribution and the function I've defined above. I mean, I know the Dirac delta distribution has a value we can adjust to change the 'spreadness' but... why use Dirac delta distribution. I can't seem to understand this "generalized" property that it has

@Azmuth it's not whatever to me. If you're gonna help, please give correct information, otherwise, don't bother.

@JingleBells Well, for a dirac distribution, $$\int \delta(x) f(x) dx = f(0)$$

@JingleBells The Dirac delta fulfills $\int_a^b \delta(x)f(x)\mathrm{d}x = f(0)$ if $0\in(a,b)$ and $\int_a^b \delta(x) f(x)\mathrm{d}x = 0$ otherwise. Your function (let's call it $d(x)$) does not have that property, it has $\int_a^b d(x)f(x)\mathrm{d}x = 0$ always.

Yes, what @ACuriousMind says

This is the defining property of $\delta(x)$ - it is not a function defined by its values on points, it is an object (a "distribution") defined by its value when integrated against other functions $f(x)$ in this way.

you can get it as the limit of some series of ever-narrower functions (search for "nascent $\delta$ function"), but the limit is again only meaningful under the integral sign

1:41 PM

@ACuriousMind Hmm, I kinda get it. So I should not try to imagine delta(x) as a graphable probability distribution but as an abstract function that has some specific value when integrated with some other function? But why in $\int_a^b \delta(x)f(x)\mathrm{d}x = f(0)$ the result is f(0)?

Yes, and that's it's definition - there is no other "why" than that (unless you define it via the nascent deltas, then this is a result of computing the limit I talk about, but then you have to wonder why you'd define it by that limit in the first place...). It's just sometimes surprisingly useful to have a distribution like this. In terms of statistics, it represents an "absolutely certain" distribution around the value 0 - all its moments like standard deviation etc. are zero.

And we write $\delta(x - x_0)$ or $\delta_{x_0}(x)$ for the same distribution around the value $x_0$ instead of $0$.

Got it. Thank you so much!!!

2:28 PM

Can we infer that since the complex plane is a separable topological space, the infinite dimensional complex Hilbert spaces of regular quantum mechanics are separable?

Since each component is therefore separable

No, there are Hilbert spaces which aren't separable

We just focus on the separable ones

user image

@skillpatrol ^

hmm ok

2:42 PM

@Charlie Careful! "separable" in the topological sense is not the same as "separable" in the Hilbert space sense.

oh no

or wait, maybe it is

sorry, I must have been thinking of some other property, the two are indeed the same

I apologize for adding to your confusion, ignore me :P

Don't worry when you already have this much confusion a little more can't hurt :P

good to hear

I consider myself a connoisseur of confusion

2:48 PM

@Stupidquestioninc you can always ask us ...

My reasoning for the above was that if we're dealing with $\Bbb C^\infty$, if each component takes values in the complex plane, and we can find a dense subset of $\Bbb C$ that is countable, we can find a dense subset of $\Bbb C^\infty$ that is countble, since each of its components are

the axiom of separability of the Hilbert spaces in QM has been bothering me for a while, maybe my book will get to it eventually

tex.stackexchange.com/help/badges/59/publicist?userid=13423 🤔🤔🤔🤔🤔

I have no recollection of that thread, or of posting a link to it

Like, at all

I'm not quite sure what the problem was but I was getting that same error trying to run the Annalen der Physik class over a newish installation of MikTeX. Related to the advice here, running initexmf --mkmaps made the problem go away. — E.P. Feb 1 '17 at 0:36

Turns out I did

I forget the example of a non-separable Hilbert space

Lemme get the Hilbert space textbook

And I still have no memory of posting that

Who remembers anything

2:53 PM

Precisely

I write all my posts high on LSD

that is how you get the physics juices flowing

Goddamn

Every example in that book is separable

26

Q: Nonseparable Hilbert spaces

Being nonseparable Banach space is in fact nothing special: one meets the first examples in the standard functional analysis course, when one learns about $\ell^p$ or $L^p[0,1]$ spaces-these spaces are nonseparable when $p=\infty$. However I spoke to one person which said to me that nonseparab...

fa.functional-analysis banach-spaces hilbert-spaces

@Charlie Don't get too worked up about it, in practice it just means that there there exists a countable Hilbert basis for the space

Non-separable Hilbert spaces are kind of that thing in math where you basically have to make up the examples yourself

Because they basically never occur

@ACuriousMind Sorry for backtracking into someone else's conversation, now this has me wondering if I remember the Dirac function correctly. "The Dirac delta fulfills $\int_a^b \delta(x)f(x)\mathrm{d}x = f(0)$ if $0\in(a,b)$ " That's basically because $\int_a^b \delta(x) \mathrm{dx} = 1$ right at $f(0)$, so the integral basically multiplies to 0 everywhere except $f(0)$, so the result winds up being $1* f(0)$?

You lose that property when the space gets "too big", e.g. if you take an infinite tensor product of infinite-dimensional spaces, see physics.stackexchange.com/a/90031/50583 and its references to Wightman/Straeter

2:59 PM

ok, it's mostly just a passing thought every so often

@JMac integrals don't work that way - $\int fg \neq \int f \cdot \int g$!

That $\int \delta = 1$ is just a direct consequence of the definition $\int \delta f = f(0)$ when you write $\int \delta = \int \delta \cdot 1 = 1$, since the constant function of value 1 has the value 1 at 0

really the main problem with the Dirac delta is that physicists write $\delta(x)$ as if it were a function :P

I've a 400 page handwritten notes on something. What's the fastest way, I can get them as a document?

Typing them will take years! :'(

@ACuriousMind It's a function for asymptotic numbers :p

Any idea/suggestion/method is appreciated during such difficult times...

Well, I guess technically not, but there exists functions infinitesimally close to the delta function in it

3:05 PM

There are some apps that let you scan @Azmuth

Also, I doubt if any software will get my handwriting...

@Charlie Not sure if those apps will get my handwriting....

@ACuriousMind But if you're multiplying two functions, and the first of them is only non-zero at a particular point, and it's area is 1 at that spot, wouldn't the area under that curve multiplied by another curve always wind up being just the second function at that particular point?

Worth a try

@Charlie I tried OCR scanners, not worth.. more time correcting things then typing them by hand...

The fastest way, I've found is to use Latex, any other faster methods?

@Charlie Is there any specific good scanner you know? The notes are written on single ruling page with no sense of straight lines/handwriting...

@JMac No, if you do this with any ordinary function that is only non-zero at one point the integral is always zero

3:08 PM

Some lines are diagonal, some are half empty, some are complex equations, some are underlined...

@JohnRennie Any idea?

I think the problem is thinking of $\delta(x)$ as a function @JMac If you take $$\int_{-\infty}^\infty G_n(x)f(x) dx,$$ where $G_n(x)$ is the Gaussian function and $n$ is the variance and take the limit as $n\rightarrow 0$ you can derive the correct result.

this is because the product $fg$ would only be different from zero at one point, but ordinary Lebesgue integration cannot detect differences between functions on a zero-measure set (a point in this case), so this is the same integral as that of the constant zero function

idk if that is helpful

the important part is that there is no function $G(x) = \lim_{n\to\infty} G_n(x)$ such that $\int G(x)f(x)\mathrm{d}x = \lim_{n\to \infty} \int G_n(x)f(x)$ (also I think that should be that their variance is $1/n$)

by talking about $\delta(x)$, we essentially define this object on the l.h.s. into existence, but it's not a function

@ACuriousMind Can an ordinary function have a non-zero area while being zero at every point but 1? Like isn't that the property of the dirac delta that actually allows the integral to be equal to $f(0)$?

3:13 PM

@JMac no ordinary function can do that, that's the point

if your function is just non-zero at one point, integrals of its product are always just 0

@ACuriousMind ... But doesn't the dirac delta directly contradict that?

I wasn't trying to imply that the logic can extend to ordinary functions, but more that the Dirac delta is kinda a special case where it only leaves $1*f(0)$ as the only place where the integral has an area, so that's the result.

You're confusing two things

@Azmuth what do you need to do with the notes. Are they an important reference of are you archiving them just in case you might want to refer to them one day?

There are functions, which map points of $\mathbb{R}^n$ to $\mathbb{R}$

ie $f(x)$

$f$ is a function, $f(x)$ is a real number for a given $x$

A distribution maps functions to $\mathbb{R}$

The delta distribution doesn't have $x$ as an argument, but $f$

The "proper" notation is $\delta[f] = f(0)$

You never evaluate $\delta$ at a point

@JMac No, the dirac delta doesn't have values "at a point" because it's not a function! You can't write $\delta(2)$ or $\delta(0)$ and ask what that value is, it just doesn't mean anything.

3:21 PM

But can you not find values of $\int \delta(x)\mathrm{d}x$ at a point?

I don't know what you mean by that

No, that's just a real number

it has no value at a point

no more that $1$ has values at a point

I thought it had the value of $1$ at $f(0)$ and $0$ everywhere else.

I think there's a lot of ways of defining it, some more formal than others, defining it as a "function" such that $f(0)=\infty$ and $f(x\neq0)=0$ is an informal definition that is sufficient for most of its usage in physics

@JohnRennie For reference (Probably), I'm not sure about that.

3:26 PM

well it "works" as long as you don't try to do anything with that definition :P

@Azmuth then scan them to a pdf. If/when you find yourself referring to them retype the page(s) that you're referring to.

If you hardly ever refer to them then you won't need to retype anything. If you do find yourself referring to them frequently then the bits you use most will soon be retyped.

@JohnRennie Well.... I guess, scanning handwritten form would not be good... Isn't there any way to get them typed?

No. OCRing handwriting is an exercise in frustration. Unless you or someone else is will to sit down and manually retype the lot you're not going to get the notes typed.

How much time will it take to manually retyping whole 200-300 pages?

A friend of mine had a similar problem and he paid a secretary to retype his handwritten notes. Unfortunately the secretary know nothing about science and the final result was full of mistakes. One that I remember was the secretary typed inorganic anion as inorganic onion.

3:32 PM

lol... haha... :)

How much time did it take?

There's your solution @Azmuth get a secretary :P

@Azmuth ages

@Charlie Welp, I can pay most 2000k USD only... Can I get a secretary for that much? :P XD

And it was expensive since secretaries have an annoying tendency to want to be paid.

@JohnRennie damn... I'll do it myself... 1 paper takes around 10 min... So, 5000 minutes in total.... that's 500 minutes per day... or 4 hours for 20 days...

Seems that'd work...

3:35 PM

You clearly have more spare time than I do.

I wake up at 4, so, By, 8, it's 4 hrs.. :)

9 to 1, study... 2 to 6, sleep, then, 7 to 10, study...

Umm.... bit hard, but I can pull this....

I notice a lack of "work" in that schedule.

@JohnRennie That's what unpaid internships and jobs that require "3 years of experience" in the field for every entry level position are for.

No work for now... busy... study time... will work once I'll be free :)

Is it worth it attempting to understand Bayesian vs frequentist probability? I've been trying for the past 15 minutes, going around the internet but I still can't understand the difference.

3:39 PM

@Azmuth Just two days ago you told us you work for a start-up, now you say you don't have any work because you're studying. Which is it?

@ACuriousMind It's startup we are flexible with dates.

@Azmuth lol, again stating random facts I see

"facts"

Ah yes, I forgot start-ups famously have no time pressure and don't require anyone to do work

So my province managed to basically avoid a major crisis with Covid (we've had like minimal cases, and only 6 currently); but this year has still managed to be a total pile of crap here. Instead of having a bad case of the virus, we had Canada's deadliest rampage; and now there are lobster fishermen setting fire to places that support First Nations lobster fishermen. This year is such a mess.

3:46 PM

feel free to remember whatever :)

Here in Bulgaria, we're expecting to move to online schooling again. Numbers have been "rising" and we're expecting schools to close after a week/s.

@JMac That sucks

also, why are "lobster fishermen" not just "lobsterermen"? (in case you were wondering how my brain is doing today :P)

@ACuriousMind I see you've learned my language

I'm (slightly) surprised there isn't like a push for a more gender neutral term than "fisherman".

I don't hear it said many other ways here :S

lobsmen

3:49 PM

@JMac fishy people

lmen

men

Pretty sure you're going in the wrong direction Jingle Bells

"lobsterperson" sounds vaguely threatening but intriguing

n

Yeah I think the problem with Lobsterperson is the implication that there's a race of lobster people waiting to take over the land.

3:52 PM

are you saying there isn't?

fear of global warming intensifies

Never. I'm just saying that the Lobsterpeople wouldn't want that. They are trying to lay low.

I'll just call them n, since it doesn't seem taken

@Charlie ugh, don't remind me

petition to rename "lobster fishermen" to "n". Your signature here plz:

3:57 PM

What y'all buying?

I just ordered a Realme A11 and Black Hoodies on this festive offer...

@Azmuth food

The dumbest part about the lobster drama is that they are mad that First Nation's people are allowed to fish a "moderate livelihood" outside of the lobster fishing season. It's stupid for two reasons, the first being that we made these agreements with them after we forcibly took the land to try to smooth things out (and instead just treated them like shit for centuries).

The really dumb part is that a corporation that fishes magnitudes more than the First Nations is allowed to fish all year round and they aren't being targeted at all. People are getting some really racist vibes.

just today saw a news article from 1990 where our government committed to cutting CO2 emissions by 1/4th in 15 years. Took them 30.

An actually useful suggestion: When someone edits their message, the added/removed part should be made green or crossed red respectively, and then fade away after 5 seconds, so the person who reads/had read the message can know what has changed without having to reread the whole message.

@ACuriousMind And you guys have done pretty well on pivoting to renewables relative to a lot of places, no?

4:03 PM

Sort of, I guess? Not the worst there, but you mustn't forget we're a major producer of cars and still mine coal in a lot of places

we can use OTEC to reduce global warming :P

okay, i'mma go for now

Can anybody help with this rather trivial thing

go for it

user image

I keep getting my Lorentz transform the wrong way around (index wise),

Am I just doing this the wrong way? It feels like it should be relatively easy

why is the position of the summed index different between $\partial_\mu$ and $\partial^\mu$?

I would have expected $\partial_\mu \mapsto {\Lambda_\mu}^\nu \partial_\nu$ and $\partial^\mu\mapsto {\Lambda^\mu}_\nu \partial^\nu$.

4:14 PM

@ACuriousMind Fair enough. Canada is still doing pretty bad IMO. We have massive oil reserves that we still are extracting... but because it's in sand it also takes quite a bit of power to extract the oil, so it's like we pollute so that we can sell products to further pollute.

user image

@JakeRose does this help? (not speaking as an authority on the subject mind you :P)

@ACuriousMind seems I copied the notes wrong and the space time derivative should transform with the inverse

@Charlie mhm. Shouldn’t the LT on the derivative be an inverse?

yes I may be missing $^{-1}$ powers in there

also, why do we have to transform the minnows ki metric if it’s just going to be the same anyway?

wait lemme think

4:25 PM

@JakeRose that makes sense since ${(\Lambda^{-1})^\mu}_\nu = {\Lambda_\nu}^\mu$

I mean the metric is still covariant right

The only restriction on the Lorentz transformations is that $\Lambda^T\eta\Lambda=\eta$

@ACuriousMind is it? Why?

@JakeRose it follows directly from the requirement that they preserve the metric, see physics.stackexchange.com/a/37953/50583

@ACuriousMind thanks for the answer!

4:45 PM

@M.N.Raia there is cutting edge (experimental) research relating to this let me know if youd like to discuss it sometime

Great. Think I figured it all out! Thanks!

Man the windows 10 quick start feature is weird. Because of it, the computer won't update when it's "Shut Down" and turned back on. You need to literally use the "Restart" feature or disable quick start.

I didn't know win 10 had a quick start feature :P

does it do anything if you're already booting from an SSD?

5:00 PM

I don't know, I literally just disabled it. I'm already booting from a SSD so I'm curious to see how boot changes or if it even takes much longer.

You might already be using Quick Start by default if you never changed it.

I'm pretty sure mine updates when it shuts down, though

Mine was being weird so I followed instructions that I conveniently found on SuperUser SE. superuser.com/a/994285/696188 I'm actually gonna restart right now and see what happens.

Okay so it took about 2 minutes to do a full shutdown and update. I see barely any difference in actual booting besides it being a tad bit slower for the first 30 seconds or so once Windows opens.

hey guys, I am thinking about the Uniqueness theorem in E&M -- if I find a solution for the potential, then it is the solution.

but then what if I add a constant potential to it?

Okay now I'm confused. Shutting down like that installed one update; but it's saying I need to restart to install the other one. Shutting down again didn't install that update :(

two potentials are still physically equal

but now it seems that I have two solutions

5:09 PM

@Shing what are your boundary conditions?

@ACuriousMind the specific case I was thinking a box with fixed potentials on walls. region of interest are the walls + inside

@Shing then adding a constant potential changes the value of the potential on the walls, hence the result of adding that is not a solution to your boundary conditions

@ACuriousMind I think I get it now, thanks!

Make_Stackexchange_Great_Again

5:33 PM

A baby hummingbird drinking the juice of a raspberry.

user image

@Make_Stackexchange_Great_Again Is it having an accident on that hand?

Make_Stackexchange_Great_Again

Yup.

The circle of life.

:-)

This is totally -> twitter.com/SrBachchan/status/1318063662280527873?s=20 @ACuriousMind 's kind of rock voice! :P

PS - I don't understand a single line of word the kid is singing...!

hehehe XD

Make_Stackexchange_Great_Again

Music isn't always about understanding.

5:48 PM

:)

Make_Stackexchange_Great_Again

:-)

6:05 PM

Since Traversable Wormholes are a quite restrict topic, can I post a question here?

sure, but no guarantees anyone knows the answer, though @Slereah is usually interested in wormholes and other assorted relativity weirdness

Thanks

0

Q: Doubt on Newman-Janis algorithm for a traversable Wormhole

Recently in a paper $[1]$ the researchers presented a rotating traversable wormhole solution using the famous Newman-Janis Algorithm $[2]$. But something is anoying me. In $[1]$ they presented the Morris-Throne metric in the following way (section 4): (...) we rewrite the spherical symmetry worm...

general-relativity spacetime differential-geometry metric-tensor kerr-newman-metric

ah, I didn't realize you meant a question you had already posted :P anyway, there's probably a minor typo - you suddenly say Newton-Penrose instead of Newton-Janis - that's all I can say about that question...

Make_Stackexchange_Great_Again

May I suggest, next time, you wait at least one hour before posting in chat :-)

Preferably, longer....

YMMV

depending on traffic conditions

@Make_Stackexchange_Great_Again thanks for the tip.

Make_Stackexchange_Great_Again

6:19 PM

np

@ACuriousMind Yes...sorry haha

6:39 PM

@M.N.Raia As far as I can tell those coefficients are still arbitrary?

They're just expressed differently

The only difference being that the radial one is only defined for $\mathbb{R}^+$, but when expressed like that you just do it for both sides of the throat

6:53 PM

@Slereah What I would like to know is: f(r) could be e^{2\phi} + r and g(r) = 1-b(r)/r + a

And the null tetrads remain the as as the article?

1 hour later…

7:53 PM

Why use covariance when correlation gives you the same thing but with normalized magnitudes that make sense?

Make_Stackexchange_Great_Again

Cross. Validation?

Not sure what that is yet Trump

@JingleBells who's saying you should use covariance instead of correlation? :P

@JingleBells The stats.SE is called Cross Validated, and they even have a question where I can't really see any of the answers making a point other than "correlation is a nicer form of covariance"

@ACuriousMind No one actually :P

@ACuriousMind Yeah, I agree. I just got confused because they introduce covariance first and then correlation but then I realized correlation is a normalized covariance :P

well if they had started with correlation you'd be here asking why we have to divide by the variances ;P

8:02 PM

xDD

exactly, so I guess I kinda wondered why covariance exists in the first place when correlation is better, but now I know

Make_Stackexchange_Great_Again

8:13 PM

Knowing is half the battle.

at the very least

engineers would round that up to 1

so you just won the battle

One way or another, the engineers always win

Make_Stackexchange_Great_Again

to them, the battle is twice as large as it needs to be

Then they'll round up to two

by induction they always win

Make_Stackexchange_Great_Again

how many engineers does it take to screw in a lightbulb?

5, one to hold the bulb, and 4 to drink until the room starts spinning

8:29 PM

It's a decent joke but how does it use the fact they're engineers at all lol

Make_Stackexchange_Great_Again

8:40 PM

if they were programmers, they wouldn't do it

because that's a hardware problem

2 hours later…

Stupid question inc

10:39 PM

@JohnRennie Thanks! Yesterday I was mentally unstable so today being stabilized I found it put myself.

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