"In theoretical physics, quantum nonlocality refers to the phenomenon by which the measurement statistics of a multipartite quantum system do not admit an interpretation in terms of a local realistic theory."

This is an example of overcomplicating simple things (something that really bugs me in today's world). It would have been enough if they said - "In theoretical physics, quantum nonlocality is the phenomenon where entangled particles can exchange information faster than the speed of light. This goes against Einstein's special relativity.

I have a theory that people are doing this because they enjoy being around smart sentences, words and twisted linguistics. It makes them feel smart. It makes others think they are smart (and therefore makes them feel good about themselves) - higher social status, higher chances of reproduction.

Is there some type of ML algorithm (neural network?) where it's somewhat trained to classify images and it sees an object and predicts that the probability of that object being a tree is 0.8. Now, can that network train itself based on it's prediction? So next time it sees the same tree, it predicts 0.9. That way the algorithm strengthens it's ability to recognize trees in general by lacking new data. I'm not particlarly sure what I'm talking about, it just popped into my head.

@NovaliumCompany No, your paraphrase is incorrect. Quantum entanglement does not allow for transfer of usable information, and only in some interpretations is it believed that there is (unusable, undetectable) communication at all

Sometimes, people use jargon because no colloquial paraphrase carries the same meaning and this is an acceptable use of jargon!

@NovaliumCompany quite often in mathematics and the more mathematical end of physics it isn't possible to define things in natural language and you have to use technical terms. This makes it sound like jargon jibberish, but it isn't. The terms used are standard terms that have a precise definition.

@JMac Community votes and an AI.

@JohnRennie I think this question is big-list type and should be closed.

@JohanLiebert I've already voted to close it

@JohnRennie Got it

morning

@NovaliumCompany that would also be wrong

explains where quantum woo came from

also as far as I'm aware, non-local theories have also been ruled out

"Bohm found a hidden assumption, that of “locality”, in Bell’s proof of his theorem, but later proofs [110] do not use any such assumptions."

@JohnRennie Hey, can you email me at [email protected]?

I better run before ACM shows me abap

@BernardoMeurer done :-)

sciencedirect.com/science/article/abs/pii/0375960187900752

Bell's theorem sans assumptions of locality

@Slereah I'm not talking about the correctness of my simplified sentence. I understand, as JohnRennie and ACuriousMind said, that some things in technical fields (physics, maths...) are better to be explained using references to the components, instead of explaining the definitions of the components themselves. It just triggers me a bit when I see someone on youtube explaining something incredibly understandable way while the classical definitions of "high-status individuals" fail to do that.

@NovaliumCompany Well you didn't seem to understand it, so maybe it's not so easy to explain after all!

@Slereah That's because the only definition I read is the one google gave me.

@NovaliumCompany you need to be a bit cautious. I've earned lots of rep on the PSE explaining things in ways that non-experts can understand, but most of those explanations are ultimately misleading because they have to oversimplify to be comprehensible.

My take on the issue is that quantum interpretation is both 1) super complicated to understand properly 2) Not actually that useful in doing actual quantum mechanics

If you wish to really study the issue, I can advise some books, but they are 1) extremely technical and 2) probably not worth it

I don't think quantum interpretations is a good thing to study as a layman and I have never seen it lead to much interesting

@JohnRennie I agree. I'm just saying, some things can be explained simpler than google definitions.

I hope you like non-Kolmogorov probabilities and Hilbert lattices!

Most people don't even seem to realize that the whole quantum argument isn't even about the probabilistic nature

If quantum mechanics' problem was just about probabilities, it wouldn't be that controversial

It's the contextuality

Plus, failure is the greatest teacher of all, so I'm ready to confuse myself along the way with oversimplified definitions if that means getting some grasp on the idea. That gives me the incentive to build on the oversimplified foundation with more details. (e.g undestand that the electron doesn't actually spin in an orbital form around the nucleus)

How will you know if you have failed

Not understanding quantum interpretations isn't like failing to juggle plates

there's no clear signal that you've done so

I will realize that what I've learned is not completely correct, or not completely finished.

@Slereah I'm not interested in QM.

There's some 90 years old physicists who still haven't learned it, if I'm to trust some other old physicists

My opinion is that oversimplified, vague, unfinished definitions, at least for me, work well.

I see learning as solving a puzzle where more parts constantly come in.

Few people even bother looking up measurement for classical theories :V

I want to shoot something about vector calculus, Sam sir are you ready?

If Sam is me then I am no sir

I am no Sam either, come to think of it

"sam" means alone in Bulgarian.

Neither am I Bulgarian

xDD

@Slereah :) I thought Sam is short form of Samuel

It is in America

I do not hail from the land of cowboys

AHHAHHAHAHHA

HAHAHHAHAH

All right, Stokes Theorem says for any vector field $\mathbf A$ and and any surface $\sigma$ with boundary curve as $s$ , we have $$ \int_{\sigma} \mathbf A \cdot d\sigma = \oint \mathbf A \cdot d\mathbf s $$

That is true

No.

What's the mistake?

For a start, $ds$ is a vector field of normal vectors to the surface considered

It is not generally going to be $(1,1,1)$

In fact, since the surface has to be closed, it can't be $(1,1,1)$ everywhere

Also $dU \neq \frac{\partial U}{\partial x} dx$

I didn't get you, why surface needs to be closed?

Because that is the theorem.

That's why the symbol you see is $\oint$

Well that's for boundary, circulation around the boundary

Yes.

But that is a symbol for "integration over a closed surface"

So, why surface need to be closed?

I mean, do you want a proof of Stokes' theorem?

Ah yes, I see

This is for surfaces

in case of surfaces, it's the boundary curve that is closed

Same comments, tho

Yes.

How?

is it $\partial U$ ?

The definition is $$dU = \sum \frac{\partial U}{\partial x_i} dx_i$$

I mean $\frac{\partial U}{\partial x} dx = \partial U $

All right! I got you.

$$\mathbf A \cdot d\mathbf s = \frac{\partial U}{\partial x} dx+ \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z} dz \\ \mathbf A \cdot d\mathbf s = dU $$

Are we right now ?

Although again, typically you don't have $$\vec{\nabla} U \cdot d\vec{s} = \frac{\partial U}{\partial x} dx + \ldots$$

It depends on the boundary you're doing

For instance, take a disk

The surface is a disk, the boundary curve is a circle

okay, then?

okay

okay

So $\vec{n} \cdot (\dot{x}, \dot{y}) = 0$

This is $$- n_x \sin(t) + n_y \cos(t) = 0$$

This is your normal vector

so see you have got the original curve in parametric form

I just took the normal dot product when I wrote $$\mathbf A \cdot d\mathbf s = \frac{\partial U}{\partial x} dx+ \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z} dz$$

Yes, the issue is that $ds$ isn't the same as $(dx, dy, dz)$

You're integrating along a curve, for a start

A curve is in one dimension

Your integral should be along a single variable

I thought we can write any vector in component form and $d\mathbf s$ is after all an infinitesimal vector only.

It's an infinitesimal vector at a location along a curve

ie it's a map $$ds : \mathbb{R} \to \mathbb{R}^3$$

well not really, but you get the idea

at a point on your curve, you associate an infinitesimal vector

So, am I wrong $$\mathbf A \cdot d\mathbf s = \frac{\partial U}{\partial x} dx+ \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z} dz$$ here?

Yes.

Why? Means $d\mathbf s \neq dx \hat i + dy \hat j + dz \hat k $ ?

Yes.

@skullpatrol You know the main thing is that you cannot reason with some people here, they argue only because they want something to talk about. They don't understand anything. So, leave it Pal!

Oh wait, actually Stokes theorem integrates wrt the tangent, not the normal

To be specific, what you have is, for a curve $\vec{x}(t)$, $$\vec{A} \cdot d\vec{s} = \vec{A} \cdot \dot{\vec{x}}(t) dt$$

This is what an integral along a curve is

I cannot see why $d\vec s$ cannot be broken down into components

Well here you have $$\vec{A} \cdot \dot{\vec{x}}(t) dt = \sum A_i \frac{dx_i}{dt}(t) dt$$

Here you can indeed use $\dot{x}_i dt = dx_i$

Can't we cancel $dt$ s ?

So I suppose you can, yes

Although then you have to switch the integral boundaries

So yes that should be about that

Can we go on without any disputes ?

sure

Since $\mathbf A \cdot d\mathbf s = dU$ ,therefore, by Stokes theorem $$ \int_{\sigma} \mathbf A \cdot d\boldsymbol{\sigma} = \oint dU$$

Yes.

As this is the integral along a loop, the boundaries should be identical

Therefore $$\oint dU = 0$$

Since $\oint dU = 0$ (I myself don't know why but this is true)

Hmmmm

Let me think

Having an integral equal to zero doesn't guarantee that the function is zero

Hey! I'm such a care-less man !

I forgot all the way to write $curl~\mathbf A$

OH MY GOD!

Stokes Theorem is $$ \int_{\sigma} curl~\mathbf A \cdot d\boldsymbol{\sigma} = \oint \mathbf A \cdot d\mathbf s$$

Is there any chance that you can forgive me and we can continue ?

Well $\mathrm{Curl}(\vec{\nabla} U) = 0$, if I remember correctly

Which should be a much shorter proof

That's only I'm striving to prove

You can do it without integrals

Yes, but as I mentionned, $\int f = 0$ doesn't necessarily imply $f = 0$

This should be $$\int_{\sigma} curl~\mathbf A \cdot d\boldsymbol{\sigma} = 0$$

Although... I think that is true if this is true for every surface $\sigma$

I have a proof of that somewhere

Lemme see

What?

Well, if $$\int_\sigma f = 0$$ for every possible choice of $\sigma$, then $f = 0$

Yes, as in our formulation $\sigma$ was arbitrary

Then yes, that would be the proof

Although proving that is a bit tricky

Why can't we prove that from propositional logic?

For a start because propositional logic deals with propositions

I meant from propositional logic we can say "since that is true for every surface $\sigma$" and "we know that integral of zero is zero" therefore "f is zero" ?

well no, because you also have that integrals of non-zero functions can be zero

there is a proof for it but it involves limits of domains of integration

All right for now let's accept that it is true, can we move on with $$ curl ~\mathbf A \cdot d\sigma = 0$$

Well I would object to using $d\sigma$ outside of an integral :p

AHA! that was my problem you know

If $$\int_{\sigma} curl~\mathbf A \cdot d\sigma = 0$$ then can we assume $$curl~\mathbf A =0$$ because doing that means giving no respect to the dot product.

I think you can show this by picking $\sigma$ a sphere of radius $R$ and then taking the limit of $R$ to zero

Or any area, really

How would that prove anything?

Well, I tried to show something similar here, apparently :

samuel-lereah.com/db/problemdb/Static%20point%20charge

I trust past me

How much past is that?

Who can tell

Time is complicated

Well you're working day and night to for Time only :)

Here's a dumb argument :

Therefore $ (\nabla \times A) \cdot n = 0$, from the previous argument, for an arbitrary vector field $n$

Which means that $\nabla \times A = 0$

Okay, now how can we remove that $\mathbf n$

$n$ is arbitrary, therefore it is unimportant

@Slereah How? It could be that curl of A is perpendicular to that n

Yes, but with the inner product, if $a \cdot b = 0$ for any $a$, then $b = 0$

A dumb proof for this : take $a$ to be all possible basis vector

Then $e_x \cdot b = 0$ and $e_y \cdot b = 0$, implying $b_x = 0$ and $b_y = 0$

I don't know linear algebra yet, so can we just call basis vector as $\hat i$

No.

I meant $\hat i$, $\hat j$ and $\hat k$

I'm aware.

So, how can we remove $\mathbf n$ from $$ curl~\mathbf A \cdot \mathbf n = 0$$ to get $$curl ~\mathbf A = 0$$

I have done so!

You may marvel at it

You can check it more intuively by picking as your surface $\sigma$ a square with normal vector pointed in the direction $e_i$

Then do it in every direction

Therefore, all components are zero

Ah! Now I see

@skullpatrol Today Johann Liebert informed me about Ron Maimon and you know I found him great! Now, I see @ACuriousMind is nothing, he is just acting the way he talks.

I'm trying to find a coordinate-free proof because coordinates are for cheaters :p

Hm, I guess the simplest proof is

put up :D

If $a \cdot b = 0$ for all $a$, then take $a = b$ implies $b \cdot b = 0$

Therefore $b$ has zero norm

Inner product is positive definite, therefore $b \cdot b = 0$ implies $b = 0$

WOW! That's one of the greatest proof

I really enjoyed that proof

It is very much not

No it was best

you see no one can object it

is there any loop holes in it?

it is fine

He he he he he he he he HAHAHHAHAHAH :)

Thank you (can I call you Sammy)

Okay I will not put up those requests again.

Thank you Slereah

@Knight Calling somebody "nothing" is incredibly insulting and rude -- I hope this is just a case where you misspoke or used the wrong word and you don't mean to call ACM (or anybody) "nothing."

@Knight Please consider the possibility that you are taking this determination on a rather incomplete set of information. RM is a complicated character and any simplistic view about him is unlikely to be well informed -- he made some great contributions, but he also had some very rough interactions with other people (both here, elsewhere on SE, and off-SE) that are no longer accessible but which did happen.

In other words, please do your research in depth before you canonize him.

You cannot serve both God and Maimon

@EmilioPisanty Yes I may agree that my research is incomplete. If it’s not against the rules can you please tell me what really happened with Ron Maimon (I mean how come he had people on his side also)

@peterh-ReinstateMonica Hello! How are you doing?

@tpg2114 Thank you for giving me a chance to clarify, when I attributed him with “nothing” I meant he is nothing sort of what he displays, he displays (at least to me) that he is rude and doesn’t consider pardon for breaking of any rules or new ideas

@tpg2114 I called ACM nothing because I saw Shog9 and our Dmckee taking that matter of Ron so calmly and I was impressed. You know I had great admiration for my friend @rob and you know I respect you a lot and so to @DavidZ (why @DavidZ is not coming to the h bar) but Shog9 and Dmckee compelled me to admire them for their cool actions and words.

Suspended!

Image not found 😀

@JohanLiebert What?

@Knight yeah he is rude to everyone!

@JohanLiebert “is” who?

@Knight It's a picture of Ron Maimon banned on Quora

@Knight Ron Maimon

Banned on Quora? Oh My God! Are there any links

?

Just search for Ron Maimon. He is quite famous like Lubos Motl both being past users of Physics.SE.

All right! Now I see Unemployment made him an evil (or to portray like an evil so that someone could offer him a job)

@Knight Thanks, well

He is too blunt.

@peterh-ReinstateMonica Is that eagle in your profile picture or a hawk?

@Knight I'm almost sure that's an owl.

@Knight it's owl.

@JohanLiebert No not too much. But we cannot deny that his knowledge was nice and he started using curse words only after suspension from SE

@Semiclassical look what I found =)

from doi.org/10.1016/S0079-6638(08)70215-4

@Knight no he was suspended because he was rude not that it made him rude.

@EmilioPisanty doing some lines?

look at those beautiful cusps on the inside =)

@Knight It seems weird to say "we cannot deny" about events that you weren't around for...

@Slereah nope, just looking for good catastrophe-theory references for a paper I'm writing =)

@Knight "he started using curse words only after suspension from SE" is basically flat-out false. The nature of moderation means that hurtful content is removed (which is good!), and that means that there is often no openly-available evidence of what happened. But let me assure you, having been there when it happened, that it did.

I have enough powers to look at Maimon's posts now, I think

@EmilioPisanty though if someone is willing to dig into this stuff then there is a lot to get the idea of what happened then

Some objective facts that can be ascertained independently using the Wayback Machine: he was suspended several times on this site. About halfway through a one-year PSE suspension, he was given a ten-year suspension by SE staff, due to behaviour elsewhere on the network. This was then extended to ~300 years at some point, for undisclosed reasons which the people involved have asserted are nontrivial.

@Slereah My impression (from reading about it, having not been here at the time) is that it often occurred in comments too, so even with the privilege to see deleted posts we likely can't see a lot of the actual things that led to problems, since deleted comments work differently.

However:

discussing this further basically entails discussing the actions and interactions of someone, in a venue where they cannot post and thus cannot defend themselves. As such, this discussion should stop more or less about now.

ah, right on cue. Hi, cavalry @rob

Greetings

@rob how's things?

@EmilioPisanty My life is so much more complicated than I wish it were. Everything is harder than it should be.

@rob I think this bug has been reported previously

I had a great weekend! I took an oil painting class!

And I injured myself.

we're still waiting for them to fix it ¯\ _(ツ)_/¯

Honestly. Who gets injured painting still lifes?

@rob I agree

How about you, @EmilioPisanty? What's new in your world?

@rob currently sprinting to finish a paper by the end of the month

It's about halfway through, so there is definitely hope

@EmilioPisanty What sort of paper?

@rob somewhat long-form, somewhat technical

I'm trying to pin down a vague concept that's been around my field for ~25 years into a single, precise, canonical form

it involves taking traditional concepts and making them complex-valued

which is always fun =)

@EmilioPisanty That sounds nifty

yeah, I'm pretty happy with how it's turning out

@Knight Good question :-) Afaik I need to check it on the google images

@Knight Owl, yeah! :)

@peterh-ReinstateMonica Hahahahaha.

@rob Come on you raver, you seer of visions, Come on you painter, you piper, you prisoner, and shine

Ahah

I think I found the appropriate theorem to prove the $\int_\Omega f = 0 \to f = 0$

$$\mu(\Omega) \inf f \leq \int_\Omega f \leq \mu(\Omega) \sup f$$

Slereah

Educate me about the history of Dirac Delta Function (please ping me when you reply, it's a humble request)

I don't know the history of the Dirac function

you may google for it

My problem is: It was Heaviside who formulated Maxwell's equations in late 19th century and Dirac came into the ring around 1928, so how Heaviside found the divergence of $\mathbf E$ without using Dirac Delta Function?

@Knight Maxwell formulated Maxwell's equations, though it was Heaviside who wrote them in the form we are familiar with today. And I don't see why you need the Dirac delta to write down the divergence of the electric field.

@JohnRennie In $$ \mathbf E (\vec r) = k \int_V \frac{\rho dV}{R^2}\hat R $$

$$\nabla \cdot \mathbf E = k \int_V \rho~dV ~\nabla \cdot \frac{\hat R}{R^2} $$

Now, $\nabla \cdot \frac{\hat R}{R^2} = \delta ^3 $

Is that the way Maxwell derived it? Presumably not since he didn't know about the Dirac delta.

That's really I want to know my sir, you always understand me :) you're so cute and charming and GREAT

I have a vague recollection that Maxwell user quaternions in his derivation. I've never studied how he did it since the equations are far more accessible in Heaviside's formulation.

How did Heaviside do it? Dirac came in 1928

@JohnRennie

@Knight Maxwell had already done it and Heaviside simply rewrote his equations.

Just because the result can be derived your way does not mean that is the only way to derive it.

okay

@JohnRennie He he he he he he he . he he eh eheh :)

But I have to concede that not having studied Maxwell's original work I don't know exactly how the result was obtained.

@JohnRennie Sir what does it mean "fixing the servers" ?

@Knight I look after getting on for a thousand servers scattered across the North West of England. My job is to check them first thing in the morning to see if any problems have developed overnight.

Any serious problems I pass on to the engineers, but if there are problems that can be easily fixed I'll do that immediately.

@JohnRennie Is that your hobby or a job?

My job.

:)

Windows servers are pretty reliable, but with a thousand of them there's usually at least one that needs some tender loving care in the morning.

I read in 10th Grade that "servers provide data upon requests"

The term server generally refers to the computer. The computer can run many different types of program that provide data.

So you fix them by sitting in your room or you have to go there?

The simplest is just file sharing i.e. the server stores user files. But other common forms are database servers like SQL Server and mail servers like Exchange.

@Knight I can log into the servers remotely over the Internet, so I do everything from my desk. If a server needs a part replaced then our engineers do that.

basically it means that you use programming (if that's the word I want) for fixing the servers

No. But it's hard to explain how I fix servers unless you already know how they work.

@JohnRennie Out of curiosity, are you guys using AMD, Intel, or neither on the hardware side of things?

We use Dell hardware, and Dell use only Intel stuff in their servers.

I see I see. Although that might change :)

@dsm the new AMD CPUs are very impressive, but outside of the specialist data centre applications CPU speed isn't that big a deal for servers and CPU price is not a large part of the server price.

I'm sure Dell are looking at the AMD CPUs - they'd be foolish not to. But I doubt any of the server manufacturers are racing to switch to AMD.

It's definitely going to be interesting to see how things pan out with the big names, especially with continued delays with Intel. Overhauls do seem pretty intense, logistically. Exciting to see the top ten CUP best sellers on Amazon right now, as far as the consumer marketplace goes

The consumer and server markets are very different.

If I was building a gaming PC right now I'd be buying AMD.

@dsm Ryzen is definitely more bang for your buck over i5/i7/i9, but that isn't server CPUs.

But with servers CPU choice is less of a big deal.

Defintely, agree with both of you 100%. Excited to see some real competition in this area for once. I've got to imagine servers are or will be a high priority for AMD going forward

The specialist workstation market is where the AMD chips are likely to have the biggest impact.

The guys who buy workstations to do rendering or finite element calculations.

Traditionally you'd buy those workstations with 28 core Xeons, probably twin, and the 32 and 64 core Ryzens and Epycs are an obvious fit there.

Servers are more like just managing data effectively, whereas workstations are more like a bunch of calculations, that's the major difference isn't it?

Yes.

Workstations are basically single user.

@JohnRennie Say that AMD provided a significantly better option to server CPUs than Intel, in whatever metric relevant to your company. How much of a bear would that overhaul/transition be?

@dsm For us none at all. We just install Windows on the server then install whatever software is needed. We really don't care what flavour the CPU is.

Or perhaps that question is moot, seeing as how it would first come from Dell

For Dell I doubt redesigning their motherboards to use AMD instead of Intel is that big a deal. My guess is they have already done that as part of testing.

But suppose Dell come to us and say "we can now offer servers with AMD". Why would we buy them? If they were cheaper I guess that's a reason, but the cost of the CPU is only a small part of the cost of a server so they wouldn't be much cheaper.

I guess it would be nice to have servers with 64 core CPUs just for bragging rights :-) But none of our servers need that much power.

I imagine the main incentive for transitioning would be increased efficiency and lower energy costs for cooling a server complex. But I'm not sure if increased power goes hand in hand with increased efficiency for servers. Does it? I'd assume not. That seems like architecture.

Speaking from a viewpoint of knowing very little about computers

Hi all. Could someone have a look at the work I did in this question: physics.stackexchange.com/questions/531833/…,

After many trials, I can still not find a mistake in my calculation and I have no clue how to show from the given statement, that the propagator is zero for x<v_s t. Thank you in advance. Have a nice day!

Hello!

I have a tablet that shows Mass (gr) and Observed Period (s) values.

For example, Mass: 50gr - Observed Period: 0.72s

Mass: 100gr - Observed Period: 0.85s

Now, here is the question about this harmonic oscillator.

Draw the squares of the periods of vibrational movements as a function of mass. There mass values given on the table don't include the mass of the spring. Find the active mass of the spring by extending the graph.

I think I will use the $\omega^2=\frac{C}{M} \Rightarrow \omega=\frac{2\pi}{T}$ equations here.

$\omega^2=\frac{4\pi^2}{T^2}=\frac{C}{M}$

$T^2=\frac{4M\pi^2}{C}$

Is this right?

Looks fine to me (except that your $\implies$ is not really an implication :P), what's the problem?

Yeah, exactly :D

Actually the original question is this:

4) Mass-Spring System/Datas

The data shown on graphs 7.3 and 7.4 are taken by observing a mass-spring system.

a) Draw the squares of the periods of vibrational movements as a function of mass. There mass values given on the table don't include the mass of the spring. Find the active mass of the spring by extending the graph.

b) Find the spring coefficient.