Let me summarize what I'm saying as such. There are two frames of reference so far: the lab frame (which is the same as the nuclei's frame), and the electron's frame.

LOL ... poor you.

very nice lecture though

Calculate in closed-form $$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

Please show me your way for my question above (my solution here is pretty elegant I'd say and hard to guess to be honest).

Well from that simple intuition you reformulate classical mechanics using the notion of expected value, and then mathematically decompose a probability distribution into wave functions, and then from the wave function you assume that it has a "quasi"-classical approximation of the form $e^{iaS}$ where $a$ is a conversion factor to make the exponent dimensionless, that $a = \frac{1}{\hbar}$, i.e. $\psi \approx e^{iS/\hbar}$

I still remember the proof of cauchy schwarz In most analysis books they just show how to do it using quadratic formula without telling you where it comes from!

@Semiclassical What...

In the lab frame, there are equal densities of electrons and nuclei. That's also true in the electron frame, since both will change by the same amount in changing the reference frame.

but your exposition was excellent @TedShifrin

Thanks.

@Semiclassical Can you check my understanding of the derivation of magnetic field from Lorentz transform?

@BalarkaSen something else unclear?

Not yet.

Let me finish my summary.

Okay

@Danu I am unclear about what the commutators mean (I am guessing covariance), as I mentioned a couple messages back. Did you reply to it? Chat's moving too fast for me today.

@BalarkaSen that is the uncertainty principle, it's not just the intuition for it, it is it if you understand it

In the lab frame, the nuclei are at rest and the electrons are moving. Consequently, the electrons comprise an electric current and so produce a magnetic field.

@Semiclassical $$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

@Jake1234: See Exercise A.2.4 in my geometry notes. (There's a solution at the back.)

The electrons in each wire experience the magnetic field produced by the other, and therefore are attracted due to the Lorentz force.

@TedShifrin do you by any chance got a electronic version if your book ?

Now, in the electron frame, the electrons are at rest and the nuclei are moving. Moreover, they're moving in the opposite direction that the electrons were in the lab frame.

@Adeek: Not for public consumption, no, sorry.

@Semiclassical I remember you used to be a fan of such questions.

@bolbteppa I don't see where you explained how $h/4\pi$ comes from except giving a link to something on the internet, to be honest - which was what my question was about.

@BalarkaSen Sorry, what do you mean by covariance?

okay

Covariance of random variables, as in statistics.

...hrm. Something is weird here.

aka, $E[X]E[Y] - E[XY]$.

The conclusion I'm being led to (which I don't entirely like) is the following:

So now you have currents due to the moving nuclei, and these again attract each other owing to the induced magnetic fields.

@robjohn posting here such questions seems a loss of time $$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2,$$ I see no interest in them and I wonder why.

@user1618033 Kind've in the middle of answering something else...

@BalarkaSen What is the problem with just understanding it as a commutator: $AB-BA$?

@LeakyNun So in both cases, there's a net Lorentz force on the wire (which is comprised of both the electrons and the nuclei).

Hence, attraction in both cases.

Sure.

@Semiclassical OK OK, you're right.

Now, what I don't entirely like about that is that the particles experiencing a net force changes depending on the reference frame.

@BalarkaSen To be honest, I was responding to your actual quote "So, I don't understand the uncertainty principle" before the rest, good luck with that so

Back to my work.

The other is that my argument about charge densities has the following rebuttal.

@Danu It's unclear to me what the commutator represents when you're working with position and momentum as your random variables, maybe?

@bolbteppa You didn't read the question, then.

I said I started with charge densities $(n_e,n_n)$ in the lab frame, and then obtained new, larger charge densities upon boosting to the electron frame.

but it seems like I could turn that around, and argue that boosting back to the lab frame should also increase both charge densities.

And that's definitely preposterous.

big word.

@BalarkaSen Random variables?? I'm working with linear operators on Hilbert spaces.

So I've definitely gone wrong somewhere in here...

You have successfully lost me, then, @Danu :)

@BalarkaSen Excuse me for responding to a direct quote :\ Since you were worrying about whether the same $h$ from $E = hv$ was in a different question as if there are many of them I can only assume you don't have the fundamentals down and go back to basics

@bolbteppa Dude, that's not what I asked. I know that $h$ and this $h$ are the same, I am asking why the hell they are the same.

@BalarkaSen Alright. So, in what sense do you regard the things I am thinking about as operators as "random variables"?

@BalarkaSen dude you literally said "I am probably asking for a -- at least moral -- derivation of the principle" so I gave you the most moral derivation of the uncertainty principle there is

@Semiclassical So can you check my understanding?

At this point, I probably can't since I don't trust my reasoning right now. Something I wrong with what I'm saying.

@bolbteppa No, you're misunderstanding. I asked for a moral derivation of the upper bound of the product of those two things being $h/4\pi$.

What the uncertainity principle means intuitively etc are just explained in my textbook, I don't want another explanation for it.

Relax, guys. No point in arguing.

on dimensional grounds, one can say that, if something like the HUP exists, it must take the form $\sigma_x \sigma_p\geq C h$ for some constant $C$

@BalarkaSen Did you see my answer regarding the numerical value?

I am annoyed about misinterpreting a question by reading just the first line and not the rest of it, especially when the question is 2 lines.

and the question is then "why $C=1/(4\pi)$ in particular"

@Semiclassical Let's say we have charge densities $(n_{eL},n_{eE},n_{nL},n_{nE})$.

Right.

We have $n_{eL}>n_{eE}$

We also have $n_{nL}<n_{nE}$

@Danu I haven't studied quantum mechanics at the level of generality you're speaking of :) I am just thinking of momentum and position being two random variables and uncertainity principle saying their standard deviation having product more than $h/4\pi$.

So everything is conserved.

@Danu Let me look back.

@Semiclassical A self-aware physicist.

In the electron's point of view, electrons are not moving.

That's what we should have, yes.

And what's the problem?

heya @PVAL

Hi @Ted

Well, when I say "that's what we should have" that's on the grounds of the two reference frames being symmetric with one another.

@BalarkaSen Yikes, okay :P

@Semiclassical And where's the asymmetry?

After all, it's just a chapter on high school chemistry, @Danu :)

@MichaelA If I do end up at SB in April, do you want to look at that $S^6$ paper?

Hang on. I need to sort my brain a bit.

@BalarkaSen Sorry, I automatically assumed you'd be thinking about the "sophisticated perspective", so to say.

@Semiclassical I'd advise you to use Quicksort to do that in $O(n\log(n))$ time.

@MikeMiller Is there any way that $S^6$ paper is correct?

@user1618033 Did you post that one? There are so many questions these days, that I don't see many that I consider interesting.

Heya @robjohn. Lovely to see you for a change!

In the meantime, can anyone tell me how it is derived that $\displaystyle\sum_{r=0}^\infty\frac{1}{\tbinom{2r}{r}}=\frac{36+2\sqrt3\pi}{27}‌​$?

I think here's where I'm bothered. On the one hand, it seems correct that $n_{eE}=\gamma n_{eL}$.

@LeakyNun That series diverges.

where $\gamma=1/\sqrt{1-v^2/c^2}$. Lorentz factor etc.

@PVAL It's not impossible. I have heard one person say it's the most credible attempt they'd seen (but still decided against reading it because if it was right eventually someone would write it better)

@TedShifrin You can see me?

@robjohn Nope. The series converges to exactly that limit.

@Semiclassical sure

Seems like it might be fun to read anyway.

the question is what $n_{nL}/n_{nE}$ should be.

@robjohn Well, electronically speaking :)

@robjohn oh, wait

@robjohn better

@Semiclassical Isn't it also $\gamma$?

Well I just remember hearing about a few errors pretty close to when it was posted which were immediately "fixed" according to the author. But when people poke holes in the argument so quickly, I find it very hard to not be skeptical of a result.

@Semiclassical Because $v_{nE}=v_{eL}$.

That's what I'd need to avoid my 'preposterous' statement earlier, yes.

@PVAL: I'm with you.

and $v_{nL}=v_{eE}=0$

wait

I remember that too.

I think R. Bryant's paper on Chern's attempt to prove nonexistence is probably a better paper.

@Danu Ah, alright. It kind of spooks me that $h$ appears in the uncertainty principle because it's not morally clear to me how to get to the uncertainty principle from the slogan "energy comes in packets".

@PVAL I've never seen that. When did it appear?

That's on the list of papers I'm going to read that I;m never going to read.

@Ted Idk if its published its on Bryant's website though.

Let's say both at once. $n_{eE}=\gamma n_{eL}$, and $n_{nE}=(\bullet)n_{nL}$.

Sure

@LeakyNun That one can be finished in many ways. I think you might like to discover alone some ways.

What should $(\bullet)$ be?

Hmm, I stopped keeping up with Robert's preprints a decade ago. Let me look :)

To avoid my preposterous conclusion, I'd say it should be $1/\gamma$.

@Semiclassical $\frac1\gamma$

@BalarkaSen After thinking about it for a little bit more, it seems to me that even a factor ~10^3 would not make a HUGE difference

arxiv.org/abs/1405.3405

@BalarkaSen Also I don't think that is very clear to most people.

I really feel like someone should just solve this already.

Right. And for some reason that really bothers me

@Ted @Mike That's the paper if your interested.

I am.

Thanks, @PVAL :)

@Danu Interesting.

@Semiclassical Why?

It'll be full of moving frames and prolongations, I'm sure.

@robjohn That is... beautiful.

It seems to be unpublished. I guess Bryant didn't want to publish someone elses work, even if he wrote it up.

For one, because it'd mean $n_{eE}+n_{nE}=\gamma n_{eL}+\frac{1}{\gamma}n_{nL}$

@LeakyNun I also have an answer that can be modified to give a generalization of that sum. Using $(2)$ and $(3)$ from this answer.

which would mean that, if you start out in an electrically neutral frame, then you can boost to one where it's not.

Hi @rob!

@J.M. Hey there! How are you doing?

@BalarkaSen I'm tempting to conjecture 10^6 could still be ok too, but I'm not sure :P

@robjohn pretty okay, all things considered. How about you?

@Danu How are you coming to these conclusions/conjectures?

@BalarkaSen Alright, just for reference - statements like "So, I don't understand the uncertainty principle" ... "I am probably asking for a -- at least moral -- derivation of the principle" ... "The same $h$ that appears in $E=hν$?" imply you have no idea what it really means, and when I respond directly to the first quote and you say that's not what you asked I can only say wtf tbh, the question you asked afterwards is directly answered here (P47/48) by a simple inequality:

books.google.ie/…

@robjohn nice

@robjohn Not on main yet.

I'll be back in 30 min.

@BalarkaSen Nothing concrete

@bolbteppa Just drop it. Don't take it personally.

When does your one physixs friend get on, @Danu?

@robjohn nice

@MikeMiller Being who?

@robjohn one can do it even simpler.

@bolbteppa I think Danu and SemiC understood my question just fine. But I appreciate your explanations. I don't claim to understand the uncertainty principle (I am no physicist), but I have been told and I have read the "physical meaning" of the principle multiple times already. It doesn't answer the question I have though, sorry.

@J.M. Doing really well, but busy offline a lot these days

Anyway, I have to go for 30 min.

@user1618033 The generalization?

The question seems to be "why $h/4\pi$ and not, say, $h/2\pi$"

@robjohn Very cool :)

@JasperLoy: proofs of transcendence are not such 'small things' imo :P

@Semiclassical Yes.

@robjohn Yeah, real life tends to get in the way of your hobbies. ;)

Hi, Jasper. I saw your message about the inscribed circles. What's the easiest proof?

Hi @Jas, you are lemon-yellow in this current incarnation. :)

@J.M. besides getting complicated in itself.

@LeakyNun I'm looking for an authoritative reference right now to clear this up, since I don't seem to be remembering it correctly.

Well I have provided answer to what the uncertainty principle is, and exactly where the specific inequality you asked for comes from, maybe you don't recognize equation (16.7) in that link because the book writes $\hbar/2\pi$ but since $\hbar = h/2\pi$ it is $h/4 \pi$?

@TedShifrin Bonsoir

Bonsoir, @JeSuis

@TedShifrin comment allez vous ?

@Semiclassical Alright, I'll deal with this tomorrow.

It's 1:36 AM here now.

oof.

get some sleep, yeah

it's 12:36 pm here, so we're on opposite sides of the globe

@JeSuis un fou

Ça va, merci, @JeSuis, et vous?

@Danu The one that understands me when I spew my math garbage and asked about $\delta^2$.

@LeakyNun lol

@JeSuis un violon, pour l'amour de Dieu

@TedShifrin je ne sais pas pourquoi j'ai dit vous, on avait passé cette étape, je vais bien, merci. What's up ?

LOL, well, I didn't want to be rude and tutoie you :P

@LeakyNun Looking at some references, it seems to be a more subtle question than I gave it credit for

due to the difference between the fields produced by a moving charge and an electric current. it's subtle.

Yeah, @Jasper, I knew all that. I guess I couldn't do it right in my sleep. I know I had this figured out years ago, because I put questions on our high school math tournament exam about it.

@LeakyNun This paper seems to give a nice presentation on the problem: citeseerx.ist.psu.edu/viewdoc/…

@Jasper: So you get $r = A/s$, where $s$ is semiperimeter. Where does Pythagorean come in?

@TedShifrin :D, no rudeness.

oh, that occurred to me in my sleep, too, but without paper it didn't work. Let me try.

@Danu I meant ACuriousMind. He's now helping me with my nonsense.

jeez, conversations in this chat get me more tired than being sick

Oh, duh, I see, @Jasper. I didn't do the last bit of algebra in my sleep :P Thanks!

there are quite a few conversations going on right now. hard to follow

I'm already senile.

Thanks, Dr. Jasper :)

It's there.

@JasperLoy that's a bit much innit? or at least, you start having an intimate relationship with porcelain.

I should go back to mute mode like I did a couple days ago.

@JasperLoy well, if you sweat it out I guess... :D

but the humidity is torture.

@TedShifrin as-tu une idée sur l'exercice suivant : On considère une application $f:\mathbb{R}^2\to \mathbb{R}^2$ telle que $|| f(y)-f(x)||\leq || x-y||$ pour tout $x,y dans \mathbb{R}^2$. Pour tout entier naturel $n$, on note $f^n$ la $n^\mathrm{e}$ itérée de $f$. Démontrez que, pour tout $x dans \mathbb{R}^2$, la suite $(f^n (x)/n)$ converge.

@J.M. It's easy to sweat out lots of water here. The humidity is very low.

@robjohn You're on the other side of the world, my friend. :) The tropics do not have that luxury.

@robjohn Last summer was quite humid. :(

@JasperLoy It must have been accidental

@JeSuis: Il me semble que la suite $\|f^n(x)\|$ est bornée?

@JasperLoy Look at my avatar... do I look angry? ;-p

@robjohn Perennially sad?

la série ?@TedShifrin

It's a very broad field, @Jasper, so it depends on one's interests/tastes.

oops, @JeSuis.

@TedShifrin Here or in the south?

Here. When I got here the end of July, it appeared I had brought the south with me. You and I concurred.

@TedShifrin Yes. It was pretty humid.

@Jasper: I'm fond of Griffiths/Harris, actually. But Ronnie Wells's book is clear. I even learned a lot from Morrow/Kodaira.

There is also DeMailly's relatively new book, @Jasper, which you can download free.

But it has two years' worth of material in it, @Jasper, although a few mistakes, too. :P

I have heard praises of Demailly from prof.

I know. I have pages of errata.

Jean Pierre Demailly ?

@Jasper: I also have a scan of the notes from the course I taught at MIT in 1980. I proved all that stuff in the course.

Oui, @JeSuis, je crois.

The price is not that bad, on account of I have a copy. I still don't understand the value of endless book discussion without combining it with endless reading.

@TedShifrin Quel livre ?

Géométrie complexe.

I hope you can start doing some mathematics at some point. I feel that would be much more rewarding than thinking about doing it, personally.

It always takes me a day to get everything working when I update TeXLive. So I think I'll skip it this time :P

@JasperLoy Yeah, it depends ONLY on you.

Well, it is a fresh install. You don't need to delete the old one, actually. But I have all sorts of extra fonts and it takes me a while to get them recognized, every time.

@JasperLoy If I considered the advice given on this chat I would have been dead mathematically. Try as much possible to depend only on your forces in all you do.

You have the last word about your success!

@JasperLoy nice colour :)

well well .. if it isn't the Imp

@r9m which institute are you in now? :D

@Sawarnik right now? .. the institution called 'home'!

@r9m wah .. and what do you plan for future?

@Sawarnik I plan to leave the plan making for future ..

@robjohn there one can also use the beta function, but I was mainly referring to the particular case. Then there is also a way related to some clever telescoping sums (which I didn't post anywhere so far).

@r9m have you seen this one? :-) $$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

@user1618033 not until now ..

I'm out to jog a few km, and then return back, but not before finishing some stuff (that is like 80% done).

@r9m Hope you like it. :-)

BBL

FuLly equipped and ready to go out but I return from the door for one thing

...

^^^ Star it!

BBL

Hello!

How do I make "n" the subject of this? n(n - 1) = x / 5 ???

@Monad You mean put it on the form $n= \cdots$?

??? Your thing has dollars signs, a backward slash, dots, and idk

There is a link to the right on how to enable the processing of it

but what I wrote was just n = ...

I'm trying to make it so n = something that doesn't contain "n"

Have you seen the quadratic formula?

Actually yeah I could use that

But... all I'm trying to do is invert a function I have... It's 5nn - 5n, the nn is n squared. And if in a sequence form 1, 2, 3, 4 = 10, 30, 60, 100, but I'm trying to reverse that.

Try writing out the quadratic formula (remember that you need the positive solution)

I used an online calculator to find the inverse function of 5nn - 5n... and I replaced the plus-minus with +, seems to work... guess I'm just going to floor(inverse) then.

Jes

But I'm doing something with programming, I was trying to find the easiest solution

and I did... so it's fine

Ik Ik it's but a re-arrangement

o/

Good nights,

Yay I made it perfect... it's so good! :)

The probability of logical perfection is 1 over infinity, some consider "their" perfection, their imprefections, so for them perfection has a probability of 1 - 1 over infinity.

aka 0.9 recurring

Weird shit happens when you mix mathematics and philosophy :T

Jasper Loy not exactly :T

Yes exactly

^

@JasperLoy Spinoza? :P

@JasperLoy Does that continue 24/7?

Ughh... this is worst than I thought... I thought my teacher meant there's hope in the world but nvm,

@Monad Glad to be of service.

@JasperLoy OK :) I find I drink between 4 and 9 liters per day.

@Monad Try to find a number between 1.000... AND 0.999... ?

Skull Petrol 0.99904

:D

how is that between them?

1.000 - 0.99904 > 0 & 0.99904 - 0.999 > 0

oh did you mean recurring?

@Monad You seem to be ignoring the dots after the number

Yes^

Those three dots mean a lot :P

Oopsie daisy... define "find"... if you're asking me to write it out then I can't the same way you couldn't express in terms of digits from 0 - 9 even though conceptually it could be.

couldn't express infinity*

@Monad there is no number between 1 and 0.999...

@Monad Give a proof of existence

(you will fail)

@Danu just like giving proof of infinity :T

@Monad infinity is not a statement, so it can neither be proven nor disproven

@Tobias oh yah there's no number, and?

@Monad and hence they are equal

a - a = 0 means a = a @Monad