The h Bar

the gif's hypnotizing me a little

yeah i have seen the explanation. iirc after the rarrangement it's a little shorter in height

and width

user228700

Ugggggh, eff that. Not able to get it to link correctly.

user228700

@BalarkaSen Yep.

The Banach–Tarski Paradox

►

That's some crazy video.

10:04 AM

@Kaumudi.H I'm not sure that works with chocolate :-)

user228700

YES. Jesus Christ, yes, that's the one.

For youtube you can just post the URL and the chat software will automatically onebox it.

user228700

@BalarkaSen It is! Took two re-watches to fully (sort of) understand everything.

user228700

@JohnRennie Unfortunately, no :'-(

Yeah, super-weird stuff. I actually don't know the proof of Banach-Tarksi.

*Tarski

10:05 AM

@JohnRennie I wish we can watch video from that small box without opening new tab

@BalarkaSen there are a number of YouTube videos that go through the details, but it's quite involved. I have to confess I lost interest partway through.

I thought the video was very well made, but stopped midway because I didn't want to learn proof of a theorem I don't know from youtube :P

I'll read it at some point eventually

@BalarkaSen what's your favorite proof in math?

Mine is the proof that all triangles are isoceles :-)

@Fawad hard to say. Most basic, but dazzling proof I remember seeing is Euler's (not Euclid's) proof of infinitude of primes.

Using $\prod (1-1/p)^{-1}$

for complicated theorems i can go on listing but...

10:10 AM

@BalarkaSen yes,I will look on internet

I like the proof of insolvability of quintics. Great, but very complicated.

I like the proof that Euclid's fifth postulate cannot be proved from the first four.

is there a constructive proof of that, or is it just by contradiction

it's a constructive proof by Gauss. he constructed hyperbolic geometry, where the first four holds and the fifth doesn't :P

thereby proving logical independency

well that is a proof by contradiction :p

@DanielSank I'm...not sure what I expected ;) If you have to jump through hoops to get it here, just leave it be and drink it yourself, I appreciate the gesture anyway :)

10:14 AM

Ah yeah good point

How many euclidian axioms even hold for general manifolds

Most of them break down if there's singulrities

@JohnRennie Obligatory denial: That's not my sock :P

@Slereah 2nd postulate breaks down if every geodesic is closed like in a sphere

1st breaks down if the manifold's incomplete i guess

@ACuriousMind I almost posted a comment to suggest you should be more discreet in choosing the usernames for your sock puppets :-)

3rd is mostly garbage

10:16 AM

@0celouvsky As a physicist, I would use it it anyway ;P

@JohnRennie I mean, if I had need for a sock and created one, I would probably choose a name like that to make it obvious it's mine

I'm now tempted to create an account called JRisGod just to see how long it lasts :-)

If you don't vote for yourself or something, it would last indefinitely :P

@Slereah Didn't @0celouvsky once spend a considerable amount of time formulating these in differential geometric language? He should know the answer

@ACuriousMind not active users aren't removed?

@Fawad No, why would they be?

For the record, chat room users, don't create sock puppet accounts. Instead hack other people's accounts and use them to vote yourself up.

2

10:21 AM

Maybe there's some automatic deletion I'm not aware of, but generally, user accounts are only deleted if they violate rules

@JohnRennie you have some free time?

@JohnRennie So, what's your account password?

3

@Fawad I'm supposed to be working on programming task due on Monday but I'm having severe motivational issues so yes I have some free time :-)

@Fawad However I prefer to chat here in the main room not in a separate chat room.

@BalarkaSen 8 asterisks - it fools everyone.

@JohnRennie so I am not going to think others disturb now. I will ask you here.

Iam looking for a text which looks something like this:

I don't think the chat is currently discussing anything of major importance :-)

10:26 AM

@BalarkaSen Impressive haxx0r skills on display there

3

@ACuriousMind it works more often than you might think

@ACuriousMind What can I say? I'm really good at it.

@BalarkaSen ********* ?

trying to think of an 8 string curse

10:28 AM

@JohnRennie that text is real,you can see it's in this person name

yikes there's too many

@Fawad I have no idea what that is. Some form of script I'd guess, but I don't think it's Arabic.

@BalarkaSen Just glue two four letter words together :P

@ACuriousMind I used to work for the KGB, I know everything about hacking.

Soviet ~~philosopher~~ spy confirmed.

10:31 AM

@BalarkaSen at age -15?

philosopher and spy are not mutually exclusive

@BalarkaSen hmm electrodynamics has more than eight letters and so does fluid mechanics ...

William Lee was a spy for the Interzone

@JohnRennie You really have quite a stock of curse words

@JohnRennie ratsauce?

Old Monty Python joke incomprehensible to anyone under the age of forty

10:33 AM

caught up on the chat and I have to say, Gibraltar? That's a bit of a controversial colony to bring up isn't it? I much prefer the less controversial ones, like the Falklands

less controversial ones, like the Falklands - you're not Argentinian then?

@Phase you break your streak :O

Sarcasm's hard to portray with text tbh

Is there a sarcasm tag in mathjax

@Fawad Whatever do you mean by that sir

There is irony punctuation in unicode, but nobody uses it :P

I didn't even know this existed, well, TIL.

Doesn't it kind of detract from Irony to make it clear that it's ironic before it's even been read?

10:37 AM

@JohnRennie you know any site where I can ask for it?

@Fawad not offhand ...

@Phase Sort of, but I'd argue that in actual conversations, you often can tell without parsing the content just by the gestures and intonation whether something is ironic, too

@JohnRennie I asked in few chat rooms.I didn't found any suitable site to ask. And no one know about that in other chat rooms.

@ACuriousMind True

I have a friend whose normal cadence is close to what irony sounds like from other people. It happens frequently that she's asked "Was that serious or were you being ironic?" :D

10:42 AM

When I'm being ironic it's pretty indistinguishable from when I'm talking normally

same with sarcasm

But that might just be from being British

There is commonly used irony punctuation nowadays

it is just emojis

:laughing crying emoji: :hand emoji: :100 emoji:

centipede emoji

When can SE chat get emojis? Should I make a Meta post about it?

that's my favorite

10:43 AM

Here is a useful ironic emoji

🤔

2

🤔What did they mean by this?🤔

@Phase Dear god please no. I much prefer the ASCII signs over these tiny yellow faces that I can't really distinguish on a large monitor.

🤔really makes you think

🤔Hmmmmm🤔

🤔that is very interesting🤔

10:44 AM

I also think that my browser renders most unicode emojis as the same faces, come to think of it

@ACuriousMind What's wrong, are you confused? 🤔

Like, I don't see a difference between any of the emojis you posted

$$🤔\int (R - \Lambda) e d^nx🤔$$

🤔In the grand scheme of things, are there differences?🤔

Jesus christ

@Slereah blank message?

10:45 AM

your computerotron probably can't display emojis

Anonymous

My browser doesn't even render the emoji's

Anonymous

I'm getting blank boxes

you can't break chat using emojis, but you can break with ASCII

$\frac{d}{d 🤔} e^{🤔} = e^{🤔}$

🤔🤔🤔

@BalarkaSen and somewhat by using zogla too

10:46 AM

Obviously @Phase does not remember that time we did the Einstein Field Equation in emojis

I don't think I was there then

Sad, I wonder what it was like 🤔🤔🤔

lol

:36934884 🤔what did he mean by this🤔

So I've just been copy pasting these, what's the actual input for them

It is unicode symbol 1f914

I was thinking of using smarturl and the emoji movie wiki to make some kind of crap joke, but I read a bit, and did you know Patrick Stewart is going to play the Poo emoji? Just let that sink in

10:49 AM

@Phase alt + f4

O jee thanks lemme just try that

Ohhhh Noooooooooooo

alt+f4 is too obvious

You should have said windows key + L

@Phase Oh my goodness, it's true!

the things we do for money

but well, you know

Tbf there are worse ways to make money than voicing an anthropomorphized poo

10:55 AM

for that money I'll be a poo in your dumb movie, too

A teaser trailer was recently released to largely negative feedback online

you can buy a lot of GR for poo money

@Slereah dumb movie? 🤔

I don't want to judge something without seeing it, but

here we are

Wow

this is

Wow

10:57 AM

I predict that the emoji movie will not be very good

Anonymous

@JohnRennie It would have been better with kid voices...

Bold move, I know

@Slereah heresy

Anonymous

The bored tone of the characters ruins it

@Phase well it's a sort of x-men in that sense

10:58 AM

@Slereah They said the same thing about Pulp fiction

could be

But I do not fear being wrong

X-Men: Days of the future poo

If it is the new citizen kane, then fine

I think the Emoji movie needs a different director

Days of future pants

10:59 AM

Like tarantino

or Michael bay

yeah i vote for tarantino

"do I look like a poo?"

or maybe lars von trier

David Lynch

"then why are you trying to flush me like one"

11:00 AM

Or John Waters

@Slereah David Lynch is fine, you know.

John Waters would make a great emoji movie

he's a good guy

I know

Although to be fair

I only like one thing of his

"Rabbits"

Oh

I've just had a terrifying premonition

Actually I'm gonna keep this to myself, at least until I patent it

I'd hate to get Einstein'd by a patent clerk [UPSIDE DOWN QUESTION MARK]

11:02 AM

@Phase is it about an emoji version of Final Destination?

Emoji version of Serbian Movie

user123733

heya

@user123733 ayeh

@BalarkaSen no but thanks for the patent

Anonymous

11:07 AM

@BalarkaSen Another question :P If $f(x)f(y)=f(x+y)$ for all real $x,y$ then which one should be correct? f(0)=0 or f(0)=1? It seems like both can be true.. [Additionally the question mentions f(5)=2 and f'(0)=3, but I don't think that's relevant here]

both can be true, yes

you're gonna need more info

for $\exp$ it is $1$, for the constant zero function it's $0$

Slereah answered it, yeah.

Anonymous

Well, do we actually need to know f(0) to find f'(5) ?

I don't think so

I can solve it with just f(5) and f'(0)

Anonymous

@BalarkaSen How would you do it? I was trying to use the limit definition of derivative

11:11 AM

@blue f(0)=1 I think. If $f(x)f(y)=f(x+y)$ means $(k^x)×(k^y)=k^{(x+y)}$

Anonymous

lim h->0 $\frac{f(5+h)-f(5)}{h}$

f(5 + h) - f(5) = f(5)f(h) - f(5).

take f(5) outside

OH. f(0) might not be 1

breaks my heart

Anonymous

@BalarkaSen Exactly

Anonymous

That's what I was thinking

The exponential function obeys that property

and $e^0$ is very much 1

Anonymous

11:12 AM

@Slereah But that's a special case...

Anonymous

Exponential function is not the only solution to the functional eqn

Anonymous

(Probably)

Yes

constant $0$ function does obey it

It does seem like you need more information. f(0) = 1 is probably needed.

Also $1$

11:13 AM

@ACuriousMind what am I alleged to know?

where to find the pirate's gold

@ACuriousMind so it doesn't hold in general, the issue being that the fundamental theorem of calculus need not hold

For absolutely continuous functions it shoul work

Anonymous

@BalarkaSen This was a AIEEE problem from 2002. They usually don't give wrong questions. But as of now it seems like "insufficient information". Thanks though

In particular loos hits ones

@blue OK, wait. f(5 + 0) = f(5)f(0).

f(5+0) = f(5) = 2

That means f(0) = 1 by cancelling

11:15 AM

@0celouvsky I think it's about a question I asked

Yeah I think it might only be 0 for the zero function

The essential info was f(5) = 2 :) That restricts the function severely; you can't be the zero function.

Anonymous

@BalarkaSen Ah ha!

Anonymous

Cool

Anonymous

I didn't see that :P

Anonymous

11:16 AM

Good question then :)

Tricky, yep.

I found a SE post that quoted from a textbook that $\sum_i \vert x_i \rangle \langle x_i \vert = 1$ is a false assumption, and was confused. Is it to do with the properties of infinite dimensional hilbert spaces?

@BalarkaSen what you think of my approach,co‌rrect?

@Phase If the $|x_i\rangle$ form a basis, it's true.

you're assuming that it's a power function

Dangerous assumption

11:18 AM

You can show that converges to the identity in the norm topology of operators.

@0celouvsky Well then that's even more confusing, it was this post physics.stackexchange.com/questions/328817/…

because now you have to prove that all functions obeying that property are power functions

Anonymous

@Fawad That's not true always...

I guess I must be misinterpreting it

@blue example?

11:19 AM

@Phase $\sum |x_i\rangle \langle x_i|$ and $\int |x\rangle\langle x|dx$ are completely different!

More on this later, I have to get going or else

@Phase There are no countably many $\lvert x_i\rangle$ for position

Ok, but that's not what I was confused about

Anonymous

@Fawad $f(x)=0$

Well I see you don't have time for this,you carry on @blue

@Fawad How do you conclude $f(x)= k^x$?

11:20 AM

and @ACuriousMind does that affect it?

The assertion is not really that the sum-identity is false, but that the sum doesn't make sense to begin with in this case

@BalarkaSen by seeing it's property how it behaves?

The integral is also of dubious validity

But you don't need it anyway

That doesn't make sense.

$f(x) = 1$ satisfies $f(x)f(y) = f(x + y)$

Anonymous

@Fawad There are many more functions which have the same property. See the example I gave.

11:21 AM

(The integral doesn't make much more sense, actually, but it works :P)

so |psi> != Sum |xi><xi|psi> ?

@BalarkaSen that's $1^k$

@Phase What are the $\lvert x_i \rangle$?

I would assume the basis vectors of position

@Slereah Ok. 0 it is then.

11:21 AM

$0^k$

And another probably heinous assumption being that they are basis vectors in an infinite dimensional hilbert space

^and orthonormal ofc

@Phase Well, then that's the point, you can't label the "basis vectors of position" with an $i$ you can sum over - there are uncountably many of them labeled by real numbers, not integers.

You mean 0^x. That's not defined for nonpositive x.

oh

@BalarkaSen Well he didn't specify the domain

11:22 AM

I see I guess

And actually, these vectors do not actually lie inside the Hilbert space (their norm is not defined) and do not form a "basis" in the proper sense at all

Is it because $\aleph_1 > \aleph_0$ that you can't represent it with integers?

Anonymous

@Slereah I said for all real x and y....

Alas, wave mechanics works mostly if you pretend that they do and just replace the sum by a formal integral sign

Oh well :p

11:23 AM

@Phase yes

Yes

Ok, thanks for the help @ACuriousMind @0celouvsky

I wonder how much hypothesis one needs to solve $f(x)f(y) = f(x + y)$ by $a^x$.

$\log f(x+y) = \log f(x) + \log f(y)$

So it's a Cauchy functional equation on $g(x) = \log f(x)$

On the positive reals, you need $\log \circ f$ to be continuous. In particular you need $f$ to be continuous.

Hello @heather

@BalarkaSen I'm pretty sure continuity is sufficient.

11:26 AM

Yeah I think so too.

So that's it folks. It's hard to write down a counterexample to what @Fawad is claiming because they are nowhere continuous.

Anonymous

@BalarkaSen Can we prove f is continuous from the given conditions f(5)=2 and f'(0)=3...?

But they exist.

Anonymous

for f(x+y)=f(x)f(y)

Anonymous

Say f(x+h)=f(x)f(h)

@blue Differentiability at 0 should mean it's every differentiable hence continuous

I think

Anonymous

11:28 AM

@BalarkaSen Wait, how?

@blue it is continuous. My example/approach wasn't a continuous function...

Hello @Secret

Anonymous

@Fawad the power function is continuous...

@BalarkaSen can you prove they exist, though

f'(x) = lim (f(x+h) - f(x))/h = lim f(x)(f(h) - 1)/h = f(x) f'(0) [because f(0) = 1 from the hypothesis f(5) = 2].

so it's everywhere diff, hence everywhere cont

Is there an ideal for addition of the real

If there was you could define a weird function defined on that ideal and defined differently outside

Anonymous

11:30 AM

@BalarkaSen Right, then I think we can derive the power function from that functional eqn

Anonymous

Given f is continuous and differentiable

Anonymous

And follows the given conditions

@Slereah yeah but only using axiom of choice

Well I guess pick that

@blue On the positive half-axis, yes, by this

11:31 AM

the function is $= 2^x$ on that ideal, $3^x$ outside

@Slereah you searched on internet if not then how you got that?

Anonymous

@Fawad differential eqn should do it

Because that's like the basic property of the exponential function

Anonymous

f(x)=f'(x)

Anonymous

y=dy/dx

Anonymous

11:33 AM

ln(y)=x+c

@Slereah That's...a question I never thought about.

@BalarkaSen There is?

That's so counterintuitive that I will require a supporting link

Ah, right, just f'(x) = f(x) f'(0) gives a differential equation and you can solve that. No need for Cauchy on particular problem

well suck it up AIEEE

@ACuriousMind Examples of nonlinear functions satisfying f(x + y) = f(x) + f(y)? It's on that wiki link.

Oh, no, not that, I thought you were referring to the ideal thing

Oh

@Slereah There is no such ideal. Suppose we have $I\subset \mathbb{R}$. Then $x + y \in I$ for all $x\in\mathbb{R},y\in I$, but for $z\in \mathbb{R}- I$, $y = z - x$ clearly makes $x +y$ lie outside $I$.

11:36 AM

So is there such an ideal?

Ah ok

I didn't read the ideal question

too bad

Can't think of any other way to build such a function then

Ideals only make sense in rings etc. where the operation is not invertible

Fields don't have any ideals except the zero ideal and the full field

The hyperreals have an ideal

Every infinitesimal is an ideal under multiplication

Oh wait, only for finite numbers

I guess the hyperreals without infinite numbers don't form a field

nevermind

Emilio Pisanty

Yo @AccidentalFourierTransform did you LooksOK this one? physics.stackexchange.com/review/first-posts/172637

I think it speaks pretty badly about the queues that that answer went through both a First Posts and a Late Answers reviews without ending up with a Not An Answer flag on it

physics.stackexchange.com/posts/328868/timeline (10k+, sorry everyone else)

11:46 AM

What are some ways to start physicist fights

There's quantum interpretations

QFT formalism

Rigor in mathematics

Quantum gravity

Maybe time assymetry in physics

Any of the question that will make Motl start to have big throbbing forehead veins

@Slereah I think I know the most controversial thing though

Signature convention?

@Slereah Careful not to blast the brains up tho

really any notation issue

If the real part of the non trivial zeroes of the Zeta function are all $\frac{🤔}{2}$

Jim

11:51 AM

Would anyone have any issues with it if we rename all $|j,m>$ states as "Jim states"?

2

@EmilioPisanty The review says "reviewed", but I can't determine what action has been taken.

Emilio Pisanty

@ACuriousMind it sat for a fair while at 'this item is no longer [not?] reviewable' on the LQP review physics.stackexchange.com/review/low-quality-posts/172641 before it was deleted

the review now says invalidated, which will be because of the moderator deletion

Everrett's thesis is pretty good

it reminds me of Rovelli's section on QM interpretation

"A, being an orthodox quantum theorist, then believes that the outcome of his measurement is undetermined and that the process is correctly described by Process 1."

Good old A

GPhys

when I'm imagining double slit experiments in my head, for electron double slit experiments it's very easy to come up with a realization of an observer that will always collapse the electron wavefunctions at the point of the double slits - a very bright lightbulb, for example

Jim

The term "Orthodox quantum theorist", to me, seems like an oxymoron

GPhys

11:59 AM

is there a similarly practical observer I can imagine for the photon double slit experiment?