The h Bar

Obliv

00:02

ozurdilerim. benim kuzenimde cok zor girdi okula. Turkiyede education system ban baskadir. kesinlikle bir yol var senin icin, sadece bulman lazim. Ingilizcegin iyidir ve cok rahat geciceksin buruya

her state degisik kanunlar var. En basda sen buruyanin nasil immigrate yapican orenmen lazim. for example: what kind of visa they offer and what is the conditions for them

kurallari ogrenmek yani

user587860

Ozur dilemene gerek yok tabii ki. Ama bence ilk kabul almak daha onemli

Relativisticcucumber

04:05

@Obliv hm is this true? never seen anything like that before in the US

Silly Goose

@Obliv maybe people like terence tao XD

PM 2Ring

06:10

@Supersymmetry & @Obliv Please don't converse here in Turkish. The language of this room is English. The mods and room owners can't easily moderate posts in languages they can't read. And it's rude to other room visitors. But feel free to create a new room for your Turkish conversation.

@Sanjana Fourier square wave sagecell.sagemath.org/…

naturallyInconsistent

06:59

@PM2Ring oh I feel sad that you didn't choose to do |x| instead. It is easy to implement the changes, though, from your nice work: change f to (pi ** 2)/8 - sum(cos(k * x * pi)/(k ** 2) and then you can also plot and print the derivative of f to get the square wave.

Ryder Rude

08:02

hi

how did ancient people measure that earth takes 365 days to revolve

constellations maybe

was an accurate measurement tho

"Aristarchus made the first calculation of the Earth-Sun distance in about 300BC. He did it by measuring the angular separation of the Sun and Moon."

Slereah

08:18

> Eudemus attributed to him the discovery of 'the fact that the period of the sun with reference to the solstices is not always the same'; the vague phrase seems to mean that he discovered the inequality of the length of the four astronomical seasons, that is, the four parts of the 'tropical' year as divided by the solstices and equinoxes.
Eudemus presumably referred to the written works by Thales On the Solstice and On the Equinoxes mentioned by Diogenes Laertius. He knew of the division of the year into 365 days, which he probably learnt from Egypt.

The basic trick to look at the length of the year is to just look at when the stars rise at the same point

It is something you with fairly low tech means

Ryder Rude

365 requires some accuracy in measuring constellation position

Slereah

You just really need a lot of measurements for it

Which according to the Greeks the Egyptians and Chaldeans had a ton of

Ryder Rude

yeah, it's not as impressive

as what Aristarchus did

Slereah

Egyptian civil calendar had 365 days

> The civil calendar was established at some early date in or before the Old Kingdom, with probable evidence of its use early in the reign of Shepseskaf (c. 2510 BC, Dynasty IV) and certain attestation during the reign of Neferirkare (mid-25th century BC, Dynasty V).

Ryder Rude

i hav no idea how i wud measure the distance to sun myself

Slereah

08:29

Apparently probably based on the observation of the rise of Sirius

Ryder Rude

inpossible without telescopes

naturallyInconsistent

@Slereah Just Egyptian prehistory alone, let alone Ancient Egypt, had thousands upon thousands of years. Just keeping records would have given some hints.

Slereah

Telescopes don't give you any more informations about stars than you can get without a telescope, you just get more precise data

Ryder Rude

the 7 day week is based on the moon cycles

Slereah

you just really need to measure angles

naturallyInconsistent

08:31

@RyderRude why do you have to be so brazenly wrong? You already quoted about Aristarchus.

Slereah

You just get data like this

Ryder Rude

@naturallyInconsistent sorry. i thought he used telescopes

Slereah

How could he

Ryder Rude

@Slereah is this ancient

Slereah

Babylonian

Ryder Rude

08:33

@Slereah i mean some ancient analogue of telescope

naturallyInconsistent

@RyderRude Telescopes started becoming a thing in 1608, as Wiki writes. Just a short while before Galileo got to work on one and turned it to the Moon.

Slereah

People used weird shit like this

Mural instrument

A mural instrument is an angle measuring instrument mounted on or built into a wall. For astronomical purposes, these walls were oriented so they lie precisely on the meridian. A mural instrument that measured angles from 0 to 90 degrees was called a mural quadrant. They were utilized as astronomical devices in ancient Egypt and ancient Greece. Edmond Halley, due to the lack of an assistant and only one vertical wire in his transit, confined himself to the use of a mural quadrant built by George Graham after its erection in 1725 at the Royal Observatory, Greenwich. Bradley's first observation with...

naturallyInconsistent

@RyderRude You mean an angle measuring tool

Ryder Rude

the 7 days is based on moon cycle, the 365 dyas is based on season cycle, the 24 hr is based on day-night cycle

and a calendar combines all these units of time

it is 7 days becuz the sale day was decided based on moon cycle

in Babylon

sorry idk what 24 is based on. it is randomly dividing the time into 24

Slereah

The cool tool of antiquity was the gnomon rly

Greek astronomers won't shut up about the gnomon

Ryder Rude

08:39

it is a nice tool. anyone can make it

the word "clockwise" has been derived from.this device

i mean the definition of clockwise

this device used to go clockwise becuz humans originated in southern hemisphere

if humana had used this device in northern hemisphere, then anti clockwise would hav been defined as clockwise

Slereah

Yes, Greece and Egypt famously in the southern hemisphere

Ryder Rude

it was before greek and egypt civilisations

see this youtu.be/P5rBJ746F1g?si=g9OTskGQiSZcULoH

ACuriousMind

you know, it's possible to not say stuff as if you were stating facts if they are not facts :P

2

Ryder Rude

sorry he says civilisation evolved in the northern hemisphere @Slereah

so clockwise has been named after the behavior of this device in the northern hemisphere

if civilisation was southern, anti clockwise wud hav been called clockwise

naturallyInconsistent

It is the whole, totally contrary to common sense, and we even challenged you, and you doubled down. Why can't you just pause a bit and process a challenge before commenting?

Slereah

08:51

He has the wisdom of Karl Pilkington

naturallyInconsistent

Wiki isn't helping there: what is Pilkington relevant to this about? Presumably something about acting hardheaded?

Slereah

Karl Pilkington is a man willing to construct entire theories based on something he vaguely remembers reading once

naturallyInconsistent

ahhhh

Ryder Rude

the theory of clockwise is correct. i just mixed the hemisphere

Slereah

Karl Pilkington Karl's Future Predictions

►

Many amazing ideas

ACuriousMind

08:55

it's the southern hemisphere where horizontal sundials go counterclockwise, but in any case there is no inherent reason to base clocks on horizontal sundials rather than vertical ones (which go the opposite direction to horizontal ones)

naturallyInconsistent

Yes, we are aware of that. It is not that bad that you made a mistake. It is the, Slereah challenged you, and you doubled down, that is the issue.

Ryder Rude

Africa is in the southern one. i said it becuz humans originated there

so i thought it was before Greece and Egypt

naturallyInconsistent

@ACuriousMind nah, putting something on the floor is much easier than hanging something up. Even walls are not something easy to work with, when we are talking about that far back in time.

@RyderRude yes, I was very much willing to give you the benefit of the doubt there. I went to check when you insisted upon it. Why won't you do some fact-checking when being challenged?

Ryder Rude

i did. thats y i saw the video again

humans still originated in Africa. it seems like Neil is talking about the origin of specific civilisations

it says the population was 5 million when humans originated

i would have expected a thousand

maybe if we go back to some previous species like homo erectus, it would be a thousand

ACuriousMind

09:49

None of these statements make any sense without being much more specific: There is generically no sharp point in time when any (sub)species emerges, the nature of evolution is gradual.

@naturallyInconsistent sure, but stranger accidents have happened in history: If the first clockmakers had for some reason had a cultural attachment to vertical sundials, they might have chosen differently, even if the horizontal ones are more common

naturallyInconsistent

@ACuriousMind I'm not disagreeing that there is some chance of it accidentally doing that, but I think it is highly contrived. It is far easier to use a make-shift thing, e.g. tree. If it were hung up somewhere, then it would be far more work to do, with a lot of chances for the thing hung up to fall down, e.g. earthquakes. I just don't think there is a serious chance of us doing it the vertical way.

ACuriousMind

@naturallyInconsistent I just recently visited a small town near here where there was a hundreds of years old vertical sundial on the old town hall

naturallyInconsistent

@ACuriousMind did anything have the inclination to walk into it? miahahaha

ACuriousMind

it was like three meters up on the wall

naturallyInconsistent

aww

Slereah

10:15

Feeling bad for the N-ray guy because he went down in history as a big dumdum

Poster boy for dumb science

nickbros123

In NR electrodynamics, Whenever Ive come across Energy in any way (say in pure electrostatics, magnetostatics, or electrodynamics- poyntings vector..), Macroscopically Ive always made sense of things as large blobs worth a 1000 molecules being moved around. this is how I justified all the formulas. Is this a good picture to keep, or should I do away with this and stick to abstract averaging processes?

in FNH robinson, in a throwaway fashion, he discusses an averaging procedure called "truncation" , where $\langle E^2\rangle =(\langle E \rangle)^2$, this got me thinking about how someone could try to average the energy density itself, but it will lead to other complications, like dielectrics etc where self energy of these "blobs" in my head, dont get accounted for in any way.

Ryder Rude

10:53

@ACuriousMind statements like this are suposd to say : "there were somewhere between 4 million and 6 million humans when humans evolved, where 4 million is the mark where we're sure there were no humans

but it's still weird becuz how can u hav 4 million non-humans and then 6 million humans a few hundred years later

we r suposd to pick two marks : one where we can classify the species as non-humans and one where we can classify them as humans

these marks would hav too be to far apart then

fqq

11:20

@RyderRude most of Africa is in the northern hemisphere

ACuriousMind

@RyderRude Where do you get this nonsense from? "For the time of speciation of Homo sapiens, some 200,000 years ago, an effective population size of the order of 10,000 to 30,000 individuals has been estimated, with an actual "census population" of early Homo sapiens of roughly 100,000 to 300,000 individuals." (from Wiki)

PM 2Ring

11:40

@RyderRude Well, 24 is nice because it has lots of factors. However, the Moon's mean daily motion (relative to the stars) is roughly 12°, and the Moon's angular diameter is around 30 arc-minutes, so it moves through its own diameter in an hour.

ACuriousMind

11:58

@PM2Ring fixed-length hours actually post-date the division of the day into 24 hours; the reason for the 24 hours is that the Egyptians used a system based on 12s (possibly because there are ~12 months in a year), cf. e.g. scientificamerican.com/article/…, hsm.stackexchange.com/a/2881/3797

PM 2Ring

@ACuriousMind Sure. But the Moon's angular size and speed are quite variable, so "Moon hours" are definitely not fixed in size. ;) FWIW, I recently posted an answer about the variations in lunar motion. astronomy.stackexchange.com/a/55112/16685

Ryder Rude

@fqq yes. sorry

PM 2Ring

In the early 2000s, I made a Sundial generating program in POV-Ray. The sundial itself was pretty easy. The pseudorandom grass was a bit tricky on an Amiga with 10 MB of RAM.

Ryder Rude

@ACuriousMind it was a website when i searched it. but it actually is talking about the number of humans in 8000BCE for some reason

idk how i keep making mistakes

i knew it had to be a few thousand

PM 2Ring

12:16

There were many different hour schemes. Some sundials could be very tricky to read because they implemented multiple schemes. mysundial.ca/tsp/hours.html

Ryder Rude

@PM2Ring is this animated??

the grass looks real

there was one conference to set the value of pi as 3

sorry it's 3.2 en.m.wikipedia.org/wiki/Indiana_Pi_Bill

PM 2Ring

12:34

@RyderRude It could easily be animated. I never bothered to make an anim with it, though. The code that calculates the Sun's position is pretty basic, but it should be accurate to within a minute or so. I have much more accurate code these days. astronomy.stackexchange.com/a/49546/16685

And of course, when I want high accuracy, I just ask JPL. ;)

user587860

I've still been failing to find a god damn left, unitary action of $\text{SL}(2,\mathbb{C})\times \text{SL}(2,\mathbb{C})$ on $\mathcal{L}^2(\text{H}^{+}_{3})$

user587860

If $G$ is a unimodular Lie group and $K$ is a compact subgroup, $G\times G$ has a unitary left-action on $\mathcal{L}^2(G)$ and merely a left-action on $\mathcal{L}^2(G/K)$. Since $G = \text{SL}(2,\mathbb{C})$ is a unimodular Lie group, it seems that $\text{SL}(2,\mathbb{C})\times \text{SL}(2,\mathbb{C})$ has a unitary left-action on $\mathcal{L}^2(\text{Fr}(\text{H}_3)) = \mathcal{L}^2(\text{SL}(2,\mathbb{C}))$. But I am not sure what's going in the direct product

PM 2Ring

Here's some Python code that calculates pi using the arithmetic-geometric mean in integer arithmetic. It converges quadratically. gist.github.com/PM2Ring/032ffa274fec417c47cc31f014dcfa5c I also have a version of that algorithm which has no non-trivial divisions: all the divisions are done using bit shifts, including in the sqrt steps.

Ryder Rude

AGM sounds interesting

have i already shared this philosophy to u plato.stanford.edu/entries/structural-realism ? @PM2Ring

PM 2Ring

12:59

AGM is pretty awesome. You can use it to calculate elliptic integrals. You can use it to calculate all the standard inverse circular & hyperbolic trig functions, and logarithms. It may not be the fastest way if you only need low precision, but for high precision, it's unbeatable (assuming you're using an efficient multiplication algorithm).

Ryder Rude

nicee

so u like Fourier transform or Taylor series more

some people say that Taylor series is also projecting onto a basis

but i dont see it

PM 2Ring

@RyderRude I don't think I've seen that page before, but I just bookmarked it.

Ryder Rude

greatt

it is a really nice philosophy of science

PM 2Ring

Fourier is good for periodic stuff. Or stuff that can be made periodic. Taylor has pretty broad scope, but the convergence of Taylor series often sucks. Fortunately, you can often transform Taylor series into Padé approximants that have a larger radius of convergence, and converge faster.

One fun thing with discrete Fourier is circular convolution. en.wikipedia.org/wiki/Circular_convolution

in Mathematics, Aug 24, 2021 at 15:47, by PM 2Ring

A few weeks ago, I was playing around with drawing a closed loop through a set of points, using cubic Bézier curves. I learned a nice algorithm that ensures that both the 1st & 2nd derivatives of the Bézier functions match at their endpoints. But for n points, it requires solving a system of n equations in n unknowns, which gets expensive for large n.

> Fortunately, these equations can be expressed as circulant matrices, so the problem can be efficiently solved using Fast Fourier transforms.

sagecell.sagemath.org/…

Ryder Rude

13:15

i too came across Pade approximations a few months ago

convolution is just matrix multiplication ,right

PM 2Ring

There's a really simple algo for making Padés from Taylor series. math.stackexchange.com/a/4823743/207316

Ryder Rude

i made this transform which is like Fourier transforms a few months ago math.stackexchange.com/questions/4756290/…

PM 2Ring

Convolution is just long multiplication. ;) en.wikipedia.org/wiki/Convolution Circular convolution is nice & symmetrical because it wraps around, which makes it an ideal match with Fourier techniques.

Ryder Rude

yes

John Zimmerman

13:54

schrodinger heat equation in 1d free particle

$+$ sign instead of $-$ sign

*Schrodinger wave equation

I want to understand how the evolution occurs

Please tell me

Slereah

14:10

lol

William Martens

I think I have said this before but I have to express my appreciation once more to this site(s) and to this chat;
**THANKS** for helping out so much as you all do! :) I have gotten so so much help, with my questions, explanations, and finding things. Not to exclude understanding things in a better way, really this site is a life saver! (okay, not literally but you get the idea)

(like, When I have asked a question, I have always gotten good answers, comments, and replies)

edit: I tried to do bold text but doesn't seem to work

Ryder Rude

14:34

@Slereah why is this wrong

Ryder Rude

15:14

hello, in infinite dimensional spaces, is $U^{\dagger}U=I$ insufficient to conclude if a transform is unitary?

if we take a boost transform, $U(f(p))=f(\Lambda p)$, then this is sometimes unitary sometimes unitary depending on the normalisation of $|p\rangle$

but $U^{\dagger} U=I$ does not depend on the normalisation of basis vectors

Jagerber48

15:38

Is $U^{\dagger}U = I$ not the definition of unitary? Even for operators on infinite dimensional spaces?

ACuriousMind

Dec 14 at 12:58, by ACuriousMind

what needs to be invariant under unitary transform is not their ill-defined "product" with each other, but their resolutions of the identity. In the relativistic case, you have $1 = \int \mathrm{d}\Lambda_p \lvert 0\rangle\langle 0\rvert$, where by $\mathrm{d}\Lambda_p$ I mean the usual Lorentz-invariant measure, and this whole thing is invariant under Lorentz transformations

you need to stop obsessing over what happens to the $\lvert p\rangle$ and start thinking about what your inner product on the $f(p)$ is with respect to which you want to call this operator "unitary"

the boosts are perfectly unitary, you simply haven't thought carefully enough about how the $\lvert p\rangle$ and their normalization are related to the Lorentz invariant inner product on the mass shell

@Jagerber48 The definition is actually $U^\dagger U = I$ and $UU^\dagger = I$, which is only equivalent to just $U^\dagger U = I$ in the finite-dimensional case, but that subtlety is not the problem here

Slereah

But what about $\overset{U^\dagger}{U} = I$ and $\overset{U}{U^\dagger} = I$

Surely it should be true in every direction

Ryder Rude

@ACuriousMind im getting an incorrect conclusion. if $U(p,p')$ is a transform and $\int dp' U(p,p') U^*(p',p")=\delta (p-p")$, then $(U f(p), U g(p))=(f(p), g(p))$ is true regardless of what measure $m(p) d^3p$ we use to define the inner product

ACuriousMind

heh

Ryder Rude

this conclusion is false, right?

the inner product is $\int d^3p m(p) f(p) g^* (p)$

ACuriousMind

15:56

First of all you need to be careful which things here are functions of 3-momentum and which are functions of 4-momentum

because when we write something like $\langle p\vert p'\rangle = f(p)\delta(p-p')$, everything is treated as a function of 4-momentum

but the relativistic wavefunctions in momentum space are only functions of 3-momentum, as they are functions on the mass shell, and the integral is also only over the 3-momenta (with a factor $m(p)$ making them Lorentz invariant)

Ryder Rude

im getting that the inner product with any $m(p)$ is invariant as long as $U(p,p')$ satisfies $\int dp' U(p,p") U^*(p",p)=\delta $. @ACuriousMind

so far im not talking about boosts. lets say we talk about general U(p,p')

everything is a function of 3-momentum in what i write

is this conclusion correct or incorrect?

depending on the answer, i will branch the discussion

i was using functions of three-momentum, to clarify

the fourth component is determined here by the 3

@ACuriousMind everything is 3 momenta. U(p,p'), f(p), \delta ^3(p-p') and $d^3p$

pls help

ACuriousMind

16:13

@RyderRude And in what sense do you think your $\int \mathrm{d}p' U(p,p')U(p',p'') = \delta(p - p'')$ for 3-momenta applies to boosts?

Ryder Rude

there must be some U(p,p') for boosts, right?

it is a linear transform from wavefunctions to wavefunctions

and the wavefunctions r on 3-momenta

ACuriousMind

The problem is not with the existence of $U(p,p')$!

it's more with its definition

how do you think this is defined

Ryder Rude

u mean an expression of U(p,p')?

ACuriousMind

and why is your measure here $\mathrm{d}p'$ instead of the Lorentz invariant meausre

Ryder Rude

it's just the general expression of any linear transform. so far, we're not talking about boosts, but boosts too have to fit into this

ACuriousMind

16:18

you just have mashed together a bunch of formulae from relativistic and non-relativistic physics into one mess

Ryder Rude

if u take any hilbert space and any linear tranaform, the action is $\int dp' U(p,p') f(p)$ ,right?

im just using a geneal concept of linear transform, and then we apply this logic to boosts

ACuriousMind

where does the $U(p,p')$ for an arbitrary linear transform come from?

Ryder Rude

i would say a linear transform is defined by some U(p,p') and the action being that integral?

so if u want to specify ur transform, u just specify ur U(p,p')

ooh a linear transform is more generally a linear function F(f(p))

from this, we can derive a U(p',p), right?

so there is a map F ---> U(p,p')

from linear functions to matrices

ACuriousMind

the $U(p,p')$ is what's called an "integral kernel" for the linear operator and what I want to you to write down is how you're actually defining it

because if you do this, the $\lvert p\rangle$ will appear somewhere in there

Ryder Rude

the defining expression is $\int dp U(p,p') f(p)= L[f(p)]$ where $L$ is the linear operator

ACuriousMind

16:25

so why is that $\mathrm{d}p$ and not the Lorentz-invariant measure? How do you actually get $U(p,p')$

Ryder Rude

ok i will just get to how im defining it for boost

in the relativistic normalised basis, a boost is $\psi (p) ---> \psi (\Lambda p) \frac{1}{\gamma}$

this is F

now we get U

we have $\int dp' \delta (\Lambda p - p') \frac{1}{\gamma} f(p')=\frac{1}{\gamma} f(\Lambda p)$. from the lhs, we fetch our U

@ACuriousMind becuz a general linear transform on any hilbert space fits into this $\int dp U(p,p') f(p')$. so we dont need a lorentz invariant measure right now

this U shud satisfy that integral which gets us delta

but my problem is, if any U satisfies that integral, then any inner product $\int dp m(p) f(p) g^*(p)$ is invariant regardless of what u choose for $m(p)$

becuz u cud just put a $U^{\dagger} U$ inside this integral as it's identity

sorry when i write $\psi (\Lambda p)$, it makes it seem like im working with four-momenta. but the fourth component is agian determined by the three components

Slereah

16:41

Reading all that history of science has taught me to be humble because you don't want to go down as that guyvin history

You don't want historians to name something like SLEREAH'S FOLLY

Mr. Feynman

It's too late

we have evidence on the hbar

Slereah

Noooo

Mr. Feynman

16:55

QPT in a nutshell: Taylor expand where you can't

Come on, how can they take $\int_{\mathbb{R}^2}d^2\vec{x} g(x)$ and expand $|\vec{x}|\to 0$

Slereah

17:08

Quantum phield theory?

Mr. Feynman

Quantum phase transitions

nickbros123

17:37

@Mr.Feynman this is just JD jackson

Or rather: Taylor expand everything you see

Mr. Feynman

Nah, I don't think Jackson does it randomly :P

nickbros123

He does it in the work done formula

Mr. Feynman

Where?

nickbros123

Which I think is a pretty random place

Mr. Feynman

When I say random I mean illicit :P

nickbros123

17:40

@Mr.Feynman I don't remember exactly where, but he does derive the Polarisation term $P\cdot E$ term in so doing, so it's not aimlessly doing it. But chapter 6 of JDJ has a definite formula: Taylor expand everything, drop !(first 2 terms)

@Mr.Feynman Taylor expansion is always allowed

Mr. Feynman

@nickbros123 We're not discussing the regurality of a function here

You can't truncate an expansion over a variable integrated over the entire plane

But after all it seems they were secretly integrating the series but decided to do so in the sloppiest way possible

nickbros123

I'm glad I haven't encountered such functions in my studies till now

Mr. Feynman

I mean, this is simply what I wrote above

47 mins ago, by Mr. Feynman

Come on, how can they take $\int_{\mathbb{R}^2}d^2\vec{x} g(x)$ and expand $|\vec{x}|\to 0$

nickbros123

are they taylor expanding that integral?

either I dont understand whats going on or are they taylor expanding a real number? (most likely the former)

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