The h Bar

Semiclassical

20:00

I think a lot of physics is reliant on metaphysics. I don't say that as a bad thing.

Balarka Sen

I mean that's a fine terminology. A completely useless one however

It's consistent, just useless

dmckee

So, the technical adjective we're looking at here is "platonist", right? Which gives you a sense of just how many BS sessions have failed to resolve the issue.

Semiclassical

eh.

there's different versions of platonism

you can suppose that math is 'real' without saying it's physical

skullpatrol

plato liked young boys

Semiclassical

that is a genuine philosophical position

...

Anonymous

20:01

Next time before debating with someone I'll keep the Oxford dictionary open and tell them to do the same :P (same edition)

Anonymous

@skullpatrol XD

Balarka Sen

If you were expecting to get a cat_transcendence_limitless.gif reaction to my explanation of why the Fermat's last theorem is a natural question to consider, I must say you should have seen disappointment coming

No scientific result on existence can be "physical" by your notion.

Anonymous

@BalarkaSen Well, I was seeing it coming. But since it was you I decided to keep listening :P

Anonymous

Anyhow, let's get back to something else now...anything else.....XD

Balarka Sen

I propose to talk about why Fermat's last theorem is interesting instead of why it's "physical", given that latter is now an established bullshit notion. :D

Semiclassical

20:09

Works for me.

skullpatrol

Me too.

Semiclassical

otherwise I'd have to talk about Kant's distinction between analytic a priori, synthetic a posteriori, and synthetic a priori propositions

and good god do i not want to do that

Balarka Sen

lol

skullpatrol

Kant is toooo dense

Semiclassical

@skullpatrol Kant as a thinker, interesting

Kant as a writer...

...

Balarka Sen

20:11

This reiterates the point I made earlier about how philosophy not having a unified language of communication makes it harder for outsiders to appreciate

skullpatrol

he makes Russell sound talkative :P

Semiclassical

lol

0celo7

@BalarkaSen wait, you actually tried to argue that FLT is physical?

Semiclassical

His lectures tend to be more readable, probably because those were adapted from notes other people took

Balarka Sen

@0celo7 I tried to argue why it's interesting and meaningful/compatible with our perception, yes.

Which is a good question and has a good answer

0celo7

20:13

how do diophantine equations matter to perception

Balarka Sen

It's unclear why anybody should give two shits about solutions of a random diophantine equation

@0celo7 Read ^^

0celo7

Where did you say anything about "perception"?

well, I've gotta prove a Noether theorem

skullpatrol

wb @ACuriousMind

Balarka Sen

A solution to FLT inevitably involves trying to generalize the unique factorization property of integers... that's why it's compatible with our intuition of the fundamentals...

(Technically only for Kummer primes, but let's ignore that. I don't know the story beyond Kummer theory)

But that's approximately the reason people got excited about it

0celo7

yawn

number theory has no motivation

Semiclassical

20:19

I think one could just boil this down to an example from antiquity

namely, the infinitude of primes

skullpatrol

numbers are the motivation

Semiclassical

Especially since the sieve of Erastothones does seem so pleasingly concrete.

Balarka Sen

Right

I think that's how I first really understood primes

Good call

Semiclassical

There is something quite striking about that, yeah

Balarka Sen

Very true

skullpatrol

20:22

Fundamental.

Semiclassical

(it's also cool how the sieve idea is essential to linking primes to the Riemann zeta function, but that's more esoteric)

Balarka Sen

That $\zeta$ has a pole at $s = 1$, yup.

0celo7

Am I damaged if prime numbers are so boring to me it hurts

Semiclassical

lol

skullpatrol

yes

Balarka Sen

20:24

I think I first understood why the Riemann hypothesis is really interesting by understanding what information $\zeta(s)$ spits out about primes as you let $\Re[s]$ vary towards $1$

If you have $s = 2$, $\zeta(2) = \pi^2/6$ by the famous Euler computation

That already says there are infinitely many primes

By writing out the product and seeing it can't be rational

Semiclassical

I don't really mean to come to specific philosophical conclusions btw. My point is more that their resolution is only self-evident if you already have conclusions of your own

skullpatrol

@0celo7 without them there would be no fundamental theorem of arithmetic :P

CooperCape

1 is a prime. Rights for 1 as a prime. 1 is a prime in the scriptures.

Balarka Sen

Near $s = 1$ you get Merten's theorem about convergence of $\sum 1/p$

near the line $\Re[s] = 1$ understanding that there are no zeroes of $\zeta$ gives you a proof of PNT

It seems obvious that going $\Re[s] < 1$ would be even more revealing about the prime numbers, really

0celo7

does anyone have a good pdf of Weinberg Vol 1?

I can only find pretty cancerous ones

skullpatrol

20:27

wb @JohnRennie

Semiclassical

ugh

boring latex task is boring

skullpatrol

the devil is in the details :P

Balarka Sen

i hate latex so much id rather use typewriters lol

0celo7

@BalarkaSen contrarian

Balarka Sen

well im about to go to bed

0celo7

20:30

I want to have the EMS TeX format

Balarka Sen

this was a hot fiery death and destruction discussion

0celo7

@Semiclassical do you think I could email someone there and get it?

Semiclassical

no clue

0celo7

"integrating by parts"

aka

"incorrect math"

0celo7

20:48

@ACuriousMind help

I'm trying to derive the conservation law for a translation invariant lagrangian

but it seems like one has to do the integration overall of spacetime

is there a more rigorous way to do this?

(maybe it's not true in general?)

bangs head

I do not understand physics

0celo7

22:03

@Slereah I finally understand Noether's theorem

bolbteppa

22:49

3 days later, hydrogen atom wave functions from absolutely nothing, with all stupid normalization conventions $[(n+l)!]^{1 \text{ or } 3}$!

Coulomb units are a nightmare, random $1/4$ conventions, stupid old definitions of Laguerre polynomials

19 pages long, will become 20!!!!

\begin{align}
\psi &= R_{nlm}(\rho) Y_{lm}(\theta,\phi) \\
&= \sqrt{ (\dfrac{2}{r_0})^3 \dfrac{(n-l-1)!}{2n(n+l)!} } e^{-\dfrac{\rho}{2}} \rho^l L_{n+l}^{2l+ 1} Y_{lm}(\theta,\phi) \\
&= \sqrt{ (\dfrac{2}{r_0})^3 \dfrac{(n-l-1)!}{2n(n+l)!} } e^{-\dfrac{\rho}{2}} \rho^l L_{n+l}^{2l+ 1} \sqrt{ \dfrac{2l+1}{4 \pi} \dfrac{(l-m)!}{(l+m)!} }e^{\pm i m \theta} P^m_l(\theta) \\
&= \sqrt{ (\dfrac{2}{r_0})^3 \dfrac{(n-l-1)!}{2n(n+l)!} } e^{-\dfrac{\rho}{2}} \rho^l \dfrac{d^{2l+1}}{d \rho^{2l+1}} L_{n+l} \sqrt{ \dfrac{2l+1}{4 \pi} \dfrac{(l-m)!}{(l+m)!} }e^{\pm i m \theta} \dfrac{1}{2^l l!} (1 - x…

0celo7

wtf...

bolbteppa

Pure insanity, no wonder I skipped this before

That's the atom son

In the non-relativistic approximation ignoring spin which immediately leads to fine structure splitting so the solution has fun problems immediately

How do you write Rodrigues in terms of trig functions nicely so I can make the solution even more logical

0celo7

we made up compact notation to avoid this insanity

bolbteppa

Imagine, some books throw in $[(n+l)!]^3$ in that coefficient instead of $(n+l)!$, just for fun

0celo7

23:04

@bolbteppa which books

bolbteppa

Griffiths QM for example

0celo7

garbage book!

bolbteppa

Then most books (e.g. Pauling) will calculate the normalization using generating functions, Landau does it I think with insane hypergeometric functions, nobody just takes this stuff on directly, except Sommerfeld's old book, which leaves lots of gaps but at least it's direct and forces you to learn it all nicely

Yes, now page 20!

Haven't even began the relativistic hydrogen atom problem, which involves coupled hypergeometric functions for spi****nors, fun times

(Ah I put the wrong Laguerre in there, looks messy!)

$$ \psi = \sqrt{ (\dfrac{2}{r_0})^3 \dfrac{(n-l-1)!}{2n(n+l)!} } e^{-\dfrac{\rho}{2}} \rho^l \dfrac{d^{2l+1}}{d \rho^{2l+1}} [\dfrac{e^{\rho}}{(n+l)!} \dfrac{d^{n+l}}{d \rho^{n+l}}( \rho^{n+l} e^{-\rho})] \sqrt{ \dfrac{2l+1}{4 \pi} \dfrac{(l-m)!}{(l+m)!} }e^{\pm i m \theta} \dfrac{1}{2^l l!} (1 - x^2)^{\frac{m}{2}} \dfrac{d^{m+l}}{dx^{m+l}} (x^2 - 1)^l $$

I knew it wouldn't fit

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