@lılostafa Ugh, depends on your convention

this should pretty much prove that SR can deal with accelerations just fine, even though in this simulation it is instantaneous or "near instantaneous" ones i was using for simplicity. I could "easily" do the same simulation with non-instantaneous acceleration

but if you can't formulate at least in principle a way to simulate it

Since the $\arcsin$ makes the $\sin(\theta)$ disappear lightly, I choose it here ;P

@ACuriousMind there's a chatroom that needs mod help

the latest set of messages is questionable.

@ACuriousMind thanks

@heather what were they?

a bunch of inappropriate-ish messages whining and insulting mods

I have zero experience with that notation (exterior product). what is exactly $d\theta {\wedge} d\phi$?

(by inappropriate-ish i mean more than just insulting mods)

\wedge : $\wedge$

you lost a $

@lılostafa You can think about it as $d\theta d\phi$, in the sense that it's the oriented surface element which you use to integrate stuff

I am not sure if you need to know more

it's the exterior product of the two one-forms

@Semiclassical Thanks :)

@BalarkaSen >:(

in more generic coordinates, it'd be $dA/(r^2\sin \theta)$ on the surface of the sphere (maybe sine phi, i'm not sure which convention you're using)

You can write it very simply in tensor notation

@BalarkaSen I want the absolute basics at least for now

$A_\mu \wedge B_\nu = A_{[\mu} B_{\nu]}$

I looked at the conversation above. That's all you need to know, @lılostafa

@Semiclassical - i would like to see a relativistic quantum simulation on the lowest level we have reached so far, with no simplifications. Just a bunch of particles interacting with each other in spacetime. Even if it was just a split of a nanosecond we could simulate accurately

I think people try to do that with stuff like lattice QCD

@Semiclassical do we understand Physics well enough to create such a simulation?

though that probably involves assumptions as well

Yes.

It's not terribly complicated

@Semiclassical Well, $r^2 \sin\phi dr \wedge d\theta \wedge d\phi$ is pullback of $dx \wedge dy \wedge dz$ by the spherical parametrization, modulo signs

@ACuriousMind He's defined $B$ as this ^ so I guess it's $\cos \theta$

actually, it definitely does

in lattice QCD you'd only be considering one kind of interaction. so you're not considering all possible physics at once.

@lılostafa right

i'm sooo far from being an expert on lattice QCD, though, and i should probably not say more

@ACuriousMind is an expert

neat.

what is it with people naming points $o$

@BalarkaSen what does $F_{\theta \phi}$ here mean exactly? which component of $F$ (in spherical coordinates) is this? (The curl of $A$ here is the $\hat r$ component of it)

@Slereah so did you see my emails to the GR guy?

I have yes

@0celo7 Haldane said in his Nobel lecture last year that many consider this discovery (Berry phase) worth a Nobel prize

I think the distinction to be made is one of mathematical content vs. physical implications

the details are Not Nice

mathematically, I could well believe that Berry phase is nothing but a fact about certain kinds of bundles

yeah, as a mathematician I find Berry phase to be bundle crap

nevertheless, it's still been a remarkably productive concept within condensed matter physics

some people like bundle crap, but I don't

and I have fresh wounds from a battle with bundles

sounds like you've got a bundle complex

@lılostafa My result contains overall factors of $1/r^3$, so it is plausible, but it looks far uglier than what these notes got

@ACuriousMind It should be a monopole in the B space

what's the most accepted theory how particles obtain their rest mass?

What I got from converting $\sin(\theta)\mathrm{d}\theta\wedge\mathrm{d}\phi$ to Cartesian space is definitely not a monopole :D

Higgs boson

She just called an ideal the algebras version of a black hole, incredible... youtube.com/watch?v=svSy58w6clw

well, it absorbs anything

sooo

What a way to frame it

@Semiclassical The Higgs boson fully accounts for all the rest mass? Quantum fluctuations do not add anything to the rest mass?

@ACuriousMind So was Berry wrong? :/

i dunno

This is what he's obtained in [his original paper](mecklenburg.bol.ucla.edu/Berry%20Berry%20Phase%201984.pdf)

@lılostafa More likely is that Tong meant a different coordinate conversion than what I did

I also think that maybe Tong didn't quite do the right thing there when computing $F$ in spherical coordinates as he did - the curl in spherical coordinates doesn't look as simple as that

Maybe he wrote down that curl without thinking about that, then wrote the expression for the magnetic field in Cartesian coordinates because he knew what should come out without actually bothering to check whether his expression converts to that.

I thought it's just a very simple coordinate conversion which I can't understand.

Now that I see it isn't that straightforward, it's much more acceptable for me that I don't understand it :)

But yeah, I'm pretty sure that his Cartesian expression is correct, but his polar expression is not - he took the curl wrongly

peer reviewing doesn't seem to be an easy job

Einstein found out the hard way

It's not a job. They just ask you do it and don't pay you anything. :P

@Slereah you still here?

maybe

@Slereah I am setting up shop by the GR books in the library

name a book with causality in it and I will check

the GR book?

Is there only one

There should be a peer review fund which is independent

there are at least 2

Check what

@pZombie when there's money involved it fails to be independent

I mean I'm sure all the books with causality in it that I know, you also know

@Slereah the strong causality proofs

@ACuriousMind OK, thanks a lot for your time! I'll try to follow other approaches (like Berry's paper)

I dunno man, Kriele?

@0celo7 how so? It should be government sponsored and the funds should go to the top scientists having the best chances of doing a successful peer review

@0celo7it is in the interest of everyone for successful theories to be accepted as early as possible

or dismissed as early as possible if they are wrong

the government doesn't have money

it is stolen from people's pockets

maybe try not to further propagate the cycle of violence that is taxation

Can we post messages that are %100 flag-worthy in binary or Morse? Would it be OK?

@lılostafa You can, and it would not be okay.

Not really. If you count the taxes average Joe pays to the government, you will find that it is just a fraction of where the government gets its money from

doesn't matter

@ACuriousMind But practically, if they get flagged I don't think people who check the flagges are going to convert them back to text

We can use more complicated encryptions like this one too enigma.louisedade.co.uk/enigma.html

@lılostafa If I see a flag on such a message, you can bet I'm going to decode it before acting on the flag either way

I imagine people will translate them

@lılostafa Well, you could also just post PGP encrypted messages. Without good reason I'd ask you to stop that, though, and move them to trash as they're only of interest to the recipient.

i was surprised myself. In Greece, even if you taxed EVERYONE as in average worker 100%, hence around 10k per year, you would get only to around 50 billions given half the population was working, which it is not, but let's say it is so. Yet, the government spends over 100 billions yearly

@Slereah Yeah Kriele just declares they exist

and that is with 100% taxes

it's the hidden axiom of general relativity

i wish there were some clear postulates of GR

just as there are for SR

The postulates of GR are pretty clear

sure, shoot then. What are they?

I'm a bit sick of them really

Depends if you go the mathematical or physical route

Mathematics is

All together

@ACuriousMind The goal is to have it semi-public, not indecipherable by all but one person. But OK.

Let M be a connected, paracompact, Hausdorff, second countable pseudo Riemannian manifold with metric $g$ of signature $-+++$ and torsion-free, metric connection $\nabla$

(and maybe assume that there is also a time orientation and a volume form)

And the metric $g$ obeys the relation $$G_{\mu\nu}[g] = T_{\mu\nu}[g, \varphi_i]$$

Slereah - that might be a postulate which leads to GR, but it is not basic enough as is "light always travels at C etc"

Well that is what it is

"light travels at c" does not give SR

@0celo7 the two postulate of SR lead to SR

and what are they?

the more physical axioms of GR are the principle of equivalence and of general covariance

@0celo7 i am sure you know them

what are ordinary and extraordinary polarization?

(this question of mine gives some context)

the united states before and after the last election? :P

@Slereah can an observer in an infinitesimal small space tell whether he is in a spaceship accelerating, at rest, free falling in a gravity field or on the surface of a planet? I don't think he could discern between any of those 4 cases if we were to consider him to be infinitesimal small

@heather A birefringent crystal typically has two axes - one called "ordinary" and the other "extraordinary", so these would be the polarizations along these two axes, I guess

huh, okay.

thanks.

@ACuriousMind How on Earth do you know that

i guess...how do you get that into some more "normal" type, like horizontal or vertical?

is there some sort of conversion method?

hey, optics is physics

@heather Well, it depends on how the crystal is oriented with respect to your "horizontal" and "vertical" axes... :P

@0celo7 I am a physicist, you know...

@heather In a biaxial crystal, the ordinary-polarized rays see a refractive index $n_o$ and the refractive index of the extraordinary ray will be in between $n_o$ and $n_e$

@Slereah And this is what bugs me about the supposed derivation of GR in many places i found. They seem to be comparing an accelerating spaceship with someone on the surface of a planet as being equivalent, then go on to compare someone free falling in a gravity field to free floating in empty space as being equivalent. Yet, as i see it, all 4 cases are equivalent when we are talking local, infinitesimal small

an infinitesimal small spring with some mass attached to it, would not register any acceleration at all, because for it to register acceleration, it would require to be extended over some space

that is, if one can even imagine an infinitesimal small spring

@Slereah idk, I hope Galloway writes back

would be nice

Also what did Galloway do

His name sounds familiar

check the biblio of BEE

he did lots

@ACuriousMind Why is that component of F called $F_{\theta \phi}$? It should be $F_r$. (I'm saying based on the component of $\nabla \times A$ in that equation)

Let me put it this way.

@lılostafa It depends on whether or not you look at the $F$ as a 2-form or as a vector (related via Hodge duality)