The h Bar

John Duffield

13:00

@skull patrol : I've read a lot of Einstein material, I know the answers. If you plot light-clock rates at different elevations your plot is curved. Hence you could say time is curved. But there isn't any actual thing called time that exists. Space exists, and motion, but time is a derived dimension. You can't move through like you can move through the space dimensions.

ACuriousMind

The only Lie-type bracket you might encounter in physics that is not a commutator is probably the Poisson bracket

@Slereah No, Lorentzian geometry is in signature (1,n-1). General (p,q) is pseudo-Riemannian.

John Duffield

@skull patrol : Hence time is emergent, so space can exist without time. See A World without Time for what Einstein says, and see what a clock does for the evidence. When you open up a clock you don't see time flowing through it. A clock doesn't literally measure the flow of time like it's some cosmic gas meter. It "clocks up" some kind of regular cyclical motion, such as a that of a vibrating crystal.

skull petrol

Ok, so time is what elapses between events, correct? @JohnDuffield

John Duffield

@skull patrol : not really. That's what we call it, but when you look closely, you find that there is no actual thing you can see or touch etc called time. What's actually happening is that light and electrochemical signals and things are all moving. A clock clocks up motion.

Work calls I'm afraid. Busy busy busy. Bye for now.

skull petrol

Bye Sir.

Secret

13:14

@Slereah
http://philsci-archive.pitt.edu/10945/1/NSP_SHPMP_Final_Version.pdf

I have finished reading the paper you referred me

It seems in the paper, the author went through 3 different criteria for non superluminal propagation:

Given a manifold $M$ with some prescribed data $\Phi$ at some point $A$

1. NSP1: Change in field values at A should only change the field values within the casual cone of A. This ensures the uniqueness of some solution at some later time given some initial data at A

skull petrol

:O

You just set the record for the longest post I've ever seen

ACuriousMind

Seriously, is this chat bug known on Meta?

skull petrol

That's not a bug, it's a monster :P

ACuriousMind

Okay, it's, as usual, the multiline that's responsible.

skull petrol

Houston we have a problem.

ACuriousMind

13:21

Hm, it's not a bug

Secret

without all those spacing, my stuff is not readable, just like those wordy journal articles in phys chem

ACuriousMind

Digging on meta says that the multiline environment disables all markdown and restrictions because it is for pasting stuff.

Secret

...except I actually type this from scratch

there's just so much I want to go through and I don't knwo how to make it even simpler and concise

skull petrol

Yes, you put a lot of effort into it.

ACuriousMind

@Secret Keeping to the message length restriction and not posting more than two or three in a row might be a useful exercise in being concise. For instance, more than half of that monster post is a summary of the paper. If Slereah referred you to it, why are you repeating back the content he presumably knows to him?

Hmm...I wonder if a mod could use the multiline environment to superping every user of SE at once... :D

Tuntuni

13:26

Hi, can someone tell me where I could find more information on arithmetic vs. algebraic sum?

My physics book mentions it and so does one yahoo answer, but I can't find any "formal" information on it.

I know the difference, but maybe it has a different "formal name"?

Secret

@ACuriousMind I am trying to check my understanding on the paper with Slereah on whether it is correct, given my physics intuition is just not good enough to not bump into wrong or misleading interpretations, especially for GR and quantum mech

Checking that I am on the right track is important to ensure the points I want to discussed about does not sound nonsense because of a misconception in my part (which I often ran into because of the tendency to overintepret what the maths told me)

ACuriousMind

@Tuntuni I had to look up "arithmetic sum". They're just sums where a constant is added every time. What is an "algebraic sum"?

Tuntuni

Hm, I can find a definition for "algebraic sum" on Wikipedia's suggestion page (although no articles), but nothing for arithmetic sum?

@ACuriousMind That's an arithmetic progression, not an arithmetic sum.

ACuriousMind

Then I know neither of those two types of sums.

skull petrol

An algebraic sum uses variables while an arithmetic sum uses numbers.

ACuriousMind

13:29

@skullpetrol Seriously?

That's a stupid distinction if I ever heard one

skull petrol

Why not?

Tuntuni

Arithmetic sum is supposed to be a summation without taking the signs of the numbers in account.

So basically just their absolute values.

While an algebraic sum does.

ACuriousMind

That makes even less sense to me, why would anyone invent those names for that?

Tuntuni

No idea, that's why I want some more information ofnit.

I thought it was a common thing, lol.

"Because of the different directions of action of the two voltages, instead of taking their arithmetic sum, we need to take their algebraic sum." (a not-so-good translation from my book)

The section talks about connecting multiple batteries but with their poles reversed.

ACuriousMind

@Tuntuni I don't think "arithmetic sum" and "algebraic sum" are all that common, and I'm not sure what more you hope to find on that.

0celo7

13:33

@ACuriousMind D:

Tuntuni

I hoped to find a serious definition in some math theory text or something.

0celo7

@skullpetrol really?

I think when he posted his whole CV he set the record

@ACuriousMind Why a mod?

ACuriousMind

@0celo7 Because non-mods can't superping

0celo7

What is superping

ACuriousMind

@0celo7 Ping users who have not been in chat in the last 2 days

0celo7

13:36

@ACuriousMind I knew that

@ACuriousMind ah

why doesn't KK come here at all anymore

certainly he's not busy 24/7, and he goes on the main site all the time

@ACuriousMind real analysis 1 final exam

anything special?

that's the measure theory free course

Slereah

and back home

0celo7

@Slereah huh?

Slereah

@ACuriousMind You're pseudo lorentzian :V

ACuriousMind

@0celo7 Not sure what kind of question that is

Slereah

Also then what is a (p,q,r) metric signature called

0celo7

13:44

@ACuriousMind is it what you'd expect from a first real analysis course

ACuriousMind

@0celo7 Pretty much

0celo7

@ACuriousMind second semester exam here

ACuriousMind

Yeah, Lebesgue integrals. We were tardy and had to finish that in the third semester

0celo7

the measure theory real analysis course does not have any final exams online, sadly

homepages.uconn.edu/~rib02005/rags100514.pdf that's the book, apparently

Slereah

Real physicists don't use no Lebesgue integral

0celo7

13:49

@ACuriousMind did go right into sigma algebras and borel sets?

it looks like we do that in the third semester

Slereah

An integral is just $\lim_{\Delta x \rightarrow 0} \sum_{x = a}^{b} f(x + \Delta x) \Delta x$!

ACuriousMind

@0celo7 No, we didn't take the measure theoretic route in my analysis class

0celo7

oh that's right, your lecturer was some old number theorist :D

@Slereah agreed

but the PDE classes have Lebesge integrals as recommended background

so...

...believes in the caloric fluid :'D

@Slereah hehe

still laughing at that

Slereah

You laugh but just wait to see how path integrals are defined with physicist

0celo7

I do know some QFT you know

Slereah

13:56

Oh yes but I've done like 6 months of working on path integrals

Like non-relativistic

Down to the nitty gritties

0celo7

just integrate over all paths

I don't see the issue

Slereah

It is pretty not rigorous :p

0celo7

manifolds are vector spaces, I don't see the issue

Slereah

I told you about the $dx^2 = dt$ thing

0celo7

all manifolds are isomorphic to R^n

I don't see the issue!

Slereah

13:57

Then you truly are a physicist!

0celo7

I bet you ppl can't provide one counterexample

@Slereah well, that's neither of my majors

Slereah

The sphere isn't isomorphic to $R^n$

0celo7

of course it is

Slereah

Hey you know what else isn't isomorphic to $R^n$?

0celo7

just map every point to R^n, done

Slereah

13:58

Two copies of $R^n$

Oh wait

Does isomorphic imply continuous maps

0celo7

what like R^2n?

Slereah

Or not

0celo7

@Slereah that's homeo

in any case

Slereah

Well I suppose every manifold has the same cardinality, yes :p

0celo7

all manifolds are vector spaces because all manifolds are homeomorphic to $\mathbb{R}^n$

Slereah

13:59

Well maybe not the empty set

ACuriousMind

@0celo7 Not true. An "isomorphism of manifolds" is always continuous.

Slereah

and discrete sets of points

0celo7

@ACuriousMind then what is a homeo

ffs these definitions

Slereah

you're a homeo

2

0celo7

that's up there with son of a bricklayer

that has to be the most creative insult ever :D

Slereah

14:00

It is so very quaint

It's right up there with "Son of a motherless goat"

0celo7

in a string theory vs. LQG debate:

You son of an engineer!

ACuriousMind

@0celo7 It's admittedly silly, but one calls an isomorphism between topological spaces a homeomorphism for some reason. So an isomorphism of manifolds is also a homeomorphism, but when they come e.g. with differentiable structures, it is more.

0celo7

@ACuriousMind in any case

all manifolds are vector spcaes

just because you can't define the vector space doesn't mean it exists

ACuriousMind

You drunk?

0celo7

no, I'm taking a stand against popmath

ACuriousMind

14:02

Ah, trolling then.

0celo7

procrastinating is more like it

and I do not troll

I challenge the accepted truths as a part of my crusade to rid mathematics of cargo cult proofs

Slereah

He's all about Gauss and the Theorems

0celo7

Gauss Bonnet is incorrect

since every manifold is homeomorphic to R^n

all manifolds are flat

bet you won't see that in a "mainstream" book

Slereah

So

The earth is flat

0celo7

yes, google on "flat earth theory"

Slereah

14:05

I've been to the forums :p

It is quite entertaining

Secret

3

Q: Is every "nice" topological vector space a manifold?

Say $V$ is a topological vector space. What conditions do you need to add on $V$ to make it a (topological, maybe infinite-dimensional) manifold? For instance, can we view the Schwartz class functions, test function spaces, or distributions as being locally homeomorphic to a Banach space?

Slereah

They work very hard to not do the simplest experiments to show that the earh is round

0celo7

Pope Pius 2 and the evidence demand it

so, time for tea

work calls!

bye

$PP^2\wedge\mathcal{E}$ is the name of the game

Slereah

@Secret : Sounds about right?

Didn't read the part about the extended cone yet tho

Been busy

0celo7

you know what else

$\lim_{n\to\infty}\int f_n=\int \lim_{n\to\infty}f_n$

this is obviously true

Slereah

14:15

Everything commutes

0celo7

dunno why cargo cult math forbids it

@Slereah exactly

$R(X,Y)Z=[\nabla_X,\nabla_Y]Z+\nabla_{[X,Y]}Z\equiv0$

the Riemann tensor is zero

Slereah

That is why the earth is flat

0celo7

you know what else

$\lim_{k\to\infty}\int\mathrm{e}^{\mathrm{i}kx}f(x)\mathrm{d}x=0$

easy

math types refuse to believe it

you know what else

$G_{\mu\nu}=T_{\mu\nu}$

but above we showed Riem=0

so $T_{\mu\nu}=0$

simple

quantum mechanics is wrong too

Slereah

Well

Everything commutes

0celo7

$[q,p]=0\ne\mathrm{i}$

Slereah

14:19

Also

$a - b = b - a$

0celo7

hamiltonian mechanics is wrong too

Slereah

All numbers are 0

Secret

wait how can $G_{\mu\nu}=T_{\mu\nu}$
Isn't $G_{\mu\nu}=T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$ from EFE?

Slereah

No?

0celo7

$\{q,p\}=0\ne1$

@Secret no?

go take a GR course, buddy

but remember it's all lies

also

Newtonian mechanics is wrong

$F=mv$

for crying out loud Archimedes was correct. Duh!

Secret

14:22

Sorry should be $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$

ACuriousMind

@0celo7 Wat

Do you mean Aristoteles?

0celo7

no

you know what else

axiom of choice is only conditionally correct

Secret

how's so?

0celo7

look at Banach Tarski

absolutely silly

so in that case it's wrong

but all vector spaces have a basis

this is obviously true

so in that case it's right

and if AoC ever increases GPD, it's right in that instance

this is simple logic, people

obe

Logic or intuition?

ACuriousMind

14:30

Not sure if you think you're funny :P

0celo7

another problem: there's no difference

@ACuriousMind this is serious

you know what else

I'm not convinced the angles in a triangle add up to 180

ACuriousMind

@0celo7 No, it isn't. You're just spamming the room with blatantly false stuff you don't even believe yourself.

0celo7

@ACuriousMind indeed. brainstorming for a paper

I should write it on "violence in mathematics: how a small child was shunned in an internet chatroom"

that would be easier to write than the moral implications of sanctions

Slereah

What is the correct way of writing the Euler Lagrange equation

0celo7

can't write an incorrect equation the "right way" :)

ACuriousMind

14:42

@Slereah What do you mean?

Slereah

Is it $\frac{\partial \mathcal{L}}{\partial \phi} - \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \partial_t \phi} = 0$ or $\frac{\delta \mathcal{L}}{\delta \phi} - \frac{d}{dt}\frac{\delta \mathcal{L}}{\delta \partial_t \phi} = 0$

0celo7

first

second is used by some people for no good reason

ACuriousMind

@Slereah $\partial$. $\delta$ is used by people who think that's a functional derivative.

Slereah

Is it not though

ACuriousMind

@Slereah No, the Lagrangian is a function that happens to get fields plugged into it, but $\phi$ and $\partial_\mu\phi$ are just variables to it.

Slereah

14:47

Alright

ACuriousMind

What is true is that the whole Euler-Lagrange equation is the expression for the functional derivative $\frac{\delta S}{\delta \phi}$ being zero.

0celo7

indeed

and then there's the Maupertuis thingie

which I don't really understand

ACuriousMind

@0celo7 What do you mean?

0celo7

@ACuriousMind I don't know how to derive it or how it's different or special

Slereah

@ACuriousMind : Wald also defines the momentum as $\pi = \frac{\delta}{\delta \dot\phi} \mathcal L$

Is it also incorrect?

ACuriousMind

14:56

@Slereah If he claims that is a functional derivative, yes. The Lagrangian is really just a function, and we do not care that $\phi$ is a field when we write down $\partial \mathcal{L}/\partial \mathcal{\phi}$.

The true functional derivative would not treat $\phi$ and $\dot{\phi}$ as independent, for one.

@0celo7 It's different because it gives trajectories for constant energy and known start- and end-points, in contrast to the usual action principles which give trajectories given known start- and end-points as well as times. Also, Maupertuis gives the trajectory as p(q) in phase space, while the usual principles give q(t) or (q(t),p(t)).

0celo7

So we take the integral $\int p\mathrm{d}q-H\mathrm{d}t$

then assume $H$ is constant

and vary the integral?

Slereah

What am I supposed to do if even Wald lies to me!

0celo7

WHAT

WHERE

Slereah

cf above

skull petrol

WHO

WHY

0celo7

15:04

which book

which page

which year

Slereah

Quantum field theory in curved spacetime and black hole thermodynamics

Or whatever it is called

0celo7

yeah

that's at home

Slereah

It's the chapter on quantum field theory in flat space

0celo7

page? I have a "legal" copy with me

Slereah

you can use your eyes

to see it

using vision

0celo7

15:05

page?

skull petrol

And light

0celo7

page, seriously

I need to see the heresy for myself

Slereah

The book is in the toilets

and I am too lazy to go get it

0celo7

@ACuriousMind I'm planning to stop by Freire's office to talk about stuff

I'll throw the Maupertuis principle in

ACuriousMind

@0celo7 I find it better this way: Usually, we vary $\int L \mathrm{d}t$. If $H$ is constant, then adding it does not change the stationary points: $\int(L+H)\mathrm{d}t$. But $L$ and $H$ are Legendre transforms of each other, so $\int p\dot{q}\mathrm{d}t$ and $\dot{q}\mathrm{d}t=\mathrm{d}q$ gives the abbreviated action for Maupertuis

0celo7

15:07

ACuriousMind

So, for constant $H$, you might as well just vary $\int p\mathrm{d}q$.

0celo7

@ACuriousMind makes sense

you make sense

Arnold does not D:

but I will think on it some more myself

or talk to Freire

Slereah

@0celo7 : p. 36

eq. 3.2.1

0celo7

see the image above...

Slereah

Well I didn't see it

0celo7

15:09

I don't see the issue

Slereah

Because I was busy GOING DOWN TO THE BATHROOM FOR YOU

Ask @ACuriousMind about it

0celo7

@ACuriousMind what is @Slereah's problem?

Secret

Slereah (discussion to be enriched later...)

0celo7

@Slereah but he doesn't

you're a liar

stop slandering Bob's name

ACuriousMind

@0celo7 Just read what we wrote - if the $\delta$ is supposed to be a functional derivative, it's wrong.

0celo7

15:12

Why?

You can't take the functional derivative wrt. $\dot\phi$?

ACuriousMind

@0celo7 Because of how a functional derivative is defined: If you have a functional $F[\phi]$, then the functional derivative is a function $\frac{\delta F}{\delta\phi}(x)$. But $\mathcal{L}(\phi,\dot{\phi},t)$ is not a functional, it is just a function in three variables.

0celo7

wtf

the action is not a functional??

ACuriousMind

You can't take the functional derivative of something that is not a functional.

0celo7

WTF IS S

ACuriousMind

@0celo7 $\mathcal{L}$ is not the action

0celo7

15:16

where is $\mathcal{L}$!?

ACuriousMind

In what @Slereah claimed Wald wrote :P

0celo7

see the image for what the book actually says

ACuriousMind

About the functional derivative w.r.t. $\dot{\phi}$ I'll have to think a bit.

0celo7

as I said, @Slereah needs to stop slandering Bob's name

@Slereah where is $\pi=\delta\mathcal{L}/\delta\dot\phi$

I need picture evidence

otoh, Zee has

ACuriousMind

@0celo7 I don't think you can take the functional derivative w.r.t. $\dot{\phi}$.

0celo7

15:23

wai

ACuriousMind

For that, you'd have to consider the action as a functional $S[\phi,\dot{\phi}]$ where $\phi$ and $\dot{\phi}$ are independent functions. But that's not what $S$ is. The action always considers $\dot{\phi}$ to be the time derivative of $\phi$.

0celo7

Stop saying things I know

The things I know don't convince me, obviously

Slereah

Wait

is the momentum as the variation of S correct

Or is it still incorrect

ACuriousMind

@0celo7 Not this again. Unless you can't define the functional derivative w.r.t. $\dot{\phi}$, it doesn't exist. The burden of proof is not on the person claiming it doesn't exist.,

@Slereah Huh? $\delta S/\delta\phi$ is just the Euler-Lagrange expression, if you mean that.

Slereah

@0celo7 @ACuriousMind This one

0celo7

15:28

What if I define it s.t. $\phi$ and $\dot\phi$ are independent

Slereah

Although wait that equation doesn't make sense either, does it?

0celo7

and then make them dependent after the variation

Slereah

Wouldn't there be an integral if that was the variation of S?

0celo7

@Slereah no

Slereah

Oh wait, is it because $\delta f(x) / \delta f(y) = \delta(x-y)$

ACuriousMind

15:29

@0celo7 Possible, but non-standard.

Slereah

or something

0celo7

@ACuriousMind good enough for me

ACuriousMind

Could be what Wald had in mind, though.

Slereah

I DON'T KNOW ANYTHING ANYMORE

0celo7

@ACuriousMind :(

I don't want burden

pretty sure $\delta S$ is an integral

but $\delta S/\delta\phi$ is a function

ACuriousMind

15:32

The variation $\delta S$ between two functions $\phi,\psi$ is given by $\delta S[\phi,\psi] = \int \frac{\delta S}{\delta \phi}(x)\psi(x)\mathrm{d}x$

0celo7

I knew that!

Slereah

but what is the variation

of love

0celo7

?

obe

Is SR the same to GR as QM is to QFT?

ACuriousMind

Haddaway - What Is Love

►

@obe No

0celo7

15:42

Aha! Freire worked at Max Plank doing GR!

obe

@ACuriousMind How?

0celo7

No wonder he teaches a pseudo-Riemannian class

ACuriousMind

@obe Why do you think it would be?

obe

Idk I guessed.

0celo7

math.utk.edu/~freire/teaching/m567f08/m568s09announcement.html

omg I need to get him to teach this

obe

15:44

Were you the one interested student?

0celo7

2009

obe

Time machine.

0celo7

his policy is 5 people need to sign up

none of the math people give a rats ass about GR

and the physics department is 90% particle and condensed matter physics

:(

math.utk.edu/~freire/symplectic.txt

why did he post a TeX code for a lecture proposal

Slereah

@obe wot

Galois in the Field

Can anyone talk to me about Fock spaces here?

ACuriousMind

15:48

They are focking awesome

::hides::

Galois in the Field

Is that also a yes? :P

Slereah

It's a bunch of Hilbert spaces glued together

0celo7

Language!

ACuriousMind

@GaloisintheField Yes, what do you want to know?

obe

@Slereah An idea...

Galois in the Field

15:50

So $\mathcal{F}_2$ is a 6-dimensional $gl(3)$-module

Slereah

$\oplus_{i = 1}^n \text{Sym}{\otimes_{k = 1}^i \mathcal{H}}$

Secret

https://i.sstatic.net/TBmH1.png
(2)

Slereah

Or something

ACuriousMind

@GaloisintheField Uh...what is $\mathcal{F}_2$?

0celo7

@Slereah its in the back of toilet book

Slereah

15:51

Probably

Galois in the Field

@ACuriousMind The fock subspace of all 2-boson states spanned by homogeneous polynomials of degree 2 in the boson creation operators acting on the physical vacuum state

Slereah

Is the Fock space itself a Hilbert space btw

I suspect it is

ACuriousMind

@Slereah Yes, silly, the tensor product and direct sums of hilbert spaces are Hilbert spaces

yuggib

@GaloisintheField Take a look at this

ams.org/journals/tran/1956-081-01/S0002-9947-1956-0076317-8/…

Slereah

@ACuriousMind : I suspected as much since particles are a lie :p

Hence only the total space of all things is the Real Hilbert Space

yuggib

15:53

@Slereah but it is complex...

ACuriousMind

@GaloisintheField Not sure why you say it is six-dimensional or a $\mathfrak{gl}(3)$-module. What physical system are you considering and how many creator/annihilator pairs do you have?

Slereah

Well yes, but I mean real in the metaphysical sense!

yuggib

@Slereah define $\text{Real}\bigr\rvert_{\text{Metaphysics}}$

ACuriousMind

@GaloisintheField Additionally, no physicist will understand you when you talk about $\mathfrak{gl}(3)$-modules. ;) Say it carries a representation of $\mathfrak{gl}(3)$

Galois in the Field

Boson creators $b_i^\dagger$ $i=1,2,3$

Oh okay haha

ACuriousMind

15:57

@GaloisintheField Ah, yes, I now agree with you it is six-dimensional and a $\mathfrak{gl}(3)$-module

yuggib

@GaloisintheField And anyways, why would it be six dimensional? It depends on what one-particle space you choose as a starting point

Galois in the Field

Anyway my question is, we have $\cal F_2$ as the direct sum of two irreducible $o(3)$-modules (carries a $o(3)$ representation?)

ACuriousMind

@yuggib There are only three creation operators, so the one-particle space is three-dimensional.

Galois in the Field

How do I find the highest weights of my $o(3)$-modules?

ACuriousMind

@GaloisintheField Do you know...Clebsch-Gordan coefficients? :D

Galois in the Field

15:59

Twp

Yep

yuggib

@ACuriousMind then it is not worth to call it a Fock space...

you may as well call it $L^2$

Slereah

I don't like Clebsch-Gordan because it sounds like someone misspelt Klein Gordon

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