The h Bar

AccidentalFourierTransform

18:00

after all, the canonical algebra is always determined by the kinetic part of the Lagrangian

free or interacting, the algebra is fixed by the form of the terms with derivatives

and those are usually the same for free and interacting theories

Slereah

Oh so it is a valid assumption?

it's just the unitary thing at $t \neq 0$ that isn't?

AccidentalFourierTransform

what do you mean by valid? :-P

IMHO, it is reasonable

Slereah

Is it mathematically objectionable

AccidentalFourierTransform

yes, like everything else

Slereah

I mean obviously yes, physically the interacting theory will be somewhat similar to the free theory

but that's a far cry from being rigorous

AccidentalFourierTransform

18:03

but you can derive Feynman diagrams directly from the CCR, so I'm willing to accept them as good enough and move on :-P

Slereah

But the only other way to get them is via the path integral, which is also not rigorous

AccidentalFourierTransform

screw path integrals

Slereah

I'm fine with not too rigorous assumptions, but I would like to at least know what those assumptions are

AccidentalFourierTransform

the assumption is that $[\phi,\frac{\partial\mathcal L}{\partial\dot\phi}]\sim\delta$

for any theory, interacting or not

Slereah

"As already mentionned, this is not possible. The field $\Phi$ must be more singular than $\Phi^0$ and this manifests itself afterwards in the need for an infinite renormalization of $\Phi$. The magic formula refers to the "unrenormalized field" $\Phi_u$. The true field is then written as $Z^{-1/2} \Phi_u$"

0celouvskyopoulo7

18:06

@JohnRennie lol

Slereah

So yes apparently the $\Phi(x,0) = \Phi^0(x,0)$ is bullshit

Oh wait, there's more!

0celouvskyopoulo7

What are you talking about?

Slereah

"Actually the justification of the last step is not possible. The vectors $e^{iH_0t} \Omega$ converge only weakly to a multiple of $\Omega_0$. The operator between the bars depends on $t$ and $t'$. As a consequence, it does not converge as $t$ or $t'$ tend to infinity."

@0celouvskyopoulo7 The interaction picture

And why it's all a lie

There are many reasons, apparently

it is the greatest lie that was ever told

0celouvskyopoulo7

You mean besides the whole "it was a virgin birth" thing?

Slereah

I don't think they assume a virgin birth at all in QFT

So I think the three big lies to make QFT work are

1) you can multiply distributions fine

2) The field operator and the free field operators are kind of the same

3) The vacuum state is gonna be similar to the free vacuum state

Those are the three lies of QFT

AccidentalFourierTransform

18:16

I really dont think you need 2&3

sure, some people use that

but they are dispensable assumptions

Slereah

what do u use instead, if you use the Hilbert space formalism

The path integral formalism doesn't assume that but it has other lies

AccidentalFourierTransform

QFT books are very bad

I have to write one of my own

Slereah

Don't we all

AccidentalFourierTransform

using only 1)

0celouvskyopoulo7

QFT is wrong

Slereah

18:18

well yes

I know

0celouvskyopoulo7

There are no lies in GR

Slereah

cf above

AccidentalFourierTransform

QFT is wrong but its fun

and many of its ingredients can be as rigorous as you want

Slereah

It's unfortunate that GR is perfect and QFT is awful when we can mostly test for QFT

0celouvskyopoulo7

But, why

How do physicists use provably false things?

How does their conscience allow it?

ACuriousMind

18:21

"provably false" != "morally false" :P

Sha Vuklia

Guys, maybe a silly question, but Newton's second law basically states that we can add forces vectorially. However, (when mass is assumed constant) force is $F=m\frac{d^2r}{dt^2}$. Can't we just say that since position vectors can be added vectorially, then velocity and acceleration vectors can be added vectorially too, and if you multiply by a scalar (i.e., mass), this can be added vectorially too?

0celouvskyopoulo7

I get a stomachache if I use something that I know isn't completely true or whatever

Or if I doubt it's true

Slereah

Maybe I should make a PSE post to check

0celouvskyopoulo7

@ShaVuklia That's not what the second law says

It just says $F=ma$

What about addition?

AccidentalFourierTransform

what if $m$ is actually a matrix?

MOND, V.2

Sha Vuklia

18:23

@0celouvskyopoulo7 well it technically says $F_{\text{res}}=ma$.. but we already defined $F=ma$ for a single force.

0celouvskyopoulo7

Random

AccidentalFourierTransform

$\boldsymbol F,\boldsymbol a,\boldsymbol v$ and $\boldsymbol r$ are all vectors

0celouvskyopoulo7

Did you have a class on that?

@AccidentalFourierTransform Jesus Christ

Italic bold?

AccidentalFourierTransform

oh no

you are one of those

0celouvskyopoulo7

@ACuriousMind Can I insult him?

ACuriousMind

18:24

@ShaVuklia In joshphysics' formulation, the point is not that you can add force vectors. The point is that the total force on an object will be the sum of the forces from the "isolated" interactions, and not something else.

0celouvskyopoulo7

@ShaVuklia Part of the definition of force is that you add forces vectorially.

That's what we mean by "force is a vector"

ACuriousMind

You have axiomatically no reason to believe that the sum of the isolated force vectors will be the total force - and that's what the second law in this formulation does.

0celouvskyopoulo7

@ACuriousMind What?

The axiom is that forces are vectors, hence you add them.

AccidentalFourierTransform

what do you call a duality with three things?

ACuriousMind

@AccidentalFourierTransform triality

0celouvskyopoulo7

18:26

Then the law says the total force (axiomatically the sum of individual forces) is the acceleration.

ACuriousMind

I've seen Witten use it, it must be a word :P

AccidentalFourierTransform

oh then Im totally using it

0celouvskyopoulo7

Yes, there's an SO(8) triality.

@ACuriousMind English to Witten dictionary takes a whole new meaning now

Sha Vuklia

but how do we know then that we can add velocity vectors vectorially?

ACuriousMind

@0celouvskyopoulo7 I think we're trying to say the same thing.

@ShaVuklia They're...vectors. Of course you can add them, that's part of the definition of a vector

Sha Vuklia

18:28

but a single force is also a vector by definition...

Sir Cumference

I need a new punny math username

0celouvskyopoulo7

And you add forces

Sir Cumference

Please help

0celouvskyopoulo7

I'm confused what the confusion is

And why this is so important to you

ACuriousMind

@ShaVuklia Yes, and you can add force vectors.

0celouvskyopoulo7

18:28

@ACuriousMind Admit it, you don't read what I write

AccidentalFourierTransform

CisIsJustAPhase

Cis (mathematics)

cis is a less commonly used mathematical notation defined by cis(x) = cos(x) + i sin(x), where cos is the cosine function, i is the imaginary unit and sin is the sine. The notation is less commonly used than Euler's formula, e i x {\displaystyle e^{ix}} , which offers an even shorter and more general notation for cos(x) + i sin(x). == Overview == The cis notation was first coined by William Rowan Hamilton in Elements of Quaternions (1866) and subsequently used by Irving Stringham in works...

0celouvskyopoulo7

@ShaVuklia Did you even look at Arnold?

Sir Cumference

@AccidentalFourierTransform That's gonna get me banned

AccidentalFourierTransform

y

Sha Vuklia

@0celouvskyopoulo7 I'm reading Goldstein

but maybe I should, though I thought it was too difficult for me

let me recheck

0celouvskyopoulo7

18:29

@ShaVuklia You need to read Arnold.

Sir Cumference

@AccidentalFourierTransform Should be pretty obvious

0celouvskyopoulo7

The abstract formulation of Newtonian mechanics begins with the Galilean group on an affine space.

ACuriousMind

@ShaVuklia The second law doesn't say "you can totally add force vectors", it says "the sum of the individual (isolated) force vectors (the "total force") is equal to mass times acceleration".

The thing isn't that you can add forces, it's that their sum is actually physically meaningful and equal to $ma$.

Sha Vuklia

ok, maybe Arnold is doable for me (i have it in front of me now)

Bernardo Meurer

Hehehe, maybe Arnold is doable

Sha Vuklia

18:32

@ACuriousMind give me one sec to let that sink in

0celouvskyopoulo7

amazon.com/Measure-Properties-Functions-Advanced-Mathematics/dp/…

huh

actually affordable

oh

it's not even that long

I'd be better off reading Federer

Slereah

0

Q: QFT and its lies

I have been trying to figure out all the erroneous assumptions of QFT (as performed in an operator theory) that allow it to function as it currently is. So far, the three big candidates I found are these : The product of distributions is allowed in some sense (this runs afoul of Schwartz's imp...

a good title

0celouvskyopoulo7

@ACuriousMind can you give me vamp powers pls

Sha Vuklia

One last try then; A force $i$ is by definition: $\vec F_i=m\frac{d\vec r_i}{dt}$. So for the vector sum of all forces $i$ on some object with mass $m$, we get: $\sum_{i}\vec F_i=\sum_i m\frac{d^2\vec r_i}{dt^2}=m\sum_i\frac{d^2\vec r_i}{dt^2}$. However, just like we can add position vectors and velocity vectors, we should be able to add acceleration vectors. So we simply get $m\frac{d^2\vec r}{dt^2}$, where $\frac{d^2\vec r}{dt^2}=\sum_i\frac{d^2\vec r_i}{dt^2}$.

0celouvskyopoulo7

what is Schwartz's theorem @Slereah

@ShaVuklia No need to put vector arrows

Sha Vuklia

18:37

ok

Slereah

Schwartz's impossibility theorem states that you cannot have a differential algebra on the space of distributions.

0celouvskyopoulo7

or use index notation if you have to

@Slereah By differential algebra you mean an algebra with an endomorphism acting according to the Leibniz rule?

Slereah

There's no algebra on distributions that satisfies the Leibniz rule for derivatives

yes

Sha Vuklia

in my eyes, we are just adding acceleration vectors, so I don't see what exactly it is that Newton's second law is giving

Slereah

The distribution algebra as they are done today get around this by having some products not be distributions

AccidentalFourierTransform

18:39

why do you always say "interactive" instead of "interacting"?

Bernardo Meurer

@ShaVuklia A headache, that's what it's giving

Sha Vuklia

sigh:P

Slereah

I'll correct it

0celouvskyopoulo7

Why do you want a differential algebra?

@AccidentalFourierTransform Probably French something

Bernardo Meurer

@0celouvskyopoulo7 Apparently, I only need to know about differentiable manifolds

ACuriousMind

18:40

@ShaVuklia The point is that you defined the $F_i$ from the accelerations in isolation. The second law is in a setting with many different objects that are not in isolation. You have, from the axioms so far, no reason to believe that the sum of the forces (or accelerations) in isolation should be the actual total force/acceleration in this situation. That it is is the content of the second law.

Slereah

@0celouvskyopoulo7 Well to solve PDEs you usually want to be able to use derivatives

AccidentalFourierTransform

derivatings*

Sha Vuklia

@ACuriousMind RIGHT, thanks. I think that answer suffices

0celouvskyopoulo7

@BernardoMeurer What?

Bernardo Meurer

@0celouvskyopoulo7 What what

0celouvskyopoulo7

18:42

You only need to know about differentiable manifolds?

Sha Vuklia

I need to remember I'm doing physics, and not just maths:P

0celouvskyopoulo7

As opposed to what?

@ShaVuklia Bah

Do math

none of this crap in math

People always explain exactly what they mean, there's no ambiguity, and no errors

Bernardo Meurer

@0celouvskyopoulo7 Yes, for Analysis II

Sha Vuklia

@0celouvskyopoulo7 that's why maths is fundamentally less exciting than physics, I think

at least for me

0celouvskyopoulo7

(i) I was joking

(ii) that's a terrible opinion if true

Sha Vuklia

18:43

haha oh sorry:P

ACuriousMind

@0celouvskyopoulo7 You should've used an irony mark there :P

Bernardo Meurer

All your fields are shit

Computer Science rocks

Sir Cumference

Pffh, astrophysics is best physics

0celouvskyopoulo7

Whatever Quavo does is the best field, tbh

AccidentalFourierTransform

the best field is $\psi$

0celouvskyopoulo7

18:44

The best field is $\Bbb C$

AccidentalFourierTransform

1+1=0

delete that

you have two flags already

Sir Cumference

What

0celouvskyopoulo7

@Slereah haha that comment

AccidentalFourierTransform

quickly

0celouvskyopoulo7

haahahahahahah

AccidentalFourierTransform

18:45

delet this

Sir Cumference

@AccidentalFourierTransform Me?

AccidentalFourierTransform

YES

Sir Cumference

Wait why

AccidentalFourierTransform

In the pertubation expansion the terms are large (infinite) and need 'renormalization'. — lalala 4 mins ago

0celouvskyopoulo7

fucking physicists, honestly

Slereah

18:46

infinite is fairly large

Sir Cumference

@AccidentalFourierTransform Why did you make me delete that...?

0celouvskyopoulo7

You tell them their theory is shit and then they do something more shit to get out of it

Bernardo Meurer

@0celouvskyopoulo7 Quavo is shit

AccidentalFourierTransform

@SirCumference trolling is a art?

Sir Cumference

@AccidentalFourierTransform ...I hate you

0celouvskyopoulo7

18:47

*an

Sir Cumference

I'd think there'd be a rule about claiming flags when there are none

0celouvskyopoulo7

@SirCumference Please delete this.

@SirCumference There is.

Sha Vuklia

@ACuriousMind would it be an equivalent second law if we said that the force vector associated with an interaction in isolation is the same force vector when there are other interactions as well?

0celouvskyopoulo7

@AccidentalFourierTransform not a troll, just showing how bad your grammar is

AccidentalFourierTransform

you're*

ACuriousMind

18:49

@AccidentalFourierTransform Do you think the content of that link is appropriate to post here?

AccidentalFourierTransform

lol, didnt read past the first paragraph

delet it pls

ACuriousMind

Maybe...read further next time :P

0celouvskyopoulo7

@AccidentalFourierTransform No, I think your is correct.

Slereah

Ahahah

I just got a downvote

rob

@0celouvskyopoulo7 You're correct: your's correct.

0celouvskyopoulo7

18:51

@rob Of course it is.

He's trying to troll me.

Slereah

Some QFT lad is mad

0celouvskyopoulo7

@AccidentalFourierTransform What is your comment supposed to mean?

ACuriousMind

@Slereah Well...when I saw the title I immediately hovered over the downvote button until I had read it :P

Slereah

The title was non-negotiable

AccidentalFourierTransform

clickbait

@0celouvskyopoulo7 idk

0celouvskyopoulo7

18:54

I really need to be studying

What am I doing with my life

Why does this chat exist?

Bernardo Meurer

@0celouvskyopoulo7 Without this chat I would spend too much time actually doing things

Which will inevitably lead to world end

ACuriousMind

@0celouvskyopoulo7 To trap people like you, apparently :P

0celouvskyopoulo7

@BernardoMeurer You still haven't told me why you "only" need to know smooth manifolds, or what you need to know about them.

ACuriousMind

Ah, yes, and to distract Bernardo from achieving total potato dominance

Bernardo Meurer

@ACuriousMind Have I told you my latest step towards complete planetary destruction?

@0celouvskyopoulo7 Because that's all my course covers apparently

ACuriousMind

18:55

@BernardoMeurer You punched your wall and bloodied your hand? :P

Bernardo Meurer

And I don't know, I learned how to prove something is a smooth manifold

0celouvskyopoulo7

That's not what smooth manifolds are about

Bernardo Meurer

@ACuriousMind Yes, no. I got an FPGA and have been doing work on it :P

0celouvskyopoulo7

What did you learn about them?

Abcd

Why does force exist?

Bernardo Meurer

18:56

@0celouvskyopoulo7 I don't know what they're useful for yet

I learned parametrization

and path gradients (?)

ACuriousMind

@BernardoMeurer I read your bragging about running linux on it, yes

@Abcd Because otherwise the world would be pretty boring.

Bernardo Meurer

@ACuriousMind Yes, I will continue to brag until I run doom on it

0celouvskyopoulo7

@Slereah Does a metric like $$g_{ij}=\left(1+\frac{\mathcal M}{|x|}\right)^4\delta_{ij} +p_{ij},\quad |x|^2|p_{ij}|+|x|^3|Dp_{ij}|+|x|^4|D^2p_{ij}|\le C$$ look like a decent enough metric on a Cauchy surface?

Abcd

@ACuriousMind Serious answer please :/

0celouvskyopoulo7

@Abcd Why should he

ACuriousMind

18:57

@0celouvskyopoulo7 No, that metric looks very indecent to me

Slereah

@0celouvskyopoulo7 wat

Abcd

@0celouvskyopoulo7 Why shouldn't he?

0celouvskyopoulo7

ugh you two

ACuriousMind

@Abcd I don't know what a serious answer would be. Why does anything exist?

0celouvskyopoulo7

@ACuriousMind Because God decided to make it.

Abcd

18:58

@0celouvskyopoulo7 Ah! that's cool.

0celouvskyopoulo7

@Slereah This is allegedly isometric to Schwartzschild when $p_{ij}\equiv 0$

is that obvious to you?

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