The h Bar

Satyajit Sen

00:30

Since I've been told that advertising on chat whenever you post a new question is a good idea, here...

0

Q: What is the difference between collimating, waveguiding, attenuating, and polarizing?

What's the difference between collimating, waveguiding, attenuating, and polarizing? Are they related on any level at all?

Looking forward to being enlightened. :)

bolbteppa

How would you explain the repeal of Net Neutrality? We did it with the Whopper. Watch the video below: https://t.co/9EWjtbenv8

Tweeted by BurgerKing on January 24, 2018 at 2:07 PM

Jake Rose

00:45

Hey is anybody free?

I can't figure out a pretty basic trig question

Cookie Toast

Man is born free, and everywhere he is in chains.

Fawad

01:08

@JakeRose just ask.if someone want to help then they will

Cookie Toast

Speaking of asking, I've got a question about the mechanical wave equation. My textbook defines $y(x,t) = A\sin (kx-\omega t)$. It then goes on to say that through the principal of superposition I can rewrite the equation to a function of $x$ multiplied by a function of $t$, in this case $\sin (kx)\cdot 2A\cos (\omega t)$. I'm not seeing how this happens...

Using the formula $\sin (a-b)$, I get $\sin (kx) \cos(\omega t) - \sin (\omega t)\cos (kx)$ but I don't see the connection here

I've also tried rewriting the equation with Euler's formula, but I'm not seeing the connection there either.

Correction, that should be $\sin(kx)\cos(\omega t) - \cos(kx)\sin(\omega t)$

bolbteppa

01:36

@CookieToast by superposition you have that $y_1(x,t) = A \sin( k x - \omega t) + A \sin(kx + \omega t)$ and so you use $sin(a+b) = 2 \sin[\frac{a+b}{2}]\cos[\frac{a-b}{2}]$

Cookie Toast

Ok, let me check my math for a moment :)

@bolbteppa is that the correct formula for $\sin (a + b)$? I've never seen it before

bolbteppa

\begin{align*}
f(x) + f(y) &= \cos(x) + \cos(y) + i[\sin(x) + \sin(y)] \\
&= \cos(x) + i \sin(x) + \cos(y) + i \sin(y) \\
&= e^{ix} + e^{iy} \\
&= e^{i \frac{x+y}{2}}[e^{i \frac{x-y}{2}} + e^{i \frac{y-x}{2}}] \\
&= [\cos(\frac{x+y}{2}) + i \sin(\frac{x+y}{2})][\cos(\frac{x-y}{2}) + i \sin(\frac{x-y}{2}) + \cos(\frac{y-x}{2}) + i \sin(\frac{y-x}{2})] \\
&= [\cos(\frac{x+y}{2}) + i \sin(\frac{x+y}{2})][\cos(\frac{x-y}{2}) + i \sin(\frac{x-y}{2}) + \cos(\frac{x-y}{2}) - i \sin(\frac{x-y}{2})] \\
&= [\cos(\frac{x+y}{2}) + i \sin(\frac{x+y}{2})]2 \cos(\frac{x-y}{2}) \\

Yup

Nightmare remembering these formulas

Cookie Toast

Oh my god. D:

We aren't even expected to know Euler's identity in my class. It's lower level community college physics

I don't know how the hell our prof expects us to do our homework

bolbteppa

01:56

2

A: How to prove $\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$ using addition theorems?

Apply to the two factors on the right the formula $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ and its twin for the cosine. My favourite explanation actually involves physics: on the left hand side we have the sound of two strings on musical instruments where one of the two i...

There is a nice explanation using sound there to try help remember it

Oh wait

Yeah, these formulae are a nightmare

'this'

is the best you can really hope for

given here

6

A: geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$

$$\begin{align} 2 \cos A \cos B &= \cos(A-B)+\cos(A+B) \\[6pt] 2 \sin A \,\sin B &= \cos(A-B)-\cos(A+B) \end{align}$$ Note. Although not labeled (yet), these identities are also evident: $$\begin{align} 2 \,\sin A \cos B &= \sin(A+B)+\sin(A-B) \\[6pt] 2 \cos A \,\sin B &= \sin(A+B)-\sin(A-B) ...

On a side-note, I found this:

8

A: Why does $A\sin{k(x+c)}=a\sin{kx}+b\cos{kx}$ imply that $A=\sqrt{a^2+b^2}$ and $\tan{c}=-b/a$?

Here's my picture-proof of the identity, with $k=1$ and @MarkBennet's suggestion to remove the negative sign from the tangent: $$p \sin(\bullet) + q \cos(\bullet) = r \sin(\bullet +\circ ), \quad\text{where}\quad r = \sqrt{p^2+q^2} \quad\text{and}\quad \tan(\circ) = \frac{q}{p}$$ I walk thro...

Amazing

Cookie Toast

@bolbteppa I have to step away from the computer, but I really appreciate the response!

Physics Meta

02:20

0

Q: Is it OK to use the "unclear" option for closing clear questions by unclear-minded users?

The original intention of the option unclear what you're asking is obvious. Nevertheless, every now and then there are questions in the closing queue that are actually clear enough, but end up receiving close votes as "unclear". Some examples: Is $Tds=dh-dp/\rho $ a valid definition for entrop...

discussion

Secret

04:18

thebulletin.org/timeline

I wonder, will we ever survive to see the clock to reach exactly midnight on the site

Fawad

04:39

I got till $T_x$ but I don’t see how to calculate equivalent thermal resistance

😢

Emilio Pisanty

@dmckee thx =)

Just needed an extra pair of eyeballs on it

There was (still is?) a nonzero chance that I came across as a condescending prick

I felt it was important to say it but it was worth minimising that chance

Cookie Toast

@fawad "From the diagram, we observe that $1+1=2$. Thus the heat flowing from the slab can be easily calculated." :P

0celo7

05:01

@BalarkaSen too late I already saw it

skullpatrol

@EmilioPisanty +1

vzn

05:19

0

Q: multivariate stochastic gradient descent reducing to linear matrix recurrence relation

the wikipedia pg on gradient descent gives the iterative (multivariate/ vector) formula $\mathbf{a}_{n+1} = \mathbf{a}_n - \gamma \nabla \mathbf{F}(\mathbf{a}_n)$ where $\mathbf{a}_n$ moves toward the minimum of $\mathbf{F(x)}$ and $\gamma$ is a step size. my question, when does this reduc...

linear-algebra matrices differential-equations recurrence-relations gradient-descent

0celo7

@EmilioPisanty lmao

@Slereah I think the shitty french math is actually genius

ACuriousMind

07:29

@SatyajitSen for the record, we generally don't encourage posting your new questions on chat unless you have reason to believe it is of particular interest to someone currently in the chat.

skullpatrol

haven't you heard, it pays to advertise? :P

...but yes; that is the purpose of the "feeds"...

Secret

07:58

some seemed to be out of their usual timezone today... hmm...

2

user228700

08:09

@JohnR: I used a mini vacuum cleaner to clean my keyboard and guess what?

John Rennie

@Kaumudi.H what?

user228700

Hardly any dust at all! :-D

John Rennie

@Kaumudi.H as news goes that isn't the most Earth-shattering I've heard :-)

Slereah

morning

user228700

>.< Well, I distinctly remember having this conversation:

user228700

08:12

Dec 9 '16 at 7:48, by user246160

@Kaumudi All the excitement and urge to maintain the device vanishes after a month and believe me, you will start eating pizza on your laptop keyboard very soon:-P (unless you are like my father who has been using the same laptop for the last ten years :-P)

user228700

AHA! Found it!

user228700

That might've been Doraemon. In any case, huzzah, it has been over one year now and it's still quite clean!

user228700

@Slereah Hey! :-) How's it hanging?

Slereah

Well it's friday

so things will soon improve

Abcd

why did doraemon leave?

user228700

08:14

Dec 9 '16 at 7:50, by John Rennie

I'm with @TheStackExchange on this one :-)

user228700

HAH! :-)

user228700

@Abcd He was banned.

user228700

@Slereah Ha, cool :-)

Abcd

@Kaumudi.H why

Secret

(cont.) ...yeah, people are definitely out of the usual timezones today. Looks like we have an anormally in australian day. This is intriguing...

user228700

08:15

@Abcd Oh, it's a terribly long story and I doubt if I should dive into it. You could search for it in the transcript if you wish.

Abcd

maybe he was a JEE guy...

user228700

Uh, yes, he was.

Captain Bohemian

I never knew there is a mini vacuum cleaner designed to clean keyboard. dust seems to accummulate on my keyboard so frequently that I can't have time to wipe it off.

Secret

(cont.) Perhaps, that might explain why my recent PhD progress is very sluggish. It seems it is either
1. My PhD goes fine and the whole world descended into chaos
2. The opposite happens

(example of an illuminate conspiracy theory grade theory)

user228700

@CaptainBohemian It wasn't specifically designed for a keyboard! :-) My mum uses it to clean many surfaces in our home.

John Rennie

08:19

@Kaumudi.H I'm kind of with you on laptop keyboards. It's a lot easier to just be careful to keep them clean than it is to try and clean them once you've got them dirty.

user228700

Yep! :-)

John Rennie

You can replace laptop keyboards fairly easily but it's a somewhat fiddly job and the keyboards are quite expensive

user228700

Hmm.

user228700

Oh, my God.

John Rennie

08:25

Looks a bit like me at that age :-)

user228700

Hahaha, really? :'-)

John Rennie

I used to get called "Joe 90" at school

user228700

I didn't know what that is, but Googled it to...understand nothing.

user228700

The Adventures of a child super-spy! Aha! x'D

John Rennie

Do an image search

user228700

08:27

Oh, BTW, has my avatar changed at all?

user228700

@JohnRennie x'D

John Rennie

@Kaumudi.H not that I can see.

user228700

Huh. Well, I changed it. Maybe in some more time...

John Rennie

If you've just changed it the change takes a while to work its way through to the chat room.

user228700

@JohnR: Any chance you looked anything like this a couple decades ago:

user228700

08:31

user228700

? :-P

John Rennie

That looks wholly unlike me at any point in my life

user228700

:-)

John Rennie

In any case the statement is true, but then you were among the stars before you left Earth :-)

user228700

:-) Ha, true!

user228700

08:33

I feel bad that you are missing this. I don't quite understand how it is you are able to dislike the whole genre!

Slereah

Alas we are not since Elvis died

but I guess the stars belong to the sky

John Rennie

Ah, that picture is a character from The Office

user228700

Yep! :-)

user228700

Dwight "Danger" Schrute.

Slereah

a much more interesting character than Jim

Which we are asked to sympathize with but is awful

user228700

08:41

@Slereah Interesting? Absolutely! x'D

user228700

I mean...

Slereah

He was also the protagonist in Super

which was a great movie

user228700

Ooh, looks interesting!

user228700

That poster doesn't do the scene any justice. His priceless expressions make it funnier than ever!

Slereah

08:43

It's a realistic superhero movie sort of thing

user228700

Ah, I see.

Slereah

just a dude going around clubbing drug dealers with a wrench

user228700

x'D Hahaha.

user228700

Oh, Ellen Paige's in it, too!

Slereah

SHUT UP CRIME!

►

great catchphrase, too

The preparation for the final fight is just him making pipe bombs in his kitchen

user228700

08:48

x'D I see.

Slereah

09:14

Hey @ACuriousMind

Is there a BRST operator corresponding to time reversal symmetry

ACuriousMind

@Slereah I don't know what that means - time reversal is not a gauge symmetry

Slereah

That paper on quantizing Polyakov mentionned that the symmetry of the action was $\Bbb Z_2 \times \mathcal F$

Reparametrization and inversion of the parameter

I assume the $\Bbb Z_2$ part is the part involved in removing the negative frequency part from the Hilbert space, since it transforms the momentum from one light cone to the other

Is it correct or not

Because the BRST operator $(p^2 + m^2)$ doesn't seem to select specifically positive frequencies

ACuriousMind

Sure, it's a Z2 symmetry but it's not gauge in the sense of the BRST formalism. Discrete symmetries have no infinitesimal generators and hence no associated constraints

Slereah

How does one remove the negative frequency part of the hilbert space, then?

Some dude also mentionned this btw

@Slereah Gupta-Bleuler reads $(\partial^\mu A_\mu)^+ |\Psi\rangle = 0$, where "$+$" means to take the positive frequency part. — AccidentalFourierTransform 11 hours ago

Is it related

bolbteppa

09:45

Apparently $\partial^{\mu} A_{\mu} |\Phi > = 0$ doesn't allow creation of photons so you have to weaken it to $\partial^{\mu} A_{\mu}^+ |\Phi> = 0$ i.e. only the annihilation operator part of the field acting on the vacuum goes to zero

Slereah

"Let us look at this problem a bit closer. A Hilbert space is described by a
set of linear operators and a vector space of states, together with a good$^1$ definition
of inner product."

At least for once it's not a nice definition

ACuriousMind

@Slereah no, the GB "positive" part is because trying to use the full operator is too strong and basically gives you no states

Slereah

Hm

How does it work for Polyakov, then?

ACuriousMind

How does what work?

bolbteppa

The BRST current for Maxwell reduces to the positive-frequency Gupta-Bleuler condition in Fourier space, and GP is an infinite number of conditions in position space also, should write this up hmm

Slereah

09:49

@ACuriousMind Doesn't the proper quantization of it remove the negative frequency part of the Hilbert space, to have a positive definite product

But I can't see what part of the process would select those states

ACuriousMind

@Slereah The result of the standard BRST procedure is a space with a proper inner product. This is guaranteed by the BRST formalism and it doesn't know anything about "negative frequencies"

bolbteppa

and BRST commutes with the Hamiltonian preserving the space of states in time, while Gupta-Bleuler wont in the non-abelian case

Slereah

Isn't the original Hilbert space $L^2(R^3)$ out of which you have to select states that obey the constraint?

ACuriousMind

You know, I've got nothing to do this Sunday, you've been asking for so long about this I might put up a self-Q&A about that

Slereah

I guess it's hard because I'm trying to link it to the naive quantization approach

Apparently the one that suffers from the negative norm is the Dirac-Fock quantization

I should read that whole particle quantization thesis really

But it's 300 pages

Whole book worth

that dude earned his PhD

Secret

10:00

Philosophy of science question: Is it always true that for any subset of phenomenon in the universe that is described by some scientific model, whenever experiments discover a phenomenon which cannot be captured by the model, there always exists a way to either generalise, modify or replace the model so that the new phenomenon as well the old phenomenon can be described in the new framework?

Or in short, is it theoretically possible for a phenomenon in the universe that there exists no models that can simutaneneously account for it and all known phenomenon?

Slereah

You can always enlarge a model to accomodate new data

It may not necessarily be a nice model

Secret

I see

Slereah

It might not satisfy the golden rules of good models

Agreeing with experiment, being predictive and being applicable to arbitrary experiments

princeton.edu/~hhalvors/teaching/phi520_f2012/putnam-craig.pdf

3

^Model with arbitrary data

Since we can only produce a finite amount of experimental data and you can always put the results of those experiments as axioms you can just construct a theory for any set of observation

A bad theory but one nonetheless

bolbteppa

One thing I've been meaning to check is why brst is just replacing $\Lambda^a$ in $\delta A^a_{\mu} = \frac{1}{g} \partial_{\mu} \Lambda^a + f^{abc} A_{\mu}^b \Lambda^c$ by $\Lambda^a = \eta^a \lambda$ for $\lambda$ a constant spinor, i.e. it factors the gauge vector into a product of spinors one of which is the ghost in your FP action, I mean it kind of makes sense

Slereah

Aren't vectors factorizable into products of spinors?

Or is it just null vectors

I forget

Hm

bolbteppa

10:09

Every 4-vector can be represented by a 2x2 matrix, and then this matrix can be factors into a direct product of two spinors

Slereah

I think I'm getting somewhere with this thesis

skullpatrol

\o @dmckee

bolbteppa

So I mean I guess that makes sense, and I guess that's how brst 'remembers' the original gauge invariance because you factored the thing and are sneakily using it...

Slereah

He goes on about how Dirac states can be given by $$\psi^D(x,t) = e^{-it\hat A}\psi(x)$$

bolbteppa

but

Slereah

10:10

For $\hat A$ some component of the constraint

So I guess it is indeed possible to select the right part of the Hilbert space in the Dirac-Fock formalism

I just need to understand how you can do that and I'm golden

I hate BRST so much

All constraints, rly

bolbteppa

Dirac constrained quantization does the same thing BRST does

Slereah

They are all awful

in their own ways

bolbteppa

There's an example for the point particle given exactly I linked you to before a while ago, he shows what you get in brst is the same as for Dirac and he set up the Dirac method, also I can give a reference which derives the GP condition from the BRST current and sets up GP earlier without the (insane) abstraction of that thesis :p I need to write them up still, need to know what's going on first

It's all awful, but apparently research papers are still coming out on Dirac stuff

Slereah

There's also a formalism where you reduce the classical phase space and then quantize it

It seems even worse

bolbteppa

google: 'best transformation'

'best gupta bleuler'

Slereah

10:18

More like worst transformation

There's a reason it ends with -rst

bolbteppa

it was the best of times it was the blurst of times

►

Slereah

A classic

Skinner invites Chalmers on steamed hams

►

bolbteppa

hmm

Slereah

How do you even reduce the phase space, anyway

Do you just quotient the phase space with the gauge group

That would be an awful manifold

bolbteppa

If you had an action which was gauge invariant, then you do the whole FP method, which is basically adding Lagrange multiplier constraints fixing the gauge and constraining changes in the gauge (the FP determinant term), then given that your new action contains new terms which are fermionic, it kind of makes sense to factor the bosonic terms into products of fermions

and then finding some overall fermion symmetry which has to be very similar to the original gauge symmetry just because the action had gauge symmetry, but the new variables allow a little freedom to massage it to make sure the new action has symmetry... Thus if you find a global symmetry, it will commute with the Hamiltonian and thus it's action on states will characterize them

Why that guarantees physical states I don't know

But the whole factoring your original symmetry into something involving fermions doesn't seem so strange since fermions arose out of constraining changes in the gauge in the first place

Slereah

10:32

IIRC the whole Grassman variable trick is just a ruse

It's just so you can use the Berezin integral to give you the appropriate constraint

bolbteppa

There's some crazy way to avoid it I think which ends up as sums over closed loops

ACuriousMind

@Slereah basically yes, it's called symplectic reduction and the resulting manifold is very nice and still symplectic

Slereah

If it's so nice how come nobody ever does it

Everyone just uses BRST

bolbteppa

Kaku does it, shows they don't affect the theory at all so are just a ruse

ACuriousMind

@Slereah explicit reduction of the phase space requires you to "solve the constraints", which is infeasible for all but the simplest cases

Slereah

10:37

What's a case where it is possible?

Is it one of those toy model cases

ACuriousMind

@Slereah It also destroys stuff like general covariance in EM - of course the two remaining do. O.f of the gauge field don't transform in any nice way under Lorentz transformations

bolbteppa

It's really crazy that if you do BRST for Yang-Mills in a way that ends with the action ending up as a total derivative, your transformations are only nilpotent when the EOM hold, however if you do BRST by doing a Hubbard-Stra... transform (i.e. adding a variable so things become more Gaussian) then they are nilpotent everywhere

Balarka Sen

@ACuriousMind Hm, I guess it could be an interesting question when a Lie group quotient of a symplectic manifold still gets a natural symplectic structure

Abcd

what does "pendulum clock loses time" mean?

Slereah

The pendulum's frequency lowers?

I would guess

ACuriousMind

10:44

If you want to know more about symplectic reduction, I think the book by Marsden (?) Is the usual reference

Slereah

Why does Haag's theorem have the Hilbert space with a unitary rep of SU(2)?

Abcd

@Slereah and what would "loses 24 seconds per day" mean?

Slereah

Why not the full Poincaré group?

Balarka Sen

@ACuriousMind Thanks, I shall have a look

Abcd

would it mean that the clock is 24 s behind the actual clock..

Slereah

10:48

Yes

Abcd

or is it 24 s ahead of the actual clock

@Slereah fine.

Slereah

When the pendulum loses time due to friction a second as measured by the clock is longer

Hence it should be longer

Wait, or is it

bolbteppa

Poincare is not connected

Reps need to be in terms of su(2)

Slereah

@bolbteppa Why not SL(2,C) then

bolbteppa

Locally you can use products of su(2) to get sl(2,C) I think

Slereah

10:50

I would agree but I think it's one of those things that makes the math people mad

bolbteppa

Honestly it's too hard to justify all this, I have given up about 20 times

Hand-waving is fine as more and more of it makes sense

Slereah

heh

bolbteppa

Why the hell is brst even nilpotent

Slereah

JibJab.com - Second Term

►

I just love early 00's political flash songs

A forgotten genre of the internet

Even Jibjab stopped making them

bolbteppa

Weinstein is in the pic around 0.37 I think, jesus

Slereah

10:55

You know where else Weinstein is?

Balarka Sen

I guess if the $G$-action on $(M, \omega)$ preserves the symplectic form then for the 1-parameter subgroup $g(t) = \exp(tv)$ for $v \in \mathfrak{g}$, $\mathcal{L}_v \omega = d/dt (g(t)^* \omega) = 0$

Slereah

The World Of David The Gnome - Opening Theme

►

Balarka Sen

But $\omega$ is closed, so Cartan says $d i_v \omega = 0$

Slereah

BRB INTERNATIONAL AND BOB & HARVEY WEINSTEIN PRESENT

Weinstein was behind David the Gnome all along

Balarka Sen

I forget if that means $v$ is a Hamiltonian vector field or a symplectic vector field

bolbteppa

10:57

Thank god I never seen that before and am not heart broken now

Balarka Sen

Ok, symplectic. (Hamiltonian is stronger, which means $i_v \omega$ is exact - symplectic just says it's closed)

bolbteppa

"Seriously, my childhood is ruined... First screen: BRB International and Bob and Harvey Weinstein Present..."

Balarka Sen

Wow, so every left-$G$-invariant vector field has to be symplectic

bolbteppa

"An important restriction for BRST transformations is to have them be nilpotent, in other words s2 = 0. This is important for transforming the measure in the path integral. Under a BRST transformation on the fields the Feynman measure changes by a Jacobian factor, det(l + s). (4.15) We have seen already that such determinants can be defined for operators, and that they satisfy det A = exp trlog A. If s is nilpotent, then det(l±s)~l±trs. "

that det A = exp trlog A is how Kaku shows they are sums over closed loops and irrelevant before even defining brst

Slereah

Is that why there are Quantum Anomalies

Is that when you get non-nilpotent BRST operators

Balarka Sen

11:03

What are vector fields $X$ on a Riemannian manifold $(M, g)$ called such that $\mathcal{L}_X g = 0$?

Killing?

Slereah

Yes

Balarka Sen

makes sense

bolbteppa

He writes an arbitrary matrix $A$ as $1 + M$ and shows it results in closed loops, hmm

For for general $A$ it becomes closed loops, but if $M$ is nilpotent then the volume element is supposed to be 1, hmm

bolbteppa

11:27

oh my god, sec. 2.1 people.phys.ethz.ch/~jshapiro/PDFs/… quantizing a particle on a circle and finding the physical space, he just sets a whole variable to zero!

Slereah

>Weinstein

Aaaaah

He's everywhere

bolbteppa

I know

Complicating things

That's insane though, simple examples show how crazy what you're studying is

'the expected result'...

I see in 2.12 something similar to your Dirac operator thing

Slereah

yeah it seems to be the same

I need to find out how that works

bolbteppa

11:43

If $M = I + s$ with $s^2 = 0$ then from $(I + s)(I - s) = I$ we have $\det(I + s) \det(I - s) = \det(I)$ i.e. $\det(I \pm s)^2 = 1$ or $\det(I \pm s) = \pm 1$ so that from $\det(I \pm s) = 1 \pm \mathrm{tr}(s)$ we have $1 = 1 + \mathrm{tr}(s)$ or $\mathrm{tr}(s) = 0$ so that the FP determinant is $\det(M) = 1 + 0 = 1$ so this is another hint at why brst is nilpotent if we think backwards I guess

Slereah

Are those operators or is $s$ a Grassman variable

or what

bolbteppa

it's just matrices, or operators in this case I guess

e.g. $\Delta_{FP} = \det(M)$

Nothing is ever easy, this is just a hint at why brst might be nilpotent before doing anything, kind of...

bolbteppa

12:03

Apparently another motivation for even wanting to do brst transformations is that the FP ghost action depends on $\zeta$ in $\frac{1}{2 \zeta}(\partial A)^2$ and so feynman diagrams will depend on $\zeta$ however because we know $\zeta$ is a ruse introduced to deal with gauge invariance we want a symmetry of the theory which eliminates it so that feynman diagrams will not be affected, and lo and behold brst gets rid of it...

Slereah

IIRC another perk of BRST is that the stress energy tensor of the action is the proper one only if you have the BRST term

Because the contribution of the gauge fixing term is cancelled out by the BRST term

bolbteppa

So I guess: If you have a random parameter in a theory attached to a boson which you know shouldn't matter, and you want to eliminate it, and you magically have new degrees of freedom which are fermions, all you can really do to eliminate the parameter is send your boson to fermion space by treating it as the direct product of fermions,

one of which is one of your original fermions and saying changes in it are parametrized versions of that in the other spinor and then defining variations in the new fermion d.o.f. to ensure it's a symmetry, I guess

So states are those coming from the charge of the current of the brst symmetry which mean the states are not affected by the parameter, which is why you'd want to use it to create a state space to begin with?

It being nilpotent is just because you want infinitesimal symmetries and you used a fermion to parametrize infinitesimals so it just seems to pop out being nilpotent by virtue of infinitesimal fermion parameters motivated by wanting to eliminate that stupid parameter

That honestly makes a lot more sense than anything else I've seen so far, and these parameters are going to arise in any FP dance you do

Slereah

12:19

Aren't the ghosts for fermions bosonic though?

bolbteppa

I'm not sure, a ghost is that thing which results from turning a determinant into a Grassmann gaussian?

Slereah

yes

bolbteppa

Maybe they have their own bosonic ghosts in a more formal perspective idk

Slereah

👻

bolbteppa

This whole eliminating parameters argument really seems to sync up with interacting theories and ensuring Feynman diagrams are pure which is how I think they were originally thinking

I don't know how the action being invariant means nothing depends on the parameter, I guess because whatever choice you pick you can change the theory by some parameter which does not affect the theory cancelling the effect of your choice of that parameter...

Another observation, that parameter in the Maxwell FP action is attached to (a squared version of) what ends up being the Gupta-Bleuler gauge condition on states, or contains it at least...

Slereah

12:38

Can't we change the website's algorithm to block any question that starts with "why"

They're almost always bad questions

physics.stackexchange.com/search?q=title%3Awhy

or duplicates

Are there any superselection rules beyond $B$, $Q$ and $(-1)^F$?

"A ray is said to be physically realizable if the projection operator onto it is an observable."

wot

What is a state that isn't physically realizable

Oh is it gonna be a state that disobeys superselection rules?

Ah, I see

In this book "observable" doesn't just mean hermitian

Emilio Pisanty

13:05

Man, SE the search is so broken

physics.stackexchange.com/search?q=ncomms

vs

bolbteppa

So it's not obvious BRST eliminating the parameter will not affect the theory, you need to do a big Taylor-Slavnov analysis it seems, but eliminating the parameter seems like the crux of it

Emilio Pisanty

data.stackexchange.com/physics/query/792179

bolbteppa

Or, you have this parameter which for different choices gives green functions, but we know it shouldn't affect physical results, so there must be some secret global symmetry, so that s-matrix elements are unaffected by the parameter at least, only way to find a global symmetry is sending vectors to their spinor counterparts via direct products etc

Emilio Pisanty

I hope you actually get the search to work before you put it in the spotlight like that. — GiantCowFilms Sep 14 '17 at 19:50

bolbteppa

This is in Bailin

books.google.fr/…

Emilio Pisanty

13:17

ah, there we go

bolbteppa

13:36

This makes way more sense than GB does

0celo7

@BalarkaSen youtube.com/watch?v=fX9jewlvadE

skullpatrol

what is that supposed to mean?

0celo7

@skullpatrol what is what supposed to mena

Slereah

Oh no

Italian bears

Somebody touched their gabagool

0celo7

@Slereah last week there was an Italian woman at the colloquium

talking about SPDEs

Slereah

13:42

SPagheDdiEs?

0celo7

lol

Slereah

What is the S for

0celo7

stochastic

Slereah

oh

Balarka Sen

@0celo7 seen it

0celo7

13:44

@BalarkaSen I know you have

it was on pdp

Balarka Sen

mhm

0celo7

i was just reminding you of the best meme

Balarka Sen

i love the SPAGET meme

Slereah

It's Spagett! | Tim and Eric Awesome Show, Great Job! | Adult Swim

►

this one?

skullpatrol

memes have no memeing :P

Slereah

13:48

I Sit Down When I Pee | Tim and Eric Awesome Show, Great Job! | Adult Swim

►

Many fine songs

Doo Dah Doo Doo! | Tim and Eric Awesome Show, Great Job! | Adult Swim

►

bolbteppa

14:06

Physical states are those states for which the current flowing in the time direction caused by changes in the fermion parameter is zero, i.e. $Q |\psi> = \int J_{BRST}^0 dV |\psi> = 0$, insane

That's reality according to math, everything is unchanging when we vary our fermions in time

Slereah

Are there any non-linear operators on Hilbert space

We always talk about linear ones

bolbteppa

Yes, in spontaneously broken symmetries

Nonlinear realization

In mathematical physics, nonlinear realization of a Lie group G possessing a Cartan subgroup H is a particular induced representation of G. In fact, it is a representation of a Lie algebra g {\displaystyle {\mathfrak {g}}} of G in a neighborhood of its origin. A nonlinear realization, when restricted to the subgroup H reduces to a linear representation. A nonlinear realization technique is part and parcel of many field theories with spontaneous symmetry breaking, e.g., chiral models, chiral symmetry breaking...

Slereah

15:00

127

A: Why are four-legged chairs so common?

The real question is why not 2 legged chairs? First, our legs are single points that contact the floor. How they reach the floor is not important. If you have $n$ legs evenly spaced, then the ratio between the radius of the legs and the length of the shortest axis is $\cos(\pi/n)$: 2 legs: 0...

"0 legged chairs are known as sitting on the floor."

12

Q: Can you have black holes in your black holes?

Inspired by Are we inside a black hole?, can you have a black hole such that other black holes are in them? In particular, the event horizon of the larger black hole should completely enclose the event horizons of the smaller black holes. If this is so, what happens when a black hole collides wi...

general-relativity black-holes singularities event-horizon

I think I predicted this question once

knzhou

15:42

@Slereah as a poster with almost 100 questions, about 20 of which start with 'why', i have to disagree!

Slereah

Sounds a bit biased to me!

0celo7

fucking normaliations

the French do good math but their conventions are wrong

Slereah

what is wrong

0celo7

@Slereah I'm working with $-a\Delta +R$ but the French work with $\Delta_\text{French}+R/a$ where $\Delta_\text{French}=-\Delta_\text{everyone else}$

Slereah

ah yes, the French operator

aka the Frenchian

0celo7

15:50

they do have a point that $-\Delta+R/a$ is better, however...

I don't really want to have to rewrite my thesis tho

Slereah

just redo it with proper notation

0celo7

well the French putting a negative in the definition of the Laplacian is ridiculous

but having $1/a$ is pretty good

Slereah

is this the conformal laplacian

0celo7

it is

Slereah

Laplace Beltrami

whatever

0celo7

15:55

those are different things

Slereah

Conformal Laplace Beltrami operator

means I

Since I usually keep Laplacian for Euclidian space

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