The h Bar

no_choice99

01:06

sup physicers

for a 1st intro to QFT. QFT for the gifted amateur or QFT in a nutshell?

user116211

05:24

Anybody wants to tell me why the downvotes, except for the angry iharob person who didn't like me? Any rational reasons? I would be interested on how to improve the answer! Thanks! — CuriousOne Jul 3 '15 at 0:45

user116211

Interesting ;P

TheGhostOfPerdition

05:40

hey @MAFIA36790

timeanddate.com/eclipse/transit/2016-may-9

@lucas what exactly dont you understand in the brachistochrone problem?

user116211

@TheGhostOfPerdition: heya

TheGhostOfPerdition

We have a solar eclipse today with mercury

@MAFIA36790

user116211

@TheGhostOfPerdition yeh... know that... but, hmmm. let it be.. I'm not so much interested in astronomy till yet ;/

TheGhostOfPerdition

@MAFIA36790 yeah me too, but its great to watch :)

user116211

@TheGhostOfPerdition I would like to witness such moments... but hmm... my local observatory.... okay, I'm not interested ;P

TheGhostOfPerdition

05:55

@MAFIA36790 :D

user116211

06:11

52

Q: Why do the French count so strangely?

Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade. English and German name their numbers like they count (with 11 and 12 as exception), but not French: # | English | German | French -...

user116211

@Slereah: ^^^^

user54412

07:14

@TheGhostOfPerdition It'll be tricky though. Mercury is pretty small, so I'm not sure a typical pinhole projection will catch it. And proper solar telescopes are pretty hard to come by.

kevin Tah N.

07:25

can someone tell me what statistical field theory really is? And what those guys do?

TheGhostOfPerdition

08:02

@ChrisWhite yeah, but solar goggles will work, easily available too, I had seen it when it happened in 2006

:)

FrancescoS

08:18

Hi guys! For a fermonic field, is there any non-vanishing anti commutation relation between the field and the derivative of it? $\{ \psi_\alpha(x),\partial_\mu\psi_\beta(x)\}$

FrancescoS

08:42

Sorry, anti commutation at the same time but different space-point $\{ \psi_\alpha(x),\partial_\mu\psi_\beta(y)\}_{ET}$

Slereah

"Note that in the two-dimensional case, the Lorentz group L is isomorphic to the abelian group R of real numbers (cf. Remark 1.15) and therefore agrees with its universal covering group."

What?

I can see the rotation group going away but what about boosts

Slereah

09:20

Hello golden man god

ACuriousMind

Heyhey

when you say two dimensions, do you mean spatial or spacetime dimensions?

Slereah

Spacetime

I'd expect it to be more like... $R^2$ maybe?

Translation and boost

Oh wait

He said Lorentz group

Not Poincaré

nvm

Anyway

From what I seem to understand of Wightman, you have a Hilbert space and the theory you work with restricts the dense subset of that space you use?

Is that correct

ACuriousMind

I'm not sure what you mean

Then again, my sum total of knowledge about rigorous QFT from the Wightman axioms is what bits I understand of Glimm/Jaffe :P

Slereah

Well from what I was able to gather, you define some operator valued distributions that all map to the same domain of the Hilbert space

ACuriousMind

Ah, yes

But that's not something special - many operators are defined on dense domains of definition even in usual QM

Just take the differentiation operator on $L^2(\mathbb{R})$ - it's obviously only defined on the differentiable functions.

Slereah

09:30

Alright

is the Haag's theorem's point that the dense subsets of different theories are disjoint?

among other things

ACuriousMind

@FrancescoS I don't think you can say something general about that

@Slereah No, Haag's theorem is that different theories do no act upon the same Hilbert space.

Slereah

Someone else here told me it was the same!

Stop your lies, stack exchange

ACuriousMind

That is, if you have inequivalent theories, you cannot find a unitary map from one space to the other that preserves the commutation relation

FrancescoS

@ACuriousMind thanks

Slereah

Hm

ACuriousMind

09:33

@Slereah Well...it might be the same in some strange way, but I don't see how

Slereah

Now to try and prove that I can get $\varphi(\Box f + m^2 f) = 0$

Wait

Is that an axiom of the theory or is it derived

ACuriousMind

@Slereah Are you maybe misremembering what yuggib said about the domains of his classical limit states?

Slereah

I was hoping maybe it was from the Heisenberg equation, but then again I don't think the Hamiltonian is well define with Wightman

@ACuriousMind Could be!

Who knows

ACuriousMind

@Slereah Well...if you want a "free field", you have to say what property you use to define "free field"

Slereah

I was hoping I could just use $\partial_t \varphi = [H, \varphi]$

But thinking back on it, that might not work

At least not in Wightman

I guess it's like Haag where the EoM is the basic equation

Also that paper seems to say that the propagator is ALSO part of the basics

Just like Haag

I guess it makes things easier to define things that way but it feels a bit like cheating

ACuriousMind

09:43

Yes, the "covariance operator" (which is basically the propagator) is something you need to give to construct the measures from it

Slereah

I mean it feels a bit like trying to construct a theory you already know

I'm sure you can squeeze that lemon for more juice with Wightman, but how are you supposed to find new theories?

it's not like you can guess the covariance operator for another theory

ACuriousMind

@Slereah Why not? It's really just the equation of motion

I don't see a difference between guessing Lagrangians and guessing equations of motion

Slereah

Well the equation of motion is fine, but the propagator is another business

I mean at least add the axiom "The propagator from the EoM is blablabla"

Wait, does that even make sense

You can't guess the propagator from the EoM if it's non-linear, right?

at least not directly

Then again I suppose that Wightman only works for linear theories

Wait, is it the propagator, actually

That's not quite the propagator yet, right

Not without knowing things about $f$ and $g$

ACuriousMind

@Slereah Um...they're supposed to be test functions, you can't "know" things about them

Slereah

Argh

yuggib

09:56

@ACuriousMind if you want to call the Sobolev space the space of differentiable functions :-D

Slereah

Right

$\mathcal S$ is rapidly decreasing functions, not tempered distributions

Also what the hell is $\mathcal{S}(\Gamma_m)$

ACuriousMind

@yuggib Sure ;)

Slereah

Apparently it is $\approx \mathcal S(R^3)$

But I still don't have a clue

Oh wait

I think it's the forward lightcone

yeah that's it

yuggib

10:14

@ACuriousMind I don't think there are any...at least that truly understand it ;-P

Slereah

10:40

So in Wightman, you have an operator valued distribution, that is... $\varphi[f] \in \mathcal O$, right?

And for any test function $f$, $\varphi[(\Box + m^2)f]$ is the 0 operator

FrancescoS

arxiv.org/abs/0907.2441, pag.33, eq. B.5. Do you think there is a typo (probably indices) in the first row?

He says that $\partial_\mu \chi^\alpha \sigma^\nu_{\alpha\dot{\alpha}}\sigma^\mu_{\beta\dot{\beta}}\partial_\nu{\chi^\d‌agger}^\dot{\beta} = \partial_\mu \chi^\alpha \sigma^\nu_{\alpha\dot{\alpha}}\sigma^\mu_{\beta\dot{\beta}}\partial_\nu\bar{\ch‌i}^\dot{\beta}+\partial_\mu \chi^\alpha \sigma^\mu_{\alpha\dot{\alpha}}\sigma^\nu_{\beta\dot{\beta}}\partial_\nu\bar{\ch‌i}^\dot{\beta} $. He add a piece with $\mu$ and $\nu$ exchanged…?????

Sorry for the long latex code :)

lucas

11:10

@TheGhostOfPerdition Which force make the particle to move?

TheGhostOfPerdition

11:27

@lucas Gravity and the constraint forces

lucas

@TheGhostOfPerdition What is the constraint force?

@TheGhostOfPerdition The only force field is gravity.

TheGhostOfPerdition

yeah, I was talking about a general case

yesm in this case its only gravity

@lucas

lucas

@TheGhostOfPerdition So, how can the particle move that path?

@TheGhostOfPerdition It must fall straightly.

@TheGhostOfPerdition Sorry. I forgot to say thanks because of your attention.

TheGhostOfPerdition

@lucas no, the problem of the brachistochrone is to find a curve that minimizes the transit time,

@lucas thats okay,, :D

lucas

@TheGhostOfPerdition Ok. But,1. least time is when particle moves on the straight line. 2. Which force make the particle to move?

TheGhostOfPerdition

11:35

for points (x1, y1) and (x2, y2) the minimum transit time would be for a straight line when x1=x2, but when x1!=x2 the min time would be for a cardiod

lucas

@TheGhostOfPerdition I couldn't get. What is "cardiod"?

TheGhostOfPerdition

@lucas least time is not always for a straight line

lucas

@TheGhostOfPerdition Which force make the particle to move?

TheGhostOfPerdition

sorry, its for a cycloid, not a cardioid

lucas

@TheGhostOfPerdition The particle is at rest. And the only force acting on it is gravity. So, how can it move on that path?

TheGhostOfPerdition

11:41

@lucas the particle wont, suppose you had a string shaped in the form of that path (cycloid), only then would it move, which minimizes the time

If you have a string on which the particle moves, then there would be constraint forces too

lucas

@TheGhostOfPerdition How can we talk about minimum time only. Minimum time under which conditions?

@TheGhostOfPerdition We know that force makes movement. Which force does make the particle's movement?

yuggib

@Slereah I don't think that is requested by Wightman's axioms...

TheGhostOfPerdition

@lucas contraint such that the acceleration in the y direction is a contant
if the particle is moving on a string then both gravity and the contraint (normal) forces make the particle move

ACuriousMind

@lucas The brachistochrone problem is the question of which shape of a frictionless track between two given points will let a particle sliding along that track arrive at the end point in the shortest time.

The particle slides along the track purely because of gravity.

TheGhostOfPerdition

@lucas and for your question on what conditions : v = sqrt(2gy)

lucas

11:48

@ACuriousMind Thank you because of your attention.But, I didn't see this in the problem? "sliding"

@TheGhostOfPerdition I think, the particle won't arrive to (x2,y2) under gravity only.

Slereah

@yuggib Well not Wightman in general

But for a free scalar field

ACuriousMind

@lucas What do you mean you didn't see this in the problem? Some people maybe don't state it clearly enough, but that is the definition of what a brachistochrone is.

yuggib

@Slereah I don't think so, no

lucas

@ACuriousMind I saw this problem in the "J.B.Marion" book. Classical Dynamics.

ACuriousMind

The usual demonstration (at least at some museums here) is that you have several tracks between the same start and end points and you can let balls roll along them. One of the tracks is the brachistochrone shape and the ball rolling along it will arrive at the end point first.

TheGhostOfPerdition

11:52

@lucas there is only gravity which is doing the work, there is a constraint forces which only guide the particle in that particular path but do no work

yuggib

it is definitely not true that for any test function $(\Box+m^2)f=0$

lucas

@TheGhostOfPerdition If we have constraint, then my question is solved.

TheGhostOfPerdition

@lucas :)

lucas

@TheGhostOfPerdition Thank you very much.

@ACuriousMind Thanks a lot.

TheGhostOfPerdition

@lucas welcome :-)

Slereah

11:59

@yuggib Why not

The link you gave me certainly seems to think so?

yuggib

@Slereah which link?

Slereah

mathematik.uni-muenchen.de/~schotten/LNP-cft-pdf/…

yuggib

yeah I see...it's the damn light cone condition

Slereah

that it is

Incidentally it's the answer to my question about "At what point does QFT define a time orientation"

Spectrum only in the forward light cone and all that

Trying to understand actually how they solve that equation currently

Not too easy

The answer being of course "Fourier transform", but it's a fancier one

yuggib

@Slereah the equation is not solved strictly speaking

it is true in a "distributional sense"

it is not $(\Box +m^2)f=0$ but more something along the lines of $''(\Box+m^2)''\Phi(x)=0$

where the quotation marks mean that it holds true when acting on rapid decrease functions, as distributions would do

Slereah

12:15

Well I guess it's something like $\langle a \vert \int (\Box f(x) + m^2 f(x)) \hat \varphi(x) dx \vert b \rangle = 0$

yuggib

i.e. on the argument

Slereah

Or something

yuggib

yes, but the point is that it is not $f$ that solves the equation, but $\Phi$, in some sense

Slereah

That much I got, yes

since this is supposed to be true for all $f$

When do you use a specific distribution, by the way?

I guess when you perform a measurement over a region of space

I guess if you measure something over a region $\Omega$ the measurement is $\langle \omega \vert \hat {\mathcal O}[f_\Omega] \vert \omega \rangle$?

Jester Tran

12:31

How do I find the number of real solutions to the equation $xe^x = k$ where $k \in \mathbb{R}$?

Slereah

Isn't that the Lambert W function

Jester Tran

@Slereah nvm, thanks

Slereah

That's the perk of knowing the weird functions :p

Jester Tran

haha

What is a quick method to find an approximation of a zero to strange functions like the Lambert W function or $f: \mathbb{R} \to \mathbb{R}:f(x) = x \sin (x)$?

Slereah

Lambert W is monotonous, IIRC

There's only one 0

$x = 0$

For the main branch, anyway

Jester Tran

12:46

Sorry, that was a bad example and poorly worded question. I should have said equations where you cannot solve it analytically.

Slereah

Not sure there's a generic method

Depends on the equation

Outside of just doing it numerically

Jester Tran

Yeah, what is are some quick numerical methods to approximate roots?

Slereah

Newton's algorithm thing?

Newton's method

In numerical analysis, Newton's method (also known as the Newton–Raphson method), named after Isaac Newton and Joseph Raphson, is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. The Newton–Raphson method in one variable is implemented as follows: The method starts with a function f defined over the real numbers x, the function's derivative f', and an initial guess x0 for a root of the function f. If the function satisfies the assumptions made in the derivation of the formula and the initial guess is close, then a better approximation...

Jester Tran

ok thanks

Bernard Meurer

@ACuriousMind Hey there

Jester Tran

12:52

do you know any other algortihms which doesn't involve differentiation?

besides fixed point iteration and halving interval method

Slereah

Newton is kind of a standard

what do you want to do, exactly

If you can't do derivatives numerically do differences

Jester Tran

@Slereah I want to approximate a solution for an equation I cannot solve analytically and differentiating this function is a bitch

Slereah

Just do $f(x+dx) - f(x) / dx$ for the derivative

with a small mesh

Jester Tran

Do I $dx \to 0$ later?

Slereah

13:07

It's numerical

Just doing a small dx should be enough

Good enough for an approximation if the function is smooth enough

Jester Tran

small $dx$ like $0.01$?

Slereah

Just write a little program and see what happens

Jester Tran

ok

Slereah

Hm

I wonder how I could write an operator-valued generalized function

I guess as a sequence of operator-valued functions

properly mollified

yuggib

13:23

what do you want to do?

Slereah

a Proper calculation of the energy

Where I can multiply the distributions without troubles

yuggib

if the world was so easy...

Slereah

Well it is not

The generalized function is still singular in that case

But it offers a very natural way to do the regularization

yuggib

so essentially you want to solve the quantization problem

Slereah

Do I

It's nothing new

There's a few papers on the topic

Not all that many, though

yuggib

13:29

do you mean that you want to do perturbation theory, and give meaning to products of distributions, or give meaning to e.g. the operator $\int :\hat{\phi}^4(x):dx$??

Slereah

The second

I assume it's not a trivial topic, considering the derth of papers on the topic

yuggib

then you're trying to solve the quantization problem, i.e. essentially a million dollars question

Slereah

i'll take the million dollars please

yuggib

good luck

Slereah

Yeah I'm guessing it's not a topic that goes on well

yuggib

13:31

it's a topic that went on well, with huge technicalities, in $1+1$ dimensions

Slereah

Everything is better in $1+1$ dimensions

yuggib

then went less well, and with humongous technicalities, in $2+1$ dimensions

Slereah

We really should leave this universe

Just go to the $1+1$ dimensional universe

yuggib

then it did not go at all in $3+1$ etc.

Slereah

Let's go there

Though really currently I don't even want to do interacting theory

I just want to see what the energy looks like in generalized functions

For a free field

i'm guessing there's some bullshit divergent term that's like $\propto g_{\mu\nu}$

And then you sweep it inside $\Lambda$ and pretend it doesn't exist

I am somewhat curious why Colombeau algebras aren't used all that much in QFT

I'm guessing because they're pretty recent

They were created in 84

www1.unipa.it/fabio.bagarello/ricerca/articolipdf/distrib2.pdf

This seems relevant

let's read it

arxiv.org/pdf/math-ph/9902008.pdf

also

1

A: Renormalization and Conway/Surreal Numbers

There are attempts to use nonstandard analysis (e.g., Albeverio) or Colombeau algebras, but these haven't been developed very far. I haven't seen anything in terms of surreal numbers, but they may probably substitute for the infinitesimals in nonstandard analysis.

"I cannot recommend this approach, hence have no motivation to look up the references. "

Not very encouraging

Slereah

14:18

mat.univie.ac.at/~neum/ms/ren.pdf

oh boy

Slereah

15:10

fortunately, today I bought an important theoretical tool

I bought some printer paper to print some more papers

Oh wait

I just realized

$\varphi(\Box f + m^2 f)$

That's $+m^2$

I get it

Then by distributional derivatives you can do $\varphi[\Box f] = -\Box \varphi[f]$

Anyway can we unban @0celo7

So I don't have to talk to myself here

Bernard Meurer

15:39

@ACuriousMind Heya

3075

15:49

@BernardMeurer I got into waterloo physics!!!

Check yours right now.

Slereah

@3075 this will be your Waterloo

3

Prepare to be Napoleon Blownapart

3075

lol

Bernard Meurer

16:05

@3075 HEY

CONGRATULATIONS MAN!

3075

Thanks :D

Bernard Meurer

Wow that was a quick response

(From them)

3075

Did you check yours?

Bernard Meurer

Just checked email and Quest, nothing yet

3075

Probably because your information form was late.

Bernard Meurer

16:09

Yep

I'm calm

I'M FUCKING CALM OKAY?

3075

xD

Bernard Meurer

WHY ARE YOU PRESSING ME

;-;

3075

lol dw you'll get in.

maybe even this week.

Bernard Meurer

If I get in I'll never go

I'll have a heart attack when I get the email

3075

e-mail comes a few days after it goes on quest.

so only check quest.

Bernard Meurer

16:13

Where will it be on Quest?

That website is horribly made

it's straight from 2002

3075

Ikr

for the "best compsci school in canada"

website is bad.

Bernard Meurer

24 in the world

ACuriousMind

@BernardMeurer heyhey

Bernard Meurer

It's something I've found, amazing schools have shitty admissions websites

@ACuriousMind Hey!

@ACuriousMind Tell me something I don't know

3075

@BernardMeurer do you see your program choices in a list on the first page?

click each one and see if it says "application" or something else.

Bernard Meurer

16:16

Ah I see

Yeah it says application

ACuriousMind

@BernardMeurer There is a centuries old manuscript, called the Voynich manuscript, that is written in an unknown alphabet and with strange drawings in it, and no one knows whether it is a language, a cipher, or just gibberish.

Bernard Meurer

@ACuriousMind That's pretty cool, where did they find it?

ACuriousMind

@BernardMeurer Someone found it in their attic and sold it to an antiquarian around 1900 I think

Bernard Meurer

@ACuriousMind I mean country-wise

ACuriousMind

Uh, well...we don't really know where it came from

Bernard Meurer

16:23

Oh, hey I have a maths problem for you

Supposed we have a mall or whatever with 20.000 people in it

person x goes to that mall twice a year

and person y goes to that mall 3 times a year

(Whenever they visit the mall they do so at 18:00)

if each of them sees more or less 70 people whenever they go there

what are the odds they'll meet?

ACuriousMind

I'm pretty bad at statistics :P

Bernard Meurer

So am I! I don't know how to solve this thing

And no, it's not a homework, I thought of it in the bus last friday :p

@Slereah Do you know?

Loong

@BernardMeurer simple classical statistics might not even be applicable, since such rare visits could be correlated; for example, if both do their Christmas shopping right before Christmas.

Bernard Meurer

@Loong Ignore that, supposed it's the day they visit is random out of the 365 days

*Suppose the day ...

DanielSank

17:39

@BernardMeurer <3

@BernardMeurer Need more assumptions. Are the 70 people unique?

Bernard Meurer

17:55

@DanielSank <3

@DanielSank Yes, 70 unique people

I think the day of the year problem can be solved by seeing this as a special case of the "Birthday Problem" but I couldn't do it

DanielSank

18:41

@BernardMeurer That makes it a slightly bigger pain in the ass, but you can still do it.

Bernard Meurer

@DanielSank I should stop taking the bus, I keep thinking of things like this

DanielSank

I am person X. Suppose I go to the mall and see one person. The probability that this is not person Y is 19,999/20,000.

Then I see a second different person.

Bernard Meurer

That 19.000 got me confused real hard hahaha

DanielSank

The probability that this is not person Y is 19,998/19,999.

Bernard Meurer

I was like "God dammit Dan this is too advanced already"

Okay yeah that makes sense

DanielSank

18:45

The probability that after 70 encounters I do not see person Y is therefore 19,930/20,000.

Right?

dmckee

@DanielSank If the probability for people to be at the mall is uniform. But the problem supposes at least two people for whom the probability is non-uniform, no?

DanielSank

@dmckee I think we're just working in the assumption that we have some bundle of 20,000 people at that two of them are X and Y.

Bernard Meurer

@DanielSank Yeah, correct

DanielSank

ok, so the probability that I don't meet Y is (19,930 / 20,000).

@BernardMeurer I think you can get it from here.

Bernard Meurer

Yeah but what about the days of the year Dan? What's the probability X and Y happen to be there the same day?

For say X goes there 2 random days a year and Y 3

DanielSank

18:51

Oh I see I had misunderstood that.

Ok so we computed the result in the case that they are at the mall at the same day.

So now you need to compute the probability that one of my visits overlaps with one of Y's visits.

Bernard Meurer

Yep, and that's where I get kaput

DanielSank

@BernardMeurer Oh.

Bernard Meurer

Birthday problem

In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday. While this makes for an amusing...

This may help, it's just as if X had 2 birthdays and Y 3 :p

DanielSank

@BernardMeurer Yes.

So imagine X's two days are plunked down on the calendar.

Now you have to plunk down three of Y's days.

("plunk" means "drop", basically)

Bernard Meurer

What are the chances I'll overlap plunks

DanielSank

18:57

Yes, but the right way is to ask "What is the probability that Y's first plunk misses X's plunks?".

Obviously it is (363/365).

Bernard Meurer

Yep, makes sense

DanielSank

ok, if you understand that then I'm 100% sure you can do the rest.

Bernard Meurer

I just add the next two probabilities now? 362/365 and 361/365

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