The h Bar

More Anonymous

7:49 AM

@fqq Cool

Hey all! I wanted to ask what does the $i$ epsilon prescription in propagators mean physically as a boundary condition?

\Like I remember it as a trick to make the integral converge

But beyond that as to what it physically meant no clue!

Nihar Karve

8:18 AM

I always think of it as enforcing the time-ordering prescription

someone said that it enforces boundary conditions on the propagator, but I don't really know how

ACuriousMind

Not quite sure, but: When time-ordering $\langle\Omega\vert T\phi(x)\phi(y)\vert\Omega\rangle$, we essentially switch from "prob. of particle being emitted at y and detected at x" to "prob. of particle being emitted at x and detected at y" the instant the time-order of x and y changes, i.e. we have the particle travelling forward in time always. Now, CPT symmetry tells us this means antiparticles travelling backward in time always. Supposing the correspondence between energy modes and (anti-)particles, this means the Feynman propagator follows the same "idea" as time-ordering. — ACuriousMind ♦ Oct 1 '14 at 20:27

see the answer and the comment thread for more discussion on the meaning of "pole shifting"

Sleb Lagnej

Can evolution be presented by some cost function C(gene1, gene2, gene3) where evolution's goal is to tweak the genes to reduce the cost function? Where cost function might be chance of death or chance of not reproducing, something like this? (I wanna represent evolution as a big gradient descent algorithm)

More Anonymous

8:33 AM

Yea I don't understand Trimok's comment:

"@DanielSank : the iϵ Feynman prescription, in your case p2−m2→p2−m2+iϵ, is done to be sure of the convergence of the QFT partition function Z=∫DΦeiS(ϕ), with S(ϕ)=−ϕ(□+m2)ϕ, so this has the effect to add a term e−ϵϕ2 in the integral, making the integral convergent. An other point of view is that, with the Feynman contour, you may do a Wick rotation (90° rotation), going to a imaginary (euclidean) energy , without crossing the poles. – Trimok"

(how do you reference a comment? @ACuriousMind )

ACuriousMind

@SlebLagnej sure, that function is called fitness (or rather, since maximal fitness is optimal, your cost function would be negative fitness)

More Anonymous

@SlebLagnej While I do love the stereotype of physicists knowing everything ... Are you sure you want to ask this here?

ACuriousMind

@MoreAnonymous the link behind the timestamp of the comment one-boxes the comment when pasting into chat

Sleb Lagnej

@MoreAnonymous ACuriousMind and JohnRennie know everything.

More Anonymous

@SlebLagnej I stand corrected :P

ACuriousMind

8:36 AM

note, however, that fitness is not a pure function of the genes themselves but also of the environment they exist in - what is fit in one environment might not be fit in another

More Anonymous

@ACuriousMind do you understand this comment?

@DanielSank : the $i\epsilon$ Feynman prescription, in your case $p^2- m^2 \to p^2- m^2 + i\epsilon$, is done to be sure of the convergence of the QFT partition function $Z = \int D\Phi e^{iS(\phi)}$, with $S(\phi) = - \phi (\square + m^2) \phi$, so this has the effect to add a term $e ^{-\epsilon \phi^2}$ in the integral, making the integral convergent. An other point of view is that, with the Feynman contour, you may do a Wick rotation ($90°$ rotation), going to a imaginary (euclidean) energy , without crossing the poles. — Trimok Sep 16 '14 at 9:26

ACuriousMind

multiline breaks the oneboxing :P

More Anonymous

@ACuriousMind fixed :P

Sleb Lagnej

@ACuriousMind so the higher the fitness function, the higher the chance of reproducing?

ACuriousMind

@SlebLagnej it's kind of defined as the chance of reproduction, so yes :P

More Anonymous

8:37 AM

@SlebLagnej Please tell me your looking into this for real world dating applications :P

ACuriousMind

@MoreAnonymous Yes, but without a more specific question I can't figure out what you don't understand about it :P

Sleb Lagnej

I don't have a girlfriend, so I'll go to the fitness to increase my fitness :))

(fitness = gym)

@MoreAnonymous reading a summary of Thinking fast and slow

More Anonymous

@ACuriousMind I don't understand the 2nd line

About wick rotation without crossing *the poles

Sleb Lagnej

How could you measure how good an environment is for some species to put in the fitness function as argument? Is this mathematically calculatable or just a concept described with math?

descriptive, conceptual mathematics

I coined that term in 2021, give me my nobel price

More Anonymous

@SlebLagnej You have my 2 cents :P

ACuriousMind

8:42 AM

@MoreAnonymous I don't see anything about "crossing energies" there? Trimok is saying that if you see Wick rotation as literal 90° rotation in the complex x-t plane, then the Feynman-like positioning of the poles allows you to rotate without the axis of integration "hitting" the poles during the rotation

@SlebLagnej obviously you can observe the rate of reproduction, can't you?

Sleb Lagnej

so if rate of reproduction can stand for environmental goodness, why can't it stand for gene goodness?

fitness(rate of reproduction) done?

@ACuriousMind that profile pic is scaring me

meh I'll just read about fitness before asking more

ACuriousMind

@SlebLagnej I think you've misunderstand what fitness is - it's the average reproductive success of a genotype in a given enviroment, i.e. it's a function of both the genotype and the environment. It doesn't measure "how good" the environment is, nor "how good" the genes are in an abstract sense - it measures how well adapted the genotype is to the environment.

Sleb Lagnej

as far as I know, a genotype is the genetic code of an organism. Supposedly it stays constant, so how do you measure the average reproductive success in a given environment when there's only one of that genotype?

More Anonymous

@ACuriousMind I don't think I can formulate my question :////

ACuriousMind

@SlebLagnej "Often though, genotype is used to refer to a single gene or set of genes, such as the genotype for eye color. "

Sleb Lagnej

8:50 AM

gotchya thx

so fitness can measure let's say the genotype of green eyes in a jungle forest? You get all organisms with green eyes in the jungle forest and record how much reproduce and how much die, and BAM - fitness?

More Anonymous

@ACuriousMind How does one rotate this integral 90 degrees?

Taken from here

quantumfieldtheory.info/Derivation%20of%20the%20Propagator.htm

Sleb Lagnej

im funny

ACuriousMind

@MoreAnonymous Wick rotation is just $t\mapsto \mathrm{i}t$. So in the complex plane where the real axis is $t$, this is a 90° rotation, no?

More Anonymous

I see so your saying integrate substitute $t'=it$ and integrate wont this affect the boundary of the (I think orginally real) integral

ACuriousMind

I'm not sure what you mean

More Anonymous

8:58 AM

@ACuriousMind ?

ACuriousMind

the claim is not that this leaves the value of the integral invariant (in fact, it will not)

Wick rotation is a general and a bit subtle procedure - if you haven't heard of it before, you should read up on it

More Anonymous

@ACuriousMind Any physics stack exchange links?

I'm not the most educated on it

ACuriousMind

we have plenty of questions discussing aspects of it (wick-rotation)

Sleb Lagnej

ACuriousMind could u pls confirm if my understanding is correct in my last question and im gone

ACuriousMind

@SlebLagnej sure

Sleb Lagnej

9:04 AM

thx ^^

More Anonymous

@ACuriousMind Too many in fact :/ Any good graduate level introductory texts instead?

ACuriousMind

it's rarely discussed in detail in intro texts because you only have to care about it when you start thinking about convergence properties or explicit computations in detail

people interested in rigor know the corresponding result - that you can get Lorentzian results by analytically continuing Euclidean ("Wick-rotated") results - as the Osterwalder-Schrader reconstruction theorem

Charlie

10:16 AM

When we say that the time and radial coordinates "switch" beyond the event horizon, we don't mean that the signature of the metric is changing right, this is just something quirky about the coordintes

Presumably just another weird artefact of using the "Schwarzschild coordinates" instead of a "better" coordinate system

ACuriousMind

@Charlie it's about whether the corresponding vector field (i.e. $\partial_t$ resp. $\partial_r$) is spacelike or timelike, it's not a property of the metric

Charlie

oh

bolbteppa

@MoreAnonymous The Feynman prescription in momentum space is 'derived' in Landau in the section on the electron propagator as if you'd never heard of it before, the idea is to go to position space and get an expression which shows what the answer has to be

ACuriousMind

outside of the horizon, $t$ is timelike and $r$ is spacelike - inside the horizon, it's the other way around

Charlie

ok that sounds like something you can't change with coordinates

More Anonymous

10:19 AM

@bolbteppa Which volume and chapter of Landau?

Charlie

for a similar reason to why you can't have a FTL coordinate system

bolbteppa

The section on the electron propagator in the qed one

ACuriousMind

@Charlie well, you could pick coordinates whose associated vector fields don't have this problem :P

More Anonymous

@bolbteppa Cool i'll I've a look

Charlie

Oh, I was under the impression that you can't turn timelike and spacelike vector fields into eachother with a coordinate change, maybe I'm just thinking of Lorentz transformations

Otherwise you could choose coordinates in which a worldline becomes spacelike, but again maybe that only holds for Lorentz coordinate changes

ACuriousMind

10:21 AM

you can't

time-like/space-like is an invariant notion

By "picking coordinates" I didn't mean that the change in coordinates wold magically make $\partial_t$ timelike again, I meant that if you don't use $t$ as your coordinate you have no reason to care about $\partial_t$

The standard Schwarzschild coordinates $t$ and $r$ are adapted to an observer at infinity - they are the time and distance such an observer will measure. Since such observers cannot look beyond the horizon, it is entirely irrelevant from a physical viewpoint that their time and space coordinate "switch roles" - the coordinates don't have any reasonable meaning beyond the horizon anyway

Charlie

ok sure, so no matter how you change coordinates, you're always to get this "flip" between timelike and lightlike for two of your coordinates

ACuriousMind

no

Charlie

D:

or rather, of the vector fields associated with them

ACuriousMind

it's perfectly possible to have coordinates that don't change their nature at the horizon

e.g Kruskal-Szekeres coordinates

Charlie

oh I guess the notion that you can't smoothly change coordinates that turn time/light-like vectors into light/time-like vectors is related to why there has to be a singularity there

ACuriousMind

10:29 AM

there is no singularity at the horizon :P

Charlie

well, coordinate singularity

if there weren't a coordinate singularity there you'd have to be able to smoothly flip timelike and lightlike vector fields

ACuriousMind

yes, but the very nature of a "coordinate singularity" is that it may not be there at all in other coordinates

look at the KS coordinates - they are constructed exactly so that $T$ and $X$ are smooth timelike and spacelike coordinates everywhere

Charlie

presumably any coordinate system that has a similar flip has to have a coordinate singularity there

yeah I think I see what you mean now

Slereah

You can have coordinate singularities in Minkowski space

By doing something dumb like $t \to t^{-1}$

on the appropriate range

which is a weird thing because IIRC Einstein was aware of this back then

and singularities were still a contentious topic

satan 29

10:49 AM

0

A: What is the proof of $C_{V} = \frac{fR}{2}$?

Not really sure if this amounts to a "proof", but, from the equipartition theorem, we know that each degree of freedom has an average KE = $KT/2$ associated with it. So if you have a system of $N$ particles, you have an average total KE of $fNKT/2=fN(R/N_{a})T/2 =fnRT/2$. (Where $n=N/N_{a}$ is th...

is my use of deltas correct in this answer?

recently i have learned that work and heat are inexact differentials, and thus their infinitesimals should be written using $\delta$ instead of $d$....

Sleb Lagnej

11:26 AM

Has anyone read Thinking fast and slow? I'm scared that the book is gonna induce doubt in me, my thoughts, and my beliefs. I don't know if I should read it or not. I don't want to walk around and constantly question everything I think. I don't want to constantly be wary of falling into a bias. I want to be a critical thinker (which is what I hope this book can give to me) but not to the point where I can't trust my beliefs and core values.

Charlie

I don't know if mixing them in the same term is correct @satan29

satan 29

@Charlie yes thats what bothering me too..

Charlie

Surely critical thinking should extend to your core beliefs and values?

ACuriousMind

if we're being rigorous about differentials, then dividing by them (like $\frac{1}{\mathrm{d}T}$) is a no-no

satan 29

@ACuriousMind then how would you describe the 1st law, using derivatives

ACuriousMind

11:36 AM

@satan29 what about $\mathrm{d}U = \delta Q - \delta W$ would require a quotient?

Yashas

11:46 AM

How do I rotate a spinor? The spinor has four components. I know how to derive the rotation operator for a spin 1/2 system but I'm not quite sure how to apply it for a spinor.

Andrew Micallef

[Does this work](https://chat.stackexchange.com/transcript/message/57656548#57656548)
Where $\Psi$ is a wavefunction satsifying schrodingers equation.

$$
\begin{align}
\int\limits_{-\infty}^{+\infty}{\left(\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi\right)}\,\mathrm{d}x
&= \left[\Psi^* \cdot \Psi\right]_{-\infty}^{+\infty} - \int\limits_{-\infty}^{+\infty}
{\Psi\frac{\partial \Psi^*}{\partial x}
}\,\mathrm{d}x\\
&= \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty}
\end{align}
$$

satan 29

@ACuriousMind I wanted to invoke the definition of $C_{v}$ as $$ \dfrac{dQ}{ndT}$$

ACuriousMind

@Yashas how can you know that a spinor has 4 components but not know how the Lorentz group acts on it? how did you arrive at the four components without considering that action?

@satan29 well, the problem is that while the notation is suggestive, $\frac{\mathrm{d}f}{\mathrm{d}x}$ does not actually mean "the differental of $f$ divided by the differental of $x$"

Andrew Micallef

@SlebLagnej It's a freaking good book

@SlebLagnej You can't escape bias, but your life will get better recognising some of them

ACuriousMind

in the language of differentials, it's more like $\mathrm{d}Q = nC_v\mathrm{d}T + \text{terms not involving } \mathrm{d}T$ but "division" is not an allowed operation

but on the other hand, physics in general and thermodynamics in particular is rarely careful about this :P

satan 29

11:54 AM

@ACuriousMind yes, i learned this the hard way..a derivative is not a fraction was my final takeaway

but then stuff like dividing by differentials and multiplying bu them etc occur all the time, so I figure it might as well work ;P

my main concern was mixing up the d with $\delta$

bolbteppa

@Yashas when you derive the Lorentz transformation of the Dirac equation by analyzing the equation one figures out how to do a Lorentz transformation of a spinor (which includes rotations)

Yashas

@ACuriousMind I haven't reached that level of mathematical sophistication yet. I am doing the rookie way. The question is to calculate the +1/2 helicity eigenspinor of an electron with momentum $\vec{p}' = (p \sin \theta, 0, p \cos \theta)$. I know the eigenspinor for momentum $\vec{p} = (0, 0, p)$ case. The $\vec{p}$' and $\vec{p}$ appear like a rotation along the y-axis. So I am trying to do something like $\psi' = U(R(\hat{y}, \theta))\psi$. 1/2

I am unsure how to compute the rotation operator for this case. $\psi$ has four components and a spin 1/2 system state vectors have two-components. 2/2

satan 29

@ACuriousMind 2 questions

1) again, shouldnt it be $\delta{q}$

2) to get the final result, we would need to divide by $dT$ but yoyu are saying its an invalid operation..

or are we actually integrating through a finite temp difference $\Delta T$, and then dividing?

Yashas

Now I'm not sure if the rotation operators from non-relativistic QM work here. Maybe my approach is completely wrong.

Slereah

@ACuriousMind Isn't it just $Q$ in the language of differentials

with $Q$ some inexact form

bolbteppa

12:04 PM

@Yashas what's wrong with the rotation in section 4.1.1 here

Yashas

@bolbteppa It's going over my head. I am at the level of this textbook (currently doing chapter 5, exercise problem 5.3)

Charlie

Just as a heads up @Yashas it is generally frowned upon to share links to pirated copies of textbooks, assuming that textbook isn't public domain

especially on a site like this where many of the members are authors of textbooks themselves

Yashas

@Charlie Ok. Thank You. I have flagged the comment asking a moderator to remove the link.

Balarka Sen

Wow, I didn't know you could delete messages without leaving a blank message with (deleted) showing up on it on the real-time chat (unlike the transcript where they never show up)

satan 29

deleted

Balarka Sen

12:19 PM

Oh it wasn't deleted, it was just edited.

satan 29

@ACuriousMind also, do you have source/sources on this that I can refer? whenever I wanted to read up on differentials, I end up seeing explanations with differential forms, and stuff which really goes over my head

ACuriousMind

@Yashas the 4-component spinor is a representation of the Lorentz group, not of the ordinary rotation group

of course the rotation group is a subset of the Lorentz group, but I'm not sure how you expect to "find" the representation here, i.e. I don't know what data about the "spinor" you actually have to derive it

@satan29 well, unfortunately the rigorous approach is via differential forms, no way around them :P

satan 29

F

what about my 2 other points regarding your statement?

ACuriousMind

12:34 PM

the $\delta$/$\mathrm{d}$ distinction isn't really universal, but yes, if you denote inexact forms by $\delta$ then some of my $\mathrm{d}$s should be $\delta$s

There is no need to "divide by $\mathrm{d}T$". When I have a function $f(x,y)$ then $f_x$ and $f_y$ are well-defined if I just write $\mathrm{d}f = f_x\mathrm{d}x + f_y\mathrm{d}y$, and you would compute them as $f_x = \frac{\partial f}{\partial x}$ and similarly for $y$. Even if $\delta f$ is not the derivative of anything, this would still be a valid definition, you just can't compute them by differentiating some $f$

but if you're not comfortable with forms this is probably not very useful to you

satan 29

$$nC_{v}dT + o(dT^2)= nfR/2 dT$+ o(dT^2)$

how do we deduce $$nC_{v}= nfR/2$$

from

$$nC_{v}dT + o(dT^2) = nfR/2 dT + o(dT^2) $$

(this is what I think your statement translates to

do I just accept it at this stage that the rigorous reasoning comes from differential forms and just move on...

ACuriousMind

12:58 PM

@satan29 it's like in linear algebra - if you have a basis $e_i$ then $\sum_i a_i e_i = \sum_i b_i e_i$ implies $a_i = b_i$, even though there is no notion of dividing by $e_i$

satan 29

oh hmmmm

Rajorshi Koyal

I need one help about an examination form fill up

opportunities.rbi.org.in/Scripts/bs_viewcontent.aspx?Id=3976

Here I need to get the link where I would apply as a Group B..I cannot find one..Could someone be of any help.

This is about an employment notification so an earnest request.

Charlie

1:12 PM

That link does not work for me

Slereah

1:34 PM

Sometimes I try to read the Real Math of thermodynamics

but then I see shit like this

Why is there an isocahedron in my ideal gas

bolbteppa

Obviously you have to throw a $G_2$ into the ideal gas every now and then

ACuriousMind

because the icosahedron is the ideal shape? :P

bolbteppa

What's it from

Charlie

2:07 PM

I haven't done thermodynamics in a long time and definitely before I did differential forms with any care, is "exact" in thermodynamics used in the same context as an exact differential form? I.e. those satisfying $\mathrm d\omega=0$?

and, now that I think about it, the notation for distinguishing exact and inexact differentials $\delta \omega$ and $\mathrm d\omega$ is literally just separating exact and inexact differential forms on whatever manifold we're doing thermodynamics on

whoops what i've written is a closed form

well, can't delete it now, I'll be shamed publicly forever

It is nice to see some value in learning differential forms more carefully, I'd never actually thought about how they're used in thermodynamics, probably because I've mentally blocked that entire subject because let's face it thermodynamics is a sinful topic that doesn't make sense

bolbteppa

You just need basic multivariable calculus differentials, I don't think many books really do any of this with forms 'properly', the biggest concern with being careless about differentials is when dealing with multivariable functions and forgetting something like an implicit dependence of one variable on another etc...

ACuriousMind

2:25 PM

@Charlie Yes. The point of writing something like $\delta W$ is that there's no function $W$ it would be the derivative of, see physics.stackexchange.com/a/516978/50583 and the answer by Joshphysics I link there

note that "exact" and "closed" are synonymous for forms on contractible domains

Sleb Lagnej

2:38 PM

Does Google handle confirmation bias? Like, if I type "Is university good?" and "Is university bad?", does google spit out different results and allow ppl to fool themselves?

Would google bring up arguments that "the person wants to hear"?

ACuriousMind

3:01 PM

that's why I google for keywords and don't ask it questions :P

but no one except Google knows how the search algorithm really selects its results

there's certainly plenty of personalization going on, it always shows you what it thinks users like you "want to see" regardless of how you phrase the question

Yashas

> In particle physics, the Dirac equation is a relativistic wave equation derived by British physicist Paul Dirac in 1928. In its free form, or including electromagnetic interactions, it describes all spin-½ massive particles such as electrons and quarks for which parity is a symmetry.

What does "parity is a symmetry" mean?

ACuriousMind

the same thing that "rotation is a symmetry" means, just for parity

are you confused about the 'symmetry' or about the 'parity' part? :P

Yashas

@ACuriousMind I haven't come across the term "parity"

ACuriousMind

have you tried Wiki?

Yashas

@ACuriousMind Yes, it said flipping a spatial coordinate. But then this sort of feels trivial. Why would the physics change by changing the direction of the axis? But if it's so trivial why would someone explicitly mention it in the Dirac equation wiki.

I think I am not understanding the significance of explicitly mentioning "parity is a symmetry" or I haven't understood what parity is

ACuriousMind

3:09 PM

@Yashas this is so trivial that in fact there are parity-violating physics, see en.wikipedia.org/wiki/Wu_experiment

parity effectively looks at a "mirrored world" (compared to rotation that looks at a "rotated world") - it turns out that it is a meaningful statement to say that most but not all physics are the same in a mirror universe

Yashas

My brain is hurt.

ACuriousMind

maybe it's important to note that you're not just switching the direction of one coordinate axis - you are really reversing the direction of all vectors in that component

it's not a coordinate change, it's a real transformation, just like rotation

(pls don't ask about passive trafos again charlie ;) )

Charlie

3:24 PM

@ACuriousMind Ah, I hadn't thought about it like that, that the function $W$ doesn't exist, that's actually really neat

presumably the manifold on which you do thermodynamics rigorously isn't something nice like $\Bbb R^3$ if it has non-trivial $H_{dR}$

ACuriousMind

why do you think it has non-trivial deRham?

Charlie

oh, just because the closed and exact forms do not coincide

at least at order 0

ACuriousMind

@Charlie hm? The $\delta W$ is neither closed nor exact!

Charlie

oh

ACuriousMind

and the thermodynamic manifold is something that has as its coordinates the state variables like $(N,V,T)$ or whatever is suitable in the current example

Charlie

3:34 PM

ugh now i've confused myself

oh I see what you mean, woops

John Rennie

4:15 PM

One of the great unanswered questions of our time

Sleb Lagnej

5:03 PM

@JohnRennie that post is so dumb it's hard to understand

i wanna go to da muun with balun

Yashas

Why does the spin flip when we reinterpret a particle as its antiparticle?

ACuriousMind

@SlebLagnej please don't call other users' contributions 'dumb' - criticism is fine, insults are not

@Yashas What does "interpret a particle as its antiparticle" mean?

Yashas

@ACuriousMind Feynman-Stueckelberg interpretation. I can see from the equations that negative energy solutions can be interpreted as positive energy solutions moving backward in time. I don't immediately see why spin has to flip.

why are the labels 3, 4 mapped to 2, 1 and not 1, 2? The textbook eventually states that the spins must flip when reinterpreting positron as an electron but I don't see this in the math.

I also find this idea of particles moving backward in time unconvincing. Is this solely a mathematical construct to simply drawing Feynman diagrams or is there something physical about moving backward in time?

Prateek Mourya

5:18 PM

Can anyone please help me how to even begin with this question

Q1

ACuriousMind

@Yashas have you tried evaluating a spin operator on these solutions?

everyone uses slightly different notations and bases for the solutions so without knowing your book we can't really tell you how to see it explicitly in your case

@Yashas all physics is mathematical constructs :P

Yashas

@ACuriousMind Yeah, sorry. I just realized it. I thought it was not trivial to compute v but it is actually equal to some u.

ACuriousMind

if you don't want to think about particles going backward in time, you don't have to, but CPT symmetry means there is actually content to this way of describing anti-particles - all physics is invariant under turning everything into anti-particles, then reversing parity and time

Sleb Lagnej

7:33 PM

@ACuriousMind understood

Yashas

Why is an extra minus sign required to account for the interchange of two identical fermions?

ACuriousMind

@Yashas because the fermionic c/a operators anti-commute

Yashas

@ACuriousMind c/a operators?

ACuriousMind

creation/annihilation

Sleb Lagnej

7:52 PM

Where does the Bayesianist get that prior information on which to base a prediction that doesn't require repetition? I'm trying to understand the difference between bayesian and frequentist

When I say that there's 50% chance that the coin will be heads, I mean that if you flip the coin infinitely many times, the amount of heads will approach 50%

is that bayesian?

ACuriousMind

@SlebLagnej no, that's precisely how frequentists think about probability

Sleb Lagnej

And how do the bayesians think?

I have no idea how else probability can be presented other than frequent repetition revealing some ratio

ACuriousMind

a bayesian would rather think about this as a personal belief - you believe that it is equally likely that the coin will come up heads as it is that it will come up tails, or in other words, you are no more confident that it will show heads than you are that it will show tails

see stats.stackexchange.com/q/31867 for more elaborate discussion

Sleb Lagnej

no matter how much I read, it's still not clear

well, it gets clearer

I need a simple explanation. Bayesians believe, frequentists repeat. Does that work?

that's maybe a bit too simple lol

it's late and my brain is tired, maybe it'll click tomorrow

Charlie

9:58 PM

What is the actual relationship between the Witt algebra and 2D spacetime? Is the Witt algebra the conformal algebra of $(1+1D)$ Minkowski space?

Ah it's the Virasoro algebra that's the 2D conformal algebra

ACuriousMind

@Charlie No, the algebra of infinitesimal conformal transformations in 2D is two copies of the Witt algebra

Charlie

oh I thought that was the virasoro algebra lol

ACuriousMind

the Virasoro algebra is a central extension of the Witt algebra that becomes only relevant in the passage to quantum physics

Charlie

actually I guess I hadn't really thought about it, the central extension isn't just that

ok fair

actually while we're on the topic, when we say "two commuting copies of the Witt algebra", we mean the Witt algebra Cartesian product'ed with itself, right? Surely if they commute they're just combined in the trivial way

ACuriousMind

yes, it's just the direct sum

Charlie

10:10 PM

ok nice

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