The h Bar

naturallyInconsistent

01:12

M I A O ~

user430580

02:00

🐱❤️

naturallyInconsistent

😽

Nairit Sahoo

03:15

Can somebody tell me whether in $Z[J=0]$ whether there is a contribution from disconnected diagrams?

naturallyInconsistent

H O N K ~

Silly Goose

h o n k

is rest mass just defined as the lorentz scalar $p^\mu p_\mu$?

I was wondering how does one derive that $p^\mu p_\mu = m^2$

Nairit Sahoo

@SillyGoose I thought starting with $p^\mu=m \frac{dx^\mu}{d \tau}=m u^\mu$ is a natural starting point (we can also derive it from the relativistic free particle Lagrangian and the expression for momenta conjugate to 4-position) and then just use $u_\mu u^\mu=-1$

Silly Goose

hm but this is sort of assuming that $m$ is lorentz invariant from the outset, right?

Nairit Sahoo

Yes.

Silly Goose

03:29

I guess I was wondering if it can be derived without that assumption

Nairit Sahoo

I see. Can you think of any more fundamental definition of inertial mass?

@SillyGoose There is ofcourse Einstein's original collision based derivations of that relation but much of what mass is kinda borrowed from non-relativistic physics. And another way would be correspondence principle: one can show that $m$ is the mass at rest by going to the rest frame and recovering non-relativistic physics.

naturallyInconsistent

03:48

@SillyGoose It is the magnitude of a conserved 4-vectorial quantity, so this is naturally chosen to be a definition.

Part of the identity of a particle

ACuriousMind

07:31

@SillyGoose Depends on your context - what do you want to start from to derive this?

naturallyInconsistent

07:58

@ACuriousMind amplituhedron?

qwerty

sniggers

ACuriousMind

@naturallyInconsistent gesundheit

naturallyInconsistent

@ACuriousMind thanks, miao miao needed that

the weather has been erratic. miao miao been sneezing at work all day

Dannyu NDos

Lemme check this: What does justify the notion of stress-energy tensor?

Why not spin-1½?

naturallyInconsistent

These are completely different questions

@DannyuNDos It had been entertained

@DannyuNDos fluid mechanics? Or GR. Tends to be easiest to motivate from them.

Dannyu NDos

08:09

GR

Gravitons having themselves as antiparticles baffle me, so

naturallyInconsistent

Pick up a copy of Schutz. Once you see a proper coverage of dust solution and so on, slowly deriving the stress energy tensor, you will see that everything makes sense.

Dannyu NDos

Uhh, is that for grad students?

naturallyInconsistent

@DannyuNDos photons and phonons are also self-anti-particles

Dannyu NDos

I know;

naturallyInconsistent

@DannyuNDos Schutz is very readable even for undergrads. You just need to have a good grasp of multivariate calculus. Preferably some tensor index manipulations at the simple level of Cartesian tensors in RHB, say.

qwerty

08:11

ACM, is part of the reason you were disenamoured with research because of how new research is kind of banal (kind of by necessity I guess when looking at ratios) when compared to centuries of progress when you learn established knowledge?

Dannyu NDos

@naturallyInconsistent Thanks.

naturallyInconsistent

@qwerty didn't ACM say before that he simply didn't like research, and particularly did not like being poor? His day job pays well?

Dannyu NDos

I guess, I'll wait till we uncover dark matter and dark energy.

qwerty

I was not present for comments on salaries.

or if I was I don't recall

naturallyInconsistent

@qwerty It was miao miao that recalled it wrong. ACM explicitly said that he did not leave for money.

Feb 22, 2020 at 9:34, by ACuriousMind

@Knight A bunch of reasons, but money was not the driving concern.

Aug 9, 2021 at 23:42, by ACuriousMind

ah, the unreasonable expectation that one devotes one's entire life to physics- one of the reasons I left academia

Sep 30 at 22:42, by ACuriousMind

well...it wasn't all just learning but if there had been a straightforward way to just stay learning and teaching instead of doing research I probably would've stayed in academia

Enough ACM lore for now. It is so easy to just search for more, though. The list is longgggg

qwerty

08:22

mhmm, yeah.

Ryder Rude

do u believe in idealism or materialism and why

Tobias Fünke

09:05

Hello everyone

qwerty

what's up?

naturallyInconsistent

M I A O ~

Ryder Rude

09:34

hello

Slereah

hey

qwerty

hey slereah rhymes

Slereah

09:55

It does not

qwerty

oops. please enlighten me to french phonetics >.<

Slereah

Oh French people can't pronounce it properly either :p

The pronounciation is /løʁea/

qwerty

I can't read IPA :(

Slereah

ipa-reader.xyz

not the best accent but it's close

oh you can change the accent

just put a French one

Céline [french] pronounces it fine

qwerty

Oh! that IPA reader is SO COOL.

I've tried to learn IPA before but found it quite hard.

naturallyInconsistent

10:08

oooo~

qwerty

/ði/ /ˈhæməɫˌtoʊnjən/ + set accent to Russian = the voice that haunts my dreams

xD. I wonder how useful this would be for learning accents

naturallyInconsistent

@qwerty very very very scaryyyyy

qwerty

@naturallyInconsistent ˈtriviəl

Slereah

10:28

I have experienced such things, my thesis advisor was a Russian dude :p

naturallyInconsistent

Miao miao had a nice Russian prof tho

qwerty

I think the current generation doesn't have as many soviet profs. they're missing out

Slereah

Well the soviet ones are getting a bit old

qwerty

yis

naturallyInconsistent

The next generations are the ones that will really find it rare, given what had just happened in Ukraine

qwerty

10:37

time to bring it back?

Slereah

Also I don't think post-collapse Russia invested as much into physics as the Soviet Union

qwerty

student: "Why did you study physics?" Russian prof, with beer in hand: "In Soviet Russia, you do physics or you do bomb."

Ryder Rude

11:05

do statisticians use the terms Bayesian and Frequentism differently from how philosophers use them?

in the philosophical usage, the terms are almost the same. The distinction is purely philosophical, i think

but it seems statisticians are using these terms differently

qwerty

@RyderRude I doubt it. SEP has an entry on it.

Ryder Rude

@qwerty what is SEP

329

Q: What's the difference between a confidence interval and a credible interval?

Joris and Srikant's exchange here got me wondering (again) if my internal explanations for the difference between confidence intervals and credible intervals were the correct ones. How you would explain the difference?

I found this post

qwerty

Stanford Encyclopaedia of Philosophy

Ryder Rude

@qwerty i think I'm familiar with the philosophical usage. it is only matter of interpretation rather than measurable consequences

but i seems the statistician usage has measurable consequences

qwerty

I highly doubt there's a difference.

Ryder Rude

11:09

in the above post, statisticians identify confidence intervals with frequentism while credible intervals with Bayesianism @qwerty

and confidence intervals and credible are measurably different things

i think statisticians use these terms to mean computational techniques rather than philosophies

e.g. whenever their technique involves Bayes' theorem, they say they are Bayesian

but in philosophical usage, Bayes' theorem can be interpreted under both frequentism and Bayesianism

the credible intervals computation involves Bayes' theorem. so they call it Bayesianism

and they call confidence intervals 'frequentism'

Frequentism and Bayesianism: What's the Big Deal? | SciPy 2014 | Jake VanderPlas

►

qwerty

...Bayes's theorem is just how you invert conditional probabilities. it's not synonymous with Bayesian inference.

Ryder Rude

this is another statistician talk in which he talks about measurable differences between Bayesianism and Frequentism

qwerty

there's no difference in usage. SEP is pretty good and will explain it.

Ryder Rude

@qwerty yes. I think statisticians are using these terms incorrectly

@qwerty statisticians claim difference in usage!

qwerty

who are these statisticians?

Ryder Rude

11:16

and they simultaneously say they r doing philosophy. i think they r just using incorrect terms

@qwerty look at the statistician in the video

i also checked out stats stackexchange. They have similar ideas

qwerty

sorry, I've got other things on right now and can't play videos.

Ryder Rude

ok then look at the threads. The previous one and also this :

329

Q: What's the difference between a confidence interval and a credible interval?

Joris and Srikant's exchange here got me wondering (again) if my internal explanations for the difference between confidence intervals and credible intervals were the correct ones. How you would explain the difference?

bayesian confidence-interval frequentist credible-interval fiducial

several answers, including the top one, claim that frequentism cannot update probabilities based on new information

i think they conflate Bayesianism with Bayes' theorem

Ryder Rude

11:41

ok so statisticians have some data D. And they want to estimate what model parameter $\theta$ could produce that data. Then the people who prefer computing $P(D|\theta)$ are labelled 'frequentists' and the people who prefer computing $P(\theta | D)$ are labelled Bayesians

this has nothing to do with Bayesian and Frequentist interpretation of probabilities in philosophy. They r conflating philosophies with estimation techniques

NOTE Book

hi

in such problems(regular polygons) we say -Q=+Q-2Q which is always true

but when can we remove the three charges and assume system to be only -2Q applying force?

I tried to find and i think it is only valid when we are talking about those points where coulomb force is zero or is there something else to it?

fqq

12:00

@RyderRude "I watched a couple of videos on YouTube and I now know that all statisticians are wrong about these statistics words"

3

qwerty

@fqq yes

Ryder Rude

@fqq i went through threads on Statisticis SE! I'm not going by one person's YouTube talk

Also, these are not statistics terms. They r philosophical interpretations of probabilities. they are just being misused in statistics

46

A: Bayesian vs frequentist Interpretations of Probability

In the frequentist approach, it is asserted that the only sense in which probabilities have meaning is as the limiting value of the number of successes in a sequence of trials, i.e. as $$p = \lim_{n\to\infty} \frac{k}{n}$$ where $k$ is the number of successes and $n$ is the number of trials. In...

here is another answer conflating the philosophical notions with Bayes' theorem

parts of the answer define the philosophical usage of the terms and other parts confuse these with the usage of Bayes' theorem

it ends with : "In the Bayesian approach one no longer talks of confidence intervals, but instead of credible intervals"

one comment calls this out : "any frequentist method that produces useful and self-consistent results can generally be given a Bayesian interpretation, and vice versa"

fqq

12:26

Lol

Mr. Feynman

13:21

Save us ACMan (CSM reference)

think_meaning_buildß

Tobias Fünke

13:40

@Mr.Feynman save us from what exactly? :d

A general question: Do you use overleaf? Do you know any alternative?

Since their latest update it is a mess to work with. The compilation time increased drastically, and only one other collaborator can edit the file (both in the standard version).

Mr. Feynman

@TobiasFünke I plead the Fitfth (not even American :P)

Relativisticcucumber

@TobiasFünke I had this issue but it did not persist when I added my university email which gives more compilation time

naturallyInconsistent

if you just wanna share a short snippet, mathb.in?

Mr. Feynman

I haven't used OL in a long time now

Relativisticcucumber

@Mr.Feynman fake physicist

Mr. Feynman

13:49

True, I'm not worthy of that label anymore :P

That being said, I've only used TexStudio recently

Relativisticcucumber

im having yet another issue that I believe is stemming from this discontinuity that the coulomb potential experiences. im trying to compute the Fourier transform of the coulomb potential. I used two methods 1) direct integration 2) poisson equation and Fourier transform of the Dirac delta function. the second method yields the correct result. however, I found a computation where someone does method 1 with the Yukawa potential and it worked. [...]

[...] when I set the parameter in the Yukawa potential to zero, I thought I should recover the proper coulomb computation, but alas that goes awry. what is wrong with the direct integration approach here?

Mr. Feynman

@Relativisticcucumber What is the "direct" approach? The Couloumb potential can't be Fourier transformed without a regularization such as the Yukawa regularization

Relativisticcucumber

@Mr.Feynman what do u mean by regularization?

Mr. Feynman

@Relativisticcucumber If you try to compute the fourier transform of $1/r$ using the definition, you'll see that this is a divergent integral

Regularization means making it finite

Relativisticcucumber

@Mr.Feynman yes I did indeed get that

so my approach was similar to this computation cupcakephysics.com/electromagnetism/math%20methods/2014/10/04/…

but I dont see any sort of "regularization" here

Mr. Feynman

14:01

How should I say? It's not that the Yukawa way is an alternate bridge you cross: you are building the bridge :P

naturallyInconsistent

That introduction of $\lambda$ there makes the integral not explode to infinity and is an example of regularisation

Relativisticcucumber

@Mr.Feynman I did not follow this statement XD

Mr. Feynman

@Relativisticcucumber I was trying to give you a metaphor to explain it better: it's not like the Yukawa trick is a slick way to compute a difficult integrals. It's a way to give meaning to an otherwise meaningless integral

Relativisticcucumber

@naturallyInconsistent oh okay I guess that would make sense. so I cannot say coulomb = doing this computation and setting lambda = 0?

naturallyInconsistent

but there is actually a trick; now that you know the solution of this Fourier transform, you can work backwards and show that assuming the Fourier transform form, you will get the correct result, and so the Fourier transform is what you assumed it to be.

@Relativisticcucumber I think everybody except pedants would accept this reasoning

Mr. Feynman

14:10

Now I'm wondering: is the regularization independence a property that comes from the theory of distributions? I mean, you could in principle use a different regularization and you can't be (in principle) sure it will yield the same thing

naturallyInconsistent

Well, I don't think in this case there is any other possibility, because the opposite direction, as miao miao already described, does not require regularisation

Slereah

messing around with copilot

If you're lazy you can mostly zone out and just ask things until it works

Good for me, lazily hacking together a script, but I am imagining another lazy developer making an important software out of it

Can't wait for it to happen

Tobias Fünke

14:37

@Relativisticcucumber yeah, I am already using my university email :/

@Mr.Feynman Good question, actually. I don't know if this was discussed already on the main site (there is a famous thread about the Coulomb potential Fourier transform) though.

related thread:

64

Q: Fourier transform of the Coulomb potential

When trying to find the Fourier transform of the Coulomb potential $$V(\mathbf{r})=-\frac{e^2}{r}$$ one is faced with the problem that the resulting integral is divergent. Usually, it is then argued to introduce a screening factor $e^{-\mu r}$ and take the limit $\lim_{\mu \to 0}$ at the end of...

mathematical-physics fourier-transform coulombs-law regularization

@Mr.Feynman but there is no option for collaboration, no?

Mr. Feynman

@TobiasFünke Not that I know. It's local.

Unless idk, you share a folder with the project?

@TobiasFünke Okay, the part about continuity of this answer should address my question.

Tobias Fünke

14:53

@Mr.Feynman yep, I think so

@Mr.Feynman nah, too complicated I guess

Mr. Feynman

At this point I'm not sure I should consider it a regularization after all

Tobias Fünke

I have the feeling that buying the premium version is the only sensible solution here...

Mr. Feynman

💀

Nairit Sahoo

15:35

This is so confusing: books are saying that connected correlator formula is $$\langle \phi(x_1) \phi(x_2) ... \phi(x_n) \rangle_C = \frac{\delta W}{\delta J(x_1)} \frac{\delta W}{\delta J(x_2)}... \frac{\delta W}{\delta J(x_n)} $$ evaluated at $J=0$ but for $n=0$ point function I am getting $\langle 1 \rangle_C=W=i \log(Z)$ which includes disconnected pieces. For $n>0$ I get only connected diagrams. Is this formula valid only for $n>0$ for some reason?

Mr. Feynman

15:55

Why are you applying the formula for $n=0$ in the first place?

Tobias Fünke

Indeed, the indices on the LHS start with 1, not 0, so I'd interpret this as $n\geq 1$... but that's just my guess

ACuriousMind

16:17

@TobiasFünke use whatever local TeX editor you like and share the file via git (preferred) or any cloud storage solution like Dropbox, Google Drive, etc or even pass the file via email? I don't think truly simultaneous editing is all that necessary to collaborate.

Mr. Feynman

Apparently he doesn't like that idea :P

ACuriousMind

@Slereah it's already happening and being interpreted as improvements in productivity :P

Mr. Feynman

Also, I sometimes edit a file in the middle of the night - waking up from bed to do so - because I remember that I formatted something wrong. Can you imagine having to upload it every time? What if your collaborators do that too?

ACuriousMind

@Mr.Feynman git push is a one-line command, if you can't manage that what are you doing conducting complicated science :P

Mr. Feynman

@NairitSahoo Anyways, if your notation is consistent with (most) literature and $W$ is the generating functional of connected GF, then it should not involve disconnected diagrams

Tobias Fünke

16:20

@ACuriousMind well, it is of course an option, but it certainly has some disadvantages...I have to think about it...

Mr. Feynman

@ACuriousMind I'm not as tech as you younglings

Tobias Fünke

actually, Overleaf is opensource

ACuriousMind

@Mr.Feynman pfft

Mr. Feynman

More specifically, your $n=0$ term should be the sum of all connected vacuum diagrams as $Z[0]=\exp{iW[0]}$ is the sum of all vacuum diagrams and exponenentiation of diagrams is there precisely to remove disconnected business @NairitSahoo

Slereah

@ACuriousMind I mean it's not bad, for me, jury rigging a web crawler to get data from internet archive

But if someone is making some airplane software like that

I would worry a bit more

Decided to get off my ass and finally do it because i need to save some stuff in case the archive goes down :p

ACuriousMind

16:35

@Slereah I agree, and I'm saying you should probably be very worried :P

Slereah

I did almost all of it without reading the results

someone somewhere is doing that with more important code

Nairit Sahoo

@Mr.Feynman But usually in the denominator there are disconnected pieces, right?

I mean in $\int \mathcal{D} \phi e^{iS}$ which forms the denominator of (not necessarily connected) correlators

Slereah

On the plus side

Currently using that code to back up all copies of Scientific American from Internet Archive

I should make a list of journals to back up

Nairit Sahoo

17:12

@Mr.Feynman Do you mean that $Z[J=0]$ has all diagrams but $log Z[J=0]$ does not have the disconnected pieces?

Mr. Feynman

17:28

$W$ is the exponential of the denominator. As far as I know it only includes connected diagrams

Nairit Sahoo

@Mr.Feynman Yes. But what about $Z[J=0]$? It has all the diagrams, right?

physics.umd.edu/courses/Phys851/Luty/notes/diagrams.pdf pg 16 equation 3.24 says that $Z[J]$ has all diagrams without vacuum subgraphs. Why can't there be vacuum subgraphs in $Z[J]$?

Mr. Feynman

@NairitSahoo That's a separate question.

Nairit Sahoo

@Mr.Feynman yes

Mr. Feynman

17:44

You may want to read further. Eq. $(3.30)$ should help you. The idea is the same as always: the denominator cancels vacuum graphs

Nairit Sahoo

@Mr.Feynman Equation 3.30 is one of the things which is confusing me

Actually not so

3.18 bothers me. I can clearly see vacuum subgraphs there, That is supposed to be $Z[J=0]$. So if $Z[J]$ does not have vacuum subgraphs, how can $Z[J=0]$ have?

Mr. Feynman

I think your notes just have a convoluted path. In principle there should be vacuum graph but since they are in $Z[0]$, which appears as a multiplication factor, you can ignore them because normalization kills them off

You should see $(3.23)-(3.24)$ as the conclusion of what's below

ACuriousMind

@NairitSahoo These notes are very confused, they mean $Z[J]/Z[0]$ has no vacuum subgraphs. The section you're complaining about switches at random between using $Z[J]$ for $Z[J]$ or for $Z[J]/Z[0]$.

Nairit Sahoo

Ohhh mannn... So please tell me if this understanding is correct: $Z[J]$ contains everything. Divide by $Z[0]$ to get rid of vacuum bubbles. Take log(Z) to get rid of disconnected pieces.

Mr. Feynman

Which is what I've been saying in a probably less efficient way

ACuriousMind

17:54

@NairitSahoo take log of the ratio $Z[J]/Z[0]$, but yes

Nairit Sahoo

@Mr.Feynman no I think you were being perfect. I was not able to match with the notes. But I think the notation in the notes is a bit ambiguous

@ACuriousMind Hmm. whooo

Last request. There is a sentence on pg 17 after equation (3.31) which says: "The exponentiation of the disconnected diagrams means that these diagrams are
trivially determined from the connected graphs". How to see this?

ACuriousMind

What's unclear about this statement?

If I have $A = \mathrm{e}^B$, then $A$ is determined by $B$.

Nairit Sahoo

@ACuriousMind Here $B$ is just the disconnected diagram which is a product of the connected diagrams?

ACuriousMind

what?

Nairit Sahoo

@ACuriousMind They are saying exponentiation of disconnected diagrams, not connected ones

I know that exponentiation of connected ones gives connected+disconnected

But what about exponentiation of disconnected?

ACuriousMind

18:01

are we really doing textual exegesis on a text that's already shown it's careless about its variables :P

The mean that you can, by the formula you've shown, determine all disconnected diagrams from knowing the connected diagrams

they mean "exponentiation of the connected diagrams"

I suggest finding a better source :P

Nairit Sahoo

Yeah sure. Thanks :)

More Anonymous

18:27

I dont see how Maxwell used central limit theorem to derive the boltzmann distribution

And nor precisely where it fails in SR

Any help?

Tobias Fünke

19:32

@Relativisticcucumber BTW: which book do you follow for your condensed matter course(s)? just out of interest... you posted some screenshots discussing the Hartree equation a few days ago

Relativisticcucumber

20:08

@TobiasFünke the course uses Ashcroft and mermin, which I do read/follow, but I also sometimes supplement with random online notes.

ok so in Hartree and Hartree Fock, these equations can be derived via the 'variational principle', which I had the pleasure of engaging with. so the starting point for this derivation was to 'stationize' (I believe that's the word) the expectation value of the hamiltonian. can I say that we do this because we want to minimize the energy?

ACuriousMind

I don't think stationize is an actual word :P

the usual phrase is extremize

Relativisticcucumber

@TobiasFünke I am very fond of this book actually. I think it's rather self contained, which is one of the major factors I appreciate in a textbook

ACuriousMind

of which minimizing is a subset, so yes, you're correct

Relativisticcucumber

@ACuriousMind ok makes sense, so its just like normal calculus

ACuriousMind

yes, calculus of variations is "just" calculus in infinitely many dimensions

Relativisticcucumber

20:14

ok small yippee. I have found a way that I can do calculus of variations. I can find a small subset of rules that I know to be true then restrict myself to applying those rules. then, I know I cannot be wrong and I avoid needing a large understanding of the subject.

hopefully this small set of rules can get me far enough.

younger me would be so disappointed in how 'physicist' ive become

I swore id never resort to this but here I am

ACuriousMind

you either become a mathematician young or you live long enough to see yourself become a physicist

7

Relativisticcucumber

true :,)

qwerty

20:35

@ACuriousMind boo. "make stationary"

ACuriousMind

@qwerty no, that's what a paper mill does

Silly Goose

can i work out lorentz factors by dimensional analysis?

ACuriousMind

if you already know the correct answer you can work everything out by dimensional analysis :P

Silly Goose

purcell is a bit dreadful and i am trying to find a nicer way of presenting how a field transforms without going full on into index notation

@ACuriousMind Lol

but even this seems a bit iffy as one has to assume the mass unit is lorentz invariant

this seems possibly ambiguous

Mr. Feynman

@ACuriousMind now it is and you're not gonna change that

qwerty

20:39

@ACuriousMind that's stationEry

Mr. Feynman

@ACuriousMind then what is the path that I've taken, master? Was there a secret third path?

Tobias Fünke

@Relativisticcucumber I see.

@Relativisticcucumber Yes, exactly. You want to find the set of one-particle states/orbitals which, when formed into a product state (or Slater determinant in the HF case), minimizes the energy.

Mr. Feynman

@ACuriousMind what about optimize? 🧠

@qwerty You haven't roasted me today yet. Just sayin'...

ACuriousMind

@SillyGoose if you don't know that mass is Lorentz invariant what are you doing thinking about fields, seems like you have elementary things to settle :P

Tobias Fünke

@ACuriousMind Hahahah. So true

Silly Goose

20:42

but i mean is it not unclear what this mass unit refers to?

it could be $\gamma m_0$

ACuriousMind

@qwerty that's the joke but I can't come up with another pun on stationery to continue the "no, that's a" template :(

qwerty

oh well we'll have to pencil in a joke to fill in later :p

ACuriousMind

=)

Tobias Fünke

@Relativisticcucumber yeah, A&M is fine!

Silly Goose

i mean the gamma factor appears in relativistic newton's second, no?

ACuriousMind

20:45

@SillyGoose again, if your mass is not Lorentz invariant you have bigger problems; dimensional analysis is a bit questionable at the best of times as an actual derivation but I don't understand this particular objection

@SillyGoose I think that's less an argument for the "mass unit" being not Lorentz invariant and more an argument that dimensional analysis is not a very good heuristic here :P

qwerty

@Mr.Feynman you're not coffee beans. I'll roast you when I like :p

speaking of which I am late for the daily grind. I wish I could be made stationary

Mr. Feynman

21:10

@qwerty do you roast coffee when you don't like it?

qwerty

tbh I have never roasted coffee beans. I'm more often a tea drinker

Mr. Feynman

Your roasting counter is higher for me than for damn coffee beans

qwerty

I recently learned that a subset of Germans, the east Friesians have one of the largest per capita tea consumption rates globally

ACuriousMind

@qwerty careful, whether or not Friesians "are" Germans is a rather fraught question - there are at least some Friesians who conceive of Friesians as a separate ethnicity

Tobias Fünke

lol

qwerty

21:22

@ACuriousMind aaahhh. noted!

apologies to any Friesians on chat

ACuriousMind

and there's also treaties that establish them as similar to an ethnic minority (although the term "minority" is itself contested in this context :P)

Tobias Fünke

@ACuriousMind are you, by chance, a friesenjung? :)

ACuriousMind

nope

Tobias Fünke

@ACuriousMind I've never heard of that. interesting

ACuriousMind

without getting into the strange subdivisions of local German cultures, the closest description of my origin is probably "Rhenish" or "Rheinlandish"

Slereah

21:35

Them's all krauts

ACuriousMind

says the frog eater

Tobias Fünke

@ACuriousMind so shouldn't you celebrate "Fastnacht" now? :d

@Slereah lol

ACuriousMind

@TobiasFünke I was never a big fan of that, but anyway I haven't lived there in over a decade. Also, it's Karneval in the Rheinland, Fastnacht is the more southern term (and here there's not a big deal around the 11th 11th :P)

Mr. Feynman

@ACuriousMind wie bitte?

Tobias Fünke

@ACuriousMind aha. good to know :D

Mr. Feynman

21:41

I think that also in northern Italy there are some frog-based dishes

ACuriousMind

@Mr.Feynman what 'wie bitte'? I countered one culinary stereotype with another!

qwerty

I just googled fastnacht, isn't it around Easter?

Mr. Feynman

Nothing. I wanted to learn all my German in one message and it was the best occasion I've had recently

qwerty

@Mr.Feynman easier to smuggle in the French through Switzerland?

Mr. Feynman

I don't know. Northern Italy ain't my business :P

ACuriousMind

21:44

@qwerty Yes - it's the celebration of the beginning of the pre-Easter fast originally, but over the years the "official" start of the Karnevalssaison (season of Karneval) has shifted and now the main centers of this kind of celebration (like Düsseldorf or Köln) celebrate its start on the 11th day of the 11th month (I don't know why :P)

qwerty

11/11 is remembrance Day here

one minute of silence at 11am and the playing of the Last Post

ACuriousMind

the main event is still the week around Ash Wednesday, though

qwerty

@ACuriousMind that's really interesting/unusual!

Tobias Fünke

@qwerty where do you live?

qwerty

Australia

Tobias Fünke

21:48

Ah okay

qwerty

we don't do carnival and Mardi Gras to most people would mean the LGBTI Mardi gras parade

I actually didn't realise the Christian origin of Mardi gras (only knowing the LGBTI version) until I briefly moved to Europe as a young adult

Mr. Feynman

I still don't understand half of Italian national holydays

But as long as the day is marked red on the calendar... :)

Tobias Fünke

I have to sleep now. Good night everyone!

ACuriousMind

@qwerty well, such remembrance days are a bit of a historically difficult topic here :P There's a "people's mourning day", but it's rather explicitly vague about who exactly is being mourned and shifts around in date as its fixed to being a certain Sunday. There is another day explicitly for the victims of the nazis but not for soldiers.

qwerty

I can imagine so :) I guess Germany is reputed to be handling such things a lot more tactfully and fairly than Japan, at any rate.

Slereah

22:02

The germans cannot celebrate remembrance day because they lost that war

qwerty

(also "explicitly vague" sounds like a funny oxymoron but makes perfect sense)

@Slereah we don't celebrate it :p

ACuriousMind

@qwerty it's impressive how much meaning there can be in not saying something :P

qwerty

@ACuriousMind accurate! :p chat.stackexchange.com/transcript/71?m=66598660#66598660

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