The h Bar

00:26

@Relativisticcucumber but it follows filling up Hydrogenic atom energy levels quite closely! It isn't complete nonsense

@Relativisticcucumber The lattice needs to be infinitely periodic in all relevant directions to have Bloch's theorem work, in the strict sense. But if you consider just a finite number of unit cells, then the k points become discrete, and their total number of k points is equal to the number of unit cells considered. Now, if your solid is actually not infinite, then, while it is not going to follow Bloch's theorem exactly, but it can be approximated so, and then your finite number of k points is meaningful

naturallyInconsistent

02:04

@imbAF seems like you worked this out with the lorentzian vegetable

Nairit Sahoo

Does $l=0$ guarantee ground state i.e. minimum energy in any spherically symmetric quantum mechanics problem?

naturallyInconsistent

@Relativisticcucumber this comes down to convention in the use of language. We dont use the term effective field theory for anything other than in QFT and string theory, an EFT is the renormalised long wavelength appropriate limiting behaviour of the underlying QFT or string theory. It is saying that some underlying QFT has a certain EFT approximate behaviour if you dont probe the short distance scales.

Mean field means many completely different things. It is usually used as a "effective background mean field" approximation of a many particle system squashed down into a single (or few) (quasi-)particle approximation, that you usually then solve self-consistently. All of the many mean field theories have nothing much in common with an EFT.

@NairitSahoo you should have been able to answer that on your own.

Nairit Sahoo

@naturallyInconsistent Wo wo wo... You expect too much of me :|

naturallyInconsistent

@NairitSahoo it should not be controversial to expect that people who read up on QFT should be able to do basic QM on their own.

Nairit Sahoo

@naturallyInconsistent I just found out that there's a 2002 paper on this. So I don't think it is "basic" QM?

naturallyInconsistent

02:30

@NairitSahoo Why would the existence of papers on a topic imply that it is not basic?

Ryder Rude

06:58

hi

123

08:19

Hello Everyone...

123

08:35

If a person standing still on the side of the road, bus traveling and airplane flying. I have three frame of references at which i can set coordinate system, if i select one frame of reference among these, is there any special name for selected frame of reference among all?

Hi @RyderRude

Mr. Feynman

09:21

What are they doing precisely going from $(10.15)$ to $(10.16)$?

I don't think you can just move $U(\beta, 0)=T\exp(-\int_0^\beta H_Idt)$ inside the time ordering, as it contains time values in $[0,\beta]$

I would get the same result if I considered the case $\tau>\tau'$, so that $T_\tau A(\tau)B(\tau')=A(\tau)B(\tau')$. Then I could combine all the operators to get $$U(\beta, \tau)\hat{A}(\tau)U(\tau,\tau')\hat{B}(\tau')U(\tau',0)$$
Observe that this is time ordered ($\beta>\tau>\tau'$ assumption) and put the time ordering back

Is there a shortcut without having to do this, though? As I said, I don't think putting $U(\beta,0)$ inside the time orderering brute force is sensible

Ryder Rude

@123 it is not required to give the selected frame a special name

123

@RyderRude How to handle the situation.

How to say about selected frame of reference?

Ryder Rude

09:46

@123 books will just writec"from the bus frame of reference" to mean that they have selected it

123

Aaaah... Okay Thanks

Ryder Rude

great:)

123

@RyderRude If i select one frame of reference among all inertial reference frame. Is it called stationary frame?

Madder

10:04

i am confused on why there is 7 crystal structures.
In my head, i made the connection, that the crystale structures are the ways you can fill space without overlap and gaps.
But mathematically, there are more ways (shapes) to do that?
So my definition ofa crystal structure is not correct.
but i am not finding a clear definition. They are vague.

naturallyInconsistent

@Madder it is not just overlap and gaps. It is also strict translational symmetry

@Mr.Feynman What was (10.10)?

Madder

Ok. So is the following statement correct.
There are only 7 shapes, which can fill by translation without overlap and gaps the three dimensional space. We call these shapes crystal structures?

Mr. Feynman

@naturallyInconsistent $$U(\tau,\tau'')U(\tau''\tau')=U(\tau,\tau')$$

This is Bruus Flensberg, incidentally

Ryder Rude

@123 i think it can be called that but I'm not sure it its standard terminology

naturallyInconsistent

@Mr.Feynman oh, nice book

@Madder Yes, but I think the name in English is lattice systems

Madder

10:11

@naturallyInconsistent Alright that is very satisfying. Chat Gpt tells me its a false statement. But i will take your word for it.

naturallyInconsistent

@Mr.Feynman You can. Look, (10.15) is saying, evolve from 0 to t', do B, then t' to t, do A, then evolve this whole mess back to zero, then evolve from 0 to beta. The evolving from t to zero and then 0 to beta will obviously cancel away to obtain t to beta. And then note that when you extend the time ordering operator to include the U(beta,0), then you will just be distributing the evolution operators into the midst of those A B operators as (10.15), so (10.16) equals (10.15)

@Madder Ashcroft and Mermin talks about how this definition omits things like screw translations, but that is by definition. It is a great book to read from.

Madder

Alright i will check it out. Thank you.

Mr. Feynman

10:39

I understand what you write. It's a better way to express what I had in mind in the procedure I described. Thank you for that, it helped me (I'm replying only now because I was afk).
That being said, also considering the way you described it, it's actually more convoluted than just "moving the evolution inside". There are several steps that balance it out so that at the end you get $(10.16)$. To make it clearer:
It's not the same as when you have $t_3>\max\{t_1, t_2\}$ and you move an operator $A(t_3)$ inside

Because it's already time ordered

So what I'm saying is that, we can move it inside, that I understand, but it is different from the equation I wrote above, that is a more direct step

Claudio

I might have found the problem, dont know how to solve it yet

but at least I found it

anybody know what all of this means and how can I apply to my case

rip

imbAF

12:10

@naturallyInconsistent No I didn't. Just quit that

Mr. Feynman

HONK

Silly Goose

h o n k

think_meaning_builds

HoNk

qwerty

h0nk!

Mr. Feynman

what the duck have I done

Silly Goose

12:13

I am quite confused about something. Bell tests always seem to be framed in classical probability theory. How is this not a problem when we have reason to believe that quantum mechanics does not follow the rules of classical probability theory?

It seems like Bell tests assume (1) experimental outcomes of the system at hand can be analyzed by classical probability theory, (2) the other usual assumptions of a Bell test.

Okay, then if you reach an inconsistency either $(1)$ or $(2)$ can be inappropriate hypotheses.

Relativisticcucumber

@naturallyInconsistent sorry to beat a dead horse but i just dont see how we get N (the number of points in the lattice) as the number of k's. i do agree that the lattice needs to be infinite to have continuous k's and that a finite number of cells gives a finite number of k's, but how to we get that this is precisely N?

qwerty

@Mr.Feynman duck .. duck .. Goose.

Slereah

@SillyGoose You should beware what is meant by "classical probability" exactly

Quantum probability does obey "classical probability" in some sense but not in others

Silly Goose

Well it seems classical probability is a subset of quantum probability (using the term subset loosely)

Slereah

It doesn't obey it in the sense that you do not have a state of your system for which there is a probability to find it in

But if you add the appropriate amount of context, it's entirely Kolmogorov

But then your space of events isn't "The system is in state X"

Relativisticcucumber

12:33

@naturallyInconsistent ok let me see if i get this correctly. in hydrogen, we have constant energy values in k space, so if i plot this "diagram", i get various lines corresponding to each k value. then, two interesting things happen when we put hydrogen atoms on a lattice. 1) the energies become a function of k. [...]

[...] this is because of being in a periodic potential and having bloch's function solutions. these two mean there are many ks that are allowed, so we now have $\varepsilon (k)$ energy dispersion that are not constant. secondly, these energies are very "collected". what i mean by this is that if we plot $\varepsilon (k)$ then we see for certain regions around a given $\varepsilon$, there will be many curves. [...]

[...] these condense into bands. this is because the atoms in the lattice act as perturbations and cause energy splitting of the individual hydrogen atom energy levels. am i getting this correct?

please be right

naturallyInconsistent

12:44

@Relativisticcucumber That's the part imbaf sent you the condition for k as integer $|v_a|\leq\frac{N_a}2$ for the number of UNIT CELLS $N_a$ in direction $a$

it is NOT number of points in the lattice. It is number of unit cells.

@Relativisticcucumber mostly

Ryder Rude

@Slereah r u familiar with derivations of Born rule in MWI? and r any of those convincing to u?

Slereah

I am not

Slereah

13:19

"So the first step to further determination in the Proceß is this"

Did @ACuriousMind write this part of nlab

123

As earth surface shows good inertial frame when smooth surface. Is it okay in 3D space if 1 or 2 dimensions shows inertial reference frame?

Relativisticcucumber

14:05

@naturallyInconsistent i think im starting to see. is the $k_{\nu}$ in that equation a reciprocal lattice vector? my problem is that i see how in this case $N_{\alpha}$ sets a limit on $k_{\nu}$, but according to that formula, $k_{\nu}$ doesnt increase. it's just the same for a given direction and lattice size.

Mr. Feynman

@naturallyInconsistent sorry for pinging you, if that bothers you I won't do it again. Did you see my reply above?
What I was trying to say is that unlike the example I provided with $A(t_3)$, the $U(\beta,0)$ outside $T$ in $(10.15)$ and the one inside in $(10.16)$ are not the same. The latter comes from $TU(\beta,\tau)U(\tau,\tau')U(\tau',0)$. That's what I meant when I said "we don't just move it inside" (unlike my $A(t_3)$)

naturallyInconsistent

@Relativisticcucumber if you only have one unit cell, then you only have the k point at the origin of the 1BZ. Two unit cells, and you get the k point at the BZB. The 3rd unit cell redistributes the k points so you have origin, and plus minus a third of the BZ, and so on and so forth

@Mr.Feynman you are correct that miao miao missed that reply. Looking now

@Mr.Feynman you are correct; it aint supposed to be trivial. The cancellation step is quite much the voodoo. I think it makes more sense to define the correct result as the whole thing time-ordered properly, and just dont have this weird thing that moves back to zero and then moves forward again from there

Ryder Rude

I was looking at some MWI papers about born rule derivations

There r papers from Carroll, Deutsch, Wallace

the thing i gather is that many worlders have a phd in circular reasoning

they all try to pretend that they can have Born rule in MWI without any world duplication

all of these papers have been called out in academia

Mr. Feynman

14:21

@naturallyInconsistent Thank you very meowch

naturallyInconsistent

meowth

Relativisticcucumber

@naturallyInconsistent i think i see that if we accept that $n$ is bound by $\pm \frac{N}{2}$, then i get it, but in this way of approaching it, i dont see why $n$ should be bound, all we require is that $kL$ is some even integer multiple of $\pi$

bc if i accept the bound, the obviously there are $N$ values of $n$ btwn $-\frac{N}{2}$ and $\frac{N}{2}$

Mr. Feynman

Incidentally, yesterday my cat booped my nose with her paw

naturallyInconsistent

@Relativisticcucumber n that lies outside the bound is just going to be a repeated k point. That is, every k point outside of the 1BZ can be mapped back into the 1BZ by a reciprocal lattice vector (the b vectors) and so do not give you new states.

You can use that to encode the band index value if you want, in which case it is called the extended zone scheme. You are now learning the reduced zone scheme

@Mr.Feynman awww

@Relativisticcucumber hmm, really dislike this argument.

Relativisticcucumber

@naturallyInconsistent but how can we be in the reduced zone scheme if we are using the entire length of the lattice?

i thought because we are using $L$ as the distance, this is the extended scheme .0.

naturallyInconsistent

14:33

@Relativisticcucumber Look, you are really getting confused. The entire length of the lattice just means how many unit cells are being considered. This number determines the gap in the k points within the 1BZ

The relationship between BZ and real-space is reciprocal!

Bigger N, smaller $\Delta k$

as long as your lattice parameter $a$ stays the same, the $b$ vectors defining the BZ's size, stays the same.

If you have very few unit cells, then your BZ will have few k points actually in there.

If you never exit 1BZ, i.e. if you consider every k point outside 1BZ to be unphysical, then you are in the reduced zone scheme. Even if you used $L$ in defining this.

There are other schemes. For example, you can use the periodic zone scheme, in which all bands and all k points are just infinitely repeated. But the most reasonable is the extended zone scheme, because then the "bands" are unfolded out and it looks like the free electron's single parabolic curve / surface, and every k point, even outside 1BZ, has a distinct meaning, and the band index reduces back exactly onto old H atom principle quantum number. Everything makes much more sense.

That is, you should be reading a textbook that covers all these Bloch stuff by starting with the completely free electron model, and the combination of H atom, before doing nearly free electron model. Then you can see how the physics is put together. Then tight binding, to see the opposite extreme, and be happy with the whole theory

Ryder Rude

15:27

mateusaraujo.info/2021/03/12/…

it is an article from Mateus Araujo

he has worked on Born rule derivations in MWI. In the article, he calls out all the circular logic involved in all these papers

the mistake that all these papers make is so trivial that it has got to be intentional

i think many worlders don't want to admit that their theory has world duplication

Everett admitted it. he simply defined a measure on the worlds

but modern many worlders come up with elaborate circular arguments instead of simply assuming Born rule as the measure of world duplication. they don't even talk about world duplication for the most part

one common feature of these papers is "probabilities are all in your head". they will never talk about statistics.

they will talk about "what credence a rational agent should assign"

123

16:11

Hi @RyderRude

Can you pls answer my above question.

The surface of the Earth as smooth plane frictionless considered as good inertial frame. But vertical axis is non-inertial. In 3D space . 1 dimension is non-inertial and 2D inertial reference frame. Is it okay to say?

If this is the case then, It means it is always possible to find 2D inertial frames perpendicular to non-inertial frame. Let say in empty space aircraft accelerating straight line. Or multiple forces act on an object then plane perpendicular to resultant force will show inertial frame.

Like object on the surface of the Earth balanced by normal force, balances non-inertial effect of gravity. Same if i balance the object in the non-inertial direction, then it will become inertial.

Mr. Feynman

16:33

artichoke

Ryder Rude

@123 why do u think so?

123

@RyderRude Because in case of Earth , we say surface of the Earth is approximated as good inertial frame. But vertical axis is non-inertial due to gravity

Ryder Rude

@123 why is gravity a "non inertial effect"

123

@RyderRude Because free fall object accelerate

Ryder Rude

i think non inertial effects on earth are centrifugal and coriolis force, not gravity. maybe u r again conflating Einstein and Newton's definition of non inertial

123

16:38

No. i am not conflating Einstein and Newton's definition. I am just considering acceleration of an object due to gravity

Ryder Rude

@123 free fall objects accelerate because they have a force on them, which is gravity. this doesn't mean earth is non inertial

123

Why gravity do not have non-inertial effect in case of Earth. And it give good inertial frame.

@RyderRude Ooooh... this is how we handle the gravity.

Ryder Rude

yeah..

123

So how can we test for inertial frame? We have a test to place an object

Ryder Rude

in an inertial frame, isolated objects would have no acceleration

123

16:42

How do we know object is isolated?

Ryder Rude

it should be far away from any other particle

or we can also check for pseudo forces. non inertial frames will have pseudo forces

123

Yes i remember this from our last discussion. But in reality this is not possible. How can we make object isolated. I think surface of Earth balance gravity, and smooth plane remove friction.

@RyderRude gravity has same effect as pseudo force.

Ryder Rude

@123 then u can do Galileo's inclined plane experiment to check if earth is (almost) inertial

123

@RyderRude Why you pointed out Galileo's inclined plane experiment for inertial frame test?

Ryder Rude

@123 it's to show that objects retain their velocities forever in the absence of friction

123

16:48

How you rate my questions? technical or newbie 😂

Ryder Rude

we have discussed the same stuff many times..

i think this is not a productive discussion... questions are all over the place

2

123

Yes but things always remains questions. When think further in the topic

But Gravity shows pseudo force. Then how Earth is inertial?

Ryder Rude

gravity is not a pseudo force in Newtonian mech. (People will get annoyed because many people have already explained this to u)

123

Thanks to clear me.

Hope no one annoy.

Ryder Rude

to see the actual reason why earth is non inertial, read about the centrifugal and coriolis force on earth..I think the books would have a good discussion..

@123 it is fine :)

123

16:54

Yes i have already learned a lot about centrifugal and coriolis in details.

I was just confused about gravity. and free fall.

Ryder Rude

oh

@123 oh

123

Thanks to clear me.

Ryder Rude

in the end, i just want to say that coriolis and centrifugal forces are the pseudo forces on earth, while gravity is not

123

Yes that's why newton always considered gravity as real force.

Ryder Rude

@123 when the pseudo force is constant in space and time, the perpendicular plane direction will behave inertial, yes

123

17:04

@RyderRude Aaahh.. Okay thanks.

Ryder Rude

as in, objects will retain their velocity component in the perpendicular plane if there's no force on them

123

@RyderRude Yes as Huygens and Galileo found horizontal component of projectile motion will remain the same

Ryder Rude

yes. but again, there are no pseudo forces in this example. the vertical component is changing because of a real force, which is gravity

123

Yes gravity is real force in Newton's sense

Thanks this discussion gave me fruitful idea and confidence, how surface of Earth is good inertial frame.

Silly Goose

18:06

Isn’t it strange that physical theories are based on rules that are consistent with physical phenomena, or sometimes idealized to even be unphysical rules?

for instance, in textbook quantum mechanics we postulate a state update rule and usually assume that the update happens instantaneously.

Relativisticcucumber

@SillyGoose hi

Ryder Rude

Is Reality an Illusion? | Gerard 't Hooft, Chiara Marletto, Christopher Timpson

►

@SillyGoose but I don't see a fundamental difference between this idealisation and other idealisation in physics

also, i wouldn't say idealisations are unphysical. they r approximate descriptions of the physical world

it is estimated that the time for decoherence is extremely small. so we say that the measurement is instantaneous

but in reality, the interaction with the measurement device and state change should take finite time

Silly Goose

@RyderRude i am meaning to call out all idealizations made

Ryder Rude

oh

Silly Goose

@Relativisticcucumber hi

Ryder Rude

18:21

@SillyGoose but which aspects of physics can not be called an idealisation? i think all of the mathematical modeling is an idealisation. we don't know if any of it corresponds to reality in the exact sense

Mr. Feynman

19:17

@SillyGoose I suppose that your argument against it being physical is along the lines of "nothing is instantaneous"

Or "nothing is truly infinite"

Just like - say - the density of a singularity in GR

My personal take - without any knowledge about the philosophy of science - on this is that we have to consider this issue on two different levels:
**1)** If the theory in question is a *new* theory, describing reality at a more fundamental level than the one we currently describe (as QM would be for classical physics), we should be ready to accept the possibility of things that we'd have considered as unphysical in the previous setting, so long as your already established theory pops out in some limit of the new one. For example, you do that for observables in QM. Before QM, the idea of an …

And now comes the second point

2) This point is more physical and less philosophical. Let's admit that this instantaneous collapse is not really instanteneous and that there is some explainable (I'm not saying deterministic) mechanism explaining it.

In such case, the physical content of "instantaneous collapse" is that for all intents and purposes in which the theory - in this case QM - can be applied, the time interval in which it happens is so much shorter than the relevant time-scale, that you can model it as instanteneous.

qwerty

20:34

I've been spending too much time on this site, it's invaded my dreams

ACuriousMind

@Slereah No, because Prozeß is incorrect by the current spelling rules - this has to come from a German who learned to spell before 1996 :P

imbAF

20:59

ACM I want to ask one thing. it might be silly not to know this, but i seem to struggle

If you have an eigenvector equation

And I take the complex conjugate in both sides

the ordering of the elements in the LHS should change right?

More Anonymous

Hermitian or complex conjugate?

imbAF

hermitian I would say

ACuriousMind

@imbAF What does "taking the complex conjugate" mean? Your equation is $Av = \lambda v$ where both sides are elements in the Hilbert space. "complex conjugation" is not an operation on vectors in a Hilbert space.

imbAF

It's this funny exercise I came across online

It drew my attention because while one can solve it in a long way

I was thinking that perhaps there are qualities of a hermitian matrix that can help one

solve it in a more intuitive way

ACuriousMind

That's not an answer to my question.

I didn't ask "why are you trying to do this?", I asked "what does this even mean?"

imbAF

21:13

what do you call $\dagger$

ACuriousMind

the (Hermitian) adjoint

imbAF

can one perform that in both sides of an eq,?

of the eigenvalue eq

in this case

ACuriousMind

yes

imbAF

I assume the order of what is in the LHS would change

should change

as the column vector becomes row

ACuriousMind

depends on your conventions for how things act on "adjoint vectors" really

Personally I would write this $A^\dagger v^\dagger = \lambda^\ast v^\dagger$. In Dirac notation you might write it $\langle v\rvert A^\dagger = \langle v \vert \lambda^\ast$

it's not really a meaningful question in isolation

your notation has to be consistent, but there are various ways to make it consistent

imbAF

21:17

It's a math problem

a 2x2 matrix (\epsilon \Delta \Delta^* -\epsilon)

ACuriousMind

yes, that was clear to me when you asked a math question :P

imbAF

Yeah but I solved it in a long way

because the math implies that you have an eigenvector of this hermitian matrix (u v*)^T to eigenvalue $\lambda$. Then one can show that $-v,u*$ is an eigenvector to eigenvalue -\lambda

And yes, you can consider the eigenvalue equation, solve it, compare the compoennts and show that

but, because it mentions that the matrix is Hermitian, call it H

I think, I don;t know, one can make use of this to showcase that in a more elegant way

and not just brute force it

is that wrong to think?

ACuriousMind

What exactly is the statement you want to show?

imbAF

If for that hermitian matrix (u v*)^T is an eigenvector to eigenvalue $\lambda$

than

$(-v,u*)^T$ should be an eigenvector to $-\lambda$

And,why do I need to go the long way

when maybe there's a more subtle way

perhaps

ACuriousMind

What does this have to do with your question about the complex conjugate/adjoint? At best, using the adjoint tells you that $\lambda^\ast$ is an eigenvalue, but since Hermitian operators have only real eigenvalues that's trivially true.

imbAF

21:27

You are right

Why would it

the fact that is hermitian already points at real eigenvalues

Mr. Feynman

Hello, right. Are you ACM's replacement?

imbAF

But then the only way is the long one

considering the eigenvalue equation, substituting both vectors

and by comparing showing that yeah, it's the case

ACuriousMind

I don't understand how you expect to get a statement about $-\lambda$ out of this without considering the components, given that your matrix is a very special Hermitian matrix and that specialness is purely defined in term of its components (because the diagonal elements are negatives of each other and not arbitrary)

qwerty

@Mr.Feynman can't be, ACM hasn't left

imbAF

@ACuriousMind I thought of doing the following

Mr. Feynman

21:29

@qwerty He's not left, he's right!

imbAF

considering:

$\epsilon u + \Delta v*=\lambda u$
$\Delta* u -\epsilonv*=\lambda v*$

complex conjugating both and then rearrange it so that I get the eigenvalue equation

in the hope that what I would get as the eigenvector, would be the required one

Like, a nice way of cutting corners

ACuriousMind

@Mr.Feynman I'm not a righty, I'm a lefty

Mr. Feynman

Calm down, ACM. It's not the moment to share political alignments

2

ACuriousMind

...I was just talking about my handedness!

Mr. Feynman

And I was working on my skills as a sit-down comedian

ACuriousMind

21:35

::canned laughter sounds::

Mr. Feynman

That's the worst humiliation I could ever receive. You ended my career as a virtual bar comedian

qwerty

@Mr.Feynman it was a low bar and you still failed

Mr. Feynman

@qwerty Me trying hard to come up with a retort and failing miserably

Old Italian typing machines had a QZERTY keyboard, not QWERTY

WHY?!

ACM should be familiar with QWERTZ, apparently that's the equivalent of QZERTY

qwerty

they would make great Scrabble words

Mr. Feynman

I'm afraid I've never played that game, only heard

qwerty

21:45

@Mr.Feynman have you heard of en.wikipedia.org/wiki/Nigel_Richards_(Scrabble_player) ?

Mr. Feynman

@qwerty No, but now I read of him

Dang, he read a French dictionary and won the French game?!

qwerty

i saw a fascinating video on him, "Richards wasn’t introduced to the game until he was 28, persuaded by his mother. “I said, ‘I know a game you’re not going to be very good at, because you can’t spell very well and you weren’t very good at English at school’”

@Mr.Feynman yes. no other human is even close to that kind of ability at all

Mr. Feynman

You clearly don't know how much of this transcript is in ACM's memory :P

Oh, so the guy has some sorts of eidetic memory

ACuriousMind

@Mr.Feynman I don't know about the Italian layout, but the German layout changes the position of Z so that typing the rather common TZ is easier and because Y is very rare in German

qwerty

apparently scrabble is also more of an area control game than a long-unusual-words game

@ACuriousMind i suppose that makes finding y hard in maths class

Mr. Feynman

21:49

@ACuriousMind The current IT layout is the same as US, except special symbols that are swapped

qwerty

apologies, @Mr.Feynman got my brain on puns

Mr. Feynman

You seem to have a thing for roasting me

qwerty

D: I'm sorry!

Mr. Feynman

No, that's fine. I chuckle when I get roasted

I mean, most of my humor is self-deprecation :P

qwerty

it's not targeted, i swear! :P

Mr. Feynman

21:51

Oh, I know that. If it were you'd have a fearsome enemy

ACuriousMind

@Mr.Feynman why, who's this fearsome person protecting you?

qwerty

@ACuriousMind clearly his cat...

Mr. Feynman

Damn it, I was late

This fearful beast

ACuriousMind

awww

Mr. Feynman

She's not fat. She's fluffy...

2

And fat

qwerty

21:54

yes, a fearsome warrior

Mr. Feynman

Hey, we share a deep bond. She booped my nose.

And then completely disregarded my existence for the rest of the day

ACuriousMind

if she's so fearsome you should be glad her attention does not fall upon you too much

Mr. Feynman

I'm a food source.

qwerty

22:21

sometimes I wonder why cats say "meow"/"mau"/"miau" in almost every language, except for japanese "nyan"

Mr. Feynman

@qwerty Well, you could argue that "meow" and "miao" (the latter is italian) are fairly different until you hear them

ACuriousMind

@qwerty why do English roosters go 'cockadoodledo' and German roosters 'kikeriki'?

qwerty

@Mr.Feynman I would pronounce them the same, or very similarly, even with just english phonetics!

Mr. Feynman

I think it's just s fortunate coincidence. For example dogs make a "woof", which in Italian would be "bau" (pronounced "bow")

qwerty

@Mr.Feynman english has the old-fashioned bow-wow-wow :P

Mr. Feynman

22:25

So, apparently many other people wonder why ニャン (nyan) is different

qwerty

I think the name for a cat is just "miao" (or some spelling variant of) in many languages

Mr. Feynman

That's probably for the same reason that other words are similar in western languages and different in Japanese. For example.

English=Umbrella
Italian=Ombrello
Japanese=かさ (kasa)

Probably that's a stupid choice

Basically it's like that only in Italian and English

ACuriousMind

no, 'miao', 'mau' etc is onomatopoeia, the umbrella is just the languages being related

Mr. Feynman

Of course it is. I'm arguing that's the same underlying mechanism

The same onomatopoeia can be transposed to many languages

ACuriousMind

but "umbrella" is not onomatopoeia

Mr. Feynman

22:29

While people without the influence of other languages will likely render it differently

qwerty

I really think cats named themselves rather clearly

except maybe in Japan :P

@Mr.Feynman you have miao in ancient egyptian, and chinese languages too...

ACuriousMind

It's just cultural accident which sound of the animal "wins" as the sound for it; I've heard plenty of cats incapable of producing anything close to a proper meow, while others more or less just say 'meow' :P

Mr. Feynman

ACM, I'm not saying that umbrella is onomatopoeia! I'm saying that just like this happens for words like umbrella, probably it happens for onomatopoeia. I suppose that the reason why we basically use the same onomatopoeia (meow et al) in different language is that there has been interaction between them (?)

I mean, if I'd never seen a cat and heard it meow, I wouldn't be so sure about which one between "meow" and "nyan" is the "correct" one

ACuriousMind

If I had had to pick the cat sound according to my childhood cat, it would have been closer to 'rrawaugh' :P

Mr. Feynman

That sounds like a cat in heat in my mind

qwerty

22:33

@Mr.Feynman Not the case! as I mentioned, e.g. chinese languages (all of the ones I know?), vietnamese, etc en.wikipedia.org/wiki/…

those are definitely not indo-european, as are many of the other languages which say some variant of meow

ACuriousMind

@Mr.Feynman nah, he was just a cranky old (neutered) man

Mr. Feynman

Mhhhh, there might be something about sound. The Japanese language is very strict with sounds. They only have 46 given syllables and that's what they get (although some are combined to make a few western sounds like "dee"). I'm sure it is possible to form a "meow" in Japanese, i.e. ミャオ but probably to Japanese people it wouldn't sound cattish

@ACuriousMind I see, so you kidnapped an old man, neutered him and pretended he was a cat

ACuriousMind

uh

Mr. Feynman

That didn't come across the way I hoped D:

And it's time to deactivate my brain

Good night to European users and goodbye to everyone else

qwerty

good night! :)

fqq

22:53

@ACuriousMind I tried researching it in the past but I was not able to find any reasoning for QZERTY, it seems to be an old typewriter thing and died out >40 years ago

Relativisticcucumber

23:57

@Mr.Feynman EEEEEEEEEK

ok i know where we all stand on cats but what's the vibe regarding dogs?

ACuriousMind

some dogs are okay

Relativisticcucumber

i love some dogs, but i really really dont like being licked by dogs. ive also found small dogs to be rather mean compared to larger dogs so i tend to prefer interacting with those. there's a therapy dog on campus that looks SO cute but i keep missing her office hour

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