The h Bar

Amit

12:00

@RyderRude "bad at philosophy" is way too broad and vague. I'm sure Feynman knew very little about all the technical jargon involved in the philosophy of consciousness for example. Does it mean he didn't have his own insights and ideas about this subject? No -- whatever he did think and perhaps even talk about was not recorded as part of the academic framework, but just being outside of the academic context doesn't make ideas worthless.

Slereah

Richard Feynman

Ryder Rude

@Amit i haven't read his philosophical talks all that much. I'm just assuming that Carroll has

Amit

@RyderRude That's funny

Ryder Rude

I read one of his philosophy talks in his lectures, and it was not a bad point

according to him, the philosophers of that era were trivialising relativity. And he corrected them

Amit

You don't need a Feynman for that insight ;)

Ryder Rude

12:02

which makes sense. Philosophers can misinterpret relativity if they don't study the math

Slereah

Plenty of people working on the philosophy of relativity were physicists

Ryder Rude

@Amit they said "ofc everything is relative"

@Slereah oh

Ryder is not Rude.

@Slereah you can hear the Italian influence in their New York accents.

Slereah

@RyderisnotRude. Well if you consider the Napolitans Italians, anyway!

Ryder is not Rude.

👍👍

Slereah

12:05

I'm told that butchering the word "Capocollo" as "gabagool" is some napolitan shit

sopranos but just gabagool (extended cold cut edition)

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Ryder Rude

Philosophers also discuss things like "do composite objects exist or do only their constituents exist"

Feynman wasn't a fan of this

Do Chairs Exist?

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It is from Vsauce

Amit

I can't find it now, perhaps it got deleted, there was a video that did the same trick as here (change "Mana Mana" to "Phenomena") only with edited Feynman bits where he speaks the word "Phenomena" in his heavy accent :D youtube.com/watch?v=Dnk0Be4a0aw

Slereah

> There is nothing here that would have been technically useful to a Russian bomb designer in 1950 or to an Iraqi bomb designer in 1990. But the primer contains much more than technical information. It conveys a powerful message that bomb designing is fun. The primer succeeds all too well in recreating the Los Alamos mystique, the picture of this brilliant group of city slickers suddenly dumped into the remotest corner of the Wild West and having the best time of their lives building bombs.
It helps to perpetuate the myth. ... This is what I mean by seduction-the myth, unfortunately contain…

@ACuriousMind Hey I have a fun idea for a project

Amit

Today we should focus on building interception missiles instead <_<

Slereah

Honestly i'd rather do homemade atom bombs than work on missile stuff

That shit is way more dangerous to work on

Amit

12:16

which is more dangerous?

Slereah

Ever seen what happens with 97% hydrogen peroxide

Amit

Why do you prefer the more dangerous project?

Slereah

I mean that missiles are way more dangerous

Amit

It's interception missiles!! Conceptually they should only be targeted at other missiles

Slereah

you're still gonna need that rocket fuel!

Hydrogen Peroxide: going all the way

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Amit

12:18

I don't want to actually build the stuff.. that's for engineers

Slereah

a noble profession

Trust me, i'm an engineer !

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Amit

Ever noticed that the common slang is a rocket scientist* rather than a rocket engineer? (which really modernly are Aeronautic engineers)

Slereah

That's because it's an expression made to be said by some podunk whitebread redneck hillbilly

Amit

lol

I gather it doesn't translate literally that way in French then

Slereah

No idea

Amit

12:22

Of course not, silly of me

Ryder Rude

Some decades ago, they had the idea to use projectile bombs. They would drop small things from space and use the kinetic energy for the explosion

imbAF

@Amit One question. Following the two rules for the index notation, is enough to do calculations, without thinking in terms of vectors and matrices ?

Amit

@imbAF What are the two rules?

imbAF

1. Free index should be the same in both sides. 2. You cannot have more then two indexes as upper and lower

I am trying to derive the expression that relates the inverse with the transverse and, yeah I will do it as a training

Amit

in 2. you mean identical indices right

repeating indices come in pairs and should be balanced is what you mean

imbAF

12:26

Yes that is what I mean

Amit

@imbAF inverse and transverse? you mean perpendicular and parallel? the acceleration one?

then yes these two rules are right

you can think about matrices and vectors if you want, you just don't have to write them out ;)

imbAF

I will post a picture because I ain't typing that now

I will try to derive this

Amit

cool

Ryder Rude

youtube.com/… it is a video series on LQG

LQG says that we have to triangulate space into tetrahedra, and then quantise the observables

Silly Goose

One day i’ll be able to apply cohomology to physics~~

Slereah

12:37

it has some pretty important applications

A lot of the differences between classical mechanics and relativistic mechanics are down to cohomology

Silly Goose

Oh how so

Ryder Rude

to each edge of the tetrahedron, it associates and element of SU(2). This is the "classical theory". And when we quantise it we get $ L^2(SU(2)^k)$ as a Hilbert space

this sounds like random ideas, but they believe it has correspondence with GR

Slereah

The action of a group on the Lagrangian are basically related to central extensions by $\mathbb{R}$, which relates to the second cohomology of the group

Silly Goose

I have only been suggested that cech cohomology could be applied to a project i was working on

oh i see

Slereah

The Poincaré group has a trivial such cohomology, while the Galilean group doesn't

That difference underpins a lot of consequences in physics

Silly Goose

12:40

well in that sense cohomology has an important application when using bargmann’s theorem to find irreps of poincare

I see

Slereah

The central extension thing is also true in QM by the way

Except this time it's by $U(1)$, the phase group

Ryder Rude

@Slereah what kind of consequences

Slereah

The difference of energy depending on mass or not, conservation of particle numbers, the localization of particles, that sort of things

Ryder Rude

oh yes, u once said that mass energy relation was rooted in cohomology

@Slereah localization of particles? Is this related to the position basis?

Slereah

12:56

it is

Ryder Rude

Do u mean more particles get created when we try to localise

Slereah

the lack of proper position basis in relativistic term is rooted in the cohomology according to Some Guy

Ryder Rude

oh

Slereah

Although I haven't found what he meant yet, pretty throwaway line

Ryder Rude

maybe he is seeing things that aren't there

Slereah

12:57

Ryder Rude

i dont know cohomology tho

Slereah

@ACuriousMind Do you know what he means here

ACuriousMind

@Slereah The action of the Gailiean group on phase space is non-trivially centrally extended, but there is no hope to have a similar action of the Poincaré group on the space of $x$ and $p$ since it does not have non-trivial extensions, see also this excellent answer by David Bar Moshe

Slereah

yeah but how does it relate to quantum localization

ACuriousMind

@Slereah note how David's answer shows that the non-rel representation on the quantum mechanics position space wavefunctions has to come from a non-trivial extension; this, by contraposition, means we can't have a position-space representation in relativistic QM with all the properties we desire, since it would have to carry a non-trivial extension, too

in a handwaving way, in order to have both position and momentum operators, you want to adjoin the position operators to the group with commutators $[x,p] = 1$ with the translation generator; for the Galilei group this is unproblematic and the resulting non-trivial central extension is the Bargmann group; for the Poincaré group, you can't do it while correctly preserving how the $x$ need to transform under all the elements of the Poincaré group

Silly Goose

13:19

formally, are bravais lattices $\mathbb{Z}$-modules?

ACuriousMind

@SillyGoose yes, you can say that

Silly Goose

are modules natural objects to represent crystal lattices by because modules share some usual properties of vector spaces sans guaranteeing the existence of a basis?

hmm actually i am confused about my original statement. In a bravais lattice, we are encoded some sort of spatially regular pattern. however, mathematically it seems i can always just write a map from said bravais lattice into $\mathbb{Z}^3$ as "spanned" by $\{1,0,0\}, ...$

so that it seems that any bravais lattice is isomorphic to this regular square lattice

well okay overall i am just curious about what mathematical structures are involved in bravais lattices

it almost seems like a discretization of a manifold because as you add finer details, you start considering a lattice at each point of the bravais lattice that encodes the finer details

Amit

13:41

@imbAF I only now realized you meant the transpose 🙃

imbAF

Yeah but I wrongfully said transverse instead of transpose

Amit

👍

Ryder Rude

What Is It Like to Be a Bat?

"What Is It Like to Be a Bat?" is a paper by American philosopher Thomas Nagel, first published in The Philosophical Review in October 1974, and later in Nagel's Mortal Questions (1979). The paper presents several difficulties posed by phenomenal consciousness, including the potential insolubility of the mind–body problem owing to "facts beyond the reach of human concepts", the limits of objectivity and reductionism, the "phenomenological features" of subjective experience, the limits of human imagination, and what it means to be a particular, conscious thing. Nagel famously asserts that "...

Amit

13:59

Mar 27, 2017 at 15:54, by ACuriousMind

@0celouvsky Related, What's it like to be a Bat?. Such impossible-to-explain subjective experiences are often called Qualia

@RyderRude I just figured someone would have linked that here ;) I'm surprised it's only once

Ryder Rude

Wow. Lemme read that discussion

There's barely anything...

Amit

But at least it proves retrocausality

Ryder Rude

Slereah hasn't changed

Mar 27, 2017 at 15:57, by Slereah

I was a bat once

@Amit lol

Amit

Philosophers of mind in general: "What's it like to be a Bat?",
Panpsychists: "What's it like to be a baseball bat?"

Ryder Rude

Based panpsychists

I had a toy model of panpsychism in which it causally participates in the universe

Chalmers says consciousness may or may not causally participate in the universe

Amit

14:08

Not very clear. Anything that exists, participates

Ryder Rude

yeah, but we usually think of consciousness as an epiphenomenon. It just rides along with matter and doesn't influence matter back

Amit

it's not always an happy phenomenon.. ;) no seriously, i don't know exactly what is the dividing line between phenomenon and epiphenomenon...

Ryder Rude

Chalmers has ideas in which the "degree of consciousness" is assigned to a system, and this causally affects matter by wavefunction collapse

This is similar to Penrose's ideas

@Amit idk exactly either. But here i meant the idea that there is no back reaction from consciousness to matter that is measurable in a scientific experiment

Amit

It doesn't sound falsifiable so what kind of experiment can you think of to show that?

Ryder Rude

It can never be shown exactly... But suppose e.g. Chalmer's model of consciousness predicts objective collapse that is measurable

you could test that

ACuriousMind

14:13

@SillyGoose As always, you have to be careful what the word "isomorphic" means in this context - sure, as $\mathbb{Z}$-modules, all 3d lattices are isomorphic to $\mathbb{Z}^3$. That just means the module structure alone is obviously not what you're interested in here

Amit

I suspect the idea of collapse will collapse

Ryder Rude

it will implode :)

Amit

or decohere

Ryder Rude

and create many worlds

If i had to pick, i don't think consciousness participates in the laws of physics

Amit

Again, I can only see it simply. If it doesn't participate, it doesn't exist

Don't go epi ;)

Ryder Rude

14:17

U r making a mistake :P physics doesn't describe all that there is

it descirbes measurements, no more, no less

Amit

well you actually tripped me up because you phrased it differently this time. previously you wrote "participates in the universe".

with regards to the statement that physics may not be able to describe the universe entirely, I do agree

Ryder Rude

sorry.. i said "causally participate" to mean consequences in physics

i wasn't clear enough

@Amit great :)

Silly Goose

i am considering some electromagnetic plane waves in vacuum $A_0e^{\eta(x^\mu)}, A^i e^{\eta(x^\mu)}$. I found a relation $A_0 = k_iA^i$. This looks quite like the Lorenz gauge condition $k_\mu A^\mu = 0$. However, I did not make any choice of gauge. Does this suggest I've done something wrong...

ACuriousMind

14:33

@SillyGoose What is $k$? The Lorenz gauge condition is $\partial_\mu A^\mu = 0$, there is no $k$ in there in position-space. Using which equations did you "find" this relation?

Silly Goose

oh i am silly

ACuriousMind

that much is known :P

Silly Goose

i should have put $e^{-k_\mu x^\mu}$

so I have $A^\mu = a^\mu e^{k_\mu x^\mu}$ where $a^\mu = (a^0, a^1, ...)$ are constants. i.e. the $k_\mu$ is the wave vector + frequency.

I used the Maxwell equations to derive that $a^0 = k_i a^i$.

but i sillilily conflated the condition i derived with the Lorenz gauge fix

i want pumpkin yum

nickbros123

16:19

@ACuriousMind sees an integral, starts looking for the real number $L$ so that for every $\varepsilon>0$ there exists $\delta$ so that for all tagged partitions....

naturallyInconsistent

18:12

@SillyGoose three appearances of $\mu$; that equation is kapoot

@ACuriousMind I can do with pretty easily with a very simple example, completely contained within a specific matrix representation, and people will immediately realise that contravariance and covariance is a necessary part of real-world physics that is never going to go away.

Ryder Rude

20:01

HULK Vs. SAITAMA Animation (Full Version) -Taming The Beast

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Hulk gets terribly angry in this

imbAF

20:20

I was attempting to derive the following:
$(\Lambda^T)_\beta^{\ \ \rho}= \eta_{\beta\delta}(\Lambda^{-1})^\delta_{\ \ \epsilon}\eta^{\epsilon\rho}$.

To do that I started from the expression: $(\Lambda^T)\eta\Lambda=\eta$. I was successful in doing so. Then I asked myself how does one derive the last mentioned expression. I asked Chatgpt about it. And it showed me that it can be derived from the invariance of the inner product between four vectors. But I had already derived that. Like this:
$\Lambda_{\ \ \beta}^{\alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$. No…

Vincent Thacker

@imbAF Why would you ask ChatGPT this kind of question?

Are you aware that it currently can't really do problems like a person can?

ACuriousMind

@imbAF 1. Don't trust ChatGPT or any other language model. No one guarantees you that anything they say is accurate. 2. Use ${\Lambda^\alpha}_\beta = {(\Lambda^T)_\beta}^\alpha$. Then the indices in your last equation are in the right order for matrix multiplication (column of $\Lambda^T$ with row of $\eta$).

imbAF

I asked it about, where does the expression in matrix notation comes from

And it said from considering the inner product between four vectors

And I said, but I have done that

And just by looking at the expression I can't tell that the transpose is present

while the matrix notation, does show that

$\Lambda_{\ \ \beta}^{\alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$. How can you tell that the transpose is present ?

ACuriousMind

I don't know why you repeated your question, but see my 2. above :P

without writing this in terms of $\Lambda^T$, the indices of the first two terms are not matched correctly (in this case row with row) to be a matrix multiplication

Vincent Thacker

@imbAF Because $\Lambda^\alpha_{\;\beta} = (\Lambda^T)_\beta^{\;\alpha}$

imbAF

20:30

contracted indices should be right after one another

?

Vincent Thacker

@imbAF No, this is not a requirement

imbAF

But that is what ACM is saying with this:
without writing this in terms of $\Lambda^T$, the indices of the first two terms are not matched correctly (in this case row with row) to be a matrix multiplication

Vincent Thacker

@imbAF I mean that it is not a requirement of index notation itself.

You are trying to make the index notation match up with the matrix multiplication notation

imbAF

requirement for what ?

Vincent Thacker

So you take the transpose in order to make it match.

imbAF

20:32

Yes, because I couldn't tell that the transpose was present until I saw the matrix notation

Vincent Thacker

Because in matrix multiplication, the second index of the first matrix is contracted with the first index of the second matrix, which is again by convention taking the first index as rows and second as columns

ACuriousMind

@imbAF they don't need to be "right after one another", but if you have a bunch of objects that can be interpreted as matrix, matrix multiplication is having the second index of the first factor contracted with the first index of the second factor

because that corresponds to the "multiply the columns by the rows" we do in the definition of matrix multiplications

any other kind of contraction of the indices on the objects is not a matrix multiplication

imbAF

I get it

So in other words

This is not wrong:
$\Lambda_{\ \ \beta}^{\alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$.
But if I want to mimic matrix multiplication, then I would need to write:

$(\Lambda^T)_{\beta}^{\ \ \alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$

Is this what you are both trying to say?

Vincent Thacker

@imbAF Yes

imbAF

But then you can't say that to mimic a certain tensor operation is not a requirement. In order for me to have been able to derive the initial expression that I wanted, I had to mimic the matrix multiplication

Unless there is a way for getting to that expression ($(\Lambda^T)_\beta^{\ \ \rho}= \eta_{\beta\delta}(\Lambda^{-1})^\delta_{\ \ \epsilon}\eta^{\epsilon\rho}$.)

that doesn't require you to mimic matrix multiplication in index notation

ACuriousMind

20:39

I don't really understand what you're trying to say, sorry

imbAF

I am saying that I had to change this:$\Lambda_{\ \ \beta}^{\alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$ into this: $(\Lambda^T)_{\beta}^{\ \ \alpha}\eta_{\alpha\gamma}\Lambda^{\gamma}_{\ \ \delta}=\eta_{\beta\delta}$ (which mimics matrix multplication) to get this: $(\Lambda^T)_\beta^{\ \ \rho}= \eta_{\beta\delta}(\Lambda^{-1})^\delta_{\ \ \epsilon}\eta^{\epsilon\rho}$

Vincent Thacker

@imbAF Index notation doesn't "mimic" matrix multiplication; it's a way of representing matrix multiplication (and is also capable of representing much more general expressions).

imbAF

represent*

Ok I got this

Thanks a lot

Silly Goose

@ACuriousMind i think people don't actually understand many of such things. it's like (at least my) high school calculus where you are taught rules to follow and you follow them. but then it's just some cousin of memory

i still don't really understand index notation :P i feel like i got to work though some chapters. taking enm again seems to be a good opportunity to finally learn it properly...

Vincent Thacker

@SillyGoose Yes, because the general education system isn't designed to turn everyone into mathematicians

It's purpose is just to teach people how to use it to do essential tasks

So the fastest way to do that is a mechanistic set of rules on how to do it

Relativisticcucumber

20:48

i met another person who went to heidelberg :D @ACuriousMind

and also a person from the lab i just left just went to do his masters there

heidelberg is taking over !

2

@SillyGoose i think the "memorize and use" is oftentimes a choice of the student as well

Amit

@SillyGoose the name of the cousin is "rote" :)

Silly Goose

@VincentThacker well i'm definitely not an advocate of learning things as mathematician. but i think people have a very individual preference for how to organize information. E.g. I think the amount of rep/Lie theory I learned (certainly not as much or at the rigor of a mathematician) has paid dividends in terms of organizing and processing power as I've done physics

imbAF

In Heidelberg they offer mathematical physics in Masters, is that right ?

Silly Goose

but in a quantum course all you'll get is here is the matrix exponential

(in the states at least)

Vincent Thacker

@SillyGoose Yes, same for me

Amit

20:52

Who isn't always a negative guy, this cousin. Sometimes some kind of a critical mass of rote coincides with a sudden insight and you realize you're understanding something more deeply than before

Silly Goose

@Relativisticcucumber indeed. i'm not sure i ever plan to know the gory details of distributions and functional analysis and etc. :P

I am not sure though. in some sense, it seems like one should not hold so tightly to the mathematical structures of existing physical theories because superseding physical theories have no reason to share such structures. though, conceivably, it is easiest to extend existing structures at times. Then, a physicist maybe benefits from holding on to more transcendental concepts.

"physical concepts" i suppose :P

ACuriousMind

@Relativisticcucumber we are legion

Relativisticcucumber

21:53

itsv my understanding that xray crystallography works by sending in xrays and looking at how they diffract to infer the composition of the material. but xrays are a form of ionizing radiation, right? doesnt ionizing radiation alter the material the radiation passes through? that seems unideal if one is trying to resolve the structure of the material

Mr. Feynman

22:34

Is Heidelberg a hive mind

naturallyInconsistent

@Relativisticcucumber small amounts of taking not-deepest-core electrons out of a material is unlikely to hurt the material enough

Silly Goose

23:18

@Relativisticcucumber material got hands

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