The h Bar

1:04 AM

@Semiclassical That could be the case, but this is just for a general inhomogeneous medium in the text, so it is not

DIRAC1930

3:53 AM

I think Dirac's objections regarding renormalization were more so about throwing away non-renormalizable theories when it's not clear what the final framework will be

E.g. imagine throwing away gravity because the naive quantization is non-renormalizable

instead of looking for a consistent framework where they coexist

Silly Goose

6:04 AM

Is anyone here familiar with density matrix renormalization group in quantum information context

A condensed matter theorist told me that it is used to minimize subsystem-subsystem entanglement, but I am not getting that impression from glancing at some review on the topic, but I don’t think he was saying incorrect things

More Anonymous

7:30 AM

@lucabtz every bad sci-fi movie ever!

:p lolz

user 85795

I find only documentaries worth watching and perhaps, quizz shows :p

More Anonymous

@user85795 man I suck at trivia

Documentaries are my go to as well

Got any recommendations?

user 85795

7:45 AM

I liked The man who knew infinity.

Ryder Rude

hi

user 85795

hello

Slereah

No movie is good at science

Doing an actual story involving good science is something too niche to interest movie studios

lucabtz

@DIRAC1930 i think the modern way now is that non renormalizable theories are perfectly fine

@Slereah imagine most of the movie spent in front of a blackboard talking and another good part spent typing latex on computer

Slereah

There's a few okay books in that domain

They tend not to be very good basis for a movie tho

Mr. Feynman

8:00 AM

@lucabtz fine as low energy theories, yes

There is also a famous line from Weinberg IIRC "Are non renormalizable theories renormalizable?"

Ryder Rude

@Slereah the movie Primer has science in it

also Interstellar

Slereah

I've seen them both and they are trash

Ryder Rude

why do u think so

Slereah

They have no more science in it than the average B-movie

If all you're looking for is simply the aesthetics of science you might as well watch the Andromeda Strain

Just a 30 minutes scene of scientists scrubbing up to go into a lab

Thrilling

Ryder Rude

Interstellar has a demonstration of gravitational time dilation like no other movie

and also of docking

Slereah

8:06 AM

Yeah and so does Dark Star

Dark Star even has a song about it

But simply having a reference to a scientific element does not a hard sci fi movie make

Might as well watch Disney's The Black Hole if that's all you wish for

Thrilling special effects from Dark Star

Ryder Rude

Dark Star looks disgusting

@Slereah wtf

Slereah

I'd watch Dark Star anyday over Interstellar rly

Ryder Rude

hard sci fi movie would feel like a lecture

Slereah

Pretty much

Like basically anything that Greg Egan wrote

Or the Planiverse

Ryder Rude

are they hard to understand

some movies become incomprehensible

the movie Tenet is somewhat hard sci fi

DIRAC1930

8:10 AM

Has anyone studies beta decay in the context of weak interactions in QFT

Ryder Rude

it's incomprehensible

but it's still pretty good

Slereah

@DIRAC1930 Somewhat?

Also remember the rules

Don't ask to ask :p

DIRAC1930

I thought you said you didn't know much QFT lol

Slereah

There's a lot of QFT

John Rennie

@Slereah Greg Egan at least knows what he's talking about, or at least he has a working knowledge of GR. Gregory Benford is the worst author i know for long, tedious, and half baked expositions in hard SF.

Slereah

8:15 AM

Never read him

John Rennie

You haven't missed much.

Slereah

Course the best way to convey scientific theme in a work of fiction remains

Mr. Feynman

@DIRAC1930 for practical like that calculations Fermi theory is enough

You can write amplitudes from the weak theory and see that in the low energy limit you get Fermi amplitudes

lucabtz

8:33 AM

@Mr.Feynman well yeah of course

@Mr.Feynman yeah, Schwartz book on QFT answers with a chapter called "Non-renormalizable theories are renormalizable"

you aint a crackpot anymore now

user 85795

How would you rate The man who knew infinity out of 10, with zero given if you didn't see it @JohnRennie

lucabtz

@user85795 zero

John Rennie

@user85795 I enjoyed it, and I think it was a good program about Ramanujan as a person. Obviously they wouldn't go into any detail about his work as 99.9% of viewers wouldn't understand it.

Relativisticcucumber

hodge podge question: firstly, i have seen before this idea that spinors are states that need to be acted upon twice to return to where they started, specifically, that they take a 720 degree rotation to return to the initial state. what is the implied representation here? because i have just read in hall that it seems we cannot even construct representations of $SO(3)$ for spinor states? specifically, what i read claims this for half-integer spin states, so im not sure how this generalizes.

i feel there is a connection to the fact that we can form a representation for SU(2), but im not really seeing explicitly how this all plays out

lucabtz

@Relativisticcucumber they are projective representations of $SO(3)$ not ordinary linear representations

Relativisticcucumber

8:39 AM

@Mr.Feynman did u have an identity crisis or smth where did crackpot go

@lucabtz sorry, im not sure what either of those terms mean let me check

lucabtz

ACM has a great Q&A

wait ill find it for you

Relativisticcucumber

@lucabtz omg yay thanks

lucabtz

98

Q: Why exactly do sometimes universal covers, and sometimes central extensions feature in the application of a symmetry group to quantum physics?

There seem to be two different things one must consider when representing a symmetry group in quantum mechanics: The universal cover: For instance, when representing the rotation group $\mathrm{SO}(3)$, it turns out that one must allow also $\mathrm{SU}(2)$ representations, since the negative s...

this is also on central extensions and so on

however the definition of projective representation is there as well as why they are important in QM

Relativisticcucumber

i see. i also found this sites.ualberta.ca/~vbouchar/MAPH464/… so ill look at this and acms answer

thanks for the suggestion

by the way are you a grad student? @lucabtz

DIRAC1930

@Mr.Feynman Ah okay thanks

lucabtz

8:44 AM

anyhow the whole point which relates $SU(2)$, $SO(3)$ and projective representations is the following: the spin representations are projective representations of $SO(3)$, however you can always lift a projective representation to an ordinary one as long as you consider a central extension. It turns out for $SO(3)$ the best extension you can get is an extension by $\mathbb{Z}_2$, which is the double-cover $SU(2)$. Thus the spin representations are ordinary representations of $SU(2)$

@Relativisticcucumber im not sure how this terminology applies to the european model of uni. I already have a bachelor and I'm graduating my master next week

have to apply to PhD this summer

Relativisticcucumber

@lucabtz ok i think this explains the general idea, thanks. ill read the post and site to get more clarity on the details. this is my first intro to projective reps yay

@lucabtz ah gotcha

lucabtz

anyhow i guess technically from next week ill be unemployed

user 85795

@JohnRennie Agreed. I don't think even the average viewer would appreciate the insistence of proofs by Professor Hardy.

John Rennie

When I first saw that programme advertised I thought it was going to be about Cantor.

user 85795

Yes, he was also known for infinity.

Ryder Rude

8:48 AM

can $dx <---> \frac{\partial}{\partial x}$ be called a canonical isomorphism relating these two?

this one does not need a metric. in some sense, these two are mapped to each other?

Slereah

It's not coordinate independent

DIRAC1930

I really liked The Imitation Game but apparatnly it's not very historically accurate

Ryder Rude

@Slereah oh

DIRAC1930

The Man Who Knew Infinity was also okay I thought

user 85795

Godel would be The Man Who Knew Proofs :P

Ryder Rude

9:07 AM

the metric can be integrated along a path

essentially a metric times two tangent vectors is a scalar, hence the integral over a path is a scalar

dows this sort of generalise the idea of "forms can be integrated"?

the metric is not a form

it's $\sqrt{g (v,v)}$ that can be integrated, actually

im trying to find a generalised notion of what can be integrated

Slereah

The form isn't the metric, it's the parametrization :p

The "dt" of a curve is the 1-form generated by the parametrization

Ryder Rude

oh

Slereah

Which you multiply by the 0-form of the norm

Ryder Rude

makes sense..

so only forms can be integrated

Slereah

For any k dimensional submanifold, you can pullback any k-form on it to define a volume form

And that's what you're integrating

Relativisticcucumber

9:15 AM

ah, i have learned Bargmann's theorem @lucabtz -- i think this provides the justification if i take the theorem to be true

bolbteppa

@DIRAC1930 he doesn't say anything about e.g. Yang-Mills, standard model etc unfortunately

Ryder Rude

yeah. we can do this pullback using the co ordinates on the sub-manifold

bolbteppa

$d \mathbf{r} = dx^{\mu} \partial_{\mu} \mathbf{r}$

Slereah

The weird trick being that the parametrization of a curve is just a coordinate defined on a line

But since it's one dimensional, it turns out that any nowhere vanishing vector field is also a frame

Relativisticcucumber

i have a general concern though about rep theory in physics. it seems everything that i have learned thus far applies to finite dimensional spaces, which seems to mean we can only apply basic rep theory to spin? is this so? because i thought that rep theory should also describe angular momentum in general, but now i am having doubts about this

Slereah

9:18 AM

And also there's no fundamental difference between a frame and its determinant

Ryder Rude

so $\sqrt{g(v,v)}$ is a scalar field on the line. and the $d\tau$ is the form cuz $\tau$ is the worldline co ordinate

Slereah

So things get a lot simpler in one dimenion but then you don't really notice the similarities with the higher dimensional case

The actual form of the length functional is like for some curve $X : L \to M$ $$l(X) = \int_L d\mu[X^* g]$$

It's the pullback of the volume form on the curve

And due to all that I said there is some unholy combination of parametrization, tangent vector, frame and volume form on a curve

But we typically all write it the same

Ryder Rude

thanks

Slereah

You can do similar things to any manifolds as well, that's what happens in string theory

You do exactly the same process for the worldsheet volume

But since this is 2D it looks a bit more complex

naturallyInconsistent

9:35 AM

uugh, now I'm screwed. I've done some really nice data analysis, but it is saying that my PN detector data of 241Am decays is having two different energy humps than one. Whyyyyy

Slereah

10:06 AM

(Another nice thing in 1D is that since the manifold is always flat, you always have the "canonical coordinates", which are just the proper time parametrizations)

More Anonymous

10:40 AM

@lucabtz noooo how come?

DIRAC1930

@bolbteppa My impression is that Dirac probably just wasn't that interested in the later stuff. He probably had stuff that he found more interesting to work on or it was in his later years when he was semi-retired

DIRAC1930

11:17 AM

Even in the 30s he was concentrating on essentially his own stuff. It just happened that at that time, his interests lined up exactly with what everyone else found to be important

More Anonymous

@DIRAC1930 reminds me a bit of coorelator vs causation or descriptor vs predictor ideas :p

bolbteppa

> When Dirac is commemorated as one of the greatest physicists ever, it is mainly because of his epoch-making contributions to quantum physics made in the second phase of his life. Given the outstanding quality of these contributions, it is only natural that he was unable to reach the same level of successful creativity later in life.

> After all, and despite his reputation, Dirac was only a human. Even Einstein’s career was divided into a highly creative early phase and much less productive later years. At the end of his life, Dirac felt that his earlier successes did not quite make up for his later failures.

> In 1983, he was addressed by Pierre Ramond, then a 40-year-old Florida theoretical physicist known for his important work in string theory. Ramond sought to engage the famous physicist in a conversation, but Dirac evaded: “I have nothing to talk about. My life has been a failure.” One understands Ramond’s surprise and even shock: “I could hardly believe that such a great man could look back on his life as a failure,” he noted. “What did that say about the rest of us?”

Mr. Feynman

@Relativisticcucumber Per correr miglior acque alza le vele omai la navicella del mio ingegno

More Anonymous

11:34 AM

What are some open fundamental problems in physics minus the rigour part?

No need to reply^

DIRAC1930

@MoreAnonymous Why the electron has that specific mass

More Anonymous

11:53 AM

@DIRAC1930 got it ... Man I'd really have to learn effective field theory to even venture near that one ...

Also I think a lot of academics just have high standards ... It's like they haven't failed enough in front of the massive problems that are there ... I've failed more than anyone for a long time so I get u ...

Ryder Rude

12:17 PM

@bolbteppa some people judge themselves harshly becuz they may think they did not realise their potential

Dirac has set too high standards for himself

and this mindset affects his life in a bad way

Slereah

12:46 PM

@MoreAnonymous How to get money for it

bolbteppa

1:08 PM

@RyderRude He should have picked up a string theory book

Relativisticcucumber

@Mr.Feynman idk i thought crackpot was peak mentality

naturallyInconsistent

B A H

Relativisticcucumber

BAHHHH

Mr. Feynman

@Relativisticcucumber Deceive the world

Deceive yourself

naturallyInconsistent

@Mr.Feynman Deceive the fabric of spacetime

Mr. Feynman

1:23 PM

Are GR people tailors of spacetime?

Relativisticcucumber

Slereah

Relativisticcucumber

@Mr.Feynman pants have an interesting topology

Ryder Rude

do u think the definition of tensors as multilinear maps is a basis independent definition

wikipedia uses weird things to make a basis independent definitions, like quotient spaces

Slereah

It is

bolbteppa

1:34 PM

If $[x^a,x^b] \neq 0$, is $\partial_a (x^b f) = \delta_a^b f + x^b \partial_a f$ even true

Slereah

Do you mean like in a Grassmann algebra?

bolbteppa

Maybe?

Slereah

You kinda have to redo analysis from scratch when you change your field completely, but I don't remember the specific rules if the multiplication isn't commutative

But I would beware yeah

apparently for Grassmann algebras for instance you have to use the graded version of the Leibniz rule, for a start :

swc-math.github.io/dls/DLSCartierCh9.pdf

bolbteppa

If $[x^a,x^b] = C^{ab}$, maybe $\partial_a (x^b f) = \delta_a^b f + x^b \partial_a f + C_a^b f$ or something

I don't think Grassmann even applies here it's more general than that

Slereah

Yeah but presumably Grassmann is a case of that

naturallyInconsistent

1:40 PM

@Relativisticcucumber Today my python matplotlib plots kept getting killed. Like, dies after saving ~600 plots. Fixed, and it even ran faster. Turns out, I had >20000 exponential decay plots to fit. Isn't that amazing?

bolbteppa

2

Q: Partial Derivatives in Noncommutative Spacetime

Will the order of taking partial derivatives matter in a noncommutative spacetime? If so, what implications will that have on the way we do gauge theory? For example, will our Lagrangian now contain new terms which would normally cancel out in a commutative spacetime? An answer in the context of...

quantum-field-theory general-relativity gauge-theory non-commutative-theory

Slereah

Since $[x^a, x^b] = 2 x^a x^b$

You should probably try to show the proof of Leibniz from scratch I guess?

It's not super complicated

bolbteppa

Okay that's a good point, in that special case you can set up some rules to deal with this stuff, but in general I have no idea

Slereah

I'm sure there's plenty of math about differential non-commutative algebras somewhere

Mr. Feynman

@Relativisticcucumber of course

bolbteppa

1:45 PM

This is interesting

Sanjana

2:22 PM

How do you find the Dynkin indices (a.k.a. Dynkin components/labels, etc.) of fundamental representation of any simple Lie algebra?

Actually I wanted to "prove" that the fundamental representation of $E_8$ is $248$ dimensional. I can simply use the Weyl dimensionality formula if I know the Dynkin indices for the representation. But I am confused about how to obtain the indices.

Also, is there any way to obtain the result from the Dykin diagram/Cartan matrix without going over to Dynkin indices?

Ryder Rude

what are some good recent movies

Ryder Rude

3:28 PM

Tao says there r three separate generalisations of 1D integration : Differential form integration, Measure theory and Differential equations

the first one generalises signed integration, the second one generalises unsigned integration, the third one generalises indefinite integration

Sanjana

3:56 PM

I think my question is simpler: I want to find the dimension of the algebra corresponding to $E_8$ where I am given only its Dynkin diagram. This boils down to finding the number of roots in the $E_8$ system and the answer is gonna be $240$ because I know that the rank of $E_8$ is $8$ and the dimension is $248$.

I got a sentence from wiki which says

> "In the so-called even coordinate system, $E_8$ is given as the set of all vectors in $\mathbb{R}^8$ with length squared equal to 2 such that coordinates are either all integers or all half-integers and the sum of the coordinates is even."

Even if I can derive this "definition" of $E_8$ from the Dynkin diagram then I can just count the roots to get to the number $240$

bolbteppa

4:14 PM

$E_8$ has a maximal subalgebra SO(16), you can see this by considering the extended Dynkin diagram and then deleting a spinor node i.e. deleting the simple root coming from the spinor, it is of dimension 120. When you add that spinor back you're adding a spinor rep of SO(16) to the algebra which is 128 dimensional, thus you get 248 total. The spinor alone is enough as all we did was delete the spinor simple root.

Ryder Rude

is it important to think of point derivations as functions from germs at $p$ to R or can we think of them as functions from functions to R

maybe we start with the latter, and then define point derivations on germs by taking equivalence classes

but germs are more convoluted to think of than functions

bolbteppa

That's a bit confusing and conflating things, if you take $E_8$ and delete a spinor node you get $D_7$, so we can start from the simple roots of $D_7$, you then add a simple root of the same length orthogonal to all but one, and you get $E_8$ via the simple roots, you then count the number of roots obtainable from this and you get 240, plus the 8 CSA generators and you get 248, this agrees with taking the $D_8$ maximal subalgebra along with its spinor irrep, again giving 248

Sanjana

4:33 PM

@bolbteppa Thanks...I will come back to this when I know what a spinor node is. But is there no other way? I mean something along the lines of deriving that line from wikipedia?

Silly Goose

@Relativisticcucumber me waiting for my horribly written python optimization code to minimize cost function

@Relativisticcucumber i think there is a theorem abt general unitary representations that says they’re completely reducible or some other nice property

But maybe finding irreps is different, but that seems to be the case anyways across even finite dimension groups of different types

like semisimple vs compact

bolbteppa

4:54 PM

In the picture of $E_8$, with the fork part on the right side, if you ignore the last node on the $A_7$ line you end up with the diagram of $D_7$, we can immediately see this just by simply looking at the $E_8$ diagram, this should be easy. Thus, to find $E_8$, we just need to add a simple root to the simple roots of $D_7$ that agrees with the rules $E_8$ imposes. The simple roots of $D_7$ are $e_1 - e_2$, ... , $e_6 - e_7$, $e_6 + e_7$, the simple root is $- \frac{1}{2} (e_1 + .. e_8)$ (check).

You can now count the total number of roots generated from these simple roots to be 240. Basically our roots are of the form $\pm e_i \pm e_j$, for $i \neq j=1,...,8$, giving $4 {8 \choose 2} = 112$, and $\frac{1}{2} \sum_{i=1}^e \epsilon_i e_i$ with $\Pi_{i=1}^8 \epsilon_i = + 1$ i.e. $2^{8-1} = 128$ more roots, so 240 total, and 248 elements.

Note this is now really working with the roots of $D_8 = SO(16)$ which is why we got the 112 = 120 - 8, and the remaining 128 have the same dimension as the spinor rep of $D_8$

The extra node forced us to add an $e_8$ which implicitly resulted in the extension from the roots of $D_7$ to the roots of $D_8$, along with more stuff which is the spinor rep

Sanjana

@bolbteppa I don't get the part why we should add $-\frac{1}{1} (e_1+e_8)$?

bolbteppa

$-\frac{1}{2}(e_1 + e_2 + .. + e_8)$ is the simple root associated to the extra node that we added to $D_7$ to find $E_8$, every node represents a simple root, this is the simple root which ensures (along with the $D_7$ simple roots) we reproduce the Cartan matrix associated to the Dynkin diagram of $E_8$

Sanjana

7:20 PM

@bolbteppa Got it. Thank you so much.

From the previous conversations with you...you seem pretty fluent with the theory of roots and weights. Which book do you recommend on this?

@bolbteppa Out of many other things, I wish to intuit how to write these expressions for simple roots in terms of these standard basis vectors. I mean how to do this same exercise for other simple Lie algebras...

Mr. Feynman

Anyone familiar with Altland&Simons?

bolbteppa

7:37 PM

@Sanjana Ramond's group book quickly discusses it at the end of ch. 7 for example, in general no one book is going to do everything but Georgi is one of the best

Mr. Feynman

Never mind, it's not relevant what book it is

The argument is purely a mathematical problem. See this integral?