The h Bar

Obliv

02:01

a priori equal probability distributions are so sus

i mean for practical purposes it's useful but feels so dirty

feels like classical mechanics was just a flavor of physics, but nothing I learned was fundamental in any sense. I guess learning how to learn and think systematically was beneficial..

Silly Goose

the principle of least action is a sort of "fundamental" tool :D

Obliv

can you explain like im 5 what the principle of least action is? we veeery briefly went over it in my mechanics class

$\delta S = 0$ means..?

The wiki page says it's the integral of the lagrangian w.r.t. time for some interval so I guess as that interval becomes infinitesimal the "functional" is 0?

Silly Goose

02:17

Consider classical mechanics and a particle system in it. Let $L(x, \dot{x})$ be the lagrangian of your system. Then, the action is the integral of $L$ over a time interval. The equation $\delta S = 0$ puts a constraint on $x(t)$ and $\dot{x}(t)$. This constraint is called the equation(s) of motion. Solving them gives you $x(t)$ and $\dot{x}(t)$, i.e., the dynamics of the system

in a course on classical mechanics, you will find under certain assumptions that $\delta S = 0 \implies \frac{\partial L}{dx} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = 0$, which is called the Euler-Lagrange equation, which is the general form of the equations of motion under these assumptions

Obliv

is the constraint that $T = V$?

Silly Goose

no

Obliv

hmm okay yeah he never derived the E-L equations for us but we did work with them a lot.

he said don't worry about it, just know that it works lol.

Silly Goose

the constraint is a differential equation in two variables: $x$ and $\dot{x}$

^ if your system consists of a single particle

Obliv

so the action is the integral/functional of the lagrangian over some interval of time, if we vary it infinitesimally, we find that when it's "stationary" i.e when $\delta S = 0$, we can derive the EoM for the system

Silly Goose

02:24

the stationary conditions gives us the EoM yes

Obliv

how come in my class we just derived the lagrangian directly based on the system, then just plugged into the E-L equations without worrying about the action?

Silly Goose

the E-L equations come from stationizing the appropriate action

Obliv

does it follow directly from $S = \int_{t_1}^{t_2}L(q(t),\dot{q}(t),t)dt$?

Silly Goose

the procedure in general would be 1) write down your action, 2) stationize it, 3) see what EoM is, 4) solve EoM. In a textbook in classical mechanics, we write down a general action (subject to some assumptions), stationize it, and find the E-L equations. So we do steps 1) and 2) in a more general setting so that we don't have to do it every time for particular systems. With E-L equations in hand, we can more immediately write down particular EoMs by just knowing the Lagrangian

@Obliv after stationizing it and assuming a couple of things, yes

Obliv

what does the process of stationizing look like

Is it a calculus theorem

I'm so noob D:

oh boy it's just the fundamental theorem of calculus isn't it

nvm maybe not

Silly Goose

02:32

en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation u might already be on here, but a derivation is written out here

Obliv

ooh ok thanks

Silly Goose

it looks mostly like like calculus, but (at least for me) it was hard to get used to what a variation is :P

Obliv

the problem (and good thing) about learning from wiki is the language is very precise so there are a lot of math/physics terms I have to look up and I get so sidetracked because of it :P

but it's good to get an overview on the different terms

definitely going to go on a wikipedia binge tomorrow

Silly Goose

if you would like a resource that would stick to physics (more than math) for a derivation of E-L equations, then John R. Taylor's classical mechanics seems pretty decent. if you want a flavor of the maths as well, then the supposed classic text is Vladimir Arnold's Mathematical Methods of Classical Mechanics

or if you want to go full maths you can read something about symplectic mechanics :P

Obliv

someone here was posting problems from taylor and unfortunately the ones he's posted had some typos/misprints so I'm wary of him now :P

but yeah I'll check those out to see what I missed out on..

I need to buff up my knowledge of hamiltonians too because we baarely covered that.

and it's supposedly very useful in qm

Silly Goose

02:43

i personally do not find classical mechanics to help with quantum mechanics. but i do (so far) find classical field theory to majorly help with quantum field theory

my personal bias is that if you would like to attack both classical and quantum mechanics at once: learn linear algebra and lie theory and representation theory :P

Obliv

I'm thinking about just following ted shifrin's multivariable mathematics

he covers a lot of useful math

I think he does linear forms & tensors towards the end

Silly Goose

also using only usual methods from calculus, if you treat $\delta S$ as taking the differential $dS$, then you can obtain the E-L equations

Obliv

so the total derivative of the lagrangian within the integral. hmm ok

Silly Goose

@Obliv right

Obliv

classical EM using maxwell's equations can be a classical field theory :P I don't know how much gauge invariance matters in deriving the equation of EM waves

other than it makes the derivation easier by choosing a specific gauge (lorenzian one i think?)

Silly Goose

02:50

Obliv

I can't imagine how complicated QFT gets without the simplicity of assuming smoothness in functions, symmetries etc.

Silly Goose

this is what it looks like in any case

in the final step if you assume that the variation of $x$ vanishes at the boundaries of the time interval, the second term vanishes.

Finally, the remaining eqn is true for all variations $\delta x$, which allows you to obtain the E-L eqns

Obliv

thank you ! I will digest this

eats the paper

Silly Goose

:D

Obliv

Just for confirmation, it's $$\delta \int_{t_1}^{t_2}dt L(x,\dot{x})=0 \implies \int_{t_1}^{t_2}dt \delta L(x,\dot{x}) = 0\implies \int_{t_1}^{t_2}dt \left[\frac{\partial L}{\partial x}\partial x + \frac{\partial L}{\partial \dot{x}}d\dot{x}\right] = 0$$

Silly Goose

02:55

right

i guess the quirk is that $dx \to \delta x$ i don't really know what it means :P, but I think of $\delta x$ as a variation in $x$

and recall that here $x$ is really a function

Obliv

uh... I probably shouldn't be thinking of the term variation to mean change

is it bad I think variation, curl, derivative, change are all synonymous :'D?

Silly Goose

probably depends on who you ask :P

at the least the curl is a particular type of derivative, so I wouldn't make it synonymous with the other terms you listed

Obliv

so I guess I'll leave it as $\delta x$

123

06:11

Hello Everyone...

Mr. Feynman

@SirCumference Indeed, Sir Crackpot has gone dormant

Mr. Feynman

08:03

@Obliv In a sense they are, but if you get technical you have a quite different meaning for each of those

They all describe the way something changes, the difference is what

Relativisticcucumber

monday is the deadline for phd commitment

BAH

Silly Goose

@Relativisticcucumber BAH

Silly Goose

10:07

what is the meaning of finite extension?

ACuriousMind

@SillyGoose that $\pi_0(G)$ is finite

Silly Goose

is it an extension in the sense that we start with $1 \to G \to \pi_0(G) \to 1$ and we are adding in another group $G_0$ with the inclusion map

I am also wondering: (1) Is (1.2) just referencing the definition of a compact topological space? I.e., that every open cover of a topological space has a finite subcover. (2) Where does (1.3) come from :P is this some sort of classification of finite covers or something?

(3) In (1.2), what is the role of the abelian group $A$? (4) Why are we writing all of this out in terms of sequences?

ACuriousMind

10:29

@SillyGoose you can't "start with" $1\to G \to \pi_0(G)\to 1$ because if that were an exact sequence, $G\cong \pi_0(G)$

the name "extension" comes from the idea that $G$ is an "extension" of $G_0$ by $\pi_0(G)$, cf. group extension

@SillyGoose no, and no

in (1.2), $\tilde{G}_0$ is the universal cover, and this uses that the fundamental group of Lie grous is Abelian

Silly Goose

@ACuriousMind hm does the torus in (1.3) not make the total space not simply connected? I am unfamiliar with how "simply connectedness" interacts with taking direct products of spaces

ACuriousMind

in (1.3), this is the classification of simply connected Lie groups derived from the classification of Lie algebras

Silly Goose

@ACuriousMind oh so we need to throw in the normal subgroup before $G$ in order to write an exact sequence and have $im f_1 = G_0 = ker f_2$

ACuriousMind

ah, sorry, this is not the universal covering

Silly Goose

instead of $\text{Im} id = \{1\} = \text{Ker} f_2$, which would happen if I left out the normal subgroup in the sequence

ACuriousMind

10:47

What's happening here is this: Since the representations of compact Lie groups are unitarizable, they are in particular completely reducible, so in particular the adjoint is completely reducible, and so the Lie algebra is reductive, i.e. it is a sum of an abelian $\mathfrak{t}$ and a semi-simple $\mathfrak{k}$.

This, in turn, means that there is a simply connected semi-simple Lie group $K$ such that $T\times K$ has the same Lie algebra, and hence $\tilde{G}_0 = T\times K$ is a cover of $G_0$

replace $T$ by $\mathbb{R}^n$ and you have the universal cover

Silly Goose

okay and so the semisimple $K$ splits into a direct product of the corresponding lie groups of the summands in the direct sum of the semisimple lie algebra?

so we end up with $T$ times a product of simply connected, simple lie groups?

nooo not again ;_;

maybe i should read the first bits of a more differential geometric lie theory text :P

is there a special meaning to the brackets around $\theta \wedge \theta$?

If $\wedge$ is an antisymmetric product, then I would think $\theta \wedge \theta = 0$, but that it is a special equation suggests that is not the case :P

er I guess maybe it is because $\theta$ is $\mathfrak{g}$-valued that it does not necessarily vanish?

ACuriousMind

11:11

@SillyGoose yes, the brackets are intended to be a reminder that you get commutator terms $[\theta_i,\theta_j]\mathrm{d}x^i\wedge\mathrm{d}x^j$

Silly Goose

okay i see

how would i write $\theta$'s codomain? from looking at other resources it sounds like $\theta$ should assign to each $g$ a map $f: T_gG \to T_1G$, so the codomain is like $\text{Hom}(T_g G, T_1 G)$..?

ACuriousMind

sure, a $V$-valued 1-form on a manifold $M$ is equivalently a map $T_x M \to V$ at every point; I don't know why you'd insist on writing $T_1 G$ instead of $\mathfrak{g}$, though :P

Silly Goose

i wasn't sure if it was safe to do so XD but okay i see

lucabtz

11:49

@ACuriousMind is this related to reductive groups?

Like the lie algebra of a reductive group is reductive

ACuriousMind

yes

lucabtz

@ACuriousMind also the opposite? Like a lie group is reductive iff it's lie algebra is?

I never understood the definition of reductive group, but this one for algebras is simpler. I guess kinda like with Borel subgroup/algebra

ACuriousMind

yes

@lucabtz "reductive group" is usually defined in a more general context of algebraic groups, where not everything is a Lie group

lucabtz

@ACuriousMind yeah I noticed

But that seems to require algebraic geometry which as you know I'm pretty scared of

I'd honestly would study some if I had the time too

and there are so many stuff id like to study better

such as theory of Riemann surfaces, algebraic topology, functional analysis are all things id like to study better

and then there is algebraic geometry which i find scary mostly because i see a lot of theory of rings is used there and i basically known nothing about them

oh also Lie theory and representation is something id like to study better right

Ryder Rude

14:29

hi

notmyrealname

15:09

there's two different notations for irreps of the Lorentz group. I'm familiar with the $(m,n)$ notation, but how do i translate the notation $\mathbf{k}$ to this? Is it the case that $\dim(\mathbf{k}) = k$? So, $\mathbf{1} = (0,0), \mathbf{2} = (1/2,0),\ \mathbf{3} = (1,0)$?

ACuriousMind

@notmyrealname the $\mathbf{k}$ notation is not specific to the Lorentz group, and yes, it just denotes the dimension of the rep

notmyrealname

okay, thanks. What other groups is it used for?

ACuriousMind

note that there are multiple reps of the same dimension, so this notation isn't unambiguous and you usually fix conventions in the context

It's common to use it for SU(N), specifically SU(2) and SU(3) in the Standard Model. But since it's just the dimension you might use it for any group

notmyrealname

On Wikipedia it's claimed that $3 \oplus 1$ is the adjoint representation of $so(3,1)$. This doesn't make sense to me: The dimension of $3 \oplus 1$ should be 4 but that of the adjoint representation should be 6, no?

Poincaré group

The Poincaré group, named after Henri Poincaré (1906), was first defined by Hermann Minkowski (1908) as the group of Minkowski spacetime isometries. It is a ten-dimensional non-abelian Lie group that is of importance as a model in our understanding of the most basic fundamentals of physics. == Overview == A Minkowski spacetime isometry has the property that the interval between events is left invariant. For example, if everything were postponed by two hours, including the two events and the path you took to go from one to the other, then the time interval between the events recorded by a stopwatch...

ACuriousMind

that section of the Wikipedia article is complete nonsense

(note also that it does not reference a single source)

notmyrealname

15:26

strange. Is it at least true that $2 \oplus \bar 2 = 3 \oplus 1$?

ACuriousMind

you mistyped a $\oplus$ for an $\otimes$, but that's where the nonsense already starts - for representations of $\mathrm{SO}(3)$ or $\mathrm{SU}(2)$ this is true, not for the Poincaré or Lorentz group.

$(1/2,0)\otimes (0,1/2) = (1/2,1/2)$ is irreducible as a representation of the Lorentz group

notmyrealname

thank you!

ACuriousMind

15:47

(just so that no one gets confused by the Wiki link above pointing to no specific section: I've submitted an edit to remove that section)

PM 2Ring

I thought we had a "No AI" policy regarding both answers and questions. When did that change? physics.stackexchange.com/help/ai-policy

3

19

Q: Do we want to implement the new AI policy banner on Physics SE?

Over on Meta they recently announced a new option for individual sites to have a "AI-generated content" policy banner, offering several options: We will initially offer two banner text options that all sites in the Stack Exchange network can opt-in to. Those options are the following: Reminder: ...

Also physics.meta.stackexchange.com/q/14281/123208

ACuriousMind

@PM2Ring note that that post argues (explicitly at the end) that we indeed don't have any specific "no AI" policy, since undisclosed usage of LLM output violates already other preexisting policies on plagiarism and academic honesty, which is also what the help center page you linked says

you can post LLM output, if properly attributed, and this would not be grounds for suspension (but you'll probably get a lot of downvotes)

PM 2Ring

Ah, ok. ;)

Loong

hm

That's surprisingly weak.

PM 2Ring

OTOH, attributing a chunk of text to some LLM doesn't tell you what the real source of the information is. It merely identifies the text as a piece of generated text. Which is better than not knowing that it's generated. But it's certainly nothing like the usual chain of references that science & mathematics is based on.

ACuriousMind

16:02

@Loong honestly, no one ever posts LLM output properly attributed - really the only people who use it are those trying to disguise it as their own words so they can farm rep from it

@PM2Ring sure - which is why if you just cite a chunk of LLM text you will probably be downvoted, and note that answers that "consist entirely of copied text" are also subject to deletion regardless of where the text is copied from, cf. the end of physics.stackexchange.com/help/referencing

user 85795

@Loong I agree.

hm

PM 2Ring

I'm also concerned about questions based on AI output. The OP has asked the LLM some question, they get an answer that looks right, but with maybe some puzzling aspects. Then they ask us to verify that answer. That can get messy, especially if the OP has some common misconceptions baked into the question.

Sure, we get that sort of thing all the time anyway, eg twin paradox, one-way speed of light. But when LLM is involved the OP can get into much deeper rabbit-holes than usual.

ACuriousMind

I mean, that's true, but again the worst of these questions are already off-topic by other rules (e.g. just asking whether some nonsense a model generated is correct is often off-topic as non-mainstream)

user 85795

and those deep rabbit-holes are full of word-salad for the rabbit

ACuriousMind

the thing that makes the question bad is the content, not necessarily the involvement of an LLM

user 85795

16:15

or rabbits :P

Loong

Nevertheless, I still don't think that all the users who upvoted
"Please don't use computer-generated text for questions or answers on Physics"
actually meant
"Generative artificial intelligence (a.k.a. GPT, LLM, generative AI, genAI) tools can be used to generate content for Physics Stack Exchange, but this content must be properly referenced"

4

PM 2Ring

^ This

ACuriousMind

I feel this is splitting hairs: What kind of answers that disclose which parts are computer-generated and that are not deletion-worthy by other pre-existing rules are you worried about here?

user 85795

@Loong why does your profile say "Member for 3 years, 4 months" but also "Moderator from 2015 "

ACuriousMind

Yes, technically there is a difference between forbidding the undisclosed use of these models and all use of these models; in practice I have not seen any disclosed usage at all except for a few questions in the vein of what PM2Ring just described

the big stink about the network policy (cf. physics.meta.stackexchange.com/q/14438/50583) was because it was interpreted as an attempt to remove moderators' ability to punish the undisclosed use of the models; I'm fairly confident that is what most people are really concerned about

PM 2Ring

16:26

I suppose so. If it's disclosed content, we can safely ignore it. ;)

Loong

@ACuriousMind I am not concerned over any particular answer. If something is clearly written in bad faith, you may act on this even if the particular case is not mentioned in the help center at all.

For me it's more a matter of principle. If you really don't want A, don't say "you may do A but you also have to do B" and hope that users will trip over B. Just say "don't do A".

PM 2Ring

Ah, but SE Inc have made it a bit tricky to moderate people doing A, unless they've disclosed they're doing A. So it's simpler to use the "trip them up with B" tactic.

user 85795

and add in C, D, E,... at their discretion :P

PM 2Ring

At least we don't have the level of AI use that happens on SO. I believe there's still a huge backlog of "suspected GenAI" flags on SO. And some of the mods are still on-strike.

Loong

@PM2Ring Okay, if that's the real problem, I don't mind a little detour.

user 85795

16:33

Wow, still?

ACuriousMind

@Loong well, but at least I personally don't know that I don't want A - maybe there is, now or in the future, some legitimate use for including LLM generated content into an answer that also contains some human-written bits. While I personally doubt it, that may be a limit of my imagination. What I know I don't want is people copy-pasting ChatGPT output into the answer box to farm rep, and that's what the current policy succeeds in preventing

user 85795

some are still waiting for Monica to return :(

PM 2Ring

stackoverflow.com/users/366904/cody-gray-on-strike stackoverflow.com/users/6296561/zoe-is-on-strike Zoe is still active on the MSO site, but I haven't seen Cody in ages.

user 85795

yeah, SO is a jungle

PM 2Ring

I wouldn't mind people using language models as a super spelling checker / grammar checker / style booster. But with current LLMs there's a huge risk that it will change the author's intentions. And the author may not realise it, if they don't have enough expertise, or aren't scrutinising the LLM output sufficiently. It's too easy to develop a false sense of security because of the indefatigable self-confident tone underlying the LLMs utterances.

PM 2Ring

17:08

I made a Horizons script for comparing angular diameters of Solar System bodies. astronomy.stackexchange.com/a/57419/16685 You can use it to see what bodies are capable of causing eclipses, etc. Eg,

Apollo 11 LRRR is the Laser Ranging RetroReflector placed by Buzz Aldrin.

More Anonymous

17:36

My kind of playlist:

https://www.youtube.com/watch?v=d-o3eB9sfls

https://www.youtube.com/watch?v=_03sfhb7Mvg

DanielSank

19:11

Hi, everybody.

I haven't been around for a while. What's new?

user 85795

wb pal

Mr. Feynman

Is there a scalar product suitable to invert the mode expansion in ST? Something like $$a_p\propto\int d^3 x e^{ikx}\overset{\leftrightarrow}{\partial_0}\phi$$ in QFT

Sorry, that's $a_k$

The mode expansion is $$X^\mu=x^\mu+\alpha' p^\mu\tau+\sum_{n\neq0}\frac{\tilde{\alpha}^\mu_n \mathrm{e}^{-in\sigma^+}+\alpha^\mu_n\mathrm{e}^{-in\sigma^-}}{n}$$

I tried $$\int_0^{2\pi} d\sigma e^{im\sigma^\pm}\overset{\leftrightarrow}{\partial_\tau}X^\mu$$

Where $\pm$ determines if we take left movers or right movers

Since both exponentials are written with the same sign, it doesn't seem to work and I should replace $\overset{\leftrightarrow}{\partial_\tau}$ with "symmetric" version with a plus

ACuriousMind

19:36

@Mr.Feynman It's just the formula for Fourier coefficients after you've split the $X$ into $X = X_L(\sigma^+) + X_R(\sigma^-)$

but it only holds on-shell

Obliv

@DanielSank not much, how is the quantum ai lab

ACuriousMind

in that way the ST expansion is closer to a true Fourier expansion than the mode expansion of the free field

Mr. Feynman

@ACuriousMind Yes, you're right. This avoids derivatives too

Yet, left and right movers are so similar to positive/negative frequencies I hoped there would be some analogy

Well, I always wondered why in the free field expansion of QFT we end up with that "weird" scalar product

ACuriousMind

it's because the "modes" are a mixture of the Fourier transform of the field and its canonical momentum, and the canonical momentum involves a time derivative

there's no canonical momentum involved in the mode expansion of ST, hence no derivative

Obliv

wait what $\log$ is a trig function?

ACuriousMind

19:50

no?

Silly Goose

Log is a piece of wood

2

Obliv

weird, maybe they meant trig function in the sense that it's taught first in a trigonometry class or something

(not that we have trigonometry classes anymore where I'm from)

Mr. Feynman

@ACuriousMind Uh, I never thought about that bit about the canonical momentum, makes sense. I always justified that product with its appearance in the Noether charge

Good insight

@SillyGoose legendary on Clash Royale

Silly Goose

XD

If it’s called topological quantum field theory why do i need all this algebra 🤔 lol jk

Obliv

are you a $\sum\limits_{n=0}^{\infty}\frac{1}{n!}$ simp or a $\lim_{n\to \infty}(1+\frac{1}{n})^n$ chad

@SillyGoose dude at that level you need everything

Silly Goose

19:57

Does anyone know of a textbook which covers chern-simons theory? or even a good text which would contain the math preliminaries to learn CS theory? I am currently trying to work through a set of notes but i feel i would benefit from the level of thoroughness that a textbook may provide :P

ACuriousMind

@SillyGoose that's the level at which you don't find "textbook", you find "monographs" :P

Silly Goose

@Obliv hopefully not analysis ;D

Blebs well if that’s the best one can do

Obliv

I think i had a nightmare I was trying to construct $\mathbb{R}$ using something called "dedekind cuts" which I don't know what they are but my subconscious tried to imagine what they were

I think I thought they were sections of $\mathbb{R}$ that you just string together with some rope

ACuriousMind

you will probably not find a more elementary discussion of CS theory than in Witten's original paper

Obliv

like a finite interval plane that you tie string around the numbers :D

Silly Goose

20:02

Is Quantum Field Theory and the Jones Polynomial this the original paper?

ACuriousMind

yes

Obliv

it makes me sad how difficult it is to draw $\mathbb{Q}$ irl

compared to the other Mbb characters

Slereah

Just write a Q and do a line on it

Obliv

like on the right side?

Slereah

I do on the left

Obliv

20:12

what's the canonical way

does the pope do it on the left or right, or maybe both?

ACuriousMind

I mean you don't really hand-write $\mathbb{C}$ etc. either, do you?

Obliv

in order to distinguish it as the set/field isn't that something people do? I rarely write in general (I just write up assignments using overleaf)

ACuriousMind

you just draw a C and then a straight line inside it, also usually to the left

i.e. the "draw a line somewhere on the left" is in fact how people write most of these symbols

Obliv

yeah $\mathbb{C}$ isn't so bad, $\mathbb{Z}$ was kinda tricky for a while but $\mathbb{Q}$ was just atrocious looking.. like a grade schooler's attempt at graffiti D:

I understand $\mathbb{C}$ is complex, $\mathbb{Z}$ is zahlen, but wtf is $\mathbb{Q}$ anyway

quotient?

ACuriousMind

it's quotients

Obliv

20:15

so $\mathbb{Q}/A$ is a quotient ring of the quotient ring

DanielSank

21:03

@Obliv It's very good. Lots of progress.

Silly Goose

there is a neat site called mathpix which turns images of latex into latex that you can copy and paste. useful for if you are typing up notes but using well known expressions that are in other resources :P (or if you're doing topology homework...)

@DanielSank cucumber and i are committing to phd programs this monday :D

Mr. Feynman

@ACuriousMind on a second thought, I'm not sure I agree on this statement. If you expand the conjugate momentum $\Pi_\mu=\frac{1}{2\pi\alpha'}\dot{X}^\mu$ you are in a similar situation. The string modes are a combination of the field and the conjugate momentum Fourier transforms too

Sure, it works using the Fourier decomposition of left movers and right movers as you mention, but it's not wrong to proceed like in QFT apparently (mutatis mutandis in the scalar product)

DIRAC1930

@SillyGoose What PhD programme have you decided on?

Silly Goose

im not sure yet :0

Obliv

21:20

you should decide on the best one

imo

DIRAC1930

Did you say you wanted to do condensed matter?

Silly Goose

Yes

i think topological matter stuff would be cool 0:

DIRAC1930

Sounds like interesting stuff

Silly Goose

21:39

In an ideal world i think id like to work on axiomatic qft or something similar:P

Obliv

you should put that on your profile

in an alternate universe you work on axiomatic QFT :D

but in this one you work at a goose company

DIRAC1930

In an alternative universe, I would work on alternative universes

Silly Goose

h o n k

I wonder what los alamos's interest in decoherence is :P

since it seems usually the theory department of a national (in the US) lab is still on like nuclear physics or materials physics or something like that

DIRAC1930

There are so many interesting areas of physics

But I always end up getting drawn back to the same things

Silly Goose

what are those things for you

Mr. Feynman

22:42

@Obliv what if in that alternate universe Silly Goose is a swan?

Silly Goose

22:55

or a penguin :D

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