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3:29 AM
Sorry @Mr.Feynman, saw the guidelines now. I didn't mean to bother. will be careful next time.
@Amit I did search on google, but most websites are oriented towards 12th grade JEE physics in India, and I couldn't find something relevent. So I thought maybe asking on SE would give help of some kind.
Clear this for me sir; are requests like these generally considered a bad practice here? I asked something like this previously also and people do not respond to such queries.
 
 
1 hour later…
4:38 AM
@ShikharChamoli You're free to make requests. Other people are free to help or to ignore such requests. Pinging people you're not in a conversation with is usually considered bad form.
 
Has the font that people's names are rendered in got smaller?
Maybe it's just early (05:30 here!) and my brain isn't functioning yet but the names look bit smaller ...
 
rob
@JohnRennie I see smaller type next to shorter messages.
 
Aha, yes!
A very long message written with the sole aim of seeing if the text next to this message is larger than the text next to shorter messages.
Hmm ...
 
5:31 AM
@ShikharChamoli example from Googling "graduate physics courses"
 
 
2 hours later…
7:17 AM
hello i am wondering how i should interpret $\frac{1}{H}\vert n \rangle$ -- is this just $\frac{1}{H\vert n \rangle}$?
 
7:28 AM
@Relativisticcucumber How do you want to divide by $\lvert n\rangle$?
 
im not sure what you mean. im having this question because theres a part of sakurai that says "the inverse operator $\frac{1}{E_n^{(0)} - H_0}$ is ill defined because it might act on $\vert n^{(0)} \rangle$" and i was assuming this meant what i guessed above but im not sure
 
You ask "is this just $\frac{1}{H\lvert n\rangle}$" but that doesn't mean anything: You can't divide by a vector in Hilbert space
 
i see -- yes that is also why i am uncomfortable with this quote in general because i dont know how to interpret the operators being in this form. so i guess my guess is wrong.
 
"the inverse operator" is literally just the inverse of an operator: For any operator $O$, the inverse $O^{-1}$ has $OO^{-1} = O^{-1}O = 1$
for any eigenstate $\lvert o\rangle$ with $O\lvert o\rangle = o\lvert o\rangle$, you then have $O^{-1}\lvert o\rangle = o^{-1}\lvert o\rangle$.
when $o = 0$, you hence have a problem - the inverse operator cannot act on the eigenstate with zero eigenvalue since $0^{-1}$ does not exist
in your case, $\lvert n^{(0)}$ is an eigenstate of $E_n^{(0)} - H_0$ with eigenvalue 0.
 
yes i see i see
okay thank you
and i have one more question. sakurai also says that even for $\lambda \neq 0$ (weight factor for perturbation) we can always add to $\vert n \rangle$ (solution to perturbed problem) $c_n \vert n^{(0)} \rangle$ (solution to unperturbed problem), and i do not see why this is the case when $\lambda \neq 0$
 
7:47 AM
I don't know what "can" means in that sentence
you can always add two vectors
 
isnt the point that $\vert n ^{(0)} \rangle$ is not a solution to the perturbed problem?
okay so if im understanding this section correctly, we are trying to find the form of $\vert n \rangle$ and we come to an expression, but this expression does not satisfy that as $\lambda \rightarrow 0$, $\vert n \rangle \rightarrow \vert n ^{(0)} \rangle$, so then he says "but we can add $c_n \vert n ^{(0)} \rangle$" and then that gives us the form of the proper eigenstate $\vert n \rangle$
 
8:08 AM
@Amit : Yes, it is indeed more difficult to see if one has already voted with the new voting buttons, especially on Phys.Meta.
 
I've seen the idea that the "increased voting" they saw during their tests (the stated reason for the redesign is that it increased voting in A/B testing by a lot) is just confused people clicking the button more than once to figure out whether they have already voted or not.
 
by the way i had a question about the resistance to this AI policy stack put out. to my understanding, the ban is against removing ai generated content on the grounds of it being ai generated. if the claim is that this ai generated content is bad or misleading, cant all the ai generated content still be removed on the grounds of being just wrong or bad? and if the content is perfect, then who cares who generated it?
so it seems it doesnt really make a difference whether its allowed or not if it can still be taken down regardless?
 
@Relativisticcucumber Historically we have held the firm position that moderators are not judges of correctness - we do not delete content for being "bad", we only delete plagiarized and harmful content. "Flags should not be used for technical inaccuracies, or an altogether wrong answer" is a default reason to decline flags. I tried to express this in the second paragraph of our meta post.
 
oh interesting. but isnt wrong content harmful?
 
@Relativisticcucumber By "harmful" I meant things like personal attacks or spam
 
8:21 AM
i guess my question then is if a user posts a question and another user answers with, theoretically, the most beautiful answer ever, but its plagiarized (lets say completely ai generated), then who does this hurt?
/what negative impact does this have
 
Hello, how is the volume of phase space exactly related to the energy of the system? Does the volume of phase space being constant mean that energy will also be constant?
 
I disagree and think we should remove at least obviously wrong content
 
@Relativisticcucumber The whole reputation system is based on the assumption that posts are original work.
And sites on academic topics like physics.SE generally have cultures based on academic integrity: Passing off someone else's work as your own is simply wrong, not because it "hurts" someone specifically but because it more generally undermines the mutual trust between scientists that science is built on.
 
i guess i think if someone can use ai to such a degree to generate this content where it racks up reputation reliably then they are effectively as useful a resource than someone who knows the stuff from their own brain. ofc this argument is not valid for ppl using ai to produce nonsense which i saw is an issue but yeah i guess i thought content could be removed or downvoted for being wrong
but i see the situation now so thanks for explaining
 
Oh, you can downvote content for being wrong - that's what downvotes are for!
 
8:29 AM
got it got it
 
We all know that doesn't always work
 
@Slereah I agree that it doesn't always work! And users with >20k rep can vote to delete posts as they see fit. My position is not so much that no one should delete posts for being wrong, but that that's not what moderators should do. Quality judgement by votes is up to the users.
@Relativisticcucumber Another response to the line of thought that "if the AI is useful why should we delete it?" is that if the users wanted a computer-generated response they could just ask the machine directly without going through an SE site as an intermediary. We're not depriving anyone of anything by taking a stance against computer-generated content: If that's what you want, you can just get it directly from the source
 
just install the AI directly on the site so that users may ask it
also give it that image :
or this maybe
 
9:17 AM
So I've managed to confuse myself again. So if I say the universe is isotropic and homogeneous usually I find the killing vectors which are spatially maximally symmetric and discover the flrw metric
My question is does it make sense to say the deviation vector should isotropic and homogeneous?
 
The action of Killing vectors on objects is defined by the Lie derivative
Just check out that $\mathcal{L}_K X = 0$
which is indeed a well defined notion
 
Thanks!
It's been a while I've revised this
 
anything you can lift a diffeomorphism to can get that treatment basically
 
 
1 hour later…
10:29 AM
@Qmechanic Right! Thanks for taking note of that too
 
10:46 AM
@Slereah
Since $u^\alpha \nabla_\alpha \eta^\beta = \eta^\alpha \nabla_\alpha u^\beta$

where $u$ is the 4 velocity of the geodesic and $\eta$ is the deviation

The lie derivative is $ 0 = L_V \eta = v^\alpha \nabla_\alpha \eta^\beta - \eta^\alpha \nabla_\alpha v^\beta$ where $v$ is the spatial displacement: Can I claim $v^i \propto u^i$? where $i$ are the $3$ components?
 
idk
 
maybe ask on PSE?
 
only you can judge
 
I thank the gods for the ability of freewill :P
 
11:02 AM
I would think it unlikely that a vector field would be isotropic though
unless it's the zero vector field
 
@Relativisticcucumber Write down the the angular momentum in terms of radius and tangential momentum. Show that, to conserve angular momentum, if the radius is small, then the tangential momentum is huge. Show then, that KE of the particle will become so large that it becomes classically forbidden.
@ACuriousMind I just answered a question on the main site, whereby the lack of hats contributed to a misunderstanding.
 
@Slereah I may not have asked it PSE if you posted this earlier
 
@RyderRude There is a whole bunch of authors who have decided that the two-state, spin-half, system is the correct intro to quantum theory. And they are somewhat correct. It is quite nice to see somewhat of a consensus forming.
 
11:58 AM
There is little more frustrating than trying to find a paper that you read like a month ago.
I've even got a citation, though it doesn't make any sense. What the hell is PRSL A? They were apparently publishing back in 1954, though I'm guessing they shut down a while back.
 
It is why I include a bibliography section on sketches of articles on my site
So that I remember where the cool articles are
 
Hah! Proceedings of the Royal Society A. Stupid Brits.
@Slereah unfortunately I just scribbled the citation in the margin of a sheet of paper.
 
That's essentially what my site is
Virtual piece of paper to scribble things
 
lol
 
12:25 PM
A simple proof of Coarea formula?
 
@Slereah who pays for the server cost?
 
me
 
Please see Across the Spiderverse. It has gr8 art
@naturallyInconsistent this approach is good too. I personally prefer Poisson brackets as the intro
And then u get infinite dimensional spaces aftr replacing Poisson with Commutator
 
12:56 PM
I think Poisson brackets are a bit too new to students. It will seem like magic.
Instead, I realised that before we can do 2 state systems, we actually have to first deal with 1 state systems.
 
Good point.
 
1:09 PM
Is there a word for turning a theory covariant
Like replacing $\delta \to g$ and $\partial \to \nabla$
Like quantization but relativization
 
People do a lot less of "promoting the equations to become applicable to curved manifolds"
 
@Slereah "minimal coupling"
at least in the context of gauge theories
 
@naturallyInconsistent 1 dimensional hilbert wud make QM look dumb
 
@naturallyInconsistent bit wordy
The one dimensional Hilbert space is $\mathbb{C}$ which is a theory of one state
and one operator which is the identity
it is a bit small
 
@RyderRude No, a detector could detect or not. Over a screen the distribution would be non-trivial.
@Slereah "The curved spacetime version is:"
 
1:16 PM
@naturallyInconsistent but that's an infinite dimensional hilbert space. There is one location 4 each detector
 
@RyderRude Hence why I said it is not so simple.
 
It's just $\psi (x)$
 
My trick is to take a step back and get the probability amplitude distribution in the classical setting, so that I'm really cutting through a lot of crud and getting to the quantum world faster.
Anyway, I gtg
 
are there coordinate transformations that the jacobian matrix cannot account for?
 
cya
 
1:20 PM
Do u think the average person can learn GR?
Or is it like for very intelligent students?
 
@Relativisticcucumber The Jacobian matrix gives you the derivative of the actual coordinate transform
 
Maybe there cud b such a thing as a non-differentiable co-ordinate transformation? @Slereah
 
There can indeed
 
Then i think Jacobian wud not work for it
 
You can also have "large" coordinate transformations
 
1:26 PM
@Relativisticcucumber What does it mean for "the Jacobian matrix" to "account for" a coordinate transformation? A Jacobian matrix is just the matrix of derivatives of a differentiable map.
 
@Slereah what does this mean
 
the classic example is en.wikipedia.org/wiki/Dehn_twist
 
Thank u
 
sorry i cannot get my expression to evaluate gimme a sec
im afraid to try again lets see
the way we perform a coordinate transformation is $x^{\mu^\prime} = \frac{\partial x^{\mu^\prime}}{\partial x^{\nu}}x^{\nu}$ right?
thank god
@ACuriousMind i thought the derivative entity in my above message is the jacobian matrix and i thought this expression is how we transform coordinates
 
@Relativisticcucumber I think you tapped into one of the classical traps of index notation: Coordinates are not vectors.
 
1:33 PM
We just do co ordinate transformations as $(x', y', z' t') =f(x, y, z, t)
It's just four differentiale functions 4 the new co-ordinates @Relativisticcucumber
The vectors transform linearly becuz that transformation is induced by this transformation
But even that is a choice. It is tru only if we choose to work with the co ordinate basis for vectors
 
@ACuriousMind what would be a precise definition for coordinate transformation then? i guess im misunderstanding something
 
For a start if this is how coordinate transformation worked, you wouldn't be able to do translations
 
@Relativisticcucumber honk
 
@Relativisticcucumber A coordinate transformation is just any diffemorphism $f: \mathbb{R}^n\to\mathbb{R}^n, x\mapsto f(x)$. Any (tangent) vector $v^\mu$ transforms under this transformation as $v^\mu \mapsto \frac{\partial f^\mu}{\partial x^\nu}v^\nu$. (Physicists for some reason love using the deeply confusing notation $x'$ for $f$ and then doing weird stuff with the prime)
 
@ACuriousMind maybe also invertible
 
1:37 PM
\o @आर्यभट्ट
 
I think learning GR is a big hurdle for me. I'm not sure anymore
 
@ACuriousMind but if i want to do something basic like go from spherical to cartesian coordinates i dont see how the first statement here gives me a means of doing that but it would seem that the second would?
 
did you learn the things you're supposed to learn before GR
 
Even though we write the individual coordinates in index notations also with a single upper index, i.e. $x^\mu$ or $x'^\mu$ or $x^{\mu'}$ or whatever, the coordinates are not (tangent) vectors, and the above formula with the Jacobian matrix does not apply to them
 
@Slereah what r those topics?
 
1:39 PM
@Slereah i thought about this. i was trying to derive all elements of the poincare group and failed to be able to get the translation elements
 
@Relativisticcucumber that is indeed the problem
 
@Slereah i've not learned statistical mech, nuclear physics and plasma theory stuff
But how can these b relevant for GR
 
@Relativisticcucumber What do you mean by "a means of doing that"? Spherical to Cartesian is just the map $(r,\theta,\phi) \mapsto (r\sin(\theta)\cos(\phi), r\sin(\theta)\sin(\phi), r\cos(\theta))$.
 
@ACuriousMind this might be an ill formed question but in general, how do i construct the diffeomorphism you mentioned above
 
Ooh I've also not learned Solid state physics
 
1:41 PM
@Relativisticcucumber Construct it from what?
 
i guess i see how i can write down specific maps but i was hoping for something general?
 
The diffeomorphism group is infinite dimensional, it may be a bit big to construct in general
 
Why are you trying to "construct" coordinate transformations?
 
You can construct most of them via Taylor series I guess
 
actually i am trying to construct the most general definition for "Coordinate transformation"
 
1:43 PM
You can literally write any diffeomorphism for $f$ up there
there's nothing to construct because there's no further restrictions on what a coordinate transformation is in general
(actually arbitrary coordinate transformations are even more general because here I pretended we're on $\mathbb{R}^n$/a single patch of the manifold)
 
Yes, in general, a we need a collection of maps for the whole manifolfd
It's called an..... atlas
 
ook and when you say "A Jacobian matrix is just the matrix of derivatives of a differentiable map." then this matrix is the derivative of the maps in the diffeomorphism you mentioned above?
 
Yes
The diffeomorphism is a map from $\mathbb{R}^n$ to $\mathbb{R}^n$
 
It's becuz non linear transformations locally behave linear, so u can think of them using a matrix @Relativisticcucumber
 
@Relativisticcucumber yes
 
1:46 PM
The Jacobian is the derivative with respect to the coordinates, so that it becomes a map from $\mathbb{R}^n$ to $\mathbb{R}^{n \times n}$
 
So the Jacobian can b thought of as what the co ordinate transformation does to the tangent space at a point @Relativisticcucumber
 
thank god
ok thank u all
 
Yes
 
If u fix the origin at that point, then the nearby co ordinates shud transform approximately according to the Jacobian, i think
 
Vectors can be considered as the derivatives of curves on the manifold, hence you can show that the transformation of coordinates act on that derivative through their own derivatives
 
1:48 PM
Becuz then the nearby co ordinates can b identified with the tangent vectors
There is some similar idea about Riemann Normal co-ordinates, which identifies the nearby co ordinates according to the parameter of geodesic evolution in the direction of a tengent vector from that point.
I read it very long ago, maybe the name cud b wrong
 
for Lie groups, is there a nice expression for transforming a vector in the tangent space of the identity to a vector in the tangent space at an arbitrary point?
 
@SillyGoose What do you mean by "transforming"?
 
@SillyGoose Most Lie groups are not flat space, so you can't really transport a vector uniquely
 
Like, there are infinitely many different maps one can imagine between two vector spaces (which is all you're asking for)
what particular properties do you want your "transformation" to have, and why?
 
I guess any two points on a Lie group are connected by a unique geodesic?
 
1:56 PM
hm well i'd like to be able to start with the data of the lie derivatives at the identity which i understand to just be a set of generators of the lie group, and then use a mapping (which i was told that perhaps conjugation by arbitrary group elements would work) to map them into all the lie derivatives at an arbitrary point in the lie group
idk if that makes sense
 
Well not necessarily
Not true on the sphere
 
@SillyGoose why do you want to do that?
 
and this is specifically for $SU(2)$ which perhaps has nice properties?
 
I think this may be related to the evolution formula $\frac{dP}{da}= [P, Q]$. It translates the generator $P$ by a parameter $a$ according to the evolution generated by $Q$
 
@ACuriousMind the usual form of the derivative of the exponential mapping is a bit intractable to use in this case so it would be nice to start with derivatives at the identity (which are easy to compute) and then have some nice mapping to take them to derivatives at arbitrary points. but bigger picture i need to differentiate something like $\exp(-i\Sigma_i \theta_i \Lambda_i)$ w.r.t. a single $\theta_i$. Here, $\Lambda_i$ are generators (the generalized gell-mann matrices)
and the directional derivative at the identity is a nice form because it will be $\frac{\partial}{\partial\theta_i} \exp(-i\theta_i\Lambda_i) \lvert_{\theta_i = 0}$, i.e. no "translation" term therefore no need to consider commutators and so on
 
2:03 PM
Could I get feedback on this question? It's marked as needing clarity, and a couple of rounds of edits haven't fixed that. I'm not really sure where's it's unclear, honestly.
https://physics.stackexchange.com/questions/754123/does-bremsstrahlung-disprove-the-classical-conceputalization-of-light-as-a-movin
 
@SillyGoose The canonical way to identify $T_g G$ with $T_e G\cong \mathfrak{g}$ is via $(L_g)_\ast : T_e G \to T_g G$ for $L_g : G\to G, h\mapsto gh$ is left translation by $g$.
 
Maybe it just falls victim to Betteridge's law of headlines.
 
@ACuriousMind is $T_g$ the lie derivative (?) at $g \in G$?
 
that is the tangent space
 
2:05 PM
@SillyGoose No, $T_x M$ is the notation for the tangent space of $M$ at $x\in M$
 
hm okay so is this accurate: given Lie group $G$, $T_eG$ the generators of $G$ form a basis for $T_eG$?
 
The generators of $\mathfrak{g}$
Oh wait I guess that's maybe correct
The infinitesimal generators I guess
 
e.g. for SU(2) an example is that the pauli matrices generate SU(2) and the pauli matrices form a basis for $\mathfrak{su}(2)$
 
@WaveInPlace Like we've been telling you with your other questions about photons and classical EM, the question doesn't really make sense because you seem to be working with a model of what a "photon" is that is not the model of mainstream QED, but there is not enough information to discern what exactly that model is, so no one knows what misconception to correct.
 
@WaveInPlace Decided to delete?
 
2:08 PM
okay i see
so let's say i have $SU(2)$ and take the pauli matrices as my infinitesimal generators. Let $g \in G$. Then will $g \cdot \sigma_x$ be an infinitesimal generator in $T_g G$?
 
@Amit, no it timed out. Life's been busy.
 
Do u guys think learning GR requires high intelligence
 
@RyderRude I don't think "high intelligence" is a useful concept
 
@WaveInPlace How does a question time out?! :)
 
and more generally do infinitesimal generators get mapped to infinitesimal generators by $(L_g)_*: T_eG \rightarrow T_gG$?
 
2:11 PM
@Amit by being deleted by the roomba
@SillyGoose Yes
 
oh, but I thought that's only for very old questions
 
excellent :D
 
@Amit as you see on that page, unanswered questions scored negatively will be deleted after 30 days already
 
@ACuriousMind, I thought this hewed pretty close to the mainstream. The genesis was me watching a youtube on the Thomson derivation of the Larmor formula, and wondering how that fit with the "changing electric field leads to changing magnetic field" adage.
 
Ok, got it.. I just thought since WaveInPlace mentioned it now, it ain't that old :)
 
2:13 PM
Can the average person learn GR assuming they're interested?
 
@WaveInPlace You are simultaneously talking about photons and definite values of the electric and magnetic fields as if these existed at the same time. But in the description where you have photons, the electric field is an operator, not a definite value. In the description where you have definite EM fields, there are no photons.
This is not specific to EM: If you try to mix the classical and quantum descriptions instead of properly thinking about the limiting processes that turn one into the other, you're going to arrive at nonsense. That's not a failure of physics.
 
@ACuriousMind hm so when would the usual way of computing the derivative of the exponential mapping come in handy? this method of translating the lie derivative seems much easier?
 
@ACuriousMind So much hate for semi-classical physics
 
@ACuriousMind I'm not trying to mix the two, am I? I'm trying to square the predictions of classical EM and the experimental outcomes.
 
@SillyGoose ::shrug::
 
2:16 PM
Basically just pointing out that the two don't match here, akin to the photoelectric effect.
 
@WaveInPlace Then what is the question? All you're saying is "the classical description cannot explain this" but we already know that the classical description can't explain everything light does!
 
The way the classical description broke seemed bigger than just bremsstrahlung
That was the question. Does this breakage also extend to the view of light as just propagating changes in electric/magnetic fields.
Because the Thomson derivation describes radially distributed, propagating changes in electric/magnetic fields, and those don't lead to light.
Not radially distributed light, anyway.
 
How is Wald's book for GR
 
it is a fine book
 
I have studied upto page 40-50, then i cudnt move ahead
 
2:28 PM
I wouldn't recommend it as a first GR textbook
 
It's not a good introduction no
But it's a fine second book probably
 
I will try Carroll's book again
 
The question also is why you couldn't move ahead, mathematical reasons or something else?
like just 'cause it's boring maybe? lol
 
It was mathematical reasons. Identities involving indices
 
you're not gonna get less indices in another GR book I fear
It's an indices jamboree in GR
 
2:31 PM
Oh, yeah so either a more gentle intro book like Carroll's, or work on your Math foundations...
 
I will try to brute force
 
I like the indices
It's the easiest part of GR for me, lol
 
Wow
I think i will need to sharpen my brain lol
 
You need to actually do some computations with indices to start to like them, in my experience
 
Indices are fine, the hard part is when you have to stop using the indices and do the actual sums
 
2:33 PM
Very few people get the first time what's happening just trying to follow along when someone else manipulates index equations
 
Having to compute a Riemann tensor by hand is not fun
 
but if you've done a few exercises and gotten a feel for how these things behave, they start to click into place
 
Thanks 4 the advice. I was only reading Wald's book last time
I will compute myself this time
Do u think GR requires a smart student?
 
24 mins ago, by ACuriousMind
@RyderRude I don't think "high intelligence" is a useful concept
you keep asking the same question, you'll keep getting the same answer :P
 
a patient student is probably more important
 
2:36 PM
Y do u think high intelligence is not a useful concept
 
@RyderRude What exactly do you mean when you call someone "smart"?
 
They can visualize stuff in their head very well and do computations fast
Like, say, board game players
I mean the masters
I have problms with visualising situations
But yeah, this is only one type of intelligence @ACuriousMind
 
The ability to visualize things in your mind (your "mind's eye") is not universal and some people seem to lack it entirely, cf. aphantasia
 
@RyderRude I can hardly pass the opportunity to refer you to this at this point
 
and why would you need to do computations fast? No one is rushing you.
the book won't mind if you take an hour to figure out a single page
 
2:40 PM
@Amit ive seen Feynman talk this stuff but i think he is just being humble
U dont win nobel without being exceptional lol
Feynman used to do exceptional stuff in this teens
 
@RyderRude Maybe -- but I was actually referring more to what he says after the being humble part, about that people have different ways of thinking.
Because you asked about the visualization...
 
@ACuriousMind yes, i also value computations less
@Amit i think there r too many types of intelligence
So I should not limit it to computations and visualisation
 
The thing is also, if you like something, you don't care whether you're smart enough to do it well or not, as long as you can do it at all. It's not like you need to pass an entrance exam to open Wald's book, or see an online lecture, or browse Physics.SE, etc. :)
 
If you want to do physics to get a Nobel prize I fear you may be disappointed
It's not even a good choice to get paid
At least enjoy it
 
@Amit exactly! I really try to not care about how good i am at physics.
 
2:53 PM
Also I know that nobody ever listens to that advice but it is generally a good idea to learn the basic fields before learning GR
Not gonna be easy to learn differential geometry if you can't even do geometry
 
Basic fields, classical fields you mean?
 
The basic disciplines
if you prefer
 
no, Slereah is saying you gotta understand triangles, man
 
I know triangles
 
2:55 PM
Oh, I see. I thought I knew triangles! But then someone posted this thing on Math.SE... long story short, the challenge was to do it without Trig. I couldn't do it. But I feel a bit better 'cause it took a week or two for someone to post a correct answer :)
 
I think i will b able to learn GR
It's just that i havent been studying much
 
@ACuriousMind is the subscript asterisk used here to mean "post-composition"?
 
is that image displaying
 
24 hours ago, by ACuriousMind
The $\ast$ is a limb: $f_\ast$ is the pushforward - the $f$ is kicking something forward with its leg. $f^\ast$ is the pullback - the $f$ is pulling something backward with its arm
 
@Slereah lmao
 
2:56 PM
@Slereah unfortunately yes :P
 
it's all true
 
lol i forgot about that meme format
 
In case you're interested - Do it without Trig! Without the Pythagorean theorem, too. Only basic Euclidean Geometry
(and mind the answer... spoilers...)
 
Do u guys watch Spiderverse
There is a sequel that is out
 
May 6 at 20:50, by ACuriousMind
I'm so bored of these superhero movies and other large franchises
 
3:00 PM
my mommy won't lemme watch TV before I finish Wald's book
 
@ACuriousMind this is non-conventional art
Pls try spiderverse
Its not like mcu and dc
 
there's a man dressed like a spider
sounds pretty like superhero stuff
 
It is a superhero movie, yes
But it has deep themes about responsibility and discovering yourself
And it is animation. That too, non conventional animation
I mostly dont watch this stuff
 
@Slereah lol, that's a good one
 
Every movie has themes
 
3:05 PM
Maybe there should be a superhero animation called Geometric legends
 
That's a pretty low bar to pass
 
Push-forward and pull-back will be evil brother masterminds
 
Its plot isnt conventional like an alien attack and sky beam
 
That can transport you to whatever dimension they like and try to trap you
 
Most superhero movies have a done to death plot
 
3:06 PM
The hero will start as a partial derivative but will discover that his ancestors were covariant
lol
 
It has friendship, funny jokes, deaths, huge character development
The first movie is called "Spiderman Into the Spiderverse"
An impactful character development is missing from most hollywood flicks
 
Do you have any affiliation to showbiz?
 
Lol. I only recommend good movies here
 
Just making sure :) It's okay
 
Ive never recommended Fast and Furious stuff to u guys
This is the song i meant a few days ago
@Slereah but what is the math field of the right-most dude called?
I've never seen that math
 
3:23 PM
Looks like algebraic topology
 
Oh. I'm sure this field and algebraic geometry will b very sacred stuff
 
@Slereah Pretty sure the $\mathcal{D}$ is supposed to be a derived category, which is in particular a triangulated category - hence "geometry is triangles"
 
They give off vibes of being very general
 
hm so is this an accurate drawing of how to compute the directional derivatives at an arbitrary point in a lie group given the data of the directional derivatives at the identity?
 
But ive read that differential geometry can b seen as a generalisation of algebraic geometry, in one sense. But in another sense, algebraic geometry is a generalusation of differential geometry
 
3:28 PM
i was a bit confused because the group element acts as i) a group element so under the binary operation and ii) as a map between tangent spaces
but i suppose there's nothing really wrong with that
 
Lie groups have a group action, and this group action defines a diffeomorphism on the manifold
You can just do a pushforward by any point of the manifold because of this
 
I also thought this is very intriguing stuff. It also relates to the Schrodinger picture and the Heisenberg picture. In the former, we transform the state vector using the group elements. But in the latter, we transform the algebra elements using the group elements.
The former application of a group is more natural to me.... as a group of transformations that act on some object like state vectors
The latter is somewhat hard to digest. But i guess you could see the "generators" as the "objects" on which the group can act on.
 
3:44 PM
Another way to see this is that if you have a curve passing through the neutral element on your Lie group, the left action will move this entite curve to the point g
And define its own vector there
 
@RyderRude EEEEEEEEEEEK
 
@Slereah hm so let's say we want to deal with $SU(2)$ and use the pauli matrices as generators. Then, let's associate "moving in a direction" with each pauli matrix. E.g. $\sigma_x$ is associated with "moving in the x direction". Then, will $g \cdot \sigma_x$ be mapped to the directional derivative at point $g$ in the $x$ direction?
 
@SillyGoose No
 
ok so if coordinates are defined as specified above, i have another question :P in carroll, he says that the lorentz boost transformations are $t' = tcosh\phi - xsinh\phi$ and $-tsinh\phi + xcosh\phi$ then he says "from this we see that the point defined by $x' = 0$ is moving and has a velocity $\frac{x}{t}$." but i do not understand what this claim means given the previous discussion of coordinates
 
or, well, what do you mean by "direction derivative in the $x$-direction"?
 
3:59 PM
i am having trouble even describing what "direcetional derivative" should mean at points not at the identity
 
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