@ACuriousMind Apologies for late response, I think you guys were talking about something important so I thought better not to disturb you all. so I was saying that I understand that the chats looks very mixed and messed up, tend to look up as a single statement, I really didn't mean that what you are thinking.

I discussed about this yesterday that I am nothing in front of you people, as you people are treating me as a friend then why I cannot to those who are really struggling with some concepts which feels easy to us. Obviously I am here to for those people who are struggling.

But anyways if I was the one who looked down for the people then I would not be on the PSE to help people for free they are struggling with. you can explore my profile I took the questions which can be done by using simple mathematics :-).

2nd thing here is that JEE is a competitive exam, which means one needs to perform better than other to win the race if you just think that there are these type of questions then there should not be any purpose of conducting a competitive exam, they will be called just as a normal exam. The concept for normal exam and a competitive exams here in India differs a lot.

(Don't know about outside india but in india you need to be much more trained so that you should not be doubtful on simple questions) we in india think that at every point of time the competition is still going on. So because of this the mindset of the exam in ones mind must be changed.

I'm egarly waiting to hear your point of view on this

how can one find the value of an airy function for a given input? i know there are built in algorithms on Mathematica, but i cant seem to find a way to calculate it by hand/from an expression

@TejasDahake may i ask what is going on ;o

@Obliv just cycle through all 10 taylor swift albums on youtube ;)

@Relativisticcucumber Ahh. Nothing intresting just, trying to explain how things work in india :-)

By the way good morning everyone :-)

@TejasDahake I don't think the JEE is a good way to train students to be good scientists.

I know why the Indian government uses the JEE. It's because the education system needs more investment, and the JEE is the best that can be done with the resources available.

Hi All..

Hello @JohnRennie Sir

Hi :-)

@JohnRennie Sir if you have time. Can i ask questions?

Yes, but let's switch to the problem solving room.

Okay sir. How to make new line in single message in latex?

@JohnRennie yes. Things are actually very depressing here. And i think may be because of a huge amount of population we have, might be a reason for increased competition as compared to other countries.

Everybody here aims for getting admitted in top technological institutions same thing goes for medical students as well. There are very less top universities, if we look towards the bulk of students preparing for.

@TejasDahake I realize this is what reality looks like but I deeply resent the idea that knowledge (or any kind of competence, really) should be a competition. We might not be able to change these competitive systems but that does not mean we need to approve of their competitiveness or that we should reproduce their structural disdain for those who score lower in our everyday speech (e.g. by taking the expectations of these exams and uncritically accepting them as reasonable expectations)

@Relativisticcucumber the Airy function is a "special function", meaning there isn't any closed-form expression for it in terms of elementary functions. So the only way to compute it is to pick one of the power series or integral formulae and start computing them to desired accuracy with some method (power series is just "pick an order up to which to compute", integral might require choosing a numerical method to compute the integral)

@ACuriousMind I know there must be a substitute for competitions, but if now every student wants to get admitted in these few institutions then what should government do? That is why I think the government had to introduce competitive exams which mainly judges you based on your intelligence.

@ACuriousMind As you know In other countries you can choose any course in your institution you want to study for, but this is not the case in india you won't believe that you can't even choose here what you want to study, your performance decide it. i.e the rank secured in that competitive exam decides which course you are going to get in these 'so called' prestigious institutions.

(E.x let's say your dream is to have a degree of computer science engineering, then sorry you can't. Rank list will decide it. If your secured rank meets the minimum requirements to join that course then you can but if not then you can't)

Just imagine. This level of competition exists here in India. You don't even have freedom to choose what your dream is.

If you really want to achieve something you really like, then your hardwork, intelligence and competitiveness decides that

In other words how much you can do better than other students.

We the students who are here also do not have any other chance to just skip these competitions.

But anyways, these types of system somehow teaches us how to be consistent in our life and how to just keep working hard :-)

It's a much larger structural problem. Of course, if you have much fewer places than applicants for something you need to introduce some selection criteria. But the root problems you then need to solve is: Why are there so few places? Why does everyone feel compelled to choose the "highest" they "rank" for?

And then you need to try to create opportunities for everyone and not just the few. Options for a good life that don't rest on some singular achievement on a test when people's brains aren't even fully formed. This isn't easy - politics, especially for a billion people, is hard - but that's no reason to think the status quo is as good as it gets.

@TejasDahake That is the myth of meritocracy that I'm concerned about. Plenty of people "work hard" and still get downtrodden by these kinds of systems. Being able to function under immense pressure is not a virtue we should demand of everyone.

What worries me is that I see so many people like you openly state that the system sucks, for everyone, and then turn around and defend it as "teaching how to work hard" or stuff like that. If the system sucks, don't defend it. Don't pretend the defense mechanisms you have to build to survive it are universal virtues.

To be clear: I'm not saying you're doing anything wrong by taking this exam, or wanting to be as good at it as you can

arxiv.org/pdf/2212.00526.pdf

what is this font

@ACuriousMind i'm not defending this system, I just told you the only one advantage it can have.

May be in the future India will plan to establish more and more such type of institutions to atleast reduce these types competitions but for now I don't think everything is just going to be changed in very less amount of time.

@RyanUnger Bodoni Egyptian Pro Medium

(courtesy of WhatTheFont)

@ACuriousMind can you just share me the link for problem solving chatroom?

I've a question here which i think need to be discussed

here: chat.stackexchange.com/rooms/54160/problem-solving-strategies

Haha lol. I think I'm only the one who is online there

I can't see anyone. May be because of some loading issues I think

I think the chatroom is not loaded properly here

I retried for sometime but it didn't help

@ACuriousMind yes, now I'm there, thanks for sharing

@ACuriousMind have you ever been to India before?

@TejasDahake No, I've never been outside of Europe

Oh, I see, but I invite you. Explore India try some new food :-)

You'll get variety of food items here

I'm sure you'll like some of those

I would also like to try German cuisine atleast for once in my life

@ACuriousMind I saw some of them on youtube and they look amazing

oh I've had plenty of Indian food - there are quite a few Indian expats around here

though of course it's not clear how authentic it is to the kind of food you're thinking of

@ACuriousMind Why don't we care about the projective rep. when considering angular momentum?

Is this just because we know that the $X$ in $\psi(X)$ transforms under the vector representation of $SO(3)$ when considering $\psi(X) = \psi'(R^{-1}X)$?

@ACuriousMind I'm halfway disappointed you identified your source rather than the mystery of identifying a random font!

@DanielUnderwood lol

@DIRAC1930 If you mean orbital angular momentum, then there you don't start with SO(3) and go look for representations. You know that orbital angular momentum is $x\times p$ (this is indeed derivable from the transformation of $x$, see physics.stackexchange.com/a/735616/50583) and so the properties of angular momentum are already completely fixed by the properties of $x$ and $p$

there's no freedom here, you just look at the standard representation of $x$ and $p$ on $L^2(\mathbb{R})$ and you find it's just integer $\ell$s

Has KvN formulation been implemented for General Relativity?

I think it'll give useful insights about the nature of probabilities in a dynamical metric

what

Since in principle you can do KvN with any classical (field) theory, why do you think it needs to be "implemented" for General relativity specifically?

I don't know.... The derivation of KvN involved the Lioville equation which pretty much came from the Poisson bracket

sure, and since GR has a Hamiltonian formulation (ADM formalism) you could straightforwardly use that

And General Relativity is weird because there may not exist a space-like peeling of spacetime

But ADM is a special case

@RyderRude in what sense?

It is a special case for when a space-like peeling exists. i. e. When you can also atrribute energy to spacetime

the assumption of global hyperbolicity (existence of space-like foliation) is a really weak assumption - you don't really want to have the weird universes where that fails, they usually have time travel or whatever

Hmm, but I really love the crazy universes :P

I think it's in the spirit of GR to allow for the general case

KvN is basically classicla probability theory applied to classical physics theories, right? So I think there should exist a KvN for GR in the general case when time is weird

And it will give us useful insights about the nature of probabilities in weird time

I have no idea what you mean by "nature of probabilities"

you can do statistical mechanics without a KvN formulation just fine, just put a probability density on the configuration space or phase space

When I think "probabilities", I think of them as being defined at a time point

This sort-of conflicts with GR. But I don't think it really does. All we need is KvN for the general case

@RyderRude The conflict with GR is that GR is not a theory about time evolution, at its heart

As in "probabilities" get their meaning from "what you would see if you measured at a time point"

@ACuriousMind i agree

you can do a Hamiltonian formulation of such reparametrization-invariant theories where "time" is not a fixed notion, but what you'll find is that the Hamiltonian is zero on-shell

i.e. the whole physical content of the theory is in the Hamiltonian constraints

Oh. So we gotta quantise the constraints. Or implement KvN on the constraints

I havent studied constrained quantisation yet

It is supposed to be different from canonical quantisation, in the form of the Poisson bracket

I have surface level knowledge. I saw one stackexchange post

we understand constrained quantization well-enough, there isn't any mystery here

Yeah. I now agree there isn't any mystery

but your root problem seems to be what the "probabilities" mean in this "no definite time" context and that has nothing to do with quantization as such or the KvN formulation

the thing is that a probability is just "given a state $s$, the probability to measure an outcome $o$ is $P(o|s)$"

Yeah, I want to know how probabilities get their meaning in such theorues

time doesn't appear there

it's just that usually we are interested in some trajectory of states over time $s(t)$ and then compute probabilities for measurements at different times

But probbailities are defined at a time. They tell you the likelihood of what happens at this moment

@RyderRude no, they aren't

there is no time in my definition, just a state and an outcome

But the state is the present state.. idk :P

no, a state is just a state

It'd really help if I cud see the KvN

As in, it'd make things explicit

I cudnt find it on google

no, it really wouldn't :P

:)

But is it true that GR has been quantised for 2+1?

you seem to have fixated on the idea that the KvN formulation is about doing probabilities in classical physics but that's just not true

you can just put a probability density on classical phase space without ever talking about Koopman or von Neumann

classical statistical physics is exactly that

Yeah, that's true too. But complex numbers are prettiee

Prettier :P

I don't get one thing. How does KvN avoid destructive interference despite having complex numbers

It just magically avoids it

And there must also be a path integral of KvN?

@ACuriousMind I agree. But again, GR has may not have space-like slices

Though probably, you can just implement classical probability theory on the "constrain formulation"?

I'm not sure why you think KvN is so important but it's really not

it's just a cute demonstration of how to take "classical physics is $\hbar = 0$" literally

I agree

Even classical probability theory will do for me

But how do u implement it on weird time GR?

Does there exist a method with the "constrain" stuff?

there's plenty of literature on doing statistical mechanics in curved spacetime

@ACuriousMind also, is it true that 2+1 GR has been quantised without any issues remaining? I think this would. put the nail in the coffin that GR isn't special

@ACuriousMind I will search

@RyderRude [citation needed]

GR is renormalizable in 1+1 (no propagating d.o.f.) but not renormalizable in any higher dimension afaik

@ACuriousMind I couldn't find it. Some stackexchnage answer. I'm 70% sure they said 2+1. Maybe they talked about some recent not-well-accepted result

@ACuriousMind and 1+1 is the trivial "topological GR" case. So there is some hope that GR is special

that's what I mean by "no propagating d.o.f."

Yeah

I neither know what "special" means here nor why you'd hope for it

the failure of GR to be consistently quantizable isn't special, it's just non-renormalizability, a fate that hits a lot of theories

"Special" as in, maybe the postulates of QM will also get modified

what's "special" is that GR is the only theory we think of as fundamental that has this property

@ACuriousMind Yeah, mostly I too think that GR will simply get quantised

It's a gauge theory. Happens to match with the spin-2 unitary representation, etc. Maybe it's just a perturbation theory issue

renormalizability is not, strictly speaking, a perturbative issue

but it is a technical issue far removed from the "postulates of QM"

Yeah. And we don't know if GR is non-perturbatively non-renormalizable

It is probably just a technical issue. One can hope though that QM gets modified

I felt bad when I read that 2+1 had been quantised just fine :P

why would you hope that QM gets modified?

@ACuriousMind I prefer crazier theories. Hopefully, QM can make room for the time loops from GR :)

Would be a fine addition to the QM crazy

@ACuriousMind When do we get the freedom to lift to a proj. rep?

@DIRAC1930 the point about spin is that it is not a function of $x$ and $p$, so the standard representation doesn't fix its behaviour and so the spin representation (which can be projective) is something you need to additionally specify

in principle there is nothing that forbids a projective representation of orbital angular momentum from the outset but since Stone-von Neumann says that the $L^2(\mathbb{R})$ representation is essentially the only representation of $x$ and $p$ and we don't have any projectiveness there, it just doesn't happen

Okay thanks

Why is it a common thing to hear that spin wasn't understood until the Dirac equation when it seems like the path to $SU(2)$ from $SO(3)$ is well understood?

@ACuriousMind thanks !!

@ACuriousMind i actually feel this chat has a very meritocratic mindset. if you notice, people really ask acm things very disproportionally to general people. ofc for physics this makes complete sense, but about ways to listen to music efficiently, views on education, etc. i mean everybody has these. so i feel like maybe people should, as acm says here, reflect on how they contribute to these systems and how they can change them instead of complaining while still giving into them. : )

just 2 cents from a random person so feel free to ignore

@DIRAC1930 you probably vastly overestimate how clear this picture was in early quantum theory

I'm pretty sure no one except perhaps for Weyl thought about quantum mechanics in terms of groups or algebras or whatever in these early days

projective representations in math were known since around 1907 apparently, so it's not as if this would've been long-established math at that point, either

@Relativisticcucumber I would not necessarily say that getting asked all the questions is a benefit in the usual meritocratic sense ;P

i dont think it is at all

im just saying, okay so i actually spend copious amounts of time doing nonprofit work for educational accessibility. it's smth i care A LOT about, and i also go to a very annoyingly elitist university in the US, so i see pretty much constantly people complain about these systems and turn around and contribute to them

Does spin come out more naturally when considering the Lorentz group? But IIRC, you still have to look at the proj, rep

needless to say it's EXTREMELY frustrating. be productive, dont embody the thing you hate, etc. it's in the small things how the everyday man or woman propagates these systemic issues

this isnt at you but at the people complaining yet embodying a meritocratic attitude in this chat

end educational philosophy spiel

I wonder why Dirac said in 1982 that spin was only understood in the context of the Lorentz group

IIRC

@Relativisticcucumber It's almost impossible not to contribute to a system so pervasive, especially for people who are not in positions of power and for whom refusal to engage with these systems would carry a lot of disadvantages (and you only get to positions of power by contributing to the system, so it's a catch-22!)

@DIRAC1930 if you start by wanting a relativistic equation for the wavefunction linear in time, you end up with the Dirac equation; there is no such equation for a scalar

this is why people say "spin comes from relativity" - because if you try to do rel. QM you almost inevitably stumble upon it

Ah of course

hmmm im not sure i agree. i do see this attitude of "if i dont contribute, im at a disadvantage", but that's just selfish and kind of tautological, at least to the extent that it's usually taken. i mean you cant expect someone to fight for you when you wont even fight for something, ya know?

i do agree you can weasel your way in, and that probably you do have to be involved in some sense, but idk there are a lot of things that can be done, and i think people can do a better job of asking themselves "what am i doing to fix this"

@Relativisticcucumber that's right - I'm just saying that participating in a system is distinct from justifying or defending it

ah then we come to hypocrisy xD

what I was complaining about in the messages you replied to is the latter, not necessarily the former

@Relativisticcucumber sure, if you want to call it that most people are hypocrites in this sense in one way or another

@ACuriousMind i see. well i think by contributing the way you do to this community, you contribute very positively and seem to do at least something to add good to this issue. just imploring others to try and do something similar. if you must participate in these systems, at least counteract it with some good somewhere. people spend too much time complaining ;) as you said solutions are hard so we should challenge ourselves to ponder them

But this view - that people are hypocrites if they take part in this system they despise - individualizes an issue that can only really be solved by collective action. Nothing changes if one person quits, they're just replaced by another more willing to be a "hypocrite". Yes, people should take action if they believe the system is bad, but that action is not necessarily quitting.

i can see that for sure. that's kind of what underlies my belief of "do net good" bc we arent perfect and we do live in these systems as you said

there can be an end to a means at times

but i think people need to try harder and be more accountable than they are on average, and i still do think that, okay say you need to take the JEE and study for it, thats all well and good, but still the meritocratic attitude in other situations can be dismantled

or at least can be worked on

just an example not to attack anyone in particular

i think we are relatively on the same page here, no? @ACuriousMind

@Feynman_00 come to observe? XD

Yes, I do expect people to do better and to not internalize e.g. harmful attitudes about "intelligence". But I also realize how hard it is to not do that when the entire world around you is structured to reinforce such beliefs.

Life is easier, in general, when you agree with the structure of the world around you :P

boy do i know it

If I were to transform the group $G= SU(2)$ by $S^{-1} G S$, the spinor would then become $S^{-1}\psi$ right? If you were to extend this idea to rel cases, is this roughly what people mean when they talk about Weyl, Majorana, Dirac representations etc.

@DIRAC1930 What do you mean by "transform the group"? If $S\in G$, then $S^{-1}GS = G$ - conjugation is an automorphism.

Yes but the elements will now represent something different physically

i.e. $T^{-1}S_z T$ is no longer the spin in the $z$ direction

that depends on whether you think about this as an "active" or "passive" transformation

if your $S$ represents just rotating your coordinate system ("passive transformation"), then $S^{-1}gS$ is still the same physical operation as $g$

if your $S$ represents transforming the physical system ("active transformation") then you're right

this is one of the problems with trying to assign unambiguous physical meaning to every mathematical manipulations - sometimes it really is just ambiguous

anyway, this doesn't have anything to do with Weyl or Majorana representations

the existence of Weyl or Majorana spinors is a statement about subrepresentations of the Dirac representation $D$

Weyl means $D$ is reducible as a representation of su(2) (or so(3,1)) and splits into $D_+\oplus D_-$ with $D_\pm$ being irreps of definite parity

Majorana means there is a real subrepresentation $M\subset D$ where $M$ is a real vector space (see also this question of mine where I try to formalize this without doing usual physics-y tricks)

in general, the wavefunction of a particle with spin $s$ is $L^2(\mathbb{R}^3)\otimes V_s$, where $V_s$ is the corresponding spin-s representation. That is, the wavefunction is valued in $V_s$ - it doesn't take elements of $V_s$ as its argument, the argument is still normal position

i.e. $\psi(x)\in V_s$ for $x\in\mathbb{R}^3$

Ah okay, I was talking about angular momentum but that's fine. Okay, but if I changed the spin rep, it would completely change every expectation value

I mean orbital angular momentum is just part of the $L^2(\mathbb{R}^3)$ - you have $L^2(\mathbb{R}^3) = L^2(\mathbb{R})\otimes (\oplus_{\ell\in\mathbb{N}} V_\ell)$, which is essentially just the decomposition into the radial part times spherical harmonics

Is a rep of $SO(3)$ acting on vectors of $(r,\phi,\theta)$ still the defining representation?

sure, you've just chosen different coordinates for $\mathbb{R}^3$, haven't you?

Yes, so when you do changes like this, the Hamiltonian has to change to spherical coordinates, and when you compute expectation values, you have to translate them to $(x,y,z)$ to compare results etc.

But people don't seem to do any of this when changing representations in anything other than condensed matter physics

once again I'm not really following you

we're not changing the representation, you're just changing how you express the vectors when you go from $(x,y,z)$ to $(r,\phi,\theta)$

Isn't something like that the same as changing from the Dirac to the Weyl representation?

@ACuriousMind The representation is the same, but the physical interpretation of the elements has changed right?

no, at least I don't understand why they would be related

@DIRAC1930 how? both $(x,y,z)$ and $(r,\phi,\theta)$ are just coordinates describing a point in 3d space (and hence also the vector pointing at that point from the origin)

there's nothing "physical" about this

$diag(-1,1,1)$ is a parity reversal in $x$ on the first one but a reveral in $r$ on the second one

I don't think the above is in $SO(3)$ but i was just using it as an arbitrary example

1. If you switch the way you describe vectors you also have to change how you describe operators! Matrices are always relative to a basis. 2. The change from $(x,y,z)$ to $(r,\theta,\phi)$ is not a basis change in the sense of linear algebra!

it's a coordinate change on $\mathbb{R}^3$ as a manifold, not a basis change in a vector space

linear operators on $\mathbb{R}^3$ are not matrices acting on $(r,\theta,\phi)$

Okay, so when I change from the Dirac to the Weyl rep, the Dirac equation changes to the Weyl equations

again, Dirac->Weyl is writing $D = D_+\oplus D_-$ in representation-theoretic terms

you get the Weyl equation by choosing a basis of $D_+$ and $D_-$

and I'm guessing the states $|D>= \hat{U} |W>$ so that expectation values are left invariant

we're not acting with any operators to do this!

It's just - you can write an abstract Dirac spinor $\psi$ as $\psi_+ + \psi_-$ where $\psi_\pm$ are in $D_\pm$

I get that you can do that (I think), I just don't know how you would get the same expectation values. If I were to go to $\psi(r,\dots)$ from $\psi(x,\dots)$ I would have to change the measure in the inner product

when you plug that into the Dirac equation and set $m=0$, it separates into the Weyl equations for $\psi_+$ and $\psi_-$ separately

@DIRAC1930 I don't know how you got to the idea that this is analogous to changing the coordinates for the wavefunction argument, but it's really not

the argument of the wavefunction has nothing to do with this

Because isn't $X^\mu$ written in a particular basis

what is $X^\mu$

The Lorentz group you see in physics texts are always defined as acting on this with a physical picture in mind

$X^\mu$ is a $4$ vector

yes, but what does it have to do with any of this?

The Dirac representation is some vector space $D$ on which we have an irrep of the Clifford algebra

The Weyl representations $D_\pm$ are representations of $\mathfrak{so}(3,1)$ such that $D = D_+\oplus D_-$

writing $\psi\in D$ as $\psi = \psi_+ + \psi_-$ turns any equation for a Dirac spinor into the corresponding equations for two Weyl spinors

Hmm maybe it's easier to think about this in the non-rel case. What I'm saying is someone has chosen to use the Pauli matricies

nothing here has anything to do with 4-vectors or choosing a basis or whatever

ah, now if we talk about matrices, then matrices are always tied to a particular basis

you will note that the word "matrix" appears nowhere in my above description

and we can't talk about this in the non-rel case because there are no Weyl spinors in non-relativistic physics :P

now I think what is confusing you is that physics texts often talk about the "Dirac representation" and the "Weyl representation" as a particular choice of basis so that the $\gamma$-matrices have a specific form

In terms of what I wrote above, this just corresponds to choosing two different bases of $D$ - and the basis you choose in the "Weyl representation" is such that the first two basis vectors belong to $D_+$ and the other two to $D_-$

it is important that this use of "representation" is not really commensurable with the mathematical notion of representation, which is why I generally try to avoid talking like that

(physics texts suck at group theory, exhibit no 1000 :P)

So a big thing is that expectation values are independent of representation. How would I show this for a simple case?

the Wiki article actually does this right by talking about different bases rather than representations (even if the section heading confusingly uses "representations")

@DIRAC1930 you mean "independent of basis"

and that's just ordinary linear algebra: The expectation value of an operator $A$ w.r.t. a state $\psi$ is just $\langle \psi,A\psi\rangle$, where $\langle -,-\rangle$ is the inner product

neither the application of an operator to a vector nor the inner product depends on the choice of basis

So when we write a Weyl spinor vs a Dirac spinor, we have a different basis?

well, if you write a single Weyl spinor, it's just a 2-d vector, right?

look again at what I wrote above: You have that a Dirac spinor is the sum of two Weyl spinors

and you can choose a basis of Dirac spinors such that the basis vectors are just pure Weyl spinors (i.e. one of the two Weyl spinors in their sum is zero)

Okay thanks, I need to study this a bit more

given a muon in the atmosphere with proper time $2E-6 s$ and velocity $v=.95c$ how would I go about determining how far it travels w.r.t an observer on earth?

does it depend on which direction it's moving w.r.t the observer?

or do I just multiply the dilated time with the velocity?

I'm not sure what "with proper time" means but no, it doesn't depend

so from the muon's POV, it's traveling at v=.95c velocity for 2E-6 seconds. From a "stationary" observer on earth these things change?

like it travels for less time

ah, that's supposed to be scientific notation

i would assume i'd have to use its relativistic velocity * dilated time to find the distance but this example problem is from the length contraction section only.

oh yeah sorry I like that instead of x10^-6

you can do this with either length contraction or time dilation

i.imgur.com/9I0YxvP.png i'm asked to find the distance it travels w.r.t. an observer on earth in a problem.

yes, this is a very standard exercise

okay so it's just time dilated * its velocity from its pov

but wouldnt its velocity be measured to be different from the observers POV?

you can do it like that, or you can compute how far the muon "thinks" it travels in its reference frame and then "uncontract" that distance to get the distance as seen from an observer on earth

so $L = L_0\sqrt{1-\frac{v^2}{c^2}}$ gets you the distance it traveled from the observer pov

$L_0$ being its distance it traveled?

i.imgur.com/kcfJvpt.png this is so trippy.

if you travel really close to the speed of light, space gets shortened and time gets really long?

@ACuriousMind What does he mean by transform the basic vectors? I understand the transformation of the coeffs, but why do I have to do both?

Never mind

Why in physics do they only transform the coordinates

Why for $SO(3)$ do people only transform the coordinates but for $SL(2)$ they suddenly transform both the coordinates and basis vectors

I am confused with the concept of stationary action referenced in Goldstein

@DIRAC1930 that's not what's happening

there is no difference between groups for this - everything that's written there except for the $\alpha\delta -\beta\gamma = 1$ holds in general for linear transformations

this is just elementary linear algebra

it's just two different ways of looking at a transformation - you can think about what it does to vectors, or you can think about what it does to coefficients in a fixed basis

Will the coeffs mean different things depending on which one you choose?

Nevermind I'm not thinking properly

I think I'm confused with the statement 'the action is stationary'

No whether or not there's only one path depends on your system

there can be zero or more than one

but that's unrelated to what you're talking about though

Infinitesimal variation leads to a path with an infinitesimal variation of the action