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12:46 AM
@Relativisticcucumber you are correct, jigglypuff. Are you asking how the energy bands look like in energy v.s. momentum plot, or are you asking how one energy state looks like in position space?
@Obliv this is related to the earlier problem. If you cannot solve that, you will be even less well placed to solve this.
1:17 AM
@naturallyInconsistent for reference i put this question on the main site here physics.stackexchange.com/questions/781778/…! and i think what im looking for is how to understand the image/situation in physical space -- im not sure if that is the same as position space? i dont think so?
1:37 AM
I suppose I should be answering there?
well you can answer wherever hehe
any answer is appreciated anywhere
what does "mix" mean here:
2:21 AM
@naturallyInconsistent i solved the earlier problem, but this one is tripping me up. I looked online and it seems like there are some very complicated methods. U can do it non lagrangian way but still ends up being a lot of work
2:36 AM
is there a formal and precise way to write the Hilbert space for a spin-1/2 system in the $z$-basis?
Wny one who who like to guess?
A: Can a mathematician create free will, rigour, etc etc?

J KusinHere’s a small sampling of math “causing”free will I’m aware of: Tegmark and his Mathematical Universe Hypothesis. Free will as a subjective belief/experience and all other subjective experiences are massively complex mathematical relations. Math (a paired down version of modern math) leads to p...

What's your top 3? In order of preference?
Is there anyone who would like to guess?
I'm pretty sure by definition if you "create" something you're determining its fate in a deterministic world, so how would that be free will
that question makes no sense also
And if you assume a non deterministic universe, how does one "create" something. definitions are vague
2:51 AM
Isn't it is who define?
Our freewill?
idk man, not too good with this stuff
i think free will exists within confines of reality. when u go to sleep u dont have as much free will as if you're awake, and if you're dead you pretty much have none imo.
but i'm just young and a student so my opinion doesn't matter
actually interesting thought is what is a choice? If we're part of a deterministic universe but we don't have all of the info, do our choices become free of determinism?
3:27 AM
I think physics actually describes equation of states
Any how do we know how to reduce
This to the physics we know?
We use an approximation
Based on clarity and relevance
Which i argue againn requirs clarity and relevance
sure but isn't there always the epistemological/ontological/metaphysics concern that the equations of state & physics we do know don't answer those questions
like say we had a perfect model of the universe and all of the matter, it still doesn't answer what consciousness is
Indeed which is why Wittgenstien saw language as a consensus
you could also go the extreme side of solipsism and only believe that you exist and the universe doesn't whatever that means
This mach vs Newton in einstien
If you truly understand mach
U can resolve this difference
ill look into it thanks
good day/night
3:36 AM
3:50 AM
@Relativisticcucumber DPT essentially reduces to NDPT if the perturbation is diagonal (commutes with the CSCO)
it's crazy how much I interpret these newton's laws problems differently from others
part b, it can't be as simple as $mgh = \frac{ky^2}{2}$ where $h=2y$
@NiharKarve sorry what is csco
4:12 AM
@Relativisticcucumber complete set of commuting operators
@NiharKarve but how does this answer what mixing means?
@Relativisticcucumber it means that the limit of zero perturbation from small perturbation solutions lead to those states already. Not some superposition "mixing"
@Relativisticcucumber if you cannot find a nice basis such that the perturbation matrix is diagonalised, the states then "mix", no?
variously called a "good basis" etc.
4:34 AM
i still dont know what mix means, sorry :( i only know very basic perturbation theory, so maybe im missing something here? i, so far, only know how to find correction terms for consecutive orders in the nondegenerate case
im trying to think of a more informative question one sec
@Relativisticcucumber It is that $\psi=a_1\psi_1+a_2\psi_2$ superposition being called mixing in this case.
@Relativisticcucumber meowed there
4:49 AM
@naturallyInconsistent i am reading now
@Obliv is this irodov?
@Obliv that's why the question warned that this would involve a lot of algebra
btw, is there a way to show that light is transverse, by use of the uniqueness theorem of maxwells equations? I am reading eugene hecht, and in that book, it is assumed from the get go, from experimentation that light is transverse in nature, and the author goes on to make use of plane waves to explain light stuff. but there are certain assertions (eg: if light travels along x direction, the x and z component are 0)
which i dont know immediately why are being applied
@nickbros123 is it not sufficient to just derive light waves from Maxwell's equations? The derived solutions are manifestly transverse
@naturallyInconsistent we derive the wave differential equation, but from there how to go about showing it is transverse? forgive me if these are stupid questions regarding stuff thats covered in em books, for i am not studying this topic from griffiths- im studying optics for the sake of an exam coming up this week
5:02 AM
@nickbros123 The part that makes them transverse is that $\vec\nabla\cdot\vec E=0=\vec\nabla\cdot\vec B$
so that in Fourier basis the wavevector $\vec k$ is perpendicular to the E and B fields
is there an assumption that the form of each component is proportional to $e^{i(kx-\omega t)}$ where $x$ is the direction of propagation?
similarly, the $\vec\nabla\times\vec B+\frac1c\frac{\partial\vec E}{\partial t}=\vec0$ shows that the maximum and zeroes of E and B are together, no phase shift between them
@nickbros123 we typically choose z to be the direction of propagation, but this is not really an assumption, since a plane wave must move in some direction, and our universe is isotropic
@naturallyInconsistent that makes sense
thank u
kms wait so
@naturallyInconsistent the states for the tight binding model are like $e^{ikna}$ and they are energy eigenstates? so in energy basis?
is it sensible to have a form of these in position basis?
it will not look so simple
yes, Bloch states are rather commonly plotted in space for visualisation purposes: it is very easy for us hoomans to see with our eyes that some states are sensible and near the atoms, and other states are ridiculous and near the boundaries of the unit cells where no atoms are, and so we can use this visual information to sort numerically discovered energy eigenstates and throw some in the bin
The simplest first Bloch band to think of can be as simple as $e^{ikR}\psi_0(r)$ for atomic ground state $\psi_0$
5:16 AM
even after your answer i am still a bit puzzled. i think i understand the energies more, but what im seeing is that we have N sites on the lattice and so we can have 2N electrons fill our model. then, we have these bands of energy. that's fine, but what i want to know is just where in the physical space of the lattice this means the electrons will be? to answer this question for hydrogen, [...]
[...] i would just solve for pdf in position basis and plot it and see the orbitals, but i cannot find an answer anywhere for the visualization of these electrons in the lattice -- like i see what the energies are but where are the associated energy eigenstates placed?
sorry if im repeating myself :P
@naturallyInconsistent so the electrons literally just live in the ground state of all of the atoms still?
like the OG ground state?
of the atoms in the lattice as if they were not in a lattice?
@Relativisticcucumber Then you needed to ask what it looks like in position space.
i have no frontal lobe
@Relativisticcucumber Such a Fourier combinations of ground states has a slightly different energy than just summing up the ground states!
@Relativisticcucumber yes
However, note that my argument is for Hydrogen, which is extremely easy. In the case where you have more electrons, it is clear that ground state alone $1s$ will be filled up. That band will not be enough to accommodate so many electrons. You will slowly fill up from the ground state, $2s,\ 2p,\ 3s$ etc and then you get a lot of interesting solid state physics. Of particular importance is semiconductors.
ok so just to triple confirm -- the energies behave very differently in these models compared to isolated atoms, but the actual locations where we can find the electron will still be the same as the isolated atoms? like the electrons can flow through the solid and hop around, but we still find them in these canonical states? @naturallyInconsistent
sorry i dont want to misunderstand this XD
and for CM and SS, what do you recommend? kittel?
@Relativisticcucumber Oh, no, they may visually look to us hoomans to be the same as isolated atoms, but these are not at all the same as isolated. These are lattice solutions. The wavefunctions are spread throughout the entire lattice, i.e. entire universe. In a sense, there is nothing to hop around because it is already everywhere---kinda like a plane wave we have been assuming works for photons, etc
@Relativisticcucumber Steven Simons Oxford Basics. It is a simplified version of Ashcroft & Mermin. Kittel's front part is unreadable.
Remember, periodic solutions yo
5:29 AM
@naturallyInconsistent oh no
@Relativisticcucumber have you not seen how solutions to molecules are first worked out by?
but the plane wave solutions i thought are localized -- like free particle and gaussian packets?
@naturallyInconsistent that was a response to this
@naturallyInconsistent no only single atoms
i should try a molecule -- i shall try H2?
@Relativisticcucumber then you kinda need to learn about H$_2^+$ molecular ion first
will this be in the book you recommended or i should search elsewhere?
@Relativisticcucumber a plane wave is everywhere in spacetime and is completely delocalised---it has perfect wavelength information and thus cannot possibly be localised anywhere
5:34 AM
@naturallyInconsistent how can a plane wave be normalizable?
@Relativisticcucumber typically this is done in a atomic and molecular textbook, maybe spectroscopy, and seen as a prerequisite to solid state. Life is horrible
wait i think i remember
@Relativisticcucumber only via box normalisation or better, Dirac delta distribution. It lives in Rigged Hilbert space, not Hilbert space
a plane wave is not normalizable
and that is what leads us to gaussian packets?
for free particle?
@naturallyInconsistent @naturallyInconsistent have you gone through crystal field theory by any chance?
5:35 AM
@naturallyInconsistent okay do you know a good atomic and molecular textbook XD
i am very interested in amo so that might be a good start for me
@Relativisticcucumber Gaussian packets is a superposition of plane waves to mimic a normalised, partially localised and partially fixed momentum state.
@naturallyInconsistent ok right makes sense
@nickbros123 yes, but that is so horrible as to be barely understood. I know some crazy shit there and know what phenomena to see but it is mostly ewww and avoid
@Relativisticcucumber I do, but they are quite thick. I used a Bransden & Joachain. One really does not need to study that in full to know enough to continue
okay i will look up some simple molecules then return to this
man what a mess
thanks for the help
@naturallyInconsistent that stuff is there in my chemistry module 💀💀 my institution is doing a social experiment as to how far we can push students before they off themselves
5:42 AM
@nickbros123 it is just a rather unenlightened wish for students to just be exposed to a lot of shit, most of which they would just not even remember was covered, and their presence means that there is insufficient class time to cover important concepts with any form of clarity.
@nickbros123 oh no
6:11 AM
@NiharKarve can you give an example of a perturbation that causes mixing?
6:22 AM
consider some density matrix $\rho$. is it true that $e^{-iUHU^\dagger} \rho e^{iUHU^\dagger}$ is equivalent to $e^{-iH} U^\dagger \rho U e^{iH}$? I feel like no...
but i am trying to understand the fact that sending every $\lvert \psi \rangle \mapsto U \lvert \psi \rangle$ is equivalent to sending each operator $H \mapsto U^\dagger H U$
@SillyGoose are you specifically not considering a time translation unitary operator that commutes with the Hamiltonian?
right $[U, H] \neq 0$ in general (is this what you are asking)?
@Relativisticcucumber imagine being in a system that started out completely respecting z-symmetry, and then now introducing a magnetic field in the x-direction. Then a spin half system would now have energy splitting in the $\left|\pm\right>=\frac{\left|\uparrow\right>\pm\left|\downarrow\right>}{\sqrt2}$ which are definitely superposition "mixtures" of $\left|\uparrow\right>$ and $\left|\downarrow\right>$ states
@naturallyInconsistent ok i think i see what you mean now
@SillyGoose but we almost always consider only the case that $\left[\hat U,\hat{\mathcal H}\right]=0$, so I wanted to ask you if you are explicitly considering the other case.
6:33 AM
well i am ocnsidering the case in which $U$ is an arbitrary element of $SU(n)$ to be more precise
by the Taylor's expansion reasoning, $e^{iUHU^\dagger}=Ue^{iH}U^\dagger$ so that you should be getting $Ue^{-iH}U^\dagger\rho Ue^{iH}U^\dagger$
meanwhile, I was so thrown off by the missing t in $e^{iHt}$
6:52 AM
(Remember that ACM said that the correct argument isn't the Taylor's expansion, but rather Borel functional calculus)
The Taylor expansion argument isn't wrong, but it only works for analytic functions (and in order to really make it work we'd have to think about what norm on the operators we're actually looking at so that the series converges)
flips table
7:08 AM
@JohnRennie only physicists can be so excited about dust
@JohnRennie more like well done OSIRIS-REx team; Bennu can't care less...
Oops, yes, I confused the name of the mission and the name of the asteroid.
Oh well :-)
@SillyGoose hi these two transformations are equivalent in the sense that they are related by a change of basis. This means that predictions are invariant. You should check for the invariance of expressions like $Tr(\rho H)$
only scalars are invariant under a change of basis. components of matrices change, so you cant check for the invariance of those
7:28 AM
for e.g. $UHU^{\dagger} |\psi \rangle \neq HU|\psi \rangle$, but the expectation value of $H$ in the state $U|\psi \rangle$ is the same as the expectation value of $U^{\dagger}HU$ in the state $|\psi \rangle$
the pairs Upsi, H and psi, U^+ H U are related by a change of basis by U^+
7:43 AM
similarly, the pairs $\rho _1, H_1 = U^{\dagger} \rho U, H$ and $\rho _2, H_2 = \rho, U^{\dagger} H U$ are related by a change of basis, so $Tr(\rho _1 H_1)= Tr(\rho _2 H_2)$
$Tr(\rho _1 H_1)= Tr(U^{\dagger} \rho U H)= Tr(UU^{\dagger} \rho U H U^{\dagger}) = Tr(\rho UHU^{\dagger}) =Tr(\rho_2 H_2)$
8:24 AM
@Relativisticcucumber that's an energy eigenstate, written in "position" basis. n is the lattice site
1 hour later…
9:26 AM
> The Relativist draws down the Veil of Isis, and says: this knowledge is forever hidden from us. The Teachers in the Eastern Schools reverently lift the veil, and say: the solution of even these most inner mysteries, by searching, thou shalt find.
Sorry guys, veil of Isis is down
@naturallyInconsistent the Oxford book is epic
Apparently Eddington was the guy who invented quantum mysticism
2 hours later…
11:47 AM
@Slereah I need. Comic source for this one
Thanks I'll read it when I have time
Eddington was a bit of a weird guy when you look a bit into his contributions
The guy who introduced quantum mysticism to the wider world seems to have been en.wikipedia.org/wiki/The_Mysterious_Universe
12:28 PM
I have my own interpretation of QM ...
I hope to introduce it to the larger audience at some time
12:46 PM
@MoreAnonymous can i get a trailer :)
it is a video of a single celled organism : youtube.com/shorts/Jj0gqQeYEBw?si=IPVYx1VjRGv-48QV
@Slereah no I do have a written version of the physisc ideas
*no but
@RyderRude I need some feedback to whom it was sent
It should help me shape things
oh. i hope it's good. Does it bring new math to the theory?
hmmm .. its more of research programme with some outputs of its own as well. Either way I think It's worth exploring ...
@MoreAnonymous do the predictions differ from standard QM in some domain of experiments?
As for the math ... Im not the right guy for rigour to judge this
@RyderRude It does have new proedictions imo (but I havent been able to test them out yet) ...
Wigner and WIgner's friend get to be coherent friends in my thesis :P
12:56 PM
then i would say it is a nice interpretation
Thanks :) me too
usually, people confuse "interpretation" with "different philosophy of the same math". This is why interpretations have a bad name.
but yours is good !
Yea anyway ... I think I'll see whatt some pro has to say when the time is right
and i would say it's wrong to call it an interpretation, as the word implies 'different philosophy of the same math". it should be called "solution to the measurement problem"
@MoreAnonymous ok
@MoreAnonymous :)
3 hours later…
3:48 PM
@ACuriousMind hey
I definitely understand the followings:

$L' - L = \frac{d}{dt}f$
$Q = \frac{\partial L}{\partial \dot q} - f$
then when you say to add total time derivative, are you adding it to $L'$ in which case, the final resulting lagrangian is $L+\frac{d}{dt}f + \frac{d}{dt}F$ ?
I'm just adding it to $L$
then, we can write it as:

$L' = L + \frac{\partial f}{\partial q} \dot q$
then what makes you put $L'$ instead of $L$ in $Q$ ?
$Q$ is derived such as when you got $L$ and $L'$, you put $L$ in it, not $L'$
But now that I look at it again this argument doesn't actually work
yep :)
do you remember my old proof ? - ibb.co/C93bDd0 and you said that "On the left-hand side you have no expression proportional to $\dot x$" is actually wrong. It's wrong only for cases when Lagrangian contains product of $x$ and $\dot x$. Then I asked the author of the proof why he assumes that $L$ can't contain product of $x$ and $\dot x$ and he answered me with this: ibb.co/B2GL2Rr @ACuriousMind what's your thoughts on this ?
4:07 PM
@GiorgiLagidze I have no idea what argument is being made here? Are you trying to argue that there are no functions that mix $\dot{q}$ and $q$?
@GiorgiLagidze hello. if you already assume the absence of such products, then the proof is much simpler and more general. in the absence of such products, you have $L=f(v)+g(x)$. The EL eqn is d/dt (df(v)/dv) = dg/dx. For this to be translationally invariant, the rhs has to be x independent and hence g(x) is linear in x. hence proved
who is this "author" you're interrogating here? Are we talking to ChatGPT?
@ACuriousMind ofc no. I hate ChatGPT more than you can imagine
basically, you noted correctly that proof is wrong, but then @RyderRude correctly detected that it's wrong for cases when L contains product of $x\dot x$. If it doesn't, the proof is correct (ibb.co/C93bDd0)
also not all functions that mix $q$ and $\dot{q}$ have to be functions of the scalar product between them e.g. $q^2\dot{q}^2$ is not a function of $q\cdot \dot{q}$
I definitely didn't as well understand the last reply here physicsforums.com/threads/…
how about you @RyderRude
4:11 PM
if you just assume the Lagrangian contains no mixed terms then you have $L = K(\dot{q}) + V(q)$ and it is straightforward that if this is translation invariant then $V(q) = 0$
After an evaluation of the available math subjects, I think I'll do measure theory. Is that unwise? D:
you don't even need any complicated argument because the terrible case of the quasi-symmetry just can't happen - translations on a pure function of $x$ like $V(x)$ cannot produce total time derivatives (a total time derivative needs to contain at least one $\dot{q}$) except constants
@ACuriousMind you are forgetting the case where V(x) is linear. that is why this is a much more general proof. it even covers the case of accelerated frames
@GiorgiLagidze just use the last proof i gave. it is much more general. it proves V(x) is linear
@RyderRude can you link the message ?
@GiorgiLagidze my first message to you today a few minutes ago
when i said hello
@GiorgiLagidze this proof proves that V(x) is linear if you assume homogeneity. if you further assume isotropy, then you can prove V(x) is constant
isotropy is also required to prove that V(x) is constant. homogeneity isnt sufficient
4:15 PM
you say rhs has to be x independent, but if it is x independent, how can g(x) be linear in x ?
@GiorgiLagidze ...the derivative of a linear function is a constant
@GiorgiLagidze i dont think this is correct because there exist physical Lagrangians which involve product of $x$ and $v$, like those in a space dependent magnetic field
@Mr.Feynman Depends - what are you hoping for, and what "kind" of measure theory are we talking about?
you say that "hence $g(x)$ is linear in $x$". if it's linear, then L contains $x$ :D
@GiorgiLagidze yes, you cant prove constancy of L by assuming homogeneity
4:17 PM
doesn't linear mean: $g = 10x$ for example ?
you cant prove something thats false !
constancy of L can be proved by homogeneity + isotropy
I seem to like the proof better - ibb.co/C93bDd0
if we assume L doesn't contain $x \dot x$ product, then ibb.co/C93bDd0 seems great
@GiorgiLagidze ugggh. the conclusion is incorrect anyway and so is the proof. why do you have to like this so much :P
@ACuriousMind I don't how how to describe other than "basic measure theory" the course ends with Fubini theorem apparently
@GiorgiLagidze even the conclusion is wrong. homogeneity implies linearity of V(x)
it does not imply constancy !
4:20 PM
@RyderRude which conclusion and what's wrong if you assume that L can't have product of $x \dot x$
because a homogeneous space does not imply the constancy of L
the proof doesn't say that
@GiorgiLagidze compute the EL eqn of 1/mv2 + 2x and tell me if it's homogenous
@Mr.Feynman I mean that's not "unwise" but be prepared that if you've seen people do stuff with integrals a lot of this will just be "basic properties of integrals, but with more technicalities" :P
@GiorgiLagidze the proof proves constancy of L using homogeneity
4:21 PM
constancy of L in $x$
i.e proves L can't contain $x$
yes, and that's incorrect!
compute the EL eqn of 1/2mv2 + 2x
2 = ...
so is it a translationally invariant law?
$2 = m\ddot x$
hence proved. homogeneity does not mean x non dependence
4:23 PM
I understand what you mean :p
just go through the simple proof with L=f(v)+g(x)
@ACuriousMind the other option was functional analysis but apparently this year will be more advanced and I remembered your words about Sobolev spaces :P
@GiorgiLagidze if you want to prove constancy, you can assume isotropy later
@GiorgiLagidze remember that this is only isotropic if the rhs is 0
but in your case, because you got $2x$, you don't have single particle (closed system)
@Mr.Feynman That I hated Sobolev spaces doesn't mean you would!
4:24 PM
not necessarily
and that's why the proof for your case fails
@GiorgiLagidze no, it can also be an accelerated frame
we only care about equations in inertial frame
besides, the proof you gave isnt assuming inertial frames anywherr
@ACuriousMind In all honesty, if I make a bad choice, i.e. something too difficult I'll be forced to take it or wait the next academic year to remove it and thus put off graduating :P
4:25 PM
yep, if you do so, then it works
but i have to understand your proof now
i really dont think that your proof works even if you assume the absence of product terms
And there are some shady rumors about the Profs at the Math department so I'd better avoid messing up D:
$L = f(v) + g(x)$. (eq 1)
$\frac{d}{dt} \frac{\partial f(v)}{\partial v} = \frac{\partial g}{\partial x}$ (eq 2)

now, if we got homogeneity, eq 2 must give the same result when we change $x$. if $g(x)$ is linear in $x$, eq 2 gives the same result. if $g(x)$ is non-linear, it changes, so you show that g(x) must be linear to achieve homogeneity
@RyderRude correct ?
i think you're proving something else though :P
you don't prove that L can't depend on $x$
which is what we set out to prove
4:33 PM
again, you cant prove something thats wrong!
i am proving the more general case where we dont assume inertial frames
what you're showing is L must be linear in $x$ to achieve homogeneity. if it contains $x^2$ or more, then homogeneity breaks
to get to constancy under x, you have to later assume inertial frames after this proof
@GiorgiLagidze yes
but we know all inertial frames are homogeneous. if so, even if $L$ contains $x^2$, space is still homogeneous
but with your proof, it breaks. i think what you're proving is homogeneity of lagrangian and not the space
what? we are not proving homogeneity of anything :P
we are assuming homogeneity
okay, imagine we're wokring in inertial frame. which means we know that space is homogeneous
4:37 PM
ok. then you also know that space is isotropic though
yep. and if L contains $x^2$, homogeneity breaks as you get different results
okay, i get your proof now :)
ok ;)
though for free particle, it doesn't prove that L can't depend on $x$.
after also assuming isotropy, you can prove that
what would you say about isotrophy that allows you to prove that
4:39 PM
homogeneity got u to ma=c, right? for what values of c is this this eqn isotropic?
ya, and whats L when the rhs is 0?
but values of $c$ are not related to isotropic
@GiorgiLagidze u just said that istropy implies c=0. that's a relation
if i had said that isotropy implies c=10, what would be wrong ?
values of c come from homoneity, not from isotropy
4:42 PM
ma=10 is not isotropic because it becomes ma=-10 if u rotate 180
not in inertial frame :p
as long as you do the experiment with same initial conditions
@GiorgiLagidze proving constancy of L requires homogeneity+isotropy + absence of product terms
but you don't seem to use inertial frame at all
in that case, you're correct
yeah, im only using homogeneity+isotropy+absence of product terms
@GiorgiLagidze Landau defines an "inertial frame" to be the case where we have homogeneity and isotropy!
there's nothing else to use
4:44 PM
yeah, i know, ryder though goes backward
@GiorgiLagidze i was reading Landau a bit and i think there is an huge jump in the explanation where he goes from L=f(v^2) to L=1/2 mv^2
he doesnt acknowledge it in the page i saw. does it explain this later
Landau starts as the following.
1) we only work in inertial frame
2) in inertial frame, space is homogeneous/isotropic
3) we observe free particle from inertial frame. to us, we know space must be homogeneous and isotropic.
if you now @RyderRude use your proof, something is wrong
because even if L depends on $x$, space is still homogeneous(you get the same equations)
and space is still isotropic as $x$ doesn't care about direction :p
of course $x$ "cares about direction"
a single $\vec x$ is not invariant under rotation
and what "isotropy" means formally is rotational invariance
then what would be the 4th bullet point ?

1) we only work in inertial frame
2) in inertial frame, space is homogeneous/isotropic
3) we observe free particle from inertial frame. to us, we know space must be homogeneous and isotropic.
I don't understand what these bullet points are supposed to be so I couldn't say :P
4:52 PM
it summarizes landau's approach. what he says after what
if L depends on $x$ linearly, you still get homogeneous space for free particle :P
while Landau says that it's not the case
he directly says that due to homogeneity ONLY, L can't depend on $x$
sure, that is, strictly speaking, wrong if you include quasi-symmetries of the Lagrangian
he doesn't use isotrophy to lay out the logic that L can't depend on x
we've been over this already
4:54 PM
then what's the point he mentioned it like that ?
if it's not fully correct
I cannot read Landau's mind
sometimes books make mistakes or phrase something oddly, you have to get over it
can we at least all agree that homogeneity of space doesn't imply L to be In-dependent of x ? @ACuriousMind if so, I'm good :P
we more or less agreed on that ages ago
I don't know how you missed that :P
yep, seems like I missed that and was digging it now :) I thought you weren't saying it, while I was saying it
even my (wrong) argument was only about how you could potentially choose a Lagrangian constant in x, not that it had to be constant
4:59 PM
I normal summon bolbteppa and end my turn
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