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2:38 AM
@Obliv things like this you should be asking an expert on the topic, and who may themselves be consulting a cosmologist. This is deep lore and you should not expect someone outside the field to be knowing everything.
@imbAF and this is why ACM had been telling you to try your hand with simpler cases before looking at the general case. Your way of learning necessarily runs into confusion.
@imbAF The opposite is true. The comparisons are supposed to be natural, only that bad notation hid the conversion. It is the same confusion that people using Imperial keep not understanding that unit conversions are something SI people always do, because the physics is clearly trying to link those concepts and SI helps the unit conversions by making them easy, whereas in Imperial land unit conversions are so difficult that nobody learnt to try them at every turn. Notations matter
@SillyGoose The central assumption in all of these computations is that the interactions are so weak that the initial estimate of "these microstates will still roughly have the same energies as before any interaction" is tolerable. Strictly speaking, it is necessary that these assumptions are strictly false, but it is more than just tolerable. It is easy to see that if you have so many closely and regularly spaced energy levels as to form a quasi-continuum, then obviously small shifts dont matter
@Amit What is DW? I tried googling Death novels DW and it was confusing.
2 hours later…
4:43 AM
@naturallyInconsistent Disc World? i.e. the Terry Pratchett books that feature his character Death?
@JohnRennie ahhhhh Pratchett is sooooo good
5:38 AM
supposedly one can approximate fast processes as being adiabatic; however, if i wanted to solve this problem more precisely, would a better approach be to assume that the process is constant pressure, compute the heat capacity at constant pressure, find the temperature difference, find the change in energy, and compute the heat given the fact that i know the work done on the system?
i guess i don't really understand how one should interpret this physical situation. it seems a little bit silly to approximate it as being an adiabatic process with constant $T$ and $P$ especially when I did the computation I just detailed above and found that the temperature does in fact change if we lax the assumption that the process is adiabatic and constant $T$
@SillyGoose no, because during that 1mm motion, the pressure is definitely not constant, and none of the intensive variables are well-defined constants of the entire system; instead, they temporarily become complicated functions of spacetime.
5:55 AM
What book is this, Wacky Fowl?
this is the soul destroying schroeder
No wonder it is full of silly questions
more wacky than this fowl
i am confused about this question still though and how one can solve this by using reason. i guess the first confusion is how can one decide if a process is precisely adiabatic (or can be safely treated as adiabatic)
We are kind of assuming that piston walls and so forth are very slow at transferring heat, i.e. they are originally meant to be insulating. In this way, at such sudden motions, there would be insufficient time, so $Q\approx0$, which is the content of (b). Since you know both (a) and (b), by 1st law, you know (c), and that is sufficient. Finally, you wait infinitely long for the re-establishment of equilibrium, and you can compute (d).
2 hours later…
7:57 AM
A formal question. In QM we have the position completeness relation $\int dx \lvert x\rangle\langle x\rvert=1$ in Schrödinger picture or, in the Heisenberg pictures with instantaneous eigenstates $\int dx \lvert x,t_0\rangle\langle x, t_0\rvert=1$. Formally, in QFT one has an analogous completeness relation but with path integrals. In this case we only consider the Heisenberg picture and the field operator spectrum $\hat{\phi}(x,t)\lvert\phi(x,t)\rangle=\phi(x,t)\lvert\phi(x,t)\rangle$,
$$\int D\phi \lvert\phi(x,t)\rangle\langle \phi(x,t)\rvert=1\tag{1}$$
Now, I was thinking: QM may be regarded as a 0+1D QFT, where the quantum fields would be the Heisenberg picture position operator $\hat{x}(t)$; so I wondered that one may formally write something like $(1)$ specialized to this case
$$\int Dx\lvert x,t\rangle\langle x, t\rvert=1$$
I've never seen anything like that, nor I can see what it would mean, though. Is that a thing?
TL;DR: In QM we have the usual completeness relations and in QFT we may extend this idea by means of path integrals. Now I wonder what happens if you apply the latter the QM regarded as a QFT
I mean, even intuitively does it make sense? If not, why?
@Mr.Feynman i would say this is indeed a thing, as there is no mathematical difference between particle QM and 0+1 QFT with a real-valued field.
the CCR and the hilbert space of both theories are the same. the only difference is in how you interpret the variables of the classical theory
but the quantum theory only cares about the hilbert space, operators, and the dynamical laws. there is no difference there.
8:14 AM
The ETCR have indeed the same form if you consider the Heisenberg picture CCR. Still, I don't have a meaning for that last alleged identity, if there is any. Maybe it's correct but just useless because we have the "standard" completeness relation.
I'd also appreciate some feedback from @ACuriousMind about it, when he reads this
@Mr.Feynman i just saw that you have written a big D here. how does it make sense to functionally integrate when the eigenstates are no longer the space of functions
@Mr.Feynman I was particularly annoyed when I pushed my prof and then he said that the correspondence between QM and 0+1D QFT is only established by the term-by-term agreement in the perturbation expansion of QHO. I'm looking for a direct proof that they are mathematically the same thing, not some tenuous deduction that might not work if it werent a QHO.
@Mr.Feynman I am particularly concerned that you think of this only in path integrals, which I do not think is true of QFT.
@Mr.Feynman i would say this formula is only formal (like you say), as there is not much meaning of functional integrals. However, we can lattici-fy the QFT and revise this to : $\int dx_1 dx_2... dx_i |x_1, x_2....x_i\rangle \langle x_1, x_2.... x_i|$. Then the QM case is a special case of this formula
@Mr.Feynman This is actually one of those small headaches that isn't properly handled in the standard presentations. A few books have claimed that the correct QFT resolution of identity involves $$\hat{\mathbb I}=\int\left|\phi\right>d\phi\left<\phi\right|+\text{two particle excitations}+\text{more particle excitations}$$ and not just the single particle excitations part, just that they drop out of further calculations. Life is very horrible in QFT.
@naturallyInconsistent The $\lvert\phi\rangle$ in my message are defined as eigenstates of the field, so I'm not sure about why you bring up definite particle states
Anyway, I stressed that I only care about this on a formal level, just to understand why that version of the completeness relation never appears
8:28 AM
@Mr.Feynman Eigenstates of what of the field?
What do you mean?
the field at a spacetime point is an operator and I'm considering the spectrum of that operator
@Mr.Feynman if it is at one spacetime point, then this operator, if interacting, will excite more than one particle. Only in the non-interacting case can we try to make it excite just one particle.
@naturallyInconsistent MrFeynman is just writing the completeness relation in the wavefunctional basis
That is kinda the whole point: you are trying to find the Hamiltonian eigenstates, and it may well not be the single-particle Hamitonian eigenstates. It can be multi-particle Hamiltonian eigenstates. QFT operators are very scary!
@RyderRude you cannot have completeness relations in QFT without the multi-particle part
no one said anything about a Hamiltonian eigenstates :P. this is just the completeness relation
8:32 AM
mhh if I understood your message, you're thinking of $\hat{\phi}(x,t)$ acting on vacuum $\lvert 0 \rangle$, which of course is not just 1-particle (wave-function renormalization). Here we are defining $\lvert \phi(x,t) \rangle$ as eigenvectors of the field, just like $\lvert x \rangle$ are eigenvectors of $\hat{x}$ in QM
@naturallyInconsistent that is only true when you write the completeness relation in the the free Hamiltonian eigenbasis
@RyderRude If it is free, then it is definitely possible to do it with single-particle. You are getting it backwards
the wavefunctional basis spans the entire space on its own. there are no "multi-particle" states left @naturallyInconsistent
@Mr.Feynman yes. i think he is confusing these states with the |x\rangle states, in which case you would need the multi-particle vectors too
Just to be clear, I think @naturallyInconsistent means that $1\neq\sum_{k}\lvert 1_k\rangle\langle 1_k\rvert$, but I'm not claiming that at all
$\lvert 1_k\rangle$ being 1-particle states
@Mr.Feynman actually, that brings us to another problem. Your Schrödinger picture and Heisenberg pictures are also wrong. In Schrödinger picture the states are evolving with time whereas the operators $\hat x$ are ostensibly constant in time. This means that the completeness relations are $\int\left|x,t\right>\mathrm dx\left<x,t\right|=\hat{\mathbb I}$ for Schrödinger picture, corresponding to $e^{-i\hat{\mathcal H}t}\left|x,0\right>\mathrm dx\left<x,0\right|e^{+i\hat{\mathcal H}t}$ in Heisenberg
8:38 AM
@Mr.Feynman oh. i thought he meant the completeness relation written using the position eigenkets in QFT, which is impossible anyway because the position eigenkets do not inner product to the delta function
@Mr.Feynman yes. in this case too, you would need the multi-particle states in the completeness relation. this is what i meant when i said naturallyinconsistent's statement was valid if we were working with the eigenspace of the free Hamiltonian
@naturallyInconsistent Wait, these are not state vectors, these are basis vectors (to use Sakurai's words). Basis vectors are just a way to rewrite operators in the spectral decomposition) and they are defined as $\lvert x,t\rangle=e^{iHt}\lvert x, 0\rangle$ (note the sign of the exponential)
They are the instantaneous eigenstates of time-dependent operator, so of course they do have a time label in Heisenberg picture
@Mr.Feynman I am also kinda saying that it is impossible to write "eigenvectors of the field" as $\left|\phi(x,t)\right>$; it is much easier to see that it is impossible to write them that way when you consider momentum representation, and $\left|\phi(p_1,p_2,t)\right>$ is not at all the same thing as $\left|\phi(p,t)\right>$
@Mr.Feynman No, you really only should have a specific time slice on which you expand the states in Heisenberg picture, not over many different time slices.
@Mr.Feynman Thanks for the sign convention; I do not claim that part to be correct, I'm just working off memory.
I have integrated over a fixed time slice, in fact
The Schrodinger picture is only the one corresponding to $t=0$
The other slices are identical copies of the completeness relation, one for each slice of time
Check e.g. this
@Mr.Feynman i think the real problem is that it doesnt make sense to functionally integrate in 0+1 QFT. so you cant write a capital D when you go to the special case of 0+1 QFT. you should first latticify a 1+1QFT, and then recover the 0+1 qft completeness relation as a special case
@Mr.Feynman you are writing a Dx even though x is no longer a function
it should just be dx for 0+1 QFT
Wait what? That just comes form eq. (1) but considering $x$ as the field over 0+1D
8:50 AM
@Mr.Feynman Dx implies functional integration, right?
@RyderRude This might be the problem and what I'm trying to understand yes
now are the eigenvalues of the field operator functions in 0+1 QFT?
@Mr.Feynman That isn't clear enough. I need to think; this thing is always superbly confusing.
@RyderRude Functions of what?
Of time for sure, the only spacetime coordinate at hand
yes, but you are considering the vectors at a fixed time slice
at a fixed time slice, does the $x$ in $|x,t\rangle$ vary over a space of functions?
no, it just varies over the real numbers. so u can just use dx, not Dx
8:53 AM
The construction is perfectly analogous, because the field version also consider at fixed spacetime slices (i.e. only the operator evaluated at $(x,t)$)
If you want, we may write using the same notation although it looks deceptive in QM $\lvert x(t)\rangle\langle x(t)\rvert$, while in field theory you have $\lvert \phi(x,t)\rangle\langle \phi(x,t)\rvert$
So the QM case is the field theory case with 0 spatial dimensions and $x(t)$ is the field
is the confusion just that, if you specialise the QFT case to 0+1 QFT, you should be getting a Dx?
this is opposed to the dx that we have in QM
or do you agree that the specialisation of the QFT case gives you a dx, agreeing with what we have in QM?
@RyderRude and yes
@RyderRude yes
Specializing you get $\int Dx \lvert x,t\rangle\langle x,t\rvert=1$ and unless anything went wrong in doing this, it' different from $\int dx \lvert x,t\rangle\langle x,t\rvert=1$ that we already have
I mean, maybe the notation sucks but in the first case we are integrating over all possible field configurations $x(t)$ (all the possible paths), in the latter we are using the ordinary completeness relation
@Mr.Feynman no, in the first case we are not integrating over any paths, just like the generalised QFT formula doesnt have any integration over time dependent paths either
the first case is just meaningless, as it doesnt make sense to functionally integrate over a real-valued variable. dx is the only thing that can be meaningful there
the QFT formula only has an integration over the space-dependent function space. so how can an integration over time dependent paths show up in the specialised case?
@Mr.Feynman while I am still thinking about the pictures, do you disagree with the fact that $\forall(p^\prime,p_1,p_2,t^\prime,t)\in\mathbb R^5\qquad\left<\phi(p^\prime,t^\prime)\vert\phi(p_1,p_2,t)\right>=0$?
9:09 AM
@Mr.Feynman in the generalised QFT case, at $t=0$, we are integrating over the simultaneous eigenvectors of all the field operators $\phi(x)$. The integration is done over all the eigenvalues, which is why it is a functional integration. in the 0+1 QFT case, there is just one operator's eigenvectors and one eigenvalue to be integrated
Also, I think you might be confused by the path integration specifics. We can do QFT with canonical quantisation, and then you should not have big D appearing.
@RyderRude So you're saying that even in the general field case $\int D\phi$ only integrates in $\phi(x,\cdot)$?
@Mr.Feynman yes, time is constant. the integration is on the set of space dependent fields
@naturallyInconsistent in fact there is not doubt that the latter is correct. I asked if the former could make any sense
@RyderRude I considered that hypothesis before asking. If that's it, the question has been answered
okay :)
9:14 AM
@Mr.Feynman IIRC, when doing the path integral in QFT, the big D notation is outside the Dyson expansion and the insertion of the resolution of identity is filled with small d integrations. Again, this was one part I had to prod the prof to make excruciatingly clear, and is part of why he gave up and realised I had a point and many treatments are just plain wrong.
@RyderRude But I have to understand why it is so now :P
Maybe there is nothing to understand, it's just that the original analogy was intended at a fixed time slice in the first place and $\int D\phi$ was over spatial configurations from the beginning by analogy, right @RyderRude?
@Mr.Feynman yes. the completeness relation is at a fixed time slice. like you said, the original formula gives completeness relations at each point of $t$
this is true even in QM
@RyderRude yes that's what I meant by "analogy"
but you can ask why the integration is over the set of space dependent functions, if you are unclear about that
but i guess it's clear
you may be familiar with wavefunctionals
I was misunderstanding the role of time here. I thought that "fixed time slice $t$" in QM would genealize to "fixed spacetime $(x,t)$" in QFT.
That's not how it should be read: it generalizes to "fixed time slice $t$ plus 3 spatial dimensions to deal with"
9:27 AM
yes. because QFT is QM with an infinite number of variables. time is still playing the special role
it's just that the variables are indexed by $R^3$ in QFT
I mean, up to the eigenvectors part it worked but it broke when integration came into play
Still, I think the notation $\int D\phi$ for spatial configurations only is kinda misleading here. For example, when you consider the generating functional $W[J]$, you have the same notation $\int D\phi$ for path integrals but there you're also integrating over time configurations
yes, i guess they could write $D\phi(x,y,z)$ to be clear
the eigenvectors are also of the form $|\phi (x,y,z), t\rangle$
for each function $\phi (x,y,z)$, there is a simultaneous eigenvector $|\phi (x,y,z)\rangle$ of all the field operators at $t=0$, and this eigenvector is what's being evolved in time.
but you could also write it as $|\phi (x,y,z,t)\rangle$ to denote an eigenvector of the Heisenberg evolved field operator
i think this is why you thought that that the eigenvectors picture is covariant. but it breaks down in the completeness relation
@RyderRude Let me invent the notation $D^3\phi$ and $D^4\phi$ for the two cases. Do you agree that for the completeness relation we have $$\int D^3\phi \lvert\phi(x,t)\rangle\langle \phi(x,t)\rvert=1$$ while for the generating functional we have $$Z[J]=\int \mathcal D^4 \phi e^{i \int d^4 x L_0 + J(x)\phi(x) }$$?
(where of course $D^3\phi$ indicates integration over spatial configurations and $D^4\phi$ over spacetime configurations)
@Mr.Feynman yes. i guess this notation is free real estate right now :P
@RyderRude I'm sorry, I don't understand this phrase D:
9:41 AM
i've never seen anyone write d^n x for repeated integration, for instance
@Mr.Feynman i mean i think this notation is not taken up for some other purpose
I'll possibly never use it again in my life, it's useful now though :P
Thank you for all @RyderRude
great :D
10:07 AM
Hey guys! I know I've been kinda an emotional wreck but don't worry about me too much ... this is my kinda vibe for now
Thanks to everyone in the chatroom
10:32 AM
Also who knew chess.com was epic?

this completements my music a lot more:

10:52 AM
Let's play our own styles of chess
*chess stackexchange
11:09 AM
Translation of the Indian song
11:32 AM
@MoreAnonymous it is a great song. i hope you enjoy life now
@MoreAnonymous we all worry about you. so you should make sure u are happy
@MoreAnonymous it's very cool
12:21 PM
is the book: "Keith Stowe: An introduction to Thermodynamics and Statistical Mechanics" usable for someone who only knows classical physics (I havent had a course in QM)
@RyderRude thanks !!! Much appreciated
12:37 PM
@nickbros123 i don't know the book but I'd expect the part about quantum statmech to require a good understanding of qm
1:01 PM
@fqq I dont knwo the first thing about quantum mech, I just want to learn some thermo, and perhaps classical stat mech. My school says: schroeder, blundel blundel, zemansky, reif
Do they say buy them all?
Zemansky would be my choice.
Have you taken physical chemistry?
@nickbros123 For you, I'd suggest reading Feynman Lectures Vol 1 on the heat parts, and then Callen, because your maths is fine
Maybe Kittel & Kroemer if you cannot find Callen. Augment with Pippard
1:17 PM
@naturallyInconsistent hmm, thanks
@naturallyInconsistent difficulty in finding books in the internet era? ;p
there are questionable ways to obtain ebooks...
In which case I'd recommend Ian Ford Entropic Approach
bye for now; party time
@user726941 nah, lol. one of these. and I have taken physical chemistry in highschool. Thermochem with the PV diagrams and so on
that was more or less what I studied in hs
That's a good start.
1:51 PM
\o @Amit
what's up
not much not much
2:05 PM
anyone want to play this game ? lichess.org/vjpMxx6J
@RyderRude hey ryder. I wanted to ask something about our previous discussions. in the constant speed moving train, where we were discussing the drop ball experiment, we said that to check homogeneity, we drop ball, then we move further in the train and drop it again(i.e ball is dropped from same height because as you said experiments must be done with same initial conditions ). Is this really correct ?
@GiorgiLagidze yes, this is the experimental interpretation but remember that to to mathematically check homogeneity, u transform the eqn
that was my next question. if what I said is correct, then i got a problem with lagrangian. Lagrangian for the ball is $L = \frac{1}{2}mv^2 - mgy$ and then we have $L' = \frac{1}{2}mv^2 - mg(y+a)$, so what I said seems incorrect as it's not that we moved, but we drop the ball from heigher
2:23 PM
@GiorgiLagidze are you imagining moving across the train vertically or horizontally?
if it's horizontal, then the mgy term doesnt transform as y is the vertical co ordinate
exactly. so what's the experimental interpretation ? is it wrong to say that we move horizontally and drop the ball again from the same height ?
the equation or the lagrangian themselves do not have any information about what initial condition you choose for your experiment
they can be used with any initial condition
ofc, it's just 2 experiments in the end must be done with the same initial conditions
2:26 PM
the point is that, if the eqn is invariant, then the same initial conditions give the same results
to my, this only works if we move horizontally
oh, so u want to check for vertical homogeneity
i get that, but in the experimental interpretation, which one is correct ? we move horizontally ? or vertically ?
yes, the frame is vertically homogenous too
@GiorgiLagidze depends on what co ordinate translation u did to the lagrangian : horizontal or vertical
@GiorgiLagidze here u did vertical
since we talk about dropping a ball, then moving horizontally is homogeneous, but since we want to check it mathematically, moving horizontally doesn't help us :
so we must do vertical i guess
2:29 PM
@GiorgiLagidze what do you mean? you can mathematically check for horizontal homogeneity of this EoM
same for vertical. both can be mathematically checked
in fact, it's homogenous under all translations
in terms of vertical, i guess, we also must move up, not just ball ?
because of same initial conditions for the experiment ?
yes. you just consider the same experiment in a translated co ordinate system
but if you also move up, then y + a is wrong as for you, it's still y that you're dropping from
the correct substitution is y=y'-a
you have to write the lagrangian in the new co ordinates
then there's someone external standing in the train as well, observing it and he stays at the same place for both experiments :D
because if you move up, to you, ball is still dropping from y
2:33 PM
hmmm why are you interpreting the variables in the lagrangian as initial conditions :P
y is a co ordinate. it's not an initial condition
what do you mean "dropping it from y"
ok lemme explain from the beginning
let's say two observers in a train. one observer is located slightly above the other
they both drop the ball from the same height relative to themselves. the initial conditions are the same, but in different co ordinate systems
we want to know if the experiments give the same results in both co ordinate systems
I think i got that part
you got a ball in your hand, you drop it, then you drop the ball from heigher, because to check homogeneity. You say that you also move up, so to you, ball gets dropped from the same height. so in this explanation, we got experiments done with the same initial conditions.
yes. it's same initial conditions but in different co ordinate systems
in terms of Lagrangian, for one observer, Lagrangian is 1/2mv^2 - mgy, but why is for second observer lagrangian as 1/2mv^2 - mg(y+a) then ?
to get the lagrangian of the other observer, you have to substitute y=y'-a in the first lagrangian
so the other lagrangian is 1/2mv2-mg(y'-a)
y' is the vertical co ordinate of the other observer
I think I know what you mean, will quickly draw the visual representation just to confirm
2:42 PM
@RyderRude ibb.co/GPnqhB1
da vinci would be proud :P
@GiorgiLagidze yes!
@GiorgiLagidze :P
tbh it's quite good
@GiorgiLagidze oh, but to be clear the co ordinate axes are named y and y'.
yes, you intended this. sorry
i think this was the whole point to name them such as so we're on the same page
yes, it is clear
now, if you solve L', you still get $y(t)$ equation. so from the observer that is in $y'$ frame, gets the equation of $y(t)$
isn't it little unpleasant ?
2:57 PM
L' gives u the EL eqn of y'(t). and L gives the one for y(t)
yeah, you could say that as well. true, since v^2 becomes \dot y' ^2
i think i was making the mistake of initial conditions being in the Lagrangian
this is just the galillean space transformation
and that's what we mean by translational invariance(i.e gallilei transformation) as galillei transformation does the same thing - transforming frame of coordinates
yes. and also you were transforming it wrong. ur transformed lagrangian should have the new co ordinate, not the old one
ah, you mean, $L = 1/2mv^2 - mg(y+a)$ is not quite correct ?
2:59 PM
@GiorgiLagidze i think Galilean transformations just include rotations and boosts normally. i havent seen translations being included under this term
@GiorgiLagidze yes, you should elimitate dy/dt and y in favor of dy'/dt and y'
so if y'=y+a. you can eliminate y from the original lagrangian by substituting y=y'-a
@RyderRude in that case, you always would get the same thing. i mean if L = 1/2my^2 - mgy^2, then transformed one would be L' = 1/2my'^2 - mgy'^2
Lagrangian is invariant under shifts by a constant
I really recommend an easier book for now, something that eases into Lagrangians
@GiorgiLagidze yes. technically, u still get that extra constant that gets thrown away
but we know when potential is mgy^2, space is non-homogeneous, no ?
@GiorgiLagidze btw y did u suddenly square y?
@GiorgiLagidze yes. in this case, u dont just get an extra constant
try substituting y=y'-a. u also get a linear term in y'
3:03 PM
@Mr.Feynman $\sum |x,t><x,t| = \sum e^{-iHt}|x><x|e^{iHt} = e^{-iHt} \sum |x><x|e^{iHt} = e^{-iHt}Ie^{iHt} = I$
you substitute y=y'-a in L ? we get $L = 1/2m\dot y'^2 - mg(y'-a)^2$ how do we get linear y' ?
@GiorgiLagidze expand the mg(y'-a)^2
mg(y'^2 - 2y'a + a^2)
so the middle term is linear :P
the EL eqn is different now
yes, i know. what i meant was if we got potential as mgy^2, with your approach, L' = 1/2my'^2 - mgy'^2 and in that case, y'(t) ends up the same as for L = 1/2my^2 - mgy^2.
even though they shouldn't.
3:06 PM
@GiorgiLagidze u r again using the wrong L'
@GiorgiLagidze it literally gave u this
include the linear term in y'
and what's wrong with using the following: $L = 1/2m\dot y^2 - mgy^2$ and $L' = 1/2m\dot y'^2 - mgy'^2 = 1/2my'^2 - mg(y+a)^2$
what's wrong is that this L' does not give u the correct dynamics of the translated co ordinates
L' and L must be related by the co ordinate transformation y'=y-a
you have a recipe to get L'. u cant use whatever u want :P
so we do vice versa ?

L' = 1/2m\dot y^2 - mgy'^2
L = 1/2m\dot \dot y' - mg(y' - a)^2
yes L and L' again dont have the same dynamics. so non homogeneoty
@GiorgiLagidze but what?
i get that but you got the same problem that since L = 1/2m\dot \dot y' - mg(y' - a)^2, you get y'(t) while you know that for L, coordinate system is y(t)
3:11 PM
i think you have misplaced some dashes
but anyway
@GiorgiLagidze i dont understand the question now
L is given in the coordinate frame of $y$. you say to write $L$ such as it contains $y'$ and not $y$. this means that you will find the equation $y'$ , but not $y$.
my question was since L is in the $y$ coordinate frame, isn't it unpleasant that we get $y'$ in $L$ and not $y$ ?
what you're saying is the same thing as we had put $y+a$ in $L'$ as substitution
problem is why do we get $y'$ equation of motion in $y$ frame :P
so you're saying that L was the lagrangian for the original co ordinate system. but we changed its variables and then claimed that L was the lagrangian for the new co ordinate system
yes, new lagrangian for $y'$ frame is $L'$ and original one for $y$ frame is $L$
isn't our point is to get $y'(t)$ and $y(t)$ separately and they must match ?
this problem is a can of worms. i dont know how to explain this one clearly
but i will try
@GiorgiLagidze wait u agree that the new lagrangian is L' and not L?
$L'$ is in $y'$ frame, while $L$ is in $y$ frame
3:17 PM
so whats the problem again?
is it "why is it justified to obtain L' by doing a change of co ordinates in L'?
if potential energy is mgy^2 by earth, then we got the followings.

$L = \frac{1}{2}m\dot y^2 - mgy^2$
agreed ?
okay, now, $L' = \frac{1}{2}m\dot y'^2 - mgy'^2$, correct ?
some books actually use the same symbol L for L(x, dot x ) and L(x', dot x')
Well, if $L'$ is in $y'$ frame and you agreed with that, why is my last statement incorrect
3:21 PM
you get the lagrangian for the new frame by substituting y=y'-a in the older frame's lagrangian. why is this unclear to you?
@bolbteppa I know that, why?
The problem was that I didn't acknowledge that in the path integral completeness relation we were considering a fixed time slice and integrating over spatial configurations only
i feel this problem is related to a notation that books use. they are meant to imply that the Lagrangian is a co-ordinate independent function. So they will use L(x, dot x) and L(x', dot x') for the same Lagrangian expressed in different co ordinates
but this notation is also quite confusing. it clashes with the notation of a mathematical function
@RyderRude We're on the same page now. it's not unclear. so we got:

$L = \frac{1}{2}m\dot y^2 - mgy^2$ (lagrangian in y frame)
$L' = \frac{1}{2}m\dot y'^2 - mg(y'-a)^2$ (lagrangian in y' frame)

agreed ?
because the functional forms of L(x, dot x ) and L(x', dot x') are different in general. and we still use the same symbol L
@GiorgiLagidze yes
ah and y(t) must match y'(t) now to be homogeneous
3:27 PM
yes. and the EL eqns are just different
makes sense. I think the big problem is everyone using different notations
and it's mindblowingly confusing. I wish every book said the same notation
yes. books will often denote Lagrangians expressed in different co ordinares with the same symbol L
in this notation, the Lagrangian is a co ordinate independent notion. so u cant take about "Lagrangian of a frame"
in feynman's book, he says that "everybody was using confusing notations, so i started using mine as a better one. but I realized that in order to be in this field, i had to get used to such confusing different notations" :D
imo, this co ordinate independent Lagrangian idea later comes in handy when u later formulate the theory on a manifold
but in intro classical mech, it is confusing to use the same symbol L for different functions of the arguments
yeah, crazy... :P
that's why I get confused most of the time and then it's easy to lose track of where i was
but yes, you did an amazing job helping me. I was wrong in using initial condition for Lagrangian and that was not book's problems, but mine
3:31 PM
does landau do the same thing? use the same symbol L for lagrangians expressed in different frames?
landau in that case is good hahaha
ok :)
Thanks so much @RyderRude. <3 appreciate more than you can imagine :P this was a big step understanding noether's theorem about shifts i believe
the transformation in Noether's theorem is not a change of co ordinates though
doesn't she use the coordinate shifts ?
3:36 PM
they just mean L(x+a)=L(x). there is no new co ordinate introduced
it's just x on both sides
i think i better finish what i'm on before jumping to noether. will let you know it wouldn't make sense <3
it's just supposes to be interpreted as the lagrangian staying constant when u make its arguments dependent on a symmetry parameter
it is not a change of variables
@GiorgiLagidze ok :)
In here :https://en.wikipedia.org/wiki/Relativistic_Breit%E2%80%93Wigner_distribution
It says about the phase space-dependence of the decay width. What is that?
4:31 PM
@GiorgiLagidze hi again. since you have drawn y and y' as measuring the distance to the same point, are you considering the motion of the same ball in both co ordinate systems?
to be clear: if you do two experiments with initial conditions: y(0)=a, dot y(0)=b and y'(0)=a, dot y'(0)=b, then if laws are homogenous, y(t) and y'(t) will be the same function of $t$
however, if you consider the motion of the same ball in both co ordinate systems, then y(0) $\neq$ y'(0), hence y(t) and y'(t) will be different
2 hours later…
6:30 PM
Would it be correct to say that the differential cross section is the probability of a scattering event per unit of time for a given solid angle ?
2 hours later…
8:52 PM
@naturallyInconsistent u must not know ACM's true power. But yeah, you're right. I gotta build up the foundation to understand the lore at some point. (I highly doubt I'll learn the relevant physics in my current applied route)

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