@123 i didn't use experimental values of a1 a2 and a3!

yes!!

but note that there will be two experimental values of each variable

lemme define them

Ookay. What about - 1 after mathematical multiplication of constants with sign. Pls elaborate

Yes. I'm getting there

Ookay

define a1 as the acceleration of 1 when it collides with 2. u can get this experimentally

a1' as the acceleration of 1 when it collides with 3

@RyderRude Yes by controlled experiment we can find a_1 = 1 when it collide with body 2 and body 3

@123 no....u just use whatever u find. No need to use 1

ok a1 and a1' are the accelerations of 1 when it collides with 2 and 3 respectively

Ookay.. In real experiment we will use real values

a2 and a2' are for 2 when it collides with 1 and 3 respectively

Ookay

a3 and a3' are for 3 when it collides with 1 and 2 respectively

Yes

these are all experimental values

Yes

now, k12= - a1/a2

k23= -a2'/a3'

Yes

k31=-a3/a1'

now, $a_1 \neq a_1'$

Yes

so, when u multiply, these two don't cancel out

similarly, nothing cancels out

@ACuriousMind you removed your statements about the structure of the action!

Ookay

so u don't get -1

and u get 1 only from experiments @123

@qwerty yes, because the "simple example" I wanted to make there didn't actually work

@RyderRude Aaah... That's nice explanation. and intuitive

@123 yes

It's however interesting to note that the Hamiltonian formulation of the second system in fact has different equations of motion. The canonical momenta are $p_1 = \dot{q}_2, p_2 = \dot{q}_1$ and the Hamiltonian is $H = p_1 p_2 + \omega^2 q_1 q_2$, the equations of motion are like $\dot{p}_1 = -\omega q_2$, which is the wrong way around for a Hamiltonian HO, i.e. the Hamiltonian formulation knows these two Lagrangians are not "equivalent"

@ACuriousMind was it all incorrect or just the example?

@RyderRude Can we multiply constants by mathematical rule. Or we need to only rely on experiment to get 1

@123 u can multiply by mathematical rule, but a1 and a1' just don't cancel out cuz they're not equal

@qwerty I don't think I made any substantial claim outside of the example - the "structure of the action" comment was just how I would express my previous statement about the $\delta S$ in words

using mathematics, u get - a1a2a3/a1'a2'a3' @123

Pls w8. sorry about that i will be back in 15min. Thanks

>.< may I please ask you to repeat yourself in that case? I hadn't quite gotten my head around what you were saying before it was removed

@ACuriousMind hi. is Hilbert space symmetrization enforced by hand in string theory, or is it mathematically constrained

i saw Susskind's lectures but he didn't address this..

i have tried googling too

@qwerty I was simply saying that the structural difference between $\delta S = E_1 \delta q_1 + E_2 \delta q_2$ and $\delta S = E_2 \delta q_1 + E_1 \delta q_2$ is not captured by talking about the equations of motion $E_i = 0$

pls help

I do think that's the key here - if we want to invoke Noether's theorem there has to be an action formulation, the whole point is that the relevant symmetries are symmetries of the action and not merely of the equations of motion, and so we should consider the equations of motion in their property as variations of the action instead of isolated differential equations

Ah, and now I have the right analogy: Consider the functions $f = \frac{1}{2}(x^2 + y^2)$ and the function $g = xy$. They have the same critical point, since $\mathrm{d}f = x\mathrm{d}x + y\mathrm{d}y$ and $\mathrm{d}g = y\mathrm{d}x + x\mathrm{d}y$ mean the equations for the critical point are $x=0,y=0$

but the functions are different, and no one would say they are "equivalent" just because they share a critical point, particularly because it's a minimum for $f$ but not for $g$

I think i agree with the first half of what you are saying, but it's my turn to be sick and I am going to need to rest and think on this... however I think the reason for raising the point is that the actual things we test are the equations of motion, not the action though, so if the set of equations of motion are the same, then physically it's the same. so i will need to think about this a bit more

the real question is: Do we believe in the action as more than just the thing that produces the correct critical points?

In classical mechanics you might be able to say yes, but in quantum mechanics this becomes more difficult, especially in the path integral formulation but also in the operator formalism (cf. e.g. topological effects from terms in the action that leave no trace in the e.o.m. at all)

@ACuriousMind yeah - I think this is what I was getting at above. it's the EoMs not the action which are "physical"

also do you ever email people about their papers? maybe this is a bit premature but I'm curious what Harvey Brown would have to say about the points you raise.

I have never emailed anyone about any paper but I don't think it's necessarily a bad thing to do :P

Sorry @RyderRude

@RyderRude Do we get -1 or +1 from mathematics after putting all the values

may be a1' , a2' and a3' are negative also

@ACuriousMind one of my great fears is that there will be a mistake in one of my papers and everyone sees it but no one can be bothered to email me about it, otoh the grim reality is no one cares and has even read my work :P

@123 it is fine

@123 the -1 is from mathematics

@RyderRude But there is possibility of getting negative of acceleration of prime also?

@123 i can show try to show without experiments that it mus the positive

Sure

@123 there is...some of the primes must be negative. Some of the non primes must be too

@RyderRude Ookay... may be then we will get +1

@123 ok imagine we did the first collision and a1= - 2, a2= 4

Ookay

@123 yes.thats exactly the reason it's positive in the end

@ACuriousMind I would like to encourage you to do so if you feel confident in your comments :D

now we collide a2 and a3, and we get a3=-2, a2=4

@qwerty I mean I don't disagree with the brute fact that these two Lagrangians "have the same equations of motion" yet have different symmetries

Ookay

notice how a2 remains the same. so it will cancel out

Yes

I just think this perhaps indicates that taking "the equations of motion are the same" to imply that we should consider the Lagrangians/actions equivalent is wrong, rather than saying the Noetherian symmetries aren't real/choice-dependent

now the final collision : to cancel out the previous values of both a1 and a3, we will require a1= -2 and a3=-2. this is impossible

cuz accelerations must be opposite

@RyderRude Pls check there may be type in mentioning in a_3 and a_2

@123 no, it is as intended

Ookay.

may be you have to add prime in a_2'

In a way this reminds me of Norton's dome (because these two Lagrangians are also very special, how sure are you can actually get a constant $\omega$ in 2d like that?): Do you interpret Norton's dome as the death of the notion of causality, or do you dismiss it as an artifact of the mathematical idealizations of the theory?

the point is that, when we do these three collisions, the accelerations of at least one particle will be in opposite directions in the two collisions that that particle appears in

so this acceleration will not cancel out, and will introduce a minus sign

@123 yes. the primes should be there

Aaaah Ookay let me check it with animation.

@ACuriousMind I dont think he's exactly saying that; he does call them inequivalent Lagrangians after all. imo the point is that the EoMs are what's physical

Also if we take acceleration beside change in velocity the experiment should not be colliding.

acceleration means continues interaction

for collisions, the acceleration time period will be tiny

but u can take the ratios of changes in velocity, instead of ratio of accelerations

yes. may be we can use tug of war on an ice slab

r u going to simulate it

@RyderRude Yes i have simulation for that. Using LOM

nice

After discussing you now i will never forget newton's law IN-SHA-ALLAH

This discussion gave me more confidence and revision what i have read before.

i also have a really nice proof. imagine that the accelerations of each particle was of the same sign in both collisions of that particle

Yes

this means that, in both collisions, the particle was incoming from the same side.

cuz if it comes from left, it must have negative acceleration upon collision

One more thing i observed in inelastic collision.

and if right, it must have positive acceleration

@123 what

@RyderRude Exactly that's the thing i want to calculate using simulation by introducing coordinate system and different situation

lemme finish the proof. So assume particle one did Left Left, particle 2 did Right Right and Particle 3 did Right Right or Left/Left

@RyderRude Let me show you first equation of inelastic collision

so in total u have 4 Lefts and 2 Rights, or 2 Lefts and 4 Rights. But this is impossible cuz it must be 3 Left incoming particles and 3 Right incoming particlea

@RyderRude Ookay. What is meant by that. Let me read your comments again. You can complete your answer

@123 i have finished it..

I am reading your comments

@RyderRude Ooh i didn't get it. Assume right side is positive direction and left negative.

@123 yes. That's what I've assumed

@qwerty But ACM's f and g example makes it clear that just considering EoM is insufficient. f has a real minimum and so its quantum field extension would make sense, whereas g has a saddle point and is unbounded below, leading to no sensible quantum field extension. Since whatever classical universe we are interested in had better come from sensible semi-classical limit of a valid quantum system, this would be an issue.

Object A and B both moving right both have positive initial velocities before collision. A particle which is behind of B must have higher speed to do a collision. Then after collision both again have positive final velocities

@123 assume particle 1 was incoming from Left in both collisions, particle 2 was incoming from right in both, and particle 3 was right in both. So, u have 2 Left particles and 4 right particlea

Ookay

@123 i am assuming opposite directions of initial velocities...

maybe this proof needs to be generalised more

: ) Okay. Normally i take right side to be positive direction

NP. i will check it further now using animation.

yes :)

Can we discuss inelastic collision

yes

Let take coefficient of restitution $C_R$ into account.

ok...

We know LOM (Law of conservation of momentum) always conserved whether it is elastic or inelastic collision

i also have a general proof. Assume each particle had both accelrations of the same sign (Either Left or Right). Then we get 2 L and 4R, or 4L and 2R. But this is impossible cuz it must be 3L and 3R for the accelerations

cuz each pair of accelerations is opposite, so we must have 3L and 3R

@123 yes

Problem is that. We made LOM to be conserved we introduced $C_R$ otherwise it creates an error in the equation. It means the final velocities reduces

If the final velocities after collision reduces there must be some loss same as loss in KEs, which we encountered by introducing $C_R$

do u mean momentum can't be conserved because of this CR

So this loss of must contribute to the change in objects shape.

@naturallyInconsistent mhmm i'm about to go to bed but I actually thought that was the oppsite of what @ACuriousMind wrote here chat.stackexchange.com/transcript/message/66403664#66403664 . or maybe I'm too tired and read it wrong...

@RyderRude Momentum can't be conserved without $C_R$

chat later :) goodnight

@123 so u mean momentum can't be conserved in elastic collisions??

@qwerty Oh, I phrased that poorly - the 'yes' is supposed to mean 'yes, in classical mechanics you might think the critical points are all that matters', not yes to the actual question I asked before that

in other words I agree with what nI said

@RyderRude No no... I am saying momentum also not a conserved quantity in inelastic collision. Because of reduction in final velocities after collision. To compensate this reduction we introduced $C_R$

@123 oh

Am i going correct way?

@123 i haven't studied collisions.... but momentum is always conserved, so u r going in the wrong direction

I'm feeling a bit lazy to check it via computations. Does anyone know why the normalization of many-bosons wave function has normalization $\frac{\prod_i (n_i)!}{N!}$ and sometimes $\frac{1}{N!\prod_i (n_i)!}$?

Okay NP. I will see the equation once again, and i will be going to correct myself.

ok :)

One question about N2L

sure

We know from mathematics about dependent and independent variable and their notion. Like y = f(x) = ax^2 + bx + c

@ACuriousMind miao miao hath made yall the knights who say nI~

In F = ma , we have two independent variables here. mass and acceleration. Can we write this equation F(m, a) = ma

sure

u will never see F(m,a) in a physics book tho. It's not useful

Yes i have also never seen that before

we make something a function of some parameters when we want to talk about variation in its parameters

I have seen self dependency of force F(x , t , v) it is behavior of force itself.

yes. This is useful

How force itself changing over time, or position or velocities

e.g. A time varying position dependent magnetic field

It means in physics we can only write self dependencies not as a function.

But we see sometimes position is the function of time. Here time is utilized as function.

x(t) is a function of t , t is the argument

@123 u r overthinking things. we just write whatever function is useful

Yes there is no single criteria or systematic way of writing equations

yes

@RyderRude No. I just wanted to know the systematic criteria. Sometime dependencies only used for self changing , sometimes as a function.

@RyderRude @123 Can you please take this discussion to another room? It's already close to 500 messages today and no one has interacted with the discussion except you two, it's making it hard to see any other messages.

we will just end this discussion...

These are two different use. But using same convention.

Yes

@123 let's end this discussion... u just need to get used to what we write as a function. Just continue reading more physics, and see what they tend to do

@ACuriousMind sure. NP. But i want to this under your supervision. Thanks to bear me

Thank you so much @RyderRude for your time and sharing knowledge.

cya

See you : )

@123 I don't know what you mean - I don't want this under my supervision, I'm explicitly asking you to move it where I'm not forced to see it.

ACM you always get annoyed on me.

Your knowledge is very precious. One might ask stupid questions like me.

@123 I told you two days ago that I don't consider the way you're asking questions productive, as evidenced by your lack of progress to anything but basic Newtonian mechanics in four years. You are free to ignore my judgement, but you can't expect me to contribute any more answers to your questions if you do so.

But you are not contributing when asked questions about LM

I asked about virtual displacement. Whatever the answer will be , whatever. It will be helpful for me. So i can decide based on your answer i need to spend time on that or not.

I need your POV.

Because of your knowledge level.

@123 You are not entitled to anything from me (or any other user), all contributions here are voluntary, and I choose to not engage with your questions any further.

Also i have read SR several times from books, youtube , and threads. But now i want to animate Minkowski Spacetime. There is no clue from any of the resources.

@ACuriousMind Pls understand that. Everyone finally asked questions to you. So your knowledge and POV is far more experienced. Your small comment helped me a lot.

I spent my most of the time of life in understanding physics. So your small comment has high position for me. I don't know better English word to say previous sentence. Based on your idea about topic i start reading further.

Have you started reading AP French.

You know how many years i have spent here. I don't have any exam. I am a job person. I have completed my education. But still curious knowing physics ideas since last 20 years.

@RyderisnotRude. Yes. I have read 10 chapters of it with every single detail.

But AP French till now didn't gave me no new knowledge which i don't know priori. I have done all these with my hand itself.

What about the problems at the end of each chapter.

What ever AP French is saying i know and understand very well. That's why i have no question on that.

@RyderisnotRude. No i didn't see any problem.

Do the problems.

Okay. I will . But solving most of the problems require solution manual.

Based on that i can understand problems better. And the right way of thinking.

Look for those online.

Ookay definitely i will.

In LM i am stuck in "virtual displacement". Because it is the basic building block of LM.

Do you have any idea?

I saw it in D'Lambert Principle and other places. Like virtual work

Virtual Force

@123 Read,iopscience.iop.org/article/10.1088/0143-0807/27/2/014

Okay Thanks

@Arjun Do you access of it. Pls share the pdf

@123 iopscience.iop.org/article/10.1088/0143-0807/27/2/014/pdf

I have seen this before. But don't have access

@123 check out the second link and let me know

Only abstract is shown

guys, ACM didn't just ask for a conversation to end, only for it to rise from the dead.

@123 the second link is not working?Did you not get a pdf consisting of 20 pages?

No

Can you pls share pdf to me.

You can share me your google drive link of that pdf

Do not ask people to do illegal stuff here.

NP. You can email me.

I just wanted to read

your email?

@ACuriousMind I don't know what to do lol!

Thanks for your support.

@123 delete it

Pls email me. Thank you very much in advance

I will no worries : )

When i saw this paper first and the title. I early wanted to learn this. Because this exactly address my problem.

@naturallyInconsistent do you think AP French is suitable for 123's level of development

I have read 10 chapters of AP French by understanding. You can ask whatever you want. Hopefully i might answer

I am very very serious in learning physics.

You said you didn't do the problems.

Yes without the problem. I have solved and simulate many of the problems of K&K

Mastery comes from problem solving.

@RyderisnotRude. no.

Yes i understand

But doing problems from anything would be an upgrade from all the words being exchanged.

Okay, what textbook then.

It doesn't matter. Any would do.

Give him a specific book.

@123 check your mail!

@Arjun Ahaaa.. Thanks i have received and opened. I am reading it.

good,spend some time digesting it

@Arjun 🫡

If i will have any question , can i ask .

after careful reading and cross checked with books and internet.

@123 Sure

😊

@Arjun you keep using "we(I) are(am) destined" again and again. I don't think it means what do you think it means :)

I used it only once lol : (

Destiny is the residue of habits.

@LuckyChouhan I find nothing wrong with the starred message, what do you think the problem with its meaning is?

@RyderisnotRude. Do you want to say anything about the video?

please don't just post random youtube videos here without any evident context

Currently assembling data on why $\mathbb{R}$ is important

Apparently it is ~exhibiting the cohesion~ in the cool topos

In that the shape modality is naught but the localization on R

you sounded less unhinged when you were rambling about length scales :P

Just trying to figure out how much the fact that the line pops up everywhere in topology is because it's Important or just because we like spaces that have properties that are very much like The Line

the nlab seems to imply that it's a property of a type of objects called line objects but they don't really seem to say if that's mostly just the actual line or if you can have very different ones

The examples do seem to mostly just be variations of the line though

The line as a manifold, the line as a scheme, etc

@qwerty I think the short paper you gave really buried the lede by just citing sciencedirect.com/science/article/abs/pii/037015739090137Q for the harmonic oscillator example; that paper actually discusses this inverse problem at length and shows that the existence of these inequivalent Lagrangians is directly related to the existence of non-Noetherian symmetries of the equations of motion

and you can generate all possible Lagrangians by applying these non-Noetherian symmetries to the starting Lagrangian

even the neuroscientist thinks that consciousness should be taken as a given fact, but Carroll keeps insisting on his "emergence" and "eliminative materliasm" stuff

@Slereah thats an interesting question. I believe the theory was developed with The Line(TM) in mind, or atleast TheLine^n

idk where physicists get this arrogance

"i know better about consciousness than a neuroscientist"

People use The Line a lot in topology

ie for path connectedness, homotopy, etc

or, the metric :)

that too

Also most topological spaces are in the smooth topos, which is just based on products of the line?

Apparently the topological spaces that tend to fall outside of the use of The Line are the bad ones, ie things that aren't locally compact, locally connected, second countable, etc etc

but maybe that's just the reason. We study those spaces because they are like physical space in that sense

spaces like $\ell_p$ or $\ell_{\infty}$, topologically speaking, seem nothing like $\mathbb{R}$ or its products. But on the other hand, these are sequences on The Line, so it is rather moot to use this as an example

@Slereah "all of topology is either studying spaces like The Line, or stamp collection"

@nickbros123 maybe they made topology to study lines.but unfortunately this other stuff also satisfied the definition

I think line^n was more motivating tbh. but who are we to judge

I mean once you have the Line, you can do all the things you want to it

Products, quotients, subsets, etc etc

I actually cannot understand you guys' POV. I studied topology to find my way around Manhattan

i hate finite set topologies

line is so cool

ooh but Jakobian told me they study some interesting non manifold stuff

lemme search the history