The h Bar

00:54

@EE18 The net displacement should be in the same direction (or opposite).

01:45

It should make sense. All other contributions would cancel out in pairs, and only the net displacement of the Fermi surface should contribute.

(although of course, your solar cells are not metallic and should not have a Fermi surface at all. But I'm arguing analogously.)

qwerty

02:27

@ACuriousMind I thought about this a tiny bit; I should possibly revise Hamiltonian mechanics first but to be clear, this means you're considering symmetries in the equations of motion directly (i.e. what the lecturer referred to as dynamical symmetries); rather than symmetries in the action? If this is the case, this seems much more powerful (and different from) Noether's theorem

naturallyInconsistent

@qwerty you should note how ACM refers to that still as Noether's theorem. He is definitely not considering them as different things.

qwerty

@naturallyInconsistent yes - I did note that; but her paper was all about invariant integrals which in physics would correspond to the action... so if you are considering symmetries in the EoM that is a bit different. I'm still learning this but I was under the impression there is a separate proof required to relate symmetries in the action to symmetries in the EoM; and that it is not always a given. So if you can relate conservation principles directly to the symmetries in the EoM that seems different

naturallyInconsistent

03:10

I think that is a missing the forest for the trees. Prior to Noether's theorem, people knew that the Hamiltonian formalism is very good for teasing out conserved quantities. That is kind of the whole point of developing it in the first place. The interesting bit of Noether's theorem is indeed to extend the symmetries game to the Lagrangian density. However, after her work, basically all kinds of such symmetries became known in her honour. So, the Hamiltonian formalism's version is equally

Noether's theorem, even if it predated her work.

qwerty

03:24

yes, the lecture pointed out that
> The most important part of Noether's 1918 paper was not Noether's First Theorem (hence NT1) -- that was well known by Lagrange, Poincare, Jacobi and others. She systematised this understanding of these ``\textbf{global symmetries}'' in an unprecedented way. However, her main interest were in ``\textbf{local symmetries}'', in particular the \textbf{diffeomorphism invariance} of the new theory of gravity due to Einstein.
however, I'm not sure how the Hamiltonian viewpoint would relate to the second theorem whereas her formulation seems to connect the two.

naturallyInconsistent

03:35

Would it not make sense to have the Hamiltonian version of her 2nd theorem? You can have local symmetries there too

qwerty

hmm okay. maybe I should look over Hamiltonian mechanics if it really is a good way of understanding these theorems but I've just never seen it done (at least explicitly). I'm almost a bit confused why it's not mentioned if it's a nicer formulation

naturallyInconsistent

I'm not sure what you are talking about. The fact that Hamiltonian formalism is easier to tease out conserved quantities is routinely emphasised. It is also routinely emphasised that the Hamiltonian formalism seems to love to destroy manifest Lorentz invariance, whereas Lagrangian formalism keeps it. i.e. use Lagrangian formalism to see the manifest Lorentz invariance, and work on the analysis of conserved quantities using Hamiltonian formalism, and so keep cha-cha-ing between the two as needed.

qwerty

In discussions on the Noether theorem, I've not seen it.

that's what I meant.

naturallyInconsistent

04:10

That's because it is usually discussed when Hamiltonian formalism comes up

That's a very main justification for introducing Hamiltonian formalism in a short way

qwerty

then why bother with Noether's theorems?

naturallyInconsistent

Because we want manifest Lorentz invariance. Having Noether's theorems means that we can always map the conserved quantities we find using Hamiltonian formalism into one in the Lagrangian formalism.

Ryder Rude

06:20

hi

@naturallyInconsistent mapping conserved quantities back to Lagrangian mech has nothing to do with Noether theorem. it is just a matter of replacing $p$ with $\dot x$

naturallyInconsistent

you didn't have to flaunt that you didn't even read the past however many comments that had been said, that made it clear that nothing is as simple as you think it is.

Ryder Rude

06:36

@qwerty hi. one can also have an action in Hamiltonian mech. See this physics.stackexchange.com/a/580558

123

Hello Everyone...

Ryder Rude

@123 hi

123

@RyderRude Hi

Ryder Rude

@qwerty but the proof that ACuriousMind mentioned does not use the phase space action

123

I have learned 10 chapters of AP French still found no single new information , which i don't know before. All these things i have already worked by my hand itself.

Ryder Rude

06:54

@123 did u understand joshphysics's answer?

123

Yes really gave a good explanation , i have learned few new things from his answer.

Ryder Rude

@123 what things?

123

He put the N2L in systematic way. So i can arranged the piece of puzzle now. Which was separate information for me .

Like how to tackle single force to multiple forces. Which i have already read it in different books but didn't realize it before.

Ryder Rude

i will check some of ur understanding. joshphysics made a comment on Newton's first law. u should try to re-phrase it here

@123 yes. it touched on vector sums near the end

123

@RyderRude Yes. Sure you can check. But my English is weak. Pls ask one by one. I will try to give best answer.

Ryder Rude

07:01

okay :) explain first law

like what joshphysics explained about it. not the statement in books

123

In my view of definition: "A particle natural state is its uniform linear motion unless net unbalanced force act on it to change."

JoshPhysics explained law firstly in terms of inertial frame.

Ryder Rude

@123 this is the book version tho...

123

Oops

Ryder Rude

hint : joshphysics says : first law : inertial frames exist

what statement in packaged in this law

123

@RyderRude Yes that's what i said above my dear friend

But it is matter of acceptance . Such a frames exist at which N1L appears holds true called inertial frames.

Ryder Rude

07:07

yes. That is correct!

@123 i was looking for an explanation like this

123

@RyderRude Nice... Thanks. Next my friend

Ryder Rude

elaborate on "appears to hold true"

what does that mean

123

Oookay...

Ryder Rude

it is important that u do not explain this in terms of forces

because joshphysics hasn't introduced forces yet

123

Ooooh i see , i am trying to explain this in terms of force

Let me try your way without the force.

Ryder Rude

07:11

Yes

123

Hint pls

I don't want to see joshphysics answer again.

Ryder Rude

hint : how could we check Newton's first law in experiment when we can't measure forces

propose an experiment

123

Aaaah.. Okay

If i drop a ball from my pocket it will stay at rest. If i throw it, it will have constant velocity. It is a experimental test of inertial frame.

Ryder Rude

but the ball will be affected by gravity?

it will accelerate

123

It is only possible in isolated system.

If you are asking about free fall. There is different explanation and test for it.

Do i need to explain free fall?

Ryder Rude

07:16

yes... so the experiment is ( u have said it, I'm just rephrasing it) : we discovered that objects, when they are isolated from others, continue moving with their initial velocity forever

using isolation, we r ruling out interactions

so u hav phrased it without forces

it's just that it holds in isolation, far away from other objects

123

@RyderRude Ookay. It means i don't have choice to use force.

Ryder Rude

yeh.. we haven't defined a force yet

to summarise, N1L is just the discovery that there exist frames of reference in which isolated objects keep moving with their initial velocity

and we call these inertial frames

123

But isolation word itself contain no force. Yes we can say object far away from any external influence

Yes : ) So i passed 1st test?

Ryder Rude

@123 take isolation to just mean "far away". so it just a statement about distance

we r not using force anywhere

123

@RyderRude Yes...

Ryder Rude

07:20

@123 yes :)

123

Next my friend

Ryder Rude

now u have to do N3L

cuz joshphysics did N3L next

again, no forces

123

Yes...

In an isolation every pair of interact each other by equal and opposite. Is that true.

Ryder Rude

no...

123

Otherwise pls hint

Ryder Rude

07:22

hint : give an experiment which produces a result that is equivalent to N3L

N3L is about forces. ur experiment should give an equivalent result without using forces

123

Can i use momentum

Ryder Rude

yes

123

Sure. Now i can explain it.

Ryder Rude

ooh but u have to remember that mass hasn't been defined yet

so if u use mass, u must also defined it

123

Yes mass hasn't defined. What i do now.

Ryder Rude

07:24

I'm not forbidding u from using mass, but u also have to define it

but also, joshphysics expressed this experiment without using mass. he defined mass later

123

Pls W8 2 min i have a call.

Ryder Rude

ok :)

123

Sorry about that

I am free now

Ryder Rude

yeah

123

First i need to define mass?

Ryder Rude

07:27

if u want to do it that way, u can

but joshphysics did it without mass

123

Okay. Let me explain it in my knowledge

What is better , define mass later or before? because if i want to define mass i need to use force. without force one can never define inertial mass.

Because in momentum we have mass.

Ryder Rude

don't use force yet. unless u can define it...

123

So i need to define mass without force?

Ryder Rude

yes..

joshphysics did it..

123

Let me try.

Ryder Rude

07:31

but joshphysics also expressed third law first and later defined mass

u can also do that

ACuriousMind

@qwerty I'm not considering symmetries "in the equations of motion", but the definition of a conserved quantity of course involves the equations of motion - how are you going to tell it's conserved if not by plugging it into the equations of motion? The trick is that in the Hamiltonian formalism the relationship between a symmetry and a conserved quantity is almost trivial - the quantity generates its own symmetry.

123

N1L said body natural state is linear uniform motion. So the question is, what characteristic of object involve to do it happen.

Am i going right way?

Ryder Rude

sorry u r trying to define mass

123

Yes

Ryder Rude

@123 it is okay... but vague right now

do u remember joshphysics's third law

123

07:33

Let me put it another using N3L

@RyderRude Yes i remember

Ryder Rude

just re phrase that one in ur own terms

123

Yes

ACuriousMind

@qwerty Everything I've said so far is for Noether's first theorem, which is what people usually mean by Noether's theorem - since the second theorem for gauge symmetries does not lead to conserved quantities. The formulation of gauge symmetries in Hamiltonian mechanics is a rich and interesting topic - they become so-called constraints. See, for instance, this answer of mine

123

When two isolated object collide each other than their is change of velocity after the collision.

qwerty

@ACuriousMind that doesn't really answer what I was getting at: what specifically are the symmetries you're talking about? or to phrase it another way; what is being left invariant by some specific group of transformations?

Ryder Rude

07:37

@123 yes. good direction

123

How can i put mass definition in it?

Ryder Rude

first express third law experiment without using mass

123

Ookay..

ACuriousMind

@qwerty The action

123

The change in velocity of individual object before and after the collision must depend on some property of object.

@RyderRude Am i going right?

Ryder Rude

07:40

@123 yes

@123 it's not that it "must" depend. we found this from an experiment

phrasing that experiment and its result is what u must do

123

@RyderRude Yes

qwerty

@ACuriousMind and you said that was independent of the choice of Lagrangian/Hamiltonian?

123

Velocity change of individual object before and after collision found to be depend on some property of object itself. If we did this experiment again with identical but different objects. Or did it different objects.

ACuriousMind

@qwerty Well, no, of course your action depends on the Lagrangian/Hamiltonian!

qwerty

yesterday, by ACuriousMind

this doesn't really depend on your "choice" of Hamiltonian/Lagrangian, except insofar as of course the condition to be "conserved" is that the time derivative is zero, and the time derivative is given by the Poisson bracket of a quantity with the Hamiltonian (that's just Hamilton's equation of motion)

Ryder Rude

07:45

@123 identical but different doesn't make sense..

ACuriousMind

What I argued was is that in the Hamiltonian formalism, it is easy to see that symmetries that are symmetries for one Hamiltonian are symmetries for all equivalent Hamiltonians, since the condition of being a symmetry $\{f,H\} = 0$ is the equations of motion, and two Hamiltonians are equivalent if they have the same equations of motion, right?

qwerty

but the point the lecturer was making that there can be Lagrangians (or Hamiltonians) which are inequivalent and can lead to the same equations of motion

123

identical means A and B. Two different objects have same material , shape and size. But both are two objects. My English weak

Ryder Rude

@123 hint... we collided objects and observed something about the ratio of their change in velocities

@123 oh

ACuriousMind

@qwerty In what sense can two Hamiltonians be "inequivalent" if they have the same equations of motion?

Ryder Rude

07:46

@123 it makes sense then

123

@RyderRude Ookay

ACuriousMind

And again, the symmetry condition $\{f,H\} = 0$ is just Hamilton's equations of motion for $f$ - if the two Hamiltonians have the "same equations of motion", in every meaningful sense I can assign to that phrase, then if one of them has $f$ as its symmetry, the other has $f$ as its symmetry, too

qwerty

@ACuriousMind if I may use the Lagrangian, the lectures pointed out that they are equivalent if they are related by a divergence

123

@RyderRude Yes the ration of change in their velocities are always same.

Ryder Rude

@123 yes!

qwerty

07:48

however, you may have two different Lagrangians, same equations of motion, but Lagrangians are not related by a divergence. then you get different symmetries

Ryder Rude

@123 we collide the same objects repeatedly and observe that this ratio is always the same

123

If the object are same. It doesn't matter at what velocities they have before and after collision. But ratio of their change in velocities are always same.

Ryder Rude

yes!

123

So the constant we called mass

Ryder Rude

it is the ratio of the masses of the two objects!

ACuriousMind

07:51

@qwerty What is an example for two Lagrangians with the same equations of motion yet not being related by a divergence or a scalar factor?

Ryder Rude

u have the intuition of mass here. if u collide two objects, the one that is "twice as massive" will change its velocity only half as much @123

123

@RyderRude Yes but for that we need to do this experiment with three different objects. Say A, B first, then A,C and then B,C

Ryder Rude

@123 yes. that is important too

that is to show that mass is transitive

123

Then We need to define one of them as a standard mass = 1 kg

Ryder Rude

yes!

and u collide a second object with it, and if it only changes its velocity half as much as this 1kg object, what do u call the mass of the second object?

ACuriousMind

07:53

@qwerty This really can't happen, see Qmechanic's answer on the same equations of motion implying that the difference between the two Lagrangians is a divergence

123

2Kg

Mr. Feynman

Of course there would be a Qmechanic answer for this :P

123

The more heavier object the less change in velocity.

Ryder Rude

yes

and one crucial point, we don't need to measure the mass of every single thing in the universe by colliding it with this 1kg thing

123

If change in velocity is half then it must be twice heavier than relative to standard mass.

Ryder Rude

07:55

yes

qwerty

@ACuriousMind I'll try to find examples: the lecturer said he had examples but it was either skipped in the lecture or maybe I missed it. or it's in the next lecture which I haven't gotten to yet.

Ryder Rude

so we hav figured out a 2kg thing. and we collide a third object with it, and its change is velocity is always 1/3 of the change of the 2kg object

so we assign 6kg to the third object

123

@RyderRude So how can we measure mass? We have different methods of measuring mass.

@RyderRude Yes

Ryder Rude

@123 for now, take this experiment as the one that assigns mass to objects. the gravitational technique comes later from another experimental fact

123

Okay...

Ryder Rude

07:58

so we were able to define mass only because of this property of changes in velocities

123

How can we measure inertial mass without colliding it. It should be the idea of force.

qwerty

@ACuriousMind philsci-archive.pitt.edu/2914/1/MolPhys04.pdf page 7 and 8 in the pdf

123

@RyderRude Yes.

Ryder Rude

@123 we havent defined force yet. but one may also use an experiment using springs

but that one uses gravity

123

But in this experiment we have noted that the time at which object A in contact with object B is always remain the same.

Ryder Rude

08:00

@123 u can collide it with different initial velocities

the observation is that the ratio is always the same

123

Yes. But they time of contact during collision always same for both objects

Ryder Rude

yes

ACuriousMind

@qwerty Invalid example, since the Lagrangian is not a 4-vector field in standard classical mechanics

123

Using above experiment we can define change in momentum by introducing the mass.

Ryder Rude

we r also making a hypothesis that this observation holds for arbitrarily small times. i.e. a1/a2 is always constant at all times @123

qwerty

08:02

@ACuriousMind yes the rebuttal of this idea by the lecturer was the point of his quote, so I'm not sure how the Hamiltonian formulation is relevant to whether or not it is valid...?

123

I forgot to mention one important point here. Change in velocities always equal in magnitude and opposite in direction.

Ryder Rude

@123 yes. The constant is negative.

qwerty

@ACuriousMind what about the 2D harmonic oscillator?

Ryder Rude

we r using its magnitude to define mass. forget about the sign rn @123

e.g. if the ratio is -2, we say one of the objects is twice as heavy @123

we forget about the sign when defining the mass

123

Yes

Ryder Rude

08:04

one crucial thing, the observation we made only holds in inertial frames

cuz momentum isn't conserved in non inertial

123

$\frac{a_1}{a_2} = \frac{m_2}{m_1}$

@RyderRude Yes

Ryder Rude

@123 yes.this is for the magnitudes

@123 so we use the frames, whose existence we found out in the previous experiment, to do a new experiment

and using this new experiment, we have defined mass

123

@RyderRude Yes the frame should be inertial.

Ryder Rude

now we will define force...

123

Ookay

Ryder Rude

08:06

u have defined mass. and u already have acceleration defined

123

Let me calculate mass first. Pls

Pls check yourself am i correct or not.

Ryder Rude

ok

123

$\frac{a_1}{a_2} = K_{12}$

Ryder Rude

yes

123

$\frac{a_2}{a_3} = K_{23}$

Ryder Rude

08:08

yes

123

$\frac{a_3}{a_1} = K_{31}$

Multiplying all three equations

Ryder Rude

sorry

@123 yeah..go on

123

$\frac{a_1}{a_2}\times\frac{a_2}{a_3}\times\frac{a_3}{a_1} = K$

Ooop it goes to 1

Ryder Rude

it does

123

What mistake i did. I have read this technique in simon mechanics

ACuriousMind

08:10

@qwerty debateable; one is a coordinate transformation of the other under $q_1 \mapsto q_1 + \mathrm{i}q_2$, $q_2 \mapsto \mathrm{i}q_1 + q_2$

qwerty

@ACuriousMind I dont know what the topological counterexamples are but they seem like they are an important caveat from the way Qmechanic was phrasing it

Ryder Rude

@123 there's no mistake. this is 1 indeed

123

K = 1, how can we define mass using this way.

What is the next step. I need to consult book again

ACuriousMind

@qwerty Qmechanic is always very precise, of course they are important in the sense that he doesn't want to state a theorem that's wrong; I don't think they are that important in practice, since the E-L equations are local differential equations

Ryder Rude

@123 mass is defined using K_12 and K_23. Not using K

123

08:12

Yes

Ryder Rude

K_12 is not 1, right?

123

$K_{12}\times K_{23} \times K_{31} = 1$

Ryder Rude

so m2 is K12 times m1

if u use m1 to define 1kg, then m2 is K12kg

123

Yes i can remember one of the mass we need to assign value of 1

Ryder Rude

and m3 is K23 times m2 = K23 x K12 kg

so we have assigned masses to all of the three objects

123

08:15

I am out of mind this time. How can i write $K_{12} = ?$

Ryder Rude

@123 u have already read this discussion in Simon

123

Yes but about 1 or 2 year ago

Ryder Rude

@123 K_12 can only be measured in an experiment

@123 oh

@123 it must be a great book to use this technique

123

@RyderRude Yes excellent book. I found K&K , symon, and taylor mechanics are great.

qwerty

I'm curious @ACuriousMind if you agreed with what naturallyInconsistent was saying earlier about the role of Lagrangian/Hamiltonian mechanics as well?

Ryder Rude

08:17

@123 i will check it out. I have Taylor rn

123

Other books never compelled me. I have more than 100 classical mechanics books in my laptop. But these only three are great

@RyderRude What is your opinion about Taylor?

Ryder Rude

@123 that's a lot :P

@123 i haven't read it...

@123 i learned Hamiltonian and Lagrangian from Shankar's qm book

123

Even AP French didn't gave me no new information

Ryder Rude

@123 i plan to read it

@123 oh

123

Simon mechanics also great book.

But i have read only NM part not LG.

Ryder Rude

08:20

oh

123

At that time it was beyond my scope. Now i will try that for LM

Ryder Rude

yeah.

@123 u can also try Shankar's QM Chapter 2

but it is a short review cuz the book is supposed to be on QM

@123 Simon is probably more detailed

ACuriousMind

@qwerty I don't particularly object to it, but my personal interest in Hamiltonian mechanics lies not in its classical applications, but in that it is the classical formulation that "looks like" quantum mechanics (canonical quantization is after all just replacing the Poisson bracket by a commutator) and the starting point for most serious discussions of quantization

123

@RyderRude Yes

qwerty

mhm I see.

123

08:24

@RyderRude Pls let me pick my kids from school. I will be online back after 1 hour. Sorry about that. I need to discuss about mass calculation from constants.

Thank you for your time and interaction.

ACuriousMind

Also I think the example with the harmonic oscillator finally made me understand what your concern is - Noether's theorem may associate the "same" conserved quantity with different symmetries depending on the action

Ryder Rude

@123 yes. I will discuss later

123

Thanks

qwerty

@ACuriousMind yes!!!

ACuriousMind

but I think the counterexample being pathologic (non-standard kinetic term) weakens it - it's well known (also e.g. in the discussion of Hamiltonian gauge theories) that by choosing non-standard kinetic terms one may produce all sorts of effects we don't "want" in our theory; the math seems to have a strong preference for Lagrangians of the form $\dot{q}^2 + V(q)$, and for those none of these problems occur

qwerty

08:29

yes, Harvey Brown did say they were contrived examples, but that they existed at all were an argument against saying that "symmetries explain conservation" rather than just "there is a correspondence..."

123

@RyderRude I have consult the book they did it beautifully. I will show . Thanks

ACuriousMind

@qwerty the philosophical question here really is if we need to consider non-standard kinetic terms part of physics proper

I don't know of any single real-world system described usefully with a non-standard kinetic term

and e.g. path integral quantization methods also need to assume standard kinetic term otherwise they fall apart completely

qwerty

interesting. also I'm wondering how this fits in (or not) with qmechanic's theorem? is it because as you mentioned it's a coordinate transformation?

naturallyInconsistent

@ACuriousMind phewww. So worried that something would be slightly wrong there

Ryder Rude

09:15

Einstein once proposed that all particles wer black holes

Einstein’s Other Theory of Everything

►

this theory predicted the wrong size of neutrons and they got rid of it

123

Hello @RyderRude

Ryder Rude

in string theory, is Hilbert space symmetrisation enforced by hand or constrained by the theory

@123 hi

123

@RyderRude Pls look at the simon proof above

Ryder Rude

I'll be back in few minutes

123

Sure

Ryder Rude

09:32

@123 this is all correct

and this book is a masterpiece. i can tell from this page

123

But i found one problem here

Ryder Rude

@123 what

123

multiplying all three constants should give - 1

why no minus sign

Ryder Rude

lemme check

can u show the previous page

123

Sure

ACuriousMind

09:36

@qwerty I didn't check the references, but I think I figured out the trick: The problem is - what do we mean by "the same equations of motion"? Yes, the full set of E-L equations for the two HO Lagrangians are the same, but if you look at the individual E-L expressions for the variables, they're different - in the "squeezed" case, the e.o.m. for $q_1$ is the e.o.m. for $q_2$ in the normal case and vice versa!

123

ACuriousMind

In that case, of course, the argument that if $L$ and $L'$ have the same e.o.m. so $L-L'$ has trivial e.o.m. since the E-L equations are linear in $L$ doesn't work

123

Ryder Rude

@123 k_{ij} is always positive, so the product can never be -1

ACuriousMind

And in the abstract variational sense, this certainly is not "the same equations of motion" - schematically, we consider $\delta S = E_1 \delta q_1 + E_2 \delta q_2 = 0$, where $E_i$ are the E-L expressions for the respective variable $q_i$, and the equations $E_1 \delta q_1 + E_2 \delta q_2 = 0$ and $E_2 \delta q_1 + E_1 \delta q_2 = 0$ are different equations, even if the "set of equations of motions" they determine is the same

123

09:41

@RyderRude Did he mention in the book?

Ryder Rude

@123 here

note that K12= -a1/a2

and a1/a2 is always negative

cuz they r always opposite

so k12 is always positive

but i do see how u got -1

123

Yes but it is not following mathematical rule of multiplication

Ryder Rude

yes. I see how u got a -1

123

Is there any strong argument at which we assert that constant is always positive can never be negative

Ryder Rude

k12 x k23 x k31 = - a1/a2 x - a2/a3 x - a3/a1= -1

123

09:45

@RyderRude Yes i did it this way

Ryder Rude

this is the wrong way to do it

123

Okay pls share correct way

Ryder Rude

@123 just before this relation, the book says that this relaiton is an experimental fact. Can u see why

if ur computation were true, this would be a mathematical identity

123

Nope.... :(

Ryder Rude

but it actually follows from experiments!!

i call it transitivity of masses. this has been discussed in the comments to joshphysics's answer

imagine u collided two objects and found that a1/a2= - k12 is always true for the two objects

now u collided 2 with 3 and found that a2/a3= - k23 is always true

does this mean that the a2 in this equation is the same as the a2 in the previous equation? No

qwerty

09:49

@ACuriousMind this is really fascinating to me. I wonder if anyone mentioned this to Harvey Brown since he wrote that paper in 2005? I'm guessing not since he cited himself in 2020 as well arxiv.org/abs/2010.10909

123

@RyderRude Yes i am reading the comments

@RyderRude Yes this is the problem in this definition. I didn't think about this

What is the answer of this problem? This is a serious problem

Ryder Rude

the point is that a1/a2= - k12 holds for arbitrary a1, a2. similarly a2/a3= - k23 holds for arbitrary a2, a3

and a3/a1=-k31 holds for arbitrary a3, a1

123

Yes that's why we have three different constants

Ryder Rude

imagine u collided 1 and 2, and 1 accelerated by a1 = -1 and a2 = 3. What is k12

it is -1/3

123

Yes

Ryder Rude

09:54

now u collided 2 and 3, and a2= -4 and a3= 8. so k23= -1/2

123

Okay

Ryder Rude

@123 first, note that a2 in the first experiment is different from the a2 in the second experiment

123

Yes

My question is that. if we put negative sign with acceleration. Do we need to put negative sign with constant as well?

Ryder Rude

now collide 3 and 1. it is theoretically possible that a3 was -9 and a1 was 3

now k31= - 3

@123 no the constant must be positive

correct my previous comments

123

Ookay

Ryder Rude

09:58

k12=1/3, k23=1/2 and k31=3

123

but your values constant doesn't make = 1

Ryder Rude

exactly! Why did this happen

123

What is the answer

Do de need to take one of the mass equal to 1?

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