Different Equations of Motion for different frames

[gionole]

I might be getting hooked up on this subject, but this is the last question I'm gonna ask about this.

When potential energy on earth is given by $mgy$, we know space is said to be homogeneous. If you're standing on the ground and in one case, ball is dropped from some height and then another ball is dropped from higher height, relative to ground frame, they will have the same E.O.M.

Just for this question, imagine that potential energy was given by $mgy^2$ in our universe instead of $mgy$. If you know drop a ball again from some height and then another ball from higher height, would they have the same e.o.m relative to ground or not ? If I try to derive the each ball's eom by newton, we have the following:

$\ddot y = -2gy$
$\ddot y = -2g(y+k)$ ($k$ is the second ball's dropping point relative to first ball)

Clearly, we have got the different e.o.m, which tells us that space in that case would be non-homogeneous and relative to ground, equations of each ball is different. Ofc, if potential energy had been $mgy$, space ends up homogeneous as both e.o.ms are $\ddot y = -g$ in which case, relative to ground equation of motion is the same for each ball.

Question 1: do you agree with this so far ?

Question 2: Now, imagine a slightly different approach to our experiment such as in first case, you're standing on the ground and ball is dropped from some height(k relative to you) and note the e.o.m relative to you. now, move up higher and drop a ball again from height (k relative to you) and note the e.o.m relative to you. Basically, we moved our frame as well as ball, but left the earth and everything at the same place. Would the e.o.ms be the same relative to you in each case ?

 

[Demystifier]

Q1: No. Since the potential energy is $mgy^2$, both balls will satisfy the same equation of motion $\ddot{y}=-2gy$. However, the two balls have different trajectories $y_1(t)$ and $y_2(t)$ because they correspond to two different solutions of the same equation of motion. The two solutions correspond to two different initial conditions $y_1(0)$ and $y_2(0)$, which differ by $k$.

 

[gionole]

@Demystifier This is interesting.

With the exact same $mgy^2$, we have the different explanation by using lagrangian technique. It's a known fact that, to check homogeneity, we do the following:

$L = \frac{1}{2}my^2 - mgy^2$
$L' = \frac{1}{2}my^2 - mg(y+k)^2$

and solving these two by using euler lagrange yield different e.o.ms which confirms that space would be non-homogeneous. It's clear that e.o.ms derived with lagrangian don't match with what you said: "both balls will satisfy the same equation of motion". How ?

 

[Demystifier]

To check homogeneity you compare $L'$ with $L$. In your case $L'$ is different from $L$, and their difference is not just an irrelevant time-independent constant, which implies that the system is not homogeneous and $L'$ is not equivalent to $L$. Since $L'$ is not equivalent to $L$, this means that you are not allowed to use $L'$ instead of $L$. Hence the equation of motion derived from $L'$ is wrong.

 

[gionole]

@Demystifier

hm, you lost me at: "Since L' is not equivalent to L, this means that you are not allowed to use $L′$ instead of $L$. Hence the equation of motion derived from $L'$ is wrong".

Just because $L'$ is not equivalent to $L$, that doesn't mean we can't use it. What I'm emphesizing with these 2 lagrangians is $L$ is lagrangian for the ball dropped from some height and $L'$ is the lagrangian dropped from higher location than previous ball. I think we're allowed to compare these lagrangians and if they don't yield the same thing, then homogeneity is broken. I'm not sure what's wrong with this logic.

 

[Demystifier]

Just because $L'$ is not equivalent to $L$, that doesn't mean we can't use it.

Yes it does. If we said that the system is described by $L$, then it is not described by $L'$.

$L'$ is the lagrangian dropped from higher location than previous ball.

No it isn't. The Lagrangian does not depend on the location from which the ball is dropped.

I think you are struggling to understand what then $L'$ describes. Let me give you a hint. Do you know a real physical system in which potential energy is proportional to $y^2$?

Let me also answer your Q2: The coordinate of the shifted frame is $y'=y-k$. From $\ddot{y}=-2gy$ it follows that $\ddot{y'}=-2g(y'+k)$, so the two equations of motion do not have the same form. But they are equivalent.

 

[gionole]

@Demystifier

Do you know a real physical system in which potential energy is proportional to $y^2$ ?

Unfortunately not, and that's why I said to imagine that potential energy in our universe by earth was given by $mgy^2$ instead of $mgy$.

Can you say why then we use the coordinate shift or whatever it's called to detect homogeneity ? for the material to be homogeneous with respect to property $f$, the following must be true: $f(x) = f(x+a)$. When we're talking for homogeneity of space and we got a lagrangian $L = \frac{1}{2}m\dot y^2 - mgy^2$, how do you go about checking whether space is homogeneous or not ? I'm sure you would plug $y+a$ instead of $y$, derive the eoms and say that they're different, hence laws of physics are different for each ball in different location in space. Thoughts about this ?

 

[Dale]

Unfortunately not

How about a spring.

 

[gionole]

I guess you mean $kx^2$. Even though, I don't get how this helps me understand what $L'$ is.

My question still stays the following:

Can you say why then we use the coordinate shift or whatever it's called to detect homogeneity ? for the material to be homogeneous with respect to property $f$, the following must be true: $f(x) = f(x+a)$. When we're talking for homogeneity of space and we got a lagrangian $L = \frac{1}{2}m\dot y^2 - mgy^2$, how do you go about checking whether space is homogeneous or not ? I'm sure what you do is plug $y+a$ instead of $y$, derive the eoms and say that they're different, hence laws of physics are different for each ball in different location in space.

What's the mistake I'm making ?

- is it that we don't use $y+a$ plugging in Lagrangian to derive e.o.m for the higher location ball(i.e higher location point of space) ? If not, how do we check homogeneity if we have a lagrangian ? for density property of material, you got density function and it's easy: just do: $f(x) == f(x+a)$ check and if it holds true, you got an answer. but struggling with lagrangian case.

 

[Demystifier]

Can you say why then we use the coordinate shift or whatever it's called to detect homogeneity ? for the material to be homogeneous with respect to property $f$, the following must be true: $f(x) = f(x+a)$. When we're talking for homogeneity of space and we got a lagrangian $L = \frac{1}{2}m\dot y^2 - mgy^2$, how do you go about checking whether space is homogeneous or not ? I'm sure you would plug $y+a$ instead of $y$, derive the eoms and say that they're different, hence laws of physics are different for each ball in different location in space. Thoughts about this ?

Yes, that's all OK. The point is that $a$ is not a shift of the initial position of the ball. Instead, you can think of it as a shift of the thing that produces the potential $V(x)$. For example, if $V(x)$ is the gravitational potential produced by Earth, then $a$ is a shift of Earth.

 

[Demystifier]

I guess you mean $kx^2$. Even though, I don't get how this helps me understand what $L'$ is.

For $kx^2$, the spring does not produce any force when the ball is at the position $x=0$. The $L'$ corresponds to the potential $k(x+a)^2$, i.e. the spring does not produce any force when the ball is at the position $x=-a$. Can you imagine how to realize such $L'$ in the laboratory?

 

[gionole]

@Demystifier

The idea is that in plain words, way I understand homogeneity of space is you do experiment in one location, then you do the same experiment in another location, and if the results(the object you did experiment on) behaves the same way in both experiments, then you have homogeneity. If you just assume for a second that for 2nd experiment, we don't move earth, and only focus on moving the ball(earth stays the same at the same location in both experiments), as far as I understand, you say that in that case, in both experiments, ball would have the same and equivalent e.o.m - so if I go and observe the ball dropped from some height and another ball dropped from higher location, you're saying that its e.o.m would be the same ? are you sure that this is what I would observe ? the reason I'm saying this is the higher ball definitely would have bigger potential energy due to $mgy^2$ (y being the bigger) (i.e would cause ball to accelerate faster than the first ball) which would cause different e.o.m for sure, but as you say, in both experiments, we get $\ddot y = -2gy$.

 

[Dale]

In a potential $mgy^2$ the position $y=0$ is special in the sense that there is no force there.

 

[gionole]

@Dale what does that have to do with what I said in #12 ?

 

[Dale]

@Dale what does that have to do with what I said in #12 ?

It just seems inhomogenous to me

 

[gionole]

Exactly my point. if potential energy was given by earth to be $mgy^2$ and not $mgy$, we got space as in-homogeneous, which clearly means that different points in space behave differently. In my dropping ball experiment, dropping it from some height and dropping it from some higher height would definitely have different e.o.ms because at the higher point, ball has bigger potential energy in a sense that forces are clearly different there and for below location ball which means that higher location ball would accelerate faster downward which means higher location dropping ball must have clearly different e.o.m than lower location dropping ball - clearly saying that $\ddot y = -2gy$ is correct for both of them seems confusing to me because sure, trajectories are still different due to initial positions, but you would end up such as $y_1(t) = 10 - cos(2\sqrt{5}t)$ and $y_2(t) = 15 - cos(2\sqrt{5}t)$ - if so, these trajectories are homogeneous, which i don't think clearly describes our potential energy($mgy^2$ scenario). @Dale

 

[Dale]

dropping it from some higher height would definitely have different e.o.ms because at the higher point, ball has bigger potential energy

In the potential $mgy$ the ball also has bigger potential energy at higher heights.

 

[gionole]

In the potential $mgy$ the ball also has bigger potential energy at higher heights.

Difference is that we know $F = -\frac{dU}{dy}$ and for $mgy$, Force is constant at different heights(assuming we're close to earth), while for $mgy^2$, force ends up $-2mgy$ which is different at different points.

 

[Dale]

Difference is that we know $F = -\frac{dU}{dy}$ and for $mgy$, Force is constant at different heights(assuming we're close to earth), while for $mgy^2$, force ends up $-2mgy$ which is different at different points.

Yes.

you would end up such as y1(t)=10−cos(25t) and y2(t)=15−cos(25t) - if so, these trajectories are homogeneous

So do you think this is still correct?

 

[gionole]

@Dale no, for $mgy^2$, eoms would be definitely different for each ball which wouldn't result for $y_1(t) = 10-cos(2\sqrt{5}t)$ and $y_2(t) = 10-cos(2\sqrt{5}t)$. I'm referring to reply $2$ by @Demystifier and it confused me, because it clearly says that eoms would be the same.

 

[gionole]

Since you clearly agree with me @Dale , on the point that for $mgy^2$, space would be in-homogeneous(dropping a ball from different locations would have different e.o.ms for sure to the ground frame). So we got an agreeing check here.

As for the lagrangians, to yield the different e.o.ms(which we know are what it should yield since from experiments, we know space would be in-homogeneous for $mgy^2$), then I say:

$L = \frac{1}{2}m\dot y^2 - mgy$
$L' = \frac{1}{2}m\dot y^2 - mg(y-k)^2$

Basically, $L$ is the lagrangian of the first ball(dropped from lower location) and $L'$ is the lagrangian for the ball dropped from higher location. both lagrangians are written/presented in the ground frame. If you solve this, we're clear that we get different e.o.ms which seems what we anyways thought of and aimed to get. One point I want to make is when I do $y-k$, you might think that I subtract $k$ from $y$, because initial condition of the 2nd ball changes in 2nd experiment, but that's not the whole reason I'm doing it. the reason of doing it is that we can imagine 2nd ball being in $y'$ frame($y = y' + k$) and lagrangian for 2nd ball in that frame would be $L' = \frac{1}{2}m\dot y'^2 - mgy'^2$, but we can substitute $y'$ to be $y-k$ which transforms lagrangian such as 2nd ball's lagrangian($L' = \frac{1}{2}m\dot y^2 - mg(y-k)^2$) now is seen in the same frame(y) as first ball). I think this logic should be correct ? don't you think ? @Dale

 

[Dale]

One point I want to make is when I do y−k, you might think that I subtract k from y, because initial condition of the 2nd ball changes in 2nd experiment, but that's not the whole reason I'm doing it. the reason of doing it is that we can imagine 2nd ball being in y′ frame(y=y′+k) and lagrangian for 2nd ball in that frame would be L′=12my˙′2−mgy′2, but we can substitute y′ to be y−k which transforms lagrangian such as 2nd ball's lagrangian(L′=12my˙2−mg(y−k)2) now is seen in the same frame(y) as first ball). I think this logic should be correct ? don't you think ?

I have a hard time following your argument. Sorry.

 

[gionole]

I have a hard time following your argument. Sorry.

What I'm saying in simple words is this:

in case of having $mgy^2$ instead of $mgy$, we know space is in-homogeneous which means 1st ball dropped from some height and another ball dropped from higher place than 1st ball - each ball will have different e.o.m in ground frame.

I want to show this with Lagrangian, i.e derive e.o.ms for each ball by Lagrangian method. I call ball that drops from lower height to be "1st ball" and "2nd ball" drops from higher location.

Lagrangian for the first ball is given by $L = {1}{2}m\dot y^2 - mgy^2$. Deriving the e.o.m for this by using Euler lagrange results in $\ddot y = -2gy$.

How do we get the e.o.m for the 2nd ball - to do this, we first need to write Lagrangian for this 2nd ball. I say that we can write $L = {1}{2}m\dot y^2 - mg(y-k)$ - yes, it's true that 2nd ball is dropped from higher location, but it's lagrangian still includes $y-k$ and not $y+k$ - this comes from the fact of using frames $y$ and $y'$ and $y = y'+k$

If you understand why that is, we're good. Just wanted to mention all this if I understand this correctly. @Dale thoughts ?

 

[Dale]

we first need to write Lagrangian for this 2nd ball

Why would we rewrite the Lagrangian for the 2nd ball?

 

[gionole]

@Dale We know that for those 2 balls, eoms are different. Thats what we agreed on.

How do you get an eom for 2nd ball by lagrangian method ? If you write the same lagrangian for 2nd ball as you wrote for the first one, you will get the same eom for it which would be wrong.

 

[Dale]

How do you get an eom for 2nd ball by lagrangian method ?

I would just use different initial conditions

 

[gionole]

@Dale Yes, I think it is the same thing. Though, note one important thing.

When you derive eom for first ball, you got: y’’=-2gy. Are you saying that you would use different initial condition here (as in plugging in y-k in the eom itself to get 2nd ball eom) or first you get a solution from the y’’=-2gy and whatever you get(trajectory), you would plugin in initial condition there ?

It is a very important difference. If you solve eom and then plugging in different initial condition, then you say that both balls have the same eom which is wrong as we agreed on. So I think what you are implying is substitute y-k in the first ball’s eom(y’’=-2gy) to get 2nd ball eom.

Am I right ?

 

[Dale]

you would plugin in initial condition there ?

Yes.

then you say that both balls have the same eom which is wrong as we agreed on.

I don’t think that is wrong.

 

[gionole]

@Dale Thanks so far for following me on this. I had to write big text to clearly explain what's very confusing about this and would appreciate if you read this clearly and find a mistake or agree with me in which case your #28# reply where you say: "i don't think is wrong" would be actually wrong statement.

hm, that's the thing I'm trying to differentiate as they are compeltely different thing.

$\ddot y = -2gy$ (for first ball)

as you just said, first you find a solution to this.

and solution is $y(t) = c_1cos(\sqrt{2g}t) + c_2sin(\sqrt{2g}t)$ and by using initial conditions of first ball which tells us $c_2 = 0$ considering, the initial speed(at drop moment $v=0$),
so $y(t) = c_1cos(\sqrt{2g}t)$.

Now, as you said, you put in different initial condition in this solution to get trajectory for 2nd ball, and you get: $y(t) = (c_1+k)cos(\sqrt{2g}t)$.

If you draw those graphs, you will realize that they're same(the only difference is initial condition). other than that, there's nothing different from those - this means that dropping a ball from higher location(1st ball) moves with the exact same behaviour as 2nd ball(dropped from lower location) and this is just wrong. We said a couple of messages ago, that behavior must be different because force is different at different points. Force formula ends up as $-2mgy$ and it depends on $y$ which means the higher the ball is, the more different force acting there is, hence behaviour of 2nd ball must be definitely different than behaviour of 1st ball. (see #19 and #20 reply).

With your way, you don't get the different behaviour for the 1st and 2nd ball but as we agreed, we must get.

Now, see my suggestion. If the eom of the first ball $\ddot y = -2gy$ and for the 2nd ball, $\ddot y = -2g(y-k)$, the solutions of these are:

$y(t) = c_1cos(\sqrt{2g}t)$(1st ball)
$y(t) = c_1cos(\sqrt{2g}t) + k$ (2nd ball trajectory - you get this if you solve $\ddot y = -2g(y-k)$)

Now, do you realize how important it is that eoms for 1st ball must be $\ddot y = -2gy$ and for the 2nd ball, it must be $\ddot y = -2g(y-k)$ ? otherwise, if you have the same e.o.m for both balls, you end up getting the exact same trajectory formula, and just by plugging in initial conditions($c_1$ and $c_1 + k$) doesn't seem to give us different behaviours of their movement.

 

[Dale]

If you draw those graphs, you will realize that they're same

They really are not. In particular, one is not a spatially translated version of the other.

this means that dropping a ball from higher location(1st ball) moves with the exact same behaviour as 2nd ball(dropped from lower location)

No, the second ball dropped from a lower location has a smaller acceleration. That happens whether you change coordinates or change initial conditions.

 

[Demystifier]

The idea is that in plain words, way I understand homogeneity of space is you do experiment in one location, then you do the same experiment in another location, and if the results(the object you did experiment on) behaves the same way in both experiments, then you have homogeneity.

Right. But if you study how the ball moves, then you study its whole trajectory, i.e. you don't study the behavior at one location. You cannot test homogeneity by studying the whole trajectory. You must study things which are defined at a location. For example, you can measure the force at different locations and find that the force depends on location $y$ as $-2mgy$. It is inhomogeneous because it depends on $y$, but it does not depend on the initial position $k$. The force at the same position $y$ is always the same, no matter what the initial position was.

 

[Demystifier]

If you draw those graphs, you will realize that they're same(the only difference is initial condition).

I really have no idea why do you think so. As @Dale told you, they are not the same, and the initial condition is not the only difference between them.

Basically you are saying that $A{\rm cos}\,t$ and $(A+k){\rm cos}\,t$ are the same functions. But they are not. If you don't see that, then your source of confusion is at some very elementary level.

 

[gionole]

@Demystifier

It's now clear that $Acost$ and $(A+k)cost$ are not the same. Thanks to both of you for helping me get this.

I want to get to one point though.

I think I have a misconception, because on another chat(which is clearly lost now), someone told me that when you got two lagrangians for two objects, and solving them yields different e.o.ms, then space is in-homogeneous and if solving them yields same e.o.ms, space is homogeneous.

Let's first do Active transformation method(i.e in the 2nd experiment, we move the ball higher and not the frame).

I think my confusion lies about how we can represent Lagrangian for each ball dropped - one ball from lower height than the other. Due to active transformation method, I see that to check homogeneity, we must write lagrangian for each ball in the same frame ($y$ frame) and I have a hard time doing this.

For first ball, I write: $L_1 = \frac{1}{2}m\dot y^2 - mgy^2$. How would you go about writing lagrangian($L_2$) of the 2nd ball in the same frame ?
- If you say that $L_2 = \frac{1}{2}m\dot y^2 - mgy^2$, then I get confused as e.o.ms are exactly the same and it means homogeneous which clearly we deduced it shouldn't be.
- if you say $L_2 = \frac{1}{2}m\dot y^2 - mg(y+k)^2$, then you're really substituting $y$ by $y+k$ and it seems to me in the lagrangian you're submitting different initial condition, which also seems wrong to me.

So how do you come up with lagrangian for 2nd ball then ?

 

[Dale]

I get confused as e.o.ms are exactly the same and it means homogeneous

Why do you think this? I don’t agree.

I see no need to transform the coordinates. I mean you can if you want, but why do you think it is necessary?

 

[Demystifier]

- If you say that $L_2 = \frac{1}{2}m\dot y^2 - mgy^2$, then I get confused as e.o.ms are exactly the same and it means homogeneous which clearly we deduced it shouldn't be.

As we both already told you, no, the same Lagrangians does not mean homogeneous.

So how do you come up with lagrangian for 2nd ball then ?

Again, they have the same Lagrangians, $L_2=L_1$.

Maybe the source of your confusion is this. The Lagrangian is not a property of the particle. The Lagrangian is a way to express the laws of nature that govern the behavior of any particle. Let me use an analogy with legal laws. The legal law of one country says that you can walk in a swim suit at the beach, but not at the street. This law is not homogeneous, because the law depends on the position (beach vs street). Yet, there is only one law book in this country, and this one law book says that "you can walk in a swim suit at the beach, but not at the street". The law book is not rewritten when you move from the beach to the street. Likewise, the Lagrangian is not rewritten when the particle moves from one position to another.