Different Equations of Motion for different frames

[Dale]

You can look right at the EOM directly $$\ddot y= -2gy$$ It is inhomogenous since the acceleration depends on position.

 

[gionole]

I think that was my confusion. in the $y$ frame, both balls have the same Lagrangian because they're both in the $y$ frame, hence $L = \frac{1}{2}m\dot y^2 - mgy^2$ is lagrangian for each ball and then in the solutions, we will plug in different initial conditions, which ends up giving such trajectories that they clearly have different behaviours for each ball which is again non-homogeneous. Good.

I think since we're using active transformation method(i.e describing each ball in the same frame), substituting $y+k$ instead of $y$ in the lagrangian is plain wrong.

In the passive transformation mode, i.e only one experiment is done(one ball dropped from some height) and how we can as well check homogeneity is observe the ball's motion from different frames($y$ and $y'$) and frames are related to each other by $y = y' + k$. This way, in one coordinate frame, the ball has $L = \frac{1}{2}m\dot y^2 - mgy^2$ and in another coordinate frame, the same ball would have $L' = \frac{1}{2}m\dot y'^2 - mgy'^2 = \frac{1}{2}m\dot y^2 - mg(y-k)^2$. Clearly, if we solve both, first lagrangian gives us $y(t) = c_1cos(\sqrt{2g}t)$ and second one gives: $y(t) = c_1cos(\sqrt{2g}t) + k$ which tells us that $L'$ couldn't be used as we know that system is described by $c_1cos(\sqrt{2g}t)$.

My question is since we got:

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

what makes us believe that space is inhomogeneous ? shouldn't we derive $y'$ and compare it to $y$ ? Though, From above, what we actually do is derive $y$ in both cases and yes, I agree, second $y$(derived from $L'$) is wrong, but I think end goal must be deriving $y'$ and comparing it to $y$. I think this question goes to you @Demystifier

 

[Demystifier]

@gionole let us try this way. Suppose that you never heard about Lagrangian formalism and that you only know the Newton equation. Would you still be confused? How would you describe your confusion in this case?

Let me describe how I would explain it. The Newton equation is
$$m\ddot{y}=F(y)$$
where, for the sake of generality, the force $F(y)$ is an arbitrary function. Now let us introduce a new coordinate (passive translation) $y'=y+a$, where $a$ is a constant. In terms of the new coordinate the Newton equation is written as
$$m\ddot{y'}=F(y'-a)$$
This describes the same physics as the first Newton equation above. If $y(t)$ is a solution of the first Newton equation, then $y'(t)=y(t)+a$ is the physically equivalent solution of the second Newton equation. But in general, the force is not homogeneous. We say that the force is homogeneous if and only if
$$F(y'-a)=F(y')$$
This only happens if the force is a constant, i.e. if $F(y)$ does not depend on $y$.

Can you now explain your confusion, if any, in terms of this?

 

[Demystifier]

but I think end goal must be deriving y′ and comparing it to y. I think this question goes to you @Demystifier

If by deriving you mean solving the Newton equation, no, this is not the end goal. To see that the system is not homogeneous, you don't need to actually solve the Newton equation. There is a much simpler way to do it.

 

[gionole]

Thanks @Demystifier for bearing with me all this time. I think, the problem of my confusion is not related to Lagrangian, so you're right and in this reply, I will exactly follow your newtonian way. Let's again note that potential energy was given as $mgy^2$.

I'm sure you know active and passive transformation. If you don't know exactly what I mean, here is an explanation so we're both on the same page - Youtube - since on the video, Susskind says that they are really the same, I want to understand homogeneity check in both cases.

Active Transformation

ball is dropped from some location in one experiment and then ball is dropped from higher location in 2nd experiment. Since in active transformation, coordinate doesn't change, in both experiments, we're in the same $y$ frame that doesn't change. Thanks to you, I now understand that for each experiment, the newton equation is the same - i.e $\ddot y = - 2gy$, we solve this, insert initial location as $c_1$ and $c_1+k$ and get: $y_1(t) = c_1cos({\sqrt{2g}t})$ and $y_2(t) = (c_1+k)cos({\sqrt{2g}t})$. We see that these trajectories don't have the same behaviour. (In the beginning, i thought they had - that was my confusion as well, but thanks to you and Dale, I now realize that these trajectories are such as behaviour of ball is definitely different). In this active transformation method, it's wrong to have said that for 2nd ball, newton equation is $\ddot y = -2g(y+k)$, because as said, for both experiments, newton equation is the same as already stated $\ddot y = - 2gy$. I think you agree with me on all of this ?

Passive Transformation

Ok, so you use $y' = y + k$ (so you shifted $y'$ frame downwards). Now, we don't need to do 2 experiments in the same sense and just 2 observers looking at the same ball dropping is enough, as if the ball has different behaviour to each observer, then we got non-homogeneity. For the $y$ frame observer, newton equation is the same: $\ddot y = -2gy$ and we yield $y_1(t) = c_1cos({\sqrt{2g}t})$. For the 2nd observer($y'$ frame), We can still write newton as: $\ddot y' = -2gy'$(I didn't replace $y'$ by $y+k$) and solution to this is the same, but ofc, initial location would be different - $y_2(t) = c_2cos({\sqrt{2g}t})$. We again got what we wanted - $c_1$ and $c_2$ is different, so each observer sees the motion with different behaviour. Niceee. I think you also agree with me on this.

Before moving on, do you agree with me everything so far ?
---

Ok, time for my confusion. In the passive transformation method, what you did was the following:

$\ddot y' = -2gy'$
$\ddot y = -2g(y+k)$ (since $y' = y+k$, we substitute $y'$ by $y+k$)

Ofc, if you solve it now, you will get $y(t)$ such as it won't match $y(t)$ that we got in the $y$ frame, but we know that whatever we got in $y$ frame, it's the correct one as that's what system is described by, but here, $\ddot y = -2g(y+k)$ definitely gives different $y(t)$, so solution to $\ddot y = -2g(y+k)$ can't be right.

My question is: what was the reason of why we plugged in $y+k$ instead of $y'$ ? Clearly, $y' = y+k$ is a correct transformation, so that substitution is done correctly, but $y$ then yields different result. I think plugging $y+k$ and seeing if it yields different thing is a quick way of checking homogeneity, but what I'm asking is if we solve $\ddot y = -2g(y+k)$ and it doesn't match $y(t)$ that we got from $\ddot y = -2gy$, why does that mean that space is non-homogeneous ? In my opinion, what it means is the $\ddot y = -2g(y+k)$ is a wrong equation as you pointed out in $4$ reply and can't be used.

 

[Demystifier]

@gionole perhaps you want to see what's the difference between the homogeneous and non-homogeneous system at the level of explicit solutions of the equations of motion. It can be seen as follows.

We consider a Lagrangian of the form
$$L=\frac{m\dot{x}}{2}-V(x)$$
and consider a passive translation (change of coordinate)
$$x'=x+a$$
where $a$ is a constant. Hence the Lagrangian can also be written as
$$L=\frac{m\dot{x}'}{2}-V(x'-a)$$
It is really the same Lagrangian and describes the same physics, but expressed in terms of a new coordinate $x'$ instead of $x$. The resulting equations of motion are the Newton equations
$$m\ddot{x}=-\frac{\partial V(x)}{\partial x}$$
$$m\ddot{x}'=-\frac{\partial V(x'-a)}{\partial x'}$$
We want to compare solutions of the two Newton equations, and see the homogeneity or non-homogeneity through the comparison of the solutions.

Consider first the case
$$V(x)=gx$$
In this case the two Newton equations are
$$m\ddot{x}=-g$$
$$m\ddot{x}'=-g$$
We study solutions for which the initial velocities are zero, so under this restriction the general solutions are
$$x(t)=x(0)-\frac{g}{2}t^2$$
$$x'(t)=x'(0)-\frac{g}{2}t^2$$
We see that the two solutions have the same algebraic form, namely the second solution is obtained by the replacement $x(0) \to x'(0)$. This is how the homogeneity of the potential $V(x)=gx$ is seen at the level of solutions.

Now consider the case
$$V(x)=\frac{k}{2}x^2$$
Now the two Newton equations are
$$m\ddot{x}=-kx$$
$$m\ddot{x}'=-k(x'-a)$$
The general solution of the first Newton equation (again under the restriction that the initial velocity is zero) is
$$x(t)=x(0) {\rm cos}\,\omega t$$
where $\omega=\sqrt{k/m}$. What is the general solution of the second Newton equation? If the system was homogeneous, in analogy with the homogeneous case above one would expect that it is
$$x'(t)=x'(0) {\rm cos}\,\omega t$$
Nevertheless, it is not the solution of the second Newton equation. Instead, the solution is
$$x'(t)=a+[x'(0)-a] {\rm cos}\,\omega t$$
It is not obtained by the replacement $x(0) \to x'(0)$ in the solution of the first Newton equation, which is how the non-homogeneity of $V(x)=\frac{k}{2}x^2$ is seen at the level of solutions.

 

[gionole]

1. Have you read my reply ? wanted to see what you don't agree in it so I can clear my doubts. I mean the reply #40.

2.

If the system was homogeneous, in analogy with the homogeneous case above one would expect that it is
$x'(t) = x'(0)coswt$

Hm, $x(0)$ and $x'(0)$ for sure are different. Didn't we say that $x(0)coswt$ and $x'(0)coswt$ are non-homogeneous ? they behave differently and are not spatially shifted functions. Even if the we had gotten the exact same solution of $x'(t)=x'(0)coswt$, it wouldn't be homogeneous. In reply #32, this is what you told me:

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 $Acoswt$ and $(A+k)coswt$ 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.

3. Are you sure when you got $x' = x+a$, you can write: $L = \frac{m\dot x^2}{2} - V(x'+a)$ ? shouldn'it be : $L = \frac{m\dot x^2}{2} - V(x' - a)$ since $x = x' - a$ ?

 

[Demystifier]

Passive Transformation

We can still write newton as: $\ddot y' = -2gy'$(I didn't replace $y'$ by $y+k$)

That's wrong. The correct Newton equation is $\ddot y' = -2g(y'-k)$. Since the force is $F=-2gy$, and since $y=y'-k$, it follows that the force is $F=-2g(y'-k)$. When you correct this, is everything resolved?

 

[Demystifier]

3. Are you sure when you got $x' = x+a$, you can write: $L = \frac{m\dot x^2}{2} - V(x'+a)$ ? shouldn'it be : $L = \frac{m\dot x^2}{2} - V(x' - a)$ since $x = x' - a$ ?

Sorry, it was a typo, now I've corrected it.

 

[Demystifier]

2. Hm, $x(0)$ and $x'(0)$ for sure are different. Didn't we say that $x(0)coswt$ and $x'(0)coswt$ are non-homogeneous ? they behave differently and are not spatially shifted functions. Even if the we had gotten the exact same solution of $x'(t)=x'(0)coswt$, it wouldn't be homogeneous. In reply #32, this is what you told me:

Read again what I told you. I said they are different functions, but I didn't say that it means non-homogeneous.

 

[gionole]

I think we're super close. I will let you know now what's my remaining points and if you could answer one by one, we can clear them sooner.

Question 1:

Read again what I told you. I said they are different functions, but I didn't say that it means non-homogeneous.

I think, you're right. So what you're saying is if system was homogeneous, 2 things must happen. E.O.M of $x'(t)$ definitely must be exact same form such as $x'(t) = x(0)coswt$ - (if not, clearly not homogeneous), but that's not enough, because even if it would have $x'(t) = x(0)coswt$, it would still be non-homogeneous. So sum up is: if eoms are different, it clearly means system is non-homogeneous and no further need to check anything else. If eoms are the same though, it still doesn't mean homogeneity and we need to look at the trajectory solutions. Correct ?

Question 2: Now, I think we're at a center point of my confusion.
I said: $\ddot y' = -2gy'$ in $y'$ frame. Why is this wrong ? the ball in the $y'$ frame has $y'$ coordinates, what makes you say that this is wrong ?

I understand why you're doing the following: $\ddot y' = -2g(y'-k)$ (You want to write equation of the ball not in $y$ coordinates, but in $y'$ coordinates, and that's clear, but I don't know what makes the following wrong: $\ddot y' = -2gy' => \ddot y = -2g(y+k)$ (maybe this one is correct as well, it's just doesn't give us $y'(t)$ which is what we need.)

 

[Demystifier]

E.O.M of $x'(t)$ definitely must be exact same form such as $x'(t) = x(0)coswt$

This doesn't make sense. The $x'(t) = x(0)coswt$ is not an EOM, it is a solution of an EOM.

If eoms are the same though, it still doesn't mean homogeneity and we need to look at the trajectory solutions. Correct ?

No, same eoms are enough.

Question 2: Now, I think we're at a center point of my confusion. [/B]
I said: $\ddot y' = -2gy'$ in $y'$ frame. Why is this wrong ? the ball in the $y'$ frame has $y'$ coordinates, what makes you say that this is wrong ?

Just because his coordinate is called $y'$ does not imply that the force is $-2gy'$. The symbol $y'$ is not an alternative name for the same coordinate. It is really a different coordinate, $y'\neq y$.

Perhaps it's useful to illustrate this on an example. Suppose that that we are interested in a position of a fixed object, say the city of London. There is only one London , i.e. London=London. However, $y'_{\rm London}\neq y_{\rm London}$.

I understand why you're doing the following: $\ddot y' = -2g(y'-k)$ (You want to write equation of the ball not in $y$ coordinates, but in $y'$ coordinates, and that's clear,

OK.

but I don't know what makes the following wrong: $\ddot y' = -2gy' => \ddot y = -2g(y+k)$ (maybe this one is correct as well, it's just doesn't give us $y'(t)$ which is what we need.)

The initial assumption $\ddot y' = -2gy'$ is wrong, hence the final result $\ddot y = -2g(y+k)$ is wrong. Indeed, the final result must be wrong because we know that in fact $\ddot y = -2gy$. It is not consistent that both $\ddot y = -2gy$ and $\ddot y = -2g(y+k)$ are true, unless $k=0$.

 

[gionole]

No, same eoms are enough.

Interesting. When we use active transformation method($y$ frame only), for each ball(since ball is dropped from some height and in 2nd experiment, it's dropped from higher location), we said that each ball would have $\ddot y = -2gy$.

1st ball - eom is $\ddot y = -2gy$.
2nd ball - eom is $\ddot y = -2gy$

You see, eoms are the same, so with what you say, the system has to be homogeneous. Well, let's solve them

1st ball - $y(t) = Acos(wt)$
2nd ball - $y(t) = Bcos(wt)$

A and B are initial conditions. So you and Dale said that because of these trajectories, system is not homogeneous, as these functions are not spatially translated version of the other. Hence, I just showed that even though eoms were the same, trajectories ended up such as they have different behaviour(i.e non-homogeneous). That's why I said that even though same eoms are given, that doesn't mean system is homogeneous. Even in passive transformation mode, let's say you got same eoms, but if you end up with $y(t) = Acos(wt)$ and $y'(t) = Bcos(wt)$, you still get non-homogeneity answer.

 

[Demystifier]

Interesting.

I should have been more precise. If the eom is $m\ddot x = F(x)$ and if $F(x)$ is the same for all $x$, then the system is homogeneous. If $F(x)$ is not the same for all $x$, then the system is not homogeneous. That's all.

 

[Dale]

we said that each ball would have $\ddot y = -2gy$.

1st ball - eom is $\ddot y = -2gy$.
2nd ball - eom is $\ddot y = -2gy$

You see, eoms are the same, so with what you say, the system has to be homogeneous.

Did you miss my post #36? Since $-2gy$ is not a constant function of $y$ the system is non-homogenous.

 

[gionole]

Yes, that's what I meant. EOM's being the same doesn't mean it's homogeneous. It only means that when eom doesn't depend on position, that's when system is homogeneous.

Thanks so much for putting blood and sweat. YOU ROCK <3 There's one point that I'm struggling to understand, but will think about it today so that at least I have given good thought and will let you know tomorrow the update.

 

[gionole]

@Demystifier Hi.

Point 1: Now I realize why you said that $\ddot y' = -2gy'$ is wrong and I agree. It seems that in the passive translation method(which you use $y' = y+a$), what's going on is we just change frames and leave everything in the space without any change. For example, You look at the same ball from different angle than me. In that case, for sure, if the potential energy by earth is given as $mgy^2$, if you're exactly looking at the ball from ground frame, yes for you, $mgy^2$ is correct, but to me, $mgy'^2$ would be wrong, because that says that for me, potential energy of the ball must be bigger in my frame, which is plain wrong as earth is at the same position, so whatever is potential energy in your frame, it has to be the same in mine.
Correct (1) ?

Point 2:

On this passive transformation method, I just want to ensure one thing: earth is at the same place right ? It will make sense why I'm asking in a sec.

I wonder, why people say that to check homogeneity, we must move all the important parts: This is what's read on one of the answers on stackexchange.

Set up an experiment. Find the result.

Move the experiment somewhere else and run it again. You will get the same result.

To do this, you must move all the important parts of the experiment. For example, if you drop a rock on earth, it falls. If you move the experiment out into space, it just floats. To do it right, you would have to move the earth too.

The results would not be exactly the same because gravity from the moon and the sun have a small effect. So you really need to move them too. And if you really get precise, everything in the universe has a small effect. So you need to move the whole universe. And if you do that, how do you know you have moved anything? It gets confusing.

In our explanations so far on this thread, when we did active translation method, we only changed ball's initial location to be dropped from and in passive transformation method, we just look at it from different frames, but everything in space stays at the same place. What does the quote talk about then ? Does it mean checking homogeneity on bigger scale ? I understand that homogeneity of space is scale-dependent, so I think that's what the quote means but for our thread discussion, I don't think that's necessary to do stuff like that since we're curious whether space is homogeneous on smaller scale(just to us being on earth and using $mgy^2$). Correct (2) ?

 

[Demystifier]

@gionole I think your (1) and (2) are both correct.

 

[gionole]

Hey @Demystifier . I didn't provide any update since I've been super busy with the programming job, but I also gave some thoughts on this matter and a couple of points struck me. You can answer whenever you're free - I already took so much time from you and I feel bad about this, but things I'm asking are not really explained anywhere as you explain it to me.

Point 1: In this huge thread, we have been discussing homogeneity and it turns out that $y' = y + k$ transformation and substituting $y' - k$ instead of $y$ confirms that we're using passive transformation method - i.e changing the frame while everything else stays the same - i.e object we're experimenting on only drops from single location - i.e one experiment, but just looking the same thing from different frames. I was wondering about one thing - does this method really provide a solid way of checking homogeneity - asking this because we're not really dropping the ball from multiple locations in space, but single point, so maybe we're not testing homogeneity well enough ? in the absolute sense, homogeneity should be tested as object being dropped from different points in space. What would be your explanation why passive transformation still gives a solid way of checking homogeneity ?

Point 2: I also am trying to understand isotropy and I guess, let's also use passive transformation method here. The ball is dropped one time from some single location. first, we look at it in $x,y$ frame and then in the rotated frame($x', y'$) frame. I understand that rotation can be described as :

$x' = xcos\theta + ysin\theta$ (1)
$y' = -xsin\theta + ycos\theta$ (2)

if we're in earth gravitational scenario, in $x,y$ frame, we have: $\ddot y = -g$. In order to get e.o.m of the same ball in $x', y'$, we first figure out what $y$ is from using (2).

$y = \frac{y' +xsin\theta}{cos\theta}$

$\frac{d^2}{dt^2}(\frac{y' +xsin\theta}{cos\theta}) = -g$
$\frac{1}{cos\theta}(\frac{d^2y'}{dt^2} + \frac{dx^2}{dt^2}sin\theta) = -g$

since $\ddot x = 0$, we end up:

$\ddot y' = -gcos\theta$

and this is actually different(it gives us $y'(t)$ such as its behaviour is definitely not the same as of $y(t)$).

I think I'm correct so far ? If yes, then could you quote this and say that I'm correct ?

okay, now let's do for $V = k(x^2 + y^2)$

Using the transformation and plugging them in directly in $V$, we see that in the rotated frame, potential ends up being the same, which confirms isotropy, but what if we do it separately for x and y axis.

in $x,y$ frame, $ma_x = -2kx => \ddot x = \frac{-2kx}{m}$, Now we can plug $x'cosa - y'sina$ instead of $x$ and we get: $\ddot x'cosa - \ddot y'sina = \frac{-2k(x'cosa - y'sina)}{m}$. From this, solution $x'(t)$ doesn't seem to match with $x(t)$ from $\ddot x = \frac{-2kx}{m}$. Any idea why ? or we should be making $y'sina = 0$, because we're working in $x$ direction only and $y'$ or $y$ is not $x$ direction for sure, but is this a valid thing to do ? Problem that I see why I can't make $y'sina$ to be 0 is that $y'sina$ is included in the coordinate of $x$, if I make it zero, then coordinate of $x$ is definitely wrong.

 

[Demystifier]

What would be your explanation why passive transformation still gives a solid way of checking homogeneity ?

A simple answer is, because passive and active transformation can be written in the same mathematical way.

 

[Demystifier]

but what if we do it separately for x and y axis.

in $x,y$ frame, $ma_x = -2kx => \ddot x = \frac{-2kx}{m}$, Now we can plug $x'cosa - y'sina$ instead of $x$ and we get: $\ddot x'cosa - \ddot y'sina = \frac{-2k(x'cosa - y'sina)}{m}$. From this, solution $x'(t)$ doesn't seem to match with $x(t)$ from $\ddot x = \frac{-2kx}{m}$. Any idea why ?

You say that the solution doesn't seem to match, but in fact you didn't write the solution explicitly. And if you try to find the solution, you will see that the solution for $x'(t)$ depends on $y'(t)$, which is not something you want. That's because you have two unknowns, $x'(t)$ and $y'(t)$, while one equation. You need two equations, but the second equation, the one involving $\ddot y$, "disappeared" because you decided to do it separately for $x$ and $y$. You need to turn the second equation back, because it is really the system of two equations for two unknowns. Or more geometrically, your particle moves in two dimensions, so a general motion in two dimensions cannot be described by one equation, you need two of them.

 

[gionole]

Amazing @Demystifier .. Catching up on this as I have also been looking into it and had some confusions.

Point 1:
If we check isotropy, here is the break down of ball dropping near earth of $mgy$ potential. Let's do active and passive ways both and use $\theta$ to be 30 degrees and ball's position in 1st experiment to be $x=10, y=20$ in 2D.

Active Transformation:

$\ddot y = -g$
$y_1(t) = 20 - \frac{gt^2}{2}$
$y_2(t) = (10sin30 + 20cos30) - \frac{gt^2}{2} = 22 - \frac{gt^2}{2}$

Passive:
$\ddot y = -g$
$\ddot y' = -gcos\theta = -0.86g$
$y_1(t) = 20 - \frac{gt^2}{2}$
$y_2(t) = (10sin30 + 20cos30) - \frac{0.86gt^2}{2} = 22 - \frac{0.86gt^2}{2}$

As it turns out, using $\theta = 30$ degrees rotation, active transformation suggests that space is isotropic while passive doesn't. I know that I didn't try every angle as isotropy check requires, but I still wanted to double check whether for $\theta=30$, I should get calculations such as in active mode, space is isotropic, while for passive, it's not. Is this correct ?

Point 2:
One question we might ask is how passive transformation is a solid way of checking homogeneity or isotropy since with that method, we don't actually change the test location of the object itself - as explained, it's only dropped from one single location/point in space one time. If we don't change the location of the object, are we really testing different points in space ? Here is my logical view on this:

When we observe the object from two different frames, we're essentially testing different points in space as we observe the object's motion from two different points in space. Our different frames are essentially differet points of space, right ? So if object's motion looks the same in each frame, then we tested different points(origin points of such frames) in a solid way. It's true that with this method, we don't test the same points as in active method, but it doesn't make difference at all.

Would you like this explanation ?

Point 3:

I have seen so many times that in the Lagrangian, they don't substitute $x'-a$ or $x'+a$ instead of $x$ and they directly substitute $x+a$ or $x-a$, Note that they don't use $'$ in the substitution. This seems wrong to me and you as well clearly did it the correct way in $41$ reply. Maybe people just don't use $x'$ instead of $x$ for simplicity ?

Point 4:

Depending on point 3, if I'm right that we should be substituting $x'-a$ instead of $x$ and not $x-a$, then am I right that this is exactly the type of transformation that Noether does/intends to ?

 

[gionole]

@Demystifier Hi,

sorry for disturbing you.

I have been thinking why passive transformation is a solid way of checking homogeneity.

on the following Link, It's said that in active transformation, our frame stays fixed, but points of space move, so after they moved, the same points are now given in the same frame as $x+a$, in passive transformation, we(our frame) moves and the points of space just stay at the same place, so those points are now given by $x+a$ in the new frame. So mathematically, they're the same. But here is what I fail to understand: While checking homogeneity in active transformation method, all we move is the object's initial location(i.e ball dropped from 10 meters is now dropped from 12 meters) and this really doesn't suggest that we move all points of space, we just move the object's initial location. What's your thoughts on this ?

 

[Demystifier]

While checking homogeneity in active transformation method, all we move is the object's initial location(i.e ball dropped from 10 meters is now dropped from 12 meters) and this really doesn't suggest that we move all points of space, we just move the object's initial location. What's your thoughts on this ?

If you are interested in the laws of physics for the motion of the ball, you move the ball. If you are interested in the laws of physics for the motion of all balls in the Universe, in principle you should move all balls in the Universe. If you are interested in the laws of physics for the whole Universe, in principle you should move the whole Universe. Obviously, only the first thing can be done in practice, experimentally. The other two things can only be done mentally, in theory. So in reality we only make actual experiments of the first type, and then extrapolate the results by assuming that similar results would be also valid in thought experiments of the other two types. This whole philosophy can also be summarized by the proverb - Think globally, act locally.

 

[gionole]

I think, before moving further to ask a question, you and me both need to be on the same page what active coordinate transformation generally is, because if we understand it differently, we will spend more time for no reason. The funny thing is everyone explains it differently.

What I'm first curious is which one is called "active transformation" - I asked this question here - Link but i don't think I will get an answer in short amount of time and yes, I understood your argument that ofc we can't move/shift the earth and everything upwards in real life, but I am asking basically what active transformation is in general - exact question is shown on the link

 

[vanhees71]

Active transformation means you move the entire system under consideration around, while passive transformation means you move your reference frame and describe the same system from this other reference frame. Mathematically at the end you describe both as transformations of some coordinates, building together a group.

 

[gionole]

Active transformation means you move the entire system under consideration around, while passive transformation means you move your reference frame and describe the same system from this other reference frame. Mathematically at the end you describe both as transformations of some coordinates, building together a group.

@vanhees71 So in terms of ball,earth example, here is what I think active transformation would be like:

I'm on the ground of earth and ball is dropped from some height. I am at the origin of my own coordinate system and I can easily observe ball's motion in that coordinate system. Now, we want to do active transformation. I stay where I was exactly(don't move) and my coordinate system stays at the same place(i still stand on the origin of my coordinate system), but everything else(earth, and anything else except me and my coordinate system) is shifted upwards by some number $a$. Is this what active transformation is ? if not, can you give an exact example in my use case of earth ball where I am wrong ?

 

[vanhees71]

I think you've described it correctly, i.e., if you consider the Earth and the ball as the system you want to describe. Then you need an arbitrary (inertial) reference frame, which you hold fixed. Of course, the equations of motion are
$$m \ddot{\vec{r}}_{\text{Ball}} = -G \frac{m_{\text{Ball}} m_{\text{Earth}}}{|\vec{r}_{\text{Ball}}-\vec{r}_{\text{Earth}}|^3} (\vec{r}_{\text{Ball}}-\vec{r}_{\text{Earth}}),$$
$$m \ddot{\vec{r}}_{\text{Earth}} = -G \frac{m_{\text{Ball}} m_{\text{Earth}}}{|\vec{r}_{\text{Ball}}-\vec{r}_{\text{Earth}}|^3} (\vec{r}_{\text{Earth}}-\vec{r}_{\text{Ball}}).$$
Then a transformation in the sense of an active transformation means you move both Earth and ball by $\vec{a}$ wrt. your coordinate system. For the equations of motion that means you just put $\vec{r}_{\text{Earth}}'=\vec{r}_{\text{Earth}}+\vec{a}$ instead of $\vec{r}_{\text{Earth}}$ and $\vec{r}_{\text{Ball}}'=\vec{r}_{\text{Ball}}+\vec{a}$ instead of $\vec{r}_{\text{Ball}}$. I guess you even don't need to write everything down to see that the equations of motion don't change at all, i.e., it's the same for the primed as for the unprimed position vectors.

 

[gionole]

@vanhees71 I think I get it now. Thanks very much. It seems like question after question comes into my head and I don't know why. Bear with me a little bit and i will for once and all finish this subject.

Question 1: After active transformation, you say that $\vec r'_{earth} = \vec r_{earth} + \vec a$. You sure ? Since we were on the ground and before active transformation, distance from us to the center of earth was $r_{earth}$ and now, earth is shifted upwards, then distance must have been reduced and not increased and this suggests $\vec r'_{earth} = \vec r_{earth} - \vec a$. My bad, I figured it out. It needs addition, yes distance gets decreased between origin and shifted earth's center, but we're writing this in vectors, so addition is required to get the correct vector.

Question 2: Just for my learning purposes, I could also do the following: before shifting, potential energy is $mgy$ and I write $m\ddot y = -mg$. After shifting upwards, potential energy is $mg(y-a)$ and $m\ddot y = -mg$ again because $F = -\frac{dU}{dy} = -\frac{d}{dy}(mg(y-a)) = -mg$. Is that also right ?

Question 3: Now, I'm trying to see how active transformation is the same thing as passive transformation. In passive, what we do is leave the earth and ball as it was and I shift my frame. Let's shift me and my coordinate system downwards. Well, with your calculations in #63 reply, passive would end up with the same results, but here is a tricky part. Is it really a good idea to use passive transformation method for checking homogeneity ? What I wonder is, do people really use passive transformation for checking homogeneity or they use active only ? even though mathematically, they are the same. The problem with what I have is with passive is since you don't move ball,earth in the space, you really not checking the different points in space even though that's the requirement for homogeneity, but maybe we use it since it exactly matches the active transformation mathematically ? other than that, it would be invalid though. Thoughts ?

Question 4: If I only considered ball as the system, then I'd only move the ball for sure, but would your #63 analysis be valid as you would only need to change $\vec r_{ball}$ by $\vec r_{ball} + \vec a$ ? not only that, I think passive and active transformations in such case wouldn't give you the same results, would it ?

 

[gionole]

Hey @vanhees71 , just in case, you missed #64(previous reply of mine). Thank you.