I Different Equations of Motion for different frames

[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.