I Different Equations of Motion for different frames

[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.