Ok, we really need to sort this out. I'll be as precise as I can. In Newtonian mechanics, the quantity \(|\vec{x_{rel}}|\equiv |\vec{X}-\vec{x}|\)is independent of choice of coordinates. In a problem like this where all motion is along a single axis, we can define the scalars X and x as the two objects' positions along the axis of motion. This gives \(x_{rel}\equiv X-x\) as the coordinate-independent quantity. If you disagree with this, we have more significant problems than the technicality this thread is about.
No, I understood the part in blue, that this was your construction. As I said, once you have designated X and x as positions in "x-space", you have assigned a reference frame. Therefore x and X are dependent on frame designation. If I come along with a new frame whose origin cuts through the midpoint of (X, x), at and angle θ, then your X and x are going to be equal in magnitude in my system, and opposite in sign. This is why you can't claim frame independence as you say.
Are you seriously getting on my case for forgetting to specify that my position coordinates are measured along the Earth<->ball axis?
No I am just letting you know your little scheme is not working.
Fine. My formula is valid for all coordinate systems in which the x-axis is parallel to the Earth<->ball separation.
Well, no, it's not going to be valid at all. We can create a special case for the kinematics being independent for any axis parallel to yours, but only for those cases in which the coordinates correspond. However, you are toast trying to map Galileo's experiment onto those parallel axes since the vector
r in the Law of Universal Gravitation will vary as you change systems. Therefore you are sunk. Therefore, as I said before, it's incorrect to claim frame independence. You need a coordinate transform (forward and inverse) to get between systems.
This doesn't change my conclusion at all, since choice of axis orientation is independent of choice of origin.
Let's just agree that you have no idea what a coordinate transformation is and move on. All of the skilled posters here took Linear Algebra and Vector Mechanics, so this is where you get to have conniption fits, go through catharsis, and get the remedial help you need. But seriously: get off the gas.
This is getting surreal. Classical Mechanics is a core course for a physics degree at every university I know of; it basically covers the dynamics of systems in the non-relativistic, non-quantum limit.
If you want to to convince readers that you've taken the course you're going to have to do better than that. But really, you've shown us your cards. (I say "us" without having read the comments of the people I would call "skilled". So whether or not they are on to your little scheme remains to be seen.)
Hamiltonian mechanics are a staple technique for such systems,
In some higher course, but not down at the level we are talking about. Nice try, though.
and it's beyond bizarre that you're accusing me of "inventing" things when you could just go to the Wikipedia page on Hamiltonian dynamics and verify all my techniques/terminology for yourself.
No, I do not need wikipedia to reinforce what I've posted thus far. That just says you're quote mining wikipedia. It has nothing to do with the structure of college curricula. Not knowing the curriculum is painting the fraud even louder and brighter than your messed up math did.
You've left little to the imagination, Fednis.
My current research involves the dynamics of pairs of interacting atoms in the absence of external forces; I literally do calculations in relative coordinates for money at least 3 times a week.
Yeah and I invented the Internet. The problem is that you can't impress a skilled person with your pseudoscience no matter how many credentials you pin on your chest.
Where did you get that definition? That's not what "inertial" means. An inertial frame is one that isn't accelerating.
Isn't accelerating: an accelerometer placed at the origin reads zero. This will not be the case for your proposed coordinate system, since you have stated that the Earth is accelerating relative to your origin. Therefore you are accelerating relative to the Earth, therefore the accelerometer reads nonzero, therefore you are not inertial.
In our problem, there is no force on the center of mass of the two objects, so the center of mass is not accelerating.
There most definitely is a force acting on your system or else it would not be moving at all.
That means the center-of-mass frame is an inertial frame.
Wrong, as follows:
The Earth is accelerating in this frame,
Correct. Ergo, QED.
but all inertial frames are in a constant state of rectilinear motion with respect to one another,
What is that supposed to mean? You just admitted your accelerometer is reading nonzero, so no need to try to gloss over it.
so any frame co-moving with the Earth frame cannot be inertial.
Gawd you're a nut. Any Earth-centered frame must be inertial, since the accelerometer placed at the origin of any such frame reads zero.
Actually, we need a separate thread posed as a contest to the poster who can come up with the most catastrophic consequences of the Earth magically switching to a non-inertial state.
To say that such a frame is inertial because it has a constant g-field is just nonsense.
I can certainly understand why you would think so. But therein lies the rub.