Heliocentrism from a relativistic standpoint

[Saw]

Physicists are very careful with nomenclature, Saw. Centrifugal force is a very specific thing: It is the fictitious force

$$-m\boldsymbol \Omega \times \mathbf r$$

where $\boldsymbol \Omega$ is the rate at which the reference frame is rotating with respect to the remote stars.

I'm studying your text. Shouldn't $\boldsymbol \Omega$ be squared?

 

[D H]

Good catch. However, it is not omega squared. It is

$$-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)$$

which reduces to

$$-m \Omega^2 r$$

in the special case of $\mathbf r$ being normal to $\boldsymbol \Omega$.

 

[Saw]

it is not omega squared. It is

$$-m\boldsymbol \Omega\times (\boldsymbol \Omega \times \mathbf r)$$

which reduces to

$$-m \Omega^2 r$$

in the special case of $\mathbf r$ being normal to $\boldsymbol \Omega$.

Ok, but according to the sources I am reading, if we make the analysis of the Earth-Sun system from a rotational frame (I talk now of acceleration instead of force):

- The acceleration due to the fictitious centrifugal force can be decomposed into two components: ω^2 * c (c being the distance between the center of mass of the Sun-Earth system and the Earth center) and ω^2 * r (r being the distance between the center of the Earth and the relevant point of the Earth where we're analysing the tidal effect).

- The component ω^2 * r cancels out somehow.

- So we are left exclusively with a "uniform" (the same size in all points of the Earth) centrifugal acceleration of intensity ω^2 * c.

Is that correct, in your opinion? Thanks.

 

[Saw]

Well, after deleting a last post with some blunder, I come back with a better understanding now.

Your analysis, DH, is clear:

Let us say M = mass of the Sun.
N and D are two points on the surface of the Earth separated from its center by the radius (r) of the Earth, N (from near) being the one facing the Sun and D (from distant) the one facing away from the Sun, at a given moment.
R = the distance from the center of the Sun to the center of the Earth.
CM = center of mass of the Sun-Earth system.
c = distance from the CM to the center of the Earth.

- In an inertial frame with origin at the CM (disregarding its acceleration due to other objects in the universe), objects at N and D would tend to be accelerated as follows:

* For N, a = - GM/(R^2 - r)
* For D, a = - GM/(R^2 + r)

Since the frame is inertial, these accelerations are due to "real" forces.

- In an accelerated frame with origin at the Earth center, we have to add to the real force a "fictitious" force pointing away from the Sun = + GM/R^2. Thus the accelerations would tend to be:

* For N, a' = - GM/(R^2 - r) + GM/R^2 = - net result, away from Earth center and toward the Sun
* For D, a' = - GM/(R^2 + r) + GM/R^2 = + net result, away from the Earth center and away from the Sun

However, all this is the analysis at a frozen moment, an ideal instant. It's not strange, hence, that the resulting net force is the same as if the Earth were falling in a straight line toward the Sun.

But here it is not that simple. To describe the kinematics from the simplest perspective, that of the inertial frame, the Earth is orbiting the CM, always with a fixed orientaton of its axes wrt the fixed stars: the center of the Earth is translating in a circle about the CM, whereas Points N and D are also translating in circles with the same radius (c), albeit different centers.

One solution could be what I tried to do in post #33. Why is that wrong? There I was taking the frame of the Sun, assuming that its motion around the CM is negligible and hence inexistent. But to be more precise, let's take the frame with origin at the CM (also disregarding its own acceleration due to gravitational interaction with the rest of the universe). Isn't the Earth's acceleration, wrt the CM and after a time lapse, the result of compounding the gravity force toward the Sun with the Earth's inertial tangential motion?

Another solution could be shifting to the perspective of a rotating frame and then transforming again back into the accelerated Earth frame or the inertial frame. I've got lost both in the rotating frame and in the way back. Does anyone know how to do it?

In any case, is the final result, wrt the Earth center, the following?

* For N, a' = - GM/(R^2 - r) + ω^2c
* For D, a' = - GM/(R^2 + r) + ω^2c