JesseM:
Saw said:
It is clear that the Sun, being more massive, accelerates (changes the direction) of the Earth more than viceversa. Because of that, centrifugal force due to interaction between the Sun and the Earth is much stronger on the Earth than on the Sun. Nevertheless, centrifugal force on the Earth exclusively due to gravitational attraction of the Sun (leaving rotation aside) is also very tiny. Is it discernible at all?
JesseM said:
The centrifugal force is a fictitious force in Newtonian mechanics that only appears in rotating frames--which rotating frames are you talking about in order to compare centrifugal forces? Perhaps a frame where both the center of the Sun and the center of the Earth are at rest vs. a frame where both the center of the Sun and the surface of the Sun are at rest? Of course in the latter case the centrifugal force experienced on the surface of the Sun would just depend on the rotation rate of the Sun, not on "interaction between the Sun and the Earth". If that's not what you meant, can you specify what two centrifugal forces you're talking about when you say the force is "much stronger on the Earth than on the Sun"?
I’ll tell you what was in the back of my mind and how I have thought of re-formulating it (though I am not sure at all it’s correct):
It is my understanding, in terms of Newtonian mechanics, that whenever the acceleration of an object is measured from a given frame, it’s because a real force has acted. A different thing is that if we measure from a certain reference frame, assuming it is motionless, we do not measure our own acceleration and thus attribute it to the object being observed, as if it were accelerated by a “fictitious” force. The term is a little of a misnomer (strictly speaking what is fictitious is not the force, but the assumption that the event causing the acceleration has acted on the observed object), but let’s maintain it for discussion purposes.
For example, talking about the gravitational interaction between the Earth and the Sun, leaving aside effects due to rotation of the Sun or the Earth about their respective axes…
If we choose the Earth as reference frame (as if it were non-accelerated), then it can be that:
(i) The observed object is the Sun. We infer that there is a force acting because the Sun is observed to orbit the Earth. If we explain this by assuming that all the force is suffered by the Sun, that force is mainly fictitious and only to a very small extent real.
(ii) The observed object is the distant stars (if we assume that they are so far away that, virtually, they do not interact with the Sun or the Earth). We infer again that there is a force acting because the stars are observed to orbit the Earth. If we explain this by assuming that all such force is suffered by the stars, all that force is fictitious.
(iii) The observed object is, for example, the waters of the ocean, that is to say, “part” of the Earth. (We leave aside the influence of the Moon, more relevant in practice). We infer again that there is a force acting because the waters tend to build up a bulge. At the surface of the Earth, closer to the Sun, that force is real (because the gravitation is stronger at that point). At the other side, that force is fictitious (actually, what happens is that the gravitation is stronger for the rest of the Earth).
If we chose the Sun as reference frame, the principles would not change, although the proportion between fictitious and real forces would be much in favour of the latter.
If we chose as reference frame the centre of mass of the Sun-Earth system, then all acceleration observed would be explained by real forces, except for the orbital movement of the distant stars.
Above I talked about measurement or observation, as reflected in a space-time system of coordinates. I suppose the expression “feeling” is reserved for noticing the acceleration in constituents of the frame from which you measure. That would be point (iii), i.e., tidal effects, wouldn’t it? These effects are not noticeable in small regions of space or time.
Then I get lost about the implications on this picture of GR.
PeterDonis: You mention that the orbit is a circle in flat ST, but elliptical accounting for the the curvature of ST. But isn't it also en ellipses in Newtonian mechanics? What does GR add? Is it that also time is curved?