# Heliocentrism from a relativistic standpoint

1. Mar 29, 2005

### deuced

This is one of those questions that I've always wanted to ask, but felt stupid for it.

If we look at celestial mechanics from a relativistic standpoint, considering that all inertial reference frames are equivalent and we may choose any arbitrary reference frame, how is it that we can say that the earth truly revolves around the sun?

Of course, it does if we choose the sun as a stationary frame of reference. But what if we choose the earth as our frame of reference (which, being our home, seems like a natural one)? Isn't it then perfectly legitimate to say that the sun goes around the earth? It seems to me that the heliocentrism/geocentrism question simply depends on the reference frame from which the system is viewed.

2. Mar 29, 2005

### mathman

For practical purposes the center of mass of the system is the most convenient origin for analysis. For the solar system, it is essentially the center of the sun, since there is relatively very little mass (in the solar system) outside the sun.

3. Mar 29, 2005

### JesseM

The sun can be considered to be moving in an almost inertial manner over normal timescales (although over large enough timescales it's orbiting the center of the galaxy), but over the course of a year the earth definitely does not move inertially. Inertial movement means not just constant speed, but also movement in a straight line--any change in direction is a type of acceleration, even if the speed is constant. An observer will know he's accelerating because he feels G-forces--he'll feel them if he changes speed or if he changes direction (in the case of moving in a circle at constant speed, he'll feel a 'centrifugal force').

Last edited: Mar 29, 2005
4. Mar 29, 2005

### pervect

Staff Emeritus
As various other people have said, the center of mass of the solar system is a true inertial frame (ignoring pertubations from the rest of the galaxy, at least, which are small).

Because the sun is so much heavier than any of its planets, it's _almost_ true that the sun is in an inertial frame of reference. But actually the position of the sun is perturbed (very slightly) by the motions of the planets.

The notion of what can be considered to be an inertial frame really depends on how big an area of space one wants to model.

In GR, it's convenient to model gravity by its tidal forces. When the tidal force is so small as to be negligible even when multipled by the distance scale of interest, one can consider that region of space to be "inertial", though one usually says instead "flat space-time", or one of the more technical equivalents.

When the tidal force (acceleration/meter) multipled by the distance scale (meters) is not negligible, the region of space can't really be considered to be "inertial" or "flat".

So we are left with the fact that to an astronaut in the international space-station, space-time will be reasonably "flat" in a region of a few tens of meters.

But when you go to a broader distance scale, of millions of kilometers, the gravity of the Earth and sun become very important - the Earth, for instance, doesn't follow a straight line path as it would in an inertial frame, but moves instead in an elliptical orbit.

5. Mar 29, 2005

### Creator

You've forgotten one thing. We are not alone. There are thousands of other stars which bear witness to our orbital motion - Its called parallax.
Even if you were unable to detect the earth's acceleration (about the sun), the semi-annual shift in our position relative to the stellar background makes earth's orbital motion about the sun indisputable.

Creator

6. Mar 29, 2005

### JesseM

But if the earth was moving inertially rather than accelerating, it would be just as valid to look at things from the perspective of a frame where the earth was at rest and all the stars were moving.

7. Sep 29, 2009

### Saw

I also wanted to ask this question. The answers here are very helpful. I just wanted to ask a little further.

This is my understanding (considering a simplified system where there is only the Sun and the Earth and none of them rotates):

- 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?

- In any case you can draw a picture of the solar system where the Earth's frame is the coordinate system, the Earth is deemed to be motionless and the Sun revolves around.

- This picture, however, would be more complicated to draw than a Sun-centered picture. Why? Because in the Sun-centered picture you can only care about gravitational interaction (considering that the effect due to the Earth's pull is negligible), while in the Earth-centered system most of your work, if you want the picture to be faithful, would be based on fictitious forces moving the Sun? For example, if a car (the Earth) stops and I want to describe the situation from the standpoint of the car's (accelerated) frame, I have to imagine that the driver is pushed forward by a fictitious force. Is this analogy correct?

- These comments are based on purely Newtonian mechanics. Does GR alter anyhow the conclusions?

Thanks.

8. Sep 29, 2009

### JesseM

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"?

9. Sep 29, 2009

### Staff: Mentor

The Earth *is* moving inertially, at least as far as its orbit about the Sun goes. It is in free fall, following a geodesic worldline. The presence of the Sun means that that worldline is "curved", but it's still inertial.

It is true that, in a frame of reference in which the line connecting the Earth and the Sun is considered "at rest"--i.e., a frame which is "revolving about the Sun along with the Earth"--the Earth would be considered to be "held up against the Sun's gravity by centrifugal force". But the Earth doesn't *feel* this force.

10. Sep 29, 2009

### JesseM

Yes, that's true from a GR standpoint, but since the original poster was talking about inertial frames I thought we were dealing with a simplified case where we ignore spacetime curvature and just treat the Earth as moving in a circle in flat SR spacetime, where only objects moving at constant speed and direction (relative to inertial frames) would qualify as inertial...I probably should have spelled this out though.

11. Sep 29, 2009

### Staff: Mentor

Per my previous post, the Earth doesn't *feel* this force. There is nothing wrong with setting up an "inertial" coordinate system centered on the Earth. In fact, there is such a system, called the "Earth Centered Inertial" system, as described http://celestrak.com/columns/v02n01/" [Broken]. It's used for calculations of the orbital mechanics of satellites. (I put "inertial" in quotes because, since spacetime is curved, there are still issues that need to be dealt with that don't arise with inertial frames in flat spacetime.)

The problem with such a frame if you're looking at the dynamics of the whole solar system, as mathman pointed out, is that the Earth is nowhere near the center of mass of the solar system, so everything will look a lot more complicated in such a frame.

Last edited by a moderator: May 4, 2017
12. Sep 29, 2009

### Staff: Mentor

I would agree, since you mentioned *feeling* G-forces as a distinguishing factor, which *would* be true for an object moving in a circle in flat spacetime, but isn't for the Earth moving in a slightly elliptical orbit in curved spacetime.

13. Oct 2, 2009

### Saw

JesseM:

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?

14. Oct 2, 2009

### DrGreg

Circles are possible in both GR and Newtonian theory. The difference arises when you have fixed ellipse in Newtonian theory which becomes in GR a "precessing ellipse", an ellipse whose axis is slowly rotating. (As famously measured in the orbit of Mercury.)

15. Oct 2, 2009

### Staff: Mentor

That isn't quite what I said. What I said was: in SR, since spacetime is flat, an inertial worldline *has* to be a *straight* line--straight in the Euclidean sense of "straight". In GR, spacetime can be curved, and in a curved spacetime, inertial worldlines can be curves other than Euclidean straight lines.

So in SR, any object traveling in a circle *must* be accelerated (i.e., must feel a G-force), because a circle isn't a Euclidean straight line. (Actually the object's worldline would be a helix, like a coiled spring, because of the time dimension being included--but its projection into a suitably chosen spacelike slice would be a circle.) This would also be true of an object moving in an ellipse in flat spacetime--I'm sorry if my previous post didn't make this clear. There's nothing special about circles; what I've said applies to any worldline in flat spacetime that isn't a Euclidean straight line.

But in GR, in a curved spacetime (for example, spacetime near the Sun), an object (like the Earth) can be moving on a worldline that isn't a Euclidean straight line (for example, a circular or elliptical orbit), but can still be moving inertially, because of the spacetime curvature. That's why the Earth doesn't feel any G-force from the Sun--it's moving on an inertial worldline, which is curved because of the curvature of spacetime.

(And actually, as Dr. Greg pointed out, the Earth's orbit--more precisely, its projection into a suitable spacelike slice--isn't actually a closed ellipse, because it precesses. The precession of Earth's orbit is too small for us to measure with our current tools, but the precession of Mercury's orbit has been measured and agrees with the GR prediction.)

16. Oct 2, 2009

### Staff: Mentor

I should also have mentioned that, as far as actual predictions of observations go, the difference between Newtonian gravity and GR, as far as the orbits of objects are concerned, is the precession that Dr. Greg mentioned, which is not predicted by Newtonian gravity.

Conceptually, however, there is a huge difference between the two theories. According to GR, there is no such thing as a "force of gravity"--what we normally think of as "gravity" is just spacetime curvature. So the Earth, for example, doesn't move in an elliptical orbit because of a "force" from the Sun; it does so because of the spacetime geometry in its immediate vicinity, which "tells" it that the orbit it is following is the best geodesic path it can follow. The only influence the Sun has, in GR, is in determining the spacetime curvature in the Earth's immediate vicinity.

In GR, the way you tell whether there is a force acting is by asking whether objects *feel* a force acting. And GR's prediction is that only objects moving on non-geodesic worldlines will feel a force acting. The Earth moves on a geodesic, so it feels no force; but you and I, standing on the Earth's surface, move on non-geodesic worldlines, so we feel a force. In GR, this force is not "gravity"--it's the force of the ground pushing up on us, preventing us from traveling along the nearest accessible geodesics. Similarly, if you look at any other such case, where an object feels a force that in Newtonian language we would be inclined to call "fictitious", you will find that in GR, the actual force--what the object *feels*--is due to whatever it is that is pushing the object on a non-geodesic path. And those forces are always "real" forces--the Earth pushing up on us, or a rocket engine pushing on a rocket, or the walls and floor of a subway car pushing on us as we lurch about.

17. Oct 2, 2009

### JesseM

Assuming you are talking about inertial frames, yes.
I would say a simpler description is that fictitious forces are used in accelerating frames, but I think that's basically what you're getting at here--that if you're an accelerating observer, if you use an accelerating frame where you are at rest, then objects may appear to accelerate in this frame even though they are really moving inertially, and their apparent acceleration can be accurately calculated using the same laws that apply to inertial frames as long as you add in a fictitious force. Note that the fictitious force trick works for calculating the acceleration of any object in an accelerating frame, though, not just objects which are really moving inertially--i.e. an object may be measured as accelerating in both inertial frames and a certain accelerating frame, and in that accelerating frame you can accurately predict the object's motion using a combination of real forces and fictitious ones.
I don't really understand what you mean here. Perhaps you are saying that even if there is no actual force acting on the object that appears to be accelerating in the accelerating frame, there still must be a force acting on the accelerating observer. I suppose this is true, but the magnitude of the fictitious force seen in the observer's accelerated frame doesn't have to match the magnitude of the real force on the observer seen in inertial frames, so it's not like the fictitious force is just a coded way of talking about the real force (for example, consider an observer being moved in a circle of fixed radius by some force, who is calculating the motion of an object at a different radius from himself...in this case the fictitious centrifugal force on the object in the observer's rotating frame is different from the actual centripetal force on the observer himself). Also, in principle you are free to calculate how things behave in an accelerated frame even if there are no actual observers or objects at rest in that frame--a frame is just a coordinate system after all--although in practice this usually isn't done.
I'm not sure what kind of frame you're talking about. Ordinarily if an object is rotating in a circle around some central point, then to consider a rotating frame where the object is at rest usually means considering a frame where both the object and the central point (in this case the Sun) are at rest (this is the type of frame where the fictitious forces can be broken down into a centrifugal force and a coriolis force, for example). Think of looking at a spinning record, and then placing a camera above it looking down, with the center of the lens directly above the center of the record, and the camera spinning around the center of its lens at the same rate that the record is spinning, so that the camera will show a motionless record with the entire room appearing to spin around it.

Perhaps you mean a frame where not only is the center of the Earth at rest, but points on its surface are at rest too, so the Sun appears to orbit around it once per day? This would be a much more complicated accelerating frame to analyze in terms of fictitious forces, but it could be done. However, even in this case I would still have trouble making sense of your comment "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." The rotation of the Earth about its own axis has nothing at all to do with any real force from the Sun, and thus nothing to do with the Sun's own mass.

Last edited: Oct 2, 2009
18. Oct 2, 2009

### Saw

Well, there are many things in your post and Dr. Greg's and PeterDonis' to learn from and comment, but let's clarify something:

Of course! In all my comments (I stated it but it may be somehow masked afterwards) I'm assuming that there's NO

as if simply that didn't exist, precisely because that spinning or rotation about the axis exists, let us say, for historical reasons, but is not connected with the gravitational interaction between Earth and Sun. I was just talking about orbital motion, revolution of the Earth and Sun about the common centre of mass, if you take that RF, or of the Earth about the Sun or vice versa, if you take as RF any of them.

19. Oct 2, 2009

### Saw

Linking with the attempt at clarification above... As noted, I am considering a non-real situation where the Earth does not rotate about its own axis. Given this, I assume that if I start on the side of the Earth that is facing the Sun at a given moment, during the course *of the year* there will be times when Sun-light comes laterally to me, times when it will not reach me at all (because it is shadowed by the other side of the Earth), times when again the Sun-light will come laterally from the other side and finally a time when I will be facing the Sun once again. Thus the overall picture, from the Earth's standpoint, would be one where the Sun completes a revolution around the Earth once a year. Is that what would be observed from the Earth if the latter did not spin around its own axis? I supposed so, on the grounds that such would be the reason why we experience the seasons every quarter of a year. Is it right? Can we consider that scenario?

Last edited: Oct 2, 2009
20. Oct 5, 2009

### Saw

I am still curious about the answer to the previous post... If neither the Earth nor the Sun spinned around their respective axes, would an Earth-observer feel

(i) like a bucket in a "noria" (you can see a picture of a "noria" in http://asterion.almadark.com/wp-content/uploads/2008/12/noria1.gif), so that from the Earth he would see the Sun orbiting around (and completing one orbit per year), or

(ii) like someone in a merry-go-round, so that he would see the Sun spinning around its own axis (and completing one spin per year)?