Why have I been debunking Professor Muller’s claim that “the Sun orbits the Earth just as much as the Earth orbits the Sun”? Understanding why he’s wrong makes it easier to appreciate some central but subtle concepts in general relativity, Einstein’s conception of gravity.

What I want to do today is look at the notion of tides. Tides take on more importance in general relativity than in Newton’s theory of gravity. They can tell you which objects are gravitationally dominant in a coordinate-independent way.

A few posts ago, some of the commenters attempting to refute Professor Muller focused on showing the Sun is gravitationally dominant over the Earth. They were on a correct path! But nobody quite completed the argument, so I’ll do it here.

### Why We Can’t Just Measure the Gravitational Force

If you take Muller’s point of view, you are led to the *(frankly silly)* notion that the Earth orbits the International Space Station [ISS] just as much as the other way round.

It’s true, though, that if you’re an astronaut in the ISS, you certainly don’t *feel* as though you’re in orbit around the Earth, the way you would feel if you were zipping around in a circle on an amusement park ride. If you were to sit at the window for a while, you’d see the Earth go around you and you wouldn’t feel any motion or any push in any direction. All you’d feel is the constant weird internal sensation of weightlessness — which again isn’t a feeling involving gravity itself but rather an unusual lack of pressure of your internal organs on their supporting muscles, a pressure which you’re used to on Earth.

It’s this sense of weightlessness which keeps you from doing what, for a Newtonian, would be obvious. Why not just measure the gravitational force pulling on you, confirm that it’s pointing toward the Earth’s center, and check that the path you travel relative to the Earth is consistent with the amount of force that the Earth is exerting on you? Done! That would make it clear that you, as an astronaut in the ISS, are orbiting the Earth.

But neither you nor anything in the ISS is responding to that force, as far as you can see… so how are you, in fact, to measure it?

The point of general relativity — and its notion that **the amount of gravity depends on your perspective** — is that if you are in a sufficiently small box of sufficiently low mass, and you and the box are moving without the effect of any non-gravitational forces, you cannot measure the effect of gravity * even in principle*. The weightlessness you feel is the same as you’d feel inside any object in

**any**gravitationally determined path… including one that’s so far from any gravitating objects that the path traverses completely or almost completely flat spacetime.

**This is because, just as for any smooth geometrical surface viewed close-up, a sufficiently small region of spacetime’s geometry will always appear flat.**So if you can’t do this, what are you to do instead? You should to take careful account of the words “sufficiently small” and figure out how, in principle, you *can* measure the effect of Earth’s gravity. This involves the tides… the non-constancy of gravitational effects.

### The Tides and Gravitational Dominance

While general relativity can hide the effects of a gravitational trajectory over a small region, **it cannot hide the stretching and squeezing caused by variations in gravity**… the fact that nearby but separated objects do not “want” to take the same gravitational trajectory. It can’t hide the fact that (in the presence of objects with mass) spacetime’s geometry viewed close-up is nearly flat * but not quite*. A sufficiently large region reveals that there’s a bit of curvature.

The fact that the Earth’s gravity varies in both strength and direction causes objects to stretch in the direction perpendicular to the Earth’s surface and to be squeezed in the directions parallel to Earth’s surface. This tidal effect is observable and cannot be hidden or mimicked by something else. So all the astronauts have to do, to confirm they orbit the Earth, is measure the tides.

Not that this is easy. Within the space station, the effect is tiny. But here’s where the small box comes in; the larger the experimental apparatus you have, the bigger the effect of the tide.

Newton would say that the gravitational force on an object of mass **m** orbiting at a distance **r** from the Earth (whose mass we’ll call **M**), is

**F**_{gravity}= GMm/r^{2}

where **G** is Newton’s constant. As just noted above, this force cannot be detected within the orbiting object. However, Newton would ** also** say that the tidal force on an object of length

**L**at a distance

**r**from the Earth’s center is roughly the variation of F

_{gravity}over the distance

**L**, which is

**F**_{tidal}= 2 GMm L/r^{3}

The larger is L relative to r, the larger the effect. In contrast to F_{gravity}, this force *can* in principle be measured within the orbiting object. And this measurement is completely coordinate independent (see below for the technical argument.)

In practice, the tidal force is often too small to measure. Not only that, it can be masked by other effects. On a human being with L = 2 meters, the tidal effects from the Earth are probably hard to distinguish from gravitational effects from the space station itself.

But when you get to L = 100 meters, which happens to be of the same order as the length of the space station, the Earth’s tidal effects are much greater than any gravitational effects of the space station. The tidal force is tiny (perhaps a few hundred millionths of a Newton per kilogram), but probably not beyond a precision measurement. Seeing its orientation — a squeezing in the plane parallel to the Earth’s surface, and a stretching along a line that points toward the Earth — would confirm the Earth as the cause. This, then, would imply that the Earth dominates the gravitational environment in the vicinity of the ISS… and that the ISS orbits the Earth gravitationally rather than the other way round.

### Geometry and the Tides

In general relativity, anything that can be measured locally and is coordinate independent should show up in a “curvature scalar” — a coordinate-invariant measure of curvature. The fact that the spacetime is nearly flat around the ISS is encoded in the fact that the the simplest two scalars *(the Ricci scalar R = g ^{ij}R_{ij} and the square of the Ricci tensor R^{ij}R_{ij})* are both zero. But the Kretschmann scalar K = R

^{ijkl}R

_{ijkl}is

**not**zero. It is proportional to 1/r

^{6}, where r is the distance from the dominant body. Specifically, around a body of mass M,

**K = 48 G**)M^{2}^{2}/(c^{4}r^{6}) = (12/c^{4}) (2GM/r^{3}^{2}

where c is the cosmic speed limit (a.k.a. speed of light.) Note **K is proportional to the square of the tidal force |F _{tidal}|^{2}, which goes as M-squared/r^{6}, and not to the cube of Newton’s force F_{gravity}, which would go as M-cubed/r^{6}!** As claimed, the Newtonian force is not coordinate-invariant and cannot be unambiguously observed, but the tides are coordinate-invariant and allow us to determine the object that dominates our gravitational environment.

If you measure tidal effects (i.e. the Kretschmann scalar) relatively near (but not too near) the Earth — about 2 million km or more away from our planet — you will find that the tidal effect is what you’d expect from the Sun. That is, the tidal effects are in size and in orientation what you would have predicted if F_{tidal} is from the Sun, not from the Earth. Only much closer to the Earth will you find tidal effects consistent with an Earthly origin. A survey using an array of artificial satellites would establish the Sun’s gravitational dominance throughout the majority of the solar system, including for objects that take paths similar to the Earth’s, such as the Stereo A and B spacecraft that I referred to in a recent post on this subject. And so we would conclude that the Earth lies in the Sun’s sphere of influence, and not the other way round… consistent with the argument that some commenters were wanting to make.

One potential point of confusion: the tides affecting the Earth’s ocean are dominantly from the Moon, although the Sun’s tidal effects are almost as large. Does that mean that the Earth’s orbit is significantly affected by the Moon? Not really: in the discussion above, I assumed that (as for the ISS) the mass **m** of the object on which the tides are acting is small. The Earth’s large mass corrects the reasoning I’ve used here.

If instead you probe our local environment (near the Earth’s surface, I mean) using a large and relatively low-mass object, such as a suspension bridge, you’ll find, of course, that the Earth is the dominant gravitational object. Its tidal forces far dominate any effect of the Moon and Sun.

Instead, the similar effects of the Moon and Sun on Earth’s ocean tides direct our attention to something else: Lagrange points. They offer another method, as noted by other readers, to show the Earth orbits the Sun. We can explore that avenue next time.

## 3 Responses

Gravity gradiometers measure tidal effects – the ESA Gravity Field and Steady-State Ocean Circulation Explorer (GOCE) is the best gradiometer yet flown in Earth orbit. The GOCE spacecraft’s primary instrumentation was a highly sensitive gravity gradiometer consisting of three pairs of accelerometers measuring the gravitational gradient along three orthogonal axes. In other words, the spacecraft center of mass was in free fall and the accelerometers measured the tidal effect in three dimensions, with a distance of 50 cm between each accelerometer pair.

GOCE had an accuracy of about 10 milli Eotvos (or 10^-11 / seconds^2 or 10^-11 (m/s^2)m), or about 3 x 10^-6 of the Earth’s main tidal effect of 3080 Eotvos.

Thanks! I am glad to learn this.

Lewis Caroll also described the equivalence of gravity and acceleration in Sylvie and Bruno published in 1898. “One can easily imagine a situation,” said Arthur, “where things would necessarily have no weight, relatively to each other, though each would have its usual weight, looked at by itself.”