Coordinate Independence, Kepler, and Planetary Orbits

Could you, merely by changing coordinates, argue that the Sun gravitationally orbits the Earth?  And could Einstein’s theory of gravity, which works equally well in all coordinate systems, allow you to do that?  

Despite some claims to the contrary — that all Copernicus really did was choose better coordinates than the ancient Greek astronomers — the answer is: No Way. 

How badly does the Sun’s path, nearly circular in Earth-centered (geocentric) coordinates, violate the Earth’s version of Kepler’s law?  (Kepler’s third law is the relation T=R3/2 between the period T of a gravitational orbit and the distance R, which is half the long axis of the ellipse that the orbit forms.)   Since the Moon takes about a month to orbit the Earth, and the Sun is about 400 = 202 times further from Earth than the Moon, the period of the Sun would be 4003/2 = 8000 times longer than the Moon’s, i.e. about 600 years, not 1 year. 

But is this statement coordinate-independent? Can it serve to prove, even in Einstein’s theory, that the Earth orbits the Sun and the Sun does not orbit the Earth? Yes, it is, and yes, it does. That’s what I claimed last time, and will argue more carefully today.

Of course the question of “Does X orbit Y?” is already complicated in Newtonian gravity.  There are many situations in which the question could be ambiguous (as when X and Y have almost equal mass), or when they form part of a cluster of large mass made from many objects of small mass (as with stars within a galaxy.)  But this kind of ambiguity is not what’s in question here.  Professor Muller of the University of California Berkeley claimed that what is uncomplicated in Newtonian gravity is ambiguous in Einsteinian gravity.  And we’ll see now that this is false.

Kepler’s Law Verified Without Coordinates

To check this, we can take an unambiguous Newtonian situation. Suppose, according to Newton, planet X orbits star Y in a roughly circular path of radius R over a time T, which we’ll call an XY-year.  Can R and T be determined in a coordinate-independent way, even in Einstein’s theory?

Not exactly, no.  And not without fine print.  But we won’t need an exact result with no fine print.

First, let’s measure the distance R between X and Y.  One way to do this is by using a clock on X. We need no coordinates on spacetime. All we need to know is the speed of light c (a.k.a. the cosmic speed limit.)  We fire a powerful pulse of radio waves from X toward Y at time t1, as measured by the clock on X. At time t2, as measured by the same clock, we receive a slight echo back on X.  We can then take the distance between X and Y at that time to be approximately half the light travel time multiplied by the speed of the waves, 1/2 c(t2 – t1) . By doing this repeatedly we can determine the minimum and maximum distance between X and Y over the orbit, and take R to be the average of that.

That doesn’t exactly give us R, though, for various reasons.  For instance, X and Y are moving relative to one another during the time that the light is in transit from X to Y and back again, so the light’s path isn’t exactly out and back (Figure 1), and if the distance between X and Y is not constant during that time, the time delay is not exactly proportional to the distance. More subtle is that if we used a clock on the surface of Y instead of the surface of X, we’d get a slightly different result, thanks to Einstein’s relativity, which warns us that gravitation can make clocks in different places run differently.  These and other issues make Kepler’s “R” slightly ambiguous.

Figure 1: As X and Y move relative to one another, a radio wave signal is sent (red line) from X, and bounces off Y, returning after a time T1 + T2 to X. This travel time, multiplied by c, gives an estimate of the distance between X and Y, though it is somewhat ambiguous.

But all these ambiguities are small and irrelevant, as long as the geometry of the spacetime is only weakly curved, and as long as all relative velocities (which can be approximately defined if the space is only weakly curved) are much smaller than the speed of light.  For an ordinary star with a few planets around it, weak curvature and low relative velocities (though not necessarily as mild as for the Earth and Sun) are guaranteed.  (If X and Y were black holes we would need to supplement this discussion a bit.) I’ll comment later on how to verify this without use of coordinates.  

Notice that in this measurement of R, we did not assume that X goes around Y, or that Y goes around X.  It would be okay if both go around something else, too.  We just made some light-travel-time measurements, each using a single clock and a single reflected radio pulse.  We don’t have to ask about the bigger picture here, and put coordinates on the whole system.

What about T?  It might be that (as for Earth) there are distant stars shining upon planet X. These stars are so distant that their positions on the sky change very slowly over an XY-year.  [If there were no such stars, we could send out a spacecraft very far away to play the same role.  It would take some time for it to reach a distance such that it could not possibly move very far across the sky during an XY-year. But once it was there, it would serve the same role as a distant star.]  

Once we have a distant, quasi-fixed object, we can define T in a coordinate-independent way.   The simplest is to choose a star S that is sometimes eclipsed by Y from X or vice versa.  On some day d1, the starlight from a chosen star S shines directly along the line from X to Y (and is eclipsed by X, from Y’s perspective.)  An observer on X would see this star is overhead at “midnight,” exactly opposite the position of the nearby star Y.  An observer on or near Y would see this star pass behind X in the sky.  

Now we wait, and on some day d2 this happens (or very nearly happens) again.  Then we know X and Y have performed a mutual orbit of some sort; an XY-year has passed, and so T = d2-d1.

Figure 2: As X and Y move relative to one another, X may eclipse a star S from Y’s perspective. This eclipse may be used to time the X-Y year. However, because light travels at a finite speed, there are subtleties. When the light from S reaches X (black circle), Y is in the position of the black disk; and when Y realizes S is in eclipse (at the location of the grey disk), X has moved to the grey circle. But these effects are small if the velocities involved, and the spacetime curvature, are small enough.

Again, no coordinates were needed here.  The only question is about the direction of travel of light waves, which follow straight lines in the spacetime geometry.  We only need the fact that light that arrives at X from S would continue on to Y if X did not block it, and identify the time when this happens on d1 and d2. (If X and Y were black holes we would again need to supplement this discussion a bit.)

There’s some ambiguity, as before, in defining d1 and d2.  What clock are we using?  Is it a clock on X, or a clock on Y?  Moreover, we have to remember that while the light is moving from X to Y, the line between X and Y is slowly changing; so when exactly do we define d1? Also, orbits precess; the fact that Einstein’s equations for gravity correctly predict the precession of Mercury’s orbit is a key test of their validity. Precession implies that the distance and relative velocity of X and Y won’t be exactly the same on day d1 as it is on day d2. Fortunately, any ambiguity is far less than an XY-year, as long as all curvatures in spacetime are small and the relative velocity between X and Y is small.  So we know T well enough.

In any realistic setting with planets stably orbiting a star, T and R are thus easily measured without any serious ambiguity. (Even if we replaced both the planet and the star with black holes of the same mass, we could still do these measurements with small ambiguities, as long as X and Y are far apart compared to their event horizons. We can place satellites around each one, orbiting well outside their event horizons, and are thus in regions of low curvature and low relative velocity. And then we can measure R and T between the satellites.)

Checking the Assumptions

In deriving all of this I assumed that the space between X and Y is only weakly curved, and that all relative velocities are small.  This can be verified by sending out a fleet of satellites in all directions, probing the spacetime geometry.  The paths of the satellites relative to X or Y can be estimated using the same methods I just described for X and Y — radial distances measured using radar reflection, and angular distances measured by comparison to stars or artificial distant objects that are “fixed” on the relevant time scales.  Measurements of these paths would confirm that the geometry is uncomplicated (in particular, that tidal effects are small), and that relative speeds can be roughly defined (using angular motion across the sky and Doppler shifts of light waves at known frequencies.)  

It’s key that precision measurements aren’t needed, so we don’t need to keep precise track of the spacetime curvatures and the various relative velocities.  For the Sun and Earth, the fact that the Sun violates Kepler’s Law for the Earth by a factor of hundreds makes things easy. Any ambiguities that arise because Earth doesn’t precisely orbit the Sun even in a Newtonian context, or from corrections due to Einstein’s relativity being more subtle than Newton’s, are completely irrelevant.  The Sun doesn’t even come close to orbiting the Earth, and switching to geocentric coordinates doesn’t change that.

Why Can’t Geocentric Coordinates Make the Sun Orbit the Earth?

As I emphasized last time, the problem with Muller’s argument is that he takes the Earth and Sun (let’s disregard the Moon) out of context. They aren’t in isolation; they are surrounded by, potential orbits of all kinds.  And these orbits reveal that Kepler’s law for the Sun works for satellites placed all around (but not too close to) the Earth. This includes satellites on gravitational orbits slightly outside of and or slightly inside of Earth’s path (as seen in Sun-centered [heliocentric] coordinates.)  The reverse is not true. Satellites placed just outside of and just inside of the Sun’s path (as seen in geocentric coordinates) will not satisfy Earth’s version of Kepler’s law.  They will satisfy the Sun’s version, too.

This experiment was partially done, by the way, though of course not for this purpose.  The STEREO A and B spacecraft, intended mainly to allow multiple views of the Sun, were placed on orbits that lie a little inside and a little outside Earth’s path as seen in heliocentric coordinates. (STEREO B malfunctioned some time ago, but STEREO A is still operating, and when combined with views from Earth gives us a stereoscopic view of the Sun’s activity.) The two craft have slightly smaller and larger R than the Earth-Sun system does, respectively, so they have slightly smaller and larger T, if they and the Earth all orbit the Sun. Kepler’s law thus predicts that, as seen in heliocentric coordinates, STEREO A will drift slowly ahead of Earth, while STEREO B slowly runs behind, with the effect that after about some years they both return quite close to the Earth.  (See Figures 3 and 4, which show a cartoon of their paths in both heliocentric and geocentric coordinates. See also the Wikipedia images, shown both in heliocentric fixed and in Earth-Sun-fixed rotating coordinates; these show, as my cartoon does not, how non-circularity of the orbits causes little hiccups in the latter coordinates.)

If instead the Sun orbited the Earth gravitationally as the Moon does, then STEREO A and B, which initially were are much closer to the Earth than the Sun, would orbit the Earth and satisfy the Earth’s version of Kepler’s law.  But they don’t. So the STEREO spacecraft give explicit evidence, without reference to the Earth-Sun system itself, that the Earth’s in a gravitational orbit of the Sun. (In fact Mars and Venus already do this, but the effect is less striking.)

Figure 3: Cartoon of the Stereo satellites (red and blue) and the Earth (green-blue) in their paths as seen in Sun-centered coordinates. The paths are very similar, reflecting that they are all gravitational orbits.
Figure 4: Cartoon of the Stereo satellites (red and blue) and the Sun (orange) in their paths as seen in Earth-centered coordinates. The Sun’s path is qualitatively different from those of the satellites, reflecting that they are not all gravitational orbits.

9 responses to “Coordinate Independence, Kepler, and Planetary Orbits

  1. kashyap vasavada

    A typo:
    1/2 (t2 – t1)/c .
    c should be multiplying (t2-t1)/2.

  2. kashyap vasavada

    Sorry! The wording
    “divided by the speed of the waves, ”
    has to be fixed also!

  3. Is R always coordinate-independent, or only in weakly curved spacetimes? I’m guessing not since strong gravitational lensing can result in multiple images of a star, each with a different path length. This would result in multiple possible distances measured by radio pulses depending on the path they take to get to the star.

    • I am most definitely using the fact that the spacetime around the solar system is nowhere strongly curved. The fact that there is no strong curvature anywhere nearby is a coordinate-independent statement, and so the technique used is self-consistent.

      To say it another way, if there were something that could generate strong lensing between a star and its planet, that would mean that, indeed, you could not treat the star and the planet as a two-body system. Whatever was doing the lensing would affect not only the paths of light but the nearby orbits, too.

  4. Thank you for doing all of these articles. It’s been great reading and fun to follow. I’m curious if you’ve received any feedback from Professor Muller?

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