6.5  Intersecting Orbits


Time is the longest distance between two places.

                                                Tennessee Williams, 1945


The lapse of proper time for moving clocks in a gravitational field is often computed by splitting the problem into separate components, one to account for the velocity effect in accord with special relativity, and another to account for the gravitational effect in accord with general relativity. However, the general theory subsumes the special theory, and it's often easier to treat such problems holistically from a purely general relativistic standpoint. (The persistent tendency to artificially bifurcate problems into "special" and "general" components is partly due to the historical accident that Einstein arrived at the final theory in two stages.)  In the vicinity of an isolated non-rotating spherical body whose Schwarzschild radius is 2m the metric has the form



where f = longitude and q = latitude (e.g., q = 0 at the North Pole and q = p/2 at the equator). Let's say our radial position r and our latitude q are constant for each path in question (treating r as the "radius" in the weak field approximation). Then the coefficients of (dt)2 and (df)2 are both constants, and the metric reduces to



If we're sitting on the Earth's surface at the North Pole, we have sin(q) = 0, so it follows that  dt  = dt  where r is the radius of the Earth.


On the other hand, in an equatorial orbit with radius r = R we have q = p/2, sin2(q) = 1, and so the coefficient of (df)2 is simply R2. Now, recall Kepler's law w2 R3 = m, which also happens to hold exactly in GR (provided that R is interpreted as the radial Schwarzschild coordinate and w is defined with respect to Schwarzschild coordinate time). Since w = df/dt we have R2 = m/(w2 R) = (dt/df)2 (m/R). Thus the path of a particle in a circular orbit satisfies



Now for each test particle, one sitting at the North Pole and one in a circular orbit of radius R, the path parameter t is the local proper time, so the ratio of the orbital proper time to the North Pole's proper time is



To isolate the difference in the two proper times, we can expand the above function into a power series in m/r to give



Multiplying through by dtearth and then subtracting this quantity from both sides, we get



The mass of the earth in geometrical units is about m = 0.00443 meters, and the radius of the earth is about r = 6.38(10)6 meters, so we can insert these values and integrate over a given lapse Dtearth of proper time measured on Earth to give the difference between this elapsed time and the corresponding elapsed time for the circular orbit



Consequently, for an orbit at the radius R = (3/2)r (about 2000 miles up) there is no difference in the lapses of proper time. For orbits lower than 3r/2 the satellite will show slightly less lapse of proper time (i.e., the above discrepancy will be negative), whereas for higher orbits it will show slightly more elapsed time than the corresponding interval at the North Pole. One might think that a further adjustment would be necessary to correlate the elapsed time to a particle at rest on the equator, to account for the rotation of the Earth, but the Earth tends to bulge due to its rotation in such a way that the it maintains roughly an equi-potential surface, meaning that the rate of proper time is approximately constant at sea level, regardless of latitude. Indeed this was the basis of Newton’s prediction of the Earth’s oblate shape, since (as he pointed out) if the surface of the seas was not equi-potential, the water would flow so as to achieve equilibrium.


In a low Earth orbit of, say, 360 miles, we have r/R = 0.917, so the proper time runs about 22.5 microseconds per day slower than a clock on the Earth. On the other hand, for a geo-synchronous orbital radius of 22,000 miles we have r/R = 0.18, so the orbit's lapse of proper time actually exceeds the corresponding lapse of proper time on Earth by about 43.7 microseconds per day. Of course, as R continues to increase the orbital velocity drops to zero and we are left with just coordinate time for the orbit, relative to which a clock on Earth is "running slow" by about 60 micro-seconds per day, due entirely to the gravitational potential of the earth. (Hence during a typical human life span the Earth's gravity stretches out our lives to cover an extra 1.57 seconds of coordinate time.) Of particular interest is the orbit of the Global Positioning System (GPS) satellites, which are located in circular orbits approximately 20,200 meters above the Earth’s surface (not a geosynchronous orbit, as is sometimes thought). Taking the mean radius of the Earth to be about r = 6378 meters, it follows that the orbital radius of the GPS satellites is R = 20200+6378 = 26578 meters, giving a ratio R/r of about 4.2. Inserting this ratio into the above formula, we find that the elapsed proper time per Earth-day for a GPS satellite is about 38.1 micro-seconds more than the elapsed time on Earth.


Incidentally, the value of torbit given by equation (2) goes to zero when the orbital radius R equals 3m, consistent with the fact that 3m is the radius of the orbit of light. This suggests that even if something prevented a massive object from collapsing within its Schwarzschild radius 2m, it would still be a very remarkable object if it was just within 3m, because then it could (theoretically) support circular light orbits, although I don't believe such orbits would be stable (even neglecting interference from in-falling matter). If neutrinos are massless there could also be neutrinos in 3m (unstable) orbits near such an object, although the evidence today indicates that neutrinos have a small positive mass.


The results of this and the previous section can be used to clarify the so-called twins paradox. In some treatments of special relativity the difference between the elapsed proper times along different paths between two fixed events is attributed to a difference in the locally "felt" accelerations along those paths. In other words, the asymmetry in the proper times is "explained" by the asymmetry in local accelerations. However, this explanation fails in the context of general relativity and gravity, because there are generally multiple free-fall (i.e., locally unaccelerated) paths of different proper lengths connecting two fixed events. This occurs, for example, with any two intersecting orbits with different eccentricities, provided they are arranged so that the clocks coincide at two intersections.


To illustrate, consider the intersections between a circular and a purely radial “orbit” in the gravitational field of a spherically symmetrical mass m. One clock follows a perfectly circular orbit of radius r, while the other follows a purely radial (up and down) trajectory, beginning at a height r, climbing to R, and falling back to r, as shown below.



We can arrange for the two clocks to initially coincide, and for the first clock to complete n circular orbits in the same (coordinate) time it takes for the second clock to rise and fall. Thus the objects coincide at two fixed events, and they are each in free-fall continuously in between those two events. Nevertheless, we will see that the elapsed proper times for these two objects are not the same.


Throughout this example, we will use dimensionless times and distances by dividing each quantity by the mass m in geometric units. For a circular orbit of radius r in Schwarzschild spacetime, Kepler's third law gives the proper time to complete n revolutions as



Applying the constant ratio of proper time to coordinate time for a circular orbit, we also have the coordinate time to complete n revolutions



For the radially moving object, the usual parametric cycloid relation (see Section 6.4) gives the total proper time for the rise and fall



where the parameter a satisfies the relation



The total elapsed coordinate time for the radial object is





In order for the objects to coincide at the two events, the coordinate times must be equal, i.e., we must have Dtcirc = Dtradial. Therefore, replacing r with q(1+cos(a)) in the expression for the coordinate time in circular orbits, we find that for any given n and q (= R/2) the parameter a must satisfy



Once we’ve determined the value of a for a given q and n, we can then determine the ratio of the elapsed proper times for the two paths from the relation



With n = 1 and fairly small value of r the ratio of proper times behaves as shown below.



Not surprisingly, the ratio goes to infinity as r drops to 3, because the proper time for a circular orbit of radius 3m is zero. (Recall that the "r" in our equations signifies r/m in normal goemetrical units.)


The a parameters and proper time ratios for some larger values of r with n = 1 are tabulated below.



To determine the asymptotic behavior we can substitute 1/u for the variable q in the equation expressing the relation between q and a, and then expand into a series in u to give



Now for any given n let an be defined such that



For large values of r the values of a will be quite close to an because the ratio of proper times for the two free-falling clocks is close to 1. Thus we can put a = an + da in equation (3) and expand into a series in da to give



To determine the asymptotic da as a function of R and n we can put a = an + da in equation (4) and expand into a series in da to give





For sufficiently large R the value of Bn is negligible, so we have



Inserting this into (6) and recalling that 2/R is essentially equal to [1+cos(an)]/r since a is nearly equal to an, we arrive at the result





So, for any given n, we can solve (5) for an and substitute into the above equation to give kn, and then the ratio of proper times for two free-falling clocks, one moving radially from r to R and back to r while the other completes n circular orbits at radius r, is given (for any value of r much greater than the mass m of the gravitating body) by equation (7). The values of an, kn, and R/r for several values of n are listed below.



As an example, consider a clock in a circular orbit at 360 miles above the Earth's surface. In this case the radius of the orbit is about (6.957)106 meters. Since the mass of the Earth in geometrical units is 0.00443 meters, we have the normalized radius r = (1.57053)109, and the total time of one orbit is approximately 5775 seconds (i.e., about 1.604 hours). In order for a radial trajectory to begin and end at this altitude and have the same elapsed coordinate time as one circular orbit at this altitude, the radial trajectory must extend up to R = (1.55)107 meters, which is about 5698 miles above the Earth's surface. Taking the value of k1 from the table, we have



and so the difference in elapsed proper times is given by



This is the amount by which the elapsed time on the radial (up-down) path would exceed the elapsed time on the circular path.


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