# Twin paradox with one twin in orbit, one in radial free-fall

Many questions on Physics SE relate to the twin paradox, but I did not find any that ask this specific question. Suppose that object A is in a circular orbit around a spherically symmetric, non-rotating planet (Schwarzschild metric). Suppose that object B is in radial free-fall, with a worldline that meets object A on the way up (passing arbitrarily closely without colliding) and again on the way down, as depicted here:

Object A executes exactly one period of its orbit between the two meetings, and both objects remain in free-fall between the two meetings.

The question: Which object ages less between the two meetings?

Here's what I already know, expressed using coordinates $$t,r,\phi$$ where the equation for an object's proper time $$\tau$$ is $$d\tau^2=\left(1-\frac{R}{r}\right)\,dt^2-\frac{dr^2}{1-\frac{R}{r}} - r^2d\phi^2, \tag{1}$$ where $$R$$ is the Schwarzschild radius. (This is in the plane that contains both objects' worldlines, so only $$3$$ coordinates are needed.) I already know that object A ages less when the coordinate-radius $$r$$ of object A's orbit is $$r=3R/2$$, because then A's worldline is lightlike and B's worldline is still timelike.

But does A age less than B for all radii $$r>3R/2$$? If not, then at what value(s) of $$r$$ do the two objects age the same amount between meetings?

• So you are just asking what is the proper time along each path? Aren't there duplicates - certainly the object in orbit has been addressed a number of times. – Rob Jeffries Dec 15 '18 at 20:32
• This particular problem is discussed at length in the following article: mathpages.com/rr/s6-05/6-05.htm – m4r35n357 Dec 15 '18 at 20:41
• For the impatient, the mathpages article doesn't seem to explicitly answer the question, but the answer seems to be implicit in the graph that comes after the words "behaves as shown below." The answer seems to be that in all cases, the radial twin ends up older than the circular twin. – Ben Crowell Dec 15 '18 at 21:01
• cool problem. It illustrates nicely how geodesics are local maxima of proper time. I think I will assign it next time I teach GR :). – Sam Gralla Dec 15 '18 at 21:40
• @Ben Crowell I think equation (8) answers the OP's question, do you not? – m4r35n357 Dec 16 '18 at 12:39

## Approach

My question asks whether or not the elapsed proper time between meetings is always less for A than it is for B, for every possible radius of A's orbit.

One way to answer this would be to calculate the ratio of the two elapsed proper times as in reference [1] cited by user m4r35n357, but my question is less demanding: I only want to know which of the two elapsed proper times is smaller. This less-demanding question can be answered using a less-demanding approach.

Start with the proper-time equation shown in the question, hereafter called equation (1), and consider any worldline of the form $$r(t) = r_A \hskip2cm \phi(t) = \omega t \tag{2}$$ where $$r_A$$ and $$\omega$$ are constants, meaning that they are independent of the coordinate $$t$$. Object A's circular orbit has the form (2) for a special value of $$\omega$$ that we won't need here except for the fact that it's not zero. Now, consider one cycle of A's orbit. The second of equations (2) says that the elapsed coordinate time during one cycle is $$\Delta t=\frac{2\pi}{\omega} \tag{3}$$ According to equation (1) in the question, the corresponding elapsed proper time is $$\Delta \tau_A = \left(1-\frac{R}{r_A}-r_A^2\omega^2\right)^{1/2} \,\Delta t. \tag{4}$$ Now let $$\Delta\tau_0$$ denote the elapsed proper time for an object that hovers at $$r=r_A$$ and $$\phi=0$$. This object meets A once per orbit. Equation (4) gives the elapsed proper time between meetings according to A. The elapsed proper time between meetings according to the hovering object is $$\Delta \tau_0 = \left(1-\frac{R}{r_A}\right)^{1/2}\,\Delta t \tag{5}$$ with the same value (3) of $$\Delta t$$. (The elapsed coordinate time depends only on the two meeting-events, not on the worldline that connects them.) Equations (4)-(5) give $$\Delta\tau_A < \Delta\tau_0. \tag{6}$$ This doesn't answer the question yet, because the hovering object is not the one we want. We want an object B that is in radial free-fall, meeting A twice with coordinate time interval (3) between the two meetings.

Let $$\Delta\tau_B$$ elapsed proper time between meetings according to the radially-falling object B. The goal is to prove that $$\Delta\tau_0 < \Delta\tau_B. \tag{7}$$ If we can prove (7), then combining the inequalities (6)-(7) will answer the question: the orbiting object A ages less than B between meetings, regardless of the orbital radius $$r_A$$.

## How to prove (7)

Object B is in free-fall (as is A). In other words, the worldline of object B is a geodesic. To prove (7), I will use this fact:

• Among worldlines with the given endpoints, a geodesic gives a local maximum of the elapsed proper time.

A geodesic doesn't necessarily give a global maximum (that's the point of the question), but it does give a local maximum. To exploit this, we can continuously deform the hovering worldline into B's worldline in such a way that the elapsed proper time continuously increases by virtue of this local-maximum principle.

Start with the hovering worldline. The coordinate-times of the two meetings with object A are $$t=0$$ and $$t=\Delta t$$, where $$\Delta t$$ is given by equation (3). Write $$[a,b]$$ for the coordinate-time interval from $$a$$ to $$b$$. Choose an intermediate time $$T$$, with $$T\in[0,\Delta t]$$, and consider the worldline $$W_T$$ defined like this:

• During the interval $$t\in[0,T]$$, the worldline $$W_T$$ is a radial geodesic that starts and ends at $$r=r_A$$ (at times $$0$$ and $$T$$, respectively).

• During the interval $$t\in[T,\Delta t]$$, the worldline $$W_T$$ is that of an object hovering at $$r=r_A$$.

The angular coordinate is $$\phi=0$$ during both segments, so the motion (if any) is purely radial. For any value of $$T$$, the worldline $$W_T$$ meets A twice, namely at times $$t=0$$ and $$t=\Delta t$$. By varying $$T$$ continuously from $$0$$ to $$\Delta t$$, the worldline $$W_T$$ varies continuously from that of an always-hovering object to that of an object that is in radial free fall during the whole interval between meetings. The latter case, namely $$W_{T=\Delta t}$$, corresponds to object B in the question.

From here, the proof is easy. The quantity $$\Delta\tau_0$$ in the inequality (6) corresponds to the worldline $$W_{T=0}$$. For any $$T$$, the worldline $$W_{T}$$ is a geodesic at times $$t\in[0, T]$$. For infinitesimal $$\delta T$$, the worldline $$W_{T-\delta T}$$ is not a geodesic during part of the interval $$[0,T]$$, and it differs only infinitesimally from $$W_T$$, which is a geodesic during all of that same interval. Therefore, the elapsed proper time for $$W_T$$ in the interval $$[0,T]$$ is greater than the elapsed proper time for $$W_{T-\delta T}$$ in the same interval $$[0,T]$$, because a geodesic locally maximizes the proper time. Since the two worldlines are identical to each other during the remaining interval $$[T,\Delta t]$$, we infer that the elapsed proper time for $$W_T$$ in the full interval $$[0,\Delta t]$$ is greater than the elapsed proper time for $$W_{T-\delta T}$$ in $$[0,\Delta t]$$. This is true for all $$T$$, so the elapsed proper time for B's worldline $$W_{\Delta t}$$ is greater than the elapsed proper time for the hovering worldline $$W_0$$. This gives the inequality (7), and combining this with (6) gives $$\Delta \tau_A<\Delta \tau_B$$, for arbitrary $$r_A$$. This answers the question.

Answer: Between meetings, the orbiting object A ages less than the radially-falling object B, regardless of the orbital radius $$r_A$$. This conclusion is consistent with the result of the more laborious calculations in [1], where the ratio of the elapsed proper times is calculated quantitatively.