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The usual simple statement of geodesics is that when moving from point $(x_o,y_o,z_o,t_o)$ to point $(x_1,y_1,z_1,t_1)$ the object will follow the path of greatest proper time experience, i.e., a clock traveling the geodesic will record more seconds than one going another path. I believe that is true even if you allow rocket thrusters or other energy addition devices.

But what if you consider a circular orbit which, after 1 year, results in going from $(0,0,0,0)$ to $(0,0,0,1)$? Wouldn't the proper time be maximized by simply sitting still, i.e., traveling only through time and not space? The orbiting clock would presumably run slower just as the orbiting clock in the international space station runs slower than one on Earth. In fact, if you want a physical model, assume a clock on a 400 km tall flag pole synchronized with the ISS during a pass-by (the ISS orbits at 400 km above the planet surface). One orbit later, wouldn't the ISS clock record less time than the flag pole clock?

Is that a violation of the concept "orbits are geodesics, for which the shortest distance between two 4D points is the path of greatest time"?

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One orbit later, wouldn't the ISS clock record less time than the flag pole clock?

Yes. The orbiting clock will record less time than a hovering clock.

Is that a violation of the concept "orbits are geodesics, for which the shortest distance between two 4D points is the path of greatest time"?

No, it is not a violation. In curved spacetime there can be multiple geodesic paths connecting a given pair of events. Geodesics extremize the proper time, but that is a local principle. That means that a geodesic has more proper time than any other “nearby” path, specifically any path that is only infinitesimally different.

The hovering path is not a geodesic and is not a local maximum. If instead of hovering the observer goes slightly up at the beginning and then back down at the end then there will less time dilation and a longer total elapsed proper time.

In contrast, for the circular orbit if you make any small deviation you will reduce the total elapsed proper time. So even though the hovering path is a longer path it is not a geodesic, and even though the circular orbit is a geodesic it is not the longest path. And that doesn’t violate the fact that a geodesic extremizes the proper time because they are not “nearby” paths.

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  • $\begingroup$ I commented to this effect yesterday and my comment is now gone, no notification, nothing. $\endgroup$
    – m4r35n357
    Apr 11, 2021 at 8:11
  • $\begingroup$ Is it not the case that you also have to compare like with like? It isn't just the starting position, but the starting velocity as well. $\endgroup$
    – ProfRob
    Apr 11, 2021 at 8:20
  • $\begingroup$ @ProfRob if you have infinitesimally different starting velocities then that is still a “nearby” trajectory. But of course you are right that the trajectories in the OP are not “nearby” in that sense either. $\endgroup$
    – Dale
    Apr 11, 2021 at 14:25
  • $\begingroup$ @m4r35n357 I have certainly had my share of moderator-deleted comments too. For that reason I recommend putting even short answers as answers and not comments $\endgroup$
    – Dale
    Apr 11, 2021 at 14:35
  • $\begingroup$ Thank you for your response. But I wonder if this is true: "The hovering path is not a geodesic and is not a local maximum. If instead of hovering the observer goes slightly up at the beginning and then back down at the end then there will less time dilation and a longer total elapsed proper time." Wouldn't there be more time dilation related to having a velocity and less proper time? It seems like sitting still would be the maximum possible proper time between (0,0,0,0) and (0,0,0,1) for any path. $\endgroup$ Apr 13, 2021 at 18:34

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