Clock tossing solution.
snoopies622 said:
This seems counter-intuitive to me, so I wondering if someone could please confirm or deny it:
I stand on the surface of the Earth holding two identical clocks, each set to the same time. I toss one straight up. When it comes down, it reads a later time than the one that stayed with me because while in flight it traveled on a geodesic, which maximizes proper time, while the one that stayed in my hand did not.
Hi Snoopies,
I think I have found the mathematical solution to the question you have posed.
The total proper time of the tossed clock to go up and come back down again is:
(Eq 1) \tau = R_{H}(a +sin(a))\sqrt{\frac{R_{H}}{2m}}
where 2m = the Schwarzschild radius and
a = cos^{-1} \left(2 \frac{ R_{L}}{R_{H}}-1 \right)
R(L) = The low radius that the clock is tossed from.
R(H) = The maximum height the clock gets to (apogee).
The total coordinate time is:
(Eq 2) t = (R_{H} + 4m)Qa + 4m LN\left(\frac{Q+tan(a/2)}{Q-tan(a/2)}\right) + R_{H}Q sin(a)
where LN is the natural logarithm and
Q = \sqrt{(R_{H}/2) -1 }
The total local time on the clock that remained at R(L) is simply (Eq 2) adjusted by the gravitational time dilation facter to give:
(Eq 3) t = \left((R_{H} + 4m)Qa + 4m LN\left(\frac{Q+tan(a/2)}{Q-tan(a/2)}\right) + R_{H} Q sin(a) \right)\sqrt{1-\frac{2m}{R_{O}}}
where R(o) is the radial location of the stationary observer, which in this case is R{L).
It turns out that the proper time that elapses on the clock that stays on the ground according to (Eq 3) is indeed less than the proper time of the tossed clock as per (Eq 1), which agrees with your assumption in the opening post of this thread.
However, if the location R(o) of the stationary observer is higher than the apogee R(H) of the tossed clock, the time according to that observer is greater than the proper time of the tossed clock.
The above equations are my interpretation of the ones given here
Ref:
http://www.mathpages.com/rr/s6-04/6-04.htm and I have doubled them to allow for the return journey.
snoopies622 said:
I guess the counter-intuitive part is that in space, a geodesic minimizes distance, but in space-time it maximizes distance. Is this correct?
First you have to define "distance". Usually in common terms I think of least distance as the route that takes the least proper time. As DrGreg pointed out there can be many geodesics between two points A and B but it useful to think there is always at least one geodesic that is the shortest possible route (in terms of proper time) between two points for a given initial velocity. In 3 space the route may appear curved but in fact it takes longer to travel the Euclidean "straight" line between A and B.
The above equations show that when the inertial clock traveling on the geodesic returns to the stationary non inertial clock, the free falling inertial clock shows the greater elapsed proper time. I agree that in some ways this is counter intuitive because in Special relativity the clock moving relative to your reference frame experiences less proper time than the clocks that are stationary relatived to you. To me, my intuitive in GR is that the clock that spends the most time in the "fast time zone" higher up experiences the most proper time even when time dilation due to velocity is taken into account, but I am not sure how that would be expressed formally or even if it is strictly true.
.