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Registered Member #61373
Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87
I am trying to calculate the time dilation that I experienced while flying on a jetliner. I have already calculated my "time loss" for Special Relativity (-700ns), but I cannot find a reliable calculator for general relativity at 36,000 ft alt. I am having trouble solving the complex equation for general relativity.
I want to find out how much time I "gained" from General Relativity (99.63% of Earth's surface gravity, or 0.37% less) from flying at 36,000 ft alt over a 44 hr total flight period (relative to the ground at sea level).
It is not necessary to fully understand general relativity to calculate gravitational effects on time in this case. Say you send a laser beam from the ground up to the airplane to tell you how fast time runs on earth. Assume the laser light has a certain frequency f on the ground. The beam consists of photons, each having an energy of h*f and a corresponding mass m = E/c^2 = h*f/c^2.
The photon will loose energy as it rises up to the airplane due to its mass. This energy loss is m*g*H, where m is the photon mass, g the gravitational acceleration (10 m/s^2) and H the altitude of the plane. Since the energy of the photon is h*f, f will have decreased somewhat once it has reached the plane. This implies, that clocks on earth seem to run slow when watched from the plane.
Putting all mentioned equations together, primed values are on the airplane:
E' = E - m*g*h = h*f - h*f/c^2 * g * H = h*f*(1 - g*H/c^2)
f' = E'/h = f*(1 - g*H/c^2)
So the factor, by which time runs slower on the ground is (1 - g*H/c^2)
The gravitational time dilation leads to a peculiar effect, when you drop something into a black hole. The drop will seem to slow down, once the object comes close to the event horizon. Actually it will never seem to enter the black hole. The color of the object will turn more and more reddish until it eventually will become invisible.
Registered Member #96
Joined: Thu Feb 09 2006, 05:37PM
Location: CI, Earth
Posts: 4062
Indeed, its called "redshift". Interestingly, you can now actually buy small atomic clock chips (CSAC) if you have $900 burning a hole in your pocket. Uses about the same power as a mobile phone but accurate to one second in 270M years or so the manufacturers claim. Compare with clock left back home to see if it drifts over a long haul flight.
Note, there can be effects with temperature but generally if it stays locked then the data is indeed valid.
Registered Member #61373
Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87
Uspring wrote ...
So the factor, by which time runs slower on the ground is (1 - g*H/c^2)
So it would be ( (1-(9.807m^2)*10972m/(299792458m/s^2) )*158400 sec ??? Are my parentheses in the right place? Im using 9.807 m/s/s for an exact gravity measure.
After 44h or 158400s you'd be about 158400s * 1.197e-12 = 190ns older than if you'd stayed on the ground. This is from gravitational effects only. There is, as you said, an extra contribution from the time dilation due to the planes speed. Its sign is opposite to that of the gravitational effect.
After 44h or 158400s you'd be about 158400s * 1.197e-12 = 190ns older than if you'd stayed on the ground. This is from gravitational effects only. There is, as you said, an extra contribution from the time dilation due to the planes speed. Its sign is opposite to that of the gravitational effect.
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