B Mick's Spooky Time Dilation Puzzle

AtoMick-u235
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Hmmm, , Does gravitational time dilation (speed up) cancel out earth orbit time dilation (slow down) for astronauts, , , it must do, to a certain extent

Hmmm, , ,Mick's been thinking = the present is a continuous but fleeting moment, that allows the future to flow into the past, , ,so does the past and future push and pull the present ?, , , SPOOKY !!
 
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AtoMick-u235 said:
Does gravitational time dilation (speed up) cancel out earth orbit time dilation (slow down) for astronauts, , ,
It depends on what altitude the astronauts are orbiting at. For low Earth orbit (such as the ISS), the slow-down due to orbital speed is greater than the speed-up due to increased altitude, so clocks on the ISS run slower than clocks on Earth.

At the altitude of the GPS satellites, however (orbital radius of 4.2 Earth radii), the opposite is true: the altitude effect outweighs the orbital speed effect so the natural rate of clocks on the GPS satellites is faster than that of Earth clocks (and so a frequency correction has to be applied to the GPS satellite clocks so that the output "tick rate" is the same as that of Earth clocks).

The break point between these two regimes is at an orbital radius of 1.5 Earth radii, or an altitude above Earth's surface of 0.5 Earth radii, or about 3200 km.
 
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No, GPS satellites need to account for -7 microseconds/day due to SR (motion) and +45 microseconds/day due to GR (gravity).

SO ... if an astronaut is in geosynchronous orbit, the answer is obviously no. You could probably find the one exact orbital path for which the difference is zero, but in general ... no.

EDIT: I see Peter beat me to it.
 
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AtoMick-u235 said:
Mick's been thinking = the present is a continuous but fleeting moment, that allows the future to flow into the past, , ,so does the past and future push and pull the present ?, , , SPOOKY !!
Please review the PF rules on personal speculation. Your initial question was fine by itself.
 
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...

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