# I Time dilation for satellites

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1. Apr 17, 2017

### Neutrinos02

Hello,

I read that satellites is effected by the time dilation caused by gravity and also by that one from special relativity. And so there is a need to prepare the onboard clock to ensure that the time is synchronized with a clock on Earth.

But why is this effect not symmetric? The satellites should see that the clock on earth is slowed down and vice versa?

Thanks.

2. Apr 17, 2017

### Staff: Mentor

The clocks are accelerating, tracing a roughly circular path. So the movements are not symmetrical.

3. Apr 17, 2017

### pervect

Staff Emeritus
What you read is correct. I don't quite understand the argument for why you think it should be symmetric in the first place.

We could go through the calculation in a bit more detail - depending on your math and physics background - but I'm not sure that would answer your question unless you can tell us your thought processes on why you think it should be symmetric.

4. Apr 17, 2017

### Aufbauwerk 2045

You may find it interesting to look into the work of Dr. Friedwardt Winterberg. He is perhaps the only surviving student of Heisenberg. One of his famous accomplishments was his proposal to test general relativity by using atomic clocks in earth orbit. His theory was confirmed by Hafele and Keating in 1971. This understanding of GR was very important in the development of GPS.

5. Apr 17, 2017

### Orodruin

Staff Emeritus
Actually, the clocks have zero proper acceleration.

6. Apr 18, 2017

### David Lewis

In either case, the main effect stems from a difference in gravitational potential. There is a smaller (opposing) effect due to relative motion between the satellite and the ground receiver. Acceleration per se has no significant bearing on the time differential that I'm aware of.

7. Apr 18, 2017

### Ibix

The reason that SR time dilation effects are symmetric is that two clocks in relative motion is a symmetric situation. That is, both clocks can describe the situation as "I am at rest and the other clock is moving at speed v". Since the descriptions are the same, the effects must be the same.

However, two clocks at different altitudes do not describe their situations the same way. One says the other clock is above it and the other says the other clock is below it. So there's no requirement that the effects each one sees should be the same. That doesn't immediately mean that they must see different things, but it's easy enough to show that gravitational redshift and hence time dilation must occur and that, as others have noted, it is related to the gravitational potential.

8. Apr 18, 2017

### PeroK

You might want to read up on the Hafele-Keating experiment.

One way to look at it is to consider a clock at rest relative to the Earth, but outside the Earth's gravity. Perhaps far enough away that the gravity is negligible.

Compared to that clock, a clock on the Earth's surface will have time dilation due to gravity (relatively strong) and circular motion (small). A clock in orbit will have time dilation due to gravity (weaker) and circular motion (larger).

When you take the two together you find, I recall, that the satellite clock runs slightly faster than the Earth clock. Both relative to the distant reference clock.

As the distance between the Earth and satellite clocks is not changing, they must observe this difference too.

9. Apr 18, 2017

### A.T.

Symmetry is guaranteed in flat space time, given symmetric proper acceleration profiles. Neither is given here.

10. Apr 18, 2017

### Arkalius

This depends on the orbit altitude. Clocks in roughly circular orbits with an altitude less than 50% the radius of the planet will tick slower than clocks on the surface (special relativity effects are stronger). However, orbits higher than that will have the clocks ticking faster (general relativity effects are stronger). More eccentric orbits will, of course, be more complex.

11. Apr 18, 2017

### A.T.

Is this what you get for Earth, or for any spherical mass?

12. Apr 18, 2017

### Arkalius

I did the math for Earth and found the value to be exactly 50% of Earth's radius. I haven't proved it but it appears that this is generally true for any spherical mass. An orbit radius of 1.5x the surface radius will have the object's time dilation between velocity and gravity balance out.

13. Apr 18, 2017

### Staff: Mentor

That can't be right, unless you're introducing some other condition as well. In both classical gravity and the Schwarzschild solution of GR, everything above the surface of a spherically symmetric gravitating mass depends just on the distance from the center; it is independent of where the surface is.

14. Apr 18, 2017

### Janus

Staff Emeritus
What you are looking for here is the situation where the time dilation just due to gravity at distance R1 is equal to the time dilation for an orbit of R2 for a given body.
So you are solving the equation for the ratio R2/R1 when

$$\frac{1}{\sqrt{1-\frac{2GM}{R_1 c^2}}} = \frac{1}{\sqrt{1- \frac{3GM}{R_2 c^2}}}$$

Which reduces to:
$$\frac{R_2}{R_1}= \frac{3}{2}$$

The one thing this does not take into account is that a clock sitting on the surface of the Earth will be rotating with the Earth and thus have its own velocity with respect to to the center of the Earth.

15. Apr 18, 2017

### Staff: Mentor

Ah - right. That's the additional condition I was looking for, that we are considering an object in orbit at the surface height, as opposed to resting on the surface and moving with the rotational speed of the planet.

16. Apr 18, 2017

### David Lewis

Clocks on GPS satellites run slow (in their own frame) on purpose to keep them in synch with ground clocks. So when discussing clocks ticking fast or slow, we can mean either with respect to the rate at which time passes in the clock's frame, or the apparent rate of ticking when viewed from a different frame.