Relative Time Effect on Orbital Comm: Satellite Freq & Audio/Video

[InquiringMind]

Given that the time is relatively slower for a moving object that a stationary object, does the relative difference have an effect on satellite communications? For instance, if a geostationary satellite, which is traveling very fast to appear stationary on Earth (and therefore experiencing a slower rate of time relative to Earth time), is transmitting at a frequency of 400Mhz, will it's signal appear at < 400Mhz on earth? Also, why don't audio and video from such a satellite appear slower, and sound lower when received on earth?

 

[Orodruin]

The time dilation effect on satellites is so miniscule that you will not be able to tell the difference. Also note that the satellite is also affected by gravitational time dilation, which has the opposite sign, so motion is not the only relevant factor.

The time dilation effect is relevant for the GPS system and if it did not correct for it you would quickly lose precision.

 

[Ibix]

There is an effect due to velocity and an opposite effect due to the difference in gravitational potential. Both are very small and their sum is even smaller.

The GPS relies on very high precision timing and its clocks are corrected for the effect, but that's the only application I'm aware of that cares. In other cases, it is lost in other sources of noise.

 

[Charles Kottler]

Further to the above replies, the broadcasts are initially transmitted from Earth with the satellite effectively operating as an amplifying reflector. Even if the effects of time dilation or gravity were significantly larger the frequency received back on Earth would be the same as when it was transmitted, only with a greater lag.

 

[Orodruin]

Further to the above replies, the broadcasts are initially transmitted from Earth with the satellite effectively operating as an amplifying reflector. Even if the effects of time dilation or gravity were significantly larger the frequency received back on Earth would be the same as when it was transmitted, only with a greater lag.

This is correct only for gravitational time dilation. For the time dilation due to motion it depends on the direction of motion of the satellite.

 

[pervect]

This is correct only for gravitational time dilation. For the time dilation due to motion it depends on the direction of motion of the satellite.

Offhand, I would think that the example of "The Centriguge and the Photon", in MTW's "Gravitation", $2.8, illustrates the principle applies to time dilation due to motion when one assumes a static geometry. In MTW's example there is a rotating ceentrifuge, and we ask what the redshift is between a signal emitted at one point on the rim of the centrifuge and received at another. The result must be zero, MTW derives this purely from geometric arguments.

For the case of the satellite, the satellite has a constant potential - this follows from the time-independence of the geometry. So I think there should be a constant doppler shift from ground to orbit, and this factor will be the inverse of the factor from orbit to ground.

 

[m4r35n357]

I hesitated to post this link in a B thread, but I have done it because there is an equation that you can plug numbers into, and some explanatory text with examples.

Look about a quarter of the way down for a paragraph beginning "Consequently" and the paragraph after that. The calculations show the combined effect of speed and gravity (for circular orbits) in one equation.

 

[PeterDonis]

I think there should be a constant doppler shift from ground to orbit

Not if the observer on the ground is at a fixed point on the ground. In that case, as Orodruin said, the Doppler shift between the ground observer and the orbiting observer will vary as the orbiting observer moves. (We are here assuming that the "ground" is not rotating; if it is, that adds a complication that is not really relevant here.)

The meaning of the static geometry in this context is that there is a family of observers who all see unchanging spacetime geometry along their worldlines, and such that the spacetime can be foliated by spacelike 3-surfaces that are all orthogonal to all the worldlines. The ground observer in this scenario is one such observer, but the orbiting observer is not. So even though the orbiting observer is at a constant altitude, and therefore has a constant potential, the deduction of "constant Doppler shift" you are trying to draw from that does not hold; it only holds between static observers, observers for whom both of the above things are true.

 

[Charles Kottler]

The original example was concerning geostationary orbit:

For instance, if a geostationary satellite, which is traveling very fast to appear stationary on Earth (and therefore experiencing a slower rate of time relative to Earth time),

This is correct only for gravitational time dilation. For the time dilation due to motion it depends on the direction of motion of the satellite.

The round trip time for a signal for a (stationary) Earth bound transmitter via a geostationary satellite to any stationary ground based observer will be constant (barring weather effects), so regardless of the relative speeds there cannot be any difference in the frequency of the received signal, only a delay before it arrives.

 

[PeterDonis]

The round trip time for a signal for a (stationary) Earth bound transmitter via a geostationary satellite to any stationary ground based observer will be constant (barring weather effects)

Yes, this is true. This brings up a way in which I overstated things somewhat in my last post. An observer on the rotating Earth and an observer in a geostationary orbit are both members of a family of observers who do in fact see unchanging spacetime geometry along their worldlines, even though they are not static (they are only stationary, using the technical meaning of that term--their worldlines are both members of the family of integral curves of the same timelike Killing vector field). The reason is that, relative to static observers, they both have the same angular velocity--i.e., their "orbital period" is the same. (The term is in quotes because the ground observer is not in a free-fall orbit--in fact, none of the observers in this family of observers are in free fall except for the one in the geostationary orbit. But that doesn't affect what I've said in the rest of this post.)

 

[Ibix]

To summarise the above for @InquiringMind, relativistic effects are tiny in Earth orbit and are overwhelmed by clock drift due to more mundane causes. The only exception to this is the GPS system whose onboard clocks are precise enough to need to account for the systematic drift from relativistic effects.

Specifically for the case of geosynchronous communication satellites, the case is even simpler because they act as simple reflectors. Because they are stationary above the Earth they add no effect due to their motion. And the gravitational time dilation that accumulates in the radio signal as it climbs to the satellite de-accumulates again on the way back down. So you can ignore relativistic effects in this case. Active emissions from the satellite (e.g. live video from a space station in geosynchronous orbit) would be blue shifted and appear to be running fast, but again the effect is far smaller than more mundane clock drift.

There is an additional effect for satellites in other orbits, which is a Doppler effect due to their motion relative to the Earth's surface (all the disagreement above stems from whether we're talking about the geosynchronous case or not). This effect is easily measurable and easily engineered around in communication applications.

I keep saying that the effects are tiny. The link @m4r35n357 provided shows how tiny. The third equation after equation 5 (just above the paragraph starting "Consequently") gives the extra time seen by a satellite orbiting at ##r_o## from the Earth's center compared to someone sitting on the Earth’s surface (radius ##r_e##) who experiences time ##\Delta\tau_\mathrm {earth}##. Some examples, as m4r35n357 pointed out, are in the two paragraphs below that equation.

Hope that's a useful summary.