B Is the Twins Paradox Equivalent to GPS Time Dilation?

[binis]

So it would be natural to assume that we are talking about the relativity of simultaneity.

Yes,this is.Doppler effect on the "ticking" rate (frequency) of the clock as it transmitted by signals.

 

[PeterDonis]

Yes,this is.Doppler effect on the "ticking" rate (frequency) of the clock as it transmitted by signals.

The Doppler effect is not the same thing as relativity of simultaneity.

 

[binis]

Would there be a red shift due to velocity? Yes. Could you normalize this out of the measured results?

I think no,because you don't know the velocity (post 43)

Could there be an additional red or blue shift due to Doppler? Yes. Could you account for this as a change in radio path length?

I don't think this is really taken into account.Please provide a link if you know.

 

[PeterDonis]

I don't think this is really taken into account.

All of the issues you are raising are taken into account in the operation of the GPS system. The people who designed and implemented it were well aware of relativity and its implications for what they were doing.

Please provide a link if you know.

See the reference given in post #42.

 

[binis]

Thanks.And what about the last query in #43? What is the satellite's angular velocity? Is it ω/t or ω/t' ?

 

[jbriggs444]

Thanks.And what about the last query in #43? What is the satellite's angular velocity? Is it ω/t or ω/t' ?

The last query in #43 has to do with speed, not angular velocity. I quote it here:

Linear velocity v is by definition the runned space s per time t or v=s/t If the time passed in a specific point on the earth`'s surface is t, the dilated time passed on satellite is t'. What is the speed? Is it s/t or s/t'? Only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.

Speed would be the magnitude of velocity. Velocity would be incremental displacement divided by elapsed time. Important is that the displacement and the time are taken from the same coordinate system. So the answer is that the speed of an object in the non-primed coordinate system is given by:$$|\frac{ds}{dt}|$$while the speed of an object in the primed coordinate system is given by:$$|\frac{ds'}{dt'}|$$

For the separate question of angular velocity, conventionally $\omega$ is already an angular velocity. Dividing it by a time interval would yield something like an angular acceleration. However, consistency demands that one pick a frame of reference and stick with it. It is normally improper to mix measured values from the primed and unprimed frame. So angular acceleration in the unprimed frame would be $$\frac{d \omega}{d t}$$while angular acceleration in the primed frame would be $$\frac{d \omega'}{d t'}$$ The quantity "rapidity" "celerity" is the most commonly used violation to the heuristic about not mixing measurements from different frames. It is computed as distance traveled in the stationary frame divided by elapsed proper time for the moving entity. Importantly, this means that "celerity" is not the same thing as velocity.

[Thank you, @DrGreg]

 

[DrGreg]

It is computed as distance traveled in the stationary frame divided by elapsed proper time for the moving entity.

On a technicality, that's actually the definition of "celerity" (also known as "proper velocity", a name that I consider to be misleading), but the general point you are making applies to both rapidity and celerity.

 

[Herman Trivilino]

In a thread I started awhile back, in the common twins paradox scenario, it was indicated to me that the actual time (paraphrasing) on Earth for any given time in the ship is basically “undefined” (as it can’t be verified) and/or time dilated (ticking slower) for the trip out and then shifted to having advanced quickly for the trip back resulting in an older twin on Earth. In other words, it will be symmetrical (time dilation as seen from either twin) but not “real” or “verifiable” until the trip is complete in which case there in an asymmetry in rate of time resulting in the Earth twin being older (due to the frame change of the space ship).

This is not a correct understanding of the symmetry of time dilation. And that misunderstanding is at the root of your misunderstanding of the twin paradox. First, you need to understand the role played by simultaneity in the symmetry of time dilation. Another thing that might help you is to draw a spacetime diagram of the situation. When you do that you immediately see that the twins take different paths through spacetime, that those paths are not of equal length, and therefore they each experience different amounts of elapsed time.

 

[vanhees71]

Another important point in the twin paradox is that they have to synchronize their clocks at the same place when one of them starts to travel and then have to compare their clocks when meeting again at the same place. Then their clocks show each of the twins' proper time. This "clock postulate" has been confirmed by precision measurements with high accuracy in many ways.

 

[binis]

Important is that the displacement and the time are taken from the same coordinate system.
It is normally improper to mix measured values from the primed and unprimed frame.

Right,but only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.And the relative velocity is one,having one value (magnitube=speed).What is the value?

 

[jbriggs444]

Right,but only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.And the relative velocity is one,having one value (magnitube=speed).What is the value?

LMGTFY

Each satellite in the GPS constellation orbits at an altitude of about 20,000 km from the ground, and has an orbital speed of about 14,000 km/hour (the orbital period is roughly 12 hours - contrary to popular belief, GPS satellites are not in geosynchronous or geostationary orbits).

 

[binis]

LMGTFY

Each satellite in the GPS constellation orbits at an altitude of about 20,000 km from the ground, and has an orbital speed of about 14,000 km/hour (the orbital period is roughly 12 hours - contrary to popular belief, GPS satellites are not in geosynchronous or geostationary orbits).

That is,by #56,(ds/dt)=(ds'/dt')=14000 km/h?

 

[jbriggs444]

That is,by #56,(ds/dt)=(ds'/dt')=14000 km/h?

Let us assume that you are using special relativity, an unprimed inertial frame where the center of the Earth is at rest and a primed inertial frame where the satellite is instantaneously at rest.

If you were trying to use general relativity then there would be no inertial frames available [though we may have some coordinate systems that are close enough to inertial for our purposes]. If you were trying to use a frame where the satellite is continuously at rest then we are not talking about inertial frames any longer and you'll need to identify a coordinate system before we can agree on what $\frac{ds'}{dt'}$ means.

Note that, in particular, using general relativity to argue that the satellite is "inertial" and then trying to use special relativity to argue that only relative velocities are significant will yield a nonsense result.

In the earth-centered inertial (ECI) frame, the satellite's velocity, $\frac{ds}{dt}$ is 14,000 km/h.
In the satellite's tangent inertial frame, the Earth's velocity, $\frac{ds'}{dt'}$ is 14,000 km/h.

Surely this was already obvious?

 

[binis]

Let us assume that you are using special relativity, an unprimed inertial frame where the center of the Earth is at rest and a primed inertial frame where the satellite is instantaneously at rest.If you were trying to use general relativity then there would be no inertial frames available

Τhe principle of relativity, is that there is no such thing as "earth motion" or "satellite motion", only relative motion.The satellite movement is equivalent of the earth's.Therefore ds=ds'

In the earth-centered inertial (ECI) frame, the satellite's velocity, $\frac{ds}{dt}$ is 14,000 km/h.

Satellite is almost immobile relative to the center of the Earth,differently to a point on the surface.

 

[A.T.]

Τhe principle of relativity,...

Τhe principle of relativity applies to inertial frames. There are no global inertial frames in curved space time.

Satellite is almost immobile relative to the center of the Earth

Not in an inertial frame.

 

[jbriggs444]

The satellite movement is equivalent of the earth's.Therefore ds=ds'

This is completely incorrect. Length contraction is a thing.

 

[Ibix]

Τhe principle of relativity, is that there is no such thing as "earth motion" or "satellite motion", only relative motion.

If both Earth and satellite were moving inertially in flat spacetime, you would have a point. But you must either model the satellite as moving non-inertially in flat spacetime (and don't ask too many questions about gravity) or use a full general relativistic solution in which there are no global inertial frames. Either approach invalidates your application of the principle of relativity.

 

[Janus]

Τhe principle of relativity, is that there is no such thing as "earth motion" or "satellite motion", only relative motion.The satellite movement is equivalent of the earth's.Therefore ds=ds' Satellite is almost immobile relative to the center of the Earth,differently to a point on the surface.

As many posters have already stated, this only applies when dealing with inertial frames. Once you start to make measurements from a non-inertial frame, this changes. If you are in the tail of an accelerating rocket, for example, you will measure a clock in the nose of the rocket ticking faster than your own, even though you have no relative motion between you and it. (conversely, someone in the nose will measure your clock as running slow). And this is not confined to clocks in the rocket or sharing your acceleration. You will measure any clock "ahead" of the accelerating* rocket as ticking fast, the further away, the faster it ticks.
.
Now take this rocket and arrange things so that its circling a clock at a constant speed, ( it is pointing its rockets outward in order to maintain a "forced orbit" around the clock). It is still accelerating, even though its speed remains constant. The acceleration is centripetal and always pointed at the central clock. As such, someone in the rocket would measure the central clock as ticking fast.
Someone at the central clock would be in an inertial frame and it would only be the rocket's circular velocity that would effect how they would measure the rocket's clock tick, and this would cause them to measure it as ticking slow. The point is that, while both the Central observer and rocket observer can lay claim to being "at rest", only the central observer can do so from an inertial frame, the rocket observer's "rest frame" would be a non-inertial rotating one, for which there are additional factors beyond just relative motion for determining relative clock rates.**
With GPS satellites, gravity provides the centripetal acceleration, and the difference in gravitational potential adds an additional complication, but the idea that the only important factor is "relative velocity alone" still does not hold.

* in this case "ahead" means in the direction of the acceleration

** You could also have the central clock rotating so that it maintains the same orientation with respect to the rocket at all times, and consider itself as being at rest within the same rotating frame as the rocket, But now it would be like the Nose clock in the first accelerating rocket example, and would still measure the rocket clock as ticking slow.

 

[binis]

Not in an inertial frame.

But "There are no global inertial frames in curved space time."

 

[A.T.]

Satellite is almost immobile relative to the center of the Earth

Not in an inertial frame.

But "There are no global inertial frames in curved space time."

Why "but"? Did you mean "because"?

 

[PeterDonis]

Satellite is almost immobile relative to the center of the Earth

No, it isn't. The 14,000 km/h speed that was given is relative to the center of the Earth.

differently to a point on the surface.

A point on the Earth's surface is moving, at most (for a point on the equator) at about 1600 km/h relative to the center of the Earth. So the satellite's speed relative to a point on the Earth's surface isn't that much different from its speed relative to the Earth's center.

But "There are no global inertial frames in curved space time."

But there are local ones, and as the reference to the Earth-Centered Inertial (ECI) frame should have told you, it is quite possible to treat a non-rotating frame centered on the Earth's center as inertial to a good enough approximation for many purposes, including the evaluation of the speeds of satellites.

You are hijacking someone else's thread with repeated incorrect comments that add nothing useful to the discussion. As a result, you have now been banned from further posting in this thread.