14 Mar 20
@humy saidYou omitted the full question.
No faith required. Just established and generally undisputed science which is the same science that says the Earth is round and not flat.At 3200km in orbit what is the time difference between the satellites and the receiver on earth's surface?What possible relevance does the exact figure have to anything we were talking about? Even if you wanted to know, whic ...[text shortened]... that it? You make no point and convince nobody here of anything other than you are a complete moron.
"Since you have so much faith that GR and SR equations are used show me you can use them. At 3200km in orbit what is the time difference between the satellites and the receiver on earth's surface?"
You have not answered the question.
At 3200km in orbit what is the time difference between the satellites and the receiver on earth's surface?
You have not answered that question either. That is because you are incapable of it. You are all show and no results. You are an incompetent showman.
@Metal-Brain
Here is the equation for that, do it yourself if you are so good:
t=(1 +GM/c^2 r of Earth)t
14 Mar 20
@sonhouse saidI already know. Apparently none of you are that good.
@Metal-Brain
Here is the equation for that, do it yourself if you are so good:
t=(1 +GM/c^2 r of Earth)t
14 Mar 20
@deepthought saidIt got somewhere when sonhouse admitted no Einstein equations were needed. It isn't my fault dogmatism persists.
I just took a quick look back to see if this discussion has gotten anywhere in the last 30 pages or so...
...apparently not.
@sonhouse saidExcept you need to take the motion of the satellite into account, you seem to have a sign error, and you've calculated the time difference for an asymptotic observer rather than a ground based one. The Schwartzschild metric in Polar coordinates is:
@Metal-Brain
Here is the equation for that, do it yourself if you are so good:
t=(1 +GM/c^2 r of Earth)t
𝕘 = c²(1 - r_s/r) dt⊕dt - dr⊕dr/(1 - r_s/r) - r²(dθ⊕dθ + sin²(θ.)dφ⊕dφ.)
We contract this with the tangent vector to the curve the particle moves along in order to get the proper time. To keep it simple we're having circular motion about the equator so sin(θ.) = 1. This means the proper time is:
(dτ/dt)² = c²(1 - r_s/r) - r²ω²
ω is the angular velocity of the satellite and ϖ the angular velocity of the Earth's rotation as reported by the asymptotic observer. Let's write the Earth's radius according to the asymptotic observer as R and the radius of the satellite's orbit as r. A small time interval as measured by the Earth bound observer is dT and the corresponding time interval as measured by the satellite is dτ, both of which correspond to the time interval dt as measured by the asymptotic observer. So we have:
(dτ/dT)² = [1 - r_s/r - r²ω²/c²]/[1 - r_s/R - R²ϖ²/c²]
ϖ is known, it is the length of the sidereal day, to find ω we need to solve the orbit. As a cheat we can just do a geostationary orbit so that ω = ϖ and look up R and r. Wikipedia gives:
r_s = 2GM/c² = Schwartzschild radius of Earth = 8.87mm
ω = ϖ = 2π/(seconds per sidereal day) = 7.29 E-5 Hz
R = 6378.1 km (Equatorial radius)
r = 35,786km + R = 4.21E7m
c ~ 3E8 m/s
Giving dτ/dT = 1 + 5.4E-10. So a clock on a geostationary satellite runs about one part in two billion faster than a clock on the equator.
16 Mar 20
@deepthought saidWhat about a satellite that is NOT geostationary?
Except you need to take the motion of the satellite into account, you seem to have a sign error, and you've calculated the time difference for an asymptotic observer rather than a ground based one. The Schwartzschild metric in Polar coordinates is:
𝕘 = c²(1 - r_s/r) dt⊕dt - dr⊕dr/(1 - r_s/r) - r²(dθ⊕dθ + sin²(θ.)dφ⊕dφ.)
We contract this with the tangent vector to t ...[text shortened]... on a geostationary satellite runs about one part in two billion faster than a clock on the equator.
@metal-brain saidI am aware that GPS satellites are generally at roughly half geostationary height above the Earth's surface despite the reasonable and natural assumption made by many people that they would be at exactly geostationary height but Deepthought didn't make that assumption as indicated by his words of "As a cheat...".
What about a satellite that is NOT geostationary?
You can recalculate using the same relativity time dilation equations for that lower height, whatever that lower height happens to be.
What's your point?
16 Mar 20
@metal-brain saidThen ω will have a different value from ϖ. We can solve the two body problem to get it. We could also work this stuff out for elliptical orbits and have the Earth bound observer at different latitudes and use the Kerr metric. For GPS they have them in medium Earth orbit (20,200 km above the Earth's surface) and the orbital period is half a sidereal day. So you can just plug in the new numbers to get a difference in clock rate of 0.446 ns per second.
What about a satellite that is NOT geostationary?
16 Mar 20
@deepthought saidAt 3200km in orbit there is no time difference from the surface of the earth.
Then ω will have a different value from ϖ. We can solve the two body problem to get it. We could also work this stuff out for elliptical orbits and have the Earth bound observer at different latitudes and use the Kerr metric. For GPS they have them in medium Earth orbit (20,200 km above the Earth's surface) and the orbital period is half a sidereal day. So you can just plug in the new numbers to get a difference in clock rate of 0.446 ns per second.
@metal-brain said-which would be completely irrelevant to whether relativity time dilation effects exist because;
At 3200km in orbit there is no time difference from the surface of the earth.
1, GPS aren't placed in that orbit but higher orbits. Time dilation is observed at those higher orbits.
2, according to relativity, as height of about 3,200 km height circular orbit around the Earth, the two opposing affects of relativity, i.e. from special and general relativity, for time dilation for that orbit, exactly cancel out.
(as correct said at https://demonstrations.wolfram.com/RelativisticEffectsOnSatelliteClockAsSeenFromEarth/ )
Thus whether any time dilation is observed at that height wouldn't be a test for time dilation as, according to relativity, none will be observed anyway and that fact does nothing to change the fact that time dilation effects are observed at lover orbits.
So you make no point.
@humy saidI never denied time dilation effects exist. Another lie from you.
-which would be completely irrelevant to whether relativity time dilation effects exist because;
1, GPS aren't placed in that orbit but lower orbits. Time dilation is observed at those lower orbits.
2, according to relativity, as height of about 35,786 km height circular orbit around the Earth, the two opposing affects of relativity, i.e. from special and general relativi ...[text shortened]... to change the fact that time dilation effects are observed at lover orbits.
So you make no point.
At 3200km in orbit there is no time difference from the surface of the earth, not 35,786 km. Your source of info is wrong.
For anyone here who is interested;
https://arxiv.org/pdf/gr-qc/0405001.pdf
"...
Generally speaking, clocks in low-orbiting satellites run slow due to the predominant time dilation effect, due to high orbital velocity and small altitude above the Earth’s surface. On the other hand, clocks in high-orbit satellites generally run faster than ground clocks because the gravitational potential effect is predominant.
..."
@metal-brain saidMy source of into says its 3200km.
At 3200km in orbit there is no time difference from the surface of the earth, not 35,786 km. Your source of info is wrong.
I made an edit mistake that I then corrected.
My source of info isn't wrong.
You should read the link; it says 3200km.
That same link also says;
"A GPS (Global Positioning System) satellite system is an example and practical proof of both of Einstein's theories".
Another edit error I corrected there is by changing the words "lower orbits" to "higher orbits".