Taylor/Wheeler GPS time dilation calculation (ProjectA)
qbit
GPS time dilation from the ProjectA chapter in Taylor/Wheeler book?
( http://www.eftaylor.com/pub/projecta.pdf )
I must admit I have seldom such a garbage description of an algorithm seen!
Yes, Dirk Van de Moortel in his postings often has used the same
reference, so maybe he has it working; but then maybe not, who knows... :-)
Eric Gisse
> Has anybody succeeded in using the GR method for calculating
> GPS time dilation from the ProjectA chapter in Taylor/Wheeler book?
http://groups.google.com/group/sci.physics.relativity/msg/72ba55f8bf3f3b56?dmode=source
Then again, I know what I'm doing. If you can't do it even with the
handholding of Taylor&Wheeler, you really are sad.
> (http://www.eftaylor.com/pub/projecta.pdf)
> I must admit I have seldom such a garbage description of an algorithm seen!
> Yes, Dirk Van de Moortel in his postings often has used the same
> reference, so maybe he has it working; but then maybe not, who knows... :-)
http://npoesslib.ipo.noaa.gov/IPOarchive/MAN/doc165.pdf
Poor baby can't handle the reality of the situation :(
Androcles
"qbit" <qb...@quantumworldz.com> wrote in message
news:59956.4B4a5.90...@quantumworldz.com...
: Has anybody succeeded in using the GR method for calculating
"As a result, measurement of the relative position of control tower and
airplane is accurate to 1 or 2 meters. This configuration of receivers
permits blind landing in any weather. Runway collisions can also be avoided
by using this system to monitor positions of aircraft on the ground (a task
impossible for the electromagnetic signals of radar).."
+/-2 metres in a kilometre is not so good. In fact even 10 cm below the
runway is like to cause a burst tyre with disastrous results, 10 cm above
and the pilot will do a go-a-around. You could probably get the jetway to
the
door within a metre, 2 metres is stretching it.
It's a strange fact that the electromagnetic signals of radar provide the
glideslope.
http://www.avionicswest.com/articles/glideslope_101.htm
"The timing accuracy required by the GPS is so great that general
relativistic effects are central to its performance."
You'd better use an atomic clock in the receiver, then.
It's fuckin' obvious that the paper is pack of lies written by an ignorant
jerk.
qbit
> On Aug 12, 3:38 am, "qbit" <q...@quantumworldz.com> wrote:
>
> > Has anybody succeeded in using the GR method for calculating
> > GPS time dilation from the ProjectA chapter in Taylor/Wheeler book?
>
> http://groups.google.com/group/sci.physics.relativity/msg/72ba55f8bf3f3b56?dmode=source
> Then again, I know what I'm doing. If you can't do it even with the
> handholding of Taylor&Wheeler, you really are sad.
I'm just interessted to _REPLICATE_ what these authors have
written in their book.
I want see whether what they write is true or not
by using the same algorithm given in the book,
to verify their own result.
I have the suspicion that something is not fitting the equation, so to say...
In your posting above you write the following:
|> Assumptions:
|>
|> Schwarzschild metric - nonrotating planet [I can input rotation in
|> later], spherically symmetric.
|> Circular orbits - it's easier that way.
|> The satellite and receiver are both on the equator. [even easier!]
|> Movement is in a plane with theta = pi/2 [equator] in the usual
|> spherical coordinate system.
|>
|> With that in mind...
|>
|> ds^2 = [c^2 -2GM/r]dt^2 - r^2 d\phi^2
What is d\phi ?
|> Factor out dt and make the abbreviation that w = angular velocity of
|> objects = d\phi/dt.
|>
|> ds^2 = dt^2 ( c^2 - 2GM/r - r^2w^2)
|>
|> We really care about proper time - ds^2 = d\tau^2/c^2
|>
|> d\tau = dt*sqrt(1 - 2GM/rc^2 - (rw/c)^2)
|>
|> Integrate [it doesn't really matter since r,w are not functions of
|> either coordinate or proper time].
What does this mean? Integrate or integrate not?
|> tau / t = sqrt(1 - 2GM/rc^2 - (rw/c)^2)
|>
|> Now, for my numbers
|>
|> [IN SI UNITS]
|>
|> G = 6.672x10^-11
|> r_orbit = 4.22x10^7 m
|> r_surface = 6.37x10^6 m
|> M = 6x10^24 kg
|> c = 3x10^8 m/s
|> w = 1/86400s
|>
|> Using this, I get a dtau = tau_surface - tau_orbit = 5.888x10^-10 dt
Can you show your solution with the numbers entered?
|> Which means the satellite clock is faster by about 51000 nanoseconds
|> per day. The true answer [via experiment] is about 38,000 nanoseconds
|> per day.
The Taylor/Wheeler book computes 39000 ns. Why is your result differing?
The topic is about the solution in that book.
Can you replicate just the exact steps of the book to get the exact same result?
John C. Polasek
wrote:
If you followed exactly you are in trouble, for there are several
errors. It's not v^2, it has to be v^2/c^2. Also M = MG/c^2, not
mentioned.
He failed to notice you can simplify the orbiting calculation by
noting that in orbit, using Eq. 9,
v^2 = MG/r.
So you can rewrite in Eq. 3 from
1 - 2M/r - v^2 to the simpler
1 - 3M/r
and drop the v^2. Then he goes on in Eq. 10 to derive the orbital
velocity from the orbital period which can bring in errors. You can't
use the triple form for the earth slowing, it's only for orbit
conditions.
John Polasek
qbit
> On Sun, 12 Aug 2007 12:38:00 +0100, "qbit" <qb...@quantumworldz.com> wrote:
>
> >Has anybody succeeded in using the GR method for calculating
> >GPS time dilation from the ProjectA chapter in Taylor/Wheeler book?
> >( http://www.eftaylor.com/pub/projecta.pdf )
> >I must admit I have seldom such a garbage description of an algorithm seen!
> >Yes, Dirk Van de Moortel in his postings often has used the same
> >reference, so maybe he has it working; but then maybe not, who knows... :-)
>
> If you followed exactly you are in trouble, for there are several errors.
Surprize surprize. Thank you for the confirmation.
For a book that is intended as an _introduction_ to RT
(the title reads "Exploring Black Holes: Introduction to General Relativity")
it contains such subtile and unexcusable errors!
This book was published in year 2000. Are we really the
first persons who have detected these errors in the book? :-)
And there is no sign of any correction avalaible at their web site
( http://www.eftaylor.com/download.html#general_relativity )
###########################################################################
A CALL TO TAYLOR (M.I.T) AND WHEELER (Princeton):
I am hereby publicly asking Edwin F. Taylor and John Archibald Wheeler
for a statement about these errors in their book,
and for an apology for these errors,
and for an official correction of the errors.
This is a fair request I would say since I,
and possibly many other readers, have lost that much time
due to these and other errors in the book!
If these authors are honest people then they should give a public statement.
###########################################################################
> It's not v^2, it has to be v^2/c^2.
> Also M = MG/c^2, not mentioned.
> He failed to notice you can simplify the orbiting calculation
> by noting that in orbit, using Eq. 9,
> v^2 = MG/r.
>
> So you can rewrite in Eq. 3 from
> 1 - 2M/r - v^2 to the simpler
> 1 - 3M/r
> and drop the v^2.
Ok, will try it out.
> Then he goes on in Eq. 10 to derive the orbital velocity
> from the orbital period which can bring in errors.
> You can't use the triple form for the earth slowing,
> it's only for orbit conditions.
What does this practically mean?
Is there a workaround possible, or is all of it completely useless
to get the documented and real result of 38 (or 39) microseconds
per day for the GPS ?
Eric Gisse
> "Eric Gisse" <jowr...@gmail.com> wrote
>
> > On Aug 12, 3:38 am, "qbit" <q...@quantumworldz.com> wrote:
>
> > > Has anybody succeeded in using the GR method for calculating
> > > GPS time dilation from the ProjectA chapter in Taylor/Wheeler book?
> > > (http://www.eftaylor.com/pub/projecta.pdf)
>
> > Then again, I know what I'm doing. If you can't do it even with the
> > handholding of Taylor&Wheeler, you really are sad.
>
> I'm just interessted to _REPLICATE_ what these authors have
> written in their book.
Then read the book. I can explain it to you but I cannot understand it
for you.
> I want see whether what they write is true or not
> by using the same algorithm given in the book,
> to verify their own result.
It isn't. Describing Earth as a point mass is only approximately true
- if you finish reading the rest of my post, you would see I mentioned
that the answer is only about 75% of the correct value. That might be
tweaked up a little if I used the exact orbital parameters, but it was
close enough for my purposes. I wasn't going for The True Value(tm),
it was more of a proof-of-concept for someone completely clueless.
> I have the suspicion that something is not fitting the equation, so to say...
Like because I said I was off by about 25%?
http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node5.html
A more correct solution would be to use the weak field limit which is
what was done in the page I linked. That way, the multipole moments of
the earth could be taken into account.
>
> In your posting above you write the following:
>
> |> Assumptions:
> |>
> |> Schwarzschild metric - nonrotating planet [I can input rotation in
> |> later], spherically symmetric.
> |> Circular orbits - it's easier that way.
> |> The satellite and receiver are both on the equator. [even easier!]
> |> Movement is in a plane with theta = pi/2 [equator] in the usual
> |> spherical coordinate system.
> |>
> |> With that in mind...
> |>
> |> ds^2 = [c^2 -2GM/r]dt^2 - r^2 d\phi^2
>
> What is d\phi ?
The quantity \phi is LaTeX notation.
>
> |> Factor out dt and make the abbreviation that w = angular velocity of
> |> objects = d\phi/dt.
> |>
> |> ds^2 = dt^2 ( c^2 - 2GM/r - r^2w^2)
> |>
> |> We really care about proper time - ds^2 = d\tau^2/c^2
> |>
> |> d\tau = dt*sqrt(1 - 2GM/rc^2 - (rw/c)^2)
Whoops, I never carried the negative sign through.
> |>
> |> Integrate [it doesn't really matter since r,w are not functions of
> |> either coordinate or proper time].
>
> What does this mean? Integrate or integrate not?
What is the integral of a constant?
Have you ever taken a calculus course?
>
> |> tau / t = sqrt(1 - 2GM/rc^2 - (rw/c)^2)
> |>
> |> Now, for my numbers
> |>
> |> [IN SI UNITS]
> |>
> |> G = 6.672x10^-11
> |> r_orbit = 4.22x10^7 m
> |> r_surface = 6.37x10^6 m
> |> M = 6x10^24 kg
> |> c = 3x10^8 m/s
> |> w = 1/86400s
> |>
> |> Using this, I get a dtau = tau_surface - tau_orbit = 5.888x10^-10 dt
>
> Can you show your solution with the numbers entered?
Yes, but I'm not going to do it for you. If you think you have the
mathematical and physical insight to battle relativity, you don't need
someone to enter the numbers for the calculations for you.
>
> |> Which means the satellite clock is faster by about 51000 nanoseconds
> |> per day. The true answer [via experiment] is about 38,000 nanoseconds
> |> per day.
>
> The Taylor/Wheeler book computes 39000 ns. Why is your result differing?
I didn't include the satellite velocity in the mix. The result was
close enough for my purposes. All I wanted to do was show an example
of how the effect is calculated.
> The topic is about the solution in that book.
> Can you replicate just the exact steps of the book to get the exact same result?
Yes, but I won't.
You have come on this newsgroup screaming about how relativity is
wrong wrong wrong and how Newtonian gravitation is utterly correct. If
you can't follow the exercises in an introductory book...what does
that tell you? The textbook you are working out of is of the caliber
assigned to high school students or college freshman. Folks who have
trouble with something like that have no right to tell physicists that
they are wrong about anything.
jpolasek
> "John C. Polasek" <jpola...@cfl.rr.com> wrote
You will find for the orbital delays that 2M/r will give 14us/day, M/r
(velocity) will give 7us/day for a total of 21 us/day that the
orbiting clock is slowed down from conditions at infinity.
The earth clock is slowed down much more, by 60 us/day = 1 - M/rearth.
So 60-21=39.
You say algorithm so you must be programming something. The us/day
given are after multiplying your algebra each time by K = 86.4x10^9 us/
day.
So in orbit use 1-sqrt(1-3M/r)xK = 1.5M/r*K = 21us/day. (Earth's
rotation about .1us/day).
I have a detailed derivation on my website http://www.dualspace.net,
the gravity paper, page 3. It shows the delay using my theory, pg 3
and also the relativistic version on pg. 4.
(I wondered who wrote the paper, since I don't see a signature or
attribution).
John Polasek
qbit
> by K = 86.4x10^9 us/day.
> So in orbit use 1-sqrt(1-3M/r)xK = 1.5M/r*K = 21us/day. (Earth's
> rotation about .1us/day).
> I have a detailed derivation on my website http://www.dualspace.net,
> the gravity paper, page 3. It shows the delay using my theory, pg 3
> and also the relativistic version on pg. 4.
This looks really much better than what I've studied so far.
Just one thing is unclear to me in your method:
How did you get the value 86400e6 ? (It is of course not 86400^-1)
Would this value be the same if we used the same GPS mechanism
in the orbit of another planet instead of the Earth whereby using
our same 24h clock system also for that planet?
And: what about using the Sideral Day seconds (= 86164 s) instead of 86400 ?
Would this not give a more correct value?
And: have you in your method considered the fact that the GPS satellites
are orbiting twice a day? I.e. their rotation speed is about 3873 m/s.
Do you then still get the 38 us/day ?
I'm not sure yet, but I have the feeling that using the RT method
the value 38 us/day can be true only if the satellite makes
just one orbit per day, and therefore the correct value must
be something different than the 38 us/day, perhaps 29 ?
> You say algorithm so you must be programming something.
Yes, I'm putting these Time Dilation methods into a program
for running simulations, and also for testing these methods/algorithms.
So far I can use these methods for a body (satellite, airplane),
moving along an equatorial orbit either eastwards or westwards,
over a rotating body (Earth); ie. GPS and Hafele&Keating like simulations/experiments.
FYI: I've one other working RT version of such time dilation methods,
and was interessted also to add the Taylor/Wheeler method into my collection,
but failed to do so so far due to the errors in the text :-(
Now, I will add also your method into my collection.
And I'm also developing and testing my own version for some weeks now,
but the biggest problem and uncertainty is of course the lack of
reliable information (ie. theoretic and real data), and the fact
that RT people and RT (re-)sources are not much reliable;
far from being exact or correct, as we have seen again :-(
> (I wondered who wrote the paper, since I don't see a signature or
> attribution).
I'm curious too, but since it is part of the book then I think it can only be
one or both of the same authors themselves, ie. Taylor and Wheeler.
John C. Polasek
wrote:
>"jpolasek" <jpol...@cfl.rr.com> wrote
60x50x24x10^6 usec/day.
>Would this value be the same if we used the same GPS mechanism
>in the orbit of another planet instead of the Earth whereby using
>our same 24h clock system also for that planet?
No. You'd have to orbit twice a day in their particular day.
>
>And: what about using the Sideral Day seconds (= 86164 s) instead of 86400 ?
>Would this not give a more correct value?
>
>And: have you in your method considered the fact that the GPS satellites
>are orbiting twice a day? I.e. their rotation speed is about 3873 m/s.
>Do you then still get the 38 us/day ?
>I'm not sure yet, but I have the feeling that using the RT method
>the value 38 us/day can be true only if the satellite makes
>just one orbit per day, and therefore the correct value must
>be something different than the 38 us/day, perhaps 29 ?
You can see best in my gravity paper page 3 at
http://www.dualspace.net that has a table for 2x/day derived from a
very simple formula:
delta T = V^2/2c^2*K usec/day
where V is either the orbital speed or equatorial speed or, for
gravity, where V is the escape velocity Vesc = sqrt(2MG/r)(11,170m/s
for earth).
That gives 4 values of clock slowing, 7, 14, for sky and 60 and .1
usec/day on ground, subtract to get 38.56 us/day.
>> You say algorithm so you must be programming something.
>
>Yes, I'm putting these Time Dilation methods into a program
>for running simulations, and also for testing these methods/algorithms.
>So far I can use these methods for a body (satellite, airplane),
>moving along an equatorial orbit either eastwards or westwards,
>over a rotating body (Earth); ie. GPS and Hafele&Keating like simulations/experiments.
>FYI: I've one other working RT version of such time dilation methods,
>and was interessted also to add the Taylor/Wheeler method into my collection,
>but failed to do so so far due to the errors in the text :-(
>Now, I will add also your method into my collection.
>And I'm also developing and testing my own version for some weeks now,
>but the biggest problem and uncertainty is of course the lack of
>reliable information (ie. theoretic and real data), and the fact
>that RT people and RT (re-)sources are not much reliable;
>far from being exact or correct, as we have seen again :-(
>
>> (I wondered who wrote the paper, since I don't see a signature or
>> attribution).
>
>I'm curious too, but since it is part of the book then I think it can only be
>one or both of the same authors themselves, ie. Taylor and Wheeler.
John Polasek
qbit
> >> rotation about .1us/day).
John, can you please check it again, because
isn't the term (1-3M/r) giving a negative value?
As you know mathematically the sqrt of a negative value is undefined...
Eric Gisse
> "John C. Polasek" <jpola...@cfl.rr.com> wrote
Uh, no.
It is actually quite well defined.
qbit
qbit
And, on page 4 of your document you write:
|> The second expression (12b) is simply the Lorentz contraction
|> for both velocities applied to clock rate.
Is equation 12b meant by this? I think there is no such equation in the doc, is it?
qbit
Some more issues:
1) The meaning and definition of Rg is missing.
2) You write in the doc "K = 86,400x10^-6 seconds per day", but in eqn 6&7 you use "8.64x10^-10".
3) It is unclear to me how you calculated the value 5,470 m/s in line 2 of your table.
qbit
Solved: it is of course the v_esc of the orbit as sqrt(2GM/(R+h))
Ok, now I got it all working. Yippieee! :-)
Now I have to test it...
Too bad, I would also see a bugfixed working version of Mr. Taylor's solution...
John C. Polasek
wrote:
I thought I mentioned somewhere that as commonly used, that M (it
would be clearer to say Mr instead of M) is a handy mass-radius of the
earth being made up of 3 constants (so you can compute it once):
M = Mr = M_earth*G/c^2 = .00443m = .174 inch
Mr/r will be very small indeed. Mr is the Schwarzschild black hole
radius of the earth.
John Polasek
John C. Polasek
wrote:
Sorry, Eq. 12b number is wrong. I was just illustrating relativity's
version in Eq. 6 where both orbital and earth's gravity are taken
together as a quotient and Eq. 7 where the velocity component takes
orbital velocity and equatorial velocity as a quotient.
I just wanted to show how relativity does it. I pointed out how
relativity's R_e really involves escape velocity (just prior to
equations) without their recognizing it.
My equations are independent of Eq. 6 & 7.
John Polasek
John C. Polasek
wrote:
>> > "John C. Polasek" <jpol...@cfl.rr.com> wrote
>> > >
>> > > >> So in orbit use 1-sqrt(1-3M/r)xK = 1.5M/r*K = 21us/day. (Earth's
>> > > >> rotation about .1us/day).
>> >
>> > John, can you please check it again, because
>> > isn't the term (1-3M/r) giving a negative value?
>> > As you know mathematically the sqrt of a negative value is undefined...
>>
>> And, on page 4 of your document you write:
>>
>> |> The second expression (12b) is simply the Lorentz contraction
>> |> for both velocities applied to clock rate.
>>
>> Is equation 12b meant by this? I think there is no such equation in the doc, is it?
>
>Some more issues:
>
>1) The meaning and definition of Rg is missing.
I dont see Rg
>2) You write in the doc "K = 86,400x10^-6 seconds per day", but in eqn 6&7 you use "8.64x10^-10"
I dont see it. Has to be K = 86,400 sec/day = 86,400x10^6usec/day =
8.64x10^10usec/day.
>3) It is unclear to me how you calculated the value 5,470 m/s in line 2 of your table.
From the given orbital radius of 2.66x10^7m we can derive two
velocities:
orbital V = sqrt(M_w*G/r) = 3868 m/s for sat. (velocity part)
escapeV =sqrt(2M_w*G/r) = 5470 m/s for sat. (for gravity part)
John Polasek
qbit
> On Tue, 14 Aug 2007 11:54:39 +0200, "qbit" <qb...@quantumworldz.com> wrote:
>
> >> > "John C. Polasek" <jpol...@cfl.rr.com> wrote
> >> > >
> >> > > >> So in orbit use 1-sqrt(1-3M/r)xK = 1.5M/r*K = 21us/day. (Earth's
> >> > > >> rotation about .1us/day).
> >> >
> >> > John, can you please check it again, because
> >> > isn't the term (1-3M/r) giving a negative value?
> >> > As you know mathematically the sqrt of a negative value is undefined...
> >>
> >> And, on page 4 of your document you write:
> >>
> >> |> The second expression (12b) is simply the Lorentz contraction
> >> |> for both velocities applied to clock rate.
> >>
> >> Is equation 12b meant by this? I think there is no such equation in the doc, is it?
> >
> >Some more issues:
> >
> >1) The meaning and definition of Rg is missing.
>
> I dont see Rg
It is in Eqn 6 in file dualrelativityfeb27.pdf
> >2) You write in the doc "K = 86,400x10^-6 seconds per day", but in eqn 6&7 you use "8.64x10^-10"
>
> I dont see it. Has to be K = 86,400 sec/day = 86,400x10^6usec/day =
> 8.64x10^10usec/day.
It is used in Eqn 6&7 in the above pdf, and also in your
previous postings you had written this:
|> The us/day given are after multiplying your algebra each time
|> by K = 86.4x10^9 us/day.
So, there are some inconsistencies due to the use of different scaling units.
But as said, I have finally mastered it by using just the base unit 86400 seconds.
adem
has sent the following public info to the newsgroup:
http://groups.google.com/group/sci.physics.relativity/msg/6ca94e59f57c1906?dmode=source