GPS corrections
Hauke Reddmann
orbiting at radius r over a planet with mass m) of the
special and general relativity corrections to t/t0?
(Especially where they cancel)
--
Hauke Reddmann <:-EX8 fc3...@uni-hamburg.de
His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
Androcles
"Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
For a London based sundial to gain 5 hours, move it to Washington DC.
The distance between Nelson's column and the Washington monument,
two very nice sundials, is 3670.74 miles.
That's 734.148 miles for each hour, or 12.2358 miles for a minute.
Move 0.20393 miles or 1076.7504 feet to gain one second, and that's
12921.0048 inches.
To gain a microsecond, move 0.0129210048 inches, and to gain
38 microseconds move 0.4909981824 ~= 0.5 inches.
So the relativists are claiming they've proven a GPS clock has to be GR
corrected by 38,000 nanoseconds, or by 0.5 inches a day to be as accurate as
a sundial.
The worst of them says:
"Amateurs look at data, professionals look at errorbars.
That page completely ignores the many modern measurements, which VASTLY
smaller errorbars, that all show the constancy of the speed of light in
many different situations. " -- Tom Roberts
Being a professional, I'll put error bars on my 0.5 inches,
it is now 0.5 inches +/- 0.25 inches, I could be up to 1000 miles off
in my estimate of the distance between London and Washington.
That may not make airlines very happy in their estimates of fuel
consumption, even if they do use GPS...
Actually I think relativists snort too much nose candy. I may try it
some day, but pepper makes me sneeze so I remain reluctant.
Androcles.
"If one studies too zealously, one easily loses his pants." --Albert
Einstein
Dirk Van de moortel
"Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
> orbiting at radius r over a planet with mass m) of the
> special and general relativity corrections to t/t0?
> (Especially where they cancel)
Look at
http://www.eftaylor.com/pub/projecta.pdf
Equation [3] gives the general formula.
See also
http://groups.google.com/group/sci.physics.relativity/msg/0ba81f80d2c9ccad
and a little correction on
http://groups.google.com/group/sci.physics.relativity/msg/2581429d0b66eb72
In this case
d ts / d ta = Schw(va,ra,vs,rs)
Schw(va,ra,vs,rs) = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vs^2)
where
c = 1 lightspeed
G = 1 gravitational constant
M = mass of Earth = 4.4 10^(-3) m
ta = time of equatorial observer
va = speed of equatorial observer = 1.544 10^(-6)
ra = Earth radius = 6.371 10^6 m
ts = satellite time
vs = orbital speed of satellite = 1.287 10^(-5)
rs = radius of satellite orbit = 26.6 10^6 m
One approximation is the so-called Kinematic time rate
Kine(va,va) = sqrt(1-va^2) / sqrt(1-vs^2)
and it depends on relative motion only. It is a result of
special relativity - valid in absence of gravity.
The va and vb are seen in an Earth centered non-rotating
inertial frame.
The other approximation is the so called Gravitational
potential time rate
Grav(ra,ra) = sqrt(1-2 M/ra) / sqrt(1-2 M/rs)
and it depends on location only. It would be valid for
non-moving objects only. It is an idealized result from
general relativity.
Fos a circular orbit you have
rs = M / vs^2
so expressing that the time rate is 1, and treating vs and
rs as unknowns, this gives
1 = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vs^2)
= (1-2 M/ra-va^2) / (1-3 M/rs)
Solving this for rs gives
rs = 3 M ra / ( 2 M + ra va^2 )
which gives with the numeric values
rs = 8.533 10^6 m = 8533 km
(unless I made an error somewhere).
You can verify this by expressing that the approximations
Kine and Grav cancel.
Dirk Vdm
Randy Poe
Androcles wrote:
> So the relativists are claiming they've proven a GPS clock has to be GR
> corrected by 38,000 nanoseconds, or by 0.5 inches a day to be as accurate as
> a sundial.
Can you go out to your garden right now and get me
the time off your sundial, to the nearest 10 nanoseconds?
Heck, I'll relax it a few orders of magnitude. Get me the time
to the nearest 0.1 millisecond.
Also, I'd appreciate it if you could tell me how to make sure
whether two sundials are aligned to be reading the same
time within within 0.1 millisecond. It seems to me that requires
the ability to measure an angle to about a billionth of a
radian, but obviously you have some insight on how to get
nanosecond accuracy out of a sundial without angle measurement.
Could you share?
- Randy
Koobee Wublee
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:L8cqf.78001$lt2.5...@phobos.telenet-ops.be...
>
> "Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
> news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
>> Can somebody post the general formula (assuming a satellite
>> orbiting at radius r over a planet with mass m) of the
>> special and general relativity corrections to t/t0?
>> (Especially where they cancel)
>
> Look at
> http://www.eftaylor.com/pub/projecta.pdf
Taylor's derivation above violates the principle of special relativity. GR
demands only one equation to explain GPS. Neil Ashby realized this.
However, he had to perform two fancy metric transformations without any
concrete justification. The bottom line is that GR has trouble to explain
GPS. GR's matching result to GPS is taken from the term (- U) (see below)
without any proper consideration to do so.
sqrt(g_00) = sqrt(1 - 2 U) = 1 - U - U^2 / 2...
Where
** U = G M / c^2 / r
To liberally use the term (- U) to explain the result of GPS is like cutting
words out of a newspaper to compose a sentence. It does not prove that GR
explains the result of GPS.
The (- U) is what is backwards compatible with Newtonian physics. Any
post-Newtonian theories with Newtonian result as limit should also be able
to explain GPS as well. GR is not special.
> [... Rest of handwaving nonsense snipped]
Tom Roberts
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:L8cqf.78001$lt2.5...@phobos.telenet-ops.be...
>> http://www.eftaylor.com/pub/projecta.pdf
>
> Taylor's derivation above violates the principle of special relativity.
I have no idea why you think that is relevant. GR itself does not obey
that principle -- GR has a far more general principle: general covariance.
[The principle of special relativity says in effect that the
laws of physics do not depend upon one's choice of inertial
frame; GR has no such dependence on inertial frames, and
its principle is that the laws of physics do not depend
on _any_ choice of coordinates.]
> GR
> demands only one equation to explain GPS. Neil Ashby realized this.
> However, he had to perform two fancy metric transformations without any
> concrete justification.
I have no idea what you are trying to say. "metric transformations" are
freely available in GR, and one needs no "justification" for them other
than it seems convenient to apply them. This is because general
covariance applies, and the underlying laws of physics do not depend on
coordinates in any way.
> The bottom line is that GR has trouble to explain
> GPS.
Nonsense.
> GR's matching result to GPS is taken from the term (- U) (see below)
> without any proper consideration to do so.
Sure there is "proper consideration" -- the steps you discuss are part
of an approximation to the equations of GR. An approximation that is
more than accurate enough in this situation.
> sqrt(g_00) = sqrt(1 - 2 U) = 1 - U - U^2 / 2...
> Where
> ** U = G M / c^2 / r
> To liberally use the term (- U) to explain the result of GPS is like cutting
> words out of a newspaper to compose a sentence. It does not prove that GR
> explains the result of GPS.
> The (- U) is what is backwards compatible with Newtonian physics.
>>[... Rest of handwaving nonsense snipped]
Your writing does not even rise to the level of handwaving. What are you
actually trying to say?
Note you are free to apply the equations of GR to the GPS. As, for
instance, its designers have done.
> Any
> post-Newtonian theories with Newtonian result as limit should also be able
> to explain GPS as well. GR is not special.
Sure, in some ways "GR is not special", and there are many other
theories of gravitation that explain the GPS as well as GR does. So what?
But GR is indeed "special" in that it is the simplest possible theory
that is both coordinate free and accounts for all the relevant
experimental observations to date.
[Yes, it's scary to say GR is "the simplest theory"....
But indeed, it is its simplicity that permitted Einstein
(and Hilbert) to discover it.]
Tom Roberts tjro...@lucent.com
Koobee Wublee
"Tom Roberts" <tjro...@lucent.com> wrote in message
news:42qqf.35029$7h7....@newssvr21.news.prodigy.com...
> Koobee Wublee wrote:
>> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
>> wrote in message news:L8cqf.78001$lt2.5...@phobos.telenet-ops.be...
>>>Look at
>>> http://www.eftaylor.com/pub/projecta.pdf
>>
>> Taylor's derivation above violates the principle of special relativity.
>
> I have no idea why you think that is relevant. GR itself does not obey
> that principle -- GR has a far more general principle: general covariance.
Apparently, you have not read and understood Taylor's paper on GPS.
> [The principle of special relativity says in effect that the
> laws of physics do not depend upon one's choice of inertial
> frame; GR has no such dependence on inertial frames, and
> its principle is that the laws of physics do not depend
> on _any_ choice of coordinates.]
Well, you can do better than that. To disprove my statement above, you have
to show me what I have not already known and understood. You have to show
Neil Ashby's attempt to derive GPS time dilation is as nutty as I am.
>> GR demands only one equation to explain GPS. Neil Ashby realized this.
>> However, he had to perform two fancy metric transformations without any
>> concrete justification.
>
> I have no idea what you are trying to say. "metric transformations" are
> freely available in GR, and one needs no "justification" for them other
> than it seems convenient to apply them. This is because general covariance
> applies, and the underlying laws of physics do not depend on coordinates
> in any way.
Not only you have not read and understood Taylor's paper, you have not done
so with Dr. Ashby.
>> The bottom line is that GR has trouble to explain GPS.
>
> Nonsense.
Just because you can cut out blocks of what GR has to offer to justify the
result of your observation, it does not mean you have solved the problem
using the laws of GR. What laws you may ask?
>> GR's matching result to GPS is taken from the term (- U) (see below)
>> without any proper consideration to do so.
>
> Sure there is "proper consideration" -- the steps you discuss are part of
> an approximation to the equations of GR. An approximation that is more
> than accurate enough in this situation.
To be as picky as you are, I have to correct what you just said. The
approximation is just adequate for the accuracy involve not more then
adequate as you seem to think.
>> sqrt(g_00) = sqrt(1 - 2 U) = 1 - U - U^2 / 2...
>> Where
>> ** U = G M / c^2 / r
>> To liberally use the term (- U) to explain the result of GPS is like
>> cutting words out of a newspaper to compose a sentence. It does not
>> prove that GR explains the result of GPS.
>> The (- U) is what is backwards compatible with Newtonian physics.
> >>[... Rest of handwaving nonsense snipped]
>
> Your writing does not even rise to the level of handwaving. What are you
> actually trying to say?
If you do not remember Schwarzschild's metric, please go back and review
them.
> Note you are free to apply the equations of GR to the GPS. As, for
> instance, its designers have done.
Not I. You can fudge what GR has to offer and pretend you have solved the
problem of GPS. As I said before, you can freely utilize the(-U) terms of
the satellites and on the ground together as well as the speed terms of the
satellites and the ground. Taylor's derivation violates the principle of
Special Relativity and thus General Relativity. Neil Ashby's derivation is
better.
>> Any post-Newtonian theories with Newtonian result as limit should also be
>> able to explain GPS as well. GR is not special.
>
> Sure, in some ways "GR is not special", and there are many other theories
> of gravitation that explain the GPS as well as GR does. So what?
That is why GR is not any special than others. You are free to beat this
point to death. <shrug>
> But GR is indeed "special" in that it is the simplest possible theory that
> is both coordinate free and accounts for all the relevant experimental
> observations to date.
I guess "specialness" is in the eyes of the beholder not just "beauty".
> [Yes, it's scary to say GR is "the simplest theory"....
> But indeed, it is its simplicity that permitted Einstein
> (and Hilbert) to discover it.]
Yes, that is the danger of falling in love with a theory. After that, you
are drowned in the passion of love. There is no more logical consideration
even after evidences do point towards the contrary. It is so true that
"love is color blind".
mlut...@wanadoo.fr
Dirk Van de moortel wrote :
You indeed made an error somewhere.
(To verify, calculate the orbital radius corresponding to 38
microseconds/day.
you should find about 26549 km).
Marcel Luttgens
Hauke Reddmann
radius would be equal to the geosynchronous radius.
But the resulting formula has much less appeal
than E=mc² :-)
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
news:doe6r3$5g0$1...@rzsun03.rrz.uni-hamburg.de...
> THX. My main goal was to design a planet where this
> radius would be equal to the geosynchronous radius.
> But the resulting formula has much less appeal
> than E=mc² :-)
Place all of Earth's mass on a ring at geosynchronous radius.
David A. Smith
Androcles
> THX. My main goal was to design a planet where this
> radius would be equal to the geosynchronous radius.
> But the resulting formula has much less appeal
> than E=mc² :-)
You are already on one.
At least I am, I'm geostationary and so is my desk and computer.
Androcles.
Dirk Van de moortel
"Koobee Wublee"
aka Australopithecus Afarensis
aka Scholarly Fungi
aka Time Traveler
aka Lordly Amoeba
aka Ibn Battuta
aka Marco Polo
<kub...@cox.net> wrote in message news:jQkqf.5269$LB5.1516@fed1read04...
>
> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
> in message news:L8cqf.78001$lt2.5...@phobos.telenet-ops.be...
> >
> > "Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
> > news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
> >> Can somebody post the general formula (assuming a satellite
> >> orbiting at radius r over a planet with mass m) of the
> >> special and general relativity corrections to t/t0?
> >> (Especially where they cancel)
> >
> > Look at
> > http://www.eftaylor.com/pub/projecta.pdf
>
> Taylor's derivation above violates the principle of special relativity.
You are a first class idiot with a second class agenda.
Original, but removed from archives:
http://groups.google.co.uk/groups?selm=bdq09.28353$Fq6.2...@news2.west.cox.net
But we still have the reply:
http://groups.google.co.uk/groups?&threadm=V1r09.661180$352.138570@sccrnsc02
| "Scholarly Fungi" <scholar...@yahoo.com> wrote in message
| news:bdq09.28353$Fq6.2...@news2.west.cox.net...
| > It is also unfortunate that most of the folks blindly embracing this
| > holohaux come from the white supremacists. I don't see what this would gain
| > for them other than trying to antagonize the Jews. However, this is
| > history. When I was in my early high school years, I independently came up
| > with what Butz was saying without knowing his existence. Hey, I am very
| > proud of my humble analytical skills.
There is more:
http://groups.google.com/group/sci.physics.relativity/msg/7dd80442f8d16bc0
Dirk Vdm
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135243873.6...@g44g2000cwa.googlegroups.com...
So that particular orbit is not an orbit where the two
approximating effects cancel.
Try to pay attention.
Dirk Vdm
Dirk Van de moortel
"Koobee Wublee" <kub...@cox.net> wrote in message news:b4rqf.9999$LB5.5377@fed1read04...
>
> "Tom Roberts" <tjro...@lucent.com> wrote in message
> news:42qqf.35029$7h7....@newssvr21.news.prodigy.com...
> > Koobee Wublee wrote:
> >> "Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com>
> >> wrote in message news:L8cqf.78001$lt2.5...@phobos.telenet-ops.be...
> >>>Look at
> >>> http://www.eftaylor.com/pub/projecta.pdf
> >>
> >> Taylor's derivation above violates the principle of special relativity.
> >
> > I have no idea why you think that is relevant. GR itself does not obey
> > that principle -- GR has a far more general principle: general covariance.
>
> Apparently, you have not read and understood Taylor's paper on GPS.
>
> > [The principle of special relativity says in effect that the
> > laws of physics do not depend upon one's choice of inertial
> > frame; GR has no such dependence on inertial frames, and
> > its principle is that the laws of physics do not depend
> > on _any_ choice of coordinates.]
>
> Well, you can do better than that. To disprove my statement above, you have
> to show me what I have not already known and understood. You have to show
> Neil Ashby's attempt to derive GPS time dilation is as nutty as I am.
You aren't merely nutty.
You are disgusting.
Original, but removed from archives:
http://groups.google.co.uk/groups?&threadm=fytJa.78487%24%2542.6441%40fed1read06
But we still have the reply:
http://groups.google.co.uk/groups?&threadm=XvKJa.100908$hd6.25327@fed1read05
| "Australopithecus Afarensis" <lu...@olduvaigorge.net> wrote in message
| news:fytJa.78487$%42.6441@fed1read06...
| > Thanks for posting all that and your own comments at the end. There are so
| > many lies after lies conjured up against the Nazis. I guess I'd better read
| > "Mein Kampf" to get it from the horse's mouth. It will be on my
| > things-to-do list for the near future.
Joe Fischer
>You aren't merely nutty.
>You are disgusting.
It looks like she is both crazy as a loon,
and dumb as a rock, has to be female and blonde. :-)
>There is more:
> http://groups.google.com/group/sci.physics.relativity/msg/7dd80442f8d16bc0
>Dirk Vdm
I hope you haven't started an east-west thread,
few people will be able to understand a number of
things about the world economy in the past and all
the problems associated with multiple borders and
immigration.
It probably is difficult for a young person today
to appreciate how valuable even hair was at that time,
it could be used in cushions and mattresses.
Chances are all these amateur experts should
be spending their time thinking about how people will
heat the homes above latitude 38 in Asia and North
America ten years from now.
Even with a negative birth rate, the US is in
big trouble, Einstein could help a lot if only utility
companies were free to build reactors.
Franceis lucky to have so many reactors,
and a GB company bought my local gas and
electric utility, and just raised natural gas prices
43 percent in one jump.
I know two people who call the computer
a "devil box", and I believe it, the devil doesn't
need to make people bad, just making them
stupid will have the same effect.
Isn't it awful they had to build so many
GPS satellites before they found the correct
frequency. :-)
Joe Fischer
mlut...@wanadoo.fr
You found 8533 km instead of 9567 km for that particular orbit.
For a mathematician, this is a rather big error!
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135289249....@g43g2000cwa.googlegroups.com...
In stead of the value of
| va = speed of equatorial observer = 1.544 10^(-6)
I used the value of
| vs = orbital speed of satellite = 1.287 10^(-5)
The former gives 9540 km.
You just deserved a credit for having payed attention.
A miracle :-)
Dirk Vdm
Androcles
<mlut...@wanadoo.fr> wrote in message
news:1135289249....@g43g2000cwa.googlegroups.com...
Nah... Who call the cunt a mathematician?
"That's not a relation. That's an algebraic equation" - Dork Van de merde.
That's not a dog. That's a peeing puppy.
Androcles.
mlut...@wanadoo.fr
I am wondering how you could get 8533 km by using vs = 1.287 10^(-5),
a velocity which is more than 8 times the speed of the equatorial
observer.
You will perhaps explain that you used 1.287 10^(-6). Then you should
also show the equation you used to get 8533 km. Otherwise, readers
will think that you don't understand the GR formulae.
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135377967.6...@g14g2000cwa.googlegroups.com...
That's what 12 hour period GPS satellites tend to have as
orbital speed.
>
> You will perhaps explain that you used 1.287 10^(-6).
No.
1.287 10^(-6) gives 9545 km
> Then you should
> also show the equation you used to get 8533 km. Otherwise, readers
> will think that you don't understand the GR formulae.
Can't you read and calculate?
> > > > > > rs = 3 M ra / ( 2 M + ra va^2 )
In stead of va = 1.544 10^(-6), I erroneously took
va = 1.287 10^(-5).
1.287 10^(-5) gives 8533 km.
1.544 10^(-6) gives 9540 km.
You can verify that by performing the activity called
"plugging in the values and making the calculations".
>
> Marcel Luttgens
>
> >
> > You just deserved a credit for having payed attention.
> > A miracle :-)
I am beginning to understand your problem: you don't
understand anything you read and you can't properly
express yourself.
I was too optimistic. You lost that credit.
Dirk Vdm
mlut...@wanadoo.fr
You could as well have taken 0 for the speed of the
equatorial observer. Then you would have obtained 9557 km.
But what has the orbital speed of 12 hour period GPS satellites
here to do? And the orbital radius of such satellites is not
8533 km, but 26650 km. You look rather confused!
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135424059....@g49g2000cwa.googlegroups.com...
Try to pay attention.
> And the orbital radius of such satellites is not
> 8533 km, but 26650 km. You look rather confused!
To an imbecile like you everyone looks confused.
For 2006 I wish you a permanent Memantine prescription.
Dirk Vdm
mlut...@wanadoo.fr
You are the imbecile, able to borrow a formula from some book,
but unable to use it without making mistakes.
Hopefully, your Alzheimer will not develop too much in 2006.
Marcel Luttgens
>
> Dirk Vdm
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135439277.8...@g49g2000cwa.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > <mlut...@wanadoo.fr> wrote in message news:1135424059....@g49g2000cwa.googlegroups.com...
> > >
> > > Dirk Van de moortel wrote:
[snip]
> > Try to pay attention.
> >
> > > And the orbital radius of such satellites is not
> > > 8533 km, but 26650 km. You look rather confused!
> >
> > To an imbecile like you everyone looks confused.
> > For 2006 I wish you a permanent Memantine prescription.
>
> You are the imbecile, able to borrow a formula from some book,
Never seen that formula before.
Derived it and didn't even know whether it was okay.
> but unable to use it without making mistakes.
Took the wrong numerical value.
> Hopefully, your Alzheimer will not develop too much in 2006.
On second thought, make that a one time prescription
of Euthasol.
Dirk Vdm
mlut...@wanadoo.fr
Anyhow, best wishes for 2006!
Marcel Luttgens
mlut...@wanadoo.fr
Orbital radius = 3/2 Earth radius.
Marcel Luttgens
Hauke Reddmann
That's cheating - a satellite at r=r(earth) would not have
v=v(earth) :-)
P.S. Merry Xmas to all flamers, trolls, cranks, netcops and
all other good denizens of the USENET - maybe I *should* have
went to the library instead :-)
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135525557....@g43g2000cwa.googlegroups.com...
> You should have looked for the GR formula before, because it reduces to
> Orbital radius = 3/2 Earth radius.
Which is merely an approximation.
Two facts wrong and faulty logic.
You made 3 mistakes in this single sentence.
Congratulations, this must be a record.
No wonder you can't communicate with anyone.
Dirk Vdm
Androcles
> Androcles <M...@Myplace.org> wrote:
>
>> "Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
>> news:doe6r3$5g0$1...@rzsun03.rrz.uni-hamburg.de...
>>> THX. My main goal was to design a planet where this
>>> radius would be equal to the geosynchronous radius.
>>> But the resulting formula has much less appeal
>>> than E=mc² :-)
>
>> You are already on one.
>> At least I am, I'm geostationary and so is my desk and computer.
>> Androcles.
>
> That's cheating - a satellite at r=r(earth) would not have
> v=v(earth) :-)
Not at all. Raise the Earth's rotation rate until you are weightless.
The atmosphere will spin off first, so design a bubble to keep it in.
Watch our for high tides from the moon.
You are the one cheating if you think you can increase the radius
without decreasing angular velocity.
> P.S. Merry Xmas to all flamers, trolls, cranks, netcops and
> all other good denizens of the USENET - maybe I *should* have
> went to the library instead :-)
As you wish, troll.
Androcles.
mlut...@wanadoo.fr
Dirk Van de moortel wrote:
Vdm, you are a good SR parrot, but you know zilch
about the GR formula.
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135591828.0...@f14g2000cwb.googlegroups.com...
Exercise.
Which errors did you make and where is the faulty logic?
Dirk Vdm
mlut...@wanadoo.fr
Dirk Van de moortel wrote:
> <mlut...@wanadoo.fr> wrote in message news:1135591828.0...@f14g2000cwb.googlegroups.com...
> >
> > Dirk Van de moortel wrote:
> > > <mlut...@wanadoo.fr> wrote in message news:1135525557....@g43g2000cwa.googlegroups.com...
> > > > You should have looked for the GR formula before, because it reduces to
> > > > Orbital radius = 3/2 Earth radius.
> > >
> > > Which is merely an approximation.
> > >
> > > Two facts wrong and faulty logic.
> > > You made 3 mistakes in this single sentence.
> > > Congratulations, this must be a record.
> > > No wonder you can't communicate with anyone.
> > >
> > > Dirk Vdm
> >
> > Vdm, you are a good SR parrot, but you know zilch
> > about the GR formula.
>
> Exercise.
> Which errors did you make and where is the faulty logic?
>
> Dirk Vdm
I didn't make errors. The formula I used is valid for an observer
situated at the North (or South, if you prefer) Pole.
The effect of the Earth rotation is compensed by the geometry
of the Earth (for instance, an oblateness of about 10 km
at the equator). As the Earth rotation is practically irrelevant,
the radius of the "chronosynchronous" orbit is 3/2 Earth radius.
You simply showed your ignorance by trying to incorporate
the Earth rotation into your GR formulas.
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135606039....@g14g2000cwa.googlegroups.com...
And you ignored it and just came up with a textbook
approximation.
In my original setting, ra is the equatorial radius so
rs = 3 M ra / ( 2 M + ra va^2 )
For a polar observer this obviously becomes
rs = 3/2 rp,
which is not 3/2 ra. But I was not talking about a polar
observer. I was talking about an equatorial one.
You had two facts wrong and your logic sucks.
You haven't paid attention.
Dirk Vdm
mlut...@wanadoo.fr
You are unable to realize that the orbital radius of
3/2 Earth radius is the same for observers situated
anywhere on Earth. You simply ignored the geometry
of the Earth. Ask the opinion of Tom Roberts
if you still disagree.
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135609566.2...@o13g2000cwo.googlegroups.com...
There is no such thing as "Earth radius".
There is polar radius and there is equatorial radius.
The time rate of a polar observer is the same as the
time rate of an equatorial one.
I was explicitly talking about an equatorial observer,
together with the values of his radius and his orbital
velocity, and you miserably failed to properly express
your rant.
> You simply ignored the geometry
> of the Earth.
That is exactly what you were doing. When pointed
to your error, you tried to bounce it back, just like
the poor loser we already have established you to be
since a long time.
> Ask the opinion of Tom Roberts
> if you still disagree.
Marcel, you and I don't agree on the fact that you have
severe comprehension and expression problems.
Don't be afraid there, I don't need anyone's opinion
on your problems. We all have learned to live with
them by now. Bottom line, if you are stupid and can't
formulate your chaotic "thoughs", you better shut up.
Dirk Vdm
mlut...@wanadoo.fr
> <mluttg...@wanadoo.fr> wrote in message news:1135609566.2...@o13g2000cwo.googlegroups.com...
> > Dirk Van de moortel wrote:
> > > <mluttg...@wanadoo.fr> wrote in message news:1135606039....@g14g2000cwa.googlegroups.com...
> > > > Dirk Van de moortel wrote:
> > > > > <mluttg...@wanadoo.fr> wrote in message news:1135591828.0...@f14g2000cwb.googlegroups.com...
> > > > > > Dirk Van de moortel wrote:
> > > > > > > <mluttg...@wanadoo.fr> wrote in message news:1135525557....@g43g2000cwa.googlegroups.com...
> > > > > > > Dirk Vdm
> > > > > Dirk Vdm
> > > Dirk Vdm
And an infinity of other radii.
> The time rate of a polar observer is the same as the
> time rate of an equatorial one.
You ignored this before, otherwise you would not have taken
the trouble to put the equatorial radius and velocity into your
formulas. Using the polar radius was much simpler. You are
looking as a liar and a bad loser.
> I was explicitly talking about an equatorial observer,
> together with the values of his radius and his orbital
> velocity, and you miserably failed to properly express
> your rant.
You are once more trying to hide your previous ignorance
of the fact that "the oblateness of the earth _exactly_ cancels
this effect, and two clocks at the altitude of the geoid anywhere
on earth will tick at the same rate (as observed by either clock,
or in the ECI frame)" (cf. Tom Roberts).
Next time somebody ask you the value of the "special"
orbit, I hope that you will present him a simpler, elegant and correct
GR formula instead of exhibiting again your previous needlessly
complicated formulas, and that your solution will not be false.
> > You simply ignored the geometry
> > of the Earth.
> That is exactly what you were doing. When pointed
> to your error, you tried to bounce it back, just like
> the poor loser we already have established you to be
> since a long time.
Liar, you couldn't point to an error that I didn't make.
The rest of your message is like you, dishonest and
vulgar.
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135623085.2...@z14g2000cwz.googlegroups.com...
Bravo :-)
And doing so, you just proved my point again.
You talk about "3/2 Earth radius", I correct you, and
you try to bounce it back by "correcting" me.
If you are stupid and can't properly express yourself,
you better shut up, Marcel...
>
> > The time rate of a polar observer is the same as the
> > time rate of an equatorial one.
>
> You ignored this before,
I did not ignore it.
I was specifically referring to the equatorial radius.
> otherwise you would not have taken
> the trouble to put the equatorial radius and velocity into your
> formulas.
You didn't pay attention, Marcel.
If you can't pay attention, it's better to shut up.
> Using the polar radius was much simpler. You are
> looking as a liar and a bad loser.
To someone like you, I'm sure that almost everyone looks
like a liar and a bad loser.
>
> > I was explicitly talking about an equatorial observer,
> > together with the values of his radius and his orbital
> > velocity, and you miserably failed to properly express
> > your rant.
>
> You are once more trying to hide your previous ignorance
> of the fact that "the oblateness of the earth _exactly_ cancels
> this effect, and two clocks at the altitude of the geoid anywhere
> on earth will tick at the same rate (as observed by either clock,
> or in the ECI frame)" (cf. Tom Roberts).
>
> Next time somebody ask you the value of the "special"
> orbit,
You have a reading comprehension problem and you
don't pay attention.
You are projecting your obsessions on others again.
> I hope that you will present him a simpler, elegant and correct
> GR formula instead of exhibiting again your previous needlessly
> complicated formulas, and that your solution will not be false.
You mean endlessly complicated like this
rs = 3 M ra / ( 2 M + ra va^2 )
which is valid for every local observer with its proper
earth 'radius' and rotational speed?
Take the pole and plug ra = rp and va = 0 and there
you have your 3/2 rp. Big deal.
But we all know that Marcel can't handle "needlessly
complicated formulas", don't we?
>
> > > You simply ignored the geometry
> > > of the Earth.
>
> > That is exactly what you were doing. When pointed
> > to your error, you tried to bounce it back, just like
> > the poor loser we already have established you to be
> > since a long time.
>
> Liar, you couldn't point to an error that I didn't make.
The idea is not that I point to an error. The exercise
I gave *you* was to find the two errors that *you* made.
I already gave one of them away, so now it's time for
the second part of the exercise. Hint - it was an error
of comprehension/expression. Go ahead, try and find it.
As soon as you have found that error, you can start
looking for the logical flaw in your message. No hints on
that one.
> The rest of your message is like you, dishonest and
> vulgar.
To someone like you, I'm sure that almost everyone looks
dishonest and vulgar :-)
Here's some news for you, Marcel: I wasn't talking to you.
You spotted a simple calculation error (my taking the wrong
value from a list of values) and you got a credit for that.
And then you started fucking up, just like you always do.
Most entertaining, Marcel - Keep'em coming.
Dirk Vdm
mlut...@wanadoo.fr
You wrote:
"You mean endlessly complicated like this
rs = 3 M ra / ( 2 M + ra va^2 )
which is valid for every local observer with its proper
earth 'radius' and rotational speed?
Take the pole and plug ra = rp and va = 0 and there
you have your 3/2 rp. Big deal."
You still don' t get the point, which proves that you are very stupid.
Indeed, only rs = 3/2 rp gives the correct result for *all* local
observers. Your formula
rs = 3 M ra / ( 2 M + ra va^2 )
necessarily leads to false results for all observers other than the
pole observer
because of the oblateness of the Earth. You should put that silly
formula into
your "immortal fumbles" .
Bye,
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1135634506.6...@g43g2000cwa.googlegroups.com...
[snip]
> You wrote:
>
> "You mean endlessly complicated like this
> rs = 3 M ra / ( 2 M + ra va^2 )
> which is valid for every local observer with its proper
> earth 'radius' and rotational speed?
> Take the pole and plug ra = rp and va = 0 and there
> you have your 3/2 rp. Big deal."
>
> You still don' t get the point, which proves that you are very stupid.
> Indeed, only rs = 3/2 rp gives the correct result for *all* local
> observers.
Fantastic!
Marcel Luttgens has just discovered that 28 is a constant
for all local observers :-))
>Your formula
> rs = 3 M ra / ( 2 M + ra va^2 )
> necessarily leads to false results for all observers other than the
> pole observer
> because of the oblateness of the Earth. You should put that silly
> formula into
> your "immortal fumbles" .
You haven't paid attention.
Before you jump into something, it can be useful to
find out what people are talking about.
But you probably can't pay attention.
Have you checked with your mother?
Did she drink when she was carrying you?
Dirk Vdm
Paul B. Andersen
I think you have got it inverted.
It should be:
d ts / d ta = sqrt(1-2 M/rs-vs^2)/sqrt(1-2 M/ra-va^2)
I simple first order approximation yields:
(assuming (2 M/r+v^2) << 1 )
d ts / d ta = 1 + (M/ra - M/rs) - (vs^2 - va^2)/2
The gravitational term gives a positive contribution
while the speed term gives a negative contribution,
as they should.
Or have I misinterpreted something?
> One approximation is the so-called Kinematic time rate
> Kine(va,va) = sqrt(1-va^2) / sqrt(1-vs^2)
> and it depends on relative motion only. It is a result of
> special relativity - valid in absence of gravity.
Which is > 1. It should be < 1.
> The va and vb are seen in an Earth centered non-rotating
> inertial frame.
>
> The other approximation is the so called Gravitational
> potential time rate
> Grav(ra,ra) = sqrt(1-2 M/ra) / sqrt(1-2 M/rs)
> and it depends on location only. It would be valid for
> non-moving objects only. It is an idealized result from
> general relativity.
Which is < 1. It should be > 1.
This doesn't affect the calculation below, though.
> Fos a circular orbit you have
> rs = M / vs^2
> so expressing that the time rate is 1, and treating vs and
> rs as unknowns, this gives
> 1 = sqrt(1-2 M/ra-va^2) / sqrt(1-2 M/rs-vs^2)
> = (1-2 M/ra-va^2) / (1-3 M/rs)
> Solving this for rs gives
> rs = 3 M ra / ( 2 M + ra va^2 )
> which gives with the numeric values
> rs = 8.533 10^6 m = 8533 km
> (unless I made an error somewhere).
>
> You can verify this by expressing that the approximations
> Kine and Grav cancel.
>
> Dirk Vdm
>
>
Paul
Dirk Van de moortel
"Paul B. Andersen" <paul.b....@hiadeletethis.no> wrote in message news:dp1ln6$1hn$1...@dolly.uninett.no...
Yes indeed, while copying from that old posting,
I made a mistake replacing the indices a/b with a/s
without checking again.
>
> I simple first order approximation yields:
> (assuming (2 M/r+v^2) << 1 )
> d ts / d ta = 1 + (M/ra - M/rs) - (vs^2 - va^2)/2
>
> The gravitational term gives a positive contribution
> while the speed term gives a negative contribution,
> as they should.
>
> Or have I misinterpreted something?
No, indeed, you are quite right.
>
> > One approximation is the so-called Kinematic time rate
> > Kine(va,va) = sqrt(1-va^2) / sqrt(1-vs^2)
> > and it depends on relative motion only. It is a result of
> > special relativity - valid in absence of gravity.
>
> Which is > 1. It should be < 1.
>
> > The va and vb are seen in an Earth centered non-rotating
> > inertial frame.
> >
> > The other approximation is the so called Gravitational
> > potential time rate
> > Grav(ra,ra) = sqrt(1-2 M/ra) / sqrt(1-2 M/rs)
> > and it depends on location only. It would be valid for
> > non-moving objects only. It is an idealized result from
> > general relativity.
>
> Which is < 1. It should be > 1.
>
> This doesn't affect the calculation below, though.
Indeed, it's a matter of having the indices in the correct
order. Rereading everything, I notice that I fucked up
just about everywhere. I also swapped the numerical
values of rs and ra. Yuck... haste brings along mistakes.
Next time, I'll be much more careful. Thanks for having
checked and for the correction. Amazing that no one
else noticed.
Have a nice end of year,
Dirk Vdm
Paul B. Andersen
> Dirk Van de moortel wrote:
>
>
>><mluttg...@wanadoo.fr> wrote in message news:1135609566.2...@o13g2000cwo.googlegroups.com...
>
>>
>>>3/2 Earth radius is the same for observers situated
>>>anywhere on Earth.
>
>
>>There is no such thing as "Earth radius".
>>There is polar radius and there is equatorial radius.
>
>
> And an infinity of other radii.
>
>
>>The time rate of a polar observer is the same as the
>>time rate of an equatorial one.
>
>
> You ignored this before, otherwise you would not have taken
> the trouble to put the equatorial radius and velocity into your
> formulas. Using the polar radius was much simpler. You are
> looking as a liar and a bad loser.
You cannot use the polar radius.
sqrt(1 - 2 M/ra-va^2) > sqrt(1 - 2 M/rp)
they should be equal if you were right.
The reason is that M/ra where ra is the polar radius
isn't a valid expression for the gravitational potential
at the pole, due to the oblateness of the Earth.
You have to use a smaller M than the mass of the Earth.
(The mass within a sphere with radius rp ?).
An exact calculation of the rate (relative to Schwartzschild time)
of a clock on the surface of the Earth isn't very easy.
I don't think the expression sqrt(1-2 M/ra-va^2) is
exactly right either, but it sure is better than sqrt(1 - M/rp).
Paul
Dirk Van de moortel
"Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
> Can somebody post the general formula (assuming a satellite
> orbiting at radius r over a planet with mass m) of the
> special and general relativity corrections to t/t0?
> (Especially where they cancel)
> Hauke Reddmann <:-EX8 fc3...@uni-hamburg.de
> His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
[ Second attempt,
correcting the sloppily placed indices,
adding the first order approximations,
using the correct values in the final calculation, and
adding a little comment (for the benefit of Marcel Luttgens)
Thanks, Paul Andersen, for pointing to the careless typos ]
Look at
http://www.eftaylor.com/pub/projecta.pdf
Equation [3] gives the general formula.
See also
http://groups.google.com/group/sci.physics.relativity/msg/0ba81f80d2c9ccad
and a little correction on
http://groups.google.com/group/sci.physics.relativity/msg/2581429d0b66eb72
In this case
d ts / d ta = Schw(vs,rs,va,ra)
Schw(vs,rs,va,ra) = sqrt(1-2 M/rs-vs^2) / sqrt(1-2 M/ra-va^2)
where
c = 1 lightspeed
G = 1 gravitational constant
M = mass of Earth = 4.4 10^(-3) m
ta = time of equatorial observer
va = speed of equatorial observer = 1.544 10^(-6)
ra = Earth radius = 6.371 10^6 m
ts = satellite time
vs = orbital speed of satellite = 1.287 10^(-5)
rs = radius of satellite orbit = 26.6 10^6 m
The vs and va are seen in an Earth centered non-rotating
inertial frame.
This rate can be approximated to first order as
Schw(vs,rs,va,ra) ~= 1 - (M/rs + 1/2 vs^2) + (M/ra + 1/2 va^2)
One approximation is the so-called Kinematic time rate
Kine(vs,va) = sqrt(1-vs^2) / sqrt(1-va^2)
and it depends on relative motion only. It is a result of
special relativity - valid in absence of gravity (M=0) or
at sufficiently large distances.
This can be approximated to first order to
Kine(vs,va) ~= 1 - 1/2 vs^2 + 1/2 va^2
The other approximation is the so called Gravitational
potential time rate
Grav(rs,ra) = sqrt(1-2 M/rs) / sqrt(1-2 M/ra)
and it depends on location only. It would be valid for
non-moving objects only (va = vs = 0). It is an idealized
result from general relativity.
This can be approximated to first order to
Grav(vs,va) ~= 1 - M/rs + M/ra
For a circular orbit you have
rs = M / vs^2
so expressing that the time rate is 1, and treating vs and
rs as unknowns, this gives
1 = sqrt(1-2 M/rs-vs^2) / sqrt(1-2 M/ra-va^2)
= (1-3 M/rs) / (1-2 M/ra-va^2)
Solving this for rs gives
rs = 3 M ra / ( 2 M + ra va^2 )
which gives with the numeric values above
rs = 9.540 10^6 m = 9540 km
You can verify this by expressing that the approximations
Kine and Grav cancel.
Additional note.
The equation
rs = 3 M ra / ( 2 M + ra va^2 )
is valid for all observers on the surface of the Earth, provided
ra represents the 'local Earth radius', and va the local rotational
speed w.r.t. the Earth centered non-rotating inertial frame.
This is directly related to the fact that all Earth bound observers
have the same time rate, and due to the non-sphericality of the Earth.
I used the values for ra and va on the equator, but I could have
used the values of any observer. The simplest case presents itself
for a polar observer, for whom the local Earth radius is the polar
radius ra = rp, and va = vp = 0. In that case the equation reduces to
rs = 3/2 rp
It should be noted that even this is still only approximately true.
See section 5 of
http://www.eftaylor.com/pub/projecta.pdf
Dirk Vdm
Henri Wilson
No one ever reads your posts.
>
>Have a nice end of year,
>Dirk Vdm
>
HW.
www.users.bigpond.com/hewn/index.htm
Dirk Van de moortel
"Henri Wilson" <HW@..> wrote in message news:pci9r15la3of0r9ra...@4ax.com...
Paul does.
Marcel does.
Ralph Rabbidge does.
Dirk Vdm
mlut...@wanadoo.fr
Dirk Van de moortel wrote:
Remember that I called your GR formula silly. When somebody asks
for help, the least that can be done is not to be sloppy.
Of course, all GR formulae are only approximate. But the following
BASIC program gives good practical results:
'Source: Gravitation and Cosmology, by Steve Weinberg,1972, pp 83 sq.
DEFDBL A-Z
Me = 5.983E+27 ' Earth mass in grams
Re = 6.371E+08 ' Earth polar radius in cm
day = 8.64E+10 'microseconds
'nu(s) = frequency from the satellite
'nu(e) = frequency received on Earth
'Satellite velocity: vs^2 = G*Me/(Re+H), where
'Re = Earth polar radius, Me = Earth mass, H = altitude of the
satellite
'Frequency ratio nu(s)/nu(e) = 1 + delta(nu)/nu, where
'delta(nu)/nu = -(3/2)*G*Me/(2*Re+H) + G*Me/Re
' = -3.47E-10 * (3*Re/(Re+H) - 2)
'NB: The effect of the rotation of the Earth is cancelled by the
' oblateness of the Earth, hence the choice of the polar radius Re
'There is a net red shift for H < Re/2 and a net blue shift for H >
Re/2.
Orb = Re + H 'Orb = orbital radius of the satellite in cm
INPUT "Orbital radius of the satellite in Km = ", Orb
Orb = Orb * 100000!
Shift = -3.47E-10 * (3 * Re / Orb - 2)
PRINT "Global SR+GR effects in microseconds/day = "; day * Shift: END
'SR and GR effects cancel each other:
Shift = 0
Orb = 1.041E-09 * Re / (6.94E-10 - Shift)
Orb = Orb / 100000! 'Orbital radius in km
PRINT Orb
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1136031160.0...@o13g2000cwo.googlegroups.com...
Don't be afraid, I don't care what a village idiot finds silly,
specially if that particular idiot has proven not to be able
to handle the most simple equations on the planet:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LutLog.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ApplyDerivation.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PlainlyWrong.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Indulging.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LuttgensComment.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrackpotAccept.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TrueCrackpots.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/MuchSimpler.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NegativeCrap.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/MoronLikeMe.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LuttRel.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/StupidLie.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SimplyWrong.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SpeedV.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/OnlyGalilean.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow2.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ArmsGrow.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/IfOnlyIf.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRSymbols.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CorrectRelations.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Forget.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRLuttgens.html
[snip 1972 BASIC program implementing my 'silly' equation]
Dirk Vdm
mlut...@wanadoo.fr
Your equation? Liar! You didn't even know that your equation was false,
as Paul B. Andersen demonstrated. You should put it into
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GR_Vdm.html !
Stick with SR, you are able to parrot it! GR is too difficult for your
IQ !
Marcel Luttgens
Dirk Van de moortel
<mlut...@wanadoo.fr> wrote in message news:1136074455.9...@o13g2000cwo.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > <mlut...@wanadoo.fr> wrote in message news:1136031160.0...@o13g2000cwo.googlegroups.com...
[snip]
Careful, you get too much involved emotionally - bad for your
heart condition.
> You didn't even know that your equation was false,
>
> as Paul B. Andersen demonstrated.
Since you don't understand how variables and equations
work to begin with, I fully understand why you say this,
and I won't try to explain anymore, because you wouldn't
even know or even want to know what the explanation
would be about.
> You should put it into
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/GR_Vdm.html !
> Stick with SR, you are able to parrot it! GR is too difficult for your
> IQ !
You don't understand how variables are used in equations,
so how would you know?
Dirk Vdm
mlut...@wanadoo.fr
mlut...@wanadoo.fr
> >
> > "Hauke Reddmann" <fc3...@uni-hamburg.de> wrote in message
> > news:dobac1$ffm$4...@rzsun03.rrz.uni-hamburg.de...
> >> Can somebody post the general formula (assuming a satellite
> >> orbiting at radius r over a planet with mass m) of the
> >> special and general relativity corrections to t/t0?
> >> (Especially where they cancel)
> >
> > http://www.eftaylor.com/pub/projecta.pdf
>
> Taylor's derivation above violates the principle of special relativity. GR
> demands only one equation to explain GPS. Neil Ashby realized this.
> However, he had to perform two fancy metric transformations without any
> concrete justification. The bottom line is that GR has trouble to explain
> GPS. GR's matching result to GPS is taken from the term (- U) (see below)
> without any proper consideration to do so.
>
> sqrt(g_00) = sqrt(1 - 2 U) = 1 - U - U^2 / 2...
>
> Where
>
> ** U = G M / c^2 / r
>
> To liberally use the term (- U) to explain the result of GPS is like cutting
> words out of a newspaper to compose a sentence. It does not prove that GR
> explains the result of GPS.
>
> The (- U) is what is backwards compatible with Newtonian physics. Any
> post-Newtonian theories with Newtonian result as limit should also be able
> to explain GPS as well. GR is not special.
>
> > [... Rest of handwaving nonsense snipped]
" Any post-Newtonian theories with Newtonian result as limit should
also be able
to explain GPS as well. GR is not special.3
You are quite right:
GR formula:
(dt(s)/dt(e))^2 = (1-2M/Rs)-v(s)^2)/(1-2M/Re-v(e)^2)
M = G*Mearth/c^2 and v(s)^2 corresponds to (v(s)/c)^2,
thus to G*Mearth/c^2*Rs or M/2Rs
N.B.:
Rs is the orbital radius of the satellite and Re is the Earth radius
After simplification, one gets
dt(s)/dt(e) = 1 - M/Rs - v(s)^2/2 + M/Re + v(e)^2/2
= 1 - M/Rs - M/2Rs + M/Re + v(e)^2/2
= 1 - 3M/2Rs + M/Re + v(e)^2/2
Ignoring v(e)^2/2 because of the Earth oblateness, one is left with
dt(s)/dt(e) = 1 + M*(1/Re - 3/2Rs) = 1 + shift
Notice that the shift is 0 when Rs = 3Re/2, meaning that a clock
situated on the satellite orbiting at such distance Rs from the Earth
center and a clock situated on Earth will tick at the same rate.
If the satellite were *at rest* at a distance Rs from the Earth
center (v(s)=0), we would have a *pure* GR effect given by
dt(s)/dt(e) = 1 + M * (1/Re - 1/Rs) = 1 + shift(GR)
Such GR formula can easily be derived from the potential energy formula
Ep = m(ph) * gm * (Rs-Re), where
gm = G * Mearth / (Rs * Re)
m(ph) = hNu(s)/c^2 (m(ph) = "mass" of a photon of initial frequency
Nu(s))
Thus Ep = (hNu(s)/c^2) * (G * Mearth / (Rs * Re) * (Rs-Re)
= hNu(s) * (GMearth/c^2) * (1/Re - 1/Rs)
As the photon adds Ep to its initial energy hNu(s), its
energy at the ground level (at the distance Re from the Earth center)
becomes
hNu(e) = hNu(s) + hNu(s) * (GMearth/c^2) * (1/Re - 1/Rs)
After simplfication, one gets
Nu(e)/Nu(s) = 1 + (G*Mearth/c^2) * (1/Re - 1/Rs), which is equivalent
to
the GR formula dt(s)/dt(e) = 1 + M*(1/Re - 1/Rs)
To take into account the SR effect due to the orbital velocity v(s) of
the
satellite, one has to use multiply Nu(e)/Nu(s) by sqrt(1-v(s)^2/c^2),
where
v(s)^2/c^2 = G*Mearth/Rs*c^2. Then Nu(e)/Nu(s) becomes, after replacing
G*Mearth/c^2 by M,
(1 + M(1/Re - 1/Rs)) * sqrt(1 - M/Rs), or about
(1 + M(1/Re - 1/Rs)) * (1 - M/2Rs) =
'1 + M(1/Re - 1/Rs) - M/2Rs - a negligible term
The "global" formula is thus
Nu(e)/Nu(s) = 1 + M(1/Re - 1/Rs - 1/2Rs)
= 1 + M(1/Re - 3/2Rs), which is equivalent to the GR
formula
dt(s)/dt(e) = 1 + M(1/Re - 3/2Rs)
Marcel Luttgens