Earth's orbital sagnac

[vuquta]

If the speed of light is independent of the source, then why is the earth's orbital sagnac missing from the GPS calculations.

 

The earth's rotational sagnac is included.

[Bignose]

Since the earth and the bodies rotating about it move as one, the rotational effects aren't needed. That is, both our satellites and the Earth are moving at the same speed in our orbit around the sun. No correction is necessary.

 

If the receiver was on the sun, and the transmitter on the Earth, then accounting for the effect would be needed.

[vuquta]

Since the earth and the bodies rotating about it move as one, the rotational effects aren't needed. That is, both our satellites and the Earth are moving at the same speed in our orbit around the sun. No correction is necessary.

 

If the receiver was on the sun, and the transmitter on the Earth, then accounting for the effect would be needed.

 

Here is the mathpages explanation of sagnac.

 

http://www.mathpages.com/rr/s2-07/2-07.htm

 

You will find your logic flawed.

 

If you need any help with the math, let me know.

 

The rotational sagnac is in fact a part of GPS and calculations must be performed in the hand held unit to compensate.

 

Yet, the orbital sagnac does not show up.

 

This implies light is consistent with Ritz's theory of light whereas the rotational sagnac implies Ritz's theory of light is false.

 

This is a problem.

[swansont]

Orbital Sagnac is ~1/365 as large as the rotational Sagnac for any individual clock. Same area enclosed by the signal, but the period is annual instead of daily. Rotational Sagnac is typically several hundred nanoseconds, so orbital would be less than a nanosecond, or tens of cm in positioning error, which is much less than the other errors in GPS, which has positioning errors of several meters. There is a fairly recent thread on this.

 

Further, the error is only present for signal direction that has a component along the orbital direction. So for many satellites, the Sagnac term is even smaller, and can be negative.

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Since the earth and the bodies rotating about it move as one, the rotational effects aren't needed. That is, both our satellites and the Earth are moving at the same speed in our orbit around the sun. No correction is necessary.

 

If the receiver was on the sun, and the transmitter on the Earth, then accounting for the effect would be needed.

 

No, there should be an effect. Two locations on the earth will experience a Sagnac shift when sending a signal to each other if the signal path has a component along the direction of motion. There should be an effect for orbit as well. The question is how big it is.

 

The rotation of the earth is about 40,000 km per day. A signal around the earth (transmitter and receiver co-located) at the speed of light will take about .133 second, in which case the location will have moved 0.0617 km along this curved path. That gives us a change of 206 ns. (which is less than 1% off from the actual value, had I not approximated)

 

But we want to know the orbital effect. If we ran our experiment for a year, there would be one extra rotation due to the orbit — i.e. if the earth were not rotating with respect to an inertial frame, it would look from the sun like we did one rotation. So it will loop ~240 million times, accumulating almost 49 seconds of phase from rotation during the experiment, to gain an extra .133 seconds from the orbit.

 

IOW, in the time it takes to do one loop, the earth has only completed only a small fraction of an orbit. It looks basically like the earth is traveling in a straight line, and for a straight line, there is no Sagnac. Basic qualitative application of the principle shows that the effect has to be small.

Edited July 11, 2010 by swansont
Consecutive posts merged.

[vuquta]

Orbital Sagnac is ~1/365 as large as the rotational Sagnac for any individual clock. Same area enclosed by the signal, but the period is annual instead of daily. Rotational Sagnac is typically several hundred nanoseconds, so orbital would be less than a nanosecond, or tens of cm in positioning error, which is much less than the other errors in GPS, which has positioning errors of several meters. There is a fairly recent thread on this.

 

Further, the error is only present for signal direction that has a component along the orbital direction. So for many satellites, the Sagnac term is even smaller, and can be negative.

 

Here is the problem I have with this analysis.

 

According to mathpages,

 

Δt = 2π R( 1/(c-v) - 1/(c+v) )

 

where R is is the radius of the loop and v is the linear velocity.

 

The radius of the orbital loop is 93 million miles and v = 18.55 miles per second.

 

So, your conclusions are not making sense.

 

And, the enclosed area is a wedge from the center of rotation. The center of rotation is from the sun for the orbital sagnac.

 

Perhaps, you could do the math and show using mathpages how you arrive at ~1/365.

[swansont]

R is not the radius of the orbital loop, unless the signal path is the orbit. R is the radius of the signal path loop. Or, if you don't have a circle, you use an equation that uses the area of the signal path loop. Using my example, the radius is that if the earth, because that's the area enclosed by the signal path.

 

So the only difference is in the angular frequency of the rotation. 1 year = ~365 days. Hence 1/365

[vuquta]

R is not the radius of the orbital loop, unless the signal path is the orbit. R is the radius of the signal path loop. Or, if you don't have a circle, you use an equation that uses the area of the signal path loop. Using my example, the radius is that if the earth, because that's the area enclosed by the signal path.

 

So the only difference is in the angular frequency of the rotation. 1 year = ~365 days. Hence 1/365

 

Yea, but if a satellite is on the horizon in the east and shoots at a hand held unit, then both conditions that the path is in line with the rotational path and the orbital path are both met.

 

So, in both cases, the signal is on the associated orbital/rotational loop. In addition, the entire path is not needed since clearly the rotational sagnac shows up in GPS and the entire rotational path is not travelled by the light beam.

 

 

So, how would you exclude the orbital sagnac for this case?

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R is not the radius of the orbital loop, unless the signal path is the orbit. R is the radius of the signal path loop. Or, if you don't have a circle, you use an equation that uses the area of the signal path loop. Using my example, the radius is that if the earth, because that's the area enclosed by the signal path.

 

So the only difference is in the angular frequency of the rotation. 1 year = ~365 days. Hence 1/365

 

 

I found this paper on the issue. Please check page 172.

 

http://qem.ee.nthu.edu.tw/f1b.pdf

 

I am not understanding the explanation.

[swansont]

Yea, but if a satellite is on the horizon in the east and shoots at a hand held unit, then both conditions that the path is in line with the rotational path and the orbital path are both met.

 

So, in both cases, the signal is on the associated orbital/rotational loop. In addition, the entire path is not needed since clearly the rotational sagnac shows up in GPS and the entire rotational path is not travelled by the light beam.

 

 

So, how would you exclude the orbital sagnac for this case?

 

If you follow less than the full rotational path, you get less of a shift. If you go 10,000 km along the earth's equator, you get 1/4 of the rotational shift, or about 50 ns. But that's a much smaller fraction of the orbital path length.

 

I found this paper on the issue. Please check page 172.

 

http://qem.ee.nthu.edu.tw/f1b.pdf

 

I am not understanding the explanation.

 

"I don't understand" is too vague. But he's saying that there's no measurable orbital effect.

[vuquta]

If you follow less than the full rotational path, you get less of a shift. If you go 10,000 km along the earth's equator, you get 1/4 of the rotational shift, or about 50 ns. But that's a much smaller fraction of the orbital path length.

 

Of course you get less of a shift if you do not follow the full path.

 

But, angular velocity is not the issue. Otherwise, lab based sagnacs which have a much greater angular velocity should be seeing a much greater shift based on your angular velocity explanation.

 

Further, the Total distance of orbit 942,477,780km.

 

Total one way orbital sagnac = 942,477,780( 1/c - 1/(c+v) ) = 942,477,780(v/(c(c+v)) = 942,477,780 ( 30/(c(c+30) ) = .31 seconds.

 

Assume the distance of the satellite to the unit along the orbital path is 10,000 km.

 

Fraction of the total path is 10,000/942,477,780 = 1.06*10-5 .

Fraction of total path times the total seconds for the fraction of the total seconds. 1.06*10-5 * .31 seconds = 3.29 * 10-6 seconds.

 

 

 

"I don't understand" is too vague. But he's saying that there's no measurable orbital effect.

Indeed he does.

 

This is an IOP article.

 

Here are some quotes.

 

This reinterpretation is fundamentally different from that based on the special relativity

 

Further, in the present common understanding, the null result is extrapolated without direct evidences to rule out the effect of the Earth’s rotation on wave propagation, as in Einstein’s original paper on special relativity where it is assumed that τ = 2Rt/c

 

Here is his explanation of the missing earth orbital sagnac.

 

The discrepancy in the unique propagation frame can be solved by the

local-ether model of wave propagation recently presented

 

http://qem.ee.nthu.edu.tw/f1b.pdf

[swansont]

Of course you get less of a shift if you do not follow the full path.

 

But, angular velocity is not the issue. Otherwise, lab based sagnacs which have a much greater angular velocity should be seeing a much greater shift based on your angular velocity explanation.

 

What lab-based Sagnacs have a larger angular velocity?

 

Further, the Total distance of orbit 942,477,780km.

 

Total one way orbital sagnac = 942,477,780( 1/c - 1/(c+v) ) = 942,477,780(v/(c(c+v)) = 942,477,780 ( 30/(c(c+30) ) = .31 seconds.

 

Assume the distance of the satellite to the unit along the orbital path is 10,000 km.

 

Fraction of the total path is 10,000/942,477,780 = 1.06*10-5 .

Fraction of total path times the total seconds for the fraction of the total seconds. 1.06*10-5 * .31 seconds = 3.29 * 10-6 seconds.

 

.31 seconds is the orbital Sagnac for a loop that goes along the orbit of the earth, which is not what we have.

 

Indeed he does.

 

This is an IOP article.

 

Here are some quotes.

 

This reinterpretation is fundamentally different from that based on the special relativity

 

Further, in the present common understanding, the null result is extrapolated without direct evidences to rule out the effect of the Earth’s rotation on wave propagation, as in Einstein’s original paper on special relativity where it is assumed that τ = 2Rt/c

 

Here is his explanation of the missing earth orbital sagnac.

 

The discrepancy in the unique propagation frame can be solved by the

local-ether model of wave propagation recently presented

 

http://qem.ee.nthu.edu.tw/f1b.pdf

 

Note well the word "reinterpretation." He's advancing an alternative physics model, a local ether one. You can't apply any of his theoretical explanations to relativity.

[vuquta]

What lab-based Sagnacs have a larger angular velocity?

 

Any of them are much larger than the earth's angular velocity.

 

They only need spin once per second for example. That should generate a huge sagnac if angular velocity is the issue.

 

 

 

.31 seconds is the orbital Sagnac for a loop that goes along the orbit of the earth, which is not what we have.

 

Yes we do have this if the light is shot along the path as I suggested. I said have the light shot east-west. Oh, I failed to mention along the equator at high noon.

 

This light path is along the earth's orbital path.

 

 

 

 

Note well the word "reinterpretation." He's advancing an alternative physics model, a local ether one. You can't apply any of his theoretical explanations to relativity.

 

I saw this but he provided a logical argument to show SR does not correctly explain this puzzle.

 

He claims the local dominating gravity creates/implements this "local aether" environment to explain why the orbital sagnac is missing.

 

The point that is clear from the article and my equations is the orbital sagnac is missing and should be there.

 

IOP published the article indicating they also agree it is missing or this article would never have gotten past peer review.

 

However, I think his explanation is silly and he does not produce equations to explain why it is missing.

 

I am simply making the clear point though, if the rotational sagnac exists, and the lab sagnac exists, then the orbital sagnac should exist and angular velocity cannot explain all three of these conditions.

[swansont]

Any of them are much larger than the earth's angular velocity.

 

They only need spin once per second for example. That should generate a huge sagnac if angular velocity is the issue.

 

You are implying that this is not the case, and yet have not provided any evidence. Show me where there are Sagnac interferometers being rotated that do not show angular velocity dependence.

 

 

 

Yes we do have this if the light is shot along the path as I suggested. I said have the light shot east-west. Oh, I failed to mention along the equator at high noon.

 

This light path is along the earth's orbital path.

 

 

First of all, note that your scenario is different from mine. I was using the earth as the loop.

 

How do you differentiate this from a straight line, in which case there is no Sagnac shift? What you need to know is the area enclosed by the signal path, and you haven't defined it.

 

 

I saw this but he provided a logical argument to show SR does not correctly explain this puzzle.

 

He claims the local dominating gravity creates/implements this "local aether" environment to explain why the orbital sagnac is missing.

 

The point that is clear from the article and my equations is the orbital sagnac is missing and should be there.

 

IOP published the article indicating they also agree it is missing or this article would never have gotten past peer review.

 

However, I think his explanation is silly and he does not produce equations to explain why it is missing.

 

I am simply making the clear point though, if the rotational sagnac exists, and the lab sagnac exists, then the orbital sagnac should exist and angular velocity cannot explain all three of these conditions.

 

I am not in a position to justify why an IOP journal would publish any given article. Passing peer review doesn't make something automatically true.

[vuquta]

You are implying that this is not the case, and yet have not provided any evidence. Show me where there are Sagnac interferometers being rotated that do not show angular velocity dependence.

I did not say it did not.

 

I am saying it actually depends on linear velocity and the radius of the loop as mathpages agree.

 

You are claiming it exclusively depends on angular velocity which is wrong.

 

 

 

 

 

First of all, note that your scenario is different from mine. I was using the earth as the loop.

 

 

How do you differentiate this from a straight line, in which case there is no Sagnac shift? What you need to know is the area enclosed by the signal path, and you haven't defined it.

 

How do you differentiate the satellite to earth as not a straight line?

 

I am saying, the path is right in line with the earth's orbit even more than a satellite to earth is to the earth's orbit. Clearly, that is a straight line and not a loop.

 

 

 

 

 

I am not in a position to justify why an IOP journal would publish any given article. Passing peer review doesn't make something automatically true.

 

Point taken. But, it doies not make it false either.

 

I guess that is true for any theory including SR, no?

[swansont]

I did not say it did not.

 

I am saying it actually depends on linear velocity and the radius of the loop as mathpages agree.

 

You are claiming it exclusively depends on angular velocity which is wrong.

 

 

Um, angular velocity depends on linear velocity and radius. v = wr, or w = v/r. Mathpages agrees that the effect is linear in w.

 

So saying it depends on angular velocity is the same as saying that it depends on linear velocity and radius. But while the linear speed is larger by a factor of 60 (comparing orbit and rotation), the radius is larger by a factor of more than 20,000

[elas]

Orbital Sagnac is ~1/365 as large as the rotational Sagnac for any individual clock. Same area enclosed by the signal, but the period is annual instead of daily. Rotational Sagnac is typically several hundred nanoseconds, so orbital would be less than a nanosecond, or tens of cm in positioning error, which is much less than the other errors in GPS, which has positioning errors of several meters. There is a fairly recent thread on this.

 

The original USnavy Satnav. signal gave position + or - 4 metres. this was deliberately distorted to about 1/2 mile for all shipping except the USN for obvious security reasons. It was only when hackers removed the distortion that the GPS system was altered to remove the distortion. As that was nearly 40 years ago, I presume there has been some improvement since. Local surveyors use satnav for building plots on the assumption that any error is less than 12 inches.

[swansont]

The original USnavy Satnav. signal gave position + or - 4 metres. this was deliberately distorted to about 1/2 mile for all shipping except the USN for obvious security reasons. It was only when hackers removed the distortion that the GPS system was altered to remove the distortion. As that was nearly 40 years ago, I presume there has been some improvement since. Local surveyors use satnav for building plots on the assumption that any error is less than 12 inches.

 

Selective availability was turned off in 2000. The "hacking" was differential GPS. GPS was not the original navy satnav system — that would be transit.

 

http://blogs.scienceforums.net/swansont/archives/5588

[vuquta]

Um, angular velocity depends on linear velocity and radius. v = wr, or w = v/r. Mathpages agrees that the effect is linear in w.

 

So saying it depends on angular velocity is the same as saying that it depends on linear velocity and radius. But while the linear speed is larger by a factor of 60 (comparing orbit and rotation), the radius is larger by a factor of more than 20,000

 

 

Well, see that was my point.

 

The orbital has a smaller angular with a faster linear velocity and larger radius.

 

According to mathpages, angular velocity can be omitted for radius and linear velocity.

 

Also according to mathpages, the orbital sagnac should exceed the rotational sagnac .

 

That is the math.

 

But experiments show the orbital does not exceed the rotational.

 

So, I am trying to figure out what is wrong.

 

I cannot find any way to make it work.

[swansont]

Well, see that was my point.

 

The orbital has a smaller angular with a faster linear velocity and larger radius.

 

According to mathpages, angular velocity can be omitted for radius and linear velocity.

 

Which makes no difference at all to the answer.

 

Also according to mathpages, the orbital sagnac should exceed the rotational sagnac .

 

That is the math.

 

But experiments show the orbital does not exceed the rotational.

 

So, I am trying to figure out what is wrong.

 

I cannot find any way to make it work.

 

The angular speed is smaller and the area encompassed by the signal is the same, so why do you expect the effect to be larger?

[vuquta]

Which makes no difference at all to the answer.

 

 

 

The angular speed is smaller and the area encompassed by the signal is the same, so why do you expect the effect to be larger?

 

It is my interpretation from mathpages that the area enclosed is based on the center of the loop which would be the center of the sun fpor the earth's orbital sagnac. It is an area of a wedge from that center with the arc being the hand held unit to the satellite.

 

Perhaps you could explain your position better.

 

Otherwise, I agree the angular for the oribt is much smaller.

 

 

 

Also, if you do not mind, do you see an error in my thing below? Obviously, this is a different issue but I did not want to start another thread.

 

 

 

 

Assume two rigid body spheres of radius r are in relatie motion and when the origins are co-located a light pulse is emitted.

 

When the moving frame views simultaneity of the two rigid body sphere, it fires a laser back to the rest frame origin.

 

In my opinion, these are the equations.

 

Let tl be the time O received the laser light.

 

t1 = time to when O' fired the laser.

 

t2 = time for the laser to travel back to O.

 

Obviously tl = t1 + t2.

 

Also, vt1 = distance the O' sphere traveled when simultaneity occured.

 

Also, ct2 = vt1 since that is the distance the laser must travel to hit O.

 

tl = t1 + t2. Substitute t2 = (v/c)t1.

 

tl = t1 + t2 = t1 + (v/c)t1.

 

tl = t1( c+v)/c.

 

Since we know tl, then t1 is the time when simultaneity occured in the O' frame.

 

t1 = tl c/(c+v).

 

Do you see an error in this calculation?

[swansont]

It is my interpretation from mathpages that the area enclosed is based on the center of the loop which would be the center of the sun fpor the earth's orbital sagnac. It is an area of a wedge from that center with the arc being the hand held unit to the satellite.

 

Perhaps you could explain your position better.

 

Otherwise, I agree the angular for the oribt is much smaller.

 

 

The radius or area is that of the travel of the satellites, for this example, i.e. the values of their orbit of the earth. The satellites are not orbiting the sun, meaning the sun is never within the circle defined by their orbit.

[vuquta]

The radius or area is that of the travel of the satellites, for this example, i.e. the values of their orbit of the earth. The satellites are not orbiting the sun, meaning the sun is never within the circle defined by their orbit.

 

This is not true. There exists a time they are on the orbital path.

 

Just take a satellite in front of the earth's orbit. That is on the path.

 

Then the satellite shoots a light.

 

That is just one possibility. This happens also.

[swansont]

This is not true. There exists a time they are on the orbital path.

 

Just take a satellite in front of the earth's orbit. That is on the path.

 

Then the satellite shoots a light.

 

That is just one possibility. This happens also.

 

Being "on the path" for an infinitesimal time is insufficient. The paths are tangential. What is impoprtant — what is used in the derivation — is the actual path, over a macroscopic distance. The satellites orbit the earth. That's what you use. If you do it that way, you get an answer that agrees with experiment. If you use something else, you get an answer that does not agree with experiment.

[vuquta]

Being "on the path" for an infinitesimal time is insufficient. The paths are tangential. What is impoprtant — what is used in the derivation — is the actual path, over a macroscopic distance. The satellites orbit the earth. That's what you use. If you do it that way, you get an answer that agrees with experiment. If you use something else, you get an answer that does not agree with experiment.

 

 

No, this does not work.

 

While east-west at noon, the light is on the oirbital path more than it is one the rotational path.

 

Clearly, the hand held unit and the satellite are not on the rotational path of the satellite.

 

But, at noon, at the equator, east-west, it is almost a perfect fit for being on the orbital path.

 

Your logic refutes the rotational sagnac more than it does the orbital sagnac for this particular experiment.

[swansont]

No, this does not work.

 

While east-west at noon, the light is on the oirbital path more than it is one the rotational path.

 

Clearly, the hand held unit and the satellite are not on the rotational path of the satellite.

 

But, at noon, at the equator, east-west, it is almost a perfect fit for being on the orbital path.

 

Your logic refutes the rotational sagnac more than it does the orbital sagnac for this particular experiment.

 

 

An orbital path around the earth is basically a circle. It has a certain radius. It is not "more like" any other path, depending on when or where you look at it. Its radius does not change from 10,000 km to 150,000,000 km just because of its location. They are not interchangeable. Wishing will not make it so.

[vuquta]

An orbital path around the earth is basically a circle. It has a certain radius. It is not "more like" any other path, depending on when or where you look at it. Its radius does not change from 10,000 km to 150,000,000 km just because of its location. They are not interchangeable. Wishing will not make it so.

 

1) It is about traversing the path of the loop.

 

2) The hand held unit and the satellite are clearly not on the same rotational loop.

 

3) Finally the hand held unit and the satellite are more likely on the orbital loop than the rotation loop than the rotational loop for the experiment I proposed. If you provide a proof my statement is false, I will show you why you are wrong.

 

So, you issue regarding radius is a red herring since you are the one that injected the sagnac path.