SR and c measurement

[vuquta]

But you keep claiming A, you were asked to show A, and now you say you have way B.

 

Please stick to a claim. You need to show us the transforms because you claim they fail. Either show it, or stop claiming it.

 

~moo

 

No, I did not claim A. You wanted it or you claimed some nonsense that I did not understand LT.

 

I provided A.

 

How many times do I need to show it for you to understand?

 

Sure, but I am also working from another angle that is legal.

 

I am using time dilation and the light sphere.

 

Here is LT. Other places I debate do not think I have a problem with LT and they are correct.

 

x' = ( x - vt )γ

 

t' = ( t - vx/c² )λ

 

The issue under consideration is t = rγ/c and x = rγ..

 

Simple plug and chug.

 

x' = r( λ² - v λ²/c)

 

Are you up to the alternative, yes or no.

[mooeypoo]

No, I did not claim A. You wanted it or you claimed some nonsense that I did not understand LT.

If you didn't understand, don't call it nonsense. Say you didn't understand.

 

You claim that:

 

But, by using (x,y,z) on the light sphere equal (r,0,0,r/c) and (x,y,z) equal (-r,0,0,r/c), we find different values for c² τ². and hence LT did not preserve the light sphere as Einstein claimed he proved.

 

That's some claim. You claim that Einstein was wrong. That's fine, but you need to show it.

 

We asked you to show it by actually doing the transforms, and showing us that Einstein was wrong and you are right. Either do that, or stop claiming it.

 

~moo

[vuquta]

If you didn't understand, don't call it nonsense. Say you didn't understand.

 

You claim that:

 

 

 

That's some claim. You claim that Einstein was wrong. That's fine, but you need to show it.

 

We asked you to show it by actually doing the transforms, and showing us that Einstein was wrong and you are right. Either do that, or stop claiming it.

 

~moo

 

OK, did you follow my arg with time dilation and the light sphere?

 

I am trying to do it but you do not seem able to understand it.

 

In all the other forums, I do not need to reduce to such simplicity.

 

Here is the arg again.

 

Now, do you folks want to continue with time dilation and the light sphere?

 

When time elapses in rγ/c at O, the clock at the origin O' elapses r/c, yes or no.

 

If the clock at O' elapses r/c, then light is a distance r in all directions in the frame of O', yes or no.

 

 

If you all do not want to talk, just say so and I will leave. I have plenty other places where I go.

 

Just let me know.

 

If you cannot handle this simple logic, then fine, I will go away.

 

I do not want to debate in a place that cannot deal with simple facts of the theory they support.

[mooeypoo]

OK, did you follow my arg with time dilation and the light sphere?

You are beating around the bush again.

I heard your argument. I don't want an argument, I want math. You claimed a mathematical claim, show it.

 

Can you??

 

~moo

[swansont]

Proving that the Lorentz transforms give conflicting answers is an exercise in pure math and is impossible, because the math is self-consistent.

 

Math theory does not model anything.

 

LT models light travel and relative inertial motion.

 

Apples and oranges.

 

Lorentz transforms would be valid math even if they did not model light travel and inertial motion. From a purely mathematical standpoint, they cannot give conflicting solutions. It's impossible. You have not brought up any physical experiment, so whether or not relativity is correct — whether Lorentz transforms describe the geometry of the real world — is not actually the topic here. If the Lorentz Transforms were shown to be inconsistent, it would invalidate relativity, but that's because it would invalidate all of science — all of math would be worthless.

 

And I've read your exchanges elsewhere, so I know you've been told this same thing by many people. So don't try and paint us as the ones who don't understand what's going on.

 

I'm tired of playing "find my error" with you. Every time I find one, you move on to a different scenario, but they all have the same fundamental flaw.

[vuquta]

You are beating around the bush again.

I heard your argument. I don't want an argument, I want math. You claimed a mathematical claim, show it.

 

Can you??

 

~moo

 

I said this.

 

Here is the arg again.

 

Now, do you folks want to continue with time dilation and the light sphere?

 

When time elapses in rγ/c at O, the clock at the origin O' elapses r/c, yes or no.

 

If the clock at O' elapses r/c, then light is a distance r in all directions in the frame of O', yes or no.

 

I ask this question. Since there exists a light sphere at the origin of O', when in the time coordinates of O does this light sphere reach a rest radius of r?

 

Anyway, I provided the math.

 

When the clock at O elapses rγ/c, the clock at O' elapses r/c by time dilation. Now, in the frame of O', when the clock at O' elapses r/c, light is a distance r in all directions. In particular, it is a distance r along the positive x-axis.

 

Yet, LT calculates light is a distance r ( γ² - v γ²/ c ).

 

Under SR, we have two different answers for the distance light is from the O' origin along the positive x-axis when the clock at O elapses rγ/c.

[Cap'n Refsmmat]

Yet, LT calculates light is a distance r ( γ² - v γ²/ c ).

Can you actually perform this calculation for us, as well as the one you say gives contradictory results?

[vuquta]

Lorentz transforms would be valid math even if they did not model light travel and inertial motion. From a purely mathematical standpoint, they cannot give conflicting solutions. It's impossible. You have not brought up any physical experiment, so whether or not relativity is correct — whether Lorentz transforms describe the geometry of the real world — is not actually the topic here. If the Lorentz Transforms were shown to be inconsistent, it would invalidate relativity, but that's because it would invalidate all of science — all of math would be worthless.

 

And I've read your exchanges elsewhere, so I know you've been told this same thing by many people. So don't try and paint us as the ones who don't understand what's going on.

 

I'm tired of playing "find my error" with you. Every time I find one, you move on to a different scenario, but they all have the same fundamental flaw.

 

 

And I've read your exchanges elsewhere, so I know you've been told this same thing by many people. So don't try and paint us as the ones who don't understand what's going on.

 

At this point, noone has stipped by alternative twins contradiction.

 

Noone has stopped this angle I am on now.

 

Noone has stopped my train as stationary analysis for R of S.

 

Noone has stopped my point where t' = t and the vector math that results in an emission speed of light < c.

 

You see I do read lots of typing the words this is wrong, but I never see any proof.

 

I would think as a scientist, this type of thing would be beyond you. It would seem, you would simply refute any of my logic outright and that would be sufficient.

 

To proclaim LT is consistent, then it must be consistent with any type of calculation.

 

My twins contradiction forced the absoluteness of the clock sync up against reciprocol time dilation causing at least one of the two LT calculations to be wrong.

 

This has not been touched by anyone anywhere since I introduced it 7 months ago.

 

In fact, the twins contradiction shows the components of SR calculate different answers, when arranged a certain way, for the same situation which is logically inconsistent. They shoiuld arrive at the same answer.

---

Merged post follows:

Consecutive posts merged

Can you actually perform this calculation for us, as well as the one you say gives contradictory results?

 

Yes,

 

This is LT

 

the clock elapses rγ/c in O. Thus, light is a distance rγ in all directions by the light sphere.

 

Now, consider onloy the positive x-axis point.

 

x' = ( x - vt )γ

 

x = rγ and t = rγ/c.

substitute the two above in the LT distance,

 

x' = ( x - vt )γ = ( rγ - v(rγ/c) )γ = r ( γ - v(γ/c) )γ = r ( γ² - vγ² /c )

 

Therefore, x' = r ( γ² - vγ² /c ).

 

Hence, LT calculates that light is a distance x' from the origin O' along the positive x-axis and that is

r ( γ² - vγ² /c ).

 

 

On the other hand, by time dilation if the clock at O elapses rγ/c, then the clock at O' elapses r/c. Now, in the context of O' stationary, when the clock at the origin elapses r/c, this implies light must be a distance r in all directions by the light sphere. Hence, light is a distance r from O' along the positive x-axis.

 

So, when the clock at O elapses rγ/c

by LT, light is a distance r ( γ² - vγ² /c ) along the positive x-axis.

 

However, by the time dilation and the light sphere, when the clock at O elapses rγ/c, light is a distance r from O' along the positive x-axis.

 

But, r ( γ² - vγ² /c ) < r and hence, O' must disagree with itself on points overtaken by the light cone, which is a contradiction.

[swansont]

 

You see I do read lots of typing the words this is wrong, but I never see any proof.

 

I would think as a scientist, this type of thing would be beyond you. It would seem, you would simply refute any of my logic outright and that would be sufficient.

 

Saying things like "this assumes absolute simultaneity" is a refutation of your "logic." You just ignore it.

 

To proclaim LT is consistent, then it must be consistent with any type of calculation.

No, it will not be consistent with an improper calculation.

 

As I stated long ago, you can find a lot of "proofs" that 1=0. But it's because of an invalid math operation somewhere in the proof. Similarly, the only way to show that Lorentz transforms disagree is to make an invalid step in the math. Because all you are doing is math.

[vuquta]

Saying things like "this assumes absolute simultaneity" is a refutation of your "logic." You just ignore it.

 

 

No, it will not be consistent with an improper calculation.

 

As I stated long ago, you can find a lot of "proofs" that 1=0. But it's because of an invalid math operation somewhere in the proof. Similarly, the only way to show that Lorentz transforms disagree is to make an invalid step in the math. Because all you are doing is math.

 

You are therefore claiming time dilation is false.

 

Time dilation is derived from LT.

 

Now what?

 

Let's now only talk with math with this problem.

 

Can you do this?

[vuquta]

I will respond to this issue here.

 

Originally Posted by vuquta

The revolution is a circular path just like Sagnac.

 

What logic causes this to be false? I do not understand this.

 

 

swansont

The light does not complete the circle along the path of the orbit. To first order, the path of the orbit is linear over that time.

 

Well, here is the Sagnac rotational correction for GPS.

 

Δt = Rv/c²

 

R is the distance to the satellite at light emission.

 

This is not a closed loop either.

 

Next, light reaches the ground unit in .05 seconds. The earth rotates 73 feet in that time. That certainly cannot be called anything like a curved path.

 

Next, the correction does not contain any curvature in Δt = Rv/c². It is a function of the speed of the GPS unit in the loop. The speed of the unit in the rotational loop is .28 miles per second. The speed of the unit in the orbital loop is 18.55 miles per second.

 

It is also a function of the distance to the satellite at light emission.

 

Given the facts above, there is nothing available logically or mathematically to rule out the Sagnac correction of the earth's orbit around the sun.

 

Yet, it does not exist.

 

There must exist some type of math to rule it out.

[swansont]

I will respond to this issue here.

 

 

 

 

 

 

Well, here is the Sagnac rotational correction for GPS.

 

Δt = Rv/c²

 

R is the distance to the satellite at light emission.

 

This is not a closed loop either.

 

Next, light reaches the ground unit in .05 seconds. The earth rotates 73 feet in that time. That certainly cannot be called anything like a curved path.

 

Next, the correction does not contain any curvature in Δt = Rv/c². It is a function of the speed of the GPS unit in the loop. The speed of the unit in the rotational loop is .28 miles per second. The speed of the unit in the orbital loop is 18.55 miles per second.

 

It is also a function of the distance to the satellite at light emission.

 

Given the facts above, there is nothing available logically or mathematically to rule out the Sagnac correction of the earth's orbit around the sun.

 

Yet, it does not exist.

 

There must exist some type of math to rule it out.

 

There certainly are logical reasons to claim the Sagnac effect of the orbit is small, but you have to actually understand the Sagnac effect, and it is abundantly clear that you do not. Taking statements applied to a specific problem and trying to make them apply in general in order to falsify them is a straw man argument. Stop it. There are a number of explanations of the Sagnac effect on the web. Read a few of them, please.

 

The path around the earth — what we were discussing — is a closed loop. If you had followed the argument in the other thread, you would have seen where I explained that the phase accumulation would occur continually around the loop. If you divided it into 200 segments, each segment would contribute ~1 nanosecond. No closed loop necessary.

 

The path around the earth is curved. So is the orbit about the sun. However, what is important is the amount of curvature, since linear paths will not contribute. The earth rotates about its axis in 1 day. It revolves around the sun in ~365 days. As such the amount of curvature, relative to a full circle, is 365 times larger for the earth. So for any Sagnac interferometer (properly oriented), the earth's rotation will give you 365 times more phase accumulation than will its revolution. The path around the sun can only contribute 2*pi of phase to the interferometer, and it would take a full year to do that. On the earth, that happens in a day.

[vuquta]

There certainly are logical reasons to claim the Sagnac effect of the orbit is small, but you have to actually understand the Sagnac effect, and it is abundantly clear that you do not. Taking statements applied to a specific problem and trying to make them apply in general in order to falsify them is a straw man argument. Stop it. There are a number of explanations of the Sagnac effect on the web. Read a few of them, please.

 

The path around the earth — what we were discussing — is a closed loop. If you had followed the argument in the other thread, you would have seen where I explained that the phase accumulation would occur continually around the loop. If you divided it into 200 segments, each segment would contribute ~1 nanosecond. No closed loop necessary.

 

The path around the earth is curved. So is the orbit about the sun. However, what is important is the amount of curvature, since linear paths will not contribute. The earth rotates about its axis in 1 day. It revolves around the sun in ~365 days. As such the amount of curvature, relative to a full circle, is 365 times larger for the earth. So for any Sagnac interferometer (properly oriented), the earth's rotation will give you 365 times more phase accumulation than will its revolution. The path around the sun can only contribute 2*pi of phase to the interferometer, and it would take a full year to do that. On the earth, that happens in a day.

 

Here are the actual equations for the phase difference in .05 seconds which is about the time it takes the signal to move from the satellite to the ground receiver.

 

p = pi

The phase shift equations is

(8pAcw/λ)/(c² – v²).

 

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

 

Where A = p r²

v = wr

v is the linear velocity.

 

Hence, wA = p r² (v/r) = p r v

 

So, the total phase difference is:

(8pc(p r v)/λ)/(c² – v²) = (8p²rvc/λ)/(c² – v²)

Earth radius - 6360 km

Earth rotational speed - 1669.8 km per hour = 0.4639 km/s

 

Earth distance to sun - 150,000,000 km

Earth orbital speed - 18.55 km/s

 

Earth rotational total phase difference for entire path.

(8p²rvc/λ)/(c² – v²)

(8p²(6360)(0.4639)c/λ)/(c² – v²)

(69886366343.874118567296/l)/89999999999.78479679

 

= 0.776515/λ

 

The rotational loop length is 2pr = 39961.057872 km

 

It take light about ,05 seconds to move from the satellite to the GPS unit.

 

The earth rotates in ,05, (0.4639)(.05) = 0.023195km

The fraction of movement to the total path is

0.023195/39961 = 5.80e-7

 

So, the total phase difference for this short path is

5.80e-7*0.776515/l = 4.5072e-7/λ

 

 

Earth orbital total phase difference for entire path.

(8p²rvc/λ)/(c² – v²)

(8p²(150,000,000)(18.55)c/λ)/(c² – v²)

(6590921594189464.728/l)/89999999655.8975

= 73232.4624/λ

 

The orbital loop length is 2pr = 942477780 km

 

In .05 seconds, the earth travels

.05*(18.55) = 0.9275km

 

The fraction of movement to the total path is

0.9275/942477780 = 9.8411e-10

 

So, the total phase difference for this short path is

 

9.8411e-10*73232.4624/λ = 7.20686581e-5/λ

 

 

Therefore, the orbital sagnac correction of

9.8411e-10*73232.4624/λ = 7.20686581e-5/λ

 

is much greater than the rotational sagnac correction of

5.80e-7*0.776515/l = 4.5072e-7/λ

 

 

But, the orbital sagnac does not exists or it is not published.

[swansont]

Why are you using the radius of the earth's orbit? That does not describe the area of the path of the light.

 

But, the orbital sagnac does not exists or it is not published.

 

It does exist. It depends on the location of the receivers, so I don't know what, exactly, you would publish.

[vuquta]

Why are you using the radius of the earth's orbit? That does not describe the area of the path of the light.

 

Oh, I was only comparing an experiment done on the earth with an emission point and a receiver both fixed to the earth. I was not using GPS which is based on the distance to the satellite and is therefore smaller. Sagnac has been seen for this type of setup for the earth's rotation.

 

 

 

It does exist. It depends on the location of the receivers, so I don't know what, exactly, you would publish.

 

Let's put them on the earth as above.

 

My calculations are for earth based points.

 

It shows the orbital correction is larger than the rotational correction.

 

But, the rotational correction shows up in GPS as a smaller value and in earth based experiments.

 

But, the orbital sagnac does not.

 

I think you believe because the angular velocity of the earth's orbit is much less than that of the rotation, you think the orbital correction is small.

 

But, as you said above, the total correction uses the area swept out by the two points.

 

Since that area starts from the sun to the earth, the orbital correction uses this very large area times a very small angular velocity such that it is a total of about 60 times that of the rotation correction between points A and B.

 

I am basing all this on the two points being at the equator east and west at noon.

 

This setup will only show a rotational correction.

 

So, perhaps you have the math as to why it does not show up.

 

My math demonstrates it should show up.

 

In fact, a satellite could be at the equator directly east on the horizon 10,000 miles away and the orbital value will still not be there.

 

In this setup, while light moves c, the point move toward the light pulse from the satellite at 30 km/s cutting down the distance for that light to travel.

[swansont]

But, as you said above, the total correction uses the area swept out by the two points.

 

Since that area starts from the sun to the earth, the orbital correction uses this very large area times a very small angular velocity such that it is a total of about 60 times that of the rotation correction between points A and B.

 

I don't know how you would draw a closed path around the earth, or from the earth to a satellite and back to earth, describing the travel of a signal, that would encompass the sun. You stated that the signal takes 0.05 seconds. How does it go around the sun in that amount of time? The sun is ~8 light-minues away.

 

Your analysis is wrong.

[vuquta]

I don't know how you would draw a closed path around the earth, or from the earth to a satellite and back to earth, describing the travel of a signal, that would encompass the sun. You stated that the signal takes 0.05 seconds. How does it go around the sun in that amount of time? The sun is ~8 light-minues away.

 

Your analysis is wrong.

 

A complete closed path is not need by experiments done on the earth with rotational sagnac and also with GPS. A partial path is all that is needed.

 

And it does not matter how long it takes light to travel from the rotational center.

 

It only matter how long light takes to travel around the light path.

 

 

Here is how it is derived.

 

 

In time t, light moves ct.

It must move the circumference of the circle plus the distance the receiver moved on the loop in time t.

 

Thus,

 

ct1 = 2πr + vt1.

 

t1 = 2πr/(c-v)

 

For the other direction, light must move the circumference of the circle plus the distance the receiver moved on the loop in time t.

 

Thus,

 

ct2 = 2πr - vt2.

 

t2 = 2πr/(c+v)

 

Now let's find the sagnac correction.

 

t1 - r2 = ∆t = 2πr/(c-v) - 2πr/(c+v) = 2πr (1/(c-v) - 1/(c+v) ) = 2πr ((c+v)/(c²-v²) - (c-v)/(c²-v²) )

 

∆t = 4πrv/(c²-v²)

 

Now, using angular velocity of ω = v/r

 

and A = πr², we find πrv = Aω

 

Hence, in another format,

 

∆t = 4Aω/(c²-v²).

 

So, it is not about coordinate changes, acceleration or angular velocity.

 

It is about the path length being the circumference of the circle plus the change in the position of the receiver on that circumference.

[swansont]

It is about the path length being the circumference of the circle plus the change in the position of the receiver on that circumference.

 

Precisely. Now all that is left is for you to apply this concept properly.

[vuquta]

Originally Posted by vuquta

It is about the path length being the circumference of the circle plus the change in the position of the receiver on that circumference.

 

Precisely. Now all that is left is for you to apply this concept properly.

 

OK.

 

Assume the satellite is in the east just off the horizon at the equator at noon.

 

While light moves at ct toward the receiver, where c is the fixed speed of light in space, ther receiver moves along the circumference of the orbit vt, where v = 18.55mps.

 

Let's assume the satellite is 10,000 miles away.

 

So, in time t, light must be 10,000 miles minus the distance the unit moves along the circumference of the orbit which is vy.

 

Hence,

 

ct = 10,000 - 18.55t.

 

So, the sagnac correction is:

t = 10,000/(c-18.55)

[swansont]

A satellite would usually be moving eastward, and we normally don't mix units like that.

 

Now, why 18.55 m/s? That seems unreasonable. The orbital speed of the satellite is determined by its distance form the center of the earth, and would be several km/s.

[vuquta]

A satellite would usually be moving eastward, and we normally don't mix units like that.

 

Now, why 18.55 m/s? That seems unreasonable. The orbital speed of the satellite is determined by its distance form the center of the earth, and would be several km/s.

 

 

Yea, I am bring unclear.

 

I see your point.

 

I was talking about the earth's orbit around the sun.

 

The satellite and the hand held unit are on the earth's orbital path at 18.55 mps.

 

And the speed I am talking about is the linear speed not the angular speed, though the two are related by wr = v.

 

I could use either, but if I use angular speed to calculate sagnac, I will need to use the area of the circle sector paths. For the earth's orbit, that area is very large.

 

 

There has got to be a logical reason why the sagnac for the earth's orbit is not being picked up but is for the rotation.

 

The math shows that it should.

[swansont]

18.55 mps? The earth's orbit is at about 30 km/s.

 

But you have not described a closed loop. If the path were linear, there would be no phase shift at all. For the orbit about the sun, the area enclosed by the light path in that time will be very small, hence a small phase shift.

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Merged post follows:

Consecutive posts merged

Light would take about 3140 sec to complete an orbit about the sun. In this time, the earth moves 94200 km (@ 30 km/s). L/c gives .314 sec of Sagnac, for a complete orbit.

 

If a satellite is 30,000 km away, that's 0.1 sec for a signal to get there, or 3 km of travel of the earth. Out of a circumference of almost 10^9 km. So that's 3 parts in 10^9 of .314 sec, or about a nanosecond of Sagnac from the orbit about the sun.

 

GPS Sagnac can be hundreds of nanoseconds. For a path around the earth (207 ns of rotational Sagnac), that's 40,000 km, so the orbital Sagnac will be ~1.3 nanoseconds.

Edited April 18, 2010 by swansont

[vuquta]

18.55 mps? The earth's orbit is at about 30 km/s.

 

But you have not described a closed loop. If the path were linear, there would be no phase shift at all. For the orbit about the sun, the area enclosed by the light path in that time will be very small, hence a small phase shift.

 

But, the closed loop is not needed. GPS shows the rotational sagnac is proven without a closed loop using only a potion of the loop.

 

 

Light would take about 3140 sec to complete an orbit about the sun. In this time, the earth moves 94200 km (@ 30 km/s). L/c gives .314 sec of Sagnac, for a complete orbit.

 

If a satellite is 30,000 km away, that's 0.1 sec for a signal to get there, or 3 km of travel of the earth. Out of a circumference of almost 10^9 km. So that's 3 parts in 10^9 of .314 sec, or about a nanosecond of Sagnac from the orbit about the sun.

 

GPS Sagnac can be hundreds of nanoseconds. For a path around the earth (207 ns of rotational Sagnac), that's 40,000 km, so the orbital Sagnac will be ~1.3 nanoseconds.

 

The problems above is you did not ratio off the rotational motion of the earth in .1 seconds like you did above.

 

That is 0.0463 km traveled in .1 seconds.

 

Now you have 0.0463 / 40,000 * 207 ns

[swansont]

The problems above is you did not ratio off the rotational motion of the earth in .1 seconds like you did above.

 

That is 0.0463 km traveled in .1 seconds.

 

Now you have 0.0463 / 40,000 * 207 ns

 

No, you don't multiply the ratio by the Sagnac shift — that's multiplying it twice.

 

Light takes about .13 seconds to go around the earth. In that time the earth moves about 0.06 km. L/c = 201 ns using those numbers. In 0.1 seconds, you get a shift of about 155 ns. Assuming you are going around the earth; satellite signals use a different path and will give you a different shift, but it will be roughly that magnitude.

[vuquta]

No, you don't multiply the ratio by the Sagnac shift — that's multiplying it twice.

 

Light takes about .13 seconds to go around the earth. In that time the earth moves about 0.06 km. L/c = 201 ns using those numbers. In 0.1 seconds, you get a shift of about 155 ns. Assuming you are going around the earth; satellite signals use a different path and will give you a different shift, but it will be roughly that magnitude.

 

Light takes about .13 seconds to go around the earth.

agreed

 

In that time the earth moves about 0.06 km

This is where you are going wrong.

We are measuring the sagnac effect for a path from A to B on the earth.

 

So, light is not going around the earth. It is going from A to B.

 

Say that distance is 10 km.

 

You will find the sagnac correction higher for the orbit vs the rotation.

 

It is not about now long light takes to move around the circumference.

 

It is about how far light moves while the earth moves.