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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. 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

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.

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)

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.

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. 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.

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.

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.

Why does the satellite view C±V and what is v. Merged post follows: Consecutive posts merged The revolution is a circular path just like Sagnac. What logic causes this to be false? I do not understand this.

I see in ther forums, they think the light will return at the same time. I am in agreement with this forum that Sagnac would dictate an unequal travel. Then you want to cut the two paths into small pieces and then claim these small pieces I suppose put end to end would be different. But that is not what MMX demonstrates and you did mention MMX. See, I am just not seeing how you are tying these two together. Then you bring up an aether wind and have not precisely defined it. In order to invent an object, you would need experimental data evidence to support this new object. Thoughts and math theory would not be sufficient in my mind. This in my view is why your paper was rejected. You invented an object your cannot verify. Finally, what is your math for the .02% timing differential?

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?

OMG, you are talking to me. Is this true?

What is aether wind? How does it have a speed? What is it? Since MMX has null results, how do you compare your logic to MMX? Are you claiming MMX is wrong or are you claiming the two directions are wrong. I am just not seeing a logical argument.

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 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.

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.

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.

Why did you go to all this trouble? There are many in agreement. One such attempt is known as the Emission Hypothesis (or the ballistic theory of light), and was developed partly by Walther Ritz (C&N p.353). According to this theory, light behaves like bullets shot from a gun, its speed with respect to the source being a universal constant and independent of any ether. This idea is consistent with the null results of the Michelson-Morley experiment and many others. http://laser.phys.ualberta.ca/~egerton/specrel3.htm Modern Physics/Michelson-Morley Experiment Walter Ritz's emitter theory (or ballistic theory), was also consistent with the results of the experiment http://en.wikibooks.org/wiki/Modern_Physics:Michelson-Morley_Experiment This rules out any conceptually coherent ballistic theory of light propagation, according to which the speed of light is the vector sum of the velocity of the source plus a vector of magnitude c. Ironically, the original Michelson-Morley experiment was consistent with the ballistic theory, but inconsistent with the naïve ether theory, whereas the Sagnac effect is consistent with the naïve ether theory but inconsistent with the ballistic theory. Of course, both results are consistent with fully relativistic theories of Lorentz and Einstein, since according to both theories light is propagated at a speed independent of the state of motion of the source. http://www.mathpages.com/rr/s2-07/2-07.htm Now, this might seem to suggest that light is a disturbance in a material medium in which the objects A,B,C just happen to be at rest, but this is ruled out by the fact that it applies regardless of the state of (uniform) motion of those objects. Naturally this implies that the flash of light propagates isotropically with respect to the inertial rest coordinates of object D as well. http://www.mathpages.com/rr/s1-04/1-04.htm

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.

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) OK. 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. Merged post follows: Consecutive posts merged 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. Well, instead of making proclamations you cannot prove, let's see where I go. Here you are claiming LT is exactly the same thing as math theory. I am betting otherwise, I must put up or shut up. I say we find out whether I can do this one way or the other. I have 3 ways. I am proceeding with the easiest.

No this does not explain it. It is simple time dilation along the positive x-axis. I am not talking about the negative x-axis right now. So, does the clock at the origin of the moving frame elapse r/c when the clock in the stationary frame elapses rγ/c, or is time dilation false?

I am not operating in the moving frame. First, I send a beam down the positive x-axis at co-location of O and O'. Then, I allow rγ/c to elapse on the O clock. Time dilation forces r/c to elapse on the clock at the origin of the O' clock. Do you see this part?

This is is true for all x = ct? I think not. It is true for one light beam only and not for all mappings. One mapping has ξ² = c² τ², but for another, τ is a different value. How is that a sphere? Also, what does the clock read at the origin of the moving light sphere? 2.7 The Sagnac Effect Quantitatively, if we let w denote the angular speed of the loop, then the circumferential tangent speed of the end point is v = wR, and the sum of the speeds of the pulses and the receiver at the "end" point is c-v in the co-rotating direction and c+v in the counter-rotating direction. Both pulses begin with an initial separation of 2pR from the end point, so the difference between the travel times is http://www.mathpages.com/rr/s2-07/2-07.htm

Yes, they come up wit two different times. That means we do not have a light sphere. If SR is consistent, it should preserve the light sphere such that the result is a light sphere. Worse the clock at the origin of the moving frame moves with the frame and beats time dilated.

that is simple algebra. (r,0,0,r/c) results in t' = r/c(√( (c+v)/(c-v))) (-r,0,0,r/c) results in t' = r/c(√( (c-v)/(c+v))) So, the stationary light sphere does not map into the moving light sphere. In other words, for LT to be consistent, it must map to a light sphere in the moving frame with t' = r/c from the stationary light sphere. LT does not do this. Further, LT does not map the origin of the light sphere frame to frame. So, under LT, the origin is not mapped and there does not exists a time in the stationary frame such that the moving frame will see a light sphere. In fact, on the time interval [r/c(√( (c-v)/(c+v))), r/c(√( (c-v)/(c+v)))] LT claims the moving frame is seeing r/c.

This does not explain anything. Have you considered something more obvious? Sagnac is measuring two one way light travels. MMX measures round trip. If it is the case that Sagnac measures basically, c+v and c-v because of light path differentials, then MMX would measure (c+v) + (c-v) /2 = c, a constant.

Yes, I meant 4-D.