My Problems With Relativity

[tomh4040]

I am understanding you so far. All this is according to relativity.
The Sagnac effect can be seen on Earth by splitting a signal and sending it round the Earth in opposite directions, as per the Wiki link. As the signal is traveling round, the Earth rotates, moving the Rx towards one signal and away from the other. I have done the calculations for this, and here they are. Taking the equatorial circumference as 40075000 M, and c to be 299792458 M/s, the equatorial speed of the surface is 463.8 M/s, and the signal (light or radio) takes 0.1336758 seconds to circumnavigate. During that time, the Earth has moved 61.9988 M, so shortening one path and lengthening the other by that amount. Looking at the Earth from above the north pole, it rotates anti-clockwise, so the signal going anti-clockwise or W to E has to travel 40075061.9988 M, which is 40075000 M plus 61.9988 M traveled during the signal travel time. The signal going E to W travels 40074938.0012 M. The rotation of the Earth has the effect of the E to W signal arriving at the Rx before the W to E signal. This can be seen from the Rx on the surface, not just by an observer at the axis of rotation. If the observer measures the two signal speeds relative to the Earth’s surface, he will find that they are not c, but are 299,792,946.8 M/s and 299,791,994.5 M/s respectively. One other point to note is that this effect is dependant on the fact (according to relativity) that the Rx moves during the transit time of the signal, and moreover, moves at 463.8 M/s. As the Rx is moving, there is a Doppler shift present at the Rx. Bear in mind that the Doppler shift is not dependant on distance, and can be seen whenever there is a difference in speed between Tx and Rx. The speed difference in this case is 463.8 M/s, giving a Doppler shift of 2,011 Hertz when the transmitter is a radio with a frequency of 130 meg. Note in the above description that there is no fixed speed for light, it is variable WRT the Earth.
There is therefore a big problem which will now be outlined.

As I live at approximately 54 degrees north, I have calculated the speed of rotation at that latitude, and it is 272.6 M/s. The experiments I have done have been at that latitude with that rotational speed. If a signal was to be sent round the Earth at this latitude, it would take 0.0785824 sec to circumnavigate. The rotational speed is 272.6 M/s, so during the transit time the Rx has moved 21.42 M, giving a Doppler shift of 118.2 Hertz at 130 meg. This is in accordance with the above paragraph.

I own various items of radio equipment: transmitters and receivers, a frequency counter, and a beat frequency oscillator. Transmitting at 130 meg, the frequency counter showed 129,999,273, sometimes fluctuating above or below that figure by 4 Hertz. This means that my transmitter (or frequency counter) is off frequency by 727 Hertz, but stable to within 8 Hertz. With the BFO connected, a tone of 727 Hertz was heard, which did not change as the orientation between the Tx and Rx was changed.

As the fluctuating frequency changes were well within 10 Hertz, the expected Doppler shift of 118 Hertz would have shown up easily if present. The fact is that there is no Doppler shift. The speed of light is a constant WRT the Earth – it is the same in all directions. To put this another way, the speed of light is constant on a rotating reference frame (RFR).
This was totally expected, and is borne out by other everyday occurrences. If the speed of the radio signal, or light, varies depending on whether it is going W to E or E to W, this would show up easily in for instance radar. At my latitude a policeman’s radar speed gun would show a brick wall speeding along at 272.6 M/s if it was east or west of him, and stationary if it was north or south of him. Air traffic control radar would be useless for the same reason. As the radar signal is constant WRT the Earth over a short distance, it must be the same over a longer distance, and if arranged to split and circumnavigate the Earth in both directions as above, both beams would arrive back at the starting point at the same time.

[QUOTE= HMHW]
The effect is much too small to be observable for an interferometer on the scale of the Michelson-Morley apparatus; besides, it is only observable in geometries enclosing a certain area. However, when they made it bigger, the effect was indeed observed.
[/QUOTE]

That reply is both interesting and incorrect. The MMX was specifically set up to look for the motion of the Earth through the aether. Michelson and Morley certainly knew what they were doing – they were looking for second order effects. If the effect is only observable in geometries enclosing a certain area (note how HMHW is being very vague again, he has not specified what or where this certain area is), then making the MMX bigger will not make the effect appear. The plain fact is however, that if such an effect did exist, it would be seen at the Rx, and by anybody who could see the Rx, no matter where they were on the planet. As pointed out above, if the speed of light is not a constant relative to the RFR of the Earth, radar would be useless.

[ZenBeam]

@tomh4040

As I live at approximately 54 degrees north, I have calculated the speed of rotation at that latitude, and it is 272.6 M/s. The experiments I have done have been at that latitude with that rotational speed. If a signal was to be sent round the Earth at this latitude, it would take 0.0785824 sec to circumnavigate. The rotational speed is 272.6 M/s, so during the transit time the Rx has moved 21.42 M, giving a Doppler shift of 118.2 Hertz at 130 meg. This is in accordance with the above paragraph.

I own various items of radio equipment: transmitters and receivers, a frequency counter, and a beat frequency oscillator. Transmitting at 130 meg, the frequency counter showed 129,999,273, sometimes fluctuating above or below that figure by 4 Hertz. This means that my transmitter (or frequency counter) is off frequency by 727 Hertz, but stable to within 8 Hertz. With the BFO connected, a tone of 727 Hertz was heard, which did not change as the orientation between the Tx and Rx was changed.

Are you claiming to have transmitted a 130 MHz signal, and to have received it after it traveled around the Earth?

[Great_Antibob]

@ZenBeam

Are you claiming to have transmitted a 130 MHz signal, and to have received it after it traveled around the Earth?

Yeah, his numbers indicate he thinks he’s transmitted a signal around the round.

Let’s fix the numbers, assuming the signal takes the more sensible path DIRECTLY to the receiver.

Assume tomh4040 has better than average resources at his disposal and can position the transmitter and receiver 1km apart. That means the receiver has moved roughly 91 microns in that time.

The obvious :smack: doesn’t do this botched thought experiment any justice.

[naita]

No, I don’t think that’s his claim. I think he’s mangling basic physics again.

A Doppler shift depends on a difference in speed between receiver and transmitter, how larger are those differences in your setup, Tom?

The Sagnac effect is the result of waves moving in opposite directions around a rotating loop, does your setup involve waves moving in opposite directions around a rotating loop, Tom?

[Half_Man_Half_Wit]

@tomh4040

As the Rx is moving, there is a Doppler shift present at the Rx.

Nope, Doppler shifts cancel exactly:
[

Also, Doppler effects cancel out, so the Sagnac effect does not involve Doppler effect. In the case of ring laser interferometry it is important to be aware of this. When the ring laser setup is rotating the counterpropagating beams undergo frequency shifts in opposite directions. This frequency shift is not related to Doppler shift.

](Sagnac effect - Wikipedia)

@tomh4040

That reply is both interesting and incorrect. The MMX was specifically set up to look for the motion of the Earth through the aether. Michelson and Morley certainly knew what they were doing – they were looking for second order effects. If the effect is only observable in geometries enclosing a certain area (note how HMHW is being very vague again, he has not specified what or where this certain area is), then making the MMX bigger will not make the effect appear. The plain fact is however, that if such an effect did exist, it would be seen at the Rx, and by anybody who could see the Rx, no matter where they were on the planet. As pointed out above, if the speed of light is not a constant relative to the RFR of the Earth, radar would be useless.

Look, at least read the references I provide. The Sagnac effect shows up if the geometry of the interferometer circumscribes a certain area – if the light goes ‘around a loop’. It’s the ‘A’ in the formula for the time difference Δt given here. This is necessary to detect the rotational acceleration. So, the Michelson-Morley experiment will not detect any difference in the speed of light. However, making the experiment bigger, and modifying it such that it does enclose an area with its arms, as the wiki article I linked you to very clearly describes, makes the effect detectable.

[tomh4040]

Zenbeam, that is a very good example of a ridiculous answer.

@Great_Antibob

Let’s fix the numbers, assuming the signal takes the more sensible path DIRECTLY to the receiver.
Assume tomh4040 has better than average resources at his disposal and can position the transmitter and receiver 1km apart. That means the receiver has moved roughly 91 microns in that time.

Bear in mind that the Doppler shift is not dependant on distance, and can be seen whenever there is a difference in speed between Tx and Rx. The speed difference in this case [at the equator] is 463.8 M/s, giving a Doppler shift of 2,011 Hertz when the transmitter is a radio with a frequency of 130 meg. Note in the above description that there is no fixed speed for light, it is variable WRT the Earth.
The above paragraph was in my last post.
It does not matter how far the Rx has moved, what causes the Doppler shift is how fast it is moving. Everybody who believes relativity believes that the Rx has moved WRT the advancing signal during the transit time of the signal. This causes a Doppler shift. I did not do a thought experiment, it was an actual hardware experiment, and although I did not have the Tx and Rx very far apart when I did the experiment at home, the same principle is used in radar, which does have a tremendous range. Please tell me why no Doppler shift shows up in a radar return when the radar reflects from a distant stationary object which is east or west of the transmitter, even though it is moving at [at my latitude] 272.6 M/s WRT the advancing signal.

naita No, I don’t think that’s his claim. I think he’s mangling basic physics again.

A Doppler shift depends on a difference in speed between receiver and transmitter, how larger are those differences in your setup, Tom?

I assume that was a typo - do you mean “how much larger”? What differences are you referring to?
I am using the rotational speed of the Earth at 54 degrees latitude, which is 272.6 M/s. The signal, on leaving the Tx, travels east or west to the Rx which relativists claim is moving at 272.6 M/s. This causes a Doppler shift. Now consider radar. Why does this Doppler shift not show on radar? Question for HMHW. What cancels the Doppler shift in radar?
For radar to work, the speed of the signal must be the same in all directions, anywhere on Earth. This of course means that a radar signal going W to E has the same speed as a radar signal going E to W relative to the Earth. If allowed or reflected to go all the way round the Earth, the two signals would arrive back together at the starting point. One would not arrive before the other. The same applies to a light signal.

[ZenBeam]

@tomh4040

Zenbeam, that is a very good example of a ridiculous answer.

Garbage in, garbage oujt.

[naita]

@tomh4040

I assume that was a typo - do you mean “how much larger”? What differences are you referring to?
I am using the rotational speed of the Earth at 54 degrees latitude, which is 272.6 M/s. The signal, on leaving the Tx, travels east or west to the Rx which relativists claim is moving at 272.6 M/s. This causes a Doppler shift. Now consider radar. Why does this Doppler shift not show on radar? Question for HMHW. What cancels the Doppler shift in radar?
For radar to work, the speed of the signal must be the same in all directions, anywhere on Earth. This of course means that a radar signal going W to E has the same speed as a radar signal going E to W relative to the Earth. If allowed or reflected to go all the way round the Earth, the two signals would arrive back together at the starting point. One would not arrive before the other. The same applies to a light signal.

No, I meant how large. Your Rx and Tx have the same speed, and nearly the same velocity as they are relatively close. Doppler shift is caused by a difference in radial velocity. How large is the difference in radial velocity between your Rx and Tx?

You’re attacking Relativity by presenting ideas that are true in nearly every theoretical framework, including Relativity and an entrenched ether.

[Great_Antibob]

@tomh4040

Please tell me why no Doppler shift shows up in a radar return when the radar reflects from a distant stationary object which is east or west of the transmitter, even though it is moving at [at my latitude] 272.6 M/s WRT the advancing signal.

As other posters have noted, Doppler shift comes from a difference in radial velocities.

So, yes, distance (North-South, at any rate) can make a difference. If they are close together, there is no reason to expect any significant Doppler shift.

Even in classical physics, this is true. Your thought experiment still demonstrates a lack of understanding of even a Newtonian physical framework.

And if you performed the actual experiment, the lack of an apparent Doppler shift is not inconsistent with physics (classical or relativistic) as it is generally accepted.

[Half_Man_Half_Wit]

@tomh4040

Question for HMHW. What cancels the Doppler shift in radar?

For a Sagnac effect to exist, as I stated, the geometry must enclose an area, which is not the case in the radar geometry.

For a linear Sagnac effect to exist, the acceleration must be parallel to the velocity; in the case of rotation on the surface of the Earth, it is perpendicular. So you can treat sender and receiver as being in an (approximately) inertial frame of reference, in which of course no Doppler shift is to be expected (otherwise, you could, if only the sender and receiver existed, determine your state of motion absolutely, violating both Galilean and Einsteinian relativity).

What you’re claiming is akin to saying that if you throw a ball towards me, which hits me in the head and bounces back to you, is different when carried out in motion than when carried out at rest: if you throw the ball at v, it bounces back at v (assuming my mass much larger than the ball’s); but when I’m moving at v’, then the ball bounces back at v + v’. But you forget that you’re moving at v’, as well, otherwise, we’d get closer to one another; so the v’ just drops out. It’s the same with radar waves bouncing back and forth.

Even if the above weren’t true, and the motion of the Earth introduced a speed difference of dv to the speed of light (it doesn’t), you wouldn’t see any effect on radar: from the sender to the target, light would propagate at a speed of c - dv if traveling in the same direction as the Earth rotates; however, from the target back to the sender, it would travel at c + dv, so it’d take the same time as if it had traveled at c.

[tomh4040]

[QUOTE=HMHW]
What you’re claiming is akin to saying that if you throw a ball towards me, which hits me in the head and bounces back to you, is different when carried out in motion than when carried out at rest: if you throw the ball at v, it bounces back at v (assuming my mass much larger than the ball’s); but when I’m moving at v’, then the ball bounces back at v + v’. But you forget that you’re moving at v’, as well, otherwise, we’d get closer to one another; so the v’ just drops out. It’s the same with radar waves bouncing back and forth.
[/quote]

You are trying to have it both ways here. When talking about light going round the Earth, you are saying that the Rx is moving at 463.8 M/s (at the equator) relative to the advancing light beam, so the light has to catch up with it (or reaches it sooner depending on which way round it is going. When talking about radar (the ball is representative of the radar), you are saying that the Rx does not move relative to the advancing radar beam (the ball).

@HMHW

Even if the above weren’t true, and the motion of the Earth introduced a speed difference of dv to the speed of light (it doesn’t), you wouldn’t see any effect on radar: from the sender to the target, light would propagate at a speed of c - dv if traveling in the same direction as the Earth rotates; however, from the target back to the sender, it would travel at c + dv, so it’d take the same time as if it had traveled at c.

Your assertion is that a light signal split into two beams and transmitted round the world both W to E and E to W would arrive back at its starting point at different times. If the light signals arrive back at the starting point at different times, which you say they do, this means that the speed over the ground was different in different directions. If the two signal speeds are measured WRT the Earth’s surface, they are not c (299,792,458 M/s), but are 299,791,994.2 M/s W to E, and 299,792,921.8 M/s E to W. As these are not the same, they can be referred to as vl WE and vl EW (velocity of light west to east and east to west).

I have done the calculations to get a definitive answer, and here it is. As the Tx and Rx are on a fixed rotating platform (the Earth), they are a fixed distance apart, which in the case of radar is just a matter of a few millimeters. The different speed of light in different directions is what causes the frequency changes.
I will use radar at the equator, operating at 10 Gig (10,000,000,000 Hz). The radar is pointed at a distant stationary target which is to the east. The speed of light (and radar) going W to E is 299,791,994.2 M/s ( vl WE ) relative to the Earth. As the transmitter, the target, and the receiver are all stationary relative to the Earth, it can also be said that the speed of light is relative to the receiver or any observer on the Earth.

vl WE = 299,791,994.2 M/s
vl EW = 299,792,921.8 M/s

wl (wavelength) = vl WE / f = 299791994.2 / 10000000000 = 0.02997919942 M

The velocity is 299,791,994.2 M/s, the frequency is 10,000,000,000 Hz, the wavelength is 0.02997919942 M.
The radar pulse hits the target and reflects back, now going E to W. It is traveling back to the source at the higher speed of 299,792,921.8 M/s. This introduces a shift in frequency due to the change of speed.

f = vl EW / wl = 299792921.8 / 0.02997919942 = 10,000,030,941.4 Hz

This pulse arrives back at the radar receiver, there is no further shift in frequency so the total frequency shift is 30,941.4 Hz.
I was wrong in a previous posting when I said that the frequency change would be equivalent to the rotational speed of the Earth. I hadn’t done the maths with a calculator. The result surprised me at first, I was not expecting double the shift.

Transmitted radar frequency 10,000,000,000 Hz, received frequency 10,000,030,941.4 Hz. The frequency change due to the change of vl is 30,941.4 Hz. This is representative of a speed of 927 M/s, and is due to the difference in velocity of the radar pulse in different directions causing a frequency shift. This shift is additive, hence twice the rotational speed is shown.

This does not happen. In reality, there is no frequency shift between transmitted and received pulse when using a fixed object as target. The speed of light is homogenous and WRT the Earth. The Earth is an RFR (rotating frame of reference). This conclusion is at odds with the relativists’ claim that the speed of light is not homogenous WRT an RFR.

All the above calculations and conclusions are correct if your assertion is correct that a beam of light which is split and sent in opposite directions round the Earth, arrives back at the origin at different times because of the Earth’s rotation.

[naita]

@tomh4040

All the above calculations and conclusions are correct if your assertion is correct that a beam of light which is split and sent in opposite directions round the Earth, arrives back at the origin at different times because of the Earth’s rotation.

No, they are not.

Are you now disputing that the sagnac effect is real? Do you have an alternative explanation to it being caused by a difference in clockwise and counter-clockwise loop length on a rotating Earth?

[BunnyTVS]

@tomh4040

This does not happen. In reality, there is no frequency shift between transmitted and received pulse when using a fixed object as target. The speed of light is homogenous and WRT the Earth. The Earth is an RFR (rotating frame of reference). This conclusion is at odds with the relativists’ claim that the speed of light is not homogenous WRT an RFR.

This site says there is a phase shift defined by the frequency of the lightwave and the distance between the objects. In fact the phase shift is based on 2r, which is what you discovered in your calculations.

It is the difference in phase between consecutive pulses that forms the basis for doppler radar, which provides velocity information, as well as range and bearing.

[tomh4040]

@naita

No, they are not.

You cannot just say no, please point out where my calculations are wrong.

@naita

Are you now disputing that the sagnac effect is real? Do you have an alternative explanation to it being caused by a difference in clockwise and counter-clockwise loop length on a rotating Earth?

The Sagnac effect is real, I have never disputed that. What is under discussion here is that if the Sagnac effect exists for a light pulse split and sent round the Earth W to E and E to W (which has never been done at the Earth’s surface), then the speed of light is different in the W to E direction than in the E to W direction WRT the Earth itself. This means that a radar pulse reflected from a distant target which is east or west of the radar station undergoes a change of speed when it is reflected. This change in the speed of the radar pulse will cause a change in frequency at the receiver which will show as a non zero speed for a fixed target on the radar screen. This does not happen. Please explain why.

[Trinopus]

@tomh4040

. . . This means that a radar pulse reflected from a distant target which is east or west of the radar station undergoes a change of speed when it is reflected. This change in the speed of the radar pulse will cause a change in frequency at the receiver which will show as a non zero speed for a fixed target on the radar screen. This does not happen. Please explain why.

I may be wrong, but I believe that radar beams do not change their velocity; it is always the speed of light. Radar beams going to or from objects which are advancing or receding will change their energy (and therefore their frequency) and that is how Doppler radar detects the radial velocity of things like thunderstorms or enemy bomber aircraft.

You’re right when you say the frequency changes. I think you are incorrect when you say the beam’s velocity changes. More expert responses should follow.

(FWIW, I only answer regarding things I know, or am pretty sure of. I wouldn’t comment on the Sagnac Effect, because I don’t know enough about it.)

[naita]

I’ll reply one more time, but I’m really going to abandon the thread this time.

@tomh4040

The Sagnac effect is real, I have never disputed that. What is under discussion here is that if the Sagnac effect exists for a light pulse split and sent round the Earth W to E and E to W (which has never been done at the Earth’s surface),

The Michelson–Gale–Pearson experiment is equivalent to this, even if it the loop doesn’t include the Earth’s pole.

then the speed of light is different in the W to E direction than in the E to W direction WRT the Earth itself.

Not really. The velocity of light is different, and you need a distance large enough for the velocity differences to matter. The speed is the same.

This means that a radar pulse reflected from a distant target which is east or west of the radar station undergoes a change of speed when it is reflected.

Dealing with rotational reference frames is a bitch, but I don’t think I’d describe it like that.

This change in the speed of the radar pulse will cause a change in frequency at the receiver which will show as a non zero speed for a fixed target on the radar screen. This does not happen. Please explain why.

What matters when looking for frequency shifts is the differences in radial velocity. Any transmitter or receiver on the Earth’s surface will obviously have the same absolute magnitude of velocity, the same speed, but no point has the same velocity. (Assuming a smooth surface, two points on the same vector normal to the axis of rotation will.)

To find the expected frequency shift we need to see if that difference in velocity translate to a difference in radial velocity.

Let’s start with a transmitter and receiver quite close to each other. They’ll be moving with the same speed, and as they are close together their velocities will only differ insignificantly. This is true in any system of reference, an inertial flyby, a co-rotating observer etc. No frequency shift observed or expected.

Now move them further apart. Let’s say they’re at the equator, for simplicity’s sake, and we move them so they are an angle θ apart as measured from the centre of rotation.

I first assumed they were not so far apart they moved significantly during the experiment, the point remains the same, but I could ignore the angular velocity’s influence on θ. Having done that and then introduced the angular velocity, I realised this is geometrically equal to a larger initial separation, so the following applies in both cases.

If we initially transmit in the direction of rotation, the transmitter has a velocity θ/2 above line of sight, and the receiver has a velocity θ/2 below the line of sight. The radial velocity difference is then zero, which is not unsurprising as they’re not moving closer or further apart. With no difference in radial velocity, there’s no frequency shift.

What’s left then to explain the Sagnac effect is the differences in path length. And show that the path in the direction of rotation is longer than one against the direction of rotation. Here’s where I preceive one of your issues lie, as you feel the arguments for the absolute nature of rotation is insufficient and that the shorter path is some sort of relativity trickery.

But the differences in path length also exist when we dont have a loop, but just a transmitter and receiver separated by the angle θ. Since their actual velocity is different you have to aim above the receiver to hit it when aiming against the direction of rotation, a longer path than direct line of sight, and below the receiver to hit it when you’re aiming in the direction of rotation, a shorter path than direct line of sight, at least for small angles, and always shorter than the path in the opposite direction. What this gives us is a difference in travel time, which gives us a fringe shift only in a closed loop.

I’d scan the geometric scribblings I used for this, but they’re not very legible and I’ve spent too much time on this post as is.

[naita]

@naita

Not really. The velocity of light is different, and you need a distance large enough for the velocity differences to matter. The speed is the same.

What matters when looking for frequency shifts is the differences in radial velocity. Any transmitter or receiver on the Earth’s surface will obviously have the same absolute magnitude of velocity, the same speed, but no point has the same velocity. (Assuming a smooth surface, two points on the same vector normal to the axis of rotation will.)

And assuming we’re not talking about points at different latitudes. :smack:

I’m so not going to mess with 3D geometry to prove this exhaustively.

[tomh4040]

@Trinopus

I may be wrong, but I believe that radar beams do not change their velocity; it is always the speed of light. Radar beams going to or from objects which are advancing or receding will change their energy (and therefore their frequency) and that is how Doppler radar detects the radial velocity of things like thunderstorms or enemy bomber aircraft.

You’re right when you say the frequency changes. I think you are incorrect when you say the beam’s velocity changes. More expert responses should follow.

(FWIW, I only answer regarding things I know, or am pretty sure of. I wouldn’t comment on the Sagnac Effect, because I don’t know enough about it.)

Thank you for an honest answer Trinopus. It has already been said on this forum that a light beam which is split and sent round the Earth in opposite directions (at the equator) will not arrive back at the starting point simultaneously because of the rotation of the Earth. The beam going W to E takes longer than the beam going E to W. This means that WRT the Earth, the two beams are traveling at different speeds, if they were traveling at the same speed, they would arrive together.
Radar is electromagnetic, the same as light. Therefore the speed of a radar pulse sent W to E is 299,791,994.2 M/s, and the speed of a radar pulse sent E to W is 299,792,921.8 M/s (taking c to be 299,792,458 M/s, ignoring the index of refraction of the atmosphere). When the radar pulse is reflected from a distant stationary object which is east or west of the transmitter, its speed is changed. This causes a corresponding change in frequency, which is interpreted by the radar as a speed. In other words, if relativity is correct, the returning radar pulse would show the stationary object to be moving. This does not happen.
See my post on 04-08-2012 05:14 AM for the maths involved.

[QUOTE=naita]

Quote from tomh4040. “This change in the speed of the radar pulse will cause a change in frequency at the receiver which will show as a non zero speed for a fixed target on the radar screen. This does not happen. Please explain why.”

What matters when looking for frequency shifts is the differences in radial velocity.
[/QUOTE]

Your sentence “What matters…” is correct, but only if you assume a constant speed of light. You cannot get round the fact that if two beams of light start off at the same time, travel round the Earth, and arrive back at different times, then their speed is different. Two cars which set off from the same starting point, go round a circular track in opposite directions and do not arrive back at their starting point together have been traveling at different (average) speeds. That is indisputable. The same is true of light.
A change in the speed of light will cause a change in frequency, just as a difference in radial velocity will.
Please show me where my maths is wrong to back up your statement “No, they are not.”

[Trinopus]

@tomh4040

Thank you for an honest answer Trinopus. It has already been said on this forum that a light beam which is split and sent round the Earth in opposite directions (at the equator) will not arrive back at the starting point simultaneously because of the rotation of the Earth. The beam going W to E takes longer than the beam going E to W. This means that WRT the Earth, the two beams are traveling at different speeds, if they were traveling at the same speed, they would arrive together.

This, I think, oversimplifies. It isn’t merely going EW or WE. It’s going in a combination of all directions. The beam of light has to be going in the same direction as the earth’s rotation, at some point, and counter to the earth’s rotation, at another point, and then across the direction of the earth’s rotation at yet other points. It has to go “around” a very large circle.

A radar beam, on the other hand, just goes straight out and straight back. The earth’s rotation is not relevant to that kind of path. The pathway may be within a rotating frame of reference, but the beam isn’t going “around” anything.

If you take a circle, and flatten it, it becomes an ellipse. Flatten it further, and it is an ellipse with very high eccentricity. The limit is an “ellipse” of infinite eccentricity, which is simply a perfectly flat line. It isn’t really an “ellipse” any more. In the same way, a degenerate triangle can appear to be a straight line.

The Sagnac Effect (from what I can tell) applies to figures that are not degenerate, but which are round, or octagonal, or even square. It doesn’t seem to apply to straight lines. Thus, the radar beam example is not appropriate.

Again, I very much hasten to add that I may be wrong. None of the Sagnac Effect articles I’ve seen show a simple “there and back” light trajectory, but always show a two dimensional figure.

(Think of it this way: if the figure is a straight line, then what is the difference in going “around” it clockwise vs. counterclockwise? The two would be exactly the same!)

[tomh4040]

@Trinopus

This, I think, oversimplifies. It isn’t merely going EW or WE. It’s going in a combination of all directions. The beam of light has to be going in the same direction as the earth’s rotation, at some point, and counter to the earth’s rotation, at another point, and then across the direction of the earth’s rotation at yet other points. It has to go “around” a very large circle.

The light is reflected by mirrors so obviously this has never been done. If it goes W to E or E to W round the equator or a line of latitude, it is going with or against the earth’s rotation, and in no other direction.

A radar beam, on the other hand, just goes straight out and straight back. The earth’s rotation is not relevant to that kind of path. The pathway may be within a rotating frame of reference, but the beam isn’t going “around” anything.

If light takes a different length of time to traverse the Earth in different directions, then its speed is different in different directions. Take a part of the path of light round the Earth and keep it to within line of sight say horizon to horizon (in the Wiki example this could be between mirrors), and it is still going faster E to W than W to E. A radar pulse will obviously be the same - it will have two different speeds.

The Sagnac Effect (from what I can tell) applies to figures that are not degenerate, but which are round, or octagonal, or even square. It doesn’t seem to apply to straight lines. Thus, the radar beam example is not appropriate.

We are not talking about the Sagnac effect or the Doppler shift now, we are talking about the speed of light being different in different directions, and the effect that that has on the transmitted frequency.
What I am saying in this section assumes the validity of relativity. Let’s recap briefly to show how we arrived here.
The Sagnac effect shows up in an interferometer, which, while it is rotating, is a rotating frame of reference or an RFR. The laser gyro works on that principle. If I hold it in my hand and rotate it left or right, it drives a motor left or right (in my application). This happens because the receiver is moving away from one beam and towards the other, causing a difference in arrival times of the beams due to the path length changing (this is treating the speed of light as c WRT any IFR). Look at those light beams from within the laser gyro. WRT the gyro, the speed of light is different in different directions. The reason for that is that the path lengths are identical WRT the gyro itself, and yet light has reached the receiver from one direction before the other. If the time of transmission is the same, and then the beam is split to go in opposite directions round identical paths, and the times of reception are different, then the speeds are different.
Now we change our focus to the Earth. The Earth is also an RFR, and must obey the same rules as any other RFR, so that all that was said above about the gyro applies to the Earth if we restrict our discussion to going round it at the equator, or round any path parallel to it (line of latitude).
The Wiki example linked to previously, shows that a beam of light split and sent round the equator W to E and E to W has different arrival times due to the fact that the Earth is rotating. We immediately see that WRT the Earth, the speed of light E to W is faster than W to E.
As HMHW pointed out, in any round trip such as radar, this difference would average out to the speed of light - c ( or to be precise c/n). What does not average out is the change of frequency cause by the change of speed. This is not a Doppler shift. Wiki quote :- “The Doppler effect (or Doppler shift), named after Austrian Christian Andreas Doppler who proposed it in 1842 in Prague, is the change in frequency of a wave for an observer moving relative to the source of the wave.” When Christian Doppler discovered this shift, he assumed that the speed of light really was a constant, so the only way this frequency change could come about (in light) was as he proposed it.

If relativity is correct, then the speed of light is faster E to W than W to E relative to the Earth. According to relativity, when the radar pulse reflects from a stationary target which is to the east or west, it speed changes. This causes a shift in frequency. This would cause a stationary target to show as moving. See my calculations in posting #941. This does not happen. We know that radar works, so the speed of light must be the same W to E as E to W.

Now we have a big problem for relativity. The laser gyro does work, so if it works due to different arrival times of the light beams when measured from an IFR (which when measured from within the LG are two different speeds), then light sent round the Earth will also arrive at different times due to being at two different speeds when measured WRT the Earth. If light has a different speed E to W than W to E, which must be the case as the Earth is also an RFR, then radar will not work. Radar does work, so the speed of light is constant WRT the Earth. How can the speed of light be constant for one RFR and not constant for another RFR?