The extra cycles are there in a very similar way to where they come from when the radar pulse reflects from a target which is approaching the radar station in an IFR. In that case, the speed of light is the same in both directions, and as the target moves closer, each successive wave peak is reflected from slightly closer and has a shorter distance to travel back to the radar station. The received frequency is higher than that which was transmitted. There are plenty of articles covering the Doppler shift.
The Earth has two speeds for light if we restrict our discussion to the equator and east west travel. As the radar pulse is reflected from the stationary target which is east of the radar station, its speed increases because the pulse is now traveling E to W. When that pulse is received back at the radar aerial, each successive peak is received at an earlier time because of the higher speed. The received frequency is higher than that which was transmitted. This is not a Doppler shift, although it is very similar. The Doppler shift applies only in an IFR to a target moving relative to the source (in this case, the radar station). The principle of relativity (in the restricted sense) applies here. When the radar pulse hits the target and is reflected, there can be no difference in the observed effect between the target moving and assuming a constant speed of light, or the target being stationary and assuming the speed of light to be changing.
This effect is observed in electro magnetics, if a coil is held stationary and a magnet passed by it, a current is induced in the coil. If the magnet is held stationary and the coil passed by it, a current is induced in the coil. In both cases the current is of the same magnitude and polarity, and it cannot be decided which action caused it.
You may have noticed my agreement with Einstein here. Some of what he said makes sense, but not all of what he said makes sense.
I have figured it out for myself, and this is the logical answer. I cannot see any flaw in my arguments. If relativity is correct, a ring laser will work, but radar will not. They both work, so something is wrong somewhere.
The target has no radial velocity relative to the transmitter. Radial velocity, the thing a Doppler radar measures. Relative, the very essence of relativity. A frequency shift is not expected in relativity.
It’s not expected without relativity either.
I knew this would be pointless, but you were so close in your post.
No, they aren’t.
In a Doppler shift, the extra cycles come from the space between the source and the target. If the target is coming closer, the number of cycles in that space will decrease. That is where the extra cycles come from. That’s not the case for a stationary target.
Suppose the target is only ten feet away from the radar. The radar can only transmit ten cycles at 1 GHz before the target starts reflecting. Leave the radar on. If the reflected frequency is higher than the transmitted frequency, eventually the number of reflected cycles will be ten more than have reached the target. That’s the same as the total cycles transmitted from the radar. After that, there will then be more reflected cycles than the number the radar has ever transmitted. This is clearly absurd.
What is wrong is your belief that the reflected frequency changes in this situation.
This is, of course, entirely possible. I urge you to take into account the possibility that you, yourself, are, likewise, not as fully educated on the matter as you might believe.
Since we don’t have Stephen Hawking or Roger Penrose here, we’ll have to continue to do the best we can.
Missing a word; the speed of light is invariant relative to any and all IFRs.
All correct, as I understand it. In earlier posts, you tried to “collapse” the ring of lasers into a straight-line apparatus, sending light due east-west and then back again due west-east. By shrinking the ring into a line, this caused the Sagnac experiment to become the Michelson-Morley experiment, and thus the Sagnac effect failed to materialize.
The Sagnac effect appears in rings, but not in straight-line EW/NS interferometers.
If all the relativists say one thing, and I say another, I am likely to be wrong.
To date, you have not shown relativity to be wrong. I’m not important. I’m just a middlin’ educated guy who is interested in the subject and who knows a few things. I can do Lorenz transform math. I can solve the “telephone pole and the barn” problem. I’d never heard of the Sagnac Effect until this thread.
You can do whatever you want to me; but since your beef is with Einstein, you really need to produce something that refutes him, not just me.
[nitpick]“Wabbit Season”[/nitpick]
FWIW, I’m so tired by now, just checking in every now and then and catching up. I thought I’d catch some nice two-way Socrates-Plato thing,
with nicely interchanging roles.
I sensed from the first page that one role player, attitudinally, should continue waiting tables at high-class restaurants.
Just jumping in to say that this thought-experiment shows pretty clearly that tomh4040 is wrong on this point, and so I think it deserves emphasis. (I am assuming you are referring to the case of a stationary source/target/receiver in the earths RFR). I am divining that tomh4040 is basing his argument on the fact that the speed of light is c1 as it arrives at the target, c2 after it is reflected, and due to the relationship c=λν, the frequency change must be (c1-c2)/λ. But this is assuming that λ stays the same, among other things. Nothing is so simple in a RFR.
BTW, can anyone enlighten me as to what tomh4040 is referring to when he says things like “If relativity is correct, a ring laser will work, but radar will not.”? It must be lost in this long train-wreck somewhere, but I can’t figure out what precisely are the two experimental setups that tomh4040 keeps referring to. Can someone (please not tomh4040) provide an executive summary of what they think he is saying?
A ring laser interferometer has a loop where light is passed from a transmitter in both directions back to a dector at the origin, and if it’s rotating you get a fringe pattern because, as viewed from any IRF, the path lengths are different.
From the reference frame of the transmitter and detector one could say that the speed of light differs in the two directions. tomh4040 thinks this should result in a doppler shift if you just shoot a radar at a mirror.
OK so he is saying that if if one were to be in the RFR somewhere along the loop of a sagnac interferometer, that one could set up a mirror along the loop, shoot a radar at the mirror, and that the frequency of the incident light would be different from the reflected light. In essence, he is giving an example meant to illustrate that the general result that a sagnac interferometer only finds interference if the loop encloses a non-zero area, is wrong. In his example, the path taken by the radar encloses zero area, and so should show zero interference. But he is creating a different experiment, one in which one must include the radar’s reflection off the mirror in the calculation. What actually happens is that as each wave pulse reflects off the mirror at a greater speed than incident, the wave pulse stretches out (the front of the wave pulse has a higher speed than the back of the wave pulse for a small stretch of time) and the increase in wavelength exactly compensates for the increase in speed such that the ratio c’/λ’=ν remains constant. In this way energy is conserved, and there is no doppler shift.
Without relativity it definitely will not happen. The speed of light will be constant WRT the Earth, the local gravitational field, or the entrained aether.
The Earth is an RFR, not an IFR, and the speed of light is not a constant relative to an RFR. This is what causes the entry/exit point to move during the transit time of light circumnavigating the globe. See my posting yesterday at 09:08 followed by ZenBeam at 10:00 .
This is part of it :- According to Wiki, “…if a system of mirrors (or other device to steer the light around the globe) is set up, and light sent round the Earth at the equator or along a line of latitude, it will arrive back at the entry/exit point at different times because the Earth has rotated during the transit time of the light.”
Also from Wiki :- “When the platform [the Sagnac interferometer] is rotating, the point of entry/exit moves during the transit time of the light. So one beam has covered less distance than the other beam. This creates the shift in the interference pattern. Therefore, the interference pattern obtained at each angular velocity of the platform features a different phase-shift particular to that angular velocity. In the above discussion, the rotation mentioned is rotation with respect to an inertial reference frame.”
As the signal is traveling round, the Earth rotates, moving the Rx towards one signal and away from the other. This is according to relativity. 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.
All this is agreed by your fellow posters. The target does have a radial velocity. If it did not, there would be no Sagnac effect.
[Quote=ZenBeam]
Originally Posted by ZenBeam
I could provide an explanation, but it would be pointless to do so.
I knew this would be pointless, but you were so close in your post.
[/quote]
If by so close, you mean so close to disproving relativity, I am not just so close, I am there.
[Quote=ZenBeam]
:
The extra cycles are there in a very similar way to where they come from when the radar pulse reflects from a target which is approaching the radar station in an IFR.
No, they aren’t.
In a Doppler shift, the extra cycles come from the space between the source and the target. If the target is coming closer, the number of cycles in that space will decrease. That is where the extra cycles come from. That’s not the case for a stationary target.
[/quote]
The Doppler shift is a measure of speed, not distance. The distance is immaterial, as long as the target is far enough away to allow for the returning pulse to activate the comparator. I only know the “block diagram” description of how radar works, I think I am correct there. A pulse is transmitted and the Tx switched off; the receiver is switched on to receive the pulse. The target has to be far enough away to allow for this switching time, that is all.
[Quote=ZenBeam]
:
Suppose the target is only ten feet away from the radar. The radar can only transmit ten cycles at 1 GHz before the target starts reflecting. Leave the radar on. If the reflected frequency is higher than the transmitted frequency, eventually the number of reflected cycles will be ten more than have reached the target. That’s the same as the total cycles transmitted from the radar. After that, there will then be more reflected cycles than the number the radar has ever transmitted. This is clearly absurd.
[/quote]
This is nonsense. If the radar (Tx) is left on, there will be any number of reflected cycles, not just ten, so I will assume this is a mistake on your part. Did you mean to say “switch the radar off”?
This is still nonsense. If ten waves are transmitted, then the Tx switched off, they hit the target and are reflected. Those ten waves will have taken 0.00000001 seconds to transmit (that is the time the Tx is on). They will then be reflected at a higher speed and will arrive back at the Rx. The time for reception of those ten waves at the RX is not 0.00000001 secs as it was when transmitted, but a shorter time while still holding ten waves. There will be no more waves received. The received frequency will be higher.
I was once a relativist, for many years as it happens, but there are too many unanswered questions about both SRT and GRT. I keep an open mind, and look at all the evidence. The evidence points to serious flaws in relativity (I use the word relativity when what we are discussing intrudes on general relativity as well as special relativity, as in the RFR or accelerated frames in general). Until I can be proved wrong on my claim that light/radar changes its frequency when changing direction from W to E to E to W on being reflected, I will claim relativity to be wrong. Relativists agree that light going round the equator does indeed change its velocity on reversing direction. If you have been keeping up with this thread, you will know why I say that radar will change its frequency (I have changed to radar because it is easier to discuss and detect the frequency changes). This is what now lies at the crux of the matter. All talk about closed loops versus open loops is a red herring, as is talk about the Sagnac effect. The Sagnac effect just led us here.
This is an example of confusion about the Sagnac effect being the cause of this frequency change. It is not. The frequency change is caused by the change of speed, it has nothing to do with relative movement. Also I have never used the NS direction. I have always used E to W and W to E.
If we restrict our discussion to travel along the equator, there are two speeds for light/radar. It is faster going E to W than it is going W to E. This is not in dispute. Some of you think that the light has circumnavigate the globe for this to happen. Wrong – it does not. The radar station sends a radar pulse to a stationary target on the horizon to the west. The radar pulse goes to the target faster than it returns. This is not in dispute by the majority of you. After reflection, the radar pulse is received at the aerial at a slower speed than on the outbound leg, so each successive wave crest is received later than it would be if the speed was the same as on the outbound leg. This shows the stationary target to be moving. The maths can be found in a previous posting.
You are also confusing the Sagnac effect with what is under discussion here. Also it is not a Doppler shift. You are nearly correct in saying “…each wave pulse reflects off the mirror at a greater speed than incident, the wave pulse stretches out (the front of the wave pulse has a higher speed than the back of the wave pulse for a small stretch of time)…” You will have to excuse me for rewriting this, and changing it to “…each wave pulse (crest?) reflects off the mirror at a greater speed than incident,” [OK so far. Perhaps I am being too pedantic, but I want to get this right.] “the wave crests stretch out (the front of the wave pulse has a higher speed than the back of the wave pulse for a small stretch of time)…” This part of the sentence does not make sense. The front of the wave pulse (crest) cannot have a different speed than the back of the wave pulse (crest), not if we are talking about the same wave. Do I assume that you are talking about the whole radar pulse, which of course contains many waves and therefore many wave crests? If that is so, then you are wrong, as the front and back of the radar pulse cannot have different speeds. This must be so, as when reflected, the whole of the radar pulse is at the same new speed. For the speed of light to be constant or invariant (which is the same thing) relative to an IFR, it cannot have two or more speeds going in the same direction round the equator, even for a small stretch of time. This is absolutely not correct in relativity, as it destroys the constancy of the speed of light (c) relative to an IFR.
You are saying that the wavelength increases and not the frequency, but there is no evidence for this. Sound in a wind increases or decreases in frequency depending on whether the wind is blowing towards or away from the listener. Ripples in a pond after a stone has been thrown in, will pass a marker, say a shadow on the water, at eg 2 Hertz. Throw the stone in a flowing stream, and the ripples will pass the marker at 1 Hertz if downstream of the stone, and at 4 Hertz if upstream of the stone.
The mechanism and explanation for the frequency change is simple – the radar pulse at the higher speed after reflection (assuming a change from W to E to E to W), impinges on the aerial faster than when it was transmitted. Each wave crest is therefore received sooner than it would have been, leading to an increase in received frequency. This is perfectly logical and consistent.
How does the wavelength change in your model? If I have rewritten your explanation incorrectly, please clarify. Please also ensure that what you are saying is correct in relativity.
It is all correct, and correctly written. Leave the radar on continuously, and tell me how many cycles have been reflected after 1 billion seconds. Assume you are at the equator.
He did not make a mistake. He is completely correct. Light travels roughly 1 foot in one nanosecond, so for a 1 GHz wave, 10 cycles will have been emitted before the first reaches (and is reflected by) the mirror. In his example, the radar is left on.
You don’t seem to understand what ‘frequency’ means. If the radar is left on, the received frequency cannot be higher than the emitted frequency, or else eventually you will have received more cycles than were emitted in the first place.
Perhaps you are misunderstanding the fact that in my example, the front of the wave pulse (or crest) is the part of the wave that has been reflected, and is going in the opposite direction. Its speed is therefore different from the part of the wave form which has not yet reached the mirror and is still travelling at a different speed. I was describing the process by which the wavelength of the waveform is changed during the reflection process.
Those are completely different examples having little relation to your proposed experiment involving reflection of radar waves off a mirror in a RFR.
Again, no, as ZenBeam correctly pointed out. It is not at all logically consistent. You are simply not taking into account the change in wavelength which cancels out the effect of the change in speed on the frequency. The process by which the wavelength changes was described above.
If it had a radial velocity relative to the transmitter, the distance between the two would change. At any moment, the velocity component along the axis connecting A and B, is equal for A and B. I did the math for this before realizing it was obvious.
A and B have different velocities, but this doesn’t matter, the only thing that matters is the velocity component in the direction the beam travels, and that is equal between the two.
If I use a speed radar from a vehicle moving at 100 mph on a vehicle moving at 100 mph, there’s no doppler shift. This is true in a linear system, and a rotating system, in Newtonian mechanics and Einsteinian.
Imagine a system rotating at one third the speed of light. The distance between the transmitter/receiver at A and reflector at B is 1000 meters along the equator, we transmit at a frequency of 25 GHz.
Transmitting with the direction of rotation the speed of light in the RRF of A is 2,0*10[sup]8[/sup] m/s. (Ignoring the curve the path gets in that RRF.)
Any wave front takes 5,0 μs to travel the 1000 m from A to B. The wavelength of the pulse is c[sub]AB[/sub]/f = (2,0*10[sup]8[/sup] m/s) / 25 GHz = 8 mm. The reflector is hit at a frequency of 25GHz.
In any IRF any wave front takes 5,0 μs to travel at c the 1500 m necessary to catch up with B. A and B have identical radial velocities. The reflector is hit at a frequency of 25 GHz.
The reflected pulse moves in the direction BA at a speed of 4,0*10[sup]8[/sup] m/s.
Any wave front takes 2,5 μs to travel the 1000 m from A to B. The wavelength of the pulse is c[sub]BA[/sub]/f = (4,0*10[sup]8[/sup] m/s) / 25 GHz = 16 mm. The reflector is hit at a frequency of 25GHz.
In any IRF any wave front takes 2,5 μs to travel at c the 500 m necessary to meet up with A. A and B have identical radial velocities. The reflector is hit at a frequency of 25 GHz.
If you want to look at wavelenghts:
Imagine A is transmitting in 5 μs pulses.
In A’s RRF at t = 5 μs, there is a pulse 125 000 wavelengths long, spanning the 1000 m between A and B. B is hit at 25 GHz
In an IRF at the same time, there are also 125 000 wavelenghts, compressed by the speed of A in the direction of transmission, but this compression is matched exactly by the speed of B in the direction opposite reception. B is hit at 25 GHz.
In A’s RRF at t = 7,5 μs, half of the 125 000 wavelength spans the 1000 m between B and A. A is hit at 25 GHz.
In an IRF at the same time, there are also 62 500 wavelengths, stretched out by the speed of B in the opposite direction of retransmission, but this stretching is matched exactly by the speed of A in the direction of reception. A is hit at 25 GHz.
Try making the necessary drawings if this doesn’t convince you.
I’m sorry, I don’t know why I said 1 billion seconds. That’s just making things difficult. Just 1 second is long enough. Don’t worry about receiving the radar. I just want to know how many cycles you believe have reflected from the target exactly 1 second after the radar turned on.
I’ll even do the calculations for you, tomh4040. All you have to do is check them yourself.
Above, you said the reflected frequency of a 10 GHz wave would be 10,000,030,941.4 Hz (your post 491). So the reflected frequency of a 1 GHz wave will be 1,000,003,094.14 Hz. Do you agree tomh4040?
So in 1 second, about 1,000,003,094 cycles would be reflected (we won’t worry about a fraction of a cycle). But it took about 0.00000001 seconds for the wave to travel the ten feet to the target, so only about 1,000,003,084 cycles would be reflected. Do you agree that 1,000,003,084 cycles will be reflected from the target 1 second after the radar is turned on, tomh4040? To be clear, I don’t agree with that number, I’m just interested in whether you believe that number is correct.
Please check the math tomh4040. I don’t want to misrepresent what you believe will happen.
There were a lot of posts in reply, and I will have to read them all before replying properly, but yours was the last, and does need some clarification.
The distance from the radar station is immaterial, as is the travel time (for purposes of measuring the frequency change). Why are you saying that ten less cycles are reflected? In 0.00000001 second ten cycles will be transmitted, but the Tx is on for at least one second so that is also immaterial.
I have a very busy period coming up, I may be away from my PC for a while.
I’m asking how many cycles have been reflected 1 second after the radar was turned on. Those ten haven’t reached the target yet, so they haven’t been reflected (yet). We’ll go ahead and stipulate that the radar turns off after exactly one second. That might make things simpler.
ZenBeam you are completely wrong in your assertion that the Doppler shift comes from the space between the source and the target. The distance between the two is completely irrelevant. Read up on the Doppler shift. Here is a passage from Wiki. Note the absence of any distance. “The relative changes in frequency can be explained as follows. When the source of the waves is moving toward the observer, each successive wave crest is emitted from a position closer to the observer than the previous wave. Therefore each wave takes slightly less time to reach the observer than the previous wave. Therefore the time between the arrival of successive wave crests at the observer is reduced, causing an increase in the frequency. While they are travelling, the distance between successive wave fronts is reduced; so the waves “bunch together”. Conversely, if the source of waves is moving away from the observer, each wave is emitted from a position farther from the observer than the previous wave, so the arrival time between successive waves is increased, reducing the frequency. The distance between successive wave fronts is increased, so the waves “spread out”.”
Please get your facts straight on this point before proceeding.
Look back on some of my posting involving the mathematical relationship between wavelength, frequency, and speed. If I did not know what frequency was, could I have written that? Are you getting so desperate that you are just throwing mud at me now without thinking what sort of light it puts you in? Whether the transmitter is left on or switched off, the mechanism and explanation for the frequency change is simple – the radar pulse at the higher speed after reflection (assuming a change from W to E to E to W), impinges on the aerial faster than when it was transmitted. Each wave crest is therefore received sooner than it would have been, leading to an increase in received frequency. This is perfectly logical and consistent.
You are posting nonsense. A lawyer friend of mine gave me this advice “In a debate, as well as in a court room, never ask a question which you do not know the answer to.” You could do well to heed that advice. In your quote above you said that you (the Rx) would eventually receive more cycles than were transmitted (talking about the frequency shift caused by the different speeds of light in different directions).
The second sentence in that reply of yours above can be applied to the Doppler shift in an IFR caused when the target is moving. You should have asked yourself that question about the frequency increase caused by the Doppler shift of a reflection from an approaching target. In an IFR the speed of light is constant (c), so for a one second pulse at 10 Gig containing 10,000,000,000 wave crests, the pulse is reflected and is still at c, but because the reflector has moved closer during that one second, the pulse is shorter than one second when received. The frequency has increased because the 10,000,000,000 wave crests have been received in a shorter time. Note how similar the Doppler shift explanation is to the frequency change with speed explanation.
You are getting your reference frames mixed up. The radar pulse is c relative to an IFR, not to the Earth. According to relativity the target is moving in a non linear fashion relative to that IFR, and is therefore moving away from or towards the radar pulse. This has the effect of making the speed of light relative to the Earth c+v or c-v, where v is the rotational speed of the Earth. The radial velocity is therefore WRT the radar pulse.
This is only true in a linear system, in other words in an IFR. Read the previous postings. This has been agreed by your fellow relativists. If you change the setup so the cars are accelerating (no longer in an IFR), this is no longer true. When the cars are accelerating, the radar pulse has to travel further to reach the (front) car because it has increased its speed during the travel time of the radar pulse (so has the rear car, but that is immaterial). According to relativity, the Earth is not an IFR, so the above description holds.
I have read your maths, and you are still confusing IFRs and RFRs.
Here you have a wave front traveling at 2*10^8 M/s from A to B in the direction of rotation. This is an RFR and is correct.
Here you have a wave front traveling at 3*10^8 M/s, and you call it an IFR. It is not an IFR, it is the same RFR as used in (1), and in an RFR the speed of light is not c.
You now have a wave front traveling at 4*10^8 M/s from A to B. This contradicts quote 1.
Here you have light at 2*10^8 M/s in an IFR. Do I really need to go on?
This is all a complete mess. Please rewrite it. Better still, go back to my example in posting #491 where I have worked it all out correctly, using actual values as found on the Earth. I have used a bit of poetic licence, using c instead of using c/n as the speed of light on the Earth relative to an IFR.
In a reflection time of 0.9999999 seconds, which is one second after the radar is turned on, 1,000,003,084 cycles will have been reflected from the target. Wait for the reflection time to reach one second and the full compliment of 1,000,003,094 cycles will have been reflected from the target. This is one second after hitting the target, not one second after transmission. You appear to be getting hung up on the Tx to target distance. This is completely immaterial, forget about it.
I’m just going to skip right to this part. tomh4040, at the time you say there have been 1,000,003,084 cycles reflected from the target, the radar has only transmitted 1,000,000,000 cycles. You are now arguing that more cycles have been reflected from the wall than have been transmitted. This is as silly as arguing that you can throw 1,000,000,000 balls at a wall, and have 1,000,003,084 balls bounce back.
