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#501
04-18-2012, 05:17 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by tomh4040 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.
By "other direction" I mean a circular "race track" shape, or other closed form. At one point it is going toward (say) Vega. At another point it is going at ninety degrees to a line toward Vega. At yet another point it is going away from Vega.

Radar just goes straight out and straight back. You need "other directions" -- such as around the earth along a line of latitude -- for a Sagnac Effect.

Quote:
 If light takes a different length of time to traverse the Earth in different directions, then its speed is different in different directions.
I believe this has all been covered by an examination of the equations of General Relativity. You get lots of weird stuff, like frame dragging.

I'm pretty good with math, but I can't do GR math. Someone else may be able to help here.

Quote:
 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.
No. A radar pulse only goes straight out and straight back. It does not enclose a shaped pathway. Thus, GR's rotating framework doesn't apply, and the speed is not in any way different in the EW or WE path.

This specific thing has been tested a million ways from Sunday, and the speed difference you assert is not observed.

Quote:
 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.
Frequency, yes. Speed, no.

Quote:
 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.
As I understand it, yes. The effect takes place inside a rotating frame of reference.

Quote:
 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).
As I understand it, yes.

Quote:
 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 I understand it, yes: the GR equations provide for this behavior in a rotating frame of reference.

Quote:
 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).
A radar beam is different than a closed loop path. A radar beam is "straight out and straight back" and falls under a different set of equations.

(I wish that some of our more knowledgeable members would help here. I am not totally confident of my statements here, just pretty darn sure.)

Quote:
 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.
The Doppler, or "red-shift," is how light obeys the law of conservation of energy, without changing speed. If I bounce my radar off of an airplane that it going away from me, the beam that comes back has lost energy, but not speed. It's still going exactly c; it just has a longer wavelength.

Quote:
 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.
Nope. That is exactly opposite of what Relativity says. It is also not what anyone observes. Radar beams never alter from the speed of light, c. Never. It is this constant speed that allows radar to be so very accurate in measuring range.

Quote:
 This causes a shift in frequency.
Yes. A shift in frequency. NOT a shift in speed.

Quote:
 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.
And, in fact, that's what you get. The speed of light is the same from W to E as from E to W.

Quote:
 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?
Wrong, sir. Wrong. The radar beam is not altered by a rotating frame of reference. Only a closed path beam, going around a race-track, or a circle, or simply going around a square, is affected. The radar beam does not go around a shaped course.

You are applying the rules involved in one experiment to another experiment which is not the same.

A feather falls at the same speed as a hammer....on the moon. Try it in your living room, and it doesn't work. You don't get to alter the circumstances of the experiment, and then demand that the results remain unchanged.
#502
04-18-2012, 06:49 PM
 Andy L Member Join Date: Oct 2000 Posts: 2,294
Quote:
 Originally Posted by Trinopus No. A radar pulse only goes straight out and straight back. It does not enclose a shaped pathway. Thus, GR's rotating framework doesn't apply, and the speed is not in any way different in the EW or WE path.
This seems to me to be analogous to the fact that on the surface of a sphere, a closed polygon will exhibit an angle excess (http://en.wikipedia.org/wiki/Angle_excess) proportional to the area enclosed, while a degenerate path (back and forth, not enclosing any area) will exhibit no such angle excess (since GR is intimately related to curved geometry, the fact that this kind of analogy exists is not that surprising).
#503
04-18-2012, 06:59 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Andy L: I think that's what's going on in this situation, but I very emphatically want to say, once again, this is stuff where I'm only able to dip my toe into fathoms and fathoms of depth. I am at risk of violating the unspoken rule of General Questions: don't answer if you don't really know. I think I know...but truth has a way of biting the presumptuous on the sternpost.
#504
04-18-2012, 07:02 PM
 Andy L Member Join Date: Oct 2000 Posts: 2,294
Quote:
 Originally Posted by Trinopus Andy L: I think that's what's going on in this situation, but I very emphatically want to say, once again, this is stuff where I'm only able to dip my toe into fathoms and fathoms of depth. I am at risk of violating the unspoken rule of General Questions: don't answer if you don't really know. I think I know...but truth has a way of biting the presumptuous on the sternpost.
Thanks, both for the quick answer and the caveat.
#505
04-18-2012, 10:04 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
Originally Posted by Trinopus
Quote:
 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.
Nope. That is exactly opposite of what Relativity says. It is also not what anyone observes. Radar beams never alter from the speed of light, c. Never. It is this constant speed that allows radar to be so very accurate in measuring range.
You are correct in an inertial reference frame. In the rotating reference frame where the Earth is stationary, the E to W and W to E speeds will be different, leading to the Sagnac effect in a loop.

Quote:
Quote:
 This causes a shift in frequency.
Yes. A shift in frequency. NOT a shift in speed.
Again, this is correct in an inertial reference frame. If you look at it in the rotating reference frame, and assuming the transmitter and "stationary" target are stationary WRT the Earth, then the frequency does not change.

So to recap, viewed from an inertial reference frame the speed of light is constant, and the frequency changes. Viewed from the rotational reference frame fixed to the Earth, the speed of light is different on the two legs, but the frequency is the same on both of those two legs. In neither of those frames do we have a speed of light that varies and a frequency that varies.

We had a thread a few months back that's relevant here, discussing how only the two-way speed of light can actually be measured, and not the one-way speed of light.

#506
04-18-2012, 10:11 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by ZenBeam You are correct in an inertial reference frame. In the rotating reference frame where the Earth is stationary, the E to W and W to E speeds will be different, leading to the Sagnac effect in a loop.
But what about the case of a "degenerate loop," a straight line EW, reflected WE again?

As far as I can tell, tomh4040 is saying that what is true in a loop is also true in a degenerate loop, and I'm pretty sure that's a problem...

Quote:
 . . . We had a thread a few months back that's relevant here, discussing how only the two-way speed of light can actually be measured, and not the one-way speed of light.
I'll put this on my homework reading list!
#507
04-19-2012, 12:24 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
In the frame of reference attached to the rotating Earth, the EW speed is different from the WE speed for paths that travel all the way around the Earth, so they must also be different for each portion of that path*. When you have a degenerate loop, those differences cancel out over the round trip**. For a loop with area (where the normal to the loop isn't perpendicular to the rotation axis of the frame of reference), those difference don't cancel out.

For a source and target both motionless with respect to the Earth, for a given frequency there will be some number of wavelengths between the source and target for the EW path, and a different number for the other path. In the frame attached to the Earth, those are two numbers because the speed of light in that rotating frame is different in the two direction. In an inertial frame of reference, where the Earth is stationary (but still rotating), you'll have those same two numbers of wavelengths, but in that case, it's because the target has moved between when the source emitted the radiation and when it reaches the target, so the distances traveled are different for the outgoing path and the return path.

* ignoring the "only the two-way speed of light can be measured." issue discussed in those links.

** which is essentially why measuring the two-way speed of light isn't sufficient to determine the one-way speed of light.
#508
04-22-2012, 06:26 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by Trinopus By "other direction" I mean a circular "race track" shape, or other closed form. At one point it is going toward (say) Vega. At another point it is going at ninety degrees to a line toward Vega. At yet another point it is going away from Vega. Radar just goes straight out and straight back. You need "other directions" -- such as around the earth along a line of latitude -- for a Sagnac Effect.
This is not the Sagnac effect. The Sagnac effect just led us here. Radar is electro-magnetic, just the same as light, so it obeys the same rules as light. If light between two mirrors has a speed of 299,792,946.8 M/s E to W and 299,791,994.5 M/s W to E, which it must have, as its overall speed in each direction is unchanged during the travel time, then radar must be the same.

Quote:
 No. A radar pulse only goes straight out and straight back. It does not enclose a shaped pathway. Thus, GR's rotating framework doesn't apply, and the speed is not in any way different in the EW or WE path.
Light only goes straight out and either back or on another path, in both cases being reflected. Look at the path of light between any two of the mirrors in the round the world travel path. The light going W to E is slower than the light going E to W as measured on Earth. If light going round the world is faster one way than the other, it must have the same speed difference between any two mirrors. This also applies to the light from the emitter going to the first mirror, and the mirror being aligned to reflect the light straight back to the receiver which is adjacent to the emitter – the light will have two different speeds. Now install a radar station next to the emitter/receiver and send a pulse out to a target adjacent to the mirror. It will reflect back, and the outbound speed will be different to the inbound speed. Light and radar are going along exactly the same paths, and to obey relativity they must both be constant (c) WRT an IFR, which means they have a different speed WRT Earth while having the same speed as each other. Therefore they must both have the same speed outbound (299,792,946.8 M/s E to W) as they have inbound (299,791,994.5 M/s W to E).
They are both electro-magnetic phenomena. If the speed of radar is constant relative to the Earth, then light must also be constant relative to the Earth, but relativists tell us it is not, light is constant relative to an (arbitrary) IFR. This must hold for all electro-magnetic phenomena.

Quote:
 This specific thing has been tested a million ways from Sunday, and the speed difference you assert is not observed.
If relativity is correct, it would have been observed if it had been looked for – I suspect that it has not. The speed of light on (or in) an RFR is relative to an IFR, so its speed in the RFR must be different in different directions.

Quote:
 The Doppler, or "red-shift," is how light obeys the law of conservation of energy, without changing speed. If I bounce my radar off of an airplane that it going away from me, the beam that comes back has lost energy, but not speed. It's still going exactly c; it just has a longer wavelength.
This is assuming a constant speed for light on an RFR.

Quote:
 Nope. That is exactly opposite of what Relativity says. It is also not what anyone observes. Radar beams never alter from the speed of light, c. Never. It is this constant speed that allows radar to be so very accurate in measuring range.
But there is a problem here too. Assuming you are at the equator, and pointing the radar east or west, you now have light at one speed, and the radar pulse at another.

Quote:
 And, in fact, that's what you get. The speed of light is the same from W to E as from E to W.
You are now contradicting yourself. Earlier you said that the speed of light was different W to E than E to W, agreeing with me.
From a previous posting of yours
Quote:
 Quote: tomh4040 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 [laser] 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). Unquote As I understand it, yes..
Quote:
 Originally Posted by ZenBeam In the frame of reference attached to the rotating Earth, the EW speed is different from the WE speed for paths that travel all the way around the Earth, so they must also be different for each portion of that path*. When you have a degenerate loop, those differences cancel out over the round trip**. For a loop with area (where the normal to the loop isn't perpendicular to the rotation axis of the frame of reference), those difference don't cancel out.
The speed differences cancel out, sure. The frequency change caused by the change of speed does not.

Quote:
 For a source and target both motionless with respect to the Earth, for a given frequency there will be some number of wavelengths between the source and target for the EW path, and a different number for the other path. In the frame attached to the Earth, those are two numbers because the speed of light in that rotating frame is different in the two direction. In an inertial frame of reference, where the Earth is stationary (but still rotating), you'll have those same two numbers of wavelengths, but in that case, it's because the target has moved between when the source emitted the radiation and when it reaches the target, so the distances traveled are different for the outgoing path and the return path.
To make this paragraph clearer, are the source and target motionless WRT the Earth on the Earth? If they are, we are in agreement. What do you mean by “In an inertial frame of reference, where the Earth is stationary (but still rotating)…”? An IFR does not rotate. If it rotates, it is an RFR. In an IFR, the speed of light is c, and the target cannot move WRT that IFR during the transit time of the light. In an RFR (as described) the speed of light is not c. Please clarify the position of the observer.
#509
04-22-2012, 08:59 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 The speed differences cancel out, sure. The frequency change caused by the change of speed does not..
There is no frequency change.
#510
04-23-2012, 12:17 AM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by tomh4040 . . . You are now contradicting yourself. Earlier you said that the speed of light was different W to E than E to W, agreeing with me. From a previous posting of yours . . .
I don't play this kind of game.
#511
04-23-2012, 05:47 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam There is no frequency change.
As the speed changes from the slower speed W to E, to the faster speed E to W when reflected, and is received back at the radar station, each successive wave hits the receiver before it otherwise would have done. The received frequency is therefore higher than it otherwise would have been.
Going the other way, it would change from faster to a slower speed, so at the radar station the received signal would be at a lower frequency.

Quote:
 Originally Posted by Trinopus Quote from tomh4040 ". . . You are now contradicting yourself. Earlier you said that the speed of light was different W to E than E to W, agreeing with me. From a previous posting of yours . . . " I don't play this kind of game.
This is from the earlier posting of mine referred to above, with your answer to it.
"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."

Your reply was - "As I understand it, yes. The effect takes place inside a rotating frame of reference."

Trinopus, I thought you were a man of integrity. What you posted is in black and white for all to see. You cannot have it both ways, if light is a constant (c) relative to an IFR, it is therefore variable WRT the Earth (or any other RFR). Radar must also be constant relative to an IFR, and variable WRT the Earth.

Last edited by tomh4040; 04-23-2012 at 05:48 PM. Reason: Missing quotation marks
#512
04-23-2012, 07:19 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 As the speed changes from the slower speed W to E, to the faster speed E to W when reflected, and is received back at the radar station, each successive wave hits the receiver before it otherwise would have done. The received frequency is therefore higher than it otherwise would have been. Going the other way, it would change from faster to a slower speed, so at the radar station the received signal would be at a lower frequency.
No, you're wrong.
#513
04-24-2012, 11:03 AM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam No, you're wrong.
As I said to a poster not so long ago - was it you? You cannot just say "you are wrong" without providing explanation or proof.

What are you saying is wrong?
If you are saying the frequency does not change, this is a direct consequence of the change of speed. When the light/radar increases (eg) its speed on being reflected, the successive wave crests reach the receiver before they otherwise would have done, so the frequency is increased. This same effect can be heard in sound. When listening to an outdoor concert in a varying wind, the pitch changes. If the wind blows towards the listener, and is sometimes faster then slower, the pitch goes up when it is faster and down again when it is slower.
If you say the speed change from W to E to E to W is wrong, then you are arguing against Einstein, as this is accepted by all relativists.
A point to note here. Trinopus, and probably others, are adamant in saying that the speed of a radar pulse is constant (c) WRT the Earth. Think about that. Einstein's second postulate says that the speed of light is constant with respect to an inertial frame of reference, and as all IFRs are equal, that means any and all IFRs. Radar and light are both electro magnetic, so they both obey the same rules. If radar has the velocity c WRT the Earth, it cannot have the velocity c WRT any IFR, and that goes against the second postulate – it just cannot happen. The speed of light is c relative to any IFR - no exceptions. This paragraph assumes the validity of Einstein’s second postulate.
#514
04-24-2012, 12:18 PM
 Telemark Charter Member Join Date: Apr 2000 Location: Hub of the sports world Posts: 13,354
Quote:
 Originally Posted by tomh4040 As I said to a poster not so long ago - was it you? You cannot just say "you are wrong" without providing explanation or proof.
Since you apparently either don't read or fail to understand the explanations, why bother?
#515
04-24-2012, 12:21 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 As I said to a poster not so long ago - was it you? You cannot just say "you are wrong" without providing explanation or proof.
Sure I can. I could provide an explanation, but it would be pointless to do so. Figure it out yourself.
#516
04-24-2012, 10:40 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by tomh4040 . . . When the light/radar increases (eg) its speed on being reflected . . .
The problem is that this doesn't happen. This is exactly what Relativity says doesn't happen -- and which experiments have verified not to happen.

If the light/radar bounces off of something that is moving away from you, the light/radar changes frequency. This is the standard Doppler effect, or Red Shift. But the observed speed of light does not change. This goes back to the Michelson/Morley experiment, which has been reproduced many thousands of times, with always-increasing accuracy.

Quote:
 This same effect can be heard in sound. When listening to an outdoor concert in a varying wind, the pitch changes. If the wind blows towards the listener, and is sometimes faster then slower, the pitch goes up when it is faster and down again when it is slower.
Yes. However, sound obeys different laws than light. Sound is a compression wave, just to begin with, whereas light behaves more like a transverse wave. You can't polarize a compression wave; there is no such thing as polarized sound.

(Hm... Actually... I don't know that for sure. There might be. Anyone know for certain?)

Quote:
 . . . A point to note here. Trinopus, and probably others, are adamant in saying that the speed of a radar pulse is constant (c) WRT the Earth. Think about that. Einstein's second postulate says that the speed of light is constant with respect to an inertial frame of reference, and as all IFRs are equal, that means any and all IFRs. Radar and light are both electro magnetic, so they both obey the same rules.
The speed of light is constant with respect to any and all inertial frames of reference. This is why you get such things as Lorentz transformations, time dilation, increase of energy, etc.

It doesn't contradict Einstein; it is exactly what Einstein said, and it has been experimentally verified more times than I've had coffee!

Quote:
 If radar has the velocity c WRT the Earth, it cannot have the velocity c WRT any IFR . . .
This claim does not have experimental support.

Quote:
 , and that goes against the second postulate – it just cannot happen. The speed of light is c relative to any IFR - no exceptions. This paragraph assumes the validity of Einstein’s second postulate.
This statement, on the other hand, does have ample experimental support.

This thread has gotten testy lately, and I acknowledge my own blame for part of this. However, I very strongly suggest that we all avoid using debate techniques, as this is not a debate forum. Ask questions, and, to the degree we are able, we will answer them.
#517
04-28-2012, 01:33 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
Originally Posted by Trinopus
Quote:
 Originally Posted by tomh4040 . . . When the light/radar increases (eg) its speed on being reflected . . .
The problem is that this doesn't happen. This is exactly what Relativity says doesn't happen -- and which experiments have verified not to happen.
The problem is that this is exactly what relativity says does happen. The Earth is an RFR, and relativity says that light is the constant c only relative to an IFR, not to an RFR. Therefore relative to the Earth light is faster going E to W than going W to E. This is exactly what causes the fringe shift in the Sagnac interferometer. If experiments have verified this not to happen, then relativity has a big problem. It has to explain why in one RFR - the Sagnac interferometer or ring laser), light is two different speeds relative to that RFR, while in another RFR - the Earth, light speed is constant relative to that RFR.

Quote:
 If the light/radar bounces off of something that is moving away from you, the light/radar changes frequency. This is the standard Doppler effect, or Red Shift. But the observed speed of light does not change. This goes back to the Michelson/Morley experiment, which has been reproduced many thousands of times, with always-increasing accuracy.
Two answers from a relativity viewpoint.
First point – correct, assuming a constant speed of light, which assumes we are in an IFR.
Second point – incorrect. As we are in an RFR, the speed of light WRT the Earth is faster E to W than W to E, so the speed changes when its direction changes. When the radar pulse reflects from a stationary object to the west, its speed decreases, which it must do to adhere to relativity. On reception, because of the speed decrease, each successive wave will reach the aerial later than it would have done had there been no speed change, causing a reduction in the received frequency.
Light is constant (c) relative to any IFR, which leads to the above point two. That has been discussed and agreed on this forum.

Quote:
 This claim “If radar has the velocity c WRT the Earth, it cannot have the velocity c WRT any IFR .” does not have experimental support.
If radar has the velocity c relative to an IFR, which to be consistent with relativity, it has to have as it is electro magnetic, it cannot have the velocity c relative to the Earth, and this is accepted by relativists. Turn it around and you get my quote above.
If A is not equal to B, then B is not equal to A..
#518
04-28-2012, 09:42 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by tomh4040 The problem is that this is exactly what relativity says does happen. . . .
That the speed of light changes? No; that is exactly what relativity says does not happen.

The constancy of the speed of light is at the core of special relativity.

We seem to be at an impasse. Can you show us any reason this thread serves any further purpose? It seems to me to have degenerated into "Rabbit Season," "Duck Season" non-communication.
#519
04-29-2012, 10:08 AM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by Trinopus That the speed of light changes? No; that is exactly what relativity says does not happen. The constancy of the speed of light is at the core of special relativity. We seem to be at an impasse. Can you show us any reason this thread serves any further purpose? It seems to me to have degenerated into "Rabbit Season," "Duck Season" non-communication.
It seems to me that you are not as well up on relativity theory as you think you are, or as other posters on this forum are. It has been said on this forum, and quite correctly if relativity is to be believed, that the speed of light is c relative to any and all IFRs. It is precisely because of the constancy of the speed of light in an IFR and not in an RFR, that the Sagnac effect exists. While the ring laser is not rotating, it is an IFR, and light which is split and sent round the loop in opposite directions will arrive back at the entry/exit point simultaneously. The light has the velocity c WRT the ring laser and all other IFRs, so there is no fringe shift. When the ring laser is rotating, it is no longer an IFR, it is an RFR. The light which is split and sent round it is c WRT any IFR, it is not c WRT the ring laser. If it was c WRT the ring laser, the light would arrive back at the entry/exit point simultaneously even when it was rotating, and there would be no fringe shift. There is a fringe shift however, and you can look it up on Wikipedia, or read on.
Because the ring laser is rotating (relative to an IFR), during the transit time of the light, the entry/exit point has moved towards one beam and away from the other. The path lengths are now unequal, so the light has two different distances to travel to reach the entry/exit point, causing one beam to arrive before the other. This causes a fringe shift.
Now look at what is going on from within (relative to) the ring laser. Within the ring laser, the path lengths are equal. Light going round in one direction has arrived back at the entry/exit point before the light going in the other direction. The light beams set off together, so the speed of light must be different in different directions relative to the ring laser. This is how the ring laser works – no problem so far.
Now take a look at the Earth, which is an RFR, just like the ring laser, but bigger.
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. This will not just show as fringe shift because the difference in arrival times of the light is easily measurable – see my previous posting.
Now look at what is going on from (relative to) the Earth. On the Earth, the path lengths are equal. Light going round in one direction has arrived back at the entry/exit point before the light going in the other direction. The light beams set off together, so the speed of light must be different W to E than E to W as measured on (relative to) the Earth. The light does not have to go all the way round the Earth for the speed difference to be there, as one poster declared, but by sending it all the way round, the path lengths can be seen to be identical and the same clock used to determine the travel times of the two beams. The speed is different W to E than E to W as said above.
Radar and light are both electro magnetic phenomena, and both obey the same rules. For radar to be c relative to an IFR, it cannot be c relative to an RFR, so on reflection from a stationary target to the east or west, the speed changes, leading to a change of frequency.

Quote:
 Originally Posted by ZenBeam You are correct in an inertial reference frame. In the rotating reference frame where the Earth is stationary, the E to W and W to E speeds will be different, leading to the Sagnac effect in a loop.
This was in reply to a previous posting of yours, Trinopus - so you are going against what the rest of the relativists are saying.
#520
04-29-2012, 11:00 AM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 It has been said on this forum, and quite correctly if relativity is to be believed, that the speed of light is c relative to any and all IFRs. It is precisely because of the constancy of the speed of light in an IFR and not in an RFR, that the Sagnac effect exists. While the ring laser is not rotating, it is an IFR, and light which is split and sent round the loop in opposite directions will arrive back at the entry/exit point simultaneously. The light has the velocity c WRT the ring laser and all other IFRs, so there is no fringe shift. When the ring laser is rotating, it is no longer an IFR, it is an RFR. The light which is split and sent round it is c WRT any IFR, it is not c WRT the ring laser. If it was c WRT the ring laser, the light would arrive back at the entry/exit point simultaneously even when it was rotating, and there would be no fringe shift. There is a fringe shift however, and you can look it up on Wikipedia, or read on. Because the ring laser is rotating (relative to an IFR), during the transit time of the light, the entry/exit point has moved towards one beam and away from the other. The path lengths are now unequal, so the light has two different distances to travel to reach the entry/exit point, causing one beam to arrive before the other. This causes a fringe shift. Now look at what is going on from within (relative to) the ring laser. Within the ring laser, the path lengths are equal. Light going round in one direction has arrived back at the entry/exit point before the light going in the other direction. The light beams set off together, so the speed of light must be different in different directions relative to the ring laser. This is how the ring laser works – no problem so far. Now take a look at the Earth, which is an RFR, just like the ring laser, but bigger. 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. [...] Now look at what is going on from (relative to) the Earth. On the Earth, the path lengths are equal. Light going round in one direction has arrived back at the entry/exit point before the light going in the other direction. The light beams set off together, so the speed of light must be different W to E than E to W as measured on (relative to) the Earth. The light does not have to go all the way round the Earth for the speed difference to be there, as one poster declared, but by sending it all the way round, the path lengths can be seen to be identical and the same clock used to determine the travel times of the two beams. The speed is different W to E than E to W as said above. Radar and light are both electro magnetic phenomena, and both obey the same rules. For radar to be c relative to an IFR, it cannot be c relative to an RFR, so on reflection from a stationary target to the east or west, the speed changes, [...]
OK, this is written clearly and is also correct.

Quote:
 [...] leading to a change of frequency.
No. There is no frequency change, since the target is stationary. If you are using a 1 GHz radar, then 1 billion cycles hit the target in a 1 second period. That means 1 billion cycles are reflected in that 1 second period. If the frequency were higher after reflection, where would the extra cycles come from?
#521
04-29-2012, 02:39 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam OK, this is written clearly and is also correct. No. There is no frequency change, since the target is stationary. If you are using a 1 GHz radar, then 1 billion cycles hit the target in a 1 second period. That means 1 billion cycles are reflected in that 1 second period. If the frequency were higher after reflection, where would the extra cycles come from?
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.

Quote:
Originally Posted by ZenBeam
Quote:
 Originally Posted by tomh4040 As I said to a poster not so long ago - was it you? You cannot just say "you are wrong" without providing explanation or proof
Sure I can. I could provide an explanation, but it would be pointless to do so. Figure it out yourself.
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.
#522
04-29-2012, 03:19 PM
 naita Guest Join Date: Jun 2002
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.
#523
04-29-2012, 04:38 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 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:
 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.

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:
 Originally Posted by tomh4040 They both work, so something is wrong somewhere.
What is wrong is your belief that the reflected frequency changes in this situation.
#524
04-29-2012, 08:08 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by tomh4040 It seems to me that you are not as well up on relativity theory as you think you are, or as other posters on this forum are.
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.

Quote:
 It has been said on this forum, and quite correctly if relativity is to be believed, that the speed of light is c relative to any and all IFRs.
Missing a word; the speed of light is invariant relative to any and all IFRs.

Quote:
 It is precisely because of the constancy of the speed of light in an IFR and not in an RFR, that the Sagnac effect exists. While the ring laser is not rotating, it is an IFR, and light which is split and sent round the loop in opposite directions will arrive back at the entry/exit point simultaneously. The light has the velocity c WRT the ring laser and all other IFRs, so there is no fringe shift. When the ring laser is rotating, it is no longer an IFR, it is an RFR. The light which is split and sent round it is c WRT any IFR, it is not c WRT the ring laser. If it was c WRT the ring laser, the light would arrive back at the entry/exit point simultaneously even when it was rotating, and there would be no fringe shift. There is a fringe shift however, and you can look it up on Wikipedia, or read on.
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.

Quote:
 This was in reply to a previous posting of yours, Trinopus - so you are going against what the rest of the relativists are saying.
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.

Last edited by Trinopus; 04-29-2012 at 08:08 PM.
#525
04-29-2012, 10:03 PM
 Leo Bloom Cares about Zombies Join Date: Jun 2009 Posts: 5,434
Quote:
 Originally Posted by Trinopus ....Can you show us any reason this thread serves any further purpose? It seems to me to have degenerated into "Rabbit Season," "Duck Season" non-communication.
[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.

Last edited by Leo Bloom; 04-29-2012 at 10:03 PM.
#526
04-29-2012, 10:25 PM
 Trinopus Member Join Date: Dec 2002 Location: San Diego, CA Posts: 7,098
Quote:
 Originally Posted by Leo Bloom [nitpick]"Wabbit Season"[/nitpick]
Of course!

Quote:
 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.
It's been worth it for me, just to learn of the Sagnac Effect.

Quote:
 I sensed from the first page that one role player, attitudinally, should continue waiting tables at high-class restaurants.
The tips offset the costs of my ISP, allowing me to participate in discussions far above my social status.
#527
04-30-2012, 12:06 AM
 Leo Bloom Cares about Zombies Join Date: Jun 2009 Posts: 5,434
Your acting was fine. You don't need to work for tips at restaurants.
#528
04-30-2012, 08:47 AM
 iamnotbatman Guest Join Date: Aug 2010
Quote:
 Originally Posted by 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.
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?
#529
04-30-2012, 10:36 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by iamnotbatman 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.
#530
04-30-2012, 11:47 AM
 iamnotbatman Guest Join Date: Aug 2010
Quote:
 Originally Posted by naita 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.
#531
05-01-2012, 11:40 AM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by naita 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.
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:
 Originally Posted by 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.
If by so close, you mean so close to disproving relativity, I am not just so close, I am there.
Quote:
 Originally Posted by 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.
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:
 Originally Posted by 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.
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.

Quote:
 Originally Posted by Trinopus 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. 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.
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.

Quote:
 Originally Posted by Trinopus 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.
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.

Quote:
 Originally Posted by iamnotbatman 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.
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.
#532
05-01-2012, 12:49 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 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.
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.
#533
05-01-2012, 12:52 PM
 iamnotbatman Guest Join Date: Aug 2010
Quote:
 Originally Posted by tomh4040 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”?
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.

Quote:
 Originally Posted by tomh4040 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.
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.
#534
05-01-2012, 01:02 PM
 iamnotbatman Guest Join Date: Aug 2010
Quote:
 Originally Posted by tomh4040 “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.
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.

Quote:
 Originally Posted by tomh4040 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.
Those are completely different examples having little relation to your proposed experiment involving reflection of radar waves off a mirror in a RFR.

Quote:
 Originally Posted by tomh4040 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.
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.
#535
05-01-2012, 01:45 PM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by tomh4040 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.
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*108 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 cAB/f = (2,0*108 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*108 m/s.

Any wave front takes 2,5 μs to travel the 1000 m from A to B. The wavelength of the pulse is cBA/f = (4,0*108 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.

Last edited by naita; 05-01-2012 at 01:49 PM.
#536
05-01-2012, 05:12 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by ZenBeam 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.
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.
#537
05-02-2012, 05:53 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam 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.
#538
05-02-2012, 08:19 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 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?
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.
#539
05-09-2012, 05:35 PM
 tomh4040 Guest Join Date: Oct 2011
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".”

Quote:
 Originally Posted by iamnotbatman 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.
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.

Quote:
 Originally Posted by naita 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.
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.
Quote:
 If I use a speed radar from a vehicle moving at 100 mph on a vehicle [in front] (added by tomh4040 for clarity) 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.
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.
Quote:
 Originally Posted by naita 1. Any wave front takes 5,0 μs to travel the 1000 m from A to B. The wavelength of the pulse is cAB/f = (2,0*108 m/s) / 25 GHz = 8 mm. The reflector is hit at a frequency of 25GHz.
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.
Quote:
 Originally Posted by naita 2. 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.
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.
Quote:
 Originally Posted by naita 3. Any wave front takes 2,5 μs to travel the 1000 m from A to B. The wavelength of the pulse is cBA/f = (4,0*108 m/s) / 25 GHz = 16 mm. The reflector is hit at a frequency of 25GHz.
You now have a wave front traveling at 4*10^8 M/s from A to B. This contradicts quote 1.
Quote:
 Originally Posted by naita 4. 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.
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.

Quote:
 Originally Posted by ZenBeam 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.
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.
#540
05-09-2012, 07:19 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
 Originally Posted by tomh4040 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.

You are clearly, unequivocally wrong here.

Goodbye.
#541
05-09-2012, 08:06 PM
 Andy L Member Join Date: Oct 2000 Posts: 2,294
Quote:
 Originally Posted by tomh4040 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.
Ok. Ready to read the quote and note the absence of distance.

Quote:
 Originally Posted by tomh4040 “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".”
I see several references to the change in distance being integral to the Doppler shift. Don't you? How does this disagree with Zenbeam's "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:
#542
05-10-2012, 03:55 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by tomh4040 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.
No, the radial velocity is the difference between the velocity component of the transmitter and target along the line connecting the point of origin at the moment of transmission and the target at the moment of impact.

Quote:
 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.
No, this is always true. If the target of my doppler radar, at the instant of impact, has the same velocity component as me along the line connecting the two of us, I will register zero speed. Acceleration may change that in the next instant, but for co-rotating points the change in direction is nulled out.

Quote:
 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.
It's the same situation described in two different reference frames. As viewed from an inertial observer, this is what happens. If you want to limit it to newtonian relativity, the observer is stationary at point A and the planet rotates away beneath him.

Quote:
 You now have a wave front traveling at 4*10^8 M/s from A to B. This contradicts quote 1.
It's a typo. I'm describing the return trip, so I meant B to A.

Quote:
 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.
That is indeed a silly mistake on my part. The return travel time should be approx 1,67 μs for a IFR speed of 3*108 and a RFR speed of 6*108.

In #491 you're confusing absolute and relative speeds as well as ignoring that doppler shifts require a difference in radial velocities.
#543
05-10-2012, 04:02 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by naita That is indeed a silly mistake on my part. The return travel time should be approx 1,67 μs for a IFR speed of 3*108 and a RFR speed of 6*108.
ETA: No, wait, I assumed they'd meet halfway again. I'll find the correct values when I'm not in the middle of teaching a class.
#544
05-10-2012, 04:51 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by naita That is indeed a silly mistake on my part. The return travel time should be approx 1,67 μs for a IFR speed of 3*108 and a RFR speed of 6*108.
I corrected the wrong part.

The travel time is indeed 2,5 μs. For the RFR we then get a speed of 4*108 m/s.

In the IFR the target moves 250 m, and the pulse moves 750 m.
#545
05-10-2012, 05:47 AM
 iamnotbatman Guest Join Date: Aug 2010
This is ridiculous. Good luck, guys.
#546
05-10-2012, 11:17 AM
 naita Guest Join Date: Jun 2002
Let’s try it this way. We follow two wavefronts from transmitter to target and back. We use a 25GHz signal. That means the two wavefronts leave the transmitter 40 ps apart. The two wavefronts travel paths of equivalent length, so they also arrive at the target 40 ps apart. They are reflected the instance they hit the target (or if there is a delay it’s equal for both), so they leave in the opposite direction 40 ps apart. They will travel a different path back, but it will have the same length for both wavefronts, so they will arrive 40 ps apart. That means we’re receiving a 25GHz signal.

In any IFR the wavefronts are moving at c at all times and the outbound and inbound paths are of different length and have constantly moving endpoints. As the paragraph above shows, this doesn’t matter to the frequency.

In the rotating reference frame of the transmitter, the wavefronts travel one curved path on the outbound leg and a different curved path on the returning leg, due to coriolis forces. Also the speed changes, which changes the wavelength. As the first paragraph of this post shows, this does not matter to the frequency.

Now please explain what you believe is wrong with the first paragraph. Wavelength times frequency equals speed, so we can change one, keep one unchanged, as long as the third changes to suit. The frequency and period however, are always the inverse of each other. How do you suggest we get a change in period when the distance between transmitter and target, and thus the one way travel time of any component of our signal, is a constant?

Don't get hung up on the preconception that a change in speed gives a doppler effect and change in frequency. Just examine the physical situation and the impossibility of changing the period of the signal without changing the distance.

If this doesn't convince you, I'm willing to illustrate the movement of these two wavefronts in detail.
#547
05-14-2012, 06:57 PM
 tomh4040 Guest Join Date: Oct 2011
Quote:
 Originally Posted by ZenBeam 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. You are clearly, unequivocally wrong here. Goodbye.
Well done. Full marks to you for tripping me up. I usually wait for a while after writing my answer to avoid mistakes like that. You wrote cycles, I read cycles per second. The two are not the same, I simply made a mistake - a bad mistake, as in the same posting, I had put the explanation quite succinctly. In your haste to score points, you did not read it. Here it is again.
"...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."
The answer is not that more cycles have been reflected than were sent. Put quite simply, the same number of cycles that were sent in a one second pulse are received in a pulse which is now shorter than one second, leading to an increase in frequency.

Quote:
 Originally Posted by Andy L I see several references to the change in distance being integral to the Doppler shift. Don't you? How does this disagree with Zenbeam's "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."
“…each wave takes slightly less time to reach the observer…” is indeed a reference to distance, but that distance can be any distance at all without affecting the Doppler shift one iota.
“..the distance between successive wave fronts is reduced (increased)…”refers to the signal, not the distance apart of target and radar station. The distance is not integral to the Doppler shift.
I can stand in front of a car coming towards me (you wish) and listen to its horn sounding. The driver states that his horn has a frequency of 1000Hz, yet I hear it at 1050Hz. It does not matter whether I am 10M away or 100M away, the frequency I hear is 1050HZ, telling me that the car is traveling at 50Kph (assuming the speed of sound to be 1000Kph). THE DISTANCE FROM ME TO THE CAR IS TOTALY IMMATERIAL TO THE DOPPLER SHIFT.

Naita, please read up on the Sagnac effect, especially the description of how a signal which is split and sent round the world (at the equator) from the entry/exit point will arrive back at the same entry/exit point at two different times. When you understand how it works, rejoin this discussion. You apparently are the only person on this forum who disagrees with me on this point. By this point, I don’t mean the point that frequency changes with a change of speed of the signal, I mean the point that the speed of light is different going W to E than it is going E to W.

Quote:
 Originally Posted by naita If I use a speed radar from a vehicle moving at 100 mph on a vehicle [in front] (added by tomh4040 for clarity) 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.
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.
Briefly put :- During the transit time of the signal, the non inertial frame of reference has increased its speed, and therefore its distance, so the signal has further to go to reach the target. From within that non IFR, which can also be called an AFR (accelerated frame of reference), if I measure the speed of light, I will find that it is not c, but a lesser value. Relativity assumes no difference between all non IFRs, leading to the speed of light as measured on Earth being faster E to W than it is W to E. The above paragraph is agreed with by all relativists. Please read up on this also. There is a point to clarify here. If the two cars are on the Earth, it might be supposed that they comprise an IFR, as they are both doing 100 mph. They do not, because the system they are in (the Earth) is an RFR, not an IFR, so when the speed of the cars is referenced back to an IFR which is away from the Earth, the rotational speed of the Earth has to be taken into account.

Quote:
 Originally Posted by naita Let’s try it this way. We follow two wavefronts from transmitter to target and back. We use a 25GHz signal. That means the two wavefronts leave the transmitter 40 ps apart. The two wavefronts travel paths of equivalent length, so they also arrive at the target 40 ps apart. They are reflected the instance they hit the target (or if there is a delay it’s equal for both), so they leave in the opposite direction 40 ps apart. They will travel a different path back, but it will have the same length for both wavefronts, so they will arrive 40 ps apart. That means we’re receiving a 25GHz signal. In any IFR the wavefronts are moving at c at all times and the outbound and inbound paths are of different length and have constantly moving endpoints. As the paragraph above shows, this doesn’t matter to the frequency.
I am assuming the target is stationary relative to the radar station and this is an IFR. The errors there are that the signals do not travel a different path back, and the end points are not constantly moving. They travel the same path both ways. In this IFR there will be no frequency change as there is no radial movement.
The Earth is an RFR not an IFR. When the signal reverses its direction upon reflection, its speed increases (assuming a change of direction from W to E to E to W) so the wavefronts are less than 40 ps apart. I find it easier to think of a one second pulse on the leg towards the target, which is reflected at a higher speed. That pulse is now less than one second in duration, so if it contained 25*10^9 wavefronts (wavecrests) in one second on the way out, it now contains 25*10^9 wavefronts in less than one second on the way back. The number of wavecrests have not increased (well done to ZenBeam again), but the time has decreased. This causes an increase in frequency. Re-read my last post. This is very similar to the Doppler shift, when after hitting a target (which is moving closer) and being reflected, the total number of cycles stays the same, but they are bunched closer together. In other words the number of wavecrests have not increased, but the reception time has decreased.

Quote:
 In the rotating reference frame of the transmitter, the wavefronts travel one curved path on the outbound leg and a different curved path on the returning leg, due to coriolis forces. Also the speed changes, which changes the wavelength. As the first paragraph of this post shows, this does not matter to the frequency.
To quote Wiki again “The Coriolis force acts in a direction perpendicular to the rotation axis and to the velocity of the body in the rotating frame and is proportional to the object's speed in the rotating frame… This force causes moving objects on the surface of the Earth to veer to the right (with respect to the direction of travel) in the northern hemisphere, and to the left in the southern.”. There are no coriolis forces along the equator.
As the speed changes, the frequency changes, but the wavelength does not. This keeps the correct ratio, so your statement “the speed changes which changes the wavelength” is incorrect.
A simple example. A pulse with a frequency of 10 Hz is traveling at 20 M/s. The wavelength is therefore 2 M.
W = V/F = 20/10 = 2 .
The pulse is reflected and in doing so changes its speed to 40 M/s. That pulse is now 0.5 secs long while still containing the same number of cycles (10). The frequency is now 20 Hz. The wavelength is still 2 M. W = V/F = 40/20 = 2 .

I have gone back to my real world (nearly) example (VL is velocity of light either WE or EW) :-
Transmitted F = 10Gig, VL WE = 299,791,994.2 M/s, and VL EW = 299,792,921.8 M/s . Rotational speed of Earth at the equator is 463.8 M/s

Here are the maths to show the relationship between W, VL, and F :-
W is meters, F is Hertz, VL is M/s, W is meters. These figures are for the transmitted pulse traveling W to E.
F = 10*10^9 (10Gig) : VL WE = 291,791,994.2 : W = 0.02917919942

W = VL WE / F = 299,791,994.2 / 10,000,000,000 = 0.02997919942 M This is the wavelength of the transmitted radar pulse.
It hits the target and is reflected, 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.
The velocity is now 299,792,921.8 M/s so the 10,000,000,000 cycles which occupied one second are now occupying less than one second.
299791994.2 / 299792921.8 = 0.999996905864 secs so the frequency is increased to :-
10,000,000,000 / 0.999996905864 = 10,000,030,941.4 Hz

W = VL EW / F = 299792921.8 / 10000030941.4 = 0.02997919942 M.
Note this is the same wavelength as the transmitted pulse on the W to E leg.
#548
05-14-2012, 08:16 PM
 Andy L Member Join Date: Oct 2000 Posts: 2,294
Quote:
 Originally Posted by tomh4040 “…each wave takes slightly less time to reach the observer…” is indeed a reference to distance, but that distance can be any distance at all without affecting the Doppler shift one iota. “..the distance between successive wave fronts is reduced (increased)…”refers to the signal, not the distance apart of target and radar station. The distance is not integral to the Doppler shift. I can stand in front of a car coming towards me (you wish) and listen to its horn sounding. The driver states that his horn has a frequency of 1000Hz, yet I hear it at 1050Hz. It does not matter whether I am 10M away or 100M away, the frequency I hear is 1050HZ, telling me that the car is traveling at 50Kph (assuming the speed of sound to be 1000Kph). THE DISTANCE FROM ME TO THE CAR IS TOTALY IMMATERIAL TO THE DOPPLER SHIFT. .
Which is what Zenbeam said ""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." Agreement at last.
#549
05-14-2012, 09:52 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,706
Quote:
It will take a full 1,000,000,000 ns (1 second) to transmit the 1,000,000,000 cycles. At 1,000,003,094 Hz, those cycles will be reflected in only 999,996,906 ns, as you apparently agree, given the bolded part of your quote. Since the source is only 10 ns away, you have those 1,000,000,000 cycles finished being reflected before they have all been transmitted.

You're violating causality. How can you expect anyone to take you seriously when you write stuff like this?

You continue to be clearly, completely, and unequivocally wrong.
#550
05-15-2012, 07:28 AM
 naita Guest Join Date: Jun 2002
Quote:
 Originally Posted by tomh4040 Naita, please read up on the Sagnac effect, especially the description of how a signal which is split and sent round the world (at the equator) from the entry/exit point will arrive back at the same entry/exit point at two different times. When you understand how it works, rejoin this discussion. You apparently are the only person on this forum who disagrees with me on this point. By this point, I don’t mean the point that frequency changes with a change of speed of the signal, I mean the point that the speed of light is different going W to E than it is going E to W.
In the RFR of Earth the speed is different, yes, as I showed in my calculations. As viewed from any IFR, they are not.

Quote:
 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. Briefly put :- During the transit time of the signal, the non inertial frame of reference has increased its speed, and therefore its distance, so the signal has further to go to reach the target.
No, it has changed its velocity. The speed is the same. If you examine the situation properly, you'll also find that the radial velocity is the same in the moment of emision and the moment of reflection.

Quote:
 The Earth is an RFR not an IFR. When the signal reverses its direction upon reflection, its speed increases (assuming a change of direction from W to E to E to W) so the wavefronts are less than 40 ps apart.
No, the speed changes, and the wavelenghts change correspondently, leaving the frequency the same.

Quote:
 As the speed changes, the frequency changes, but the wavelength does not. This keeps the correct ratio, so your statement “the speed changes which changes the wavelength” is incorrect. A simple example. A pulse with a frequency of 10 Hz is traveling at 20 M/s. The wavelength is therefore 2 M. W = V/F = 20/10 = 2 . The pulse is reflected and in doing so changes its speed to 40 M/s. That pulse is now 0.5 secs long while still containing the same number of cycles (10). The frequency is now 20 Hz. The wavelength is still 2 M. W = V/F = 40/20 = 2 .
For that to make sense your reflector would emit twice the pulses it received. The only way frequency changes is if the distance between emitter and receiver changes.

Your 10 Hz transmitter is sending out 10 pulses every second, the reflector is sending out 20 pulses every second. The distance between them is constant.

The number of pulses transmitted is 10*t where t is elapsed time.
The number of pulses reflected is 20*(t-tt) where tt is travel time between transmitter and reflector. (Valid from t=tt and onwards.)

Say the travel time is 10 seconds. After 10 seconds the transmitter has sent out 100 pulses, and none have yet been reflected.

After 20 seconds the transmitter has sent out 200 pulses, and 200 pulses have been reflected, which is odd, since 100 pulses are en route between transmitter and reflector.

After 40 seconds the transmitter has sent out 400 pulses, and 600 pulses have been reflected, which is frankly mindboggling.

Quote:
 I have gone back to my real world (nearly) example (VL is velocity of light either WE or EW) :- Transmitted F = 10Gig, VL WE = 299,791,994.2 M/s, and VL EW = 299,792,921.8 M/s . Rotational speed of Earth at the equator is 463.8 M/s Here are the maths to show the relationship between W, VL, and F :- W is meters, F is Hertz, VL is M/s, W is meters. These figures are for the transmitted pulse traveling W to E. F = 10*10^9 (10Gig) : VL WE = 291,791,994.2 : W = 0.02917919942 W = VL WE / F = 299,791,994.2 / 10,000,000,000 = 0.02997919942 M This is the wavelength of the transmitted radar pulse. It hits the target and is reflected, 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.
No, it doesn't, it results in a change in wavelength. That's what happens in all other situations where the speed of a wave changes.
Frequency change requires the distance between source and receiver to change.

If the simple calculations based on your 10 Hz example don't convince you you're wrong I'll draw and explain this visually later.

Last edited by naita; 05-15-2012 at 07:31 AM. Reason: Left a bit out.

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