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.
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.
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.
Frequency, yes. Speed, no.
As I understand it, yes. The effect takes place inside a rotating frame of reference.
As I understand it, yes.
As I understand it, yes: the GR equations provide for this behavior in a rotating frame of reference.
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.)
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.
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.
Yes. A shift in frequency. NOT a shift in speed.
And, in fact, that’s what you get. The speed of light is the same from W to E as from E to W.
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.
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 (Spherical trigonometry - Wikipedia) 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).
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.
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.
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.
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.
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.
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.
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.
This is assuming a constant speed for light on an RFR.
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.
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
The speed differences cancel out, sure. The frequency change caused by the change of speed does not.
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.
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=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.
[/QUOTE]
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.
When the light circumnavigates the Earth to be received back at its origin, relativity says that the receiver has moved during the transit time of the light, so the beam going W to E arrives later than the beam going E to W. The light does not have to go all the way round. If it is received at a receiver which is within line of sight of the transmitter, the receiver has still moved during the transit time of the light. Obviously it has not moved as far – in fact the movement will have been very small. If a receiver is placed W of the transmitter, and one placed the same distance E of the transmitter, the receiver which is E of the transmitter will receive the signal later than the receiver which is W of the transmitter, but this will be impossible to verify, which is why the light has to go all the way round the Earth (which is impossible to do). As the radar return is modified by how fast the target is moving, not how far it has moved or how far away it is, the radar return will show the same speed difference no matter how far away the target is. ( A policeman with his radar gun will clock your speed at eg 100Kph whether you are 10 meters away or 1000 meters away, or whether you are doing 100Kph for 10 seconds or 10 minutes.)
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.
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.
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?)
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!
This claim does not have experimental support.
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.
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.
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.
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..
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.
This was in reply to a previous posting of yours, Trinopus - so you are going against what the rest of the relativists are saying.
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?