Ok. Ready to read the quote and note the absence of distance.
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.”
Good advice for all
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.
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.
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.
It’s a typo. I’m describing the return trip, so I meant B to A.
That is indeed a silly mistake on my part. The return travel time should be approx 1,67 μs for a IFR speed of 310[sup]8[/sup] and a RFR speed of 610[sup]8[/sup].
In #491 you’re confusing absolute and relative speeds as well as ignoring that doppler shifts require a difference in radial velocities.
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.
I corrected the wrong part. 
The travel time is indeed 2,5 μs. For the RFR we then get a speed of 4*10[sup]8[/sup] m/s.
In the IFR the target moves 250 m, and the pulse moves 750 m.
This is ridiculous. Good luck, guys.
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.
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.
“…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=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.
[/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. 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.
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 2510^9 wavefronts (wavecrests) in one second on the way out, it now contains 2510^9 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.
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.
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.
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.
In the RFR of Earth the speed is different, yes, as I showed in my calculations. As viewed from any IFR, they are not.
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.
No, the speed changes, and the wavelenghts change correspondently, leaving the frequency the same.
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 10t where t is elapsed time.
The number of pulses reflected is 20(t-t[sub]t[/sub]) where t[sub]t[/sub] is travel time between transmitter and reflector. (Valid from t=t[sub]t[/sub] 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.
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.
If you can’t come up with anything better than that, you have already accepted that you have lost the argument.
The travel time to the target is irrelevant. The one second pulse hits the target. After reflection the pulse will be less than one second.
Goodbye.
Tom, I encourage you to take another look at Bunny’s post a few pages ago:
Whenever someone points out an obvious flaw in your understanding of introductory physics, you either 1)dismiss the error as a minor technical slip-up of yours or 2)just ignore them. Off the top of my head, the three most obvious errors repeatedly pointed out to you but which you have entirely ignored were 1)
This is unambiguously exactly the opposite of very basic physics (and intuition to boot).
error 2)
This betrays all sorts of misunderstanding. Again, as others have pointed out, what you mean by “turning it on its side” is to introduce another force, namely, the earth’s gravitational force. Of course the model would play out differently if we add in new forces! This observation does not undermine the accuracy of the diagram whatsoever.
error 3), in my mind the most basic
Again, even intuitively, this is clearly wrong. Just think about it - how can the velocity at the equator and at the pole be the same?
I am very much a layman, so I apologize for any errors I may have made myself. My point is that even someone with a very rudimentary knowledge of basic physics can see the gaping holes in your reasoning. Your unwillingness to address these errors (among many others) proves either that you 1) aren’t arguing in good faith, or 2) are suffering from a serious case of the Dunning-Kruger effect, as has already been mentioned, or perhaps some combination of the two.
I’ll be surprised and impressed if you actually address the errors in the quotations I posted.
Painted yourself into a corner, I see.
We agree on this point.
I was talking about a linearly accelerated frame in that passage, with the light/radar pulse being in the same direction as the line of travel.
Read the explanation and the maths again. If the radar increases its velocity, then the same number of wavecrests are now in a shorter duration pulse. Velocity and frequency have changed, so the wavelength stays the same. This can be seen in any wave phenomena.
There are two ways the frequency can change. You have mentioned one. The other one is if the velocity changes.
The reflector does not reflect 20 cycles in a second. If you remember, I fell into that trap with ZenBeam. The reflector reflects 10 cycles, yes. But it reflects them in half a second not one second, so the frequency is doubled.
This is almost word for word what ZenBeam put, and the answer is the same. Paragraph two - correct.
Paragraph three - incorrect. 200 pulses have not been reflected, only 100 have been reflected, but they have been reflected in half the time.
Paragraph four - same as three. You are confusing pulses (cycles - period) with cycles per second, which is Hertz.
In all other situations; sound or waves on water for example, the frequency changes. Listen to an outdoor orchestra in a varying wind, throw pebbles in a pond and a flowing river. You will hear/see for yourself.
That is one of the mechanisms for a change of frequency, the other is a change of speed.
Please do it now. You, and all the other posters have seen my maths. Let us all see yours.
Of course I will address the errors (?), but in return, you will have to answer my questions about the apparent incompatibility of the Sagnac effect with radar in my last few postings.
Point 1. The beach ball. I was using poetic licence. A beach ball has a very low weight to size ratio, the characteristics are unknown, and air resistance plays a big part. Substitute for a ball of known characteristics, and use that. The characteristics of this ball are that is heavy, so air resistance can be (nearly) ignored, and it reforms at 10kph. When this ball is hit by a bat and therefore compressed, it will compress to half its size in the direction it is hit, and then reform again. The bat moving at 10kph (WRT the ground) hits the ball, which compresses while still in contact with the bat. The bat is still moving at 10kph and the ball is fully compressed. Now bat and ball are both moving at 10kph WRT the ground. The ball reforms to its original shape while the surface of the ball is still in contact with the bat and is attempting to move towards the bat at 10kph (WRT the bat). As the two surfaces are in contact, this means that the centre of mass of the ball is pushed away from the bat at 10kph. The bat is still moving at 10kph WRT the ground, so the ball is now moving at 20kph WRT the ground.
We have moved on since then. Now you tell me why I am wrong in saying the Sagnac effect is incompatible with radar. See my posting #547.
Let’s again place the transmitter at A and the reflector at B.
Let’s nail down our description of the rotating reference frame co-rotating with A. In this reference frame neither A or B moves, and there’s a constant distance between them. The speed of light differs in the AB and BA direction due to it being a rotating reference frame.
Let’s look at just two wavefronts, one cycle, emitted from A at 25GHz, i.e. 40 ps apart.
If we start transmitting at t=0, the second wavefront will be emitted at t=40 ps.
We can call travel time from A to B t[sub]AB[/sub].
The first wavefront will hit B at t[sub]AB[/sub], the second wavefront will hit B at t[sub]AB[/sub] + 40 ps
As they hit they are reflected. The first wavefront is reflected at t[sub]AB[/sub], the second wavefront is reflected at t[sub]AB[/sub] + 40 ps
Travel time from B to A is different from the travel time from A to B. We can call travel time from B to A t[sub]BA[/sub].
The first wavefront will return to A at t[sub]AB[/sub] + t[sub]BA[/sub].
The second wavefront will return to B at t[sub]AB[/sub] + t[sub]BA[/sub] + 40 ps
They’re received at A on their return 40 ps apart, which is 25GHz.
Now I suspect you’ll disagree that A and B are not moving, but in the reference frame co-rotating with A they are still. That’s what’s meant by a co-rotating reference frame.
One can pick an infinity of other reference frames, both rotating, accelerated and inertial, where A and B do move, but the only one that makes sense to examine (the other’s will give equivalent results, just with horrible math) is an inertial one co-moving with A and B’s center of rotation. In that frame A and B do have acceleration and velocity, but as in all other inertial frames of reference, c is constant and the radial velocity between co-rotating points A and B is 0.
Now the following is just an animation I made for my own amusement which shows “photons” moving with or against the direction of rotation, it’s not meant to really enlighten anyone in this thread. But it’s as good a place to share it as any: http://www.geogebratube.org/student/m9696 (requires up to date Java, and in my case, use of my second choice browser)
I’m sorry, Tom, but you are clearly being disingenuous. You only “explained” one of your obvious errors. Is this
poetic license? Give me a break.
Again, the lack of knowledge exhibited in that quotation means you just don’t have the ability (at this time) to make sense of something like relativity.
Lastly, a few times throughout the discussion you’ve expressed interest in converting others to your view. If you really want the ether theory to gain traction I’d encourage you yourself to stop talking about it, as you are supremely unconvincing.
I answered only one point, and I notice you did not disagree with the explanation. Now I want you to answer mine.
This is ridiculous. I’ve already admitted I’m a layman, and certainly you can admit that your question isn’t basic physics in the same way that the definition of velocity is. So you might say that our positions in this matter lack symmetry.
Again: I know what velocity is. You seem to not. If you don’t understand velocity, you couldn’t possibly understand the physics you’re asking about.
Isn’t this the core of the issue? According to SR, this doesn’t happen. It’s also never been observed to happen.
Your posting contains another mistake, but reading it made me realize that there is a mistake in my work also. I will get to that later, but now to yours.
Not a quote, but you said “On the outward journey, to the reflector, the radar pulse is traveling at say 300,000,000 M/s, with a time between wavecrests of 40 ps. On the return journey, at say 200,000,000 M/s, the wavecrests are still 40 ps apart.” That is what you put ( I may have the velocity/direction wrong, but the maths will still be correct). You also said that the frequency is 25 Gig in both directions.
For the outward pulse, the wavecrests are 40 ps apart in time, so they have moved 0.003 M in that time.
This corresponds to a frequency of : F = V / W = 3e8 / 0.012 = 25 Gig
They are reflected also at 40 ps apart. Now the velocity is not 3e8 but is 2e8 (the return pulse). The wavecrests have moved 0.002 M in that time.
This corresponds to a frequency of : F = V / W = 2e8 / 0.012 = 16,666,666,666.66 Gig
This took me by surprise, and I looked harder at my previous postings. There is a mistake there, but as I am about to leave for a well earned holiday in Crete, you will have to wait until my return for my announcement. Of course you are all allowed to speculate in my absence.
A quick answer to Trinopus. Relativity says that the speed of light is a constant (c) in any and all IFRs. To be constant in an IFR, it cannot be constant in an RFR. It is for that very reason that a ring laser (Sagnac interferometer) works. Read up on that.
To John Cazale. You consider yourself a layman, but that does not stop you disagreeing with the points I have made in this discussion. You cannot hide behind a layman’s tag when it suits you not to answer a question, and yet come out from behind it to disagree with me. Read back through these posts, you will find that I do know the difference between speed and velocity. You may be making that remark based on the observation that I sometimes use the word speed, and at other times use the word velocity. Sometimes they are interchangeable, sometimes they are not.