What is the relationship between classical and relativistic Doppler formulas?

[Albertgauss]

Hi all,

This is a follow-up to post 140 on April 9, 2013, the Triplets20.jpg of Triplet's Paradox.

First of all, fantastic diagram.

Is the following logic right as can be gleaned off that excellent diagram of Triplets20.jpg right?

Both ships turn around (earth proper time 20 months) and approach each for a long time before either ship knows the other ship has turned around. In fact, it won’t be until Earth proper time 35 that each ship realizes the other has turned around. In fact, when they do realize the other ship has turned around, and did so awhile ago, the Doppler shift will go very fast from red to blue, hitting all the colors in between in sequence.

 

[ghwellsjr]

In a sense, the answer to your question is yes, but it will really go from way beyond red to way beyond blue. However, when we are considering spacetime diagrams like this, we like to ignore those kinds of details because unless the observers on the ships have some sophisticated electronic equipment, they will never be able to see the other ships. We like to focus instead on the timing of the signals. Fortunately, the real problems in seeing such ships will never occur since there is no way we could enact this scenario for a multitude of physical reasons.

Here is the image re-uploaded:

 

[Albertgauss]

"way beyond red to way beyond blue". My guess is that, in reality, it could go from radio to gamma. I know red and blue are a very small sub-spectrum (less than 0.1%) of what humans see. Is this what was meant here? If so, I'm fine here because I know red and blue shift pertain to the zillions of colors humans can't see.

Is the sophisticated equipment needed to hit the remainder of the spectrum that is normally invisible to human eyes (outside the rainbow)? Is it just an issue of things happening not only in other colors (radio, gamma, etc) but the change-over from red to blue shift being too quick (micro, nano, etc. seconds on timescales too short too perceive) or something like that?

Timing of the signals is pretty well-handled in the other posts, except for smooth acceleration in a turnaround which is normally left as a discontinuous junction for simplicity. But combining the results of the numerous other posts, that led me to guess what should happen here as the Doppler changes over when the turnaround is perceived by the receiving ship. It seems some kind of changeover must be possible and something observable, even if I don't know what is observable yet, I realize now.

Look's like I opened up another post.
"Fortunately, the real problems in seeing such ships will never occur since there is no way we could enact this scenario for a multitude of physical reasons."

Where I can read about the physical reasons for this? Is it the issue of rigidity under acceleration where g's are too high for ships to turn around or something like that? But, at lightspeed, its conceivable ships could turn around. How would their turnaround effects not be seen eventually when their light makes it observers farther away?

 

[sweet springs]

Where I can read about the physical reasons for this? Is it the issue of rigidity under acceleration where g's are too high for ships to turn around or something like that? But, at lightspeed, its conceivable ships could turn around. How would their turnaround effects not be seen eventually when their light makes it observers farther away?

You do not have to turn around the ship. You just jump off the going ship and board on the returning ship when they cross. Do not forget to wear a shock absorbing suit to save your life.

 

[ghwellsjr]

This thread is pertaining to this old thread:

https://www.physicsforums.com/threads/triplet-paradox.644948/

"way beyond red to way beyond blue". My guess is that, in reality, it could go from radio to gamma. I know red and blue are a very small sub-spectrum (less than 0.1%) of what humans see. Is this what was meant here? If so, I'm fine here because I know red and blue shift pertain to the zillions of colors humans can't see.

Yes, that is all correct.

Is the sophisticated equipment needed to hit the remainder of the spectrum that is normally invisible to human eyes (outside the rainbow)? Is it just an issue of things happening not only in other colors (radio, gamma, etc) but the change-over from red to blue shift being too quick (micro, nano, etc. seconds on timescales too short too perceive) or something like that?

That is all correct too.

Timing of the signals is pretty well-handled in the other posts, except for smooth acceleration in a turnaround which is normally left as a discontinuous junction for simplicity. But combining the results of the numerous other posts, that led me to guess what should happen here as the Doppler changes over when the turnaround is perceived by the receiving ship. It seems some kind of changeover must be possible and something observable, even if I don't know what is observable yet, I realize now.

Just like there are night-vision goggles that convert infrared into visible light and just like there are telescopes that see light in other spectra that present images that we can see, we can imagine similar instruments available to high speed astronauts. Of course, when an astronaut accelerates, as they both did at the Coordinate Time of 20 months, there will be an immediate Doppler shift so their instruments will need to "retune", or if they finally get the image of the distant astronaut accelerating as they did at the Coordinate Time of 35 months, they will also have to "retune". The difference is that in the first case it is predictable whereas in the second case it is not.

Look's like I opened up another post.
"Fortunately, the real problems in seeing such ships will never occur since there is no way we could enact this scenario for a multitude of physical reasons."

Where I can read about the physical reasons for this? Is it the issue of rigidity under acceleration where g's are too high for ships to turn around or something like that?

That would be one issue for very rapid turn around but the main problem is the energy requirements. It takes an enormous amount of energy to accelerate tiny particles to near lightspeed. Accelerating a spaceship to a speed where relativistic effects would be as dramatic as those that we describe in our scenarios is out of the question.

But, at lightspeed, its conceivable ships could turn around.

I don't know what you mean by this. First off, no spaceship or even a particle can be accelerated to lightspeed. And the closer to light speed a spaceship is going, the more energy it takes to turn it around, unless we consider a slingshot approach around a massive object.

How would their turnaround effects not be seen eventually when their light makes it observers farther away?

They can be seen with the right instrumentation.

There is another way to get around all these practical problems which is to image that each spaceship is emitting a very broadband, periodic flash of electromagnet energy which the other spaceships keep track of. Then no matter how much Doppler shift there is, there would be some energy in the visible range. So in my example above, we would have each ship emit a flash every month according to their own Proper Time.

 

[Albertgauss]

Hi all,

I think I got it with those replies, but I wanted to comment.

@sweetsprings

What you propose is a nice way to understand other effects of the twin paradox and make them easier to understand, but, if people could go near lightspeed, they wouldn't just hop between ships, they would turn the ship with thrusters. Yes, I am assuming a future where mankind can power ships to go near lightspeed. But its understanding the turnaround itself is a part of what I've been trying to understand these last few days.

"there will be an immediate Doppler shift so their instruments will need to "retune","

The "retuning" must be because of the change in reference frames. The ship begins in one inertial frame, progresses through a series of them until it ends up in a final inertial reference frame. Is this what was meant by that?

I know about the energy requirements of lightspeed and the impossibility of today's technology to put such a ship. I believe that mankind will find a way someway to get there, though, and that's okay to speculate about. In these posts, I just assume the energy requirements are met, even though there's certainly no "engine" yet, probably not for centuries.

"First off, no spaceship or even a particle can be accelerated to lightspeed"

I know, I just got lazy and rounded up. Meant to say "near lightspeed".

"emitting a very broadband, periodic flash of electromagnet energy..."

That's really clever all that goes with this line. I hadn't thought about that.

 

[sweet springs]

Hi

"there will be an immediate Doppler shift so their instruments will need to "retune","

Thanks for your comment. When you make U turn, you are in rotating system during you draw a semi circle. It's a littler bit complicated. Or you make a retrorocket until you get same speed but inverse direction to go home feeling acceleration toward the "go" direction. It's easier to handle so I will explain this case. The acceleration is gravity and the gravity well causes time delay. The Earth is far upward of the gravity well, so Earth time goes by very rapid and explains the story of the Earth aged brother.

Doppler shift does not appear hear. Maybe I have to read your posts carefully how Doppler appears and plays here. Best regards.

 

[ghwellsjr]

"there will be an immediate Doppler shift so their instruments will need to "retune","

The "retuning" must be because of the change in reference frames. The ship begins in one inertial frame, progresses through a series of them until it ends up in a final inertial reference frame. Is this what was meant by that?

There is no need to think in terms of a change in reference frames. Any single Inertial Reference Frame (IRF) can handle any scenario in Special Relativity. Look again at the diagram that I copied from the other thread in post #2. Now I have to admit that I didn't really look at it carefully myself and I see that I have made some wrong statements concerning how the Doppler changes for each ship. During the first leg of their trips, both ships see the other one red-shifted but at the Coordinate Time of 20 months (their Proper Times of 16 months), the Doppler shift goes away and they each begin to see the other ship normally because they are both traveling at the same speed in the same direction. Then at the Coordinate Time of 35 months (their Proper Times of 28 months), they each begin to see a blue Doppler shift of the other ship.

But these effects are not dependent on any particular IRF. All IRF's will indicate the same Doppler effects. Furthermore, even when people refer to an accelerated object (ship) as changing reference frames, they generally mean rest frames and in the above IRF, neither ship is ever at rest.

To illustrate, I'm going to upload another copy of the above IRF diagram but it will be a little smaller:

I want you to notice the ratio of the thin black lines intersecting the red dots along the thick red line representing the ship that is on the left. For the first leg of the trip, there are 4 red dots per thin black line. This means that the Doppler ratio is 0.25. The red ship sees the black ship's timing as 1/4 of its own.

Then at the red ship's turnaround, at its Proper Time of 16 months, there is a one to one relationship between the black lines and the red dots which continues until the red ship's Proper Time of 28 months. At that time the red ship sees the black ship turn around and he begins to see the timing of the black ship at 4 times his own which continues until all three observers meet up again.

Now you might be tempted to think (at least I was) that if we transformed to the rest frame of the red ship during the second half of the trip, we would also see the black ship at rest during at least part of that time interval:

Well it was a nice idea but it didn't pan out. Nevertheless, some of the signals that were sent while the black ship was at rest in this IRF arrive at the red ship while it is at rest.

But you should also notice that the three different Doppler ratios that we observed in the first IRF are present in this IRF in exactly the same way. Look at the Proper Times to see that the changes in Doppler ratios occur at the same times as far as each observer is concerned. Can you see that?

Finally, for completeness sake, here is the rest frame for the red ship during the first half of the trip:

Again, note the same Doppler ratios at the same Proper Times.

I know about the energy requirements of lightspeed and the impossibility of today's technology to put such a ship. I believe that mankind will find a way someway to get there, though, and that's okay to speculate about. In these posts, I just assume the energy requirements are met, even though there's certainly no "engine" yet, probably not for centuries.

Probably not for ever. Have you tried calculating the energy requirements?

 

[ghwellsjr]

Hi

Thanks for your comment. When you make U turn, you are in rotating system during you draw a semi circle. It's a littler bit complicated. Or you make a retrorocket until you get same speed but inverse direction to go home feeling acceleration toward the "go" direction. It's easier to handle so I will explain this case. The acceleration is gravity and the gravity well causes time delay. The Earth is far upward of the gravity well, so Earth time goes by very rapid and explains the story of the Earth aged brother.

We're not talking about the Earth aged brother, we're talking about a traveling triplet who is on the other side of Earth so I don't think your comments apply.

Even if we were talking about the Earth aged brother, we don't need to explain anything in this scenario by invoking gravity or a gravity well. This is not a General Relativity problem, it is merely Special Relativity which does not incorporate gravity.

Doppler shift does not appear hear. Maybe I have to read your posts carefully how Doppler appears and plays here. Best regards.

Yes, that would be a good idea. Read my posts carefully. We're talking about Doppler which always appears in all scenarios, at least conceptually

 

[sweet springs]

Thanks for #9 I am now considering Doppler effect.
Say there are lighted clocks on outer wall of the triplet's rockets. The triplets have telescopes to keep watching the clocks of the other two.
Doppler effect changes colors of images, analogue needle or digital number display. Not Doppler effect but relativistic time delay between IRs and the null world line relation between the rockets explain paces of the clocks observed. Am I right?

 

[PeterDonis]

Not Doppler effect but relativistic time delay between IRs and the null world line relation between the rockets explain paces of the clocks observed. Am I right?

Not necessarily. What "explains" what depends on how you want to conceptualize the scenario. For example, if you work it out you will see that the "paces of the clocks observed" exactly correlates with the Doppler shift observed; so each twin can, if he wants, "explain" everything he observes in terms of the observed Doppler shift.

Similar remarks apply to your explanation of the ordinary twin paradox in terms of a gravity well, and ghwellsjr's objection:

we don't need to explain anything in this scenario by invoking gravity or a gravity well.

This is correct, we don't *need* to invoke gravity or a gravity well, but it is not necessarily *wrong* to do so either; it's just another way of conceptualizing the scenario. (You do need to be aware, though, that this "gravity well" is not quite the same as a "real" gravity well produced by a gravitating body; it has zero tidal effects.)

The Usenet Physics FAQ article on the twin paradox goes into all this:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[ghwellsjr]

Thanks for #9 I am now considering Doppler effect.
Say there are lighted clocks on outer wall of the triplet's rockets. The triplets have telescopes to keep watching the clocks of the other two.
Doppler effect changes colors of images, analogue needle or digital number display. Not Doppler effect but relativistic time delay between IRs and the null world line relation between the rockets explain paces of the clocks observed. Am I right?

The Doppler effect is the problem that Special Relativity solves. When two inertial objects have motion between them, they both observe the same Doppler effect in the other one. This was a rather surprising result when it was discovered and especially when light was believed to travel at a constant speed relative to a fixed medium. For other Doppler effects such as sound in air or waves on the surface of water, the Doppler effect is not the same for both objects depending on their motions through the medium. The way light behaves presented a challenge that was simply solved by Einstein postulating that the light travels at a constant speed relative to any inertial frame and not just to one inertial frame related to a fixed medium.

As a result of that postulate, which cannot be observed, and his first postulate that includes the reciprocal Doppler effect, which is observed, Special Relativity solves the reciprocal Doppler effect by establishing that time slows down for objects that are moving according to the Inertial Reference Frame in which light is traveling at a constant speed. SR also derives the Lorentz Transformation so that when you create a scenario according to one IRF, you can transform all the coordinates for the events in that IRF to another IRF moving with respect to the first IRF and all the observables, including the Doppler effects, will remain intact in the second IRF. That's what my spacetime diagrams depict.

As you have noted, each IRF has a different [coordinate] time delay for the light to travel and a different [coordinate] time delay between [Proper Time] ticks of the clocks along each worldline. In addition, there are different [coordinate] distances that the light travels and different [coordinate] distances that the objects travel and different [coordinate] speeds that the objects travel, but the same [coordinate] speed for the light to travel. And throughout it all, the Doppler effects remain the same in each IRF.

 

[ghwellsjr]

Not necessarily. What "explains" what depends on how you want to conceptualize the scenario. For example, if you work it out you will see that the "paces of the clocks observed" exactly correlates with the Doppler shift observed; so each twin can, if he wants, "explain" everything he observes in terms of the observed Doppler shift.

Similar remarks apply to your explanation of the ordinary twin paradox in terms of a gravity well, and ghwellsjr's objection:

we don't need to explain anything in this scenario by invoking gravity or a gravity well.

This is correct, we don't *need* to invoke gravity or a gravity well, but it is not necessarily *wrong* to do so either; it's just another way of conceptualizing the scenario. (You do need to be aware, though, that this "gravity well" is not quite the same as a "real" gravity well produced by a gravitating body; it has zero tidal effects.)

The Usenet Physics FAQ article on the twin paradox goes into all this:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

The link specifically applies the gravity well to the twin paradox. Can it be made to work for the rest frame of the red ship explaining what happens to the black ship?

I don't see the radar method described in the Usenet link. I think it is the best one of all as I described in post #144 of the original Triplet thread:

https://www.physicsforums.com/threads/triplet-paradox.644948/page-8#post-4508402

Unfortunately, until they fix the links to the images, you will have to click on the thumbnail to see the diagram.

But here I have uploaded it again:

 

[PeterDonis]

The link specifically applies the gravity well to the twin paradox. Can it be made to work for the rest frame of the red ship explaining what happens to the black ship?

Yes (assuming you mean the red ship in this thread's scenario), but it would be a different "gravity well" (in the sense of a different non-inertial frame--see below) than if we were using it in the rest frame of the black ship. The "gravity well" being invoked is frame-dependent; it only exists in a non-inertial frame in which a particular ship (red or black) is always at rest, and only while that ship is firing its rockets--and the specific non-inertial frame will be different for each ship, so the specific "gravity well", and with it things like which ship is "higher up" in the gravity well, will be different. If you want to object that such a "gravity well" is fictitious and not "real", I sympathize (and an invariant way of stating this would be that this "gravity well" has zero tidal gravity, and therefore spacetime is still flat, not curved), but bear in mind that the "gravity well" is an interpretation--it's a way of making intuitive sense of what the math is telling you when you construct the non-inertial frame in question.

Also, since the black ship and the red ship are both non-inertial, each one's motion in the "gravity well" of the other is more complicated than in the simpler case discussed in the Usenet article. (But the motion of the blue ship in each case will be the same as the motion of Terence in the Usenet article.)

(Note that saying "the" non-inertial frame in which a given ship is at rest is a misnomer: there is no such single unique non-inertial frame. The one the FAQ article is implicitly using is probably the most "natural" one to use, but there are other possible ones, and not all of them would support a "gravity well" interpretation, at least not in the simple form in which the article gives it.)

I don't see the radar method described in the Usenet link.

You're right, it isn't. I think that's because the radar method is a different way of constructing a non-inertial frame than the one the article implicitly uses in its "gravity well" analysis.

 

[sweet springs]

Hi. I am getting to know Doppler effect you mean.

Proper frequencies of light or light frequencies observed in the rest frames are common knowledge of the triplet so that they calculate the relative speeds.

Powers of light source are also common knowledge in order to know by telescope brightness observation the distances or how long time ago the relative speeds above took such values.

Is that correct?

 

[ghwellsjr]

Hi. I am getting to know Doppler effect you mean.

Good, I'm glad to hear that.

Proper frequencies of light or light frequencies observed in the rest frames are common knowledge of the triplet so that they calculate the relative speeds.

Powers of light source are also common knowledge in order to know by telescope brightness observation the distances or how long time ago the relative speeds above took such values.

Is that correct?

I think your comments are really more appropriate for the cosmology forum. I think these issues are more complex than the simple answers we would get from just Special Relativity.

 

[ghwellsjr]

..The "gravity well" being invoked is frame-dependent; it only exists in a non-inertial frame in which a particular ship (red or black) is always at rest, and only while that ship is firing its rockets...

Are you saying that it takes three different frames to explain the entire scenario: one before the ship fires its rockets, a second while it fires its rockets and a third afterwards?

Is it possible to draw a spacetime diagram to depict this interpretation? Or does it take three?

 

[sweet springs]

The formula of relativistic Doppler effect,
$$\frac{\nu}{\nu_0}=\sqrt{\frac{1+\beta}{1-\beta}}$$,
gives red shift to leaving bodies and blue shift to approaching bodies.
SR gives same time dilation to both the leaving and the approaching bodies.
How observed Doppler effect and SR law coincide?

 

[m4r35n357]

The formula of relativistic Doppler effect,
$$\frac{\nu}{\nu_0}=\sqrt{\frac{1+\beta}{1-\beta}}$$,
gives red shift to leaving bodies and blue shift to approaching bodies.
SR gives same time dilation to both the leaving and the approaching bodies.
How observed Doppler effect and SR law coincide?

The doppler effect describes what you would see with your eyes or a measuring instrument. Time dilation is a propery of the underlying model (Minkowski spacetime), and you would not (normally) see it directly.

OK, so that's probably not very helpful, but it's important to know. I would suggest reading up on the relativistic doppler effect, because the equation you have given is only one-dimensional. The full equation (which gives the doppler shift for two or more dimensions) and lots of other information, including the way the doppler effect relates to time dilation(gamma) is given here: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

There is another effect that you should probably know about at some point, which is the aberration of light. http://mathpages.com/rr/s2-05/2-05.htm This is the key to what you would really see in relativistic situations. You might not be ready for it now, but your line of questions is likely to end up here eventually. Incidentally, the Wikipedia link is not very good which is why I linked to a better one.

OK, enough hard sums, to get an idea of what this all looks like you can download an interactive demo of SR visualization at: http://realtimerelativity.org/

There is also a game demo using these principles at http://gamelab.mit.edu/games/a-slower-speed-of-light/

Or look at this thread https://www.physicsforums.com/threads/my-twin-paradox-youtube-channel.772796/ for my home-made videos.

 

[sweet springs]

PS We see images of Doppler light, and feel as though such and such happens now as we see..
Not now due to propagation delay. Not as seen because Doppler effect including direction of motion exaggerate or dissipate what is really happening.

 

[sweet springs]

Thanks for #19.
Doppler image sequence during the trip coincides with SR, of course.
It does not mean classical Doppler image sequence causes or proves SR effects including triplet paradox, does it?

 

[ghwellsjr]

The formula of relativistic Doppler effect,
$$\frac{\nu}{\nu_0}=\sqrt{\frac{1+\beta}{1-\beta}}$$,
gives red shift to leaving bodies and blue shift to approaching bodies.
SR gives same time dilation to both the leaving and the approaching bodies.
How observed Doppler effect and SR law coincide?

I showed examples of this in post #8. The observed Doppler shifts are the same in all IRF's but the Time Dilations are different.

 

[ghwellsjr]

Thanks for #19.
Doppler image sequence during the trip coincides with SR, of course.
It does not mean classical Doppler image sequence causes or proves SR effects including triplet paradox, does it?

Classical Doppler doesn't work for light so I don't see why you would expect it to cause or prove SR.

 

[ghwellsjr]

PS We see images of Doppler light, and feel as though such and such happens now as we see..
Not now due to propagation delay. Not as seen because Doppler effect including direction of motion exaggerate or dissipate what is really happening.

What observers see is what is really happening but no one should jump to the conclusion that what they see of a distant object is happening at the same time they see it.

 

[sweet springs]

What observers see is what is really happening.

So you say leaving red-shift clock work slow and aproaching blue-shift clock work fast as the observer see?

 

[sweet springs]

Classical Doppler doesn't work for light so I don't see why you would expect it to cause or prove SR.

Relativistic, OK. Relativistic Doppler effect is derived from SR. I think relativistic Doppler effect give us nothing new to us who already know SR.

 

[ghwellsjr]

So you say leaving red-shift clock work slow and aproaching blue-shift clock work fast as the observer see?

No, clocks don't work any slower or faster just because of the way some remote person is looking at them.

Consider two inertial observers with clocks approaching each other from a great distance. They will each see the others clock working faster than their own. This is a blue Doppler shift. As they pass each other, what they see suddenly changes but the clocks are not working any differently. As they recede away from each other, they each see the others clock working slower than their own. Now it is a red Doppler shift. Each of their observations is what they each actually and really see but they don't jump to the conclusion that just because they first see the other ones clock going faster and then slower after they pass that the clocks have actually and really changed how fast they are working.

 

[ghwellsjr]

Relativistic, OK. Relativistic Doppler effect is derived from SR. I think relativistic Doppler effect give us nothing new to us who already know SR.

I would not say that the Relativistic Doppler effect is derived from SR. I would say that it is one of the many aspects of the first postulate of SR (the Principle of Relativity), independent of the second postulate of SR (the constancy of light propagation). Of course, SR has to demonstrate the authenticity of Relativistic Doppler in order to comport with reality. So in that sense I would say that SR is derived from Relativistic Doppler (among a lot of other things).

 

[sweet springs]

Now it is a red Doppler shift. Each of their observations is what they each actually and really see but they don't jump to the conclusion that just because they first see the other ones clock going faster and then slower after they pass that the clocks have actually and really changed how fast they are working.

They do not judge speed of the clocks by observed Doppler shift, thanks.

 

[sweet springs]

I would not say that the Relativistic Doppler effect is derived from SR. I would say that it is one of the many aspects of the first postulate of SR (the Principle of Relativity), independent of the second postulate of SR (the constancy of light propagation).

Classical Doppler coeficient $$1-\beta$$ multiplied by $$\gamma$$ becomes relativistic Doppler $$\sqrt{\frac{1-\beta}{1+\beta}}$$. Is this independent of the second postulate?

 

[ghwellsjr]

Classical Doppler coeficient $$1-\beta$$ multiplied by $$\gamma$$ becomes relativistic Doppler $$\sqrt{\frac{1-\beta}{1+\beta}}$$. Is this independent of the second postulate?

Where did you get that formulation for the classical Doppler coefficient? It has only one speed in it. It needs to have two, one for the source relative to the medium and one for the receiver relative to the medium. Here's the formula from wikipedia:

 

[sweet springs]

Any receiver can regard that the medium of light rests with him. Put $$v_r=0$$ for IFR of the receiver. It is beyond a pure Newtonian view, I admit. Further, relativistic time dilation factor $$\gamma$$ should be multiplied. I emphasized this factor.

 

[ghwellsjr]

Any receiver can regard that the medium of light rests with him.

If that is true, then isn't it also true that any receiver can regard that the medium of light rests with the source? Or that the medium of light is traveling at any speed with regard to the receiver?

Put $$v_r=0$$ for IFR of the receiver.

If I do that, I don't get your Classical Doppler coefficient that you posted in #30.

It is beyond a pure Newtonian view, I admit.

Then why are you pursuing this?

Further, relativistic time dilation factor $$\gamma$$ should be multiplied.

Why? Where is it declared that you should multiply by gamma instead of dividing or doing something else?

I emphasized this factor.

Did you come up with this argument on your own or can you provide a reference for it?

If you want to see how Special Relativity calculates the Relativistic Doppler factor, you can look in Einstein's 1905 paper in section 7:

http://www.fourmilab.ch/etexts/einstein/specrel/www/

I think you are misunderstanding what I am saying. Please go back and read my post #12 and see if there is anything in it that you disagree with or don't understand.

 

[sweet springs]

Thanks. I should be more careful.

Following the classical way obeserver at rest in media observe light from the source moving with speed beta (positive for leaving) as

$$\nu\ at\ rest=\frac{1}{1+\beta}\nu_0\ in\ motion$$

Observer moving in media observe light from the source at rest is

$$\nu\ in \ motion=(1-\beta)\nu_0\ at\ rest$$

In order these two coincide

$$\nu_0\ in\ motion=\sqrt{1-\beta^2}\nu_0\ at\ rest$$
$$\nu\ in\ motion=\sqrt{1-\beta^2}\nu\ at\ rest$$

Now we know there is no media. Choose one, at rest or in mortion, and delete the other.

$$\nu=\sqrt{\frac{1-\beta}{1+\beta}}\nu_0$$

Improved, I hope.

 

[sweet springs]

PS FIrst order approximation for both the formula is $$\nu=(1-\beta)\nu_0$$. I confused it be classical Doppler effect of light.