Apparent position and light travel time

[gonegahgah]

I was a little lost Randall when you initially gave my animation the okay. I had in my mind some more animations to do up to explore the notions further; but were they now redundant?

In some respects it is better now because I can again look at doing those animations and see where they take me.

For the moment I have redone the animation now in a way that I thought might be acceptable to yourself and Doc. Can I get your comment, and yours too Doc if you would kindly, on the present form as to whether they look acceptable to both of you.

What I have done is that I have redone the animations where the emission points of both objects now pass through the same point. When they both pass through this point they both emit a photon which travels towards the Earth along the same path. I have shown the photons as traveling together and reaching the receiver at the same time.

Just some caveats...

I have had to rotate the bluish object a smidgeon more clockwise around the photon emitter point to keep it facing the photon which travels out in a shower from it; than in the original animation. I have done the same with the bluish photons as well to keep them pointing directly away from the bluish object. I hope that makes sense.

I would just note that the emitted photon from the bluish object - although still perpendicular to the surface of that object - is no longer perpendicular to the path of the bluish object - as it was in the original animation. The emitted photon for the reddish object remains perpendicular to both the surface and the path of the reddish object as in the original animation.

Ignoring those last caveats does the attached animation now look okay?

Edit:
I added longer and coloured tails to the photons Randall to show more clearly where they face away and originate from. I hope this is in line with what you suggested? Thanks for the suggestion; it does look better.

Also I did notice as you mention that "the actual direction of the objects makes no difference at all in tracking these two photons". The telescope still has to be pointed at the same angle for both of them. I am still wanting to explore further animations and ideas in relation to this and also the ephemeris and related notes I am looking into. Hopefully this will become clearer for me as I do so; either way it turns out. But I certainly do acknowledge this.

 

[RandallB]

I don't understand the new plot. With the two pohtons produced at the same place & time I don't see why the rain drop points would be on the same vector, are you trying to point them at the future positions of the emitting objects for some reason?
I think you need make clear all four referance frames including the "preferred" CBR. It looks to me that all three of these red blue & Earth are traveling mostly to the left against the CBR frame.
The prior digagram was better showing the two photos traveling at different speeds in that Earth frame. Here they may as well be the same photon, makes it hard to see it ia not traveling at c in this frame diagram.

 

[RandallB]

What you are describing is not possible. You have four reference frames you are dealing with.
Earth, red & blue plus CMB.
Any one of the first three might be the same frame as CMB but each of those three cases is a new and different problem.

And as I said before the effect I think you were try to achieve in post 47 requires all three frames moving relative to each other to have their most significant movement be relative to the CMB frame (as the ‘preferred’ frame) and strongly to the left. Then the blue rain drop point would vector be about 2:00 O’clock and red at about 2:30 O’clock where you still show them in an opposite perspective. I think this angle perspective as it would exist while the two photons meet at Earth in true simultaneity should be same in all frames. That is red would never be above blue.

To build a record the way you want in each frame requires that you commit to a single preferred frame. You are not doing something that can fit with traditional relativity that allows any frame to serve as the preferred frame.

 

[gonegahgah]

Everything seemed to be okay until I brought in the fourth actor - dam I wish I had never hired him - but is that the case? I'll go back and recheck this with you Randall.

Every area on the Earth is passing through photon showers that come from various directions, many of which arrive at the same apparent locations as each other. Some objects will disappear fully or partly behind other objects - that they are behind or as in an eclipse - as far as it appears to us from our vantage point. So some photons attempt to travel to us via the same apparent paths but are blocked on the way.

In truth it is unusual for an object to approach another in the manner I have depicted in the last animation.

Despite that we should be able to depict what it should look like if such an event occurred. This is what I am wanting to do with your help Randall. I am trying to depict how you would have me depict it but there must be more accuracy to it than one retreating from "about 2:00 o'clock" and the other from "about 2:30 o'clock". Also it should be possible to depict anything from different perspectives with a different actor made to stand still.

But first may we recheck ideas from the previous animations that we covered to see if they are okay still?

The attached animation depicts only the reddish object and Earth. In the left frame, from the red object's perspective, it considers itself to be standing still and it emits a shower of photons in all directions from around its entire surface. The Earth passes through this shower of photons. One photon is emitted from one point and continues outwards until the Earth collides with it.

The photon is shown as retreating at the perpendicular to the tangent because it is a single photon traveling out as part of that face of the planet.

As previously discussed with Doc that he was then happy with we see:
1. the Sun not where it actually is in the sky but back where it was 8 minutes ago
2. but we see the face that is 8 minutes of arc further forward around from where we were 8 minutes ago
3. as that face looked 8 minutes ago.

In other words we see the face of the Sun that is currently facing us; not the face that was facing us when the photons began their journey; although as that current face looked 8 minutes ago. I'll just check: is this okay with you Randall?

The same can be said for the red object. The Earth passes through the shower that comes from the face that is currently facing it; not the face that was facing it when the photons first emitted; although the Earth does see an older picture of the current face. So this is why I depict the photons as retreating perpendicular from the face of objects (as per the space platform example).

So any depiction should show the photon retreating centrally perpendicular from the face it emitted from or as the face itself traveling outwards along the perpendicular.

The right frame depicts things from the Earth's perspective which considers itself to be standing still (all normal rotation is removed for simplification). As required by Doc, and which I agree with instead of the rotating translation I originally did, I have reciprocated the motion of objects to change this perspective. So the Earth moves at the same speed but in the opposite direction while keeping the red object still and vice-versa in the opposite frame. I have also maintained the exact position of the photon with respect to the other actors in each frame for each position of travel.

Any attempts by me to make, what I would have thought of as SR corrections for the photon between the frames, have been stamped upon so I have always returned to having this constant relationship between all the actors including the photon for both frames: the Earth, the Sun, the photon in the original animations; the Earth, the red object, the blue object, & their respective photon in the latest animations.

So, if we may, can we look at the current simplified animation attached. Is it okay or are there problems with the previous 'two actor one photon' animations Randall?

 

[RandallB]

...

The right frame depicts things from the Earth's perspective which considers itself to be standing still (all normal rotation is removed for simplification). As required by Doc, and which I agree with instead of the rotating translation I originally did, I have reciprocated the motion of objects to change this perspective. So the Earth moves at the same speed but in the opposite direction while keeping the red object still and vice-versa in the opposite frame. I have also maintained the exact position of the photon with respect to the other actors in each frame for each position of travel.

I am taking it that you are using Earth not as rotating or orbiting the sun but an object in space with location to be defined if we can.

Using both these two new diagrams only. Based on the orientation of the photon “raindrop” and how it moves from the perspective of both the red object and ‘Earth’ I can attempt to draw conclusions about the orientation of both with respect to a preferred reference frame defined by the CBR.

I would conclude that the red object is stationary wrt the CBR.
And Earth was not rotating but moving to the left very fast wrt that preferred frame.

More:
In order to achieve plots of unobservable “instantaneous” positions as I understand your trying to do, I still do not see any alternative but to define each of your frames wrt a preferred frame. And the CBR frame for a large area of local space is the only one would suggest.

I just don’t see how you will be able to define a photon with the raindrop shape you want without committing to a preferred frame.

 

[gonegahgah]

My apologise. I am just treating the Earth as another non-orbiting body.

Apparantly from our discussions it is not possible to see the same event from different points-of-view so I will cease that line of reasoning.

I will return to my going away and studying the ephemeris and come back later with other diagrams that I wanted to create and explore. Maybe they will be more practical?

 

[RandallB]

Apparantly from our discussions it is not possible to see the same event from different points-of-view so I will cease that line of reasoning.

NO not at all.
What your approach is revealing is that you cannot allow both POV to assume they are in a preferred frame of reference. Because the preferred frame of reference is going to define how the photon “raindrop shape” will appear in all reference frames. And only one can do that and still remain consistant.

The method you are using to plot photon paths into or away from an object at any angle will have that raindrop vector inline with the photon path for that objects reference frame only if in is using “The Preferred Frame”.

e.g. When an object or the Earth is not stationary with that CBR but moving in a straight line wrt the CBR one single line going through it will have the raindrop vectors in alignment with it. Of course the speed of the photons as plotted on that line cannot be appear to travel at “c”, the shadow thing again.

It took awhile to understand just what you were plotting with your technique. And I ‘m not sure what your objective in making them is. But I am sure if you do not account for this need to set a common preferred frame (presumably CBR based) I suspect you will only find confusion and nothing useful.

 

[gonegahgah]

Thanks Randall.

Doc, I apologise that I haven't still looked into the ephemeris question yet. There are too many things in this world to do. It is still something I want to get a better grasp of.

Another diagram occurred to me. Could I get your opinion on this diagram.

It depicts two bulbs and an eye. One of the bulbs is stationary relative to the eye and the other is moving at .6c across the vision of the eye.

The bulbs make contact in front of the eye completing a circuit which causes them to blink which is broken as soon as they pass.

The eye is at a level between where the bulbs contact so the distances at the time of contact are the same from the eye to each bulb.

Although one bulb is traveling at .6c they are effectively at the same point relative to the eye at the time of the blink so should the light take the same time to travel to the eye for both bulbs? If not could you provide a little explanation of why?

 

[Doc Al]

Although one bulb is traveling at .6c they are effectively at the same point relative to the eye at the time of the blink so should the light take the same time to travel to the eye for both bulbs?

That's what I would say.

 

[RandallB]

Could I get your opinion on this diagram.

It depicts two bulbs and an eye. One of the bulbs is stationary relative to the eye and the other is moving at .6c across the vision of the eye.

The bulbs make contact in front of the eye completing a circuit which causes them to blink which is broken as soon as they pass.

The eye is at a level between where the bulbs contact so the distances at the time of contact are the same from the eye to each bulb.

Although one bulb is traveling at .6c they are effectively at the same point relative to the eye at the time of the blink so should the light take the same time to travel to the eye for both bulbs?

Of course yes, but a small modification should make your example much more illustrative.

First, for clarity I’m sure when you say “across the vision of the eye” (observer) you mean to say that a line from that spot along the path of travel drawn to the observer would be perpendicular to that line of travel.
Now instead of requiring contact cause the light flash have both remain on emitting a monochrome color of yellow. Simply use a couple blocking screens about a mile long on either side of the perpendicular creating a slot that allows a blink of light to come to the observer from the traveling source. The local yellow light just glows. And as you suspected the each wave of light from both sources travel the perpendicular line at the same speed.
The question is how far apart will the separated waves of light be??
Or what color will that blink of light be? Blue Yellow Or Red

I’m sure you know that for the light visible from the traveler before the first screen starts blocking the view on the incoming approach will be seen as BLUE. A Doppler shift directly related to c being constant in all frames. Likewise I’m sure you know the light seen from the traveling source moving away after passing the second screen will be RED for the same Doppler shift reasons.
To guess the color of the light at passing do you split the diff and assume it will be Yellow?
OR might you consider the rate of time being used to source the light as compared to the rate of time of the observer?

There is a name for this example “transverse something” I don’t recall;
I suspect Doc Al likely knows and may have a link to a good explanation of it.

 

[Doc Al]

There is a name for this example “transverse something” I don’t recall;
I suspect Doc Al likely knows and may have a link to a good explanation of it.

This is called the transverse Doppler effect. I'll see if I can find a decent link describing it.

Edit: http://mysite.du.edu/~jcalvert/phys/doppler.htm"

 

[1effect]

This is called the transverse Doppler effect. I'll see if I can find a decent link describing it.

See

here:

and

here:

 

[gonegahgah]

Thanks All. So the answer is provided as "yes, the blink will arrive at the eye from both bulbs at the same time." I will explore this more here later.

Randall, good clarifications. You mention the doppler shift which I agree with.

Taking the following diagram at Wikipedia that 1effect kindly provided:
http://upload.wikimedia.org/wikipedia/en/e/e0/XYCoordinates.gif"

One thing it depicts is that for the observer who is moving is that the further objects are away to the side (that are stationary) the more they appear red shifted as the observer accelerates.

In my diagram the bulb is moving instead of the observer but that is just a relative consideration. We could consider the moving bulb to be stationary and that the eye is moving instead; it is all the same. In either respect I would expect as the Wikipedia diagram shows that the moving bulb would appear to do its one blink red shifted to the eye; while the stationary bulb would blink yellow.

That is correct; is it not?

 

[gonegahgah]

Doc, Randall, is that correct?

The following diagram I've drawn shows this doesn't it? It depicts the bulb with the eye moving by it at .6c. As in the first example the bulbs blink at point of contact which is when the eye is directly in line with the bulbs as per the bulb that it stays stationary relative to. The eye continues moving to the left relative to the moving bulb.

However as you said Randall the time the light takes to travel needs to be taken into account. By the time the light reaches the eye, the eye has now reached a position where it is moving away from the bulb. So the blink should be yellow for the bulb that remains stationary with respect to the eye; but should blink redder for the other bulb in the example that is moving in relation to the eye.

Doc and Randall, can you tell me whether that is correct or not?

 

[RandallB]

The following diagram I've drawn shows this doesn't it? ...

However as you said Randall the time the light takes to travel needs to be taken into account. ...
... tell me whether that is correct or not?

No not quite, (I'd like your diagram better if it compared just before and after reaching the perpendicular point).

I prefer the link from Dr Al, from that you should have picked up that the travel time for light was not the issue, nor what I had said. My comment was “might you consider the rate of time being used to source the light".

The point is you no longer have a Doppler Effect as if you were track side listening to a train and its whistle go by; It sounds different than the one parked next to you as it comes towards you or away from you, but for the instant it passes by it sounds the same. Why shouldn’t light behave the same way?

The source of light passing by relative to your observer is experiencing time at a slower rate thus as yellow light is cycling at a slower rate than yellow light generated in the observer frame and thus measured in the observer frame as having a longer (red shifted) wavelength.
Of course any observer moving with the source would experience the same ‘slow time’ and contracted lengths that define how the light is generated and will see the light as yellow.

The reason your observer sees red is because of the time rate difference between the two frames, not the Doppler Effect or any change in the “travel time for light”.

 

[gonegahgah]

I was going to get to sound in my next post. The main difference between sound and light of course is that sound requires a medium to travel through and light does not. I'll explore this difference tonight and ask for your feedback.

But ultimately the answer you have given is "yes, the observer will see the moving bulb do its one blink as redder than the stationary bulb" in this example. That is what I read in your reply. That is correct isn't it?

 

[RandallB]

Yes your result is correct
The issue is the cause of the "Transverse Doppler effect" :

A relative difference in time and how distance is measured between the two frames in the case of the light problem. (Again not the traditional Doppler Effect or any change in the “travel time for light”)

There is no significant difference in how time and distance is measured between frames with speed differences that apply to sound problems – hence there is no such thing as a "Transverse Doppler effect" for a sound example AFAIK.

 

[gonegahgah]

Thanks Randall.

Onto sound as I mentioned.

I've drawn some diagrams to depict the movement of sound through the air.

Each of the diagrams shows an observer (or listener), an emitter (a horn) and the transmission of the sound. It is an amazing horn because it produces a pure harmonic wave sound.
To the left is depicted what I expect the listener to hear.

Each of the diagrams shows a few things:
- the pitch of the wave relative to the air it moves through
- how long the wave takes to reach the listener (either 3 or 6 units of time)
- the relative movement of each of the actors either stationary or half speed of sound.

The actors include the listener, the horn, the sound and also the air mass through which the sound moves.

The 1st diagram depicts the listener, horn and air mass as all stationary. This is the base diagram for comparison.
The 2nd diagram depicts only the listener moving. As can be seen the sound gets to the listener in half the time and is twice the pitch.
The 3rd diagram depicts only the horn moving. As can be seen the sound takes the normal time to reach the listener but is twice the pitch.
The 4th diagram depicts only the air mass moving. This is the interesting as you can see the sound takes half the time to reach the listener but they hear a normal pitch.
The 5th diagram depicts the person and horn moving together. The listener hears the sound in half the time but at normal pitch.

Hopefully Randall (hopefully you will too Doc) you will agree with all these diagrams.

What they show is that the speed of sound is dependent on the movement of the air mass relative to the receiver (or vice versa if measuring from the receiver).

The diagrams should basically agree with real life experience.
1st) Whether an emitter is moving or not the sound will take the same amount of time to reach us.
2nd) If we move towards an emitter the sound will take less time to reach us.
3rd) As per 1st but if the emitter is moving it will change the pitch we hear.
5th) If you are in a car following another car the car in front will sound normal just as the last diagram shows (though the sound will reach you quicker)...
4th) and this is basically the same as the 5th diagram. If an air mass is traveling toward you then the sound it carries will travel towards you with its added speed.

From the diagrams it can be seen that the pitch you hear for sound is determined by the relative speed of emitter to you (I'll add after time delay and direction effects are taken into account but we can discuss that later), but the time taken for the sound to reach you is affected by the relative speed of the air masses around you.

Can you examine this and tell me what you think Randall & Doc.

 

[gonegahgah]

I should add the horn only makes one honk. The separate waves represent the honk at each time interval ie after 1 time unit, after 2 time unit, 3 time unit, 4 time unit, 5 time unit, 6 time unit.

The dotted actors represent where the actors were when the horn honked and the solid actors represent where they are when the listener hears the sound.

The sound wave lengths shown can also be considered as their lengths if time were frozen as well as representing their frequency relative to the air mass as mentioned.

 

[RandallB]

You might include a code for the wave form to indicate pitch heard.
N Normal, H High, L Low.
I interpret the shape of the wave forms in each left side ear diagram to be from top down as N, H, H, N, N.
With a speed near “S” or higher for the source you create a standing wave holding more and more energy (a sonic boom). Not sure how you want to show that and of course light does not do that.

(edit added comment)
So your notes are reasonable, but incomplete as they do not define how in some cases the volume is significantly affected as well. Example the horn can be made to move at a speed of 1.1s - faster than sound moves in the air; so upon reaching the observer the newer sounds will be heard very shortly before the older sounds. Not sure how you match up the phases of those – the energy contained in those sounds would not be canceled out by out of phase conditions. Glass breaking sonic booms from real tests show that energy remains very real.

I suppose these could demonstrate characteristics sound has that light does not display; in contrast to the "Transverse Doppler effect" light displays which sound does not.

So Yes your notes look ok, just be alert to faster than sound considerations

 

[gonegahgah]

That is correct in my opinion Randall; light doesn't use a medium to travel so it can not have the exact equivalent of a sonic boom.

It also tends to mean that light has a greater range on its frequency:

"It's thought that the short wavelength limit is the vicinity of the Planck length, and the long wavelength limit is the size of the universe itself (see physical cosmology), although in principle the spectrum is infinite and continuous." http://en.wikipedia.org/wiki/Electromagnetic_spectrum"

Putting N Normal & H High is a good idea to reduce confusion. You are correct about which ones are which.

I guess you are fairly happy with the results I depict?

The results show that if the sound emitter moves relative to the air mass then the air mass will draw the note along relative to it but the note gets stretched and lowered in pitch or shortened and increased in pitch due to being pulled away or pushed towards the air mass as it emits.

In the same respect the listener will change the pitch of the sound by moving towards or away from the air mass so that the note shortens and increases pitch as the listener moves towards and through it and it lengthens and decreases pitch as the listener pulls away from the note as it moves through them.

This is all good I hope?

 

[gonegahgah]

My apologies Randall; we must have crossed in our answers. Thanks for that reply.

Here are more diagrams showing sound when we are moving in a path that passes in front of the emitter - the path being perpendicular to a direct line to the emitter.

The leftmost diagram (1) shows the horn doing a single honk while the listener moves along the path at half the speed of sound. Although the listener was directly in front of the horn when it honked it takes longer to receive the sound ie. 4 units of time.

The middle diagram (2) shows the horn moving this time while the listener stands still. This time - although the horn is moving - the listener hears the honk in the normal length of time ie. 3 units of time, because the air mass relative to the listener determines how long it takes for them to hear the sound.

The right diagram (3) shows both the horn and air mass moving while the listener is standing still. The sound takes longer to reach the listener ie. 4 units of time, because it is carried with the air mass.

All diagrams depict that the listener hears the same change in pitch. This is because the relative movement between the horn and the user is the same ie. the horn is moving away from the listener at the same angle (or vice versa; it is all relative). The movement of the air mass does not change the pitch.

Again these examples are for sound. Are these diagrams still okay?

 

[RandallB]

Again these examples are for sound. Are these diagrams still okay?

No not for me
Let's take into account your statement (call them premises)

“because the relative movement between the horn and the user is the same …..(it is all relative). The movement of the air mass does not change the pitch.

In First diagram using the same “relative movement between the horn and the user” before the user past the horn it would be higher not lower. Same relative movement different pitch from different locations. You already said as much in prior diagrams.

Diagram two: Appears to display a transverse perpendicular transfer of sound from horn to user.
I’ve already stated my opinion that sound would not exhibit a "Transverse Doppler effect" and the diagram does not change my opinion of no change in pitch.

Diagram three: I consider evaluated incorrectly – had the horn remained stationary the pitch would have been lower because of the movement of the air contrary to your above premise.
Since the horn is moving away with the wind blowing as shown the pitch would be even lower.

As to: “the horn is moving away from the listener at the same angle” Don’t know you’re driving at; but that angle would never remain the same – and if angle has something significant to do with it (not important to me) these diagrams will not show it.

 

[gonegahgah]

"user"! Sorry Randall, I'm doing to much programming, lol.

I have to head off shortly so I will look at these further tonight hopefully.

I'm not 100% happy with the diagrams; they do need to be drawn better to show how things would more physically occur.

I just want to quickly look at diagram 2. What will actually happen as the horn moves in its path from the perpendicular point - where the honk begins - is that the honk wave will be dragged out by the horn's sideways movement; just like in the 3rd diagram of the previous 5 direct line examples - except that in that diagram the sound was squashed as the horn and listener are moving directly towards each other and in our new examples they are all moving transversely away. So the wave length would be altered to a lower pitch by the moving horn.

Does that sound better?

What I meant by "angle" is that in each of the three diagrams the horn and listener experience the same relative movement to each other - horn to left in 1st, horn to right in 2nd & 3rd agreed - but still in all the movement is from the perpedicular point and then moves away from that point at the same angle. In my opinion they will all hear the same sound because of this. I will look to diagram and explain this better tonight.

added:
Sorry should mention that the honk occurs at the point where the listener or horn is dotted and is heard where they are solid.

 

[gonegahgah]

I might go through the diagrams one by one Randall as they take a little time to do.

Here is the leftmost diagram tonight. You can see that I have modified it to more accurately reflect actual physical relativities.

Now it shows the sound wave as it spreads out in three directions. These co-incide with where the listener's ear is when each respective crest of the whole sound wave reaches the listener's ear. The green wave shows the relativity of the front crest of the sound wave, the yellow wave the middle crest, and the red wave the tail crest. Of course sound waves travel as compressions so the listener traveling across the sound will travel across the compressions making them longer or shorter depending upon how they are moving through them.

In our example the listener is moving away from through the compressions so the sound waves are longer to them and lower in pitch. This coincides with where they are as the sound reaches them and their relative direction of movement in relation to the sound.

It appears to me that the resultant wavelength is an addition of the original wave length with how far the listener has traveled between one crest and the next reaching them which will reduce or increase this wavelength.

In our example, where the listener passed beside the horn when it honked, the wavelength continues to get longer as the journey unfolds though the rate slows down as the listener heads off into infinity. This isn't shown super clearly in the diagram but it agrees with the minute measurements from the diagram.

Is this diagram correct now Randall? Can I do anything to improve it?

edit:
Should clarify that "rate slows down" means that the wavelength gets longer more slowly as the journey to infinity continues. The wave length will approach - but not reach - the original wavelength plus the distance traveled for one unit of time at half the speed of sound when the listener reaches towards infinity.

 

[RandallB]

New Comments & Diagram inculding angle comments still don't make sense to me.

 

[gonegahgah]

Hi Randall, sorry for my absence. Its taken me some time to nut out the following diagram and other things have been keeping me busy also.

I've looked at my second diagram where the horn is moving (and not the air nor listener) and have seen that it is wrong.

Here is a new diagram. It has three parts: a key, a diagram showing just the horn moving, a diagram showing just the listener and air mass moving.

I've dispensed with showing sine waves on the diagram as I've found it hard to represent the wave showing how the sound compressions move towards the point of hearing. Instead I've replaced it with expanding circles to represent particular crests of the expanding sound compressions. The green expanding circle is the leading compression (or crest), the yellow expanding circle is the middle compression, and the red expanding circle is the rear compression.

I'll explain the key first.
- The dotted figures represent where the actors (horn, listener, air) are at the time the middle crest is emitted ie. t(me) - time middle crest emitted.
- The solid figures represent where the actors are at the time the middle crest is heard ie. t(mh) - time middle crest heard.
Note: where an actor remains stationary the dotted form is hidden by the solid form.
Note: the air is left as invisible so you have to imagine it stationary and moving.
- The green dot shows the point where the lead crest is emitted from and heard at.
- The yellow dot shows the point where the middle crest is emitted from and heard at.
- The red dot shows the point where the rear crest is emitted from and heard at.
Note: where the point of emission or hearing is the same for each crest the green and yellow dot are hidden behind the red dot.
- The little rainbow indicates the early rainbow on the diagrams which show where all the crests are at time t(me) + time for one wavelength to travel ie. t(me) + λ / v(s).
Note: λ is wavelength, v(s) is velocity of sound.
- The crossed green, yellow, red lines indicate where on the diagrams each crest actually meets the ear (at different times of course).

From what I could work out the time each crest would be heard is:
t(mh) = perpendicular distance / v(s)
t(lh) = sqrt( t(mh)^2 + (λ / v(s))^2 ) - λ / v(s)
t(rh) = sqrt( t(mh)^2 + (λ / v(s))^2 ) + λ / v(s)
Note: lh - lead crest heard, mh - middle crest heard, rh - rear crest heard.

t(mh) is fairly obvious because the sound travels at the speed of sound directly along the perpendicular over the distance.
t(lh) & t(rh) both adjust the time of travel by the increased distance of traveling along the hypotenuse from the point of emission to the point of being heard.
Note: the hypotenuse is formed by the time traveled for one wavelength ie. λ / v(s) against the time to travel perpendicular giving: hyp = sqrt( t(mh)^2 + (λ / v(s))^2 ).
t(lh) emitted one wavelength earlier so that time has to be subtracted ie. - λ / v(s).
t(rh) emitted one wavelength later so that times has to be added ie. + λ / v(s).

This means t(lh) will be greater than t(mh) by amount z less difference of emission time,
and t(rh) will be greater than t(mh) by amount z add difference of emission time.
The result is that t(mh) - t(lh) < t(rh) - t(mh) but (t(rh) - t(lh)) / 2 = the normal wavelength.

It means that the sound from before the perpendicular will be higher pitch and the sound from after the perpendicular will be lower pitch.
Note: the amount will be barely noticeable in our example due to the small arc covered. The sound should be fairly normal at the perpendicular for this example.

That is completely different to what I said I know.

With the right diagram I have instead shown the air mass and person moving both at .5v(s) and the horn stationary. In all respects this diagram is basically just a change of perspective upon the first diagram. ie The horn moving and you and the air mass not moving can equally be considered to be you and the air mass moving and the horn not moving. So in all respects the results heard would be the same.

I note that the red and green crest overlap in the right diagram and almost match the yellow crest which is just slightly shorter. If you were to take the three overlapped hearing points (green, yellow, red) from the left diagram, kept them with their matching heard crests and separated the dots by the distance moved in the right diagram this would match how they look in the right diagram. So this would appear to be correct.

Can you study this for me Randall and make sure I have done everything correctly.

 

[RandallB]

I believe in our prior discussions I had advised that the Light version of your approach demanded that a preferred frame be established and followed by all frames. The traditional SR approach of any reference frame could be used as a preferred frame would not produce the same results. That is you could create descriptions from the view of any of the three reference frames with identical results only if each of those views took only one of the three frames as preferred for defining the correct “Apparent position” of each photon.

Now redoing this same “Apparent position” type of plotting effort for sound requires the same thing.
Only here it should be easier to identify which of the three potential reference frames should correctly be considered “preferred”. As we understand the behavior of sound fairly well, it is well accepted that air is the “ether” for transmitting sound, thus the preferred frame would need to be air.

Your second diagram in the horn frame needs to use “tear drop” shapes to show the correct direction of sound approaching the listener is not the same as the “Apparent path angle” as show by the “Apparent position” plot created for the Horn in its own reference frame.

Said another way the Aberration Angle would depend on the preferred reference frame. That angle of sound impacting the listener as shown in the first diagram needs to be shown as the same in the second diagram. The tear drops would be in-line with the path in the first Air based reference frame. But the tear drops would not be in-line with the path described in your “Apparent position” view as observed by the horn.

IMO the “tear drop” indicator is needed here as well, if your “Apparent Position” method is to be useful. It would be the part of the wave crest that is following the apparent path to the listener.

This would also indicate that apparent paths in-line with tear drops to be “correct” paths in the preferred frame.

As to the wide arch of other sounds going to different locations other than the listener. The depiction shown in the Air Frame seems OK, but what you have displayed in the second view does not appear to be a correct “Apparent position” path for them based on your definition of apparent position.

I’m a little uncertain of what value such a plot might be, but making a proper plot of it might be reveal something useful I’m not thinking of off the cuff right now. You may need a series of three listeners simultaneously working out each straight line case to figure what the apparent position wave front must look like in the Horn reference frame.