Apparent position and light travel time

[RandallB]

gonegahgah

I think I see the reference frame you are using for earth. It has no rotation or tidal locking with the Sun and is rotationally fixed with the distant stars or CMB. Thus the moon fixed on this frame from the Sun POV would appear to orbit the Earth once a year, but not at all from this Earth frame POV remaining fixed in one constalation. With Earth origin (0,0) orbiting the sun using that POV you are plotting the instantaneous positions of the photons as the appear to move relative to the (0,0) origin of that Earth reference frame.
Looks ok to me as long as you remember it is plotting apparent positions of some photons that never will be observed at Earth directly. Therefore some rules like FTL can be seen as being violated just like shadows can move FTL.

What are you using to make your animations, seems you build reasonably quick. Does it have an option to allow viewer to step animation frame by frame?

 

[gonegahgah]

I'm just using my old workhorse Corel Suite 12 to do my animations Randall. In reality it is a pain staking process of adding each piece in Corel Draw, converting it to bitmap, and adding it frame by frame in Corel Photo Paint.

The only way I can think to allow anyone interested to look at it frame by frame is for you to save the animation from the web page and load it into some other program. I can save it as a .mov file but this forum will not load that format.

You got the orientations I'm using in one. I did that to simplify the model as there are of course other factors to complicate things such as Earth rotation and wobble, Earth's dance with the moon, Earth's elliptic orbit, different mediums that the light travels through such as air, possibly the Sun's rotation and wobble (I want to look at Sun rotation shortly), the gravity wells, the observer jumping up and down, and probably other factors too.

I have just tried to maintain the same spatial relationships between all the actors Doc. All the distances between things are the same in both pictures. It is only because of the arc nature of Earth's orbit that this curve is translated to the right diagram. If the orbit were shown as flat rather than circular then the translation would be a straight line.

Is that wrong? Should the photon instead be shown as taking a different relative path when considered from Earth's perspective than the path it takes when considered from the Sun's perspective?

I thought at first that maybe a consideration of GR would straighten the line but from what I can tell it would actually increase the curve instead; not decrease the resultant curve. The photon when it leaves the Sun is in a deeper part of the well than the Earth is at any point in its orbit. Am I correct that under GR that paths are longer the deeper you are in a gravity well; even though under an X,Y,Z co-ordinate system we still see them as being the same length? If that is correct then this will make the journey the light takes out to the Earth longer increasing the amount of arc the Earth will transcribe before the light reaches it? Is what I just said rubbish?

Anyhow, principally we are looking at a simplified model. I had just thought GR might help to straighten the line hence the detour.

If I have a straight line for the photon in both diagrams then this will describe a different relative path for the photon. Is that thinking wrong? Would making it a straight line in both make it correct for you?

 

[gonegahgah]

Is the following animation okay Doc now that I have redrawn the two perspectives to have a straight line from emitter to receiver (even though the two pale yellow backed paths are now not spatially identical with each other against the surrounding actors between the perspectives nor the two grey backed paths with each other)?

The left attachment is the animation showing my attempted correction from the last animation and the right attachment is a static image showing each of the time positions superimposed for closer comparison.

I have also adjusted the speed of the photons so that for the Sun's perspective they travel at the speed of light relative to the sun and for the Earth's perspective they travel at the speed of light relative to the Earth. Is this okay?

So, as far as the Sun is concerned they travel directly to where the Earth will be and take that amount of time and distance to travel; yet as far as the Earth is concerned they travel directly also to where the Earth will be but instead take the amount of time and distance to travel to where the Earth was; not where it ends up being. Is this correct?

Have I taken one step too many; or was this a good extra step to take?

Randall, I just thought I should clear up that the grey backed path is not meant to be a shadow. Instead it co-incides with the face that was actually facing the Earth when the photon and face, represented by the pale yellow back path, were sent that actually end up colliding with the Earth.

So the diagram is meant to currently show that the face that was facing the Earth where it was will not be the face that is seen but that a face further around the Sun will be the one that collides with the Earth. This is something I will check with Doc shortly as well?

 

[RandallB]

Randall, I just thought I should clear up that the grey backed path is not meant to be a shadow. Instead it co-incides with the face that was actually facing the Earth when the photon and face, represented by the pale yellow back path, were sent that actually end up colliding with the Earth.

I actually think your first diagrams were a more accurate representation of the reference frames you were describing. The curves you had shown should not be plotted as straight lines in the Earth Frame. I’m not sure what reference frame these new plots show.

Remember, the plots you had made was for where in that Earth frame the photons were for given intervals of Earth frame time. Of course they plot as a curve in that frame but the photon is still directly observed as going straight, which you had also shown in your original diagram by depicting the photon in a rain drop shape showing the direction of travel when observed, and only when observed. I don’t see the need to eliminate the curve, you seem to be correcting something that is already correct IMO.

Also I did not say your plot was a “shadow” but was like a “shadow” and you must retain that and understand what it means. For example you substitute two high speed objects moving at 0.999c East and West measured by Earth Frame observers. Collecting the observations will clearly show a FTL separation of the observations recorded. Often this is miss-identified as FTL by is “Like The Shadow” FTL effect. Picking either object and using SR nothing is FTL. Just like Hubble expansion produces FTL measurements without FTL local events. (note you cannot calculate the solution using real photons traveling at c dividing by zero problem etc.)

So IMO you need to retain your curves, and expect apparent FTL & STL photon measurements in some cases. Because if you Warp the frame to account for the needed space-time curve to produce a straight line, it can only by done for one of the two photons your discribing since they are different vectors in both frames.

 

[gonegahgah]

I would tend to agree with you Randall. I would also like to know what are your thoughts on this Doc?

I've attached the unchanged animation from the previous post again and this time also added a second attachment of a static image showing each of the time positions from the animation superimposed for closer comparison. This is the best I can do I think where I can't provide a means to step through the animation frame by frame.

I still have the question about depicting the photon rate of movement. From the Sun's perspective the distance between where the photon emits and where it hits the Earth is shorter than for these two events in the Earth's perspective. Under the SR rules shouldn't I be depicting the photon as taking longer to reach the Earth in Earth's perspective; and less time to reach the Earth in the Sun's perspective?

This isn't a pivotal question for the diagrams that I am looking to explore but while we are here I would just like to know if that is the correct answer under SR.

 

[Doc Al]

I actually think your first diagrams were a more accurate representation of the reference frames you were describing. The curves you had shown should not be plotted as straight lines in the Earth Frame. I’m not sure what reference frame these new plots show.

On second thought, I agree. You were treating the Earth as an accelerating frame, which is OK.

 

[gonegahgah]

That term 'accelerating frame' sounds good; thanks Doc.

I did use the rain drop shape for the photons as you say Randall because, one reason being that the rain analogy is often used, but I did want to depict their orientation from the Sun and to the Earth that you mentioned.

I've attached another animation and composite static image. They are not much different to the last pair. This time they additionally not only depict the photon that reaches the Earth but also the face of the Sun that reaches the Earth.

They show the same face that the photon (the one that reached the Earth) was a part of is the one that reaches the Earth and not the face that was facing the Earth at the time both photons were emitted.

At the moment by this animation, where as you mentioned Randall that everything maintains orientation to the background stars (or CMB), we would end up seeing the current orientation (face) of the Sun that we are currently in front of but what it looked like approximately 8 minutes ago.

So is that still okay Doc? That is - taking into account Wikipedia - that 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.
Step 1 & 3 should be in line with Wikipedia so I'm just mainly wanting to confirm principally no. 2 at the moment. Does this all look okay from what you have been explaining to me? Is the attached animation okay?

 

[gonegahgah]

I've carried across our animations to a scenario where we have a space platform that has two super big TVs and there is a vessel that is passing by this space platform. TV A is showing Animal Planet while TV B is showing BBC.

I have eliminated the accelerated frame that we had in our previous orbit scenario Doc in the transition so you can see in the attached animation and composite image that both perspectives now have straight line travel for the photon and TV picture.

It also eliminates any GR considerations we would have to take into account if we were wanting to recognise all factors for the orbit example.

What I did want to bring across mainly was the idea that the photons poor out from the emitter like a shower and other moving objects pass through that shower; like the planet passing through the sun's shower and a car passing through a rain shower.

In that respect I have shown two TVs pointing in different directions. I have also put shutters out from the edges of the TVs so that the viewing angle is limited to directly out in front of each TV (the shutters need to be longer so please pretend that they are).

Again, as in the orbit example, the vessel and platform are non-rotating with respect to the background stars (or CMB).

What I am considering is what will we view when when the ephemeris data tells us that the platform is directly below us.

As I mentioned, Animal Planet is showing on TV A and BBC is showing on TV B. So when the vessel is directly above the platform - by the ephemeris data - Doc then is it correct - reflecting the nature of photon showers - that the vessel should only receive Animal planet and not be able to see BBC from where they are? In both perspectives?

 

[RandallB]

I have eliminated the accelerated frame that we had in our previous orbit scenario Doc in the transition so you can see in the attached animation and composite image that both perspectives now have straight line travel for the photon and TV picture. …. with respect to the background stars (or CMB)

Just remember again as graphed you can show “apparent FTL” affects, as those are caused by unadjusted relativistic views. However, since you have everything in straight lines you can apply SR rules. Resetting the graph using Lorentz factors to show the vessel frame as having longer distances with faster time (or shorter distances with shorter time will work just as well) relative to the TV Platform should correct any “apparent FTL” affects, simultaneity issues correctly considered of course. Pure SR says you can choose either frame as the preferred frame, BUT since you consider the CBR, Cosmology (contrary to the rules of SR) typically does consider the CMB as a marker for our local preferred reference frame for such calculations. You just need to define motions relative to the CMB as in is one of them stationary wrt CMB.

What I am considering is what will we view when the ephemeris data tells us that the platform is directly below us.

Ephemeris data tables are normally designed to tell us what is actually expected to be viewed directly. I think you are trying to define ephemeris data showing where things “instantaneously are” in a local frame but unobservable due to the delay of light travel.

 

[Doc Al]

So is that still okay Doc? That is - taking into account Wikipedia - that 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.
Step 1 & 3 should be in line with Wikipedia so I'm just mainly wanting to confirm principally no. 2 at the moment. Does this all look okay from what you have been explaining to me? Is the attached animation okay?

Seems reasonable to me.

As I mentioned, Animal Planet is showing on TV A and BBC is showing on TV B. So when the vessel is directly above the platform - by the ephemeris data - Doc then is it correct - reflecting the nature of photon showers - that the vessel should only receive Animal planet and not be able to see BBC from where they are? In both perspectives?

Assuming the programs are broadcast out in a narrow beam (much like a laser), I would agree with your statements.

 

[gonegahgah]

We will need to get to the measured time aspects of the receiver Randall but it hasn't been vital just yet for the ideas that I have been principally wanting to explore and verify in the current diagrams. I still have some more diagrams - currently in embryo form in my mind - that I would like to get to to see if we can cement these initial ideas further.

When we get to the measured time aspects, through other diagrams that I am hoping to provide, then I am looking to use examples that explain themselves. Again this is embryonic in my mind. To be honest I don't know what the result will be yet but I am thankful that you two are helping me to explore these multiple minute aspects.

I agree with you that only one object can be considered stationary with respect to the CMB at a time and the other must be considered as moving against it. Sorry for that error.

Mainly in the diagrams we have just looked at I am wanting to consider the emitter and receiver as having consistent orientation and no amount of rotation. So you have been explaining my examples as having no rotation with respect to the CMB. That is a much better way of saying it which I understand now. Thanks. Also the last diagram I am relying on inertia and the lack of large gravitational bodies to maintain the emitter and receiver in straight lines against X,Y,Z space co-ordinates.

I will have to take a further look into the ephemeris data because of the explanation you have provided. I will get back on this as soon as I can study it further. I want to be fully cognisant on it before I proceed to more diagrams.

Those televisions would be assumed in the examples to be narrow beam broadcasting as you say Doc.

I will get back later about the ephemeris data.

 

[gonegahgah]

Randall & Doc, I am going to be busy now until after the New Year so I would ask if I could continue to explore this more fully then.

I will be studying the ephemeris ideas when I get more chance. I have taken a little time to look at it and a few related topics a few times now since I last posted but I still need to study this further to get a fuller understanding of it. It is all interesting stuff.

Tonight I grabbed a little time to dabble with an animation. I would like to get both your comments on it if you could and when you have a chance. (As usual I've made an animation and a superimposed static image).

It shows the Earth and two objects moving in straight lines relative to the Earth which is stationary. One is moving tangential to the Earth and the other is moving at an angle. This angle brings it towards the Earth at an angle for most of the animation.

I have shown photons traveling towards the Earth; one from each object. The photons are following the same path and arrive at the same time at the same point on the Earth. However as you can see in the animation they leave from different points.

Could I get your comments (probably on the wrongness of this) please.

 

[RandallB]

I don't have any problem with it, as it shows what I've been referring to as "shadow" observations of the photos moving at different speeds. And in this case the Blue photon 'shadow' going through the Red object, which only a "shadow" could do.

You might want to enhance the raindrop image of the blue photon striking the Earth so it is more clear that both photon vectors have the same real length giving the same value of c over the 5 time steps.

Notice the actual direction of the objects makes no difference at all in tracking these two photons.

edit: Actually that is wrong
In order to achieve the effect your are defining of both photons arriving together while plotting paths not equal to speeds of “c” in this Earth frame. It will require Earth moving to the left against the astronomical preferred frame established by the CMB. Draw an equilateral triangle from the two objects at ime stamp two, with the point reaching a horizontal line level with Earth far to the left, to define the correct angle for the photon tear drop points with blue pointing up at higher angle the red. You have those angle reversed. With out getting the Earth frame to move against the preferred CMB frame I don’t think you can create the affect you intend.

 

[andrewj]

Yes. If you want to know where they are now, you have to take into account light travel time and work it out.

do you think light slows down over time

 

[Doc Al]

Yes. If you want to know where they are now, you have to take into account light travel time and work it out.

do you think light slows down over time

No. I have no reason to think so.

 

[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