Can Light Emission Be Explained Through Newtonian Physics?

[Austin0]

The tube rotates in the moving frame , this was established in the another thread on the Thomas-Wigner rotation. In fact, the Terrell-Penrose effect is nothing but another facet of the Thomas effect.
The light beam also appears inclined in the moving frame due to the well-known effect of aberration. For the particular case of light traveling along the y-axis in frame S, it will travel at an angle cos(\theta')=\frac{cos(\theta)+\beta}{1+\beta cos(\theta)}=\beta with axis Ox' in frame S'. This is exactly the same angle the tube rotates as well, so both light beam and the tube rotate by \theta'=arccos(\beta)=arctan(\frac{1}{\beta \gamma}). So the photon does not hit the sides of the tube in either frame S or S', it travels perfectly centered along the axis of the tube in both frames.

Does any of this relate to the case where the tube is orthogonally aligned to motion of the emitter but in another observing frame??
It appears to be related to the case where the tube is at rest in the emitter frame. Yes?

 

[JesseM]

How can this stated possibility be definitively false. Especially given the equivalence of mass and energy

Because that very equivalence between mass and energy implies that E = gamma*mc^2 for massive particles, where m is the rest mass and gamma is the usual 1/sqrt(1 - v^2/c^2)...in the limit as v approaches c, gamma would approach infinity, and so would E for any particle with nonzero rest mass m! But obviously photons only have finite energy, given by the quantum formula E=hf (where f is frequency and h is Planck's constant).

Incidentally the equation E=gamma*mc^2 can also be written as E^2 = m^2c^4 + p^2c^2, where p is momentum. You can see that for a particle with m=0, this reduces to E=pc, so a photon's does have momentum (combining with the quantum formula E=hf, the momentum of a photon is p=hf/c)

In this instance it is not the general law under discussion but a specific limited aspect.
The constancy of the speed of light itself means that emissions in the direction of motion cannot conserve the forward motion of the source as would be the case with unlimited speed and simple additive velocities. No?

What do you mean by "conserve the forward motion of the source"? If the source is pointing perpendicular to its direction of motion, then the photon has the same horizontal speed s_H as the source (so it always remains directly above the current position of the source), and its vertical speed s_V is such that s_V^2 + s_H^2 = c^2 (i.e. the total velocity vector has a magnitude of c)

Given the picture I have so far it would seem impossible for any photon to have a straight linear path orthogonal to the emitter motion as observed in any other frame.
That if there is a light detector at the end of a long light absorbtive tube aligned directly perpendicular to the source motion but in another frame, that it would seem impossible for a photon to reach the detector??

Does the light detector have the same velocity as the source, so it always remains directly above the source? Or is the light detector at rest in the frame where the source is moving horizontally, so the source is only directly underneath the detector at the moment it emits the light? In the first case the light will hit the detector, in the second case it won't.

 

[Austin0]

How can this stated possibility be definitively false. Especially given the equivalence of mass and energy

Because that very equivalence between mass and energy implies that E = gamma*mc^2 for massive particles, where m is the rest mass and gamma is the usual 1/sqrt(1 - v^2/c^2)...in the limit as v approaches c, gamma would approach infinity, and so would E for any particle with nonzero rest mass m! But obviously photons only have finite energy, given by the quantum formula E=hf (where f is frequency and h is Planck's constant).

Incidentally the equation E=gamma*mc^2 can also be written as E^2 = m^2c^4 + p^2c^2, where p is momentum. You can see that for a particle with m=0, this reduces to E=pc, so a photon's does have momentum (combining with the quantum formula E=hf, the momentum of a photon is p=hf/c).

Well I certainly wouldn't argue with any of the above. ANd in fact don't have any investment in massive photon stock, but coincidentally, in an unrelated thread I just encountered a paper talking about the calculated upper limit of possible photon mass so it is not a totally absurd topic of speculation. I think approaching limits of any kind leads to a certain fuzziness. I understand the concept of infinite energy needed to accelerate any massive particle to c ,,,but it would not bust my capacity for universal strangeness if photons were exempt because they can only move at c and are not accelerating from a lower velocity. Buts that just my feeling.

In this instance it is not the general law under discussion but a specific limited aspect.
The constancy of the speed of light itself means that emissions in the direction of motion cannot conserve the forward motion of the source as would be the case with unlimited speed and simple additive velocities. No??.

What do you mean by "conserve the forward motion of the source"? If the source is pointing perpendicular to its direction of motion,((1)) then the photon has the same horizontal speed s_H as the source (so it always remains directly above the current position of the source), and its vertical speed s_V is such that s_V^2 + s_H^2 = c^2 (i.e. the total velocity vector has a magnitude of c).

What I meant was your ((1)) above. Equally true of massive particles yes??
That with emissions in the direction of motion the s_H component is not conserved at all i.e. does not contribute to the velocity vector c

Does the light detector have the same velocity as the source, so it always remains directly above the source? Or is the light detector at rest in the frame where the source is moving horizontally, so the source is only directly underneath the detector at the moment it emits the light? In the first case the light will hit the detector, in the second case it won't.

The second case.
Question?? Can a photon emitted at any angle from the moving source possibly enter the tube and reach the detector. It seems clear that there would not be but at this point I want to check everything rather than make assumptions
Don't you find it somewhat [read extremely] curious that photons are not totally independent of the motion of the source and do act like massive particles wrt transverse emissions?

ANother related question: What happens in the case of an emitter, stationary in the lab and aligned orthogonally to a moving frame if the photon is reflected off a metal reflector on the moving frame?

from previous post
3) With regard to a massive particle say a bullet that is rotating around the axis of motion: in a relative frame the forward momentum would result in a sideways drift relative to the spin??
Wrt a photon this is not such a comfortable picture. If we consider a photon as a traveling waveform this would mean it would be out of phase along an orthogonal front relative to its linear motion, yes?
It would seem to be more realistic to view the angle derived from the vector sum of conserved momentum as an actual propagation angle with the wave front in phase, relative to travel, does this make any sense??

Thanks

 

[starthaus]

SO in the actual context of the question are you saying that a photon does not conserve the forward momentum of the source motion??

I am not saying anything of the kind, I simply pointed out that your whole post is a collection of errors.

 

[yuiop]

Thanks Bob S
It appears to me that kev is talking about a different thing.
As I understand your diagram the angled tube is stationary in the lab frame.

If the tube was angled and stationery in the lab frame, then it would not be stationary in the co-moving (CM) frame. In this case the photon does not reach the end of the tube in either reference frame (if we do not allow reflections off the side the tube by e.g. lining the internal tube surface with absorptive material).

In kevs diagram the tube is stationary in the emitter frame. In that case there has never been a question about it reaching the detector.

Yep, in this case the photon reaches the detector in both reference frames and the tube is measured to be orthogonal to the line of motion in both frames.

So am I correct in thinking that if the tube is orthogonal in the lab frame no photon emitted at any angle from the moving emitter can reach the detector at the end of the tube?

No, this is not true. See the diagram I attached to my last post. The tube has to be orthogonal to the line of motion in both frames in order for the photon to reach the detector in both frames, even though the tube does not appear to be parallel to the photon path in the frame where the tube and emitter are not at rest.

I think the Terrell effect as perceived in the emitter frame would make the tube in the lab appear tilted forward on approach and even more forward [away from the observer] on recession. And likewise from the lab if the tube is in the emitter frame.

This is sort of correct but is very dependent on the location of the observer relative to the rod and what exactly you mean by "tilted forward".

Let us say we had a very long wire and a hollow tube threaded onto the wire and the tube is moving at relativistic speeds relative to the wire and an observer that is rest with the wire and standing to one side of the wire. In this case the observer sees the circular cross section at the back of the tube when the centre of the tube is opposite the observer as an elliptical cross section. The appearance of the tube is the same as the silhouette of the tube when it is rotated to an angle wrt the observer, but it is the shape of the tube that appears to change rather than the orientation of the tube wrt the observer. The long axis of the tube still appears to be parallel to the long wire that is it threaded onto. In this case there is no rotation of the long axis of the tube, but visually at first glance there appears to be. Basically Terell rotation is an optical illusion due to light travels times and the exact position of the observer, which is an aspect that is normally factored out in relativity calculations.

See these cool 3d optical illusions that come about as a result of your assumptions about the shape the object you are looking at:

http://www.maniacworld.com/Crazy-House-Optical-Illusion.html



http://www.youtube.com/watch?v=laty3vXKRek&feature=related

http://www.youtube.com/watch?v=br8dZXG1tKE&feature=related

http://www.youtube.com/watch?v=TK4qdJ8nAXM&feature=fvw

http://www.youtube.com/watch?v=Nru145ftJD4&feature=related

Sorry for all the links, but I could not find exactly the optical illusion effect I was looking for, but the anamorphic street art is probably the closest. They are fun to look at anyway.

 

[yuiop]

The tube rotates in the moving frame , this was established in the another thread on the Thomas-Wigner rotation.

We established in the other thread that the tube rotates if it is not exactly parallel or orthogonal to the motion. In the case considered in this thread, the tube is orthogonal to the motion and does not rotate.

In fact, the Terrell-Penrose effect is nothing but another facet of the Thomas effect.

It is not. It is an independent effect. Thomas rotation is a physical rotation that is measured by a grid of observers that all at rest in a given reference frame and light travel times are not a factor. There is effectively an observer at each location of the tube. Terrell-Penrose rotation is an optical illusion observed by a single observer where light transmission times cause the illusion of the object rotating and changing shape.

The light beam also appears inclined in the moving frame due to the well-known effect of aberration. For the particular case of light traveling along the y-axis in frame S, it will travel at an angle cos(\theta')=\frac{cos(\theta)+\beta}{1+\beta cos(\theta)}=\beta with axis Ox' in frame S'.

Correct. This is the rotation of the light path as observed in the frame that sees the emitter as moving.

This is exactly the same angle the tube rotates as well, so both light beam and the tube rotate by \theta'=arccos(\beta)=arctan(\frac{1}{\beta \gamma}). So the photon does not hit the sides of the tube in either frame S or S', it travels perfectly centered along the axis of the tube in both frames.

This is wrong. The tube does not rotate as well. As I said before, consider the case of a light clock with the source and detector at one end of the tube and the mirror fixed to the other end of the tube. Imagine the tube light clock is at rest in frame S and aligned with the y axis. To an observer in another reference frame S' that is moving in the x direction relative to S (and all the axes of the two frame are parallel) the tube is still aligned with the y' axis while the photon path is at an angle. If the tube rotated in the S' frame it would have to rotate one way while the photon was on its outward leg and the other way when the photon was on its return leg, which is obviously silly.

 

[starthaus]

We established in the other thread that the tube rotates if it is not exactly parallel or orthogonal to the motion. In the case considered in this thread, the tube is orthogonal to the motion and does not rotate.

Err, this is incorrect. You may want to read on the Terrell effect. It does rotate, exactly by the angle \theta'=arccos(\beta)

It is not. It is an independent effect. Thomas rotation is a physical rotation that is measured by a grid of observers that all at rest in a given reference frame and light travel times are not a factor. There is effectively an observer at each location of the tube. Terrell-Penrose rotation is an optical illusion observed by a single observer where light transmission times cause the illusion of the object rotating and changing shape.

Err, this is incorrect as well, the math is exactly the same, it has to do with marking endpoints of a rod simultaneously. See my blog on the Thomas rotation.

This is wrong. The tube does not rotate as well.

The calculations show the opposite. Please do some reading on the Terrell effect.
The tube rotates by exactly the same amount the light beam is aberrated. This explains why in both the frame of the tube+emitter and in the frame of the relatively moving observer , the light beam hits the detector at the other end of the tube (because it moves in alignment with the axis of the tube).

As I said before, consider the case of a light clock with the source and detector at one end of the tube and the mirror fixed to the other end of the tube. Imagine the tube light clock is at rest in frame S and aligned with the y axis. To an observer in another reference frame S' that is moving in the x direction relative to S (and all the axes of the two frame are parallel) the tube is still aligned with the y' axis while the photon path is at an angle. If the tube rotated in the S' frame it would have to rotate one way while the photon was on its outward leg and the other way when the photon was on its return leg, which is obviously silly.

Interesting argument but fallacious. On the return path, the light is aberrated by \pi-\theta'. Interestingly, the Terrelll effect says that the tube appears tilted by the same exact angle. This explains why, in the frame of the observer, the light never hits the walls of the tube. If it did, you would have managed to design an experiment that falsifies PoR :-)

 

[yuiop]

The tube rotates in the moving frame , this was established in the another thread on the Thomas-Wigner rotation.

We established in the other thread that the tube rotates if it is not exactly parallel or orthogonal to the motion. In the case considered in this thread, the tube is orthogonal to the motion and does not rotate.

Err, this is incorrect. You may want to read on the Terrell effect. It does rotate, exactly by the angle \theta'=arccos(\beta)

In the "other thread" we were talking about Thomas rotation and now you are getting it confused with Terrell rotation. The Terrell rotation depends on the location and velocity of the object relative to the observer and the original orientation of the object relative to the line of motion, so the equation for the Terrell rotation has a lot more factors than the equation you have given.

Err, this is incorrect as well, the math is exactly the same, it has to do with marking endpoints of a rod simultaneously. See my blog on the Thomas rotation.

The math for Thomas rotation and Terrel rotation is not the same. Thomas rotation is to do with marking the endpoints of the rod simultaneously in the reference frame of the observer. Terrell rotation on the other hand is to do with the simultaneous arrival of the light signals at the film of the camera or the back of the eye and what the observer sees is not a depiction of where the the end points of the rod are simultaneously in the observers reference frame. In the Terrell effect, it is because the observer sees a composition of events on the object that are not simultaneous, that he sees an apparent optical distortion or rotation of the object.

The calculations show the opposite. Please do some reading on the Terrell effect.
The tube rotates by exactly the same amount the light beam is aberrated. This explains why in both the frame of the tube+emitter and in the frame of the relatively moving observer , the light beam hits the detector at the other end of the tube (because it moves in alignment with the axis of the tube).

Simply wrong.

Interesting argument but fallacious. On the return path, the light is aberrated by \pi-\theta'. Interestingly, the Terrelll effect says that the tube appears tilted by the same exact angle. This explains why, in the frame of the observer, the light never hits the walls of the tube. If it did, you would have managed to design an experiment that falsifies PoR :-)

Simply wrong again. PoR is preserved and the light arrives at the detector or mirror at the end of the tube in both reference frames, despite the fact the tube does not rotate while the path of the photon does appear to rotate.

 

[Austin0]

Thanks Bob S It appears to me that kev is talking about a different thing.
As I understand your diagram the angled tube is stationary in the lab frame .

If the tube was angled and stationery in the lab frame, then it would not be stationary in the co-moving (CM) frame. In this case the photon does not reach the end of the tube in either reference frame (if we do not allow reflections off the side the tube by e.g. lining the internal tube surface with absorptive material).

I think you are confused here. In Bob's example the tube, which is stationary in the lab frame , where we are also, is aligned at the proper angle to admit the photon

In kevs diagram the tube is stationary in the emitter frame. In that case there has never been a question about it reaching the detector.

Yep, in this case the photon reaches the detector in both reference frames and the tube is measured to be orthogonal to the line of motion in both frames.

As I said this was not in question , I specifically mentioned a light clock in a previous post which is essentially what your diagram depicts if there is a mirror at both ends

So am I correct in thinking that if the tube is orthogonal in the lab frame no photon emitted at any angle from the moving emitter can reach the detector at the end of the tube?

No, this is not true. See the diagram I attached to my last post. The tube has to be orthogonal to the line of motion in both frames in order for the photon to reach the detector in both frames, even though the tube does not appear to be parallel to the photon path in the frame where the tube and emitter are not at rest.

Your diagram does not depict the subject of this enquiry. The tube and the emitter are not in the same frame and in this question the emitter is in motion not the tube or either is in motion but not in the same frame.

I think the Terrell effect as perceived in the emitter frame would make the tube in the lab appear tilted forward on approach and even more forward [away from the observer] on reccession. And likewise from the lab if the tube is in the emitter frame.
Thanks for your helpful time

This is sort of correct but is very dependent on the location of the observer relative to the rod and what exactly you mean by "tilted forward".

Let us say we had a very long wire and a hollow tube threaded onto the wire and the tube is moving at relativistic speeds relative to the wire and an observer that is rest with the wire and standing to one side of the wire. In this case the observer sees the circular cross section at the back of the tube when the centre of the tube is opposite the observer as an elliptical cross section. The appearance of the tube is the same as the silhouette of the tube when it is rotated to an angle wrt the observer, but it is the shape of the tube that appears to change rather than the orientation of the tube wrt the observer. The long axis of the tube still appears to be parallel to the long wire that is it threaded onto. In this case there is no rotation of the long axis of the tube, but visually at first glance there appears to be. Basically Terell rotation is an optical illusion due to light travels times and the exact position of the observer, which is an aspect that is normally factored out in relativity calculations.

By tilted forward I meant toward the moving observer in the emitter frame as he approached the tube assuming his POV was in line with the tube and away from him as he passed it and it fell behind.
The assumption being that the tube is horizontal and the observer's POV is aligned with it.
And I think it would be an apparent rotation in every optical sense. A tube is the ideal object for the effect with the right symmetry to produce a perfect illusion.

Thanks for the links ,,very interesting

 

[starthaus]

In the "other thread" we were talking about Thomas rotation and now you are getting it confused with Terrell[/b] rotation.

If you ever did the computations , you would have found that they are the same. Since you never completed the computations, there is no way for you to find out.

The Terrell rotation depends on the location and velocity of the object relative to the observer and the original orientation of the object relative to the line of motion, so the equation for the Terrell rotation has a lot more factors than the equation you have given.

No, I have given you everything that you needed to complete the computations. I simply did not do it for you, you need to do some work in order to learn. If you went back to the respective thread and if you tried to calculate the effect on a rod aligned with say, the y axis, as seen from the perspective of an observer moving along te x axis, you'd find out, to your surprise (and contrary to your repeated claims) that, in the frame S' of the observer, the rod is rotated away from the y' axis. You need to complete these calculations.

The math for Thomas rotation and Terrel rotation is not the same. Thomas ration is to do with marking the endpoints of the rod simultaneously in the reference frame of the observer.

Precisely.

Terrell rotation on the other hand is to do with the simultaneous arrival of the light signals at the film of the camera or the back of the eye and what the observer sees is not a depiction of where the the end points of the rod are simultaneously in the observers reference frame.

Err, wrong. The simultaneous "arrival" of light beams to the observer's film means exactly marking all the points of the rod simultaneously in the frame of the observer. This is the practical way of marking endpoints simultaneously in the case of a moving object. How else would you do it?

Simply wrong again. PoR is preserved and the light arrives at the detector or mirror at the end of the tube in both reference frames, despite the fact the tube does not rotate while the path of the photon does appear to rotate.

If you say so :-)

 

[JesseM]

Well I certainly wouldn't argue with any of the above. ANd in fact don't have any investment in massive photon stock, but coincidentally, in an unrelated thread I just encountered a paper talking about the calculated upper limit of possible photon mass so it is not a totally absurd topic of speculation.

It is absurd if you are both postulating that photons move at c and that they have mass. Any paper speculating about photon mass is presumably speculating that photons actually move slower than c (where c is defined to be the fundamental constant that appears in the Lorentz transformation/time dilation equation/etc., not defined as the speed of photons). I think approaching limits of any kind leads to a certain fuzziness. I understand the concept of infinite energy needed to accelerate any massive particle to c ,,,but it would not bust my capacity for universal strangeness if photons were exempt because they can only move at c and are not accelerating from a lower velocity. Buts that just my feeling.

What I meant was your ((1)) above. Equally true of massive particles yes??

Yes, it's equally true of massive particles that if the emitter is pointed perpendicular to its direction of motion, then the particle's horizontal speed s_H will be the same as the emitter.

That with emissions in the direction of motion the s_H component is not conserved at all i.e. does not contribute to the velocity vector c

What do you mean "does not contribute to the velocity vector c"? Do you disagree that in this case s_H^2 + s_V^2 = c^2? That looks like a "contribution" to the total speed of c to me. For example, we might have s_H=0.8c and s_V=0.6c, so then it'd be true that (0.8c)^2 + (0.6c)^2 = (0.64 + 0.36)*c^2 = c^2.

Question?? Can a photon emitted at any angle from the moving source possibly enter the tube and reach the detector. It seems clear that there would not be but at this point I want to check everything rather than make assumptions

It would depend on how wide and deep the tube was, I suppose. An angled beam might still enter one side of the tube and hit the detector on the other side.

Don't you find it somewhat [read extremely] curious that photons are not totally independent of the motion of the source and do act like massive particles wrt transverse emissions?

No. If the photon direction was independent of the source, that would falsify relativity, since there would be a single preferred frame where a photon emitted by a tube pointing in one direction would actually travel in the same direction. The first postulate of relativity says that the laws of physics should work the same way in all inertial frames, which means that if experimenters in inertial windowless ships moving at different velocities all perform the same experiment, they should all get the same result--the first postulate would be violated if an experimenter in one frame found that light exited the emitter parallel to the orientation of the emitter, but an experimenter in a different frame found that it exited at an angle relative to the orientation of the emitter.

Are you familiar with the http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_basics/index.html#Light1 thought-experiment? This thought-experiment actually depends on the fact that light always travels parallel to the emitter in the emitter's rest frame in order to derive the time dilation equation.

ANother related question: What happens in the case of an emitter, stationary in the lab and aligned orthogonally to a moving frame if the photon is reflected off a metal reflector on the moving frame?

It doesn't matter, reflection only depends on the orientation of the reflective surface at the moment the light hits it, not on the velocity of the reflective surface. If the reflector is horizontal and the beam is vertical in the lab frame, the light will bounce back down vertically in the lab frame.

from previous post
3) With regard to a massive particle say a bullet that is rotating around the axis of motion: in a relative frame the forward momentum would result in a sideways drift relative to the spin??

I don't know what you mean by "sideways drift relative to the spin". The linear momentum of the center of mass is conserved as long as no forces are acting, so in zero gravity a spinning bullet will travel in a straight line just like a nonspinning bullet.

Wrt a photon this is not such a comfortable picture. If we consider a photon as a traveling waveform this would mean it would be out of phase along an orthogonal front relative to its linear motion, yes?

Don't understand this sentence. "Out of phase" with what other wave, exactly?

 

[yuiop]

I think you are confused here. In Bob's example the tube, which is stationary in the lab frame , where we are also, is aligned at the proper angle to admit the photon

There is room for confusion here, because Bob did not indicate in which frame the tube is stationary, in his drawings. However, if you assume the angled rod is at rest in the lab frame, then it is moving in the emitter (CM) frame and still at an non-orthogonal angle. This is not what Bob drew in his diagrams. In his diagrams, the rod is orthogonal in the CM frame.

As I said this was not in question , I specifically mentioned a light clock in a previous post which is essentially what your diagram depicts if there is a mirror at both ends.

Yes, my diagram is essentially a classic Einstein light clock.

Your diagram does not depict the subject of this enquiry. The tube and the emitter are not in the same frame and in this question the emitter is in motion not the tube or either is in motion but not in the same frame.

You are right that my drawings are for a emitter and tube both at rest wrt each other and are correct in that context. Bob's drawing is not correct for the scenario you describe, or for any other scenario, because the tube is never parallel to the photon path in both frames, if the photon path is is not parallel to the relative motion of the frames.

I have attached a new diagram showing the correct relative motion and path orientations of the photon, tube, emitter and receptor for the scenario you describe in the attached drawing. The photon is the red blob and its path is shown by the red velocity vector. The rest of the objects are labelled in the diagrams.

 

[yuiop]

Err, wrong. The simultaneous "arrival" of light beams to the observer's film means exactly marking all the points of the rod simultaneously in the frame of the observer. This is the practical way of marking endpoints simultaneously in the case of a moving object.

You are exposing a fundamental lack of understanding of the meaning of simultaneity in a reference frame here.

How else would you do it?

You would do it by placing observers that are all at rest wrt each other at all the events that are being measured. These observers synchronise clocks with each other using Einstein clock synchronisation convention. If an infinite number of observers is required, then so be it. To measure the length of a moving rod, one observer that happens to be at one end of the moving rod at time t, compares his coordinates with another observer who happens to be at the other end of the rod at the same time t. We know these times are simultaneous in the mutual rest frame of the observers, because their clocks are synchronised. This gives the correct coordinate measurement of the rod in their rest frame. If on the other hand a single observer located at one end of the rod tries to measure the length of the moving rod by comparing the coordinate of the nearest end of the rod with the coordinate of the far end of the rod at the time the light arrives from the far end, he will get an incorrect coordinate length measurement, because the light from the far end was emitted at an earlier time and not simultaneously with the light from the near end. The method of using a multitude of observers (and synchronised clocks) that are all at rest with respect to each other, is fundamental to the definition of a reference frame and removes all ambiguities due to light transmission times from distant events. Terrell rotation comes about because of differences in light travel times creating an optical distortion, while Thomas rotation (and time dilation and length contraction) is what is left when all distortions due to light travel times have been removed. It is essential to understand this basic fact, very early on in any study of relativity.

 

[starthaus]

You would do it by placing observers that are all at rest wrt each other at all the events that are being measured.

This is impractical. The method that I gave you is practical.

If an infinite number of observers is required, then so be it.

What did I just tell you?

To measure the length of a moving rod, one observer that happens to be at one end of the moving rod at time t, compares his coordinates with another observer who happens to be at the other end of the rod at the same time t.

Given the fact that I wrote up the exact mathematical description of the process, I do not think that the above is necessary, nor that it can ever be implemented.

If on the other hand a single observer located at one end of the rod tries to measure the length of the moving rod by comparing the coordinate of the nearest end of the rod with the coordinate of the far end of the rod at the time the light arrives from the far end, he will get an incorrect coordinate length measurement, because the light from the far end was emitted at an earlier time and not simultaneously with the light from the near end.

This is not how it's done. Nor is this what goes on in the Terrell paper. I can explain that to you but we are digressing. What I pointed out is that , contrary to your beliefs, the tube is tilted by the same angle as the beam of light. I even gave you the tools how to calculate the tilting angle, a computation that you never completed.

 

[starthaus]

I have attached a new diagram showing the correct relative motion and path orientations of the photon, tube, emitter and receptor for the scenario you describe in the attached drawing. The photon is the red blob and its path is shown by the red velocity vector. The rest of the objects are labelled in the diagrams.

It is good to see that you corrected your errors from post #23.
The tube rotates by the same angle as the aberration angle of the light beam. Now, your new pictures do include any derivation of the rotation of either the light beam or the tube, so, how did you derive the amount rotation? Did you simply accept my proof?

 

[Austin0]

This is not how it's done. Nor is this what goes on in the Terrell paper. I can explain that to you but we are digressing. What I pointed out is that , contrary to your beliefs, the tube is tilted by the same angle as the beam of light. I even gave you the tools how to calculate the tilting angle, a computation that you never completed.

So what does go on in the Terrell paper?

 

[JesseM]

Um, isn't Terrell-Penrose rotation a purely optical effect, affecting only the visual appearance of moving objects? So if you want to just calculate the coordinate position of different parts of a tube at a particular coordinate time in some inertial frame, I think it'd be irrelevant. On the other hand, Thomas precession is said in the wikipedia article to be a kinematic effect, and kinematics concerns coordinate positions and coordinate times. This article on the relativity of simultaneity has a simple example of how a horizontal rod moving vertically in one frame will be slanted in the frame of an observer moving horizontally in the other direction (see section 4 of the article, 'An Application: The Rotation of Bodies in Transverse Motion')--would this be an example of Thomas rotation? Based on Rasalhague's comment about deriving Thomas rotation from the relativity of simultaneity in post #133 here, I would guess so.

 

[starthaus]

Um, isn't Terrell-Penrose rotation a purely optical effect, affecting only the visual appearance of moving objects? So if you want to just calculate the coordinate position of different parts of a tube at a particular coordinate time in some inertial frame, I think it'd be irrelevant.

No, the calculations are the same since they involve figuring out where the points of an object are located at the same moment of time as viewed by a moving observer.
Either way, I gave the exact mathematical formalism of how this is calculated in another thread. I left a few calculations for kev to finish.

On the other hand, Thomas precession is said in the wikipedia article to be a kinematic effect, and kinematics concerns coordinate positions and coordinate times. This article on the relativity of simultaneity has a simple example of how a horizontal rod moving vertically in one frame will be slanted in the frame of an observer moving horizontally in the other direction (see section 4 of the article, 'An Application: The Rotation of Bodies in Transverse Motion')--would this be an example of Thomas rotation?

Yes. The point was that kev , in post 23, missed the fact that the tube is rotated by the same amount as the angle of aberration of the light beam traveling through it. Thus, the light beam does not hit the walls of the tube as incorrectly shown in kev's post 23.

 

[yuiop]

Yes. The point was that kev , in post 23, missed the fact that the tube is rotated by the same amount as the angle of aberration of the light beam traveling through it. Thus, the light beam does not hit the walls of the tube as incorrectly shown in kev's post 23.

You are wrong. If you understand the diagram in post 23 correctly, the photon does not hit the walls of the tube in any of the reference frames. The diagram is correct in the context that it was given in. Call this scenario 1. In this scenario, the emitter and the tube are at rest in the CM frame and the tube and photon path are parallel to the y-axis the CM frame. In the lab frame that is moving along the x-axis of the CM frame, both the emitter and tube are moving in this frame and the photon path is at an angle to the y-axis while the tube remains parallel to the y axis. The tube and photon path are not parallel in the lab frame in this scenario, but the photon does not hit the side of the tube in either reference frame.

It is good to see that you corrected your errors from post #23.
The tube rotates by the same angle as the aberration angle of the light beam. Now, your new pictures do include any derivation of the rotation of either the light beam or the tube, so, how did you derive the amount rotation? Did you simply accept my proof?

The new diagram I posted in #42 is for a different scenario. (Call it scenario 2). It is not a correction of post 23. In scenario 2 the tube is at rest in the lab frame while the emitter is at rest in the CM frame. In this scenario the lab operators manually align the tube with the photon path to ensure the photon passes through the tube. In the transformation from the lab frame to the CM frame, the tube and photon path rotate by different amounts (contrary to your claims) and the tube and photon path are not parallel in the transformed frame. In both scenarios 1 and 2, the photon path is necessarily parallel to the tube in the rest frame of the tube by design, but in both scenarios when we transform to another reference frame the photon path and tube are no longer parallel. Both scenarios are consistent about this and both scenarios/diagrams contradict your claim that the rotation angle of the photon path (aberration) is the same as the rotation angle of the tube. This is simply not true.

Derivation of formulas for second diagram:

If the photon path is parallel to the y-axis in the CM frame then we know if the lab frame is moving in the negative x direction, that the x component of the lights velocity vector is equal to the \beta_x. We also know that the speed of light is c in all reference frames so we can say c = 1 = \sqrt{\beta_x^2+\beta_y^2} = \sqrt{\beta_x^2' + \beta_y^2'}
This implies \beta_y' = \sqrt{1 -\beta_x^2'. The angle of the photon path in lab frame (S') is \theta' = atan(\beta_y'/\beta_x' ) and by substitution of the y' component we obtain \theta' = atan \left(\sqrt{(1-\beta_x^2')}/\beta_x'\right) = atan\left( \frac{1}{\gamma \beta_x'} \right).

This is the equation given by BobS earlier. We can also say by symmetry, that \theta' = atan(y'/x') where (x',y') is a coordinate point on the photon path in frame S'.

The angle of the statioanry tube is the same as the angle of the photon path in the lab frame, because it is manually aligned by the lab operators to allow the photon to pass through. The angle of the moving tube in the S (CM) frame is \theta = atan(y/ x) = atan(y' *\gamma/x') = atan(tan(\theta')* \gamma). This is effectively due to the Thomas rotation of the tube (which is the difference between \theta and \theta') and is not the same as the aberration of the photon path. Thomas rotation derived in two lines! Therefore the photon path and tube are not parallel when we transform to the emitter frame and this is what is depicted in the diagram attached to post 42. You agree the diagram in post 42 is correct, but it contradicts your claims that the photon path and tube rotate by the same angle under transformation.

It seems to me that you are saying the diagram posted by BobS in post #22 where the tube and photon path are parallel in both reference frames is correct. Is that the case? I think the diagram posted by BobS is incorrect and I do not mean to "harp on" about Bob's error because with all due respect he acknowledges he might be mistaken (and he deserves credit for being the first to post the correct equation for the photon path rotation that we all agree with), but I have to bring it up because you seem to be adhering to what his diagram is showing. What is you position? Which diagram do you think is correct? If you think it the one I posted in #42 then you will have withdraw your claim that the photon path and tube rotate by the same amount under a transformation to a different reference frame, because that is not what it shows. If the tube and photon path are the parallel in one frame and if as you suggest the photon path and tube rotate by the same amount under a transformation, this implies that the photon path and tube is parallel in all reference frames which simply in not true.

 

[yuiop]

To measure the length of a moving rod, one observer that happens to be at one end of the moving rod at time t, compares his coordinates with another observer who happens to be at the other end of the rod at the same time t.

This is impractical. The method that I gave you is practical.
...
Given the fact that I wrote up the exact mathematical description of the process, I do not think that the above is necessary, nor that it can ever be implemented.

Seeing as it would only require two observers to measure the length of a moving rod, I would not say it is impractical. It is also essentially the method used by Einstein in his original papers.

The Terrell rotation you keep going on about, is just an optical illusion as JesseM said. A train moving along a track appears to rotate relative to the track as seen by a single observer at the side of the track, but it is not a physical rotation as can readily be seen by noting that the wheels at the front and back of the train remain in contact with the track and are therefore at the same distance from the observer.

[PLAIN]http://www.anu.edu.au/physics/Searle/Tram_warp.jpg

The apparent illusion of being rotated is not even good. The part of train that is rotated away from the observer is the same height as the part of the train that is rotated toward the observer, so there is depth perception of being rotated. All that happens is if you assume all points of light that arrive simultaneously at the eye or film back were emitted simultaneously, is that the train is shortened and sheered giving a vague impression of being rotated.

The attached image is a screenshot from the educational video about Terrell rotation (See the "BACKLIGHT.rm" video on this website http://www.anu.edu.au/physics/Searle/ ) . The yellow outline is the composite model of the tram that is made when it is incorrectly assumed that all points of light from the tram were emitted simultaneously. I have added two white lines to indicate the location of the track. The composite model (of points that are simultaneous) is still parallel to the track and just looks superficially like it is rotated. Is a model is correctly composed of points on the train that are all simultaneous in the observers frame, the train is rectangular, parallel to the track and length contracted. Note that while length contraction is not visible in a Terrell rotated sphere, it is visible in a Terrell rotated rectangular object.

 

[JesseM]

No, the calculations are the same since they involve figuring out where the points of an object are located at the same moment of time as viewed by a moving observer.

What do you mean "the calculations are the same"? Are you suggesting that, if I have a grid of rulers defining my coordinate system, then the markings on my ruler which visually appear to coincide with different points on the object at a single instant actually correspond to the position-coordinates the object occupied at a single moment in my frame? (i.e., the positions which different points on the object occupied simultaneously according to my rest frame's definition of simultaneity) If so that's clearly incorrect--the Terrell-Penrose effect depends on the fact that the light rays from different points on a moving object which reach my eyes at the same instant were actually emitted at different time-coordinates in my frame. For example, one simple case of the Terrell-Penrose effect is that for an object traveling straight towards me, if a light ray from the back of the object reaches my eyes at the same moment of a light ray from the front, the light ray from the back must actually have been emitted at an earlier time-coordinate in my frame, which causes the object to visually appear stretched to a length greater than its coordinate length in my frame.

For example, suppose a ship with a rest length of 10 light-seconds is traveling towards me at 0.8c, traveling in the +x direction. If the front reaches my own position x=0 at t=100, then the front has x(t) = 0.8c*t - 80 (when you plug in t=100 you get x=0). Since the ship is length-contracted to a length of 6 light-seconds in my frame, that suggest that at t=100 the back must have been at position x=-6, so the back must have x(t) = 0.8c*t - 86. Now, suppose at t=50 the front emits a light signal from position x=0.8*50 - 80 = -40, which reaches my eyes 40 seconds later at t=90. Also suppose at t=20 the back end emits a light signal from position x=0.8*20 - 86 = 16 - 86 = -70, which reaches my eyes 70 seconds later, also at t=90. That means that at t=90, I am seeing the front end lined up with the x=-40 mark on my ruler, and also seeing the back end lined up with the x=-70 mark, so visually the object appears to be 30 light-seconds long at that moment, five times greater than its "true" length of 6 light-seconds in my frame.

 

[starthaus]

the photon path is at an angle to the y-axis

You got this part right. This is due to aberration.

while the tube remains parallel to the y axis.

This is the part you got wrong. This is due to the fact that you never completed the calculations from the other thread, despite a lot of encouragement I've given you.

The tube and photon path are not parallel in both reference frames in this scenario,

You are repeating the same error as above.

 

[starthaus]

the Terrell-Penrose effect depends on the fact that the light rays from different points on a moving object which reach my eyes at the same instant were actually emitted at different time-coordinates in my frame.

I understand the Terrell effect extremely well. I will do a writeup for you and place it in my blog. Before I do that, you need to consider the following: the image of any object is generated by all the light rays that arrive to the camera within the very narrow time frame when the aperture is open. For objects moving at very high speeds, practice tells us that the aperture must be very short, otherwise motion blurr occurs. This means that , in practice, the image of the object in the Terrell effect is generated by all the light rays that originate in the object and arrive at the camera "objective" at the same time. For a moving object this means that the "photograph" is really a composite of different parts of the object taken at different positions. This explains why we can see the side of the cube that , under normal conditions, should be hidden from our view by the front side. The reason is that, while the apperture was open, the cube in the Terrell paper, has moved a little, allowing for light rays originating from the side face of the cube to be "unblocked" . A very good description can be found in R. Skinner "Relativity" pp.69-72. I will do a detailed explanation in my blog in a few days.

 

[yuiop]

The tube and photon path are not parallel in both reference frames in this scenario,

You are repeating the same error as above.

You are right that the above sentence is not correct and obviously contradicts my statement a couple of sentences earlier where I said "the tube and photon path are parallel to the y-axis the CM frame". I have corrected the typo in the original post.

the photon path is at an angle to the y-axis ...
...
while the tube remains parallel to the y axis.

This is the part you got wrong. This is due to the fact that you never completed the calculations from the other thread, despite a lot of encouragement I've given you.

I stand by this statement and it is supported by my diagram in post 42 that you said is correct. If you are talking about calculations that relate to Terrel rotation, then they have no bearing on whether the photon passes through the tube or not, because as everyone keeps telling you, the Terrel effect is just an optical illusion and has no physical significance in this context.

You have still not stated which diagram/s by myself or BobS best reflect your stance on the subject now that you have had time to reflect on it. So which is it, or do you not want to commit yourself at this moment in time?

 

[JesseM]

I understand the Terrell effect extremely well. I will do a writeup for you and place it in my blog.

Well, no need to do so unless it's relevant to the question of the light passing through the tube. Can you address the question I asked you before?

What do you mean "the calculations are the same"? Are you suggesting that, if I have a grid of rulers defining my coordinate system, then the markings on my ruler which visually appear to coincide with different points on the object at a single instant actually correspond to the position-coordinates the object occupied at a single moment in my frame? (i.e., the positions which different points on the object occupied simultaneously according to my rest frame's definition of simultaneity)

 

[starthaus]

Well, no need to do so unless it's relevant to the question of the light passing through the tube. Can you address the question I asked you before?

It means that the calculations involved in finding the Terrell rotation are the same as the calculations involved in finding the angle of rotation of an object as viewed from a moving frame. I went ahead and I uploaded a tutorial on Terrell. You can find it under RollingObjects in my blog. You can see the complete set of computations for a complex case (translation + rotation).

 

[Austin0]

You are wrong. If you understand the diagram in post 23 correctly, the photon does not hit the walls of the tube in any of the reference frames. The diagram is correct in the context that it was given in. Call this scenario 1. In this scenario, the emitter and the tube are at rest in the CM frame and the tube and photon path are parallel to the y-axis the CM frame. In the lab frame that is moving along the x-axis of the CM frame, both the emitter and tube are moving in this frame and the photon path is at an angle to the y-axis while the tube remains parallel to the y axis. The tube and photon path are not parallel in the lab frame in this scenario, but the photon does not hit the side of the tube in either reference frame.

The new diagram I posted in #42 is for a different scenario. (Call it scenario 2). It is not a correction of post 23. In scenario 2 the tube is at rest in the lab frame while the emitter is at rest in the CM frame. In this scenario the lab operators manually align the tube with the photon path to ensure the photon passes through the tube. In the transformation from the lab frame to the CM frame, the tube and photon path rotate by different amounts (contrary to your claims) and the tube and photon path are not parallel in the transformed frame. In both scenarios 1 and 2, the photon path is necessarily parallel to the tube in the rest frame of the tube by design, but in both scenarios when we transform to another reference frame the photon path and tube are no longer parallel. Both scenarios are consistent about this and both scenarios/diagrams contradict your claim that the rotation angle of the photon path (aberration) is the same as the rotation angle of the tube. This is simply not true.

Derivation of formulas for second diagram:

If the photon path is parallel to the y-axis in the CM frame then we know if the lab frame is moving in the negative x direction, that the x component of the lights velocity vector is equal to the \beta_x. We also know that the speed of light is c in all reference frames so we can say c = 1 = \sqrt{\beta_x^2+\beta_y^2} = \sqrt{\beta_x^2' + \beta_y^2'}
This implies \beta_y' = \sqrt{1 -\beta_x^2'. The angle of the photon path in lab frame (S') is \theta' = atan(\beta_y'/\beta_x' ) and by substitution of the y' component we obtain \theta' = atan \left(\sqrt{(1-\beta_x^2')}/\beta_x'\right) = atan\left( \frac{1}{\gamma \beta_x'} \right).

This is the equation given by BobS earlier. We can also say by symmetry, that \theta' = atan(y'/x') where (x',y') is a coordinate point on the photon path in frame S'.

The angle of the statioanry tube is the same as the angle of the photon path in the lab frame, because it is manually aligned by the lab operators to allow the photon to pass through. The angle of the moving tube in the S (CM) frame is \theta = atan(y/ x) = atan(y' *\gamma/x') = atan(tan(\theta')* \gamma). This is effectively due to the Thomas rotation of the tube (which is the difference between \theta and \theta') and is not the same as the aberration of the photon path. Thomas rotation derived in two lines! Therefore the photon path and tube are not parallel when we transform to the emitter frame and this is what is depicted in the diagram attached to post 42. You agree the diagram in post 42 is correct, but it contradicts your claims that the photon path and tube rotate by the same angle under transformation.

It seems to me that you are saying the diagram posted by BobS in post #22 where the tube and photon path are parallel in both reference frames is correct. Is that the case? I think the diagram posted by BobS is incorrect and I do not mean to "harp on" about Bob's error because with all due respect he acknowledges he might be mistaken (and he deserves credit for being the first to post the correct equation for the photon path rotation that we all agree with), but I have to bring it up because you seem to be adhering to what his diagram is showing. What is you position? Which diagram do you think is correct? If you think it the one I posted in #42 then you will have withdraw your claim that the photon path and tube rotate by the same amount under a transformation to a different reference frame, because that is not what it shows. If the tube and photon path are the parallel in one frame and if as you suggest the photon path and tube rotate by the same amount under a transformation, this implies that the photon path and tube is parallel in all reference frames which simply in not true.

I think you will find that Bon S's diagram is absolutely correct in the context.
He actually understood the parameters of my question. In his diagram he was simply illustrating the necessary conditions in each frame for the photon to make it through.
Not two pictures of a singular sequence of events.

If the tube is stationary in the lab frame, then it has to be aligned at an angle θ_lab = tan^-1(1/βγ) relative to the direction of motion.. If the tube is stationary in the rest frame of the emitter, it has to be aligned orthogonal to the direction of motion.

Bob S

 

[starthaus]

You are right that the above sentence is not correct and obviously contradicts my statement a couple of sentences earlier where I said "the tube and photon path are parallel to the y-axis the CM frame". I have corrected the typo in the original post.

It isn't just a typo, it reflects your lack of understanding of the issue. Every time I point out one of your errors you come back claiming that it was a typo.

.

You have still not stated which diagram/s by myself best reflect your stance on the subject

I don't do physics with "pictures". You can find the correct mathematical description in post #29. If you ever get around to performing the calculations I showed you in the other thread, you will understand why the tube gets inclined by the same angle as the angle of light aberration.

 

[Austin0]

From these posts I seem to understand that my initial grasp was basically correct as far as the behavior if not on any implications of mass.

it is not a totally absurd topic of speculation.

It is absurd if you are both postulating that photons move at c and that they have mass. Any paper speculating about photon mass is presumably speculating that photons actually move slower than c (where c is defined to be the fundamental constant that appears in the Lorentz transformation/time dilation equation/etc., not defined as the speed of photons).

As per early post above I was not postulating anything , just musing in print.
I quickly let it go completely as it wasn't really the point.
On the other hand having been pressed it is not really without some justification.
It seems to me that whether photons move exactly at the upper possible limit or not is not important. The derived value c for photons as it stands, appears to be correct for all EM interactions which also appears fundamental to everything else too.

I mean that a photon emitted transversely in a moving system and reflected within that system bounces straight up and down just like a Newtonian ball. It retains the forward motion of the system. i.e. conserved forward momentum. It is not independent of the motion of the source except with regard to emissions along the path of motion.

That with emissions in the direction of motion the s_H component is not conserved at all i.e. does not contribute to the velocity vector c

What do you mean "does not contribute to the velocity vector c"? Do you disagree that in this case s_H^2 + s_V^2 = c^2? That looks like a "contribution" to the total speed of c to me. For example, we might have s_H=0.8c and s_V=0.6c, so then it'd be true that (0.8c)^2 + (0.6c)^2 = (0.64 + 0.36)*c^2 = c^2.

Emmissions along motion path are c regardless of the velocity of the emitter no?
SO the x or H component of the source does not contribute.
Wrt a photon emitted directly to the rear or at an acute angle to the rear what contribution to the resulting velocity vector does the forward x component of the emitter make??

Don't you find it somewhat [read extremely] curious that photons are not totally independent of the motion of the source and do act like massive particles wrt transverse emissions?

No. If the photon direction was independent of the source, that would falsify relativity, since there would be a single preferred frame where a photon emitted by a tube pointing in one direction would actually travel in the same direction. The first postulate of relativity says that the laws of physics should work the same way in all inertial frames, which means that if experimenters in inertial windowless ships moving at different velocities all perform the same experiment, they should all get the same result--the first postulate would be violated if an experimenter in one frame found that light exited the emitter parallel to the orientation of the emitter, but an experimenter in a different frame found that it exited at an angle relative to the orientation of the emitter.

Are you familiar with the http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_basics/index.html#Light1 thought-experiment? This thought-experiment actually depends on the fact that light always travels parallel to the emitter in the emitter's rest frame in order to derive the time dilation equation.

Hi Bob S ...I am afraid I was not clear enough in my post.
The tube and detector are both at rest in the lab frame. There is clearly no problem if the tube is in the emitter frame otherwise a light clock could not work.

You are looking at things from the perspective of consistency with SR , I wasnt questioning this but simply looking at the physics and finding it strange that a massless wave would act like a massive particle.

With regard to a massive particle say a bullet that is rotating around the axis of motion: in a relative frame the forward momentum would result in a sideways drift relative to the spin??

I don't know what you mean by "sideways drift relative to the spin". The linear momentum of the center of mass is conserved as long as no forces are acting, so in zero gravity a spinning bullet will travel in a straight line just like a nonspinning bullet.

I meant that in the lab frame the bullet spin would remain parallel to the y-axis so the conserved x motion would be sideways relative to the spin.
Or conversely it would be tilted relative to the linear path it was traveling in the lab.

Wrt a photon this is not such a comfortable picture. If we consider a photon as a traveling waveform this would mean it would be out of phase along an orthogonal front relative to its linear motion, yes?
It would seem to be more realistic to view the angle derived from the vector sum of conserved momentum as an actual propagation angle with the wave front in phase, relative to travel,

Don't understand this sentence. "Out of phase" with what other wave, exactly?

Out of phase with itself , tilted relative to the path of motion .
Since that post I have followed a link Bob S posted to another thread and another link where on the last page (13 ), last paragraph of a paper I found the following which is related to Terrell rotation but also seems to answer my question here

""""One can do these same calculations from any selected observation angle and
nds similar results. The image (eye or photographic) appears to be a cube rotated
by the aberration angle.
The key issue is that one is observing with light emitted from the object (cube
in our example). In relativity light propagates with constant speed c independent of
the observer's or source speed and the key point here is that the wave front always
remains perpendicular to the direction of propagation. The only thing that changes
is the direction of propagation (and thus wavefront angle) which is what we call
relativistic aberration. Thus an image in one frame remains an image in the other
and only the angle of observation changes.
This statement is true for the case of a small object which subtends a small
solid angle. As one goes to larger angles, the aberration changes and a larger solid
angle object would be rotated and distorted by the variation in aberration angle
across the object being viewed.

 

[JesseM]

Emmissions along motion path are c regardless of the velocity of the emitter no?

Yes, but in the frame where the emitter is moving, the "motion path" is neither horizontal nor vertical, it's on a diagonal. By the pythagorean theorem, the speed on this diagonal path can be decomposed into the sum of squares of vertical and horizontal speed, just as in my example where horizontal speed was 0.8c and vertical speed was 0.6c.

SO the x or H component of the source does not contribute.

Again I am not clear what you mean by "contribute"--it seems to me that if the total speed is equal to \sqrt{(0.8c)^2 + (0.6c)^2} then both are making a contribution to the total speed! Perhaps by "contribute" you actually mean "raise above c?"

Wrt a photon emitted directly to the rear or at an acute angle to the rear what contribution to the resulting velocity vector does the forward x component of the emitter make??

Same as always, the total speed of the photon is \sqrt{s_H^2 + s_V^2}, so as long as both horizontal and vertical speed are nonzero they are both part of the calculation of the total speed. Do you disagree?

You are looking at things from the perspective of consistency with SR , I wasnt questioning this but simply looking at the physics and finding it strange that a massless wave would act like a massive particle.

Well, massless particles act like massive particles in a number of ways--for example, they both travel in straight lines at constant speed if undisturbed (and ignoring quantum randomness, which in any case says the same thing about their average expected motion). If you're just saying you "find it strange" on some intuitive level that you can't really justify rationally that's fine, but personally I don't share that intuition, so unless you do have some argument as to why it's problematic in some way I don't really see what there is to discuss here.

With regard to a massive particle say a bullet that is rotating around the axis of motion: in a relative frame the forward momentum would result in a sideways drift relative to the spin??

I don't know what you mean by "sideways drift relative to the spin". The linear momentum of the center of mass is conserved as long as no forces are acting, so in zero gravity a spinning bullet will travel in a straight line just like a nonspinning bullet.

I meant that in the lab frame the bullet spin would remain parallel to the y-axis so the conserved x motion would be sideways relative to the spin.
Or conversely it would be tilted relative to the linear path it was traveling in the lab.

You're still not providing a clear scenario to explain what you mean, for example this is the first you mentioned of a "lab frame" or of the bullet's spin being parallel to the y axis. Do you mean that the bullet's axis of rotation is parallel to the y' axis of its own rest frame, so it's still parallel to the y-axis of the lab frame? How is it moving in the lab frame--purely in the x-direction or in a combination of x and y directions? Most importantly, you didn't explain what you mean by "sideways drift"--"drift" usually denotes the idea that the values of two quantities are drifting apart (i.e. the difference between them is increasing), but here although the bullet's spin axis is different than its axis of motion in the lab frame, the difference between the two angles remains constant as the bullet travels.

Out of phase with itself , tilted relative to the path of motion .

A wave can't be out of phase with itself--if two waves are out of phase this means the peaks of the two waves don't line up (a phase difference), this doesn't make sense for a single wave which only has one set of peaks. As for the second part, I don't understand what it would mean for an electromagnetic wave to be "tilted relative to the path of motion"--what exactly would be tilted? The electric and magnetic field vectors? On this page you can see a simple animation of an electromagnetic wave created by a charge bobbing up and down on the z axis, showing how the electric and magnetic field vectors vary at different points along the x-axis (the vectors at points not along the x-axis aren't shown but they would be similar). You can see these vectors are actually perpendicular to the direction of motion--does that match what you meant by "tilted relative to the path of motion", or are you talking about something different?

Since that post I have followed a link Bob S posted to another thread and another link where on the last page (13 ), last paragraph of a paper I found the following which is related to Terrell rotation but also seems to answer my question here

""""One can do these same calculations from any selected observation angle and
nds similar results. The image (eye or photographic) appears to be a cube rotated
by the aberration angle.
The key issue is that one is observing with light emitted from the object (cube
in our example). In relativity light propagates with constant speed c independent of
the observer's or source speed and the key point here is that the wave front always
remains perpendicular to the direction of propagation. The only thing that changes
is the direction of propagation (and thus wavefront angle) which is what we call
relativistic aberration. Thus an image in one frame remains an image in the other
and only the angle of observation changes.
This statement is true for the case of a small object which subtends a small
solid angle. As one goes to larger angles, the aberration changes and a larger solid
angle object would be rotated and distorted by the variation in aberration angle
across the object being viewed.

Why do you think the statement wouldn't be true for a large object which subtends a larger solid angle? The image of the object would be distorted, and to understand this you don't have to think about waves and wavefronts, you can just imagine each point on the object emitting little particles moving at c in straight lines in all directions, so no matter where the observer is positioned one of these particles will have been emitted at the correct angle to reach his eyes (assuming the light wasn't blocked by an absorber). The Terrell-Penrose effect just has to do with the fact that the light from different parts of an object which is reaching your eyes at a single instant was not actually all emitted simultaneously in any inertial frame, which means you get weird distortions in the object's appearance.