Can Light Emission Be Explained Through Newtonian Physics?

[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 independant 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 independant 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.

 

[yuiop]

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.

I stand corrected. I missed the context of Bob's diagram that was given in an earlier post. Apologies to Bob. Normally these diagrams are given in the context of a transformation from one frame to another, i.e. the different points of view of the same set of events, of observers with motion relative to each other. In Bob's example, they manually rotated the tube in one of the frames between experiments and they do not represent the the same set of events. Anyway, in the context that it was given in, Bob diagram is correct. peh! I usually criticize "other" people for not reading the context! "kicks self".

 

[JesseM]

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.

I understand that the Penrose-Terrell effect means the image you see at a given instant "is really a composite of different parts of the object taken at different positions" (and different coordinate times), that was the whole point I was making in post #51. But you still haven't explained what relevance this optical effect is to the coordinate description of the tube in the frame where it's moving, or what you meant when you said in post #48 that "the calculations are the same"? Suppose we have a tube which is oriented perpendicular to the floor in its own frame, but it's moving horizontally (parallel to the floor) in the lab frame, are you claiming that at any given t-coordinate in the lab frame, the position of the bottom of the tube will not be directly underneath the position of the top, i.e. the tube's orientation is not perpendicular to the floor in this frame?

 

[starthaus]

I understand that the Penrose-Terrell effect means the image you see at a given instant "is really a composite of different parts of the object taken at different positions" (and different coordinate times), that was the whole point I was making in post #51.

Good, I hope you have read my explanation of the effect.

But you still haven't explained what relevance this optical effect is to the coordinate description of the tube in the frame where it's moving, or what you meant when you said in post #48 that "the calculations are the same"?

If you read the explanation, you would have seen that it employs the determination of the locus of the points for dt'=0 in the frame of the moving observer S'. So, when you apply the condition dt'=0, you get that the object gets rotated even for the case when the object and the observer move at 90 degrees from each other, a case denied by kev.

Suppose we have a tube which is oriented perpendicular to the floor in its own frame, but it's moving horizontally (parallel to the floor) in the lab frame, are you claiming that at any given t-coordinate in the lab frame, the position of the bottom of the tube will not be directly underneath the position of the top, i.e. the tube's orientation is not perpendicular to the floor in this frame?

See above. I can see that you share this misconception with kev.

 

[JesseM]

Good, I hope you have read my explanation of the effect.

My post #51 was before your own explanation, so you can see that I already understood that the Penrose-Terrell effect comes from the fact that you are seeing light emitted from different points on the object at different times, with the object having moved between those times. So, your explanation told me nothing new.

If you read the explanation, you would have seen that it employs the determination of the locus of the points for dt'=0 in the frame of the moving observer S'. So, when you apply the condition dt'=0, you get that the object gets rotated even for the case when the object and the observer move at 90 degrees from each other, a case denied by kev.

But you're just talking about visual appearances. If we want to know whether the light makes it from the front of the tube to the back or if it hits the side of the tube, visual appearances are irrelevant, we only have to worry about the coordinate positions occupied by the tube at different times, and the coordinate position of the light beam at different times. If coordinate positions of the walls of the tube are always vertical (perpendicular to the floor) in the rest frame of the tube, but the light beam is traveling at an angle in this frame, then if the tube is narrow the light entering the front will hit the side rather than reach the detector at the back. So again, do you disagree that if we have two frames moving "horizontally" along the x-axis relative to one another, then if the tube is oriented "vertically" parallel to the z-axis in one frame (in terms of the coordinate positions it occupies at any given t-coordinate in that frame), according to the Lorentz transformation for frames in standard configuration (with all their spatial axes parallel), then the tube will also be oriented "vertically" parallel to the z'-axis of the other frame (again in terms of the coordinate positions it occupies at a given t'-coordinate in the second frame)?

Suppose we have a tube which is oriented perpendicular to the floor in its own frame, but it's moving horizontally (parallel to the floor) in the lab frame, are you claiming that at any given t-coordinate in the lab frame, the position of the bottom of the tube will not be directly underneath the position of the top, i.e. the tube's orientation is not perpendicular to the floor in this frame?

See above. I can see that you share this misconception with kev.

Does calling it a "misconception" imply you disagree? Again, remember I am not talking about visual appearances, but rather about the position coordinates occupied by the tube at a single time coordinate. If you do disagree, I can give you some simple math to show that you're incorrect.

 

[starthaus]

But you're just talking about visual appearances.

No, I am not.

then if the tube is narrow the light entering the front will hit the side rather than reach the detector at the back.

If this were true (it isn't) then you have a very serious problem because in the frame of the tube , the light does not hit the walls and reaches the detector whereas, according to your above claim, in the frame of the moving observer the light hits the wall before reaching the detector (if the observer is moving fast enough wrt the tube).

 

[JesseM]

No, I am not.

Well, the Terrell-Penrose effect is concerned with visual appearances, not actual coordinate positions at a single instant of coordinate time.

If this were true (it isn't) then you have a very serious problem because in the frame of the tube , the light does not hit the walls and reaches the detector whereas, according to your above claim, in the frame of the moving observer the light hits the wall before reaching the detector (if the observer is moving fast enough wrt the tube).

You are forgetting that the tube is moving horizontally in the frame where the light is traveling up vertically (the tube is moving in the x direction while the light is traveling up parallel to the z axis in my example, so the light beam has a constant x coordinate as its z coordinate increases), so even if the tube is also oriented vertically in this frame (its sides are also parallel to the z axis at any given t coordinate), the light can still hit the side because the tube is moving sideways as the light travels from bottom to top (the x coordinate of each side is changing, so even if the light is halfway between the x coordinates of each side when it first enters the bottom of the tube, one side's x coordinate may reach the x coordinate of the light before the light has had time to reach the top). Meanwhile in the frame where the tube is at rest, the light is traveling at an angle, so it also hits the side in that frame.

Once and for all, do you disagree with my claim that if the tube is parallel to the z axis at a single t coordinate in one frame, then it is also parallel to the z' axis at a single t' coordinate in the other frame, assuming the two frames are moving parallel to each other's x and x' axes as in the Lorentz transformation for frames in standard configuration? If you do disagree I can easily provide the math to show you're wrong, but I'm not going to bother if you just want to be evasive and refuse to give me a straight answer to whether you are disputing this or not.

 

[starthaus]

Well, the Terrell-Penrose effect is concerned with visual appearances, not actual coordinate positions at a single instant of coordinate time.

Yet, if you read (and understood) the writeup I did, you would understand why the formalism
used by the Terrell effect applies perfectly to the exercise in the OP.

You are forgetting that the tube is moving horizontally in the frame where the light is traveling up vertically

No, I am not forgetting anything, I reacted to your incorrect statement that the light would hit the walls of the tube in the frame of the observer. I pointed out that this cannot be true since the same experiment would have different outcomes in two different inertial frames. Let me try to explain to you my explanation of the experiment.

1. In the frame of the tube+emitter+detector the light travels alongside the tube in an "up-down" fashion. The detector always detects the light unimpended.

2. In the frame of a moving observer, things are a little more complicated. Light is aberrated "to the right" by an angle \theta=arrcos(\beta). We know that from Einstein's light aberration formula. Now, while the light travels an infinitesimal length "diagonally" , at speed c and angle \theta wrt the horizontal axis, the tube moves "to the right" . It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally, such that while the light beam moves by cdt diagonally, the tube element moves vdt to "the right". So, the tube looks like it was re-assembled diaonally, in order to "allow" the light to reach the detector at the other end. This explains why the light hits the detector in the observer frame. Now, the elementary angle of tube slanting is ...cos(\phi)=\frac{vdt}{cdt}=\beta. So, to anybody's surprise, \phi=\theta=arccos (\beta). This also "happens" to be the angle of the Terrell rotation, so the formalism developed for solving the Terrell rotation works perfectly for solving this type of problem. Since you objected to using it, I gave you a different solution.

 

[JesseM]

No, I am not forgetting anything, I reacted to your incorrect statement that the light would hit the walls of the tube in the frame of the observer.

Nope, never said anything about it hitting in "the frame of the observer", of course it's a frame-independent fact that if it hits the wall when analyzed in one frame, it will hit the wall when analyzed in any other frame.

1. In the frame of the tube+emitter+detector the light travels alongside the tube in an "up-down" fashion. The detector always detects the light unimpended.

Austin0's thought-experiment involved a tube that was moving relative to the emitter, so there was no "frame of the tube+emitter+detector", the emitter (which shot the beam straight up in its rest frame) was in a different frame than the tube, and this was the situation I was talking about when I said the light would hit the wall.

However, even if we choose to analyze your altered thought-experiment, your explanation doesn't make much sense to me. This first part makes sense:

2. In the frame of a moving observer, things are a little more complicated. Light is aberrated "to the right" by an angle \theta=arrcos(\beta). We know that from Einstein's light aberration formula.

If the light was emitted vertically in the rest frame of the emitter/tube, so it has x(t)=0 and z(t) = c*t, then in the frame of the observer moving at speed v in the +x direction, the light has x'(t') = -vt' and z'(t') = sqrt(c^2 - v^2)*t'. So, at any given time t' the light has traveled a distance of \sqrt{x'(t')^2 + z'(t')^2} = \sqrt{v^2 * t'^2 + (c^2 - v^2)*t'^2 } = ct'. So since cos(angle) = adjacent/hypotenuse, for the angle theta between the light's path and the z-axis we have cos(theta) = sqrt(c^2 - v^2)*t'/ct' = sqrt(1 - v^2/c^2) = Beta. So yes, I agree the angle theta in this frame would be arccos(Beta).

But then you go on to say:

Now, while the light travels an infinitesimal length "diagonally" , at speed c and angle \theta wrt the horizontal axis, the tube moves "to the right" . It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally, such that while the light beam moves by cdt diagonally, the tube element moves vdt to "the right". So, the tube looks like it was re-assembled diaonally, in order to "allow" the light to reach the detector at the other end.

By "re-assembled diagonally" are you suggesting that in the frame of the moving observer, the tube is actually at an angle of arccos(Beta) relative to the z' axis at any single time t' in this frame? You're not just claiming it's a visual effect? Because once again, if you are suggesting this you're just wrong, and I can easily show you why if you like. Anyway, it's easy to see why the light doesn't hit the side in the frame of the moving observer despite the fact that the tube is still oriented parallel to the z' axis in this frame--as I mentioned above, the light has a horizontal velocity of -v along the x' axis, and so does the tube! So if the photon starts out midway between the two walls, it stays midway between them since both the photon and the walls are moving in the x' direction at the same speed.

 

[starthaus]

If coordinate positions of the walls of the tube are always vertical (perpendicular to the floor) in the rest frame of the tube, but the light beam is traveling at an angle in this frame,

It isn't traveling at an angle in the tube frame, it is traveling at an angle in the observer frame. I have explained the key role played by the aberration effect in explaining the effect several times already.

then if the tube is narrow the light entering the front will hit the side rather than reach the detector at the back.

I have already explained in my latest post why this is not the case.I have also alerted you several times already that this effect cannot happen since it would result into different outcomes of the experiment in the frame of the observer vs. the frame of the tube.

 

[starthaus]

Austin0's thought-experiment involved a tube that was moving relative to the emitter, so there was no "frame of the tube+emitter+detector",

Then you did not understand his post, so we have been talking about different things all along. This is getting really boring, do you agree that the light beam is aberrated by the angle arccos(\beta) in the observer frame? Do you agree that the light beam does not touch the walls in either frame? Do you agree that the light beam hits the detector at the end of the tube in both frames?

 

[JesseM]

Then you did not understand his post, so we have been talking about different things all along.

The post you link to said nothing about a tube, he was there talking about a simpler scenario with no tube involved. He introduced the tube in post #7 where he said:

4) 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.
That if there is a light detector at the end of a long light absorbtive tube aligned directly perpendicular to the source motion that it would seem impossible for a photon to reach the detector??

If the tube is "perpendicular to the source motion", I take that to mean that the source is moving relative to the tube. But just to make sure, I asked Austin0 about this directly in post #32, and he replied in post #33 (the bolding was his):

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.

So you see, Austin0 was clearly talking about a case where the source was moving relative to the tube, and that's the case I was considering when I talked about the light hitting the side of the tube.

Like I said, though, I'm happy to consider your alternate scenario where the source was at rest relative to the tube, and both the light beam and the tube were oriented vertically in the rest frame of the tube/source. In this case, I agreed with you that in the frame of an observer traveling in the horizontal direction, the light beam is moving at an angle of arccos(Beta) from the vertical axis. I also agree that in this scenario, the light never hits the wall of the tube, regardless of what frame we use to analyze the situation. However, if your odd comment "It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally" amounts to a claim that the tube itself is not still oriented vertically in the frame of this observer (in terms of the coordinate positions it occupies at a given time in the observer's frame, not in terms of visual appearances), then as I already said this is complete nonsense. I can show why it's nonsense with some basic algebra if you or anyone else actually believes this, but if you won't confirm that this is what you're arguing I won't bother.

edit: I see you added the following comment:

This is getting really boring, do you agree that the light beam is aberrated by the angle arccos(\beta) in the observer frame? Do you agree that the light beam does not touch the walls in either frame? Do you agree that the light beam hits the detector at the end of the tube in both frames?

In your altered scenario (not Austin0's scenario), of course I agree with all these things. In my previous post #68 I already made the following comments about your altered scenario:

So since cos(angle) = adjacent/hypotenuse, for the angle theta between the light's path and the z-axis we have cos(theta) = sqrt(c^2 - v^2)*t'/ct' = sqrt(1 - v^2/c^2) = Beta. So yes, I agree the angle theta in this frame would be arccos(Beta).

Anyway, it's easy to see why the light doesn't hit the side in the frame of the moving observer

 

[starthaus]

If the tube is "perpendicular to the source motion", I take that to mean that the source is moving relative to the tube.

If this is what you think, then the problem reduces to the case of light aberration and, in order to make sure that the light hits the detector, one needs to actively incline the tube by arcos(\beta).

So you see, Austin0 was clearly talking about a case where the source was moving relative to the tube, and that's the case I was considering when I talked about the light hitting the side of the tube.

Then the problem reduces to the case of light aberration and, in order to make sure that the light hits the detector, one needs to actively incline the tube by arcos(\beta).

Like I said, though, I'm happy to consider your alternate scenario where the source was at rest relative to the tube, and both the light beam and the tube were oriented vertically in the rest frame of the tube/source. In this case, I agreed with you that in the frame of an observer traveling in the horizontal direction, the light beam is moving at an angle of arccos(Beta) from the vertical axis. I also agree that in this scenario, the light never hits the wall of the tube, regardless of what frame we use to analyze the situation.

Good, then, for this case we agree. You need to revisit the case when the tube is moving wrt the source.

edit: I see you added the following comment:

In your altered scenario (not Austin0's scenario), of course I agree with all these things. In my previous post #68 I already made the following comments about your altered scenario:

Good. You only need to revisit the case when the tube is moving wrt the source. The tube needs to be inclined.

 

[Austin0]

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.

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.

Here I was talking about a photon emitted along motion path,,,not a diagonal because there is no y or z component. That in this case the photon was independant of the motion of the emitter because the x component of the emitter velocity made no contribution to the resulting vector.
Maybe the communication problem stems from my having an inaccurate understanding of conservation of momentum in this context.Confined to these conditions of orthogonal motion in S' I think that it is a transformation from a moving frame into the lab frame. With Newtonian velocities the Pythagorean relationship holds strictly with the y' velocity component in S' being equal to the y component in S so the summed vector magnitude would of course be greater in S with the addition of the x component. Therefore the momentum magnitude would also be greater.
With the photon in this scenario the summed x,y velocity vector in S cannot be greater than the y' velocity vector in S' yet the spatial y,y' metric is untransformed so that distance in y is equal in both frames. It is only with the time component included that this makes sense.
The elapsed time in S being greater resulting in lower y velocity in S.
So in your quote above the velocities are already transformed and of course would be consistent with measurements in S SO the real x contribution in this case would be in increased momentum represented by blueshift in S.
This raises a question ; the reduced y' velocity in S... y velocity, requires clocks in S to be running faster than in S' So how does this work with the blueshift??
If the time in S' is dilated , then the emitted photon would be redshifted relative to S.
So does the blueshift resulting from the angle of aberration raise the frequency more than the dilation redshift factor to realize a net blueshift or am I astray somewhere here??

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??

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?"

Well partly. It seems that purely on a conservation basis a photon emitted directly astern would have a v < c if the forward motion of the emitter was conserved in the normal sense.
That a photon emitted at an angle astern conserves the y' component (with the time factor transformation) to derive a directional vector but the magnitude of that (-x,y) vector would not seem to reflect the +x motion of the emitter at all.
Isn't this exactly what the addition of velocities equation indicates?

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?

As above ; I of course don't disagree but these are the transformed coordinate speeds aren't they? The velocity of the emitter does not enter into it or does it?

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 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.

Sorry if I wasn't clear. I assumed from the context of this discussion we were talking about a bullet orthogonal to its rest frame as observed in the frame where the gun is in motion. Drift in this context was meant as motion in x comparable to wind drift with actual bullets,, meaning sideways motion . No implication of wind involved in this case of course :-)

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??

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?

If you will notice after all that you have in the end both confirmed and agreed with what I said in the first place.
The graphical representation in that clip does not really apply. If you consider only the electric component viewed as a packet of parallel wave fluctuations then the bullet analogy would see them as parallel but tilted relative to the linear direction of travel. With various part of a single cycle not simultaneously in phase with other parts.
As I originally said this does not track, so it might be better to look at the aberration angle, not neccessarily as a vector sum of conserved velocities, but as an actual emission angle.
Perhaps??

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?

Sorry again. It was late and I didn't notice the paragraph separations . The only relevant part was the part I bolded as it pertained to the question above, about wave orientation.
I am interested in Terrell rotation but have already said in this thread it is not relevant to the question of the thread. I was making no comment regarding T rotation at all.

Regarding the question I asked earlier ; COuld an emitted photon at any angle make it through a perpendicular tube in the frame where the emitter was in motion??
Looking at it now there seems like there should be some angle in the moving frame where an emitted photon would travel orthogonally in the lab frame. WHere the forward momentum would be exactly equivalent and opposite to the -x component deriving from the negative angle it was emitted at. What do you think?
Thanks

 

[JesseM]

If this is what you think, then the problem reduces to the case of light aberration and, in order to make sure that the light hits the detector, one needs to actively incline the tube by arcos(\beta).

Ah, so you are finally coming out and saying that the tube is inclined? Like I said, the idea that the tube would become inclined in the observer's frame is just wrong, and it's easy to show why using the basic math of the Lorentz transformation. Suppose that in the rest frame of the tube, we pick four points on the inside of the tube--one on the left inner side at the bottom (LB) at the tube's "lip", one at the right inner bottom (RB), one at the left inner side at the top (LT), and one at the right inner side at the top (RT). And suppose in this frame the width of the opening is 10 light-seconds while the vertical length of the tube is 100 light-seconds. Then for the (constant) positions of each point in the tube's rest frame, we could have LB: x=-5 light-seconds, z=0 light-seconds, RB: x=5 light-seconds, z=0 light-seconds, LT: x=-5 light-seconds, z=100 light-seconds, RT: x=5 light-seconds, z=100 light-seconds (you can see from this that in the tube's rest frame, LT is directly about LB, and RT is directly about RB). Suppose also that the photon is moving in the +z direction at x=0, and that it passes z=0 (first enters the tube) at t=0. So, the photon's position as a function of time in this frame is x(t)=0 and z(t)=c*t.

Now consider the frame of an observer moving at 0.6c in the +x direction relative to the first frame. As I said before in post #68, if the photon had x(t)=0 and z(t)=c*t in the tube rest frame, in this observer's frame the photon will have x'(t')=-v*t' and z'(t')=sqrt(c^2 - v^2)*t', and I showed in post #68 why this does mean that the photon's path has an angle of arccos(1 - v^2/c^2) = arccos(Beta) relative to the z' axis. Hopefully you don't disagree that these would be the correct expressions for the path of the photon in the observer's frame (if you do I can give a full derivation). Anyway, for this example where the observer is moving at 0.6c, the path of the photon would be given by x'(t') = -0.6ct' and z'(t') = sqrt(c^2 - (0.6c)^2)*t' = sqrt(1 - 0.36)*ct' = 0.8ct'.

Now, consider what x'(t') would be for RB and RT in the observer's frame. For RB, in the tube's rest frame we had x=5, z=0. So, if we substitute in the Lorentz transformation equations x=1.25*(x' + 0.6ct') and z=z', this means:

1.25*(x' + 0.6ct')=5 -> x' + 0.6ct' = 5/1.25 -> x' = 4 - 0.6ct'
z'=0

You can see that x'(t') = 4 - 0.6ct' and z'(t')=0 are the correct expressions for the path of RB since if you find the x' and z' for any value of t', and then plug them into the Lorentz transformation to find the corresponding x and z in the tube rest frame, you always get back x=5 and z=0. For example, at t'=20 we would have x'(t') = 4 - 0.6*20 = 4 - 12 = -8 and z'=0. So plugging t'=20, x'=-8 and z'=0 into the Lorentz transformation we have:

x=1.25*(x' + 0.6*t')=1.25*(-8 + 0.6*20)=1.25*4=5
z=z'=0

Likewise, at t'=30 we have x(t') = 4 - 0.6*30 = 4 - 18 = -14 and z'=0. So plugging t'=30, x'=-14 and z'=0 into the Lorentz transformation we have:

x=1.25*(-14 + 0.6*30) = 1.25*4 =5
z=0

So, you can clearly see that the correct expression for RB in the observer's frame is x'(t')=4 - 0.6c*t' and z'(t')=0 -- no disagreement here I trust?

Now let's find the expression for RT in the observer's frame. In the tube's rest frame the constant coordinates were x=5, z=100, so if we substitute in the Lorentz transformation equations x=1.25*(x' + 0.6ct') and z=z', this means:

1.25*(x' + 0.6ct')=5 -> x' + 0.6ct' = 5/1.25 -> x' = 4 - 0.6ct'
z'=100

Again you can verify that the correct expression by picking any value of t', finding the corresponding x' and z', and using the Lorentz transformation to show that these convert back to x=5, z=100 in the tube's rest frame.

So, the correct expressions for RT in the observer's rest frame are x'(t') = 4 - 0.6ct' and z'(t')=100. Since the expressions for RB were x'(t') = 4 - 0.6ct' and z'(t')=0, you can see that at any given value of t', RB and RT still have the same x' coordinate in the observer's frame, and are 100 light-seconds apart on the z'-axis, exactly as was true in the tube's rest frame. Since RB and RT represent two points on the top and bottom of the inner right side of the tube, this shows that the inner right side of the tube is still vertical in the frame of the observer.

Likewise, for LT we would have x'(t') = -4 - 0.6ct' and z'(t') = 100, and for LB we would have x'(t') = -4 - 0.6ct' and z'(t') = 0. I can derive these if you wish but the method should be clear from above. So, in the observer's frame the left inner side is also oriented vertically, at any given t' coordinate any point on the left inner side has x' coordinate x'(t') = -4 - 0.6ct'.

It's also easy to see from this why there is no danger of the photon ever hitting either side of the tube in the observer's frame. Remember, the photon's own x' coordinate in the observer's frame was just x' = -0.6ct'. So if the right inner side has x'(t') = -0.6ct' + 4, that means at any given value of t' the right inner side will be 4 light-seconds to the right of the photon's current position, and if the left inner side has x'(t') = -0.6ct' - 4, that means at any given value of t' the left inner side will be 4 light-seconds to the left of the photon's current position. So, the photon will always be halfway between the two inner sides as it travels, and will continue to be until it hits the detector at the top of the tube.

Good. You only need to revisit the case when the tube is moving wrt the source. The tube needs to be inclined.

Nope, it doesn't. In the case where the tube is moving relative to the source, suppose that in the tube's rest frame the source is moving in the +x direction at 0.6c. Assume exactly the same things about the positions of LB, LT, RB, RT in the tube's rest frame as above, but in this case the photon's path in the tube rest frame is not parallel to the z-axis, instead it's given by x(t)=0.6c*t and z(t)=0.8c*t. Since every point on the right inner side has x(t)=5 in the tube's rest frame, the photon will hit the right inner side when 0.6c*t = 5, or when t=8.333... seconds. This will happen when the photon has climbed to a height of z(t)=0.8c*8.333... = 6.666... light-seconds.

Now we can analyze the same situation in the rest frame of the source, where the photon's path is just given by x'(t')=0 and z'(t') = c*t. In this frame, the formulas for LB, LT, RB and RT are the same as in the observer's frame in the first scenario (since the observer in the first scenario was moving at 0.6c in the +x direction relative to the rest frame of the tube, and so is the source in this scenario). So, we know that every point on the right inner side has x'(t') = 4 - 0.6c*t'. Thus, the right inner side will have the same x' coordinate as the photon when 4 - 0.6c*t' = 0, or when t' = 4/0.6c = 6.666... seconds. At this moment the photon has climbed to a height of z'(t') = c*6.666... = 6.666... light-seconds. So, both frames agree that the photon is absorbed by the right inner wall of the tube when it has reached a vertical height of 6.666... light-seconds, although they disagree on the time-coordinate of this event (it's natural they should agree on the vertical height since the Lorentz transformation says z=z')

 

[starthaus]

Ah, so you are finally coming out and saying that the tube is inclined?

If the source is NOT in the same frame as the tube, i.e. if the tube is MOVING wrt the source.
This is basic observation of stellar light aberration. Do you even read what I write?

 

[JesseM]

If the source is NOT in the same frame as the tube, i.e. if the tube is MOVING wrt the source.

The coordinate orientation of the tube in a frame moving relative to the tube can be derived from the coordinates of points on the tube in its rest frame plus the Lorentz transformation, they don't depend on anything else like whether the source is at rest or in motion relative to the tube. I analyzed the case of the source moving relative to the tube at the end of my last post, the x'(t') and z'(t') for the four points LB, LT, RB, and RT in the frame moving relative to the tube would be the same regardless of whether the frame is understood to be that of the source or the frame of an observer moving relative to both the source and the tube; either way, LB and LT have the same x' coordinate at any given value of t' (likewise with RB and RT), so the tube is still oriented perfectly vertically in the frame where it's moving, just as it is in its rest frame.

This is basic observation of stellar light aberration.

My example already agreed that if light moves vertically in one frame, then it moves at an angle of arccos(Beta) relative to the vertical (z axis) in the other frame. This is true regardless of whether the light moves vertically in the tube's rest frame (the scenario analyzed in the first part of my previous post), or if we assume that the source is moving relative to the tube and the light only moves vertically in the rest frame of the source (the scenario analyzed in the last two paragraphs of my previous post, where in the tube's rest frame the light had x(t)=0.6c*t and z(t)=0.8c*t, so the cosine of the angle it makes with the z axis must be adjacent/hypotenuse = 0.8c*t/c*t = 0.8 = Beta). But a vertically-oriented tube will not change its orientation in the frame moving horizontally relative to it (regardless of whether that frame is the frame of the source or just an observer), if the tube was oriented vertically in its own rest frame it will still be oriented vertically in the frame moving relative to it.

Of course we could manually re-orient the tube to match the aberrated angle of light from a source in motion relative to it, to ensure that the light actually would reach the detector, but then it will be non-vertical in both the moving frame and the tube's own rest frame. Perhaps I misunderstood and you were actually talking about such a manual reorientation when you used the word "actively" in your statement from post #72 "in order to make sure that the light hits the detector, one needs to actively incline the tube by arccos(Beta)". However, I think it's pretty obvious that your earlier posts were not talking about a manual re-orientation, that you were in fact saying that a mere shift of frames would cause a tube which was vertically aligned in one frame to be oriented at an angle in another, for example in post #67:

1. In the frame of the tube+emitter+detector the light travels alongside the tube in an "up-down" fashion. The detector always detects the light unimpended.

2. In the frame of a moving observer, things are a little more complicated. Light is aberrated "to the right" by an angle \theta=arrcos(\beta). We know that from Einstein's light aberration formula. Now, while the light travels an infinitesimal length "diagonally" , at speed c and angle \theta wrt the horizontal axis, the tube moves "to the right" . It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally, such that while the light beam moves by cdt diagonally, the tube element moves vdt to "the right". So, the tube looks like it was re-assembled diaonally, in order to "allow" the light to reach the detector at the other end. This explains why the light hits the detector in the observer frame. Now, the elementary angle of tube slanting is ...cos(\phi)=\frac{vdt}{cdt}=\beta. So, to anybody's surprise, \phi=\theta=arccos (\beta). This also "happens" to be the angle of the Terrell rotation, so the formalism developed for solving the Terrell rotation works perfectly for solving this type of problem. Since you objected to using it, I gave you a different solution.

"It is as if the tube has been cut up into infinitelly small transervsal sections and the sections have been re-arranged diagonally ... This explains why the light hits the detector in the observer frame". So, it doesn't sound like you were talking about something as simple as manually changing the angle of the tube, rather you seem to have thought that if both the light and the tube were oriented purely "up-down" in the "frame of the tube+emitter+detector", then a frame shift would show that the tube is oriented at an angle in the frame of the moving observer, and that this angle in the observer's frame "also 'happens' to be the angle of the Terrell rotation", which was why you kept talking as though the Terrell rotation was somehow relevant to a coordinate-based analysis of the tube and light beam (even though Terrell rotation is purely an optical effect, as I pointed out from the start). All of this is clearly wrong, as I've shown--if the tube is oriented in an "up-down" fashion in its own rest frame, it is also oriented in an "up-down" fashion in the frame of any observer moving perpendicular to the up-down axis.

Do you even read what I write?

Yes, but even with a careful reading your meaning is often ambiguous. In any case you don't have much ground to complain that I am not reading what you write carefully enough if you petulantly refuse to answer simple requests for clarifications about what you are saying, like the question I kept asking you over and over again in various posts about whether you disagreed with the idea that if the tube is vertical in its own rest frame, it is also vertical in the frame moving relative to it:

Suppose we have a tube which is oriented perpendicular to the floor in its own frame, but it's moving horizontally (parallel to the floor) in the lab frame, are you claiming that at any given t-coordinate in the lab frame, the position of the bottom of the tube will not be directly underneath the position of the top, i.e. the tube's orientation is not perpendicular to the floor in this frame?

So again, do you disagree that if we have two frames moving "horizontally" along the x-axis relative to one another, then if the tube is oriented "vertically" parallel to the z-axis in one frame (in terms of the coordinate positions it occupies at any given t-coordinate in that frame), according to the Lorentz transformation for frames in standard configuration (with all their spatial axes parallel), then the tube will also be oriented "vertically" parallel to the z'-axis of the other frame (again in terms of the coordinate positions it occupies at a given t'-coordinate in the second frame)?

Once and for all, do you disagree with my claim that if the tube is parallel to the z axis at a single t coordinate in one frame, then it is also parallel to the z' axis at a single t' coordinate in the other frame, assuming the two frames are moving parallel to each other's x and x' axes as in the Lorentz transformation for frames in standard configuration? If you do disagree I can easily provide the math to show you're wrong, but I'm not going to bother if you just want to be evasive and refuse to give me a straight answer to whether you are disputing this or not.

By "re-assembled diagonally" are you suggesting that in the frame of the moving observer, the tube is actually at an angle of arccos(Beta) relative to the z' axis at any single time t' in this frame? You're not just claiming it's a visual effect? Because once again, if you are suggesting this you're just wrong, and I can easily show you why if you like.

However, if your odd comment "It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally" amounts to a claim that the tube itself is not still oriented vertically in the frame of this observer (in terms of the coordinate positions it occupies at a given time in the observer's frame, not in terms of visual appearances), then as I already said this is complete nonsense. I can show why it's nonsense with some basic algebra if you or anyone else actually believes this, but if you won't confirm that this is what you're arguing I won't bother.

Maybe if you acted like a mature adult engaged in rational debate, and gave straight answers to simple questions about what you are arguing (as I always do with your questions), rather than playing these ridiculous games where you are seemingly trying to express your contempt for me by pointedly ignoring questions I ask you even when I repeat them several times, these discussions with you would get to the point more quickly rather than going on interminably as they always seem to do.

 

[yuiop]

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?

As above ; I of course don't disagree but these are the transformed coordinate speeds aren't they? The velocity of the emitter does not enter into it or does it?

There is some relevance to the velocity of the emitter. If the horizontal component of the photon's velocity ({s_H}, is equal to the emitter's velocity in a given frame, then they are equal in all reference frames. For example, if in the comoving frame the photon path is parallel to the y axis, so that the x component is zero, and the emitter is at rest in that frame, then in another frame moving at v=-0.8c relative to the first frame, then the velocity of the emitter in the new frame is +0.8c and the x component of the photon in the new frame is also +0.8c.

I am interested in Terrell rotation but have already said in this thread it is not relevant to the question of the thread.

Could you try and explain this to Starthaus. JesseM and myself are having trouble getting this across to him.

Regarding the question I asked earlier ; Could an emitted photon at any angle make it through a perpendicular tube in the frame where the emitter was in motion??

Yes, there is a specific angle where this is possible.

Looking at it now there seems like there should be some angle in the moving frame where an emitted photon would travel orthogonally in the lab frame. Where the forward momentum would be exactly equivalent and opposite to the -x component deriving from the negative angle it was emitted at.

Spot on.

 

[JesseM]

Here I was talking about a photon emitted along motion path,,,not a diagonal because there is no y or z component.

OK, I guess I figured that "emissions along motion path" was just a shorthand for "the component of the light's velocity parallel to the motion path", since we had just been talking about s_H and s_V for a photon emitted vertically in the source's rest frame. Again, it would help if you would be more descriptive instead of just using these short compact phrases, you could say something like "for light emitted in a direction parallel to the source's motion, the speed is independent of the source's speed."

That in this case the photon was independant of the motion of the emitter because the x component of the emitter velocity made no contribution to the resulting vector.

Sure, that's true for the magnitude of the velocity vector in this case, although the momentum of the photons would still depend on the velocity of the source since momentum for light is dependent on frequency, and the source's velocity does affect the frequency via the Doppler shift.

Maybe the communication problem stems from my having an inaccurate understanding of conservation of momentum in this context.Confined to these conditions of orthogonal motion in S' I think that it is a transformation from a moving frame into the lab frame. With Newtonian velocities the Pythagorean relationship holds strictly with the y' velocity component in S' being equal to the y component in S so the summed vector magnitude would of course be greater in S with the addition of the x component.

Yes, if a projectile is sent out purely in the y' direction in the rest frame S' of the emitter, then if we have another frame S moving in the x' direction of S', in this frame the projectile will have the same velocity in the y direction but will have the same velocity as the emitter in the x direction, so the total magnitude of the velocity will be greater. In relativity it is no longer true that the velocity in the y direction of S is equal to the velocity in the y' direction of S' (it's not even true for slower-than-light projectiles, although it's approximately true if the relative velocity of the two frames is small compared to the speed of light).

With the photon in this scenario the summed x,y velocity vector in S cannot be greater than the y' velocity vector in S' yet the spatial y,y' metric is untransformed so that distance in y is equal in both frames. It is only with the time component included that this makes sense.

Again you really need to provide a lot more detail to be understood. I guess by "the spatial y,y' metric is untransformed" you just mean that according to the Lorentz transformation y=y', although if so this is a poor way of saying it since it makes it sound like there are a bunch of separate metrics (the "spatial y,y' metric" in contrast to "the spatial x,x' metric" or "the temporal t,t' metric") when in fact there is only one metric which deals with all coordinates, and in any case the Lorentz transformation equations do not define the "metric", the metric would be ds^2 = dx^2 + dy^2 + dz^2 - c^2*dt^2 in the unprimed frame and ds^2 = dx'^2 + dy'^2 + dz'^2 - c^2*dt'^2 in the primed frame. Many of your idiosyncratic ways of saying things suggest you aren't very clear on the correct use of technical terms such as "metric" and "phase", so it'd be better to try to avoid technical shorthand and spell out clearly what you mean, like "according to the Lorentz transformation the y-coordinate of any point on the light beam's path would be the same as the y'-coordinate". If this is indeed what you meant, then your comment "It is only with the time component included that this makes sense" is understandable, since a single event on the photon's worldline with the same y and y' coordinate will have different t and t' coordinates in the two frames, which explains how the two frames can disagree on the velocity of the photon in the y/y' direction despite the fact that y=y' according to the Lorentz transformation.

This raises a question ; the reduced y' velocity in S... y velocity, requires clocks in S to be running faster than in S' So how does this work with the blueshift??
If the time in S' is dilated , then the emitted photon would be redshifted relative to S.
So does the blueshift resulting from the angle of aberration raise the frequency more than the dilation redshift factor to realize a net blueshift or am I astray somewhere here??

Time dilation always causes the light to be more redshifted in frequency than we'd expect in classical physics if the light was a vibration in a classical medium (the luminiferous aether) and the receiver was at rest relative to this medium (so in the frame of the receiver light waves would always travel at c, though not in the frame of the emitter in such a classical universe). In fact the relativistic frequency measured by the receiver is just sqrt(1 - v^2/c^2) times the redshifted/blueshifted frequency you'd predict in the classical scenario, see my comments in post #15 and post #18 on this thread. In the example I analyzed in post #15 for a source sending out signals at a frequency of 1/(20 seconds) and traveling straight towards the receiver at 0.6c, the classical prediction would be for the frequency to be blueshifted to 1/(8 seconds), but the relativistic prediction would be for a frequency of 1/8 times sqrt(1 - (0.6c)^2) = 0.8 = 8/10, i.e. a frequency of (1/8)*(8/10) = 1/(10 seconds).

Well partly. It seems that purely on a conservation basis a photon emitted directly astern would have a v < c if the forward motion of the emitter was conserved in the normal sense.

"Conserved"? Again you're using technical terms in a way that doesn't really make sense. If you mean that the velocity of the emitter in the observer's frame should add to the velocity of the photon in the emitter's frame to produce the velocity of the photon in the observer's frame (i.e. Galilean addition of velocities), then say something like that, but "conserved" means that some quantity doesn't change with time in a particular frame, so if you say "the forward motion of the emitter was conserved" the normal interpretation of that would be that the forward velocity/momentum of the emitter doesn't change with time, it would have nothing to do with the velocity of the photon.

That a photon emitted at an angle astern conserves the y' component (with the time factor transformation) to derive a directional vector but the magnitude of that (-x,y) vector would not seem to reflect the +x motion of the emitter at all.

Do you just mean that if the emitter is traveling in the +x direction and it sends out a photon in the -x direction, the speed of the photon in that direction will always be c regardless of the speed of the emitter? If so, yes. Do you see some problem with this, though?

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?

As above ; I of course don't disagree but these are the transformed coordinate speeds aren't they? The velocity of the emitter does not enter into it or does it?

It enters into it in the sense that if the photon was emitted in a purely vertical direction in the frame of the emitter, then in the frame where the emitter is moving horizontally at speed v, s_H = v for the photon as well. Of course if the photon was not emitted vertically in the frame of the emitter this wouldn't be true, but this was the scenario we were originally talking about.

Sorry if I wasn't clear. I assumed from the context of this discussion we were talking about a bullet orthogonal to its rest frame as observed in the frame where the gun is in motion.

"Bullet orthogonal to its rest frame" is totally unclear, a "rest frame" is just an x,y,z,t coordinate system, it doesn't have a direction for anything to be "orthogonal to", nor do I know what it means for an irregular object like a bullet to be "orthogonal to" anything. Perhaps you mean that the bullet's spin axis is orthogonal to the axis which the observer's frame is moving relative to the bullet's rest frame? For example, if the bullet's rest frame S' has axes x' and y', and the origin of the observer's frame S is moving along the +x' axis of S', then the bullet's spin axis would be parallel to the y' axis, orthogonal to the x' axis?

Drift in this context was meant as motion in x comparable to wind drift with actual bullets,, meaning sideways motion . No implication of wind involved in this case of course :-)

Why is "motion in x" called "sideways"? Was the bullet not fired in the x direction in the observer's frame? Again lots and lots of explicit detail is needed if you want these scenarios to be understood, you have to stop being so laconic in your descriptions!

Anyway, are you claiming that the center of mass of a nonspinning bullet would follow a different trajectory than a spinning bullet, if both were fired with the same initial velocity (speed and direction) and both were traveling in a vacuum with no forces acting on them?

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?

If you will notice after all that you have in the end both confirmed and agreed with what I said in the first place.

How? Why? Please spell it out! It is completely incorrect to talk about a wave being out of phase with itself for the reason I gave above, if you think the last part of my comment somehow shows that it makes sense then you are obviously misunderstanding something about the meaning of "out of phase", but I can't identify your confusion unless you explain your thinking.

The graphical representation in that clip does not really apply. If you consider only the electric component viewed as a packet of parallel wave fluctuations then the bullet analogy would see them as parallel but tilted relative to the linear direction of travel.

What is a "packet of parallel wave fluctuations"? Please don't use pseudo-technical language, try to talk in as plain English as possible since your attempts to use technical vocabulary tend to be more confusing than helpful. For example, are you talking about the fact that the little red lines (which represent the electric field vectors at different points along the x-axis) are all parallel to each other, or parallel to the y-axis? Or are you talking about some different entities (not the little red lines) which you think would be "parallel" to something else? Try to use the most dumbed-down nontechnical language you can to describe what you're picturing (better yet would be if you could actually draw a diagram and post it here, but that might be hard if you don't have a scanner or good paint program).

Also, in the animation the little red lines are always orthogonal to the x-axis which is the direction of travel, when you say "tilted relative to the linear direction of travel" are you imagining a type of electromagnetic wave where this is not true? If so I don't believe that's possible according to classical electromagnetism, the electric and magnetic field vectors always point orthogonally to the direction the wave is propogating. You can have cases where the vectors are rotating in the plane that's at right angles to the direction of propagation though, see the discussion of circular polarized light halfway down the page here (or see the animation here where the position of the blue coil at an give point along the axis parallel to the coil shows the direction of the electric field vector at that point on the axis)

With various part of a single cycle not simultaneously in phase with other parts.

Again, "phase" refers only to entire waves, not "parts" of that wave. By "various parts" do you mean the little red lines or something else?

As I originally said this does not track, so it might be better to look at the aberration angle, not neccessarily as a vector sum of conserved velocities, but as an actual emission angle.
Perhaps??

"Aberration" happens when you are comparing the direction of propagation in the emitter frame with the direction of propagation in some frame moving relative to the emitter, you didn't say anything about analyzing the same wave in two different frames in the part of the discussion which started with the comment "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". Again you need to be waaaaaay more explicit and detailed about what scenario you want to analyze and what frames you want to analyze it from, because I see no connection between the various cryptic comments you are making about waves and phase.

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.

...

Sorry again. It was late and I didn't notice the paragraph separations . The only relevant part was the part I bolded as it pertained to the question above, about wave orientation.

A plane wave of the type usually analyzed in electromagnetism doesn't have a "wave front", it just goes on forever in either direction...do you just mean that the electric field vectors at any point along the wave (the red lines in the animation) are always perpendicular to the direction the wave is propagating? If so, that's true, but I don't know what this has to do with aberration (perhaps you are concerned that if the red lines are perpendicular to the direction of motion of the wave in one frame, they would not be perpendicular in a different frame? That wouldn't be the case, although explaining why would require a careful analysis of how electric and magnetic field vectors transform between frames)

Regarding the question I asked earlier ; COuld an emitted photon at any angle make it through a perpendicular tube in the frame where the emitter was in motion??
Looking at it now there seems like there should be some angle in the moving frame where an emitted photon would travel orthogonally in the lab frame. WHere the forward momentum would be exactly equivalent and opposite to the -x component deriving from the negative angle it was emitted at. What do you think?

I think I understand what you're saying here--you mean that the emitter's frame is unprimed and the photon is emitted at an angle so it has a nonzero speed on the x-axis, and the direction of the photon's motion is in the -x direction in this frame? And that in the tube's frame which we could call the primed frame, the emitter is moving in the +x' direction? If so, yes, if the photon has a speed of v in the -x direction in the emitter frame, and meanwhile the emitter has a speed of v in the +x' direction in the tube frame, then in the tube frame the photon will travel straight up vertically.

 

[Austin0]

Originally Posted by Austin0
I am interested in Terrell rotation but have already said in this thread it is not relevant to the question of the thread.

Could you try and explain this to Starthaus. JesseM and myself are having trouble getting this across to him.

Hah! LOL Good one kev

 

[Austin0]

Hi JesseM ...Man you guys are tough. After giving me endless flack in the beginning because I used simplistic descriptions and not accepted terminology I have made a concerted effort to assimilate the right terms in order to be understood. Now you'retelling me to forget that and go back to simplistic descriptions :-)
But I do understand your point. Part of my laconic approach, as you put it, was a measure of respect on my part in not wanting to state to you, who have knowledge, the self evident.
And an assumption of your ability to derive interpretation from context. I will try to be more explicit and comprehensive in the future

Well partly. It seems that purely on a conservation basis a photon emitted directly astern would have a v < c if the forward motion of the emitter was conserved in the normal sense.

"Conserved"? Again you're using technical terms in a way that doesn't really make sense. If you mean that the velocity of the emitter in the observer's frame should add to the velocity of the photon in the emitter's frame to produce the velocity of the photon in the observer's frame (i.e. Galilean addition of velocities), then say something like that, but "conserved" means that some quantity doesn't change with time in a particular frame, so if you say "the forward motion of the emitter was conserved" the normal interpretation of that would be that the forward velocity/momentum of the emitter doesn't change with time, it would have nothing to do with the velocity of the photon.

Isn't Galilean addition of velocities an expression derived from the conservation of the motion of the source?
In a Newtonian context, where both mass and time are invariant , doesn't conservation of momentum reduce to conservation of velocities i.e. v of the frame, emitter, launcher etc. + the imparted velocity of the projectile etc. (v_f+v_p) and then *mass to derive momentum?? So doesn't conservation in this context mean preservation through the transformation??
Understand I am not arguing here , I am interested in getting a firmer grasp on the meanings.

Sorry if I wasn't clear. I assumed from the context of this discussion we were talking about a bullet orthogonal to its rest frame as observed in the frame where the gun is in motion.

"Bullet orthogonal to its rest frame" is totally unclear, a "rest frame" is just an x,y,z,t coordinate system, it doesn't have a direction for anything to be "orthogonal to", nor do I know what it means for an irregular object like a bullet to be "orthogonal to" anything. Perhaps you mean that the bullet's spin axis is orthogonal to the axis which the observer's frame is moving relative to the bullet's rest frame? For example, if the bullet's rest frame S' has axes x' and y', and the origin of the observer's frame S is moving along the +x' axis of S', then the bullet's spin axis would be parallel to the y' axis, orthogonal to the x' axis?

Drift in this context was meant as motion in x comparable to wind drift with actual bullets,, meaning sideways motion . No implication of wind involved in this case of course :-)

Why is "motion in x" called "sideways"? Was the bullet not fired in the x direction in the observer's frame? Again lots and lots of explicit detail is needed if you want these scenarios to be understood, you have to stop being so laconic in your descriptions!

The meaning was simply using a bullet in place of a photon in the original conditions, i.e. SHot perpendicular to the direction of motion ,parallel to the y axis. Spinning on the y axis
the gun moving along the x-axis and the bullet of course likewise moving in both x + y directions

Anyway, are you claiming that the center of mass of a nonspinning bullet would follow a different trajectory than a spinning bullet, if both were fired with the same initial velocity (speed and direction) and both were traveling in a vacuum with no forces acting on them?

Not at all. The only point was that the movement in the x direction would be sideways relative to the spin. Thats all.

A wave can't be out of phase with itself

If you will notice after all that you have in the end both confirmed and agreed with what I said in the first place.

How? Why? Please spell it out! It is completely incorrect to talk about a wave being out of phase with itself for the reason I gave above, if you think the last part of my comment somehow shows that it makes sense then you are obviously misunderstanding something about the meaning of "out of phase", but I can't identify your confusion unless you explain your thinking.

If you look at what I said; it was statement of the idea that a wave must be in phase with itself. That it did not make sense to think of a wave tilted relative to the motion path as would be the case with the bullet under the same conditions.
Why possibly, can we not talk obout a wave out of phase with itself? To propose the concept as a viable physical reality is another matter, but that is not what I was saying.
Do you think it is impossible to define or describe a wave that is out of phase??

The graphical representation in that clip does not really apply. If you consider only the electric component viewed as a packet of parallel wave fluctuations then the bullet analogy would see them as parallel but tilted relative to the linear direction of travel.

What is a "packet of parallel wave fluctuations"? Please don't use pseudo-technical language, try to talk in as plain English as possible since your attempts to use technical vocabulary tend to be more confusing than helpful. For example, are you talking about the fact that the little red lines (which represent the electric field vectors at different points along the x-axis) are all parallel to each other, or parallel to the y-axis? Or are you talking about some different entities (not the little red lines) which you think would be "parallel" to something else? Try to use the most dumbed-down nontechnical language you can to describe what you're picturing (better yet would be if you could actually draw a diagram and post it here, but that might be hard if you don't have a scanner or good paint program).

Isn't the photon as a wave packet a common description? The graph you referred to was a pair of orthogonal slices of what we envision as extending in space in both directions [if motion is x then in both y,-y and z,-z] yes?? So each succeeding peak of electric and magnetic fluctuation is both parallel and orthogonal to the axis of travel and extends outward from the shared axis in both directions, yes?
We can describe such a wave where the electric and magnetic extended peaks are parallel but not orthogonal to the motion axis , yes?? WIth no supposition of reality, just like we can talk about tachyons.
That is what I was describing as being untenable.
If we imagine two liquid surfaces orthogonal to each other with patches of waves on each,
traveling together , then the waves could be parallel and perpendicular to the travel or parallel but at an angle, at least conceptually. yes?? If at an angle then wouldn't it be meaningful to say that each individual wave crest was occurring not simultaneously across the crest but was happening at different times along that crest and so was out of phase with itself?
If this is incomprehensible don't worry as we both agree there aren't many waves like this.
To avoid ambiguity make that none. :-)

Also, in the animation the little red lines are always orthogonal to the x-axis which is the direction of travel, when you say "tilted relative to the linear direction of travel" are you imagining a type of electromagnetic wave where this is not true? If so I don't believe that's possible according to classical electromagnetism, the electric and magnetic field vectors always point orthogonally to the direction the wave is propogating. You can have cases where the vectors are rotating in the plane that's at right angles to the direction of propagation though, see the discussion of circular polarized light halfway down the page here (or see the animation here where the position of the blue coil at an give point along the axis parallel to the coil shows the direction of the electric field vector at that point on the axis)

Yep I didnt think it was possible either.

With various part of a single cycle not simultaneously in phase with other parts.

Again, "phase" refers only to entire waves, not "parts" of that wave. By "various parts" do you mean the little red lines or something else?

As I originally said this does not track, so it might be better to look at the aberration angle, not neccessarily as a vector sum of conserved velocities, but as an actual emission angle.
Perhaps??

"Aberration" happens when you are comparing the direction of propagation in the emitter frame with the direction of propagation in some frame moving relative to the emitter, you didn't say anything about analyzing the same wave in two different frames in the part of the discussion which started with the comment "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". Again you need to be waaaaaay more explicit and detailed about what scenario you want to analyze and what frames you want to analyze it from, because I see no connection between the various cryptic comments you are making about waves and phase.

But that of course is exactly the situation of the original scenario. Comparing the direction of travel in the lab frame [the aberrated angle] when it is emitted orthogonally in the moving frame of the emitter.
That being said I will try to be more complete and unambiguous in the future

A plane wave of the type usually analyzed in electromagnetism doesn't have a "wave front", it just goes on forever in either direction...do you just mean that the electric field vectors at any point along the wave (the red lines in the animation) are always perpendicular to the direction the wave is propagating? If so, that's true, but I don't know what this has to do with aberration (perhaps you are concerned that if the red lines are perpendicular to the direction of motion of the wave in one frame, they would not be perpendicular in a different frame? That wouldn't be the case, although explaining why would require a careful analysis of how electric and magnetic field vectors transform between frames)

It has nothing to do with aberration per se that was just a descriptive term
Dont we consider a photon as being somewhat localized , not extending indefintely in space in any direction??
SO thanks for your patience with my slow progress in learning the language.

 

[starthaus]

The coordinate orientation of the tube in a frame moving relative to the tube can be derived from the coordinates of points on the tube in its rest frame plus the Lorentz transformation, they don't depend on anything else like whether the source is at rest or in motion relative to the tube.

This is incorrect, I have already pointed out a very good source that clearly contradicts your above statememt. Here it is again. The orientation of the tube depends on multiple factors, like the relative speed between the tube and the source. This explains why we change the orientation of our telecopes as function of the time of the year.

 

[JesseM]

This is incorrect, I have already pointed out a very good source that clearly contradicts your above statememt. Here it is again.

That wikipedia article deals with the aberration of light, not of tubes. The telescopic tubes in the diagram are only parallel to the light beams because they've been manually aimed that way (astronomers want to point their telescopes in the direction of the apparent position of the star they're looking at after all). If you think something in the article besides the diagram implies that a tube which is vertical in its rest frame should be non-vertical in another frame moving horizontally relative to it, please quote the specific section of the article that you think suggests this. Meanwhile, if you really believe that, you might try looking again at my post #74 where I gave a simple derivation using the math of the Lorentz transformation to show that a tube which is vertical in its rest frame (parallel to the z axis) is still vertical at any given moment in the frame of an observer moving horizontally (in the x direction) relative to the rest frame. The math is pretty easy, if there's an error there it shouldn't be hard to spot it!

The orientation of the tube depends on multiple factors, like the relative speed between the tube and the source. This explains why we change the orientation of our telecopes as function of the time of the year.

Yes, we change the orientation of the telescopes over time by applying some force to them, it isn't a matter of the orientation being different in two different inertial frames (at least not if the two frames are such that the telescope is at rest and vertical in one, and the second frame is moving horizontally relative to the first). If you don't understand the difference between 1) the idea that you can apply a force to a telescope to change its orientation from one time to another, as measured in any single frame, and 2) the idea that an inertial telescope's orientation in one inertial frame might be different in a second inertial frame (which could be the case if the telescope's orientation was not parallel to or perpendicular to the axis of relative motion between the frames, but would not be the case if its orientation was perpendicular to the axis of motion, as I showed in post #74), then you really are hopelessly confused!

 

[starthaus]

That wikipedia article deals with the aberration of light, not of tubes. The telescopic tubes in the diagram are only parallel to the light beams because they've been manually aimed that way (astronomers want to point their telescopes in the direction of the apparent position of the star they're looking at after all).

This is what I have already been telling you repeatedly, if the tube is moving wrt the light source, you need to incline it in the direction of motion.

If you think something in the article besides the diagram implies that a tube which is vertical in its rest frame should be non-vertical in another frame moving horizontally relative to it,

The tbe cannot be vertical in the rest frame since it needs to be inclined, a fact that you have incorrectly argued with previously.

Yes, we change the orientation of the telescopes over time by applying some force to them,

It is good to see that you are finally admitting your error, albeit grudgiingly. This is what I have been telling you all along.

it isn't a matter of the orientation being different in two different inertial frames (at least not if the two frames are such that the telescope is at rest and vertical in one, and the second frame is moving horizontally relative to the first). If you don't understand the difference between 1) the idea that you can apply a force to a telescope to change its orientation from one time to another, as measured in any single frame, and 2) the idea that an inertial telescope's orientation in one inertial frame might be different in a second inertial frame (which could be the case if the telescope's orientation was not parallel to or perpendicular to the axis of relative motion between the frames, but would not be the case if its orientation was perpendicular to the axis of motion, as I showed in post #74), then you really are hopelessly confused!

The only situation when the telecope (tube) does not need to be inclined is the OTHER situation, when the tube and the light source are moving TOGETHER, as a UNIT. I have already separated the two situations many posts ago. In this particular case , the ltube moves by vdt while the light moves cdt. I have already explained the math of this situation.

 

[JesseM]

This is what I have already been telling you repeatedly, if the tube is moving wrt the light source, you need to incline it in the direction of motion.

Are you seriously trying to tell me you were talking about manually reorienting the telescope in post #67? Your words, again:

1. In the frame of the tube+emitter+detector the light travels alongside the tube in an "up-down" fashion. The detector always detects the light unimpended.

2. In the frame of a moving observer, things are a little more complicated. Light is aberrated "to the right" by an angle \theta=arrcos(\beta). We know that from Einstein's light aberration formula. Now, while the light travels an infinitesimal length "diagonally" , at speed c and angle \theta wrt the horizontal axis, the tube moves "to the right" . It is as if the tube has been cut up into infinitelly small transversal sections and the sections have been re-arranged diagonally, such that while the light beam moves by cdt diagonally, the tube element moves vdt to "the right". So, the tube looks like it was re-assembled diaonally, in order to "allow" the light to reach the detector at the other end. This explains why the light hits the detector in the observer frame. Now, the elementary angle of tube slanting is ...cos(\phi)=\frac{vdt}{cdt}=\beta. So, to anybody's surprise, \phi=\theta=arccos (\beta). This also "happens" to be the angle of the Terrell rotation, so the formalism developed for solving the Terrell rotation works perfectly for solving this type of problem. Since you objected to using it, I gave you a different solution.

"It is as if the tube has been cut up into infintelly small transversal sections and the sections have been re-arranged diagonally" would be an awfully funny way to express the idea that some human manually pushed on the tube with their finger to change its angle. And in 1) you indicated you were thinking of the scenario where the source was at rest relative to the tube and directly below it, in this scenario the same tube wouldn't need to be inclined in the observer's frame where it was moving at the same speed as the source.

Likewise, in your post #29 which was the first one where you talked about the tube rotating, you said:

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 tube rotates in the moving frame" sounds like a claim about a natural difference in how the tube looks in two frames, not a claim about someone having to push the tube with their finger to rotate it. Especially since you add "this was established in another thread on the Thomas-Wigner rotation", since Thomas-Wigner rotation is a coordinate effect which causes an object that's oriented at some angle in one frame to be oriented at a different angle in another frame (as illustrated in section 4 of this article)

You have an annoying habit of changing your argument in midstream when some previous criticism was shown to be in error, and not even acknowledging you've changed it or acknowledging the error in what you were saying previously. A good example of this was our recent exchange on this thread, where you insisted that the "t" given by the Rindler transformation was not a time coordinate but a dimensionless number, and when I showed what was wrong with that you completely "forgot" we had been talking about t and started talking about something else in post #23. And of course you changed your argument about what was wrong with kev's derivation of time dilation for a clock moving in a circular path many times on this thread, never once answering questions about whether you agreed some previous criticism had been spurious. So, it's a bit hard for me to believe your claims about what you were "already telling me repeatedly". If you really insist that you have all along been talking about how a telescope needs to be manually inclined rather than talking about some natural effect where a vertical object in one frame is at a different angle in another frame, then a good start to convincing people would be to explain how this jibes with earlier comments like the ones from posts #67 and #29 above. I don't expect you will give any explanation since you seem to be allergic to answering requests for clarification, but in that case hopefully any other readers of this thread can see through your tactics.

 

[starthaus]

Are you seriously trying to tell me you were talking about manually reorienting the telescope in post #67?

Not in post 67. In this case the telescope is comoving with the light source. I must have explained to you the difference between comoving and non-comoving light source and telescope several times , yet you continue to comingle the two DIFFERENT cases.

Your words, again:

"It is as if the tube has been cut up into infintelly small transversal sections and the sections have been re-arranged diagonally" would be an awfully funny way to express the idea that some human manually pushed on the tube with their finger to change its angle. And in 1) you indicated you were thinking of the scenario where the source was at rest relative to the tube and directly below it, in this scenario the same tube wouldn't need to be inclined in the observer's frame where it was moving at the same speed as the source.

"As if" . Means a way of explaining things. Does not mean "is". The math in the post is un-ambigous, it tells you that while the light moves diagonally cdt at an angle \theta=arrcos(\beta) the tube moves laterally vdt such that light hits the detector in both the rest frame of light source+tube+detector and in the frame of the moving observer. In this case , the tube is not inclined, it is "as if" it were inclined.

This is not the case when the tube is moving wrt the light source, which is the subject of the latest posts.

 

[JesseM]

Not in post 67. In this case the telescope is comoving with the light source. I must have explained to you the difference between comoving and non-comoving light source and telescope several times , yet you continue to comingle the two DIFFERENT cases.

I don't comingle them, I just ask you to account for your previous statements that the tube would be rotated in a way that didn't sound like a manual rotation (which seems to include all the posts where you claimed the Terrell-Penrose rotation was somehow relevant to the angle of the tube). In fact I specifically pointed out in my last post that in your post #67 you seemed to be talking about the scenario where the source was at rest relative to the tube (i.e. my comment 'And in 1) you indicated you were thinking of the scenario where the source was at rest relative to the tube and directly below it, in this scenario the same tube wouldn't need to be inclined in the observer's frame where it was moving at the same speed as the source').

"As if" . Means a way of explaining things. Does not mean "is".

So you agree that in this scenario, the tube is completely vertical in the observer's frame? I don't see how in this scenario it is "as if" the tube is rotated in the observer's frame, even as a "way of explaining things". Do you also claim you were just saying it was "as if" the tube was rotated (even though it really isn't) when you said in post #29:

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 Thomas-Wigner rotation is a genuine rotation from one frame to another, so it sure doesn't sound like you were talking about it being "as if" the tube was rotated in the moving frame even though it really remains completely vertical in the coordinates of the moving frame.

The math in the post is un-ambigous, it tells you that while the light moves diagonally cdt at an angle \theta=arrcos(\beta) the tube moves laterally vdt such that light hits the detector in both the rest frame of light source+tube+detector and in the frame of the moving observer. In this case , the tube is not inclined, it is "as if" it were inclined.

Speaking of the unambiguous math in post #67, you also said:

Now, the elementary angle of tube slanting is ...cos(\phi)=\frac{vdt}{cdt}=\beta. So, to anybody's surprise, \phi=\theta=arccos (\beta).

So here you weren't just talking about the angle of the light, but of the tube. "Elementary angle of tube slanting" sounds like a technical claim about the orientation of the tube in the moving frame, not some purely metaphorical "as if" statement.

 

[starthaus]

So here you weren't just talking about the angle of the light, but of the tube. "Elementary angle of tube slanting" sounds like a technical claim about the orientation of the tube in the moving frame, not some purely metaphorical "as if" statement.

Of course I was talking about the angle of the light through the tube, the cdt vs. vdt are evidence of that. We have been over this several times, I thought that we have long moved on to correcting your misunderstandings about the fact that the tube has to be inclined in the other scenario, when the tube is moving wrt the light source.

In order to put an end to this sterile discussion, let's see if we can agree on the following:

1. Tube comoving with the light source, the light get aberrated, the tube does not (remains vertical)

2. Tube moving wrt light source, light gets aberrated and tube must be inclined in wrt its path in order to keep the light soulce in the objective.

can you agree on the above so we can put an end to this?

 

[JesseM]

Of course I was talking about the angle of the light through the tube, the cdt vs. vdt are evidence of that.

Your comment about the "elementary angle of tube slanting" included no cdt term, are you suggesting that comment was solely about "the angle of the light through the tube" and was totally unrelated to the previous comments in the same post where you talked about the tube being diagonal, like "So, the tube looks like it was re-assembled diagonally"? (you now claim these comments were meant in a purely 'as if' sense, though it looks to me like the words 'as if' originally referred to the idea of the tube being 'cut up' into sections and 're-arranged', not to the basic notion that the tube would be diagonal in the observer's frame)

Of course you also haven't even tried to explain how your comments in post #29 are compatible with what you now say you were arguing all along. And that's just one example, many of your other early posts pretty clearly indicated you were talking about some physical effect where the tube itself would be rotated in different frames, like this comment from post #37:

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&#039;=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).

Are you going to try to claim that in these adamant (and patronizing) dismissals of kev's statements, when you said things like "the tube rotates" you really just meant that we have to rotate it with our finger if we want the light to reach the back, or that you were just saying it is "as if" it rotates even though you knew it really doesn't? Well, no, I predict you will just ignore all these earlier comments and not even try to explain them, while continuing to act as though your argument has been consistent all along.

We have been over this several times, I thought that we have long moved on to correcting your misunderstandings about the fact that the tube has to be inclined in the other scenario, when the tube is moving wrt the light source.

What "misunderstanding" would that be? I was just talking about what would happen if the tube's orientation was perpendicular to the path of the light detector in the tube's rest frame, which was the scenario Austin0 was asking about when he originally brought up the tube in post #7:

That if there is a light detector at the end of a long light absorbtive tube aligned directly perpendicular to the source motion that it would seem impossible for a photon to reach the detector??

He never asked about what angle the tube would "have to be" pushed to in order to make sure that the light would reach the detector at the top, he was just asking if the light would reach the detector at the top given a vertical tube. That was the issue I was addressing in my posts about the scenario where the source is in motion relative to the tube.

In order to put an end to this sterile discussion, let's see if we can agree on the following:

1. Tube comoving with the light source, the light get aberrated, the tube does not (remains vertical)

Yes, although pretty clearly you didn't understand this back in earlier posts like #37 (where you apparently were talking about the 'tube comoving with the light source' scenario since you referred to the 'frame of the tube+emitter').

2. Tube moving wrt light source, light gets aberrated and tube must be inclined in wrt its path in order to keep the light soulce in the objective.

"Must be"? There is no moral or physical imperative that the light reach the detector at the top of the tube, and Austin0's original question was about whether it would reach the top if the tube was vertical in its rest frame while the source was moving horizontally in that frame. But sure, if we want to ensure that the light reaches the detector at the top, then we should use our hands to rotate the tube to the same angle as the light beam's aberrated angle in the tube rest frame, a trivial fact I have never disputed.