Can Light Emission Be Explained Through Newtonian Physics?

[Austin0]

From what I have learned so far, it appears that a light emission orthogonal to motion acts exactly like a Newtonian massive particle with conserved longitudinal momentum. Is this correct?
If so it would seem to be a cosmic coincidence of monumental proportions.
Having given it some thought all I have arrived at is: Possibly Newton was right and photons do have mass, Then the massless staus would appear to be a conventional calibration of the metric and a lack of rest photons.
In this case the independence of light from the motion of the emitter would seem to only strictly apply wrt emissions parallel to motion where the physical cap of c would prevent momentum conservation while all other angles of emission would retain coordinate speed of c but would have different velocity vectors
Is there any conceptual mechanism to explain this??
Am I totally misinformed regarding beaming?
Thanks

 

[jtbell]

From what I have learned so far, it appears that a light emission orthogonal to motion acts exactly like a Newtonian massive particle with conserved longitudinal momentum. Is this correct?

What do you mean by "conserved longitudinal momentum" here?

For what it's worth, the energy-momentum four-vector (E/c, p_x, p_y, p_z) transforms from one frame to another (via a Lorentz transformation, of course) in exactly the same way as the time-position four-vector (ct, x, y, z). This is true for both massive and massless particles.

In particular, for a relative velocity in the x-direction (only), the Lorentz transformation affects only p_x, leaving p_y and p_z unchanged, just as it affects only x, leaving y and z unchanged.

 

[Austin0]

What do you mean by "conserved longitudinal momentum" here?

For what it's worth, the energy-momentum four-vector (E/c, p_x, p_y, p_z) transforms from one frame to another (via a Lorentz transformation, of course) in exactly the same way as the time-position four-vector (ct, x, y, z). This is true for both massive and massless particles.

In particular, for a relative velocity in the x-direction (only), the Lorentz transformation affects only p_x, leaving p_y and p_z unchanged, just as it affects only x, leaving y and z unchanged.

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.
As I understand it, this is true as observed from other inertial frames as well. Photon beaming.
So if the frame is moving in x and the photon is emitted with a pure y vector relative to that frame , as observed in other frames the photon would have an x velocity component as well. Is this incorrect?
Unless I am misunderstanding what I have been reading.

 

[Bob S]

Consider a hydrogen atom with the electron emitting a 2p-1s transition photon (~10.2 eV) in the y direction transverse to its motion with velocity βc in the x direction. What is the energy, direction, and mass of the photon in the moving frame?

Use Lorentz transform (in thumbnail)

In hydrogen atom rest frame, the photon is:

E = 10.2 eV =p_yc, and p_xc = p_zc = 0

The photon mass is M² = E² - (p_yc)² = 0

In moving frame:

E^* = γE

p_xc^*=- βγE

p_yc^* = p_yc = E
p_zc^* = 0

Photon mass in moving frame is

(M^*)² =(E^*)² - [(p_xc^*)² +(p_yc^*)²]

= (γE)² -[(βγE)² + E² ] = γ²(E²-β²E²) - E² = 0
Photon has no mass in moving frame.

angle in moving frame is θ = tan^-1(p_yc*/p_xc^*) = tan^-1(1/βγ)

Bob S

 

[Mentz114]

Thanks, Bob S. Good answer.

 

[DrGreg]

Austin0

1. The scalar speed of light is invariant (the same in all frames) but the vector velocity of light is not. It can change direction when you change frames.
2. Photon momentum behaves like any other kind of momentum. The total momentum of a system is conserved (remains unchanged over time when measured in the same frame throughout).
3. Momentum is conserved but not invariant.
4. Momentum of a massive particle relative to a frame is given by p = \gamma mv.
5. Momentum of a photon relative to a frame is given by p = E / c = h \nu / c = h / \lambda

 

[Austin0]

Austin0

1. The scalar speed of light is invariant (the same in all frames) but the vector velocity of light is not. It can change direction when you change frames.
2. Photon momentum behaves like any other kind of momentum. The total momentum of a system is conserved (remains unchanged over time when measured in the same frame throughout).
3. Momentum is conserved but not invariant.
4. Momentum of a massive particle relative to a frame is given by p = \gamma mv.
5. Momentum of a photon relative to a frame is given by p = E / c = h \nu / c = h / \lambda

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.
As a self test:
1) A massive particle accelerated transversely changes vector magnitude as well as direction relative to an observing frame for both velocity and momentum??
2) A photon emitted transversely changes direction of the velocity vector but not the magnitude??
Changes both the direction and magnitude of the momentum vector with the magnitude decreased relative to the magnitude in the emitting frame??

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

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??
Whats wrong with this picture?

Thanks to all for your help in elucidating this question

 

[Mentz114]

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??
Whats wrong with this picture?

My first thought on this is that from the point of view of the moving emitter, the tube will appear to be tilted towards the emitter, giving a point on its trajectory when a photon can be fired down the tube.

I'll try the calculation tomorrow unless someone (please ) beats me to it.

[edit]
The scenario is shown in the picture. The tube has end coordinates (0,0) and the (0,L). The emitter travels along (V t, y_E ). We can define the tube in 4 dimensions by imagining a light beam emitted at the bottom end and sent to the top end. This gives a 4D position vector of the event when the light reaches the top ( L/c, 0, L ). From the emitter frame this transforms

( L/c,0,L) \rightarrow ( \gamma L/c, \gamma \beta L/c, L)

The x-coord of the top moves left, tilting the tube. I've cheated by choosing axes so the bottom doesn't move, but it shows the effect.
The important thing is that the bottom and top of the tube have different relative velocities with the emitter, and so the change in the x-coords will be greater at the top than the bottom, giving the tilt. As the emitter passes over the tube, it appears to the emitter that the tube becomes vertical then starts tilting the other way.

Because every point of the tube has a different relative velocity with the emitter frame, the tube will appear bent and tilted.

I could have got this wrong but it seems right.

 

[Bob S]

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??

Look at it this way. Consider an emitter in the CM system emitting a photon in the direction perpendicular to its direction of motion at velocity βc with respect to an observer in the Lab system. The photon is moving down a long straight absorptive tube, at rest in the CM system. Regardless of the velocity of the emitter and the absorptive tube relative to the observer, the photon will always exit the end of the tube, independent of the emitter's velocity in the Lab system. The observer sees the photon exiting the absorptive tube, at an angle θ_lab = tan^-1(1/βγ) relative to its direction of motion. (see Post #4).

So why does the absorptive tube appear to be rotated in the Lab system? It is due to the Terrell effect:

http://en.wikipedia.org/wiki/Terrell_rotation



Also see visual effects in

http://www.anu.edu.au/physics/Searle/

also see Post #9 in thread https://www.physicsforums.com/showthread.php?t=19860&highlight=terrell+relativistic

Bob S

 

[Mentz114]

This is embarassing, I came back to delete my post #8 but I'm too late.

 

[yuiop]

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??
Whats wrong with this picture?

If in the rest frame of the emitter, the photon reaches a detector at the end of the tube, then it follows that in any other frame the same must be true, in relativity and in Newtonian physics. Considering a vertical tube in a train. A ball is dropped in the top of the tube by an observer on the train and the ball appears to follow a path straight down to the bottom of the tube. To an observer outside the train, the ball follows a diagonal path that is combination of the balls vertical path due to gravity and trains horizontal motion. In the Newtonian context, the ball follows a diagonal path while staying within the confines of the vertical tube at all times. In the relativistic case, it is pretty much the same. If the tube is exactly orthogonal to the motion, then it does not rotate or bend. If the tube is tilted at some angle other than orthogonal or parallel to the motion, the tube angle relative to the motion appears different in the different frames and appears to be tilted differently (but not bent). Either way the ball (or a photon) stays inside the tube in both frames and hits the target or detector in both frames. I gave an equation for calculating the velocity of a particle or photon that is emitted at an angle to the relative motion of two frames, in a similar thread: https://www.physicsforums.com/showthread.php?t=423338

 

[Austin0]

Look at it this way. Consider an emitter in the CM system emitting a photon in the direction perpendicular to its direction of motion at velocity βc with respect to an observer in the Lab system. The photon is moving down a long straight absorptive tube, at rest in the CM system. Regardless of the velocity of the emitter and the absorptive tube relative to the observer, the photon will always exit the end of the tube, independent of the emitter's velocity in the Lab system. The observer sees the photon exiting the absorptive tube, at an angle θ_lab = tan^-1(1/βγ) relative to its direction of motion. (see Post #4).

So why does the absorptive tube appear to be rotated in the Lab system? It is due to the Terrell effect:

http://en.wikipedia.org/wiki/Terrell_rotation



Also see visual effects in

http://www.anu.edu.au/physics/Searle/

also see Post #9 in thread https://www.physicsforums.com/showthread.php?t=19860&highlight=terrell+relativistic

Bob S

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.
As for Terrell rotation I am very interested , even if it is an optical effect that would not really apply to this situation. I read about the effect years ago and can't remember the source , From what I do remember the effect only involved length contraction and the optical effects resulting from light propagation time from vatious parts of the object.
I have searched for more detailed links other than the one you cited but without success.
I will try the thread you referenced , thanks.
BYW thanks for your post wrt momentum and mass. Some of the notation was unfamiliar to me but I think I got the gist.

 

[Austin0]

If in the rest frame of the emitter, the photon reaches a detector at the end of the tube, then it follows that in any other frame the same must be true, in relativity and in Newtonian physics. Considering a vertical tube in a train. A ball is dropped in the top of the tube by an observer on the train and the ball appears to follow a path straight down to the bottom of the tube. To an observer outside the train, the ball follows a diagonal path that is combination of the balls vertical path due to gravity and trains horizontal motion. In the Newtonian context, the ball follows a diagonal path while staying within the confines of the vertical tube at all times. In the relativistic case, it is pretty much the same. If the tube is exactly orthogonal to the motion, then it does not rotate or bend. If the tube is tilted at some angle other than orthogonal or parallel to the motion, the tube angle relative to the motion appears different in the different frames and appears to be tilted differently (but not bent). Either way the ball (or a photon) stays inside the tube in both frames and hits the target or detector in both frames. I gave an equation for calculating the velocity of a particle or photon that is emitted at an angle to the relative motion of two frames, in a similar thread: https://www.physicsforums.com/showthread.php?t=423338

Ooops I was too ambiguous . In this case it is dropping a ball from the train down a vertical tube in the track frame. Thanks

 

[Austin0]

This is embarassing, I came back to delete my post #8 but I'm too late.

Looking at your post I don't exactly get the source of any embarrassment.
Perhaps you could explain in detail just how you screwed up

 

[Austin0]

My first thought on this is that from the point of view of the moving emitter, the tube will appear to be tilted towards the emitter, giving a point on its trajectory when a photon can be fired down the tube.

I'll try the calculation tomorrow unless someone (please ) beats me to it.

[edit]
The scenario is shown in the picture. The tube has end coordinates (0,0) and the (0,L). The emitter travels along (V t, y_E ). We can define the tube in 4 dimensions by imagining a light beam emitted at the bottom end and sent to the top end. This gives a 4D position vector of the event when the light reaches the top ( L/c, 0, L ). From the emitter frame this transforms

( L/c,0,L) \rightarrow ( \gamma L/c, \gamma \beta L/c, L)

The x-coord of the top moves left, tilting the tube. I've cheated by choosing axes so the bottom doesn't move, but it shows the effect.
The important thing is that the bottom and top of the tube have different relative velocities with the emitter, and so the change in the x-coords will be greater at the top than the bottom, giving the tilt. As the emitter passes over the tube, it appears to the emitter that the tube becomes vertical then starts tilting the other way.

Because every point of the tube has a different relative velocity with the emitter frame, the tube will appear bent and tilted.

I could have got this wrong but it seems right.

Is it perhaps the different relative velocities wrt the emitter that you are unhappy with?

 

[Bob S]

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.

If the tube and detector are both in the Lab (observer's) frame, then the tube tilt angle is

θ_lab = tan^-1(1/βγ)

as shown in my earlier post #4. The tube is not bent in the lab frame, nor is the observed trajectory of the emitted photon. If the photon reaches the end of the tube and the detector in the lab frame, then it will also, when observed in the CM frame.

Bob S

 

[Austin0]

If the tube and detector are both in the Lab (observer's) frame, then the tube tilt angle is

θ_lab = tan^-1(1/βγ)

as shown in my earlier post #4. The tube is not bent in the lab frame, nor is the observed trajectory of the emitted photon. If the photon reaches the end of the tube and the detector in the lab frame, then it will also, when observed in the CM frame.

Bob S

Hi Bob I understood that the angle in post #4 was the vector sum i.e. direction of the photon relative to the lab frame.
If the tube (in the lab) is orthogonal to the path of the emitter where does the tilt come from? The terrell effect is purely optical as far as I know.
SO the question is; can a photon follow a path directly perpendicular to the motion of the emitter?
If not then how could it travel through the tube?
Thanks for your responce

 

[Bob S]

Hi Bob I understood that the angle in post #4 was the vector sum i.e. direction of the photon relative to the lab frame.
If the tube (in the lab) is orthogonal to the path of the emitter where does the tilt come from? The terrell effect is purely optical as far as I know.
SO the question is; can a photon follow a path directly perpendicular to the motion of the emitter?
If not then how could it travel through the tube?
Thanks for your responce

The tube in the lab cannot be orthogonal to the direction of motion; it has to be at an angle

θ_lab = tan^-1(1/βγ).

Bob S

 

[Austin0]

The tube in the lab cannot be orthogonal to the direction of motion; it has to be at an angle

θ_lab = tan^-1(1/βγ).

Bob S

What do yu mean "can't" As in " if its not at an angle the photon won't make it through "

or that somehow it is not possible to align a tube orthogonal to the motion of the emitter?
Thanks

 

[Bob S]

What do yu mean "can't" As in " if its not at an angle the photon won't make it through "

or that somehow it is not possible to align a tube orthogonal to the motion of the emitter?
Thanks

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

 

[Austin0]

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

Hi Bob S I am sorry but I am still unsure as to your meaning.

1) It has to be aligned at angle tan^-1(1/βγ) in order for the photon to make it through??

2) If it is orthogonal to the motion of the emitter in the lab frame , the photon will not get through to the detector??

So are both of these true?
Thanks for your patience

 

[Bob S]

Hi Bob S I am sorry but I am still unsure as to your meaning.

1) It has to be aligned at angle tan^-1(1/βγ) in order for the photon to make it through??

2) If it is orthogonal to the motion of the emitter in the lab frame , the photon will not get through to the detector??

So are both of these true?
Thanks for your patience

(1) is always true: "aligned at angle tan^-1(1/βγ) in the lab frame".
(2) is always true, unless the emitter is at rest in the lab frame.

[added] See thumbnail

Bob S

 

[yuiop]

(1) is always true: "aligned at angle tan^-1(1/βγ) in the lab frame".
(2) is always true, unless the emitter is at rest in the lab frame.

[added] See thumbnail

Bob S

The equation you quote is correct for the angle of the photon's trajectory relative to the motion in the lab frame, but the tube does not rotate in the way your diagram suggests. See the attached modified diagram for what actually happens.

 

[Bob S]

The equation you quote is correct for the angle of the photon's trajectory relative to the motion in the lab frame, but the tube does not rotate in the way your diagram suggests. See the attached modified diagram for what actually happens.

If the tube is aligned orthogonal to the direction of motion in the CM system, and a photodetector at the end of the tube verifies that the emitted photon reaches the photodetector, then any observer in any moving reference frame will also observe that the photon does indeed reach the photodetector. So how can you say that an observer in a moving reference frame will see the photon blocked by a misaligned tube? Or are the laws of physics different in different reference frames?

Because the photon velocity is the same in all reference frames, the observer sees the photon at different locations at different times depending on time of flight, and her distance from the location of the photon. The same holds for the absorbing tube. So maybe the Terrell effect is relevant. In any case, all observers will agree that the photon hits the detector.

Bob S

 

[yuiop]

If the tube is aligned orthogonal to the direction of motion in the CM system, and a photodetector at the end of the tube verifies that the emitted photon reaches the photodetector, then any observer in any moving reference frame will also observe that the photon does indeed reach the photodetector. So how can you say that an observer in a moving reference frame will see the photon blocked by a misaligned tube? Or are the laws of physics different in different reference frames?

Because the photon velocity is the same in all reference frames, the observer sees the photon at different locations at different times depending on time of flight, and her distance from the location of the photon. The same holds for the absorbing tube. So maybe the Terrell effect is relevant. In any case, all observers will agree that the photon hits the detector.

Bob S

I agree that all observers will agree that the photon hits the detector. I did not say the the photon is blocked by a misaligned tube. It is not blocked despite being misaligned! You can see this if you actually look at the diagram I uploaded (and I assume you have not, because the image has had zero views so far). The photon travels diagonally in the lab frame while at the same time traveling up the length of the moving tube and eventually arriving at any detector located on the far end of the tube.

Think of the famous light clock example. Let us say the mirror is at the top of the tube as the tube goes from left to right in the lab frame. If the tube had to tilt to accommodate the diagonal light path in the lab frame, it would have to tilt one way as the light pulse went on its outward journey and tilt the other way when the photon is on its return journey. This obviously does not happen.

 

[Bob S]

I agree that all observers will agree that the photon hits the detector. I did not say the the photon is blocked by a misaligned tube. It is not blocked despite being misaligned! You can see this if you actually look at the diagram I uploaded (and I assume you have not, because the image has had zero views so far). The photon travels diagonally in the lab frame while at the same time traveling up the length of the moving tube and eventually arriving at any detector located on the far end of the tube.

I did look at your thumbnail before, and I just did again. I now understand what you are saying, and I think you are right. But this is not what the observer in the lab frame actually sees, however, because according to the Terrell rotation effect, the tube will appear rotated to the observer. This simple movie of a relativistic tram going from right to left shows the observer's view of the tram apparently rotating, so the observer sees the back side of the tram as it passes. Without Terrell rotation, the observer would not see the back at all.

http://www.anu.edu.au/physics/Searle/Tram_fast.mpg

Bob S

 

[Austin0]

(1) is always true: "aligned at angle tan^-1(1/βγ) in the lab frame".
(2) is always true, unless the emitter is at rest in the lab frame.

[added] See thumbnail

Bob S

Thanks Bob S It appears to me that kev is talking about a different thing.
As I understand your diagram the angled tube is stationary in the lab frame .
In kevs diagram the tube is stationary in the emitter frame. In that case there has never been a question about it reaching the detector.

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

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

 

[starthaus]

From what I have learned so far, it appears that a light emission orthogonal to motion acts exactly like a Newtonian massive particle

This is false, light does not act like a "Newtonian" and most definitely photons are not "massive"

Having given it some thought all I have arrived at is: Possibly Newton was right and photons do have mass,

No, this is also false.

Then the massless staus would appear to be a conventional calibration of the metric and a lack of rest photons.

This is incomprehensible.

In this case the independence of light from the motion of the emitter would seem to only strictly apply wrt emissions parallel to motion

This is also false since there is a whole class of experiments on this subject that falsifies the above statement.

where the physical cap of c would prevent momentum conservation

No, the "physical cap of c" does not prevent conservation of momentum. Conservation of momentum, as explained by DrGreg, is a general law.

while all other angles of emission would retain coordinate speed of c but would have different velocity vectors

What is this supposed to mean?

Am I totally misinformed regarding beaming?

Yes.

 

[starthaus]

The equation you quote is correct for the angle of the photon's trajectory relative to the motion in the lab frame, but the tube does not rotate in the way your diagram suggests. See the attached modified diagram for what actually happens.

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

 

[Austin0]

From what I have learned so far, it appears that a light emission orthogonal to motion acts exactly like a Newtonian massive particle with conserved longitudinal momentum. Is this correct?

This is false, light does not act like a "Newtonian" and most definitely photons are not "massive"

SO in the actual context of the question are you saying that a photon does not conserve the forward momentum of the source motion??
In this context how is the motion path of a photon different from a Newtonian massive particle?

Having given it some thought all I have arrived at is: Possibly Newton was right and photons do have mass,

No, this is also false.

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

Then the massless staus would appear to be a conventional calibration of the metric and a lack of rest photons.

This is incomprehensible.

No rest photons= no rest mass to measure. But electrons and atoms react to the imparted momentum so it is not completely off the wall to consider there might be equivalent mass , in which case it is a matter of convention that they are regarded as massless.

In this case the independence of light from the motion of the emitter would seem to only strictly apply wrt emissions parallel to motion where the physical cap of c would prevent momentum conservation

This is also false since there is a whole class of experiments on this subject that falsifies the above statement.

I think these experiments were related to light parallel to motion. My comment was not made in reference to light speed independence from the motion of the source but wrt to the conservation of the source foward motion wrt photons emitted transversely.

No, the "physical cap of c" does not prevent conservation of momentum. Conservation of momentum, as explained by DrGreg, is a general law.

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

while all other angles of emission would retain coordinate speed of c but would have different velocity vectors

What is this supposed to mean?

See below

2) A photon emitted transversely changes direction of the velocity vector but not the magnitude??

Am I totally misinformed regarding beaming?
Thanks

Yes.

So far you have said nothing constructive to add to my understanding.

Perhaps you might have some insight on the question below??

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

 

[Austin0]

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

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

 

[JesseM]

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

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

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

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

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

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

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

 

[Austin0]

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

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

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

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

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

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

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

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

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

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

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

Thanks

 

[starthaus]

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

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

 

[yuiop]

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

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

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

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

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

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

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

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

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

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

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



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

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

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

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

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

 

[yuiop]

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

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

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

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

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

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

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

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

 

[starthaus]

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

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

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

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

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

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

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

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

 

[yuiop]

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

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

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

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

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

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

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

Simply wrong.

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

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

 

[Austin0]

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

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

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

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

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

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

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

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

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

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

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

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

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

Thanks for the links ,,very interesting

 

[starthaus]

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

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

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

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

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

Precisely.

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

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

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

If you say so :-)

 

[JesseM]

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

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

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

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

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

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

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

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

Don't you find it somewhat [read extremely] curious that photons are not totally 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.

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

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

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

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

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

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

 

[yuiop]

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

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

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

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

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

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

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

 

[yuiop]

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

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

How else would you do it?

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

 

[starthaus]

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

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

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

What did I just tell you?

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

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

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

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

 

[starthaus]

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

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

 

[Austin0]

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

So what does go on in the Terrell paper?

 

[JesseM]

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

 

[starthaus]

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

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

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

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

 

[yuiop]

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

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

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

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

Derivation of formulas for second diagram:

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

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

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

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

 

[yuiop]

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

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

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

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

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

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

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