Is the behavior of light that remarkable?

[DB Katzin]

This analysis seems elementary but I haven't seen it anywhere:


1) In special relativity, we are amazed that a beam of light emitted by a spaceship moving at .99c and measured by either a stationary observer (relative to the space ship) or by an observer on the moving ship has velocity c and not as we might expect 1.99c as would be the case were this a projectile fired from a conventional terrestrial vessel. Why is this so amazing? Do we not observe the same phenomenon when a sound is emitted by a moving train whose velocity, for the purpose of analogy, is less than the speed of sound in air? Both an observer on the train or on the platform measures the speed of sound in air as having the same velocity since the velocity of sound in air is independent of the velocity of the train. What differs for the two observers is the perceived frequency of the sound. The difference is of course the familiar Doppler effect. Similarly, the two observers of the beam of light will perceive the frequency of the emitted beam of light very differently. The stationary observer will see the light's frequency shifted due to the Doppler effect, the moving observer will not. The analogy breaks down because sound is purely a wave phenomenon and light is not. There are no “auditons” emitted by the train whistle. In fact nothing “new” is emitted by the whistle, it simply acts on the air molecules to create alternate regions of compression and rarefaction which propagate through space with wave-like properties at the speed of sound. This won’t do for light since light is an actual something moving wave-like through space. So the analogy with sound ends there and we demand that since it has a “particle-like nature” as well as a “wave-like nature” it should resemble
a projectile fired from a conventional vessel whose velocity appears different to different observers depending on their point of view.

Now we know the velocities of the light beam cannot be additive since c, and only c, is the velocity of light in a vacuum or in space and in this way it is wave-like and behaves like sound. But consider the momentum or "energy" of a photon, which for a classical projectile would differ for the two observers as expected: The observer on the ship sees the projectile’s momentum as a function of the projectile’s velocity alone, while the stationary observer measures the momentum as a function of both the velocity of the vessel and the velocity of the projectile. How does this sleight of hand trick help? Since we know the momentum/energy of a photon is proportionate to Planck’s constant times its frequency and since the frequency is shifted, according to the stationary observer, by the Doppler effect, he will measure the energy of the emitted photons as being a function of both the speed of light and the speed of the spaceship while the moving observer will measure the momentum/energy of the emitted photons as being a function of the speed of light alone. So it seems once we sort out the nature of light a bit thanks to the work of professors Planck and Einstein, the predictions of special relativity regarding light's peculiar behavior with respect to observers in 2 frames of reference that are in uniform rectilinear motion are really quite similar to other more familiar terrestrial phenomena.

 

[ZikZak]

Do we not observe the same phenomenon when a sound is emitted by a moving train whose velocity, for the purpose of analogy, is less than the speed of sound in air?

No. We do not.

Both an observer on the train or on the platform measures the speed of sound in air as having the same velocity since the velocity of sound in air is independent of the velocity of the train.

That they agree on the speed of sound in air is irrelevant. The relevant quantities are the speed of the sound wave in the two different reference frames, train and platform. Because the train is moving rapidly through the air at speed $v$, the sound waves move ahead of it at speed $c_s - v$ in its reference frame, NOT at $c_s$. Similarly, the sound waves directed backwards move at $c_s + v$. Etc.

we demand that since it has a “particle-like nature” as well as a “wave-like nature” it should resemble a projectile fired from a conventional vessel whose velocity appears different to different observers depending on their point of view.

No we don't.

 

[Janus]

When we say that light has the same speed for all observers, we mean as measured relative to the observer, by the observer.

IOW, the observer on the train shining a light forward will measure the light's speed as 299,792,458 m/s relative to the train, while the observer on the platform will measure it as being 299,792,458 m/s relative to the ground.

With sound however, both measure the speed to be the same relative to the medium it is traveling through. Thus, assuming a windless day, the observer on the platform will measure the speed of sound as being 340.29 m/s relative to the ground, while the train observer will measure it as being, relative to the train, 340.29 m/s - (the velocity of the train). They get different values for the speed of sound relative to themselves.

 

[DB Katzin]

When we say that light has the same speed for all observers, we mean as measured relative to the observer, by the observer.

IOW, the observer on the train shining a light forward will measure the light's speed as 299,792,458 m/s relative to the train, while the observer on the platform will measure it as being 299,792,458 m/s relative to the ground.

With sound however, both measure the speed to be the same relative to the medium it is traveling through. Thus, assuming a windless day, the observer on the platform will measure the speed of sound as being 340.29 m/s relative to the ground, while the train observer will measure it as being, relative to the train, 340.29 m/s - (the velocity of the train). They get different values for the speed of sound relative to themselves.

Thank you. If you have another moment, please continue the critique. I am deeply appreciative. The stationary observer and the moving observer will measure the emitted light as having different frequencies, will they not? As the train moves away from the stationary observer, light shined towards him from the train will appear red-shifted and the energy a function of c-v(train) while the moving observer will measure the energy as a function of c alone. Or is this also a complete misunderstanding of relatitivity.

 

[DB Katzin]

No. We do not.



That they agree on the speed of sound in air is irrelevant. The relevant quantities are the speed of the sound wave in the two different reference frames, train and platform. Because the train is moving rapidly through the air at speed $v$, the sound waves move ahead of it at speed $c_s - v$ in its reference frame, NOT at $c_s$. Similarly, the sound waves directed backwards move at $c_s + v$. Etc.



No we don't.

Thanks for the criticism, I clearly misunderstood a fundamental point. In saying "we demand," etc. this was in the context of making classical comparisons. BTW what about the frequency analysis part of the posting. I asked the question in detail in response to Janus' post. I realize the tone of my post may have sounded glib--it was originally for a different venue--but the question is serious. Thanks again.

 

[ZikZak]

The critical point is that sound has a certain speed in one particular frame: the rest frame of the air through which it travels. Light in vacuum travels at c in ALL reference frames.

 

[DB Katzin]

The critical point is that sound has a certain speed in one particular frame: the rest frame of the air through which it travels. Light in vacuum travels at c in ALL reference frames.

Yes, I see that the analogy was poorly chosen. Not to defend it, but I was looking at if from the point of view of a projectile fired from the moving vehicle whose velocity would be seen from the train as v(proj) while the stationary observer sees the velocity as v(proj) + v(train). This is the opposite of how sound would be observed. Nevertheless, the fact that the velocity of light is always measured as c by all observers, relative to their frame of reference suggests that the velocity of light's propagation through space is independent of the velocity of the moving emitter and is rather a reflection of light's relationship to space itself. In this I am proposing it is more like the propagation of sound in air with the added stipulation that it obey relativity than it is to a projectile. Thanks again.

 

[pesto]

Thank you. If you have another moment, please continue the critique. I am deeply appreciative. The stationary observer and the moving observer will measure the emitted light as having different frequencies, will they not? As the train moves away from the stationary observer, light shined towards him from the train will appear red-shifted and the energy a function of c-v(train) while the moving observer will measure the energy as a function of c alone. Or is this also a complete misunderstanding of relatitivity.

Do I have this right?

The observers will measure different frequencies, but they will also measure different wavelengths. However, the wavelength * frequency will always = c. This is not the case for sound, which, as Janus pointed out, would equal the speed of sound - the speed of the train to the train observer.

 

[DB Katzin]

Do I have this right?

The observers will measure different frequencies, but they will also measure different wavelengths. However, the wavelength * frequency will always = c. This is not the case for sound, which, as Janus pointed out, would equal the speed of sound - the speed of the train to the train observer.

What about the special case in which the sound is emitted inside one of the cars of the moving train? Both then measure the speed of sound the same in contrast to the case of a tennis ball thrown inside the train where the mover measures the velocity of the ball, but a stationary observer measures it as the sum of the 2 velocities train + ball? How does this stipulation affect the analysis?

Intuitively we "expect" the speed of the moving train/ship to impact some aspect of the light beam emitted from it when analyzed by a stationary observer. I submit this does show up in the measurement of frequency/wavelength of the emitted light. The stationary observer can determine the velocity of the train by measuring red shift of the emitted light (this assumes he knows the know the light source of the emitted light, eg lyman-alpha line.)

 

[Al68]

Yes, I see that the analogy was poorly chosen. Not to defend it, but I was looking at if from the point of view of a projectile fired from the moving vehicle whose velocity would be seen from the train as v(proj) while the stationary observer sees the velocity as v(proj) + v(train). This is the opposite of how sound would be observed. Nevertheless, the fact that the velocity of light is always measured as c by all observers, relative to their frame of reference suggests that the velocity of light's propagation through space is independent of the velocity of the moving emitter and is rather a reflection of light's relationship to space itself. In this I am proposing it is more like the propagation of sound in air with the added stipulation that it obey relativity than it is to a projectile. Thanks again.

The remarkable thing about light is that, unlike sound, it's speed is c relative to any observer. The speed of sound is just constant relative to its medium, as expected.

It's remarkable that light from a single source will have speed c relative to two different observers moving at different speeds relative to the source. It's almost like each photon "knows" which detector will eventually absorb it, and adjusts its speed accordingly. This is why physicists were so surprised by these observations at the turn of the century.

 

[D H]

Nevertheless, the fact that the velocity of light is always measured as c by all observers, relative to their frame of reference suggests that the velocity of light's propagation through space is independent of the velocity of the moving emitter and is rather a reflection of light's relationship to space itself. In this I am proposing it is more like the propagation of sound in air with the added stipulation that it obey relativity than it is to a projectile.

There's nothing new in this claim; it is a direct consequence of Maxwell's equations. Maxwell's equations describe light as a pair of oscillating electrical and magnetic waves. Coupled together, the electromagnetic wave propagates at a speed of $c=1/\varepsilon_0\mu_0$, where $\varepsilon_0$ is the (electrical) permittivity of free space and $\mu_0$ is the (magnetic) permeability of free space. There is no room in Maxwell's equations for either the speed of the emitter or observer.

 

[ZikZak]

What about the special case in which the sound is emitted inside one of the cars of the moving train? Both then measure the speed of sound the same in contrast to the case of a tennis ball thrown inside the train where the mover measures the velocity of the ball, but a stationary observer measures it as the sum of the 2 velocities train + ball? How does this stipulation affect the analysis?

No! Again: The sound wave moves at the speed of sound in the reference frame of the air. The "stationary" observer absolutely, positively DOES measure it to travel at the speed of train + sound, because the air within the train moves with it.

 

[DB Katzin]

No! Again: The sound wave moves at the speed of sound in the reference frame of the air. The "stationary" observer absolutely, positively DOES measure it to travel at the speed of train + sound, because the air within the train moves with it.

Relative to the stationary observer the speed of sound in air IS EXACTLY the speed of sound in air no matter how fast the moving source is going as long as it is below the speed of sound and for the people on the train, the speed of sound IN THE TRAIN is independent of the speed of the train. The constancy of the speed of sound in air when emitted by a moving source is what causes the doppler effect observed by the stationary observer, but not the moving observer. This is belaboring the point which is not a good one any way.

As Al68 has pointed out, the extraordinary and unexpected finding about light is that it is always measured as having velocity C by all observers even those moving at different velocities relative to the source. In discussions of relativity to the interested but untrained public, the discussion usually ends here. And we take away from it that light is nothing at all like classical things, period. In other words there is nothing about the measurement of any aspect of light that would indicate to a stationary observer whether the source of the light was moving or stationary relative to that stationary observer. I thought the properties of sound might offer a point of comparison. Clearly it doesn't. But the stationary observer can tell whether the light source is moving or not relative to himself by observing the doppler shift of the light, assuming he knows what the emitted frequency/wavelength is in its own frame of reference, eg a Lyman-alpha line. In part this occurs because relative to a stationary observer and independent of the velocity of the light source, the speed of light in a vacuum is always EXACTLY the speed of light in a vacuum, never more, never less. This is unlike solid projectiles shot or thrown from a moving source, but in some ways similar to sound in air which always moves EXACTLY the speed of sound in air relative to a stationary observer and is measured as such by the stationary observer who can also use the doppler effect to compute the velocity of the source. That's all.

I am attempting to find analogies to help others less scientifically inclined than myself, and believe me there are millions, understand some of the findings of physics. I am an MD by training with a PHD in Physiology. I did my undergraduate work in P. Chem at the University of Chicago, Enrico Fermi Institute. I obtained my BS in 1967 and graduated magna cum laude in chemisty. I began graduate work at Cal Tech that same year and had the good fortune to attend a few graduate seminars with the late Professor Richard Feynman. I was forced to quit after a year because of the Vietnam War, and returned to complete the above referenced education in 1969 at UCLA. I may be a bit rusty, but I'm not completely dim, and I deeply appreciate all thoughtful criticism. Thanks again, DB Katzin.

 

[DB Katzin]

There's nothing new in this claim; it is a direct consequence of Maxwell's equations. Maxwell's equations describe light as a pair of oscillating electrical and magnetic waves. Coupled together, the electromagnetic wave propagates at a speed of $c=1/\varepsilon_0\mu_0$, where $\varepsilon_0$ is the (electrical) permittivity of free space and $\mu_0$ is the (magnetic) permeability of free space. There is no room in Maxwell's equations for either the speed of the emitter or observer.

Excellent, thank you. This is not intended as a new observation as I discuss below, just an attempt to find everyday analogies to help in the education of the "general public." Continuing the analysis, the stationary observer can deduce the velocity of the source by observing the doppler shift of the emitted light, if any, can he not? This is also discussed below.

Regarding the movement of light through space, bear with me here I am only trying to create imagery that may help understanding, I imagine these alternating magnetic and electric fields transfering energy up and back "like" the two legs of an ice skater. The image of a photon skating through space has something pleasing about it. Alternatively, a bit like a "slinky"--for those old enought to remember--that walks down stairs by alternating the potential energy of the coiled spring with the kinetic energy of the moving spring which is ultimately derived from gravity. Please work with me on this, if possible. I realize that light is nothing like this, but by way of analogy. Thanks.

 

[DB Katzin]

No! Again: The sound wave moves at the speed of sound in the reference frame of the air. The "stationary" observer absolutely, positively DOES measure it to travel at the speed of train + sound, because the air within the train moves with it.

According to your analysis the speed of the sound of the engines of a commercial jet flying overhead at mach .75 will be mach 1.75 (v(sound) + v(jet)) and would not be heard by a stationary observer as the customary roar, but as a sonic boom. In this case your analysis is not correct.

 

[ZikZak]

According to your analysis the speed of the sound of the engines of a commercial jet flying overhead at mach .75 will be mach 1.75 (v(sound) + v(jet)) and would not be heard by a stationary observer as the customary roar, but as a sonic boom. In this case your analysis is not correct.

You specifically set the example of sound inside a train, which carries its own air along with it. Now you want to move the goalposts to a situation where the air is at rest relative to the ground.

 

[HallsofIvy]

What ZikZak is saying is that if a person inside that jet airplane, traveling at mach .75, speaks to another person inside it, sound will travel, relative to them at "the speed of sound". If a person standing on the ground were able to see the sound waves, he would see them traveling at, relative to the ground, mach 1.75.

That is a very different situation to the "sound of the engines of a commercial jet flying overhead at mach .75" which is traveling at the speed of sound in the air outside the jet which is NOT moving at mach .75.

 

[Al68]

There's nothing new in this claim; it is a direct consequence of Maxwell's equations. Maxwell's equations describe light as a pair of oscillating electrical and magnetic waves. Coupled together, the electromagnetic wave propagates at a speed of $c=1/\varepsilon_0\mu_0$, where $\varepsilon_0$ is the (electrical) permittivity of free space and $\mu_0$ is the (magnetic) permeability of free space. There is no room in Maxwell's equations for either the speed of the emitter or observer.

I think you're technically correct, but your last sentence implies that the speed of the observer might not be zero in the reference frame that those values are measured. The speed of the observer is always zero in the frame he measures the speed of light to be c. So the observer's speed is in a way inherent in Maxwell's equations, since they assume it to be zero relative to the reference frame that $\varepsilon_0$ and $u_0$ are measured. Of course the relative speed between the observer and other objects (like the earth) is irrelevant.

 

[Al68]

I thought the properties of sound might offer a point of comparison. Clearly it doesn't. But the stationary observer can tell whether the light source is moving or not relative to himself by observing the doppler shift of the light, assuming he knows what the emitted frequency/wavelength is in its own frame of reference, eg a Lyman-alpha line.

Yeah, I'd say sound is a good analogy for light's doppler effect, but not for it's propagation speed. Of course the equations are different for the doppler effect, since light speed is invariant while the speed of sound is different for different observers (at different speeds relative to its medium), but it's still a good analogy.

 

[DB Katzin]

You specifically set the example of sound inside a train, which carries its own air along with it. Now you want to move the goalposts to a situation where the air is at rest relative to the ground.

Well said, but without getting into "seeing sound sound waves" as per HallsofIvy--as opposed to IvyCoveredHalls?- if I turn off the jets and let the plane glide, open the windows and a stationary, let's call him listener, listens for the captain to make his landing announcement, what does he hear and why? If the cone of the speaker is in the aft bulkhead facing the cockpit, the forward movement of the cone will be mach 1+ relative to the ground, does the listener hear sound or only sonic booms from that part of the loudspeaker's motion? I'm really not sure at this point--I guess that's progress.

 

[Al68]

Well said, but without getting into "seeing sound sound waves" as per HallsofIvy--as opposed to IvyCoveredHalls?- if I turn off the jets and let the plane glide, open the windows and a stationary, let's call him listener, listens for the captain to make his landing announcement, what does he hear and why? If the cone of the speaker is in the aft bulkhead facing the cockpit, the forward movement of the cone will be mach 1+ relative to the ground, does the listener hear sound or only sonic booms from that part of the loudspeaker's motion? I'm really not sure at this point--I guess that's progress.

He would not hear a sonic boom. A sonic boom isn't the result of the sound waves exceeding mach 1, it's a result of the sound waves being compressed together so they are received in a much shorter time period, due to the source moving faster than the sound wave relative to the observer. The observer receives all of the sound previously made by the jet after the jet passes him.

The sound waves in your example will, relative to the ground observer, travel above mach 1 until they leave the jet, then travel at mach 1 the rest (most) of the way. Relative to the pilot, they travel at mach 1 until he receives them. Sound waves always travel at mach one relative to the air they are propagating through. All we have to do for a given observer is add (or subtract) mach 1 to the observer's velocity relative to the air the sound propagated through.

So the relative velocity of sound is s + v (or s - v), where s is the velocity of sound relative to its medium (mach 1 in air), and v is the velocity of the observer relative to the medium (air). This was assumed to be the same equation for light in classical physics, using the hypothetical aether as its medium. That is until those "remarkable and surprising" observations showed otherwise.

 

[DB Katzin]

He would not hear a sonic boom. A sonic boom isn't the result of the sound waves exceeding mach 1, it's a result of the sound waves being compressed together so they are received in a much shorter time period, due to the source moving faster than the sound wave relative to the observer. The observer receives all of the sound previously made by the jet after the jet passes him.

The sound waves in your example will, relative to the ground observer, travel above mach 1 until they leave the jet, then travel at mach 1 the rest (most) of the way. Relative to the pilot, they travel at mach 1 until he receives them. Sound waves always travel at mach one relative to the air they are propagating through. All we have to do for a given observer is add (or subtract) mach 1 to the observer's velocity relative to the air the sound propagated through.

So the relative velocity of sound is s + v (or s - v), where s is the velocity of sound relative to its medium (mach 1 in air), and v is the velocity of the observer relative to the medium (air). This was assumed to be the same equation for light in classical physics, using the hypothetical aether as its medium. That is until those "remarkable and surprising" observations showed otherwise.

Clear and will written, thanks.

 

[DB Katzin]

The remarkable thing about light is that, unlike sound, it's speed is c relative to any observer. The speed of sound is just constant relative to its medium, as expected.

It's remarkable that light from a single source will have speed c relative to two different observers moving at different speeds relative to the source. It's almost like each photon "knows" which detector will eventually absorb it, and adjusts its speed accordingly. This is why physicists were so surprised by these observations at the turn of the century.

Then for a vessel receding from a stationary observer at relative velocity .9c, light emitted by the stationary observer, and measured by him as moving through the vacuum of space at light velocity c, is also measured by the moving observer at light velocity c relative to his moving ship, despite the fact that he is receding from the light source at .9c and is aware of this fact. The moving observer will however see this light red-shifted. In this case the moving observer finds light is non-classical since in the classical case he would measure light moving at only .1c relative to him. Does this result from the "time dilation" effect or are there other factors?

Light emitted from the moving vessel towards the stationary observer will appear to both observers as moving at speed c relative to their frame of reference, even though the light source is receding from the stationary observer at .9c and both are aware of this fact. For the stationary observer the velocity of light is independent of the relative motion of the source and is a function of "empty space" alone. This does not require a relativistic explanation and is similar to the velocity of sound in air, does it? He will however see the light as red-shifted. For the moving observer light also appears to be moving away from him at velocity c. This is not the result expected from classical wave motion analysis where the relative velocities of the vessel and the receding sound wave front are additive. I believe in this case the measurement anomaly is due to the apparent shortening of space in the direction of motion of the moving light source. This apparent deformation is such that whatever the velocity of the vessel light emitted by it will always appear to be receding or advancing with velocity c.

As for light emitted within the moving vessel it is clear that the moving observer will measure the light as having velocity c for in this frame of reference he is stationary with respect to the emitted light and is unaware of time dilation which occurs relative to an external and stationary observer. The stationary observer will also see this light moving with velocity c with respect to him. This is not the expectation from classical wave motion, I now understand, thank you, because when the enclosed medium is also moving relative to the observer, the apparent relative velocity of the wave front is the sum, or difference, of the two relevant velocities.

Is this phenomenon due to the fact that the space enclosed in the moving vessel acts as though it were stationary from the point of view of the propagation of light? Unlike material things which move at the same velocity as the enclosing vessel they are within, space from the point of view of light, does not move with it. Space is always at rest with regards to the propagation of light and light will always move with velocity c in this space and will always be measured as such by a stationary observer independent of the velocity of the vessel emitting the light or where in the vessel the light is emitted. I realize this is wordy and lacks mathematical precision, but is it on the right track, so to speak? Thanks in advance and please excuse typos.

 

[George Jones]

Does this result from the "time dilation" effect or are there other factors?

There are other factors.

https://www.physicsforums.com/showthread.php?p=2206179#post2206179

Are you familiar with Lorentz transformations between inertial coordinate systems? If you are, then you should have a look at the standard derivation of how speeds are combined relativistically.

 

[jwk]

He would not hear a sonic boom. A sonic boom isn't the result of the sound waves exceeding mach 1, it's a result of the sound waves being compressed together so they are received in a much shorter time period, due to the source moving faster than the sound wave relative to the observer. The observer receives all of the sound previously made by the jet after the jet passes him.

Which is analogous to the "sonic boom" caused by electrons moving at relativistic speeds in a medium such as heavy water. This is Cherenkov radiation and is a pretty blue.

 

[Edi]

Hi, i am new here...
So, i can understand that speeding observer will measure the light coming from back as c because of time dilation, but cant, jet, figure why he would measure light going away from ship as c - i just can't address that to time dilation.

Another one - I am in a ship traveling at .9 c and another ship traveling at .99 c fly's by me.
Would i measure its speed as .99c or .09 c ? I would still have time dilation effect so it should look like it travels by at .99 c , but in that case i would calculate his time of arrival to the target (say form Earth to Mars) much much sooner than i get there, but will actually arrive not THAT soon...

I am confused!

 

[Al68]

Then for a vessel receding from a stationary observer at relative velocity .9c, light emitted by the stationary observer, and measured by him as moving through the vacuum of space at light velocity c, is also measured by the moving observer at light velocity c relative to his moving ship, despite the fact that he is receding from the light source at .9c and is aware of this fact. The moving observer will however see this light red-shifted. In this case the moving observer finds light is non-classical since in the classical case he would measure light moving at only .1c relative to him. Does this result from the "time dilation" effect or are there other factors?

Light emitted from the moving vessel towards the stationary observer will appear to both observers as moving at speed c relative to their frame of reference, even though the light source is receding from the stationary observer at .9c and both are aware of this fact. For the stationary observer the velocity of light is independent of the relative motion of the source and is a function of "empty space" alone. This does not require a relativistic explanation and is similar to the velocity of sound in air, does it? He will however see the light as red-shifted. For the moving observer light also appears to be moving away from him at velocity c. This is not the result expected from classical wave motion analysis where the relative velocities of the vessel and the receding sound wave front are additive. I believe in this case the measurement anomaly is due to the apparent shortening, or, in the receding case, lengthening of space in the direction of motion of the moving light source. This apparent deformation is such that whatever the velocity of the vessel light emitted by it will always appear to be receding or advancing with velocity c.

As for light emitted within the moving vessel it is clear that the moving observer will measure the light as having velocity c for in this frame of reference he is stationary with respect to the emitted light and is unaware of time dilation which occurs relative to an external and stationary observer. The stationary observer will also see this light moving with velocity c with respect to him. This is not the expectation from classical wave motion, I now understand, thank you, because when the enclosed medium is also moving relative to the observer, the apparent relative velocity of the wave front is the sum, or difference, of the two relevant velocities.

Is this phenomenon due to the fact that the space enclosed in the moving vessel acts as though it were stationary from the point of view of the propagation of light? Unlike material things which move at the same velocity as the enclosing vessel they are within, space from the point of view of light, does not move with it. Space is always at rest with regards to the propagation of light and light will always move with velocity c in this space and will always be measured as such by a stationary observer independent of the velocity of the vessel emitting the light or where in the vessel the light is emitted. I realize this is wordy and lacks mathematical precision, but is it on the right track, so to speak? Thanks in advance and please excuse typos.

To date there is no evidence that "space" has any property that would act as a medium or reference frame for light propagation. Light travels at c relative to any inertial observer, not relative to "space". In your example, there is nothing wrong with calling one observer "stationary" and the other "moving", but you could just as easily switch them around and the result would be the same.

In the case of sound it does matter which observer is referred to as "moving", since although each is moving relative to the other, one is moving relative to sound's medium (air) between them.