http://www.geocities.com/bpstudios/sciforums/photon_trajectory.PNG

A laser is moving quickly to the right relative to us, the observers. It is emitting a beam perpendicular to the line of its motion.

Case A and Case B show two different interpretations of possible light beam paths. Case A has the photons totally independant of the motion of the light source, while Case B shows a form of dependancy. This might be interpreted as the light source imparting some of its momentum to the photons, but I am not stipulating that as the cause. The angle shown in Case A is not important; only that there is some angle in Case A, and there is no angle in Case B.

I have my own concepts of which case would be the better approximation for various applications, but I am more interested in whether anyone would be kind enough to offer their interpretation. I just thought it might be fun to think about. Thank you.

...but I am more interested in whether anyone would be kind enough to offer their interpretation.

And you have any doubts about this? :D

Something to think about:

Which reference frame are your pictures supposed to apply to? Are they drawn from the point of view of somebody watching the emitter fly past, or from the point of view of somebody sitting on the emitter? Or is one of them for one case and one for the other?

Something to think about:

Which reference frame are your pictures supposed to apply to? Are they drawn from the point of view of somebody watching the emitter fly past, or from the point of view of somebody sitting on the emitter? Or is one of them for one case and one for the other?

I stipulated that "A laser is moving quickly to the right relative to us, the observers," but are you implying that both cases could be correct for a single laser beam?

I think that, if the light beam is being sent out at 90 degrees to the direction of motion in the observer frame, then B would be the correct picture.

I think that, if the light beam is being sent out at 90 degrees to the direction of motion in the observer frame, then B would be the correct picture.
James, would you care to elaborate?

PS:
I will edit the original post to include the 90 degree stipulation.

This is like throwing a ball out the window of a moving car. The ball has two components to its velocity: one in the direction it is thrown, and one which it picks up from the car's motion.

This is like throwing a ball out the window of a moving car. The ball has two components to its velocity: one in the direction it is thrown, and one which it picks up from the car's motion.

James,
You make it so clear to understand, thank you. Perhaps this concept could be a teaching tool, because it helps me. My scepticism of SR initially stemmed from this very misunderstanding.

Now I will share my concept of Case A and Case B:

Case A would be appropriate for accelerated frames.
Case B would be appropriate for inertial frames.


Fairly neat, would you agree? I have heard of Case A referred to as light 'curving', under acceleration, but here we see that it remains a straight line, afterall.

James,
You make it so clear to understand, thank you. Perhaps this concept could be a teaching tool, because it helps me. My scepticism of SR initially stemmed from this very misunderstanding.

Now I will share my concept of Case A and Case B:

Case A would be appropriate for accelerated frames.
Case B would be appropriate for inertial frames.


Fairly neat, would you agree? I have heard of Case A referred to as light 'curving', under acceleration, but here we see that it remains a straight line, afterall.

Now all you have to do is justify why light should have an invariance to the motion of its source in the projected beam vector but have motion induced laterally to the projected beam by the motion of the source. :D

It is "A" in either case, constant velocity, acceleration, or whatever. The light will continue to travel perpenticular to the motion of the emitter (it does not travel at 30 degrees as the picture makes it appear), but a "beam of light" is not a solid. What has been emitted (the photons) will appear to come from an emitter at the location the emitter was in when the photon was emitted. Without this being true time would not exist, information would travel instantaniously and the universe would have been fried by the amount of photons that take up every point in space.

- KitNyx

Now all you have to do is justify why light should have an invariance to the motion of its source in the projected beam vector but have motion induced laterally to the projected beam by the motion of the source. :D

Hi Mac,
You imply that you prefer Case A. I can appreciate that, and maybe even for UNIKEF, this might be an helpful teaching tool. Every theory needs clear, incontravertable ways of establishing its claims.

It is "A" in either case, constant velocity, acceleration, or whatever.

Your confidence is most impressive. I had thought about that for a long time, but now I will have to reconsider the whole thing.


The light will continue to travel perpenticular to the motion of the emitter (it does not travel at 30 degrees as the picture makes it appear),
It is already stipulated that the angle does not matter -- any non-zero angle would qualify.


but a "beam of light" is not a solid. What has been emitted (the photons) will appear to come from an emitter at the location the emitter was in when the photon was emitted.
Yes, you have it exactly right. The path of the photon will also determine where the source appears to be for anyone who happens to perceive the light.


Without this being true time would not exist, information would travel instantaniously and the universe would have been fried by the amount of photons that take up every point in space.

- KitNyx
I'm not sure if that last part is true, but it has got to be the coolest response I ever could have hoped for. Thanks :) .

Does this turn anyones crank?

http://www.laserwireless.net/Diagrams/photons.gif

Does this turn anyones crank?

http://www.laserwireless.net/Diagrams/photons.gif

It is beautiful. SuperL, could you please remind me where I can down load that animation program?

Aside from that, can you elaborate?

Here you go Neddy.

http://www.freeserifsoftware.com/Software/DrawPlus/download.asp

Well, it's just a box that pops out a photon at regular intervals.

Well, it's just a box that pops out a photon at regular intervals.
Are you implying that you subscribe to Case A?

Thank you for the link!

It definitely illustrates my point. Thank you.

- KitNyx

Is there someone here who does not subscribe to case A? Just curious...sometimes it is hard to tell exactly what people are trying to say (including myself I am sure).

- KitNyx

I think the drawings are misleading as they present light "rays" appearing instantaneously.

What do you think?

One argument.

If case B was true, how would a distant observer ever form an image of the source as it moved?

Ned Bate - I apologize for missing your response, I was working in another thread. I agree with you - "Yes, you have it exactly right. The path of the photon will also determine where the source appears to be for anyone who happens to perceive the light." - At the time the light was emitted...tell me, when you look at the sun (not wise) are you seeing the sun where and as it is exactly at that moment? or are you seeing it where and as it was 8 minutes ago? - KitNyx

Or is this the case:

http://www.laserwireless.net/Diagrams/photons2.gif

I think the drawings are misleading as they present light "rays" appearing instantaneously.

What do you think?
Well, I think that the rays should propagate at the finite rate of c, but I see your point. Give me time to learn how to make an animation

Case C. The beams lead not follow as in A. Aberation of starlight.

Thought experiment:

A tube, one light MINUTE long and one light SECOND wide, is moving uniformly at 0.5c similar to the animation arrangement above. A pulse of light 1 second long is emitted. What happens?

Use SR to show what must happen.

(you may have guessed that I think I know the answer to this whole thing... I'm very interested in everyones answers.)

Thought experiment:

A tube, one light MINUTE long and one light SECOND wide, is moving uniformly at 0.5c similar to the animation arrangement above. A pulse of light 1 second long is emitted. What happens?

Use SR to show what must happen.

(you may have guessed that I think I know the answer to this whole thing... I'm very interested in everyones answers.)
Who's second are we using for the pulse emission time? The formulation suggests the tube frame second, so I'll assume that, below.

We'll assume time 0 for the time of emmision for both the tube and embankment frame.

From the tube frame: The pulse definitely hits the detector, because the emitter is pointed straight at it, neither are moving, and light propagation is isotropic. The pulse is emitted for one second in the tube frame, so it is detected as being 1 second long. It hits the detector at tube time 60 sec. to 61 sec.

In the embankment frame: The tube and embankment frame agree on events happening, so the pulse definitely hits the detector. That means that in the enbankment frame the pulse is travelling at an angle so as to hit the detector.

Why is this so? Doesn't the isotropy of light propagation imply that photons move in the manner of your (i.e. superluminal's) first picture? Well, no. Both observers must agree on the events that happen, so if light strikes the sides in one frame, the it must also strike the sides in another frame. But light striking the sides is inconsistent with the assumption that light propagation is isotropic in all frames, which implies that the light must hit the detector in the tube frame. And if it didn't, it would imply an absolute reference frame.

For fun, let's calculate some times.

The tube is seen as length contracted in the direction of travel, but the tube's long side is perpendicular to its direction of travel, and thus still 60 lightseconds long. Hence, length contraction is not a factor in this thought experiment.

Let the time for the light to get from the emitter to the detector be T. The path of the pulse is the hypotenuse in a right angled triangle with sides 60 lightseconds and ½c * T long, the hypotenuse itself being of length cT. Setting c = 1 (which factors out, anyway), we have (Pythagoras)

T^2 = (½T)^2 + 60^2

which solves to approx. T = 69.28 sec. Note that this gives us gamma = 1.15 (ca.) directly because both observers agree the light hits the detector at tube time 60 to 61 sec. (And this gamma factor is (whew) the same as if we calculate it directly by the standard formula)

Now, the pulse is emitted for exactly 1 second on the tube clock, which is time dilated by the factor gamma in the enbankment frame, which we've measured directly. So the pulse is emitted for 1.1547 enbankment seconds, hence the detector is hit by the pulse at enbankment time 69.28 sec. to 70.43 sec. (ca.)

Yes, I know that's probably not quite what you had in mind, but it was fun, nonetheless.


I'd say it is case B, and the isotropy of light propagation proves it.

Is there someone here who does not subscribe to case A? Just curious...sometimes it is hard to tell exactly what people are trying to say (including myself I am sure).

- KitNyx
Yes, me, by the reasoning above (in case people find that too long-winded.)

Both the cases are wrong. JamesR was partly right, and I too was inclined to agree with him at first. I'm sure what he meant was what I'm going to say, from his ball-throwing analogy. The 'B' figure is partly correct, in that the ends of the rays will be as shown. But the actual paths will be as in the figure below.

<P>
<Img src= "http://www.geocities.com/virusmakermad/laser.PNG"></Img>

<P>
I'll attempt to make this clear, and perhaps, also open a venue for arguments from MacM. Imagine that laser device also has a cap, as show in the section in the bottom half of the diagram. Section A shows what is seen by an observer who is moving along with the laser. In this frame, the beam travels directly upwards and reaches the cap. Section B shows our frame of reference (in which the laser is moving). In this frame also, the beam has to reach the cap. But by the time the beam (or the tip of it, actually) reaches the top, the laser, and the cap, will have moved away. So the beam actually has to take a diagonal path in order to reach the cap. Simple reasoning shows that the beam will take this path even if there is no cap. Thus solved.

<P>
Now, MacM, I suppose will want to know why the beam <I>has to</I> reach the cap in both frames of reference. Apart from the argument that this is what is happens according to experience, there are also reasons which occur to one's common sense. I'm in no mood to elaborate them right now.

No...Figure B shows a "beam" of light that MOVES with the emitter. I agree that light contains information about its source, but that is the EXTEND of their conection once the photons are emitted. The photons will continue to travel in a straight line until the energy is absorbed/ relected/ refracted etc. Anyways, this is why we see objects not where they are at this moment, but as they were at the moment of releasing the observed/translated photons. If the answer were ANYTHING other than Figure A then we would have a violation of causality and relativity.

- KitNyx

Rosnet, I agree with you that the paths of the individual photons will look like your diagram makes out. But how does the beam (consisting of many sequentially outputted photons) look? Your path analysis would lead to the beam looking like case B, wouldn't it?

KitNyx, consider superluminal's thought experiment. Do you honestly think the pulse doesn't hit the detector, as case A suggests? Wouldn't that be a blatant contradiction of isotropic light propagation?

Ok. From the rest frame of the moving tube, in which we sit sipping tea at one end, the light "beam" must appear to be completely normal, straight and true. SR tells us that in this inertial frame, all of physics behaves normally. The detector goes of and all is well.

From the rest frame of the embankment the detector still goes off (it is an event that happened after all). however, the tube has moved 1/2 light minute to the right (wrt us) since the first photons were emitted.

So from the tube the beam must look like A and from the embankment it must look like B:

http://www.laserwireless.net/Diagrams/beam1.gif
Notice how cleverly I even drew the emitter and detector in figure B as length-contracted in the direction of motion. Spacetime itself is skewed wrt the embankment due to relative speed alone.

Rosnet, I agree with you that the paths of the individual photons will look like your diagram makes out. But how does the beam (consisting of many sequentially outputted photons) look? Your path analysis would lead to the beam looking like case B, wouldn't it?
Funkstar makes an important distiction between "path" and "beam". It is my fault for using both terms in the title of the thread.

Both Rosnet and Superluminal have drawn diagrams that show the light path on an angle in the same direction as the motion of the laser. However, I do not think that they really intended to claim an angle with respect to the laser. I believe these diagrams are just another way of choosing Case B.

My original question was whether there was an angle or not, and by drawing the diagrams this way, (although they are technically more correct when considering our rest frame where we are not allowed to even turn our heads)it becomes unclear as to whether the angle is being claimed to exist.

Neddy,

From the embankment the angle must exist. The photons travel in the skewed spacetime of the tube as seen from the embankment and must follow the path of the beam (any "Gunslinger" fans out there?) as drawn. It's ka.

Neddy,

From the embankment the angle must exist. The photons travel in the skewed spacetime of the tube as seen from the embankment and must follow the path of the beam (any "Gunslinger" fans out there?) as drawn. It's ka.
I agree that we see a forward angle from our rest frame, but unfortunately my original question (and drawings) were whether or not we would see the beam as being on some angle relative to the laser (or tube in your case).

I may have accidentally put us into the laser's frame, but all I meant us to do was to follow the moving laser with our eyes. We were supposed to detect whether there was an angle (backward) or whether the beam stays vertical with respect to the moving laser. I think that Case B is the same as your and Roswell's drawings, if you consider the location of the laser in each of my original drawings.

Consider my claim that Case A represents acceleration. In that case, drawing it as you and Rosnet have, you could have a vertical line that actually represents the beam hitting the sides of your tube. Technically, it is probably more correct to fix our rest frame as rigidly as you have, but it complicates the matter, in my opinion. I find it easier to just allow my eyes to follow the moving laser.

Yes, well, must not this be the case:

http://www.laserwireless.net/Diagrams/emitter1.gif

?????????????

No. Consider what it would look like if there were a mirror at the end of that beam...

Hmmmm...

Remember, this is a snapshot in time...

Yes, well, must not this be the case:

http://www.laserwireless.net/Diagrams/emitter1.gif

?????????????

Wow, is that how you envision the tube to look also?

Well, I'm not sure. I'm working on it...

Now, MacM, I suppose will want to know why the beam <I>has to</I> reach the cap in both frames of reference. Apart from the argument that this is what is happens according to experience, there are also reasons which occur to one's common sense. I'm in no mood to elaborate them right now.

You speculate to much. I do not function from theoretical basis. I depend more on emperical data. While I find no good reason for it to do so, evidence seems to indicate that indeed light, having left the source carries the momentum from the source and will indeed strike the moving cap when fired traversly in a moving vehicle.

Ok. Try this. In the tube (emitter/detector) frame, you place a giant L square (you know, a right angle square) with one leg against the front of the laser (emitter) and the other parallel to the beam, and of course, you measure the beam to be at 90^o to the front of the laser. From the embankment we already know that the beam makes the angle (wrt us) as drawn. If we could see the L square it would appear skewed just as the emitter I drew.

http://www.laserwireless.net/Diagrams/emitter2.gif

Ok. Try this. In the tube (emitter/detector) frame, you place a giant L square (you know, a right angle square) with one leg against the front of the laser (emitter) and the other parallel to the beam, and of course, you measure the beam to be at 90^o to the front of the laser. From the embankment we already know that the beam makes the angle (wrt us) as drawn. If we could see the L square it would appear skewed just as the emitter I drew.

http://www.laserwireless.net/Diagrams/emitter2.gif
I like it. I might not agree with it, but I like it. I am glad this thread has generated such speculative thinking.

I believe you are taking the rest frame to be 'at rest' so literally that you are reluctant to avert your eyes from a fixed point in space. My drawings are free from the distortion that you have introduced, yet they convey the same information that you are presenting. If you go back and look at Case B, I think you might detect how this apparent 'forward leaning' angle is created without violating the parallelism of the beam and the tube.

Furthermore, your square and the distortion of the system will have to shift toward the opposite side once the light beam reflects down from the detector at the top of the tube.

Neddy,

Are you referring to the picture Rosnet posted?

PS: I believe it would shift, from the embankment perspective as it passed by.

PPS: I'm still not convinced of my own reasoning.

Or is this the case:

http://www.laserwireless.net/Diagrams/photons2.gif

This is exactly what I was trying to depict by Case B. I did not have the animation program, so it was broken up into movie-like frames.

The 'forward leaning angle' that you and Rosnet have added is technically correct (I can see it at work here), but it does not seem to require distortion of an L-square, and it does not violate the parallelism of the beam and the tube. I believe both of your drawings were equivalent to my Case B, up to the point where you started experimenting with distortion of the L-square.

Here is chapter two:

Light beam from a rotating light source

http://www.geocities.com/bpstudios/sciforums/photon_trajectory_2.PNG

Figure A has the beam moving at an angle relative to the laser. Figure B has the beam moving parallel to the laser, as the laser was when each photon was emitted.

I thought the original post went pretty well, so I'll just throw this out there for general consumption. :)

Neddy,

Did you ever see the movie "Mars Attacks" and what happens to the martians brains when they hear Slim Whitman’s “Indian Love Call”?

Ouch! :eek:

Neddy,

Did you ever see the movie "Mars Attacks" and what happens to the martians brains when they hear Slim Whitman’s “Indian Love Call”?

Ouch! :eek:
That movie was pretty good! I get the subtlties of what your saying too. Don't worry, be happy. ;)

http://www.geocities.com/bpstudios/sciforums/photon_trajectory.PNG

A laser is moving quickly to the right relative to us, the observers. It is emitting a beam perpendicular to the line of its motion.

Case A and Case B show two different interpretations of possible light beam paths. Case A has the photons totally independant of the motion of the light source, while Case B shows a form of dependancy. This might be interpreted as the light source imparting some of its momentum to the photons, but I am not stipulating that as the cause. The angle shown in Case A is not important; only that there is some angle in Case A, and there is no angle in Case B.

I have my own concepts of which case would be the better approximation for various applications, but I am more interested in whether anyone would be kind enough to offer their interpretation. I just thought it might be fun to think about. Thank you.

Neddy Bate, Will this help any?
http://ourworld.cs.com/Sandgeist/absvel/aneddy.GIF

Geistkiesel

Geist:

2. What does moving isotropically, {moving in a straight line) mean?

Neddy, I'll give you a hint -

Since, to most of us 'isotropic' means invariant with respect to the direction in space (not 'moving in a straight line - that's 'linearity'), what it means is that G dosen't know what he is talking about.

Oops. Sorry! I gave it away...

It is "A" in either case, constant velocity, acceleration, or whatever. The light will continue to travel perpenticular to the motion of the emitter (it does not travel at 30 degrees as the picture makes it appear), but a "beam of light" is not a solid. What has been emitted (the photons) will appear to come from an emitter at the location the emitter was in when the photon was emitted. Without this being true time would not exist, information would travel instantaniously and the universe would have been fried by the amount of photons that take up every point in space.

- KitNyx
KitNyx,
My sentiments exactly.You have just been elected president of the absolute velocity = zero frame of reference club.

The fact that the lights do not inherit the sideways motion of the emitter means the lines of photons, the generated trajectories are invariant in space. This means the lines do not drift, until acted on by an out side force. Therefore, we may consider any point on the line an invariant physically generated velocity = 0 reference frame. Or for easier comprehension project two laser pulses at 90 degrees from each other. Isn't the crossing point invariant in space, meaning the point where the lasers cross is not moving, period?


Good work, KitNyx.

Did you know Michael? I will have to ask my basketball coach about this I suppose.
Geistkiesel

Neddy, I'll give you a hint -

Since, to most of us 'isotropic' means invariant with respect to the direction in space (not 'moving in a straight line - that's 'linearity'), what it means is that G dosen't know what he is talking about.

Oops. Sorry! I gave it away...
"invariant with respect to the direction is space" is the same thing as a straight line. Any variantion to a direction in space implies other than a straight line. I think it was the simplicity that threw you off here, that and your obsession to trash anything anti-SRT.

Why don't you give us a little treat here and explain your disagreement: like why straight line, or linearity is different from "invariant with respect to the direction is space".
Geistkiesel

http://www.geocities.com/bpstudios/sciforums/photon_trajectory.PNG

A laser is moving quickly to the right relative to us, the observers. It is emitting a beam perpendicular to the line of its motion.

Case A and Case B show two different interpretations of possible light beam paths. Case A has the photons totally independant of the motion of the light source, while Case B shows a form of dependancy. This might be interpreted as the light source imparting some of its momentum to the photons, but I am not stipulating that as the cause. The angle shown in Case A is not important; only that there is some angle in Case A, and there is no angle in Case B.

I have my own concepts of which case would be the better approximation for various applications, but I am more interested in whether anyone would be kind enough to offer their interpretation. I just thought it might be fun to think about. Thank you.
What you claim as observed and what is occurring is two different things. If a momentum impulse was transmitted to a photon emitted 90 degrees to the direction of motion of a frame moving say .9999c, use c to make it simpler, would not the velocity of the light then be c2 + c2 = C2 making C = sqrt2 (c)?
Gesitkiesel

I like it. I might not agree with it, but I like it. I am glad this thread has generated such speculative thinking.

I believe you are taking the rest frame to be 'at rest' so literally that you are reluctant to avert your eyes from a fixed point in space. My drawings are free from the distortion that you have introduced, yet they convey the same information that you are presenting. If you go back and look at Case B, I think you might detect how this apparent 'forward leaning' angle is created without violating the parallelism of the beam and the tube.

Furthermore, your square and the distortion of the system will have to shift toward the opposite side once the light beam reflects down from the detector at the top of the tube.
Another view here Neddy Bate. I do not understand how an observer can see what SL has drawn (or your A pic). How does he, the observer, "see" the angled trajectory? Even if standing from afar it appears that the light moves straight out from the laser housing. This is what an observer "sees".


What you said about reflecting, did you mean that the light would return on the same trajectory track which it used in the out bound motion? This I would agree to which is consistent with the three postulates of light I mentioned in the over worked figure above.
http://ourworld.cs.com/Sandgeist/absvel/aneddy.GIF

Geistkiesel

This is exactly what I was trying to depict by Case B. I did not have the animation program, so it was broken up into movie-like frames.

The 'forward leaning angle' that you and Rosnet have added is technically correct (I can see it at work here), but it does not seem to require distortion of an L-square, and it does not violate the parallelism of the beam and the tube. I believe both of your drawings were equivalent to my Case B, up to the point where you started experimenting with distortion of the L-square.
Superluminal has just added a velocity component orthogonal to the motion of the source. If this is so then we must also add the velocity increase componet to arrive at a measured velocity of light greater than c as well as trashing the postulate of light that is guaranteeing the light moves independently of he source of the light. We will also have to revise the postulates that the light moves isotropically, that is in a straight line (linear mode) and of course we dump the constancy of light (with the added momentum and all imposed by the motion of the source of the light).

Now if the light is given a momentum component parallel to the motion of the source, does this momentum component only come into play for light moving orthogonal to the motion of the source? What if the light is emitted at 89 degrees? Any lateral momentum component? what about 47 degrees?
How about 1 degree, any added momentum component? You understand that at this angle a velocity component via the frame-to-photon force structure will impose a definite velocity addition on the photon vector? At least as to speed, if not both speed and direction?.

Ask your observer what she saw one more time will you Neddy Bate? Just one more time?
Geistkiesel

Figure A has the beam moving at an angle relative to the laser. Figure B has the beam moving parallel to the laser, as the laser was when each photon was emitted.


What path would you follow if you <I>fall</I> off a merry-go-round? You would fall along the tangent to the circle. What would happen if you ran along the radius of the merry-go-round and jumped off? Your velocity would have a component in the direction of the tangent, as in the previous case, plus, there would also be a component in the direction of the radius. That is, you would fall diagonally. This is so with light also, as far as it's direction is concerned, but it's speed would still be 'c'. Both your figures are incorrect. Add a component to the beams in figure B in the direction of the tangent to the circle.

A clarification of my previous post: I was showing the paths of the photons. One source of confusion in these posts is the one between paths and beams.

Here is chapter two:

Light beam from a rotating light source

Figure A has the beam moving at an angle relative to the laser. Figure B has the beam moving parallel to the laser, as the laser was when each photon was emitted.

I thought the original post went pretty well, so I'll just throw this out there for general consumption. :)

I think the answer to Chapter two is like this. This time, I've given both the path and the
beam (which is a strange one).
<P>
<Img src= "http://www.geocities.com/virusmakermad/laser2.bmp"></Img>
<P>
Here the 'beam' is curved. The beam is only the set of all the photons that have been emitted
until this instant. It makes sense. Imagine holding a fire-hose and swinging it around. The
water drops would make almost the same pattern.

Neddy, I'll give you a hint -

Since, to most of us 'isotropic' means invariant with respect to the direction in space (not 'moving in a straight line - that's 'linearity'), what it means is that G dosen't know what he is talking about.

Oops. Sorry! I gave it away...
A straight line means just that invariant with respect to direction in space.

Likewise, light speed is independent of the motion of the source. If you measure the speed of light wrt to the frame at rest wrt the embankment the trajectoy is clearly a straight line. If the frame is now in motion and you insist that the photons will be dragged along by the moving frame then the observer on the embankment will see the same thing as the observer on the moving frame.

If the motion of the source drags the light along with it then the motion of the light has a component of velocity parallel to the motion of the frame. As it has a motion component orthogonal to the motion of the frame also the net photon velocity iis,kid ((Vp)² + (Vo)²)^1/2 = C'. When the frame is at rest wrt the embankment Vp (the parallel velocity component) is zero. And C = Vo.

Using the Michelson-Morely experiment as a model the arm of the interferometer oriented 90 degrees form the light source is shown in the literature of having teh zig - zag trajectory. This suggests a component of momentum parallel to the motion of the source. So is the speed of light measured along the tilted trajectory or do we see a mix of orthogonal and parallel components of light?

Using the same logic why does not the light moving parallel to the source not receive lesser a component of momentum decreasing the speed of light parallel to the direction of motion of the ineterferometer?

Finally, if the speed of light is independent of the speed of the source of light how can the speed of light be linked to the speed of the source in any way? And especially how is the link produced through the exchange of momentum?

Geistkiesel :cool:

A clarification of my previous post: I was showing the paths of the photons. One source of confusion in these posts is the one between paths and beams.
This post addressed to Superluminal, Rosnet, Neddy Bate,James R and MacM et al
who seem to hold to the proposition that the moving frame will always impose a momentum component in the direction of the frame motion.
http://www.msnusers.com/Savadance4me/Documents/collective.GIF

Geistkiesel :cool: