Light beam path from a moving light source

[superluminal]

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

 

[MacM]

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.

 

[superluminal]

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.

 

[Neddy Bate]

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.

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.

 

[superluminal]

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.

 

[Neddy Bate]

Or is this the case:

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.

 

[Neddy Bate]

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.

 

[superluminal]

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!

 

[Neddy Bate]

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!

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

 

[geistkiesel]

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?

Geistkiesel

 

[superluminal]

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...

 

[geistkiesel]

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​

 

[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​

 

[geistkiesel]

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 c[sup]2[/sup] + c[sup]2[/sup] = C[sup]2[/sup] making C = sqrt2 (c)?
Gesitkiesel​

 

[geistkiesel]

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.

Geistkiesel​

 

[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​

 

[Rosnet]

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.

 

[Rosnet]

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.

 

[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...

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 ​

 

[geistkiesel]

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.

Geistkiesel ​