There have been a number if comments complaining about my assertion that absolute zero velocity can be easily measured. What is lost in the conversation is the fact that absolute velocity = 0 is obtained, measured at the will of the navigator on the space ship or in the high school physics lab (gymasium).

Beside the fact that this runs counter to the claims of relativity theory few, if any, have grasped the simplicity of the condition of absolute velocity = 0.
Starting with the postulate of light that the motion of light is independent of the motion of the source, we see that a photon emitted in a moving frame will continue in a straight line while the frame moves with respect to the invariant trajectory of the light.

we start from the condition that an emitted photon is directed directly at a photon absorber on a photon emitter/absorber pair (peap) whwere the peap is attached topt hje frame. When the frame motion is zero the photon trajectory of all the emitted photons is identical, in absolute space, and the peap is the same for all photon emissions and absorptions. Now if the frame moves the photon in mid-flight will not be absorbed by the same absorber on he same peap. Rather another absorber, attached to the frame ill move into the photon approaching perpendicular to the x motion say. (3 dimensional devices cover all motion.). The emission point describes absolute position in space, and the lines or light trajectoriesl describe the invariant frame of reference.

There isn’t any new physics here, except for a violation of SR that denies the very concept of an absolute velocity = 0, or any absolute motion for that matter
The physics is harmless and should not frighten any SR theorists.

[deleted]. Stop posting a new thread on this everytime someone points out your mistakes.

Persol is right, geistkeisel. Please don't post a new thread every time you have a new idea. It means snippets of the same conversation are spread across multiple threads.

If you don't become a little more discriminating, I might decide to simply merge all new threads you start into a single thread.

geistkeisel,

Could you please put your argument more clearly?

Now if the frame moves the photon in mid-flight will not be absorbed by the same absorber on he same peap. Rather another absorber, attached to the frame ill move into the photon approaching perpendicular to the x motion say.

Are you talking about accelerating your absorbers while the photon is in flight? If so, why would it surprise you that the photon didn't land where it was aimed?

I also don't see what this has to do with absolute zero velocity.

Well let me educate you. If the frame were moving at constant velocity and the the photons are invariant in their trajectories, and they are, then the motion of the frame will pull the peaps away from the photons thereby forcing the photons to strike another absorber emitter which can be monitored and used as a count of dx/dt.

Do you iunderstand yet?

If the frame is accelerating or moving with uniform motion the same result. As the peaps only measure the time the photon strikes a particlular absorber.

Understand?

Can you provide a diagram which illustrates what you're talking about?

This is spam.
Can this stuff be consolidated or deleted?

Can you provide a diagram which illustrates what you're talking about?

Au - --------
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
| ||||||||
Ad - --------


The Au, (Aup) and Ad ( A down) are a typical peap unit. The vertical lines reperesent a straight-line photon motion between the components of the peaps. With zero velocity, or zero motion, a photon emitted by the Au foir instance, will always use the same exact trajectory every time. The upward emitted photon will likewsie use the same trajectory as the downward moving photon. The tightly packed peaps represent a peap system of a rectangular (or circular) grid.

We know the general condition is Vn > 0, ie there is motion of the frame more oftern than not.
As the hardware, the peaps, are physically attached to the Ve, any photon in transit between up/down surfaces will always strike a different unit when completing the trip when the frame is moving. The photon, moving perpendicular to the velocty (in this simplified version) will always arrive at another unit that has moved into the place of the unit that hs moved away.

Here is a way to demonstrate the simplicity of what is occuring. Take a plate or saucer and move it underneath a table lamp back and forth a few times. The plate is the frame moving with respect to the invariant motion of the photon (table lamp).

Extend yiour imaginatiion to seeing the whole lamp as a planar peap system.As the rear of the plate that is about to enter the light, can be considered an appraoching wall, the peaps used in the syatem are transferred from those closest to the approaching wall to those farthest away from the wall. Thus a constant supply of peaps are always provided as emitters and absorbers of the invariant photons. This is equivalent to moving the lamp. As we are only measuring velocity, the fact that new photons are emitted form different locations on he grid is insignificant.

For another points of view, which should indelibly imprint this on your mind, consider the fact that the peaps use in the Michelson-Morely experiment were not properly calibarated to provide the proper return and arrival trajectory, The bottom mirror was necessarily adjusted in order to obtain an interference pattern which was unnecessary and effectivley made it near impossible to read the motion they had originally intended to scrutinize. The distance added due to the trianguklar trajectory is a distortion of experimental parameters and effectielvy swamped out the relevant data, most of it that is.

Reflecting the transverse (to motion) moving photon perpendiculaly, where the total optical length is 64 meters, the distance the frame can be expected to move is a manageable distance in 2.3 x 10^-7 seconds, about, for a frame moving minimum to maximum published velocitoes of 29.8 to 208 km/sec we get 6,357x10^-3 meter to a maximum 4.42 x 10^-2 meters, or 6.357 mm to 4.436 cm.


Remember: The motion of light is independent of the motion of the source. The photons here know nothing of peaps and moving frames, or frames at rest. When a photon has arrived at a different unit that has moved into the spatial location of the previous unit having moved out. this difference is detected by the units connected to the computer monitoring the whole grid process. Hence motion is measured, absolute motion.

Three dimensional grid systems give the appropriate motions in all significant directions; or even mutliple grid planes at a variety of angles to increase the resolution of measurement can be used.

The crucial point being here that the photons motions are independent of the frame. There is no sidways "inertial component" of velocity of the photons.

If the frame were about to crash into the photons in flight the compuiter system simply resets another starting point(s) for the next emission of photons. The more peaps and the more dense the peaps, the higher the resolution of measurement.

Thanks for that. I see where your misconception is now, geistkiesel.

Imagine a straight tube oriented vertically, with a photon source at the bottom and a receiver at the topic. When the tube is stationary, photons travel in a straight line along the tube. They have no component of velocity in the horizontal direction.

You are worried that if we start moving the tube horizontally, along with its attached emitter and receiver, the path of the light won't change relative to a stationary reference frame, and therefore the light won't hit the receiver any more. It will instead hit the side of the tube.

Your are incorrect, though. Your wrong assumption is that the direction that the light is emitted does not change when the tube moves. It does. From the point of view of a stationary observer, when the tube is moving horizontally, the light emitted "inherits" some of that horizontal motion, and so travels diagonally through space, according to a stationary observer. This means that the stationary observer still sees the light hit the receiver, even though the receiver has moved from where it was when the light was emitted.

Your error is simply in assuming that the direction of the light doesn't change when the emitter starts moving.

Reflecting the transverse (to motion) moving photon perpendiculaly, where the total optical length is 64 meters, the distance the frame can be expected to move is a manageable distance in 2.3 x 10^-7 seconds, about, for a frame moving minimum to maximum published velocitoes of 29.8 to 208 km/sec we get 6,357x10^-3 meter to a maximum 4.42 x 10^-2 meters, or 6.357 mm to 4.436 cm.

geistkiesel :

Lets first educate ourselves.

The only thing you know is the optical length 64m and velocity of light c.


In an inertial frame the velocity of photon is constant c for all points in that frame whether the frame is moving or not. If you think that the photon hits, after 64/c sec, other than the point just opposite to the source due to frames unperciveable motion, then your basic premise of the invariance of velocity of photon is made invalid by yourself; because in your example the photon velocity is not c in respect of the source or target or any other point. You are violating your basic premise of the invariance of velocity of photon. So your whole calculation, involving such changing c to show the displacement, is not correct.

You cannot disprove relativity, that is to find the 'absolute velocity' of an inertial frame, this way by conveniently changing its fundmentals.


Maintain your original assumption that velocity of the photon is constant c for ALL the points in that frame ; your displacement figure would be 0.

geistkiesel :

Lets first educate ourselves.

The only thing you know is the optical length 64m and velocity of light c.


In an inertial frame the velocity of photon is constant c for all points in that frame whether the frame is moving or not. If you think that the photon hits, after 64/c sec, other than the point just opposite to the source due to frames unperciveable motion, then your basic premise of the invariance of velocity of photon is made invalid by yourself; because in your example the photon velocity is not c in respect of the source or target or any other point. You are violating your basic premise of the invariance of velocity of photon. So your whole calculation, involving such changing c to show the displacement, is not correct.

Everneo, you aren't seeing that transverse motion of the frame relative to the to the straight line motion of the photons does not create any measuremnt of light conditions. You don't see it por you ignorfe it and introduce anmbuguity into the discussion. The up/down velocithy is alweays constant, buit this does not prohibit a part of the frame moving out of the way of a photon that is instantaneously 3000 meters away when it begins its motion to the absorvber.

I really cannot see how this simple model could get so garbled that you invoke the speed of light postulates of relativity theory when not necesary for any purpose excepet to blather.. which I may add are merely assertions of physical fact and are unproved, they are mental postulates. Here the constancy of the speed of light in the context it is usually discussed is not perturbed by my model, so please do not try to distort what is stated andtrhe speed of light is not in iIssue.

"If you think that the photon hits, after 64/c sec, other than the point just opposite to the source due to frames unperciveable motion, then your basic premise of the invariance of velocity of photon is made invalid by yourself; "

This statement, as I read it, is a craftily constructed piece oif bullshit.
You put the 64 meters into the design, you made the "unperceivable motion" statemen see below. I slavishly maintained my model in strict observance to the independence of the motion of light.

Are you saying catagorically that a point on a frame cannot move out of the way of a photon emitted 3000 meters away when the frame is moving at 1 km/sec?

Why do you want to distort the matter? Stupid question isn't it. We both know why you are here. Don't insult me with it is from personal interest. I haven't seen any other of of your posts. Do you always use th Everneo handle? Is this your job? To make every effort to confuse the issues, get me engaged in soem usless rhetoric? and win a propaganda battle?


You cannot disprove relativity, that is to find the 'absolute velocity' of an inertial frame, this way by conveniently changing its fundmentals.

What do you mean I can't dispriove Special Relativity by conveniently inserting the laws iof physics into the discussion? I( can do anything I want to do.

Watch me yoyo.


Maintain your original assumption that velocity of the photon is constant c for ALL the points in that frame ; your displacement figure would be 0.

What is this supposed to mean?. please explain. You
just aren't talking physics, but then that isn't your job is it?

The photon velocity is really not an issue here Let us start with a single photon we find zipping along in space. This is an example of of photon motion moving in a straight line. That straight line will continue until acted upon by an outside force. Now let us trigger the photon release and absorption using one pair of peaps. Bear with me fior a tad here, as I want to say that as long as the sources are not in motiion the photons will always be absorbed and reemitted by the same peaps.

Now, let us carefull wrap a protective shield around the peaps and carefully notice the same peap emit and absorb the photons. Now the the peap units on either end of the trajectory are pulsing regularly. We install a few thousand more peaps in a grid system for a reason I will discuss. If the optical path length of the peap trajectory is 1000 meters the time for one way flight is 3.33 x 10^-6 seconds. For a frame moving 3500km/hr = 1 km /sec the frame moves a distance 3.33 x 10^-6 km 3.33 x 10^-6 km or 3.33 x 10^-3 m. approximately3.33 mm a workable distance.

Now the photons are moving up/down with respect to the frame,
the peaps, and with no frame motion (rare) each peap will emit and absorb as a unit. Now let us have our photon just emitted, when the frame begins to move perpendicular to the straight line motion of the photon trajectory. Now you can theorize all over the place if you will, but the photon that is headed down knows nothiing about sidways motion, as there is not any sideways motion. Straight line means straight line. Now as the frame to which the peap harware is attached moves ,while the photon is headed straight. Before the photon covers the exact same distance the photons always covered arrives at the target absorber, another has slid into place and absorbs the photon that has not had one micron of its lateral or linenear motion perturbed in the slightest with the single exception of the absorption of the photons at the ends of the trajectory, as another absorber has moved into place while the straight line moving particle was moving downward.

I always have the photn moving at C between absorber/emitter pairs
. I am saying it is the sideways motion that is zero. The straight line motion is invariant the motion is invariant. I don't remember putting a restriciton of 64 meters as an effective 1000 or 3000 meter system can be devised with proper reflecting mirrors not unlike the Michelson-Morely experiment. In any event with todays semiconductor technology advancing to exposing the photographic-lithographic produced circuits using X-ray wave lengths to further cram circuits together. this system has few if any picinic canceling design parameters to over come.
Just careerist SR theorists.

Everneo, you aren't seeing that transverse motion of the frame relative to the to the straight line motion of the photons does not create any measuremnt of light conditions. You don't see it por you ignorfe it and introduce anmbuguity into the discussion. The up/down velocithy is alweays constant, buit this does not prohibit a part of the frame moving out of the way of a photon that is instantaneously 3000 meters away when it begins its motion to the absorvber.Nice try. According to you the Earth is V0. A train moving releases light at speed c in front of it. IN ALL FRAMES OF REFERERENCE.

Your assumption that you'll observe a different in light you emit is VERY wrong. If you disagree then show us an experiment... cause every experiment so far disagrees with you on the matter.

Beside the fact that this runs counter to the claims of relativity theory few, if any, have grasped the simplicity of the condition of absolute velocity = 0.

With respect to who is this absolute velocity which is zero?

With respect to who is this absolute velocity which is zero?
1100f, we meet again. Assume the truth of the independence of the motion of light and the source of light. We may consider an emitted photon an object moving in a straight line until acted upon by an outside force.. Nothing particlularly novel so far. Taking the postulate at it most strict nmeaning photons do not vary their straight line motion.

Also, assume two or more photon are emitted simultaneously. The point of emission is an absolute point in space, though to maintain observation of that point is the technical responsibility of the experimenter the straight line photons are invariant in any motion. So to answer your question, the reference frame is the photons. Caveat Emptor: to grasp the sheer simplicity of the model, do not start rushing into checking out speed of light measurements or similar activity, or other special relativity subjects. This is not a substitute for Special relativity , well not by intention.

I use the word absolute velocity in regard to the photon system as from the driving force of the piostulate that the motion of the photon s are invaiaint until acted upon by an out side force. If this development runs counter to modern understanding of the words, so be it. It is easier and less confusing for all when I simplify the model, from my reference point.
So , simply, photons do not variy their motion by their own design.
Therefore any isotropic motion of light must be from a source external to the photon motion, therefore photons, once moving, are absolute in space and and by temporal definition, and as we know the velocity along any straight line all those restrictions placed on the mesuremnt process by SR theory, such as ignoring the reality of a frame's actual motion especially when measuring the speed of light with respect to any inetial frame disappears , and the corrections are applied autiomatically to the errors of SR. especially through the
planet earth, which is physically indistinguishable, by measurment, from straight line motion motion (rotational velocity for the earth frame is < 10^-8 degrees per second. All inertial frames, real and potential other than Ve, the earth, must necessarily accelerate with respect to the Ve in order that any relative motion is observed.and
BY physical implication, differentiated from theoretical implication.
I found the Ve, earth frame is preferred as this frame simply does not accelerate and the motion is measurably indistinguishble from straight line motion. .Virtually all other inertial frames, earth bound, at the very minimum must accelerate with respect to Ve before any relative velocvity is observed between Ve and any Vn. This places physical restricitons on obsevers in moving frames denying the justification of arbitrarily assuming a condition of rest with respect to Ve, a physical impossible condition to achjieve.
Therefore, the preferred frame is not a matter of mental choice, it is a reality forced upon the motion of all matter of the universe.

The point of emission is an absolute point in space,No, it's not. Different frames will see it in different places at different times.
the straight line photons are invariant in any motion.No, it's not. Different frames will see the light going in different directions.
# I found the Ve, earth frame is preferred as this frame simply does not accelerateNo, it does. We've explained this to you OVER and OVER and OVER.
the motion is measurably indistinguishble from straight line motionNo, it is. Yet another thing we've explained. Hell, your very own proposed 'mumps' testing presented a method for determining it. There are plenty of other methods for detecting it.
Virtually all other inertial frames, earth bound, at the very minimum must accelerate with respect to Ve before any relative velocvity is observed between Ve and any Vn.Who cares?
# Therefore, the preferred frame is not a matter of mental choice, it is a reality forced upon the motion of all matter of the universe.Yes it is. That's what makes it a 'reference frame'. You can choose your reference.

So, when are you going to actually address all the problems pointed out with your theory?

Why do you just answer with SR bullshit? Light moves invariantly and idependently of the moption of the light source, and light always has a measurable relative velocity with respct to a moving frame with Vn - Ve > 0, and alll of your conbined and singular incompetence is not going to change that. Sieg Heil !

Thanks for that. I see where your misconception is now, geistkiesel.

Imagine a straight tube oriented vertically, with a photon source at the bottom and a receiver at the topic. When the tube is stationary, photons travel in a straight line along the tube. They have no component of velocity in the horizontal direction.

You are worried that if we start moving the tube horizontally, along with its attached emitter and receiver, the path of the light won't change relative to a stationary reference frame, and therefore the light won't hit the receiver any more. It will instead hit the side of the tube.

Your are incorrect, though. Your wrong assumption is that the direction that the light is emitted does not change when the tube moves. It does. From the point of view of a stationary observer, when the tube is moving horizontally, the light emitted "inherits" some of that horizontal motion, and so travels diagonally through space, according to a stationary observer. This means that the stationary observer still sees the light hit the receiver, even though the receiver has moved from where it was when the light was emitted.

Your error is simply in assuming that the direction of the light doesn't change when the emitter starts moving.
More kangaroo physics. THE MOTION OF LIGHT IS INDEPENDENT OF THE MOTION OF THE SOURCE OF THE LIGHT.

Of course you wouldn't make such a statement unless you could prove it noW would you? Prove it JAmes R or leave me alsone. I'M TIRED OF YOUR DOUBLE TALKING MAKE IT UP AS YOU GO ALONG KANGAROO CRAP.


Give me a reference that proves what you just saidd. Is this the kind of bullshit the kangaroos pay you to educate their children with?.

Caveat emptor Mr. 'Roo, you better watch what the teach your kids in schoo.

THE MOTION OF LIGHT IS INDEPENDENT OF THE MOTION OF THE SOURCE OF THE LIGHT.

Yes, once the light has been emitted and is on its way.

As explained previously, different observers do not necessarily agree on which direction the light was initially emitted.

As far as proof goes, why the double standard? You have presented no proof of your claim.

Lets first educate ourselves.

The only thing you know is the optical length 64m and velocity of light c.


In an inertial frame the velocity of photon is constant c for all points in that frame whether the frame is moving or not. If you think that the photon hits, after 64/c sec, other than the point just opposite to the source due to frames unperciveable motion, then your basic premise of the invariance of velocity of photon is made invalid by yourself; because in your example the photon velocity is not c in respect of the source or target or any other point. You are violating your basic premise of the invariance of velocity of photon. So your whole calculation, involving such changing c to show the displacement, is not correct.

Everneo, you aren't seeing that transverse motion of the frame relative to the to the straight line motion of the photons does not create any measuremnt of light conditions. You don't see it por you ignorfe it and introduce anmbuguity into the discussion. The up/down velocithy is alweays constant, buit this does not prohibit a part of the frame moving out of the way of a photon that is instantaneously 3000 meters away when it begins its motion to the absorvber.

I really cannot see how this simple model could get so garbled that you invoke the speed of light postulates of relativity theory when not necesary for any purpose excepet to blather.. which I may add are merely assertions of physical fact and are unproved, they are mental postulates. Here the constancy of the speed of light in the context it is usually discussed is not perturbed by my model, so please do not try to distort what is stated andtrhe speed of light is not in iIssue.



Your assumed, absolute transverse motion of the frame that is only 'relative' to the 'straight line trajectory' of the photons, is wrong for it contradicts invariance of light - straight line trajectory of the photon with velocity c - from any point in an inertial frame such as yours. If the absorber does not belong to the emitter's pair, for that absorber the velocity of light is more than c. For that matter, velocity of the photon is not c for any other point in this spaceship like inertial frame. Your assumption of absolute velocity of the frame is a bad joke on invariance of light. Is that clear ?

Are you saying catagorically that a point on a frame cannot move out of the way of a photon emitted 3000 meters away when the frame is moving at 1 km/sec?

You are saying,

If the frame is not moving, photon emitted by E1 hits the absorber A1 in that 'peap' pair at D1 meters from E1 in t secs.

If the frame is moving, photon emitted by E1 hits the absorber Ax in a different 'peap' pair at Dy meters from E1 in t secs.

So the 'absolute' motion makes photon to change its velocity from D1/t to Dy/t for an observer at absorber Ax ?

Thats why i called your 'assumption' a bad joke. First prove, in reality, that the velocity of the photon is not invariant for an observer at Ax ( or any other point in this frame) then freely claim the cause being the absolute motion.



Why do you want to distort the matter? Stupid question isn't it. We both know why you are here. Don't insult me with it is from personal interest. I haven't seen any other of of your posts. Do you always use th Everneo handle? Is this your job? To make every effort to confuse the issues, get me engaged in soem usless rhetoric? and win a propaganda battle?
What are you talking ? who is insulting whom now ?


What do you mean I can't dispriove Special Relativity by conveniently inserting the laws iof physics into the discussion? I( can do anything I want to do.

Watch me yoyo.
You are not inserting the laws of physics, you are distorting the SR postulate by your own assumptions.

Yeah, we are all watching your rape of SR to prove that it is not 'chaste' anymore.


Maintain your original assumption that velocity of the photon is constant c for ALL the points in that frame ; your displacement figure would be 0.

What is this supposed to mean?. please explain. You
just aren't talking physics, but then that isn't your job is it?

Read above, and try to understand that your assumption of an 'absolute motion' of the frame directly changes the velocity of photon for all observers at all the points in this frame.

All you have done is bad assumption, nothing is demonstrated by you in reality.

You are saying,

If the frame is not moving, photon emitted by E1 hits the absorber A1 in that 'peap' pair at D1 meters from E1 in t secs.

If the frame is moving, photon emitted by E1 hits the absorber Ax in a different 'peap' pair at Dy meters from E1 in t secs.

The photon when emitted is independent of the motion of the origin of the photon. The photon doesn't know there is frame. Do you understand about "independence" ? The photon is moving in an absolute straight-line. The photon defines its trajectory by whatever laws govern photon motion. Therefore, an object that was once located where the lower absorber was located, with respect to the already defined line of motion of the previous photons (calibrated with as close to physical zero velocity as possible - say the south pole with the 2 devices, one oriented parallel and perpendicular to the sun-earth radius), has moved before the next photon arrives.

So the 'absolute' motion makes photon to change its velocity from D1/t to Dy/t for an observer at absorber Ax ?

No. You are imposing some garbled physics into a real simple situation. The rape of SR must have you confused. The new absorber being located on a plane of absorbers which includsed the first substituites for he one that moved due ti the frame motion. The assumption that the motion is peprpendiclular is included, thugh if the new absorber taking the place of the one that moved is farther from the original emitter, then the time of arrival will change accordingly. There being three orthogal planes of peaps, three dinelsions motion can be monitored. The rame computer is a dummy, programmed somthing robotic like an SR theorist, itmerely records thetime of arrival and sends the information off to the processor, with its identification (grid location).

Thats why i called your 'assumption' a bad joke. First prove, in reality, that the velocity of the photon is not invariant for an observer at Ax ( or any other point in this frame) then freely claim the cause being the absolute motion.

I assume the postulate of light that the motion of light is independent of the motion of the source. You have heard of this have you not? It is basic to SR theory. No independent photon motion, no SR, got it?

If you are a "mainstream" SRist, then a simplified version of the absolute velocity detector can be assembled and tested economically. But SRist don't conduct experiments, or even discuss them in public, if they are designed to violate SR do they? There is a world of difference between voluntary and involuntary absorbtion of information. Did someone break your info-cherry when you weren't looking?

LOL at your joke, politely that is, as I didn't detect any humor. Smirking I saw, humor, no.

You are saying,

If the frame is not moving, photon emitted by E1 hits the absorber A1 in that 'peap' pair at D1 meters from E1 in t secs.

If the frame is moving, photon emitted by E1 hits the absorber Ax in a different 'peap' pair at Dy meters from E1 in t secs.

The photon when emitted is independent of the motion of the origin of the photon. The photon doesn't know there is frame. Do you understand about "independence" ? The photon is moving in an absolute straight-line. The photon defines its trajectory by whatever laws govern photon motion. Therefore, an object that was once located where the lower absorber was located, with respect to the already defined line of motion of the previous photons (calibrated with as close to physical zero velocity as possible - say the south pole with the 2 devices, one oriented parallel and perpendicular to the sun-earth radius), has moved before the next photon arrives.


So you affirm that due to 'absolute' motion of the frame the other absorber Ax 'arrived' at the point where absorber A1 was there earlier to receive the photon emitted by the emitter E1.



So the 'absolute' motion makes photon to change its velocity from D1/t to Dy/t for an observer at absorber Ax ?
No. You are imposing some garbled physics into a real simple situation.....

No ?

Ax receives the photon from E1 in t secs ; so for an observer 'moving' alongwith absorber Ax the photon from its source E1 travels the distance Dy (> the distance D1 between E1 & A1) in same t secs. For him/her the velocity of the photon is not c but > c. This is NOT what is called invariance of photon velocity observed from all the points in all the inertial frames. Your assumption of absolute motion of the frame directly screws up the SR postulate by default. You are invalidating the SR postulate by your mere assumption, not by physical proof.

I repeat :

Thats why i called your 'assumption' a bad joke. First prove, in reality, that the velocity of the photon is not invariant for an observer at Ax ( or any other point in this frame) then freely claim the cause being the absolute motion.

Thats why i called your 'assumption' a bad joke. First prove, in reality, that the [1]velocity of the photon is not invariant for an observer at Ax ( or any other point in this frame) then [b] freely claim the cause being the absolute motion.

[a]The photon's motion is independent of the motion of the source attached to the frame..
[b] the cause of what exactly? I am not being cute what are you talking about?

This is as simple as I can get it. There is a photon, emitted at A-up, in space, heading to an object A-down, where the A-down target trajectory is defined by the predicted straight-line motion of the photon. Before the photon arrives at the object A-down, A-down moves out of the way and object B-down is now in the exact position A-down was in before A-down moved. This is all. The y distances, A-up to A-down, are the same for motion exactly perpendicular to the frame motion.

Look at it like an emitted photon and its emitter and absorber located to receive and send to the other is slowly wrapped into the frame and the peaps are attached to the frame, then while a photon is in motion afrom A-up to A-down, the photon knows nothing of this and knows nothing of the B-down substitution for A-down.

All the conmputer on the frame sees is the identification and time of arrival of the photon, here at B-down. Then the time of emission of the photon from B-down is recorded on the frame computer.

[a]The photon's motion is independent of the motion of the source attached to the frame..Please stop being an idiot. You are wrong. Experiments have been done, and they disagree with you. Your only argument is 'i think it is true'. Either stop making false claims, or at least except the corrections which people make. Don't just keep repeating your mistake.

Statement by Geistkiesel:
“ [a]The photon's motion is independent of the motion of the source attached to the frame.. ”

QUOTE=Persol]Please stop being an idiot. You are wrong. Experiments have been done, and they disagree with you. Your only argument is 'i think it is true'. Either stop making false claims, or at least except the corrections which people make. Don't just keep repeating your mistake.

OK I am starting to get your point, ore than you may know. So I may educate myself more properly could you please provide me some guidance to the published material that shows experiments disagreeing wih the statement that the photon motion is independent of the motion of the source? I had analyzed the problem and determined that the photon was independent of the source.

I am starting to see the wisdom breaking throuigh the curtain. Yeah, your correct I did "think it was true". You know if you hadn't been so persistent and adamant I wouldn't have even considered the possibility that I could be wrong, but I saw that anyone who believes in something as srtongly as you, well tghey must be correct, or at least more to the center of the truth than some like myself who is admittedly shamelfully casual about scientific matters (actually if I can confess, this stuff really bores me), and I haven't the neccessary passion of curiosity, and need to examine things in nature as do you.

You know I just have too much time on my hands and yes, I guess that deep down inside I did wan't to crash special relativity theory, so my ego and could be satisfied by recognition from the rest of the world.

It is kind of silly isn't it to think that a photon wouldn't be affected by motion of the source. The frame is everything, I just came to realize, everything!

http://www-astronomy.mps.ohio-state.edu/~frogel/Ast161/outline161_a00_part12.html

Please stop being an idiot. You are wrong. Experiments have been done, and they disagree with you. Your only argument is 'i think it is true'. Either stop making false claims, or at least except the corrections which people make. Don't just keep repeating your mistake.
This is a continuation of a discussion of an absolute motion measuring device. It is based on the fundamental postulate that the motion of the emitted photon is independent of the motion of the source of the photon. The schematic is shown at three instantaneous times as the photon continues downward, while the frame continues to the right where constant velocity is assumed.

Just in case there is any misunderstanding about where I have been coming from here is a brief sketch of the peap machine. The two horizontal lines show a grid (one dimension of a xy plane grid) where all activity is in the x direction. Here the photon has just been emitted from a peap unit UA10 (upper A # 10)

T1
UA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-
o


DA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-


T2
UA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-


o

DA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-


T3
UA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-



o
DA 1 2 3 4 5 6 7 8 9 10 11 12
-||--||--||--||--||--||--||--||--||--||--||--||-
The schematic shows three consecutive times from when the photon was emitted from the peap UA10, at T1. At T2 the photon is about ½ the distance to the lower part of the parallel planar grid. At T3 the photon is directly above the DA3 peap.

The schematic drawn to show the peap grids (which are attached to the moving reference frame) are moving to the right while the photon, o, continues in a straight line down. There has been no change of direction by frame or photon.

The individual peaps on the upper and lower planes are shown for the x-axis only. This grid plane is coordinated with the y and x-axis on the frame to detect all possible directions of motions. Therefore if there is an upward motion of the frame the time difference between photon travel from a pure x direction only will be indicated by the decrease in time for the photon travel between lev els of the three dimensional two tiered planar grid (3DTTPG).

It is based on the fundamental postulate that the motion of the emitted photon is independent of the motion of the source of the photon.Yes, but this is wrong. Only the speed is independent.

It is based on the fundamental postulate that the motion of the emitted photon is independent of the motion of the source of the photon.Yes, but this is wrong. Only the speed is independent. For the person moving it is moving straight down. For the stationary person, it is moving at an angle.

Yes, but this is wrong. Only the speed is independent. For the person moving it is moving straight down. For the stationary person, it is moving at an angle.
The device is a mechanical assembly. It does not need a person viewing the device. Remeber the photons are aimed directly at an abcorb er/emitter located diorectly across form itself as are all the peaps in the grid system. All the device does is: each absorbed photon is sent to the central c0omputer with it onw identification tage, thereby locating the absorber on the grid system and the time of arrival which is compared to the time of emission to determine velocity wrt the invaiant, spatial invariant, trajectory, and determining up or down motion if the time pf arrival is calculated to be > or < than the time for a motion for transmission across the peap space that is straightline in space, wrt the photon frame, which would indicate no variation off the the plane of the XY plane, which is one of three planes: YZ and ZX.

However, this up/down motion is treated with, or by, the other two grids in the 3DTTPG.

Take a quick look at the schematic a few posts back. The frame shows X direction motion, the photons Y direction motion. Looking at the photon continuing in a straight line can be seen by the alignment of the photon in all three time periods.

Your schematic is only correct in one frame of reference. In others it is incorrect.

Your schematic is only correct in one frame of reference. In others it is incorrect.
It is only meant to be correct in the frame if reference in which it is attached. The three different time periods described in the schematic are all in the same frame.

I don't understand the idea of "absolute velocity" because velocity only makes sense relative to something. IMO, arguing about absolute velocity is a basically arguing about "my reference point is better than your reference point". Isn't it?

Ugh... which means once again you are saying nothing at all about absolute velocity.

But regardless, it isn't correct in the frame it is attached. In the frame it is attached the light will hit the opposite wall in the same place as if the whole thing was 'at rest'.

If you agree, then I'm not sure what the whole point of your thought experiment was.

I don't understand the idea of "absolute velocity" because velocity only makes sense relative to something. IMO, arguing about absolute velocity is a basically arguing about "my reference point is better than your reference point". Isn't it?
Here is what I mean by "absolute velocity" in the context that I have used in this thread: The photon motion is indeoendent of the motion of the source of the photon. I take this to mean what it says :on the "four corners of the paper". A photon once emitted adopts a trajectoy that is an invariant straight-line motion. I see that a single photon trajectory may be conceptually incomplete without a "reference' to some other object.

OK, let us take two photons emitted by two independent sources in the near neighborhood of each other. The closest point of the two lines, once the photons are emitted, is invariant. The distance between he trajectoies, even zero distance, is invariant. Now take themitters and mhave them scatter photon trajectories all over the place. All relative positions of the photons trajectories is invariant.

Now take youself a huge distance away from our super-galactic conglomeration pf reality with a fleet of space ships covering a huge flat plane. The ships (or more economically, "probes") emit light in the direction of the mass of universe that we have observed. The trajectories do not have to be oriented in some perfect rectangular array, but once the photons are released and begin their more or less simultaneous motion back to the region just left by the probes, the photons, say a million of them (hey a million space probes is nothing if it woud benefit the advancement of science) will all be invariant in their relative distance and orientattion from each other. Assume a perfect emission of a million parallel beams of light. All the mass activity in the observed universe, galaxies, stars etc., will all be observed moving through this grid of photon beams that are invariant in distance with respect to each other.

This is what I mean by use of the phrase, "absolute motion". This is my "definition".

Here is what I mean by "absolute velocity" in the context that I have used in this thread: The photon motion is indeoendent of the motion of the source of the photon. I take this to mean what it says :on the "four corners of the paper". A photon once emitted adopts a trajectoy that is an invariant straight-line motion. I see that a single photon trajectory may be conceptually incomplete without a "reference' to some other object.

OK, let us take two photons emitted by two independent sources in the near neighborhood of each other. The closest point of the two lines, once the photons are emitted, is invariant. The distance between he trajectoies, even zero distance, is invariant. Now take themitters and mhave them scatter photon trajectories all over the place. All relative positions of the photons trajectories is invariant.

Now take youself a huge distance away from our super-galactic conglomeration pf reality with a fleet of space ships covering a huge flat plane. The ships (or more economically, "probes") emit light in the direction of the mass of universe that we have observed. The trajectories do not have to be oriented in some perfect rectangular array, but once the photons are released and begin their more or less simultaneous motion back to the region just left by the probes, the photons, say a million of them (hey a million space probes is nothing if it woud benefit the advancement of science) will all be invariant in their relative distance and orientattion from each other. Assume a perfect emission of a million parallel beams of light. All the mass activity in the observed universe, galaxies, stars etc., will all be observed moving through this grid of photon beams that are invariant in distance with respect to each other.

This is what I mean by use of the phrase, "absolute motion". This is my "definition".

A damn long one. :D

A damn long one. :D
Did you undertsand the long defintion Pete?

Did you undertsand the long defintion Pete?

I don't know if Pete understood, but I don't. Do you understand your own long definition for "absolute velocity"? If you could shorten the definition and simplify it so that even a kid would understand, then I know you understand.

He basically defined a frame of reference... with the problem being that he is going to get an infinite number of different results depending on movement/location of the emiiter/observer.

Back to square one.

I don't know if Pete understood, but I don't. Do you understand your own long definition for "absolute velocity"? If you could shorten the definition and simplify it so that even a kid would understand, then I know you understand.
Light motion is invariant with respect to the source of the light. A photon trajectory is at absolute rest, ain a state of absolute zero velocity in free space. Therefore, any measurement of motion with respect to the invaiant trajectory iis a measure of absolute velocity measured from a frame of reference that is at rest, absolutely with respect to to all objects in the universe.

Light motion is invariant with respect to the source of the light. No... light SPEED is invariant. Motion is not. The trajectory is also not invariant.

We've told you this at least 10 times.

No... light SPEED is invariant. Motion is not. The trajectory is also not invariant.

We've told you this at least 10 times.
You've "told" me nothing about this nor have you offered anything but statements that are not verified by any kind of reference. Are you suggesting that evrything you "tell" me i should accept as the "truth" without?

I noticed again that you have not found it within your power to answer the post directly, even assuming you are correct about "dependent motion" of the photons wrt the source. Frame and photons are moving parallel wrt each other, the photons are emitted parallel and anti-parallel to the moving source of the light and there are no properties or conditions of the experiment that would deviate the photons from the straight line trajectory motion.

If motion is not "independant" of the motion of the source, then must we add the source velocity as a speed vector, momentum of the photon, to the straight-line motion of the light, given a result C + C' > C?

Where exactly do you distinguish motion and speed in the context of your statement above?

Who is "we've"? Are you working out of some kind of coordinated committee?
Where did you tell me this "at least 10 times"?

The trajectory is not invariant is going to be very difficult to prove. Ar you suggesting that the trajectory of light is circular, or rectangular, or what?


Geistkiesel?

I don't know if Pete understood, but I don't. Do you understand your own long definition for "absolute velocity"? If you could shorten the definition and simplify it so that even a kid would understand, then I know you understand.
Excuse the reference to "Pete" Paul T.

Do you find any difficulty with the statement that "the motion of light is independent of the motion of the soure of light?" I understand this to mean that there has been no anisotropic motion of light that can be ascribed to any motion of the source of light. Is this your understanding also?

I also incorporate the statement that the trajectory of a light photon is invariant until acted upon by an outside force. Is this your understanding?

You can use this as the definition you asked of me.

geistkiesel

You've "told" me nothing about this nor have you offered anything but statements that are not verified by any kind of reference.Try doing some of your own research... I'm tired of finding links for kooks. I'm just going to make fun of you. Are you suggesting that evrything you "tell" me i should accept as the "truth" without? Yes. Yes I am.
Frame and photons are moving parallel wrt each other, the photons are emitted parallel and anti-parallel to the moving source of the light and there are no properties or conditions of the experiment that would deviate the photons from the straight line trajectory motion. [deleted]... I did answer the question. Your statement about 'no conditions that would deviate the trajectory' is simply wrong.
If motion is not "independant" of the motion of the source, then must we add the source velocity as a speed vectorMY GOD! Didn't I just say speed was constant?
Where exactly do you distinguish motion and speed in the context of your statement above?Motion is speed + direction... aka velocity. Speed is... speed and is always constant.
Who is "we've"? Are you working out of some kind of coordinated committee?Why yes, yes I am.
The trajectory is not invariant is going to be very difficult to prove. Ar you suggesting that the trajectory of light is circular, or rectangular, or what?I'm telling you that the direction depends on your frame of reference.

If motion is not "independant" of the motion of the source, then must we add the source velocity as a speed vector, momentum of the photon, to the straight-line motion of the light, given a result C + C' > C?
Motion of the source does not add/subract to velocity of light. Only red/blue shift in the photon frequency is observed if the source is moving. It is a proven fact. Better please rely on observed facts instead of your mental arithmatic.

MY GOD! Didn't I just say speed was constant?
Motion is speed + direction... aka velocity. Speed is... speed and is always constant.
Why yes, yes I am.
I'm telling you that the direction depends on your frame of reference.

I suggest you look again at the statement that the motion of light is independent of the motion of the source of the light. For you to say that the speed is always constant, with reference to the reference frame, places a constraint on the independent motion of light with respect to the source of light.

Likewise, as soon as you say the direcdtion depends on your frame of reference, means you have also constrained the direction of motion of the independent motion of light. Some freedom you place on inependent moving photons. Thankful we all must be you aren't a politician.

What deviates the light?

If SR is going to use postulates of light, SR should be thoroughly and strictly consistent.

Try doing some of your own research... I'm tired of finding links for kooks. I'm just going to make fun of you.Yes. Yes I am.
[deleted]... I did answer the question. Your statement about 'no conditions that would deviate the trajectory' is simply wrong.

Here is the full statement regarding deviation:

"Frame and photons are moving parallel wrt each other, the photons are emitted parallel and anti-parallel to the moving source of the light and there are no properties or conditions of the experiment that would deviate the photons from the straight line trajectory motion."

The mirrors are flat (parallel) with respect to each other.
The experiment With the frame at rest the frame in motion differ only by the small amount of additional time attributed to the the moving condition, t' = t(2v)/(c - V), but this is not a deviation of the photon motion, trajectory, speed.

The speed of light is invariant. The trajectory of light is invariant, otherwise the simultaneous arrival of the photon back at the resituated midpoint of the reflectors would not have occured.

If, as you say, there is something that can deviate the light, it didn't show up in the experiment described to you in my post did it? Is this external force periodoc, random, irregular and of so whjat is the nature of these, or this external force, that you say will, "deviate the trajectory"?

Hamlet says to "poor Yorick" [Y's skull] at the end of H's soliloquy in the castle's graveyard at Elsinore:

"Not one to mock your grinning? Quite Chapfallen?
Off with you now to my ladies chamber,
tell her, let her paint an thick,
to this favor she must come.

Make her laugh at that!".

Motion of the source does not add/subract to velocity of light. Only red/blue shift in the photon frequency is observed if the source is moving. It is a proven fact. Better please rely on observed facts instead of your mental arithmatic.
Tis poor form to merely cite dogma unread dear sir. I assume you are referring to the blue shift, for one, observed in the light bearing down on an observer in the earth frame, for instance?

Tell be how this can occur, the "compression" and therefore, the bluing of the emitted light, when the source acts as a constant compressor to the wave-length of the emitted light?

Lets say the source velocity of a stellar object moves at 500 km/sec. Light moves at 3x10^5 km/sec. protortionately this amounts to 1.6 x 10^-3. The compressor is moving 1/1000 of the velocity of the photon and "compresses " the photon wave length? Or the extender is moving away from the emitted photon and is stretching the wavelength?

Try adding the observer's motion in the experiment, where the frequency is measured by counting the number of full, peak-to-peak weave-lengths that pass through the plane of the observers eye, with the wavelength left untouched. SR tells you not how to measure wave-lenth , SR gives a formula that effectively subtracts the observers motion from the experiment: "Better data through axiom, and cheaper too!"

requency and w Doppler shift being attriburable to the compression and extension of the photon wave-lenth is a myth.

If there was a comression against the wave-length, that which is observed as the physical manifestation of the existence of the photon would necessarily give a momentum pulse, a specific impulse, yo momma, a velocity addition to the photon. And extensions would stretch the string of wave-lengths and produce an _______ in velocity?
Fill in the blank and win a prize..

No, this is not what happen.

Likewise, as soon as you say the direcdtion depends on your frame of reference, means you have also constrained the direction of motion of the independent motion of light.No, the direciton is not constrained and changes deppending on your frame of reference. I'm done trying to explain stuff with you, and done calling you a dumbass.

Just look up the experiment yourself.

Try doing some of your own research... I'm tired of finding links for kooks. I'm just going to make fun of you.Yes. Yes I am.
[deleted]... I did answer the question. Your statement about 'no conditions that would deviate the trajectory' is simply wrong.

In the example we were discussing the light was travelling in perfect straightline trajectories that were reflected in straightline trajectories. So if you claim it is "simply wrong" please provide the reference that would explain the error you claim is in the statement.



MY GOD! Didn't I just say speed was constant?

Quoted by geistkiesel:
" If motion is not "independant" of the motion of the source, then must we add the source velocity as a speed vector?"
You were answering this question when you stated the above. I guess you didn't realize you were speaking to a piece of paper.


Motion is speed + direction... aka velocity. Speed is... speed and is always constant.

I'm telling you that the direction depends on your frame of reference.

Your statement is amabiguous and does not attach to the definition of the independence of the speed of light wrt to the source of light. If some one be in a moving frame, or at rest, is looking at a beam of light what does a frame of reference have to do with the direction of the light beam? The beam doen's "know" it is being watched. The beam of light continues in a straight line until acted upon by an outside force, at least this is how it woprks where I come from.

If you are in a space ship that can zoom here and there, zip, left, then zip right, change direction with aplomb, stop on a dime, pick it up and put it in your pocket, accelerate, decelerate and do double reverse immellmans
with the ease of a sleek killer hawk and you spotted a photon beam in space with the source far beyond the capability of your measuring devices to locate, then what must be the conditions of your frame that would make the direction of the beam "depend on your frame of reference.", without applying a force to the beam naturally?

If some one be in a moving frame, or at rest, is looking at a beam of light what does a frame of reference have to do with the direction of the light beam?The direction is different depending on the frame of reference.

http://landau1.phys.virginia.edu/classes/109/lectures/srelwhat.html

Geistkiesel,

I think you know I would like to be on your side here but I can't. You seem to be saying that light projected orthogonal to a moving frame doesn't mimic the forward momentum of the frame. It does.

Think for a moment about placing a flashlight horizontal on a lab bench and placing a target on a geosynchronous satellite (23,000 miles above earth) = 37,024,390 = 3.7E⁷ meters or apprx 1/8 of a light second distance.

Now only do your measurement when the satellite is in full eclipse of the earth from the sun and disregard the earth's rotation for a brief test.

The earth moves around the sun in its orbit at apprx 1E^-4c or 0.0001c. If you now project a 0.125 second long beam of light it will be in contact with both your target and the source at each end.

The target is not missed even though the target moved 2.2125 miles forward during the time between when you started the projection and before the projection hit the target.

Why? Because the light beam has forward momentum associated with the motion of the source from which it was launched. It is like throwing a golf ball out a car window (ignoring air resistance) the ball moves away from the car but stays along side the car. It doesn't fall behind.

The more interesting issue you raise is why does it display this forward momentum unless it possess some mass. This view of a photon having some mass can also be equated to what happens if you do consider the earth's rotation. It does miss the target!.

That is the photon moves out radially in a straight line just like a golf ball on
the end of a string that when spun over head until the string breaks.

In this case the photon misses the target since during the 0.125 seconds the target and source have rotated 1.85 seconds of a degree and at that distance will have moved 113.5 meters around the orbit circumference from the line of sight and projection of the beam at its start.

Of course they claim photons have no rest mass since that would cause Relativity to fail and photons do not exist at rest.

Geistkiesel,

I think you know I would like to be on your side here but I can't. You seem to be saying that light projected orthogonal to a moving frame doesn't mimic the forward momentum of the frame. It does.

Think for a moment about placing a flashlight horizontal on a lab bench and placing a target on a geosynchronous satellite (23,000 miles above earth) = 37,024,390 = 3.7E⁷ meters or apprx 1/8 of a light second distance.

Now only do your measurement when the satellite is in full eclipse of the earth from the sun and disregard the earth's rotation for a brief test.

The earth moves around the sun in its orbit at apprx 1E^-4c or 0.0001c. If you now project a 0.125 second long beam of light it will be in contact with both your target and the source at each end.

The target is not missed even though the target moved 2.2125 miles forward during the time between when you started the projection and before the projection hit the target.

Why? Because the light beam has forward momentum associated with the motion of the source from which it was launched. It is like throwing a golf ball out a car window (ignoring air resistance) the ball moves away from the car but stays along side the car. It doesn't fall behind.

I'm a little fuzzy about the geometry of the synchronous orbit setup , but I see your golf ball arrangement. I was, or have been slavishly following the postulate of light declaring that the photon motion is independent of the motion of the source, which makes sense, otherwise wioukd niot we be forced to add the component of motion that has assumed the momentum of the source. I realizde it seems instinctive to view the situation as your golf ball analogy describes.

Finally, is their literature on the geosynch orbit targeting process.

Let's jack up the speed a bit and have the light focussed perpendicular to the moving frame at .9999c. Wouldn't the light now have an actual velocity of the sqrt(2)c? I recognize that all the discussions, literally, regarding the speed of light are confined to the straightline motion of the photon. Not that I'm particlularly stunned by your assertion that the potientially extra source of "possible" speed is as commonplace as you stated. I use 'possible' in the context that the motion you say is momentum induced from motion of the source may be of a different nature than regular old fashioned dx/dt motion. The slavish obedience to the independence postulate then must be discarded by all, should it not?

For short distance on the surface of the planet I see it would be diofficult to test as the source is always physically locked on to the target and th relative nearness of source and target would not easily pick up the momentum component, or eliminate it.

I have been claiming in a number of posts that the Micheleson-Morley experiment was flawed by the experimenters implicitly assuming what you stated in your post. I don't know what M &M had to say about it ,but others have paid virtually no attention to the question which is usually handled by a "the mirror drags the photon along with the interferometer" arm wave, while I have been suggesting that the experimentors biased their experiment by focusing the transverse mirror in such a way that an interference pattern was produced.

I'll have to sit on this one for while.


The more interesting issue you raise is why does it display this forward momentum unless it possess some mass. This view of a photon having some mass can also be equated to what happens if you do consider the earth's rotation. It does miss the target!.

That is the photon moves out radially in a straight line just like a golf ball on
the end of a string that when spun over head until the string breaks.

In this case the photon misses the target since during the 0.125 seconds the target and source have rotated 1.85 seconds of a degree and at that distance will have moved 113.5 meters around the orbit circumference from the line of sight and projection of the beam at its start.

Let us look at this for a second or two. Dayton Miller (who doggedly replicated the MM experimetns for 30 years) made some intersting claims regarding the measurement of the "absolute" motion of the earth and he sun (therefore the solar system) where the sun is moving at a whizz bang 208 km/sec , basically due "south". Now if one adds the thre velocity vectors of rotation, orbit motion and linear motion of the sun (whose radius or orbit is a huge distance away) whay does this say about the instantaneous direction of the nmotion of the planet earth? The rotational velocity gets swamped by the higher velocities and would this not have an additive affect on the rotational velocity vector being a mere sub-vector (.49 km/sec) to the overwhelming higher speed of the sun (two orders of magnitude)?



Of course they claim photons have no rest mass since that would cause Relativity to fail and photons do not exist at rest.
Relativity has already failed, its the srists who still keep mumbling from the grave.

When an accelerated electron oscillating around the tungsten fillament in my desk lamp emits a photon I don't reckon the electron gets a photon mass out of his bucket and sends it on its way. However it is done, producing the photon, I mean, something, a thing, an object, that effectivley moves from here to there, leaves the filament. Assuming that the electron we associate with as the source of the photon, the final product is much more than a glob of randomly oscillating energy, as the photon has all the complexities as any other subatomic particle. It can be polarized, depolarized, defracted etc. Some authors describe the ambiguous nature of "vertical" and "horizontal" polarization modes by avoiding the tough questions by declaring that there are two kinds of photons from the get go and that Mother Nature provided us with an equal amount of each, which seems to me a ridculous model. It certainly isn't what I taught her! Feynman published the "two types of light" scenario. I've not seen any contradictions to Richard F's view, except my own.

Being of a committed "prove it to me" jerk, I am forced to mull over your model. I don't surrender easily (mostly there is a lot of kicking and scratching), unless I've proved it like when I catch myself avoiding looking at myself in the mirror. I am not stressed when taking losses early, and sneering as I do at "reputation" as a quotable referenced source of scientific value, or worth (a practice I abandioned long ago) it allows me more focus on the problem of current interest. Blurting out once in a while isn't all that painful a process, unless one is a committed fanatic, no names mentioned, but this forum has their fair share. Building houses of cards may be fun to idle away time with but I don't reccomend the practice be taken to the work place.

There is really no need for a disclaimer about liking to be on my side as I know it is not in your nature to hold back any scientific opinion or belief for any reason. That wouldn't be MacM, would it? But I must say that I sincerely appreciate the statement for what it means. Thank you.

gGeistkiesel

Geistkiesel ,

You are welcome. Let me emphasize, that I am not explicitedly claiming you have been proven false on this issue but that your claims were being based on a concept of independance which in my opinion ran contrary to the generally accepted view.

The only independance I have ever seen stipulted is in the direction of propagation. That independance seems at conflict with the notion that you promote and the issue of light in a rotating system implies regarding mass.

So I have been trying to think of some natural or man made test which actually determined this issue. At the moment I cannot think of one.

Hope that clarifies my position.

.

Geistkiesel ,

You are welcome. Let me emphasize, that I am not explicitedly claiming you have been proven false on this issue but that your claims were being based on a concept of independance which in my opinion ran contrary to the generally accepted view.

The only independance I have ever seen stipulted is in the direction of propagation. That independance seems at conflict with the notion that you promote and the issue of light in a rotating system implies regarding mass.

So I have been trying to think of some natural or man made test which actually determined this issue. At the moment I cannot think of one.

Hope that clarifies my position.

I feel a strange sense of having been reprieved.

Would you please take a look here at an outline of another SOL question, which is parallel, overlapping parallel, to the one we have been discussiing. Doppler tells us that incoming photons and sources produce compressed wavelengths and receding sources and in coming photons are extended into the red, and thatDoppler is a relativity event. We may assume that the compression requires a force on the light as the wavelength cannot compress without a reaction, nor will it stretch with out tension. This being the logic, Doppler dynamics should either accelerate light Vc > C or Vc < O, neither of which is physically possible, says I, say SR and the world I would imagine.

I wrote ths out in more detail here which is only a few paragraphs. This is even off the wall for me, but if the SR industry wants to play with the logical consistency of their theory they remind us of fairly regularly, then they should be obliged.

I am being rushed by some company here so I must close abruptly. This is one I have been timidily avoiding laying on the table because it is off the wall even for me, and I don't believe a word of it, but what the heck, you live, I don't know how many times, but when you're doing it, just do it.

Geistkiesel

I feel a strange sense of having been reprieved.

Let me clarify. I am only saying that I don't see that you have been "Proven" wrong. That doesn't mean you are right, nor does it mean that my opinion carries any weight. :D

Let me clarify. I am only saying that I don't see that you have been "Proven" wrong. That doesn't mean you are right, nor does it mean that my opinion carries any weight. :D
But your opinion does carry weight, otherwise there wouldn't such a concerted and even organized effort by your opponents to break your legs as you tread the threads.

Try here to add a little weight to your frame.