A point of departure for special relativity theorists and the rest of us.

A point of departure for special relativity theorists, and the rest of us.

1. Observers in space ships who consider themselves at rest, after reading Einsteins’s “Relativity”, of course, like to conduct tests that demonstrate being able to consider his inertial frame being at rest. He will place reflectors at opposite ends of his ship and release two photons simultaneously from the midpoint of the two reflectors. Of course if he is at rest as he was when the Ve – Vn = 0 ,while rehearsing later tests in free space.
2. On Ve (the embankment) the photons reflect back at the origin simultaneously and are proven to have arrived at their respective reflectors simultaneously, also. The moving observer will tell us the photons will also arrive back at the origin simultaneously, which he then demonstrates, after telling us, “I am at rest also”.
   
3. Lets see what is happening. The left moving reflector meets the oncoming photon before the right moving photon arrives at the right moving reflector. At the instant the left moving photon arrives at the left reflector the right photon is looking across a space gap as seen here:
   
   
   Code:

   -----Ct------>| vt  | vt | 

4. The photon has just moved a distance ct, while the frame and the right reflector is located at the right most vertical line. In order to catch the reflector the frame move a distance “|||” which the photon must catch after passing through pass through 2vt as shown, during which time the frame, and ergo reflector, has moved a distance vt2 or “|||”.
   
   
   Code:

           ----------->after passing through the 2vt + v t2  gap.
          |vt  | vt |||
          |vt  | vt ||| 
   (now)|<----------- 

5. after moving a full ct distance, the same as the left photon which has traveled in the same distance in the same time frame. The difference, of course, is that the left and right photons strike the reflectors in different times, in all frames, which synchronized clocks at the left and right reflectors will attest to after review of the clock data.
   
6. This is, in most instances, a small but very real distance. For you see that little bit of Dx defines a space coming up. The t2 defines how wide. The photons both spend the same amount of time, cover the same amount of space during the total process, yet that little bit of Dx·Dt is the single piece of data in this whole little caper that guarantees the frame is in motion.
   
7. The significance of the Dx·Dt is seen best in the closing of the loop, so to speak. In order that the photon and frame arrive with simultaneity, the frame must adjust to the natural motion of the photons and therefore the frame adds that last little Dx·Dt to the “2vt” position of the current position of the midpoint just as the photons close in on M and where the disjointed right photon moving to the right is just barely, but perfectly so, catching up and adding the final balance to the books, so to speak the numbers refere tot he three steps in which this process was conducrted (the numbvers refers to natural sections of the process for instructional purposes. 0 is when the pulses were enmitted, 3 is when the meet again at the tanslated midpoint..
8. Code:

                 0      1     2                 2
            --  >|      |     ||| |      |      |<--
                                3
                       |------->|<--------| arriving at midpoint as moved.

9. both photons move equal distances, as always in these final closing periods of time. If we use this schematic as a prediction of the position of the arrival of the frame midpoint as here the photons, not the photon sources, maintain a running track of the photons’ midpoint.
   
10. Here the is a simple predefined midpoint of colliding photons whatever the frame velocity, yet one doesn’t need time dilation and frame contraction to explain this, or anything else. Some real fancy photon/frame coordination, isn’t?
    
    
11. But for that little space ||| we would be back at the stationary embankment, AKA Ve. A frame that reflects photons, and no ||| is produced as a result, is guaranteed to be at rest the existence of a ‘|||’ space on our mother nature’s process is a slam-dunk guarantee of a moving frame. The observer-analyst-special–relativity-theorist who subtracts out real motion poops on the laws of physics. You don’t believe me? Check it out. [Say, have you read grounded’s post lately? His original post I mean?] t2 is shown as,
    
    c t2. = vt + vt + vt2.
    
12. Meaning, of course, that t2. = t1 (2vt)/(C-v) where all parameters are measured in the moving frame.
    
13. Can you dig it? The Dx·Dt = 0 means a resting frame i.e. in terms of relative motion Ve – Vn = 0, or Ve = Vn = 0, while a nonzero product proves motion! Wow! I wonder what frame the observer was pretending he was at rest with respect to?
14. Remember the final position of the midpoint is measured with respect to the absolute and totally unique x, y, z, t in all the universe, the whole Oz Thing by, guess what? Anyone, right, the photons. Of course these same photons defined the point of simultaneous arrival of ‘frame’ and photon, where ‘frame’ is taken to mean the midpoint of the position of the reflectors with respect to ß||à, long before ‘the event’ of simultaneous arrival at reflector or midpoint, [again], if you get my drift. Another hint? OK, OK, -->|||<--.
    
15. All of this coordinated frame and photon stuff doesn’t boggle the mind until some real velocity is ignored in order that it becomes necessary to provide a mathematical correction for the original error of ignoring the observer’s motion in SR equations.
    
16. How to prove this is to have synchronized computers at the Left and Right reflectors that measure the time difference, on the moving frame, of when the photons arrived at the reflectors.
    
17. Assuming L is the distance from midpoint to the reflectors, then the right photon moves ct such that ct = L – vt so the left moving photon arrives at the left reflector at time t. The right moving photon arrives at the right reflector at t + t2, where t2 is the time required for the photon to pass through the |vt | vt ||| distance and arrive at the right reflector.
    
18. Timing this reflection in the stationary embankment frame, Ve, the photons arrive at the reflectors at the same instant. In the moving frame, the right moving photon arrives late at the right reflector.
    
19. The moving observer knows the photons were emitted simultaneously, he emitted the photons, and that L is the same on both sides of the midpoint, and that the only rational conclusion is that either, (1) the clocks aren’t synchronized, in which case the experiment can be repeated until JamesR completes installation of his universal clock sycnchronization device, or (2) that the observer’s frame is actually moving, contrary to his “perceptions” justified by some theory, or another.
    
20. Genesis 11:4-5 “… you positively will not die…your eyes are bound to be opened and you are bound to be like God . . .”

Geistkiesel

 

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point of departure for Special relativity theorists and the rest of us.

:m:

Too verbal for you James R? I can understand. It was a tad messy for me when i looked at it again.

This is about an experiment that you are completely familiar with.

1. A photon pair is emitted from the midpoint of two photon emitter/absorber pairs (peaps) located at both ends of the Vn inertial frame of reference at L and R, which to Ve is moving to the right with velocity v.
2. The location of the midpoint with respect to Ve, the embankment, is defined by the location of M as measured by the time and place the photons were emitted at M.
3. The Ve distance of L M = R M is ct’ in the Ve frame and as the frame is moving the L photon meets the L peap a distance vt shy of the Ve reference frame distance.

I am trying to be careful here and I will claim that the times I use in this analysis are taken from the clocks on either of the frames without ambiguity.

1. As Vn is moving v to the right as the L photon moves to the L peap, the R photon has moved ct to the right, yet still 2vt shy of the R peap. This is the positions of the photons at the instant the L photon reaches the L peap.
2. After the L photon has been emitted and traveled the same distance it traveled from the original midpoint to the L peap, the L photon has returned to the point the photons were originally emitted. The L photon has traveled a total distance 2ct at this instant, and likewise the R photon has traveled 2t at this same instant.
3. The R photon, however, had to travel an extra vt2 after covering the distance of 2vt when heading for the R peap, plus a distance = vt2, the distance the frame moved when the R photon was covering he distance 2vt
   
4. Therefore at this time the computer data bases in L and R peap differ by an amount = t2 , the time required for the R photon to catch the R peap from 2vt behind.
5. Now the two photons are located exactly ½ the distance each from the final position of M at the instant the photons arrive simultaneously at M.

The computer at M counts the time of flight from the instant the photons were emitted and from the Vn clock at M,

1. or is 2t + t2 and
2. the same times are recorded on the L and R peaps, which
3. is timed to shut off after counting 2t + t.
4. Therefore the time on the peaps computes is 2t + t2. + t,
5. where the final t is subtracted, leaving the total time for time of flight = 2t + t2.

In looking at the data from the L and R peaps, the time analyzer at M notices that the difference in arrival time

1. of the L and R photons at the L and R peaps.
2. Therefore, the experiment is repeated 1000 times more with the same result, or rhat
3. the R peap measured the arrival of the R photon t2 after the L peap measured the arrival of the L peap.

There are two possibilities concludes the analyst of the data,

1. (1) the R peap was out of synchronization in all repeated tests even though the clocks were synchronized before starting any of the experiments, or
2. (2) the frame is moving to the right and t2. marks the time the R photon took to catch up to the R peap after moving an extra vt2..
3. The (2) option is the only reasonable option

.

The total traveled distance is

1. Ct2. = 2vt + vt2 and is the distance the R photon moves to catch the R peap and is also the time the distance the L photon moved after having moved a distance 2ct .
2. After a bit of algebra we see that t2. = t(2v)/(c – v).
3. Now if t2 were zero, then v must be zero as t and C are definitely not zero.

Therefore, and this is the point of departure,

1. that the t2 > 0 proves the motion of the frame even though any Vn observer may have originally “considered” himself at rest with respect to Ve.
2. As the stationary times for the test are derived from 2ct' = 2ct + 2vt or ct' = ct + vt and t' = t(C + v)/C which means
3. t' = t if v = 0.
4. where the only difference is the t2 time, t2 proves motion to the right.
5. Time dilation and frame contraction does not explain away the t2. Anyone wanting to interject SR times at this point, which is really before SR, must be persuasive in locating any errors of physics or logic in the analysis above
6. In other words, t2. cannot be explained away by time dilation as,
7. the L and R peap are symmetrical with respect to each other and to the midpoint at M and equal time dilation and frame contraction does not zero out t2.

Those attempting to use an “equivalence” of inertial; frames argument must first overcome

1. the physical reality that the mass inertia of Ve (6 x 10^27 grams) is infinitely larger than that of Vn, that
2. the Vn necessarily accelerated with respect to Ve in order that any relative velocity of Vn - Ve > 0 be observed, that
3. Ve is measurably indistinguishable from straight line uniform motion, and
4. Ve is never observed acceleratiing and providing the velociy for an observed |Vn - Vc| > 0
5. that the only way Vn can equal zero, the rest frame velocity of Ve, is to decelerate from its current Vn > 0 to Vn = 0.

geistkiesel
11-18-2004

 

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• A shorter version and the whole point of departure​
  
  
  the relative motion of frame and photon​
  ​
  .
• A some point in the reflection process the photon moving in the direction of V, the moving frame, is located a distance 2vt from the reflector (or midpoint) of the frame that it is heading oward. Here the distance 2ct is 2x the distance from the source to the reflector that is moving in the opposite direction of the frame, or one round trip for the Left moving photon (that meets the L reflector moving to the right). Here is the arrangement just before the emission of the photons:
• Code:

  [COLOR=Red]|L_________________M__________________R_|[/COLOR]

• the L and R reflectors and the midpoin M the source of the emitted photons at t =0. The frame moves right and therfore the L photon will meet the L refletor before the R photon meets the R reflector.
  
• The R photon has to cross 2vt to get to the reflector, plus a bit more that the frame has moved while the photon was crossing 2vt. Or, said another way for t2 is the time the photon had to cross 2vt and little more, arrive at the reflector, or Ct2 = 2vt + vt2. After some manipulation,
  
  t2. = t(2v)/(C – v)
  
• Here is the picture of the photons looking across the 2vt gap:
• Code:

  [COLOR=Red]|----->|    vt      |    vt      |||[/COLOR]

• This is the same condiion the L photon sees just after reflecting back to the original midpoint and is looking at the midpoint instead of a reflector, as in below.
• Code:

  [COLOR=Red]|vt|<----ct------|----------->  |  ||   
                  0 1  2
      |<----------|-|-|--------->  | ||   
      |---------->      | | <---------|[/COLOR]

• The two vertiical bars are the distance the frame moves during the photon journey across the 2vt gap, or t2. The R photon sees this after the L photon has reflected back to the original origin while the frame has moved 2vt and now the L photon is looking at the same 2vt gap. The R photon is located such the the midpoint of both photons is where the frame will arrive simultaneously with the photons.
  
• The R photon is moving (C - (-v) )t2 = t2(C+v)t2, where v is colliding with C, while the L photon moves t2(C + (-v)) = t2(C -v) or:
  t2(C - v), both frame and photon moving right.
  and
  t2(C + v) " " " " opposite to ech other.
• Or, said another way, if t2= 0 there is no frame motion,
• otherwise for t2 > 0 the frame is moving, not with respect to Ve necessarily (Ve the embankment, planet earth). The time effect is physically real, a measure of motion in the stationary and moving frames and has nothing to do with an obsevers's perception of a frame's motion;
• rather the focus is with respect to the invariant coordinate system defined by the invariant location of the emission point of the photons.
• The time difference between the measured moving and stationary frame is due to the extra bit of motion of the frame that is required to process the motion of the reflecting photons. This effect has nothing to do with time dilation and frame contraction.
• it has everything to do with the relative motion of frame and photon.

 

If light actually moved in this Newtonian way, there would be no relativity theory. Neither would there have been a negative result in the Michelson-Morley experiment. Say you're travelling in a spaceship at 0.9c relative to the universe at large. By Newton's theory you would expect that light you fired off in the direction of travel would move at 29,979 km/s. But it doesn't - it's still travelling at 299,792km/s - and it's speed can be measured at that rate in the direction of travel, perpendicular to the direction of travel and against the direction of travel.

 

Silas I agree with your first statement, but am leery about the 2nd as there are some unanswered questions in the experiments. We’ve all considered the reality that theories are not eternal, as history has taught us. If, as you say, that the truth of the analysis means the falsity of SR, well, maybe now is the time to do just that?

However, let me suggest that I did not use the addition of velocities as you stated. Quite the converse as I assumed the independence of frame and photon, and was using that independence to advantage by measuring the time it took for the photon to move from point to point. Which is another flaw in SR that does break the indenpendence rule by linking the motion of light with the inertial frame. The relative moption of frame and photon is always measured at C, right? This hardly independence. The photon doesn't know of the frames presence.

You measure a light beam with respect to Ve the embankment and measure C = 3 x10^8 m/s. You then measure my space ship going by at .9c. When aIget a ways away I emit a pulse of light and measure the time it takes for the light to reach the end of my 4000 meter light speed measuring range. The light takes 4000/3x10^8 = 1.33x10^-5 sec. In the meantime the frame has moved ahead 2.7x10^8 x 1.33 x 10^-5 = 3591 meters. This means that a pulse of light will move Ct2 meters during the time the end of the frame has moved ahead vt2. Ct2 = Ct1 + v t2, or t2. = Ct1/(c-v). t2. = 3x10^8 x 1.33 x10^-5/(.3x10^8) = 1.33 x 10 ^-4 sec.
Therefore, Ct2 = 3x10^8 x 1.33 x10-4 = 39900 meters.

Question, other than your statement that SR is false if the anaklysis is true, can you find any fault in the process as described?

I to this day cannot conceive of any rational basis for omitting the velocity of the inertial frame when making a relative velocity measurement. The way some SR theorist reacts when suggesting that a relative velocity between frame and photon one would think they equate the concept to a dirty word.

Just from the condition of inertia alone is the embankment, the planet earth with mass = 6 x 10^ 27 grams, equivalent to conceivable inertial frame Vn, n = 1, 2 . . . …infinity? I think not. Think about the reality that the Ve planet earth is measurably indistinguishable from straight-line motion. The so-called “turning rate” is on the order of 10^-8 degrees/sec.

More later.
geistkiesel

 

Silas,

Maybe the speed of the light is linked to the gravitational field of the spaceship. So even though it may be emmited at a speed that's higher or lower than c relative to the spaceship, the spaceship's gravitational field quickly accelerates, or decelerates, the light to 299,792 km/s.

 

Silas was too brief. He should have said if you measure the relative motion of the photon, Vc, and the space ship, Vn, realtive to Ve = 0, the embankment, the photon/frame relative motion would be measured as 29,979 as he actually said. Measure Vc and Vn the light and ship velocity with respect to Ve, the embankment and the SR goes away.

Prosoothus,
Silas got it wrong, slightly, and his statement is not pertinent. Your statement violates the postulate that has never been observed to have been violated: The motion of light is independent of the motion of the source of the light.

All Silas was responding to was the familiar experiment where photons are emitted (<|>) at the midpoint (|) of reflectors on a frame moving to the right.

Code:

L|________________<|>_________________|R

Analyze the problem purely from the photons point of view with the assumptions that

• a photon will continue moving in a straight line,
• until acted upon by an outside force
• ignoring any references to the frame's motion
• In other words the reflectors and the midpoint detectors are merely outside forces acting on the photon motion as the photons move into these obstructions.

The sequence of events on the moving frame any velocity) are:

1. photons emmitted left and right simultaneously
2. after both photons move a disance ct
   • the left photon arrives at the left reflector
   • while the right photon is still short of the right reflector [that has moved to the right].
3. from these positions
   • the left photon has reflected now moving right and returned to the point of the original emission point, defined by the photon's point of emission.
   • the right photon has been reflected after the left photon was reflected) and moves a total disance ct.
4. Both photons position's can now be considered as having just been emitted and therefore, their mutual midpoint, i.e. the point where they are going to meet is defined before they arrive).
5. The problem now has the location of the original midpoint of the reflectors closer to the left photon, moving to the right into the oncoming photon from the right.
6. Both photons arrive simultaneously with the arrival of the position of the original midpoint that has been moving at a constant velocity, any velocity
7. The problem can be separated into two photon reflector problems if we back up to the point after the left photon has reflected and returned to the photon defined midpoint. Both photons have been reflected and are heading toward each other we see the following, where the left photon was emitted before the right photon from their reflection points now considerd as source points:
   
   Code:

   |-------->       [U]|[/U] |          <------|  (->frame mtotio)

8. When the photons and the observer (|)all meet simultaneously, the observer, knowing the photons were emitted from sources on his frame located equidistant from him concludes:
   • He is stationary and the photons were emitted simultaneously, (not irrational) or,
   • He is moving and the rear photon was emmited before the forward photon. (the correct guess) or,
   • He is stationary and the embankment is moiving to his rear, and concludes the forward photon was emitted before the rear photon.
9. The 3rd option is erroneous and physically impossible.

geistkiesel
10-21-2004

 

geistkiesel,

I would say that postulate is only true if the source is stationairy in a gravitational field. Since I believe that the speed of light is only equal to c relative to the gravitational field that it is passing through at any given moment, if the source of the light is moving through a gravitational field when the light is emitted, the light will not be travelling a c relative to the source.

The experiment that you have illustrated has never been performed while moving through a gravitational field. So to assume that the photons would reach their targets at the same time, under all inertial circumstances, would be a leap of faith.

 

Your leap, I presume? Hell I jumped across a long time ago.

Interesting as is your reply you really didn't touch the postulate, you merely substituted your own thinking about the matter. I read you as saying the postulate describing the independence of the motion of light and the source is true except on the planet earth, for instance. When you say that light is moving erroneously at C with respect to the source, you imply, arbitrarily, the dependence of source and light motion. The postulate has to do with independence of motion, your model say oh yes it does have a dependence, on earth that is.

Your biggest leap that you have to make, however, has to do with what the answer to, what is a gravity field? How do you know, for example, it isn't all in angular momentum conservation? You know the sun has the biggest chunk of the solar system mass, but contributes a mere percent or two to the total angular momentum of the solar system? You knew this I am sure?

If this is your thesis I need a lot more If I am to gain any real interest.
Geistkiesel

 

geistkiesel,

I wouldn't want to hijack a thread with my theory, but since this is your thread, and you asked, I will give a brief explanation of my theory.

First, I believe that a photon has a dipolar, or non-uniform, gravitational field with a positive gravitational field in the front of the photon, and a negative gravitational field in the back of the photon (kind of like a magnet). When a photon is placed in a external standard (positive) uniform gravitational field, the external field attracts the front of the photon (since same gravitational fields attract) and pushes the back of the photon (since opposite gravitational fields repel). These forces cause the photon to accelerate in the external field until its speed reaches c (which is the speed of the gravitational interaction) relative to the external field.

This would mean that the speed of light is not equal to c for all inertial observers, but only for observers that are stationairy in a gravitational field. So, for example, an observer that is stationairy on the surface of the Earth (which would mean that the observer is stationairy in the Earth's gravitational field) would always measure the speed of light to be equal to c (disregarding the small effects that the Sun's and Moon's gravity have on the light). On the other hand, an observer that is moving through a gravitational field will measure the speed of light to change, where the change is equal to the speed the observer is moving relative to the field.

And, finally, the speed of reactions in an object that is moving through a gravitational field will decrease not because of time dilation, but because the speed of light changed inside that object, and the speed of reactions in that object are directly related to the speed of light inside that object.

As for your question about what is a gravity field, my answer is I don't know. I do think that it is an extension of matter rather than the result of an exchange of particles (gravitons). Also, I believe that the negative gravitational field, which I introduced in my model, could explain dark energy.

 

And I thought I was "outre". Interesting concepts, most difficult to prove and of course it isn't going to make any SR theorists come knocking. Just from instinct if nothing else, I see a weakness in the use of gravity, which isn't all that well defined, or measured. We see the affects of course,of what we call gravity, but are we talking about body to body forces such as mass attraction? Who knows?

When I consider the mass of the planets as approximately 10^-3 that of the sun, yet the sun contributes only 1 percent, or so, of total angular momentum and I meditate on the conservation of angular momentum principle, I cannot simply relax and get on a common gravity wagon, though yours certainly isn't common, either for gravity and for photon motion.
If it weren't for the introduction of such radical constructs it would be more palatable, to me at least.

The row I am hoeing isn't nearly as long as yours I suspect.

 

geistkiesel,

Can I ask you why you feel light has velocity in the first place?

 

You are asking why I feel lght has velocity? Well, for one many people have published expeimental accounting of measuring the speed of light. The chap who measured, or detected the different times it took light to reach earth from Jupiter's moons. The fact that nothing in my reality that is observable occurs instantaneously, which is not a denial of nonlocal activity.

Since you asked, consider the velocity of a distant star moving at 500 kim/sec for instance, emitting a single photon moving 3 x 10^5 km/sec. If the star and photon are moving in the same direction how can the snail like velocity of the star compress a photon when the star is moving at .0016 of the photon motion?, or worse yet, if the star is receding from the photon motion, how the receding stretches the wavelength to a longer length.?

This model has photons oozing out of sources like little increments of tooth paste oozing from the toothpast tube.

This model also ignores the pistulate of light that tells us the motion of the source of light and the motion of the light are independent.

 

geistkiesel,

I think Quantun Quack was asking why light moves at c instead of acting like other particles, like electrons or protons, that will remain stationairy unless pushed be an external force.

I suggested that light uses gravity to move. If you believe, unlike relativists, that at any point in a photon's life it, may be travelling at a speed higher or lower than c, then there must be something that pushes or pulls the light to c. What do you believe is the photon's source of propulsion?

 

Actually Guys I just thought that seeing as you are all "Out there" so to speak why no go the ful hog and ditch the notion that llight has to travel at all.

Light is deemed to have velocity only because it's effect is interpreted that way.

"light could be an instantaneous effect which is delayed adn gives teh impression of velocity simply becasue the reflector takes time, equivalent to the distance of separation to change it's resonance and reflect the light. There are no particles or waves just simply graviotational harmonics at work.

Light being a form of alternating inverse gravity."

The distance between two objects in a vacuum is actually zero. Light has no distance to travel. The reflector takes time to reflect the light due to the reduction in light intensity caused by distance. But the distance in a physical sense is zero.

Any way , as we are going out there I thought I'd ask why not go the full thing?

 

Your hypothesis can be easily broken by just looking for other way to measure the speed of light. One example is, the one done by Romer (involving Jupiter's moon eclipse), which showed that light doesn't propagate instantaneously. Since this measurement doesn't involve mirror reflection, you need to think of other explanation. May be the Jupiter's moon also "hold" the sun light for awhile. But, again, this cannot be true. How do you explain that the delay appear to depend on Jupitar-Earth distance, not on the Jupiter's moon properties? At this point, you should immediately give up the idea until you find a better explanation than the "reflector takes time".

 

PaulT I already have, there are many flawes tothis hypothesis, this is true, however I may add that the receptors in the eyes are also reflectors that take time to reflect the light.
However My point was not the validity of the hypothesis, but that if we are going to confront conventional thinking on some points why be selective and just go all the way and confront it all.

The fact as to whether light travels of not may be a major issue in the development of theories for gravity and ex-nihlo creation of the universe but as you have pointed out in concept it fails to make full sense or be worthy of any following. It is however worth noting that ex-nihlo creation also doesn't make a hell of a lot of sense either. The big bang or similar ideas suggesting an enourmous problem for our logical assumptions of conservation laws and thermaldynamics.

On one hand science will state absolutely that energy can not be created and yet the universes coming in to existence proves otherwise. Also the observations being made of spontaneous creations of matter and decreations of matter also makes a lie of this "energy can not be created" rule.

With this light velocity thing I am suggesting that the velocity figure is actually the change in the velocity of the reflectors suface, that the distance between source and reflector is extrapolated in the reflectors vibratory rate in direct relationship with distance. Exactly how this happens is yet to be determined as is exnihlo creation.

Any sampling of the light at any distance will give you lights invariance but only because of a fundamental relationship between reflective mass and distance between those masses and not the velocity of some fictional mass less, wave like particle.

Again though this is purely hypothetical as is most of this thread, and well pink elephants can fly too I guess......

 

I can ditch the speed of light on a standing wave theory only. But initially the light has to get from there to here.

 

it gets all confused I reckon because on one hand we declare that there is no medium between objects for light to propagate in and if there is no medium then there is no real distance. The space between objects being nothing so therefore light has nothing to travel though, thus distance is an illusion of space only.
any way sorry for interupting your thread.......I'll say no more

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