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

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. 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”.

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:

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

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 “|||”.

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

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.

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

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?


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.

Meaning, of course, that t2. = t1 (2vt)/(C-v) where all parameters are measured in the moving frame.

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? 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, -->|||<--.

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.

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.

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.

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.

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.

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

I don't understand your diagrams.

:m: I don't understand your diagrams.
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.
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. 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. 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.
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. 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. 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

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. 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, or is 2t + t2 and the same times are recorded on the L and R peaps, which is timed to shut off after counting 2t + t. Therefore the time on the peaps computes is 2t + t2. + t, 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 of the L and R photons at the L and R peaps. Therefore, the experiment is repeated 1000 times more with the same result, or rhat 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) 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) 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.. The (2) option is the only reasonable option.

The total traveled distance is 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 . After a bit of algebra we see that t2. = t(2v)/(c – v). 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, 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. 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 t' = t if v = 0. where the only difference is the t2 time, t2 proves motion to the right. 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 In other words, t2. cannot be explained away by time dilation as, 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 the physical reality that the mass inertia of Ve (6 x 10^27 grams) is infinitely larger than that of Vn, that the Vn necessarily accelerated with respect to Ve in order that any relative velocity of Vn - Ve > 0 be observed, that Ve is measurably indistinguishable from straight line uniform motion, and Ve is never observed acceleratiing and providing the velociy for an observed |Vn - Vc| > 0 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

I don't understand your diagrams.
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:
|L_________________M__________________R_|
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:
|----->| vt | vt |||
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.
|vt|<----ct------|-----------> | ||
0 1 2
|<----------|-|-|---------> | ||
|----------> | | <---------|

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.

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,

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.

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 responding to geistkiesel}

“ 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. ”

[Prosoothus responds to 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.

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:
photons emmitted left and right simultaneously
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].
from these positionsthe 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.
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).
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.
Both photons arrive simultaneously with the arrival of the position of the original midpoint that has been moving at a constant velocity, any velocity
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:
|--------> | | <------| (->frame mtotio)
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.
The 3rd option is erroneous and physically impossible.
geistkiesel
10-21-2004

geistkiesel,

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.

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.

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?

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?

"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.


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".

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......

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?


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 :)

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 :)
No problemo. When a baseball disappears over the fence what do we ascribe to the medium in which the ball travels? There isn't any medium, the ball is a unit of mass. The photon is a unit of mass. Light motion and all the rest has a confusing history written by the confused, but light isn't cnfused.

If light had mass then I would have to agree but light is massless therefore has no inertia ro for that matter substance. If it had mass according to relativity it would take an infinite amount of energy to get it to travel at 'c'.

So what is our photon? I guess it could be anything you want it to be....hey?

If it has mass you would be able to see it in transit and not just in reflection.....therefore light is 2 dimensional and only discerned by effect and not substance.

If light had mass then I would have to agree but light is massless therefore has no inertia ro for that matter substance. If it had mass according to relativity it would take an infinite amount of energy to get it to travel at 'c'.

So what is our photon? I guess it could be anything you want it to be....hey?

If it has mass you would be able to see it in transit and not just in reflection.....therefore light is 2 dimensional and only discerned by effect and not substance.
The claim that "light has no mass" is a mouthful and usually expressed differently in that light is said to have "no rest mass". However, there are sufficient experimetnal results measuring the momentum transfers of photons to material bodies. Momentum without mass is unthinkable. Even though the no rest mass seems consitent to some analyses, no one has been able to capture a photon of EM radiation where Vp = 0. And remember, all the physics to date is not guaranteed to be perfectly described by the popular understanding of the physics industry.

The claim that "light has no mass" is a mouthful and usually expressed differently in that light is said to have "no rest mass". However, there are sufficient experimetnal results measuring the momentum transfers of photons to material bodies. Momentum without mass is unthinkable. Even though the no rest mass seems consitent to some analyses, no one has been able to capture a photon of EM radiation where Vp = 0. And remember, all the physics to date is not guaranteed to be perfectly described by the popular understanding of the physics industry.

Please prove that there cannot be momentum without mass.
The relation between light's momentum and energy is E = Pc, where E is the energy, P its momentum and c the speed of light.
Since the relation between energy and momentum is E^2 = P^2c^2 + m^4, (here m is the mass), it follows that m = 0 for light.

Please prove that there cannot be momentum without mass.
The relation between light's momentum and energy is E = Pc, where E is the energy, P its momentum and c the speed of light.
Since the relation between energy and momentum is E^2 = P^2c^2 + m^4, (here m is the mass), it follows that m = 0 for light.
Your mathematics is interesting. Since you introduced the mathematical expression, is it not your obligation to show that there is zero mass in the
P^2c^2 (did you eman (2Pc)^2?) term.
Is this request unfair?

Since I have your attention 1100f, and as I truly appreciate your bluntness, can you briefly explain the difference in the folloiwng 2 conditions as Einstein decsribed them in his "relativity" book.
1. A raven flying with uniform motion in a straight line is observed to have different relative velocities and directions wrt the various frames, but all will see straight line motion and accurate relative velocities, while
2. making the same observations of "light motion in vacuo", all frames will measure the velocity of like as C, and that all relative velocity will also be mesured as C (they are the same measurement).
For the purposes of this esoteric physics discussion, what physical attributes distinguish photons from ravens or ducks (I had a duck as a young child)?
What if my duck were flying by at .9999c. Would the ordinary physicist be able to detect the difference in velocities of duck and photon, .ooo1C or 3 x 10 = 30 km/sec, which is equivalent to the earth sun orbit velocity?
Asked another way, what is it about ducks and photons that deserve such bizaar treatment?
Finally, what are the implications of theoretically imposed discarding of the frame velocity when measuring the relative (or direct) speed of light? To you I ask this question, no one else.
I see that discarding essential data such as an observer's speed is not a universally recoginized physical practice.
If two automobiles are moving in opposite directions and they both make mesurements of the relative velocity of their combined motion, would the calculations of observers in either, or both frames, be acceptable to you as credible scentific evidence if offered as a showing of relative velocity, knowing that the author(s) specifically excluded the inclusuion of his own velocity?
No you would not consider the data credible, or scientific. Well then, at what velocity does SR kick in, where discarding the observer's motion is theoretically imposed axiomatically? Is this at all scientifically credible?


Thank you 1100f, I appreciate your response to my posts,

Geistkiesel

* 1. A raven flying with uniform motion in a straight line is observed to have different relative velocities and directions wrt the various frames, but all will see straight line motion and accurate relative velocities, while
* 2. making the same observations of "light motion in vacuo", all frames will measure the velocity of like as C, and that all relative velocity will also be mesured as C (they are the same measurement).

Again, for light different frames will measure different directions for the light, although the speed will be c in all frames.

* For the purposes of this esoteric physics discussion, what physical attributes distinguish photons from ravens or ducks (I had a duck as a young child)?

Photons have no feathers. Ducks have mass. Should I go on? It will be a very long list.

* What if my duck were flying by at .9999c. Would the ordinary physicist be able to detect the difference in velocities of duck and photon, .ooo1C or 3 x 10 = 30 km/sec, which is equivalent to the earth sun orbit velocity?

Depends on how the detection was done.

* Asked another way, what is it about ducks and photons that deserve such bizaar treatment?

What's bizarre? All the equations treat ducks and photons the same way, taking into account the physical differences between the two (e.g. the difference in mass).

* Finally, what are the implications of theoretically imposed discarding of the frame velocity when measuring the relative (or direct) speed of light? To you I ask this question, no one else.

It seems I'm not allowed to answer this, then.

* I see that discarding essential data such as an observer's speed is not a universally recoginized physical practice.

As I said, the observer's velocity is quite important in determining the velocity of a light beam.

* No you would not consider the data credible, or scientific. Well then, at what velocity does SR kick in, where discarding the observer's motion is theoretically imposed axiomatically? Is this at all scientifically credible?

SR applies at all velocities. It doesn't "kick in" at some specific velocity. Relativity has replaced Newtonian physics, although Newtonian physics is still a reasonable approximation for velocities much less than the speed of light.

Again, for light different frames will measure different directions for the light, although the speed will be c in all frames.
Before I reply I must "share" an observation. Your response is basically what I had just described.
Read thefollowing:


Geistkiesel States : as follows:

* 1. A raven flying with uniform motion in a straight line is observed to have different relative velocities and directions wrt the various frames, but all will see straight line motion and accurate relative velocities, while
* 2. making the same observations of "light motion in vacuo", all frames will measure the velocity of light as C, and that all relative velocity will also be mesured as C (they are the same measurement)." [end of Geistkiesel quote]
”
Your statement that light will exhibit different directions, but will have the same velocity, isn't to me a response. It is just a line from the many postulates of special relativity. Now I, expecting more in the response such as some indication of the physics of the condition that theoretically prevents, and it does effectively because only committed SRist would never dare make a contrary measurement, from measuring the relative motion of frame and photon. This is what makes me conjecture that we aen't talking about the same planet, the one with the blue sky.

Even though my "expectations" weren't met, so what maybe the relation of SR and the kind of "Newtonian" physics I have been accused of indulging in, is the kind reasonably described in yours and my posts on the subject matter in the past few weeks. I look, and look, and like Einstrein, looking and looking to resolve apparent contradictions in relativity and the motion of light, eventually designing SR theory. Similalry I am looking for a way that would allow what you say, and to what I say, that wouild link the differences into some mutual tit-for-tat.

Alas, I am not up to that task, which is really understanding in an SR sense why photons are different than ducks in the measurement of their respective relative velocities, where the ducks are measured with respect to the moving frame Vn with the Vn > 0., while the photon gets measured the same way, but the relative motion will always be measured as C, as if the frame velocity Vn = 0, which it artificially is set to zero for SR conveneince..

So ducks have feathers, and photons do not, so why this difference at the measurement spot? What do feathers, and not feathers, have to do with it? Why is this difference a physical reason to treat the objects differently? I've read the story of the vacuo, where the photon is always measured C. I then wonder what theoretical affect the measurement would bring if the last bytes of data resulted in: "That's not a photon you're measuring, that's a duck".


[quote=James R]
Photons have no feathers. Ducks have mass. Should I go on? It will be a very long list.
You can make the list as long as you like, but Iwould prefer, and it would be less work for you, if you would do other than just list differences in the phsical chaacteristics oif the the duck and the photon. Specifically, what does a duck's feather have to do with distinguishing the mesurement of the duck and the light, the frame and the photon? Feathers, what do feathers have to do with the systematic measuring differences regarding photon and frame?

I dispute the claim that the photon has no mass, but beside that, assuming the photon does have mass, what is to do with the systematic measuring differences between frame and photon?

Can you present you answer in a manner that is logically and scientifically accurate, coherent, such that an SR novice could take your response and anlayze on the basis of the infroamtion contained in the four corners of that response, the scientific worth, or value witjion your post?

Statement and question from geistkiesel: "What if my duck were flying by at .9999c. Would the ordinary physicist be able to detect the difference in velocities of duck and photon, .ooo1C or 3 x 10 = 30 km/sec, which is equivalent to the earth sun orbit velocity?"


Depends on how the detection was done.

What would be a factor in how the detection was done? How could there be a mistake? What are the conditions for a mistake in observation erroneously predciting duck versus photon?



Geistkiesel asks James R:

"Asked another way, what is it about ducks and photons that deserve such
bizarre treatment?"

What's bizarre? All the equations treat ducks and photons the same way, taking into account the physical differences between the two (e.g. the difference in mass).

So how do the equations account for the physical differences in mass? Which equations decide the issue? Any one equation to scrutinize would be welcome.


Geistkiesel asks:"Finally, what are the implications of theoretically imposed discarding of the frame velocity when measuring the relative (or direct) speed of light? To you I ask this question, no one else.
”

It seems I'm not allowed to answer this, then.


"Not allowed"? This isn't one of your 'roo generated answers is it? If you are seeking permission to speak, well I see your arm raised so by all means, speak freely man, tell us of the implications.



Geistkiesel states to James R: "I see that discarding essential data such as an observer's speed is not a universally recoginized physical practice."

As I said, the observer's velocity is quite important in determining the velocity of a light beam.

Sure it is, the motion is very important. Why is it important that you assume the moving frame Vn = 0, when measuring the speed of light and the relative speed of frame and photon?. What is the frames velocity's importance when assumed to be zero, or at rest with respect to Vew, of which you write, when measuring the realitive velocity of frame and photon, specifically, please?

All I read is "the observer on them moving frame, not knowing if she is moving assumes, she is "justified" in the assumption that she is at rest." Withjout this SR has nothing.



Geistkiesel states [the "blue" section was not quoted by James R, in his response in this post].
I see that discarding essential data such as an observer's speed is not a universally recoginized physical practice.

* If two automobiles are moving in opposite directions and they both make measurements of the relative velocity of their combined motion, would
* the calculations of observers in either, or both frames, be acceptable to you as credible scentific evidence if,
* offered as a showing of relative velocity, knowing that the author(s) specifically excluded the inclusuion of his own velocity?

No you would not consider the data credible, or scientific. Well then, at what velocity does SR kick in, where discarding the observer's motion as theoretically imposed axiomatically? Is this at all scientifically credible?

SR applies at all velocities. It doesn't "kick in" at some specific velocity. Relativity has replaced Newtonian physics, although Newtonian physics is still a reasonable approximation for velocities much less than the speed of light.
You mean to say that Relativity has replaced Newtonian physics in 'your mind,' don't you? You remind me somewhat of Rush Limbaugh that proudly broadcasts to us that to maintain a level playing field that he has "half his brain tied behind his back". Let me warn you James R, ou too Rush, you had better untie that other half of your brain, because at this moment on, you need the full use of both of your brain cells.

Are you asserting that Relativity is proved by your physical velocity discarding protocols? We [SRists] can't demonstrate Relativity, because the affect is so slight that we can ignore it, right? So we do ignore it because it cannot be proved. So therefore you might as well go along with including the observer's velocity in the measuring of automobile's relative velocities, right? A form of showing professional magnanimity to the the ignorant, the mal and mis informed, those lesser bodies undeucated and unschooled in and have accepted the irrational and physically inane model of, whatever is claimed by the proponents protecting their PSR.

Perhaps, "kick in" was inappropriately definitive, I should have asked at what velocity does one of the automobile velocities become justifiably discarded? Answer: You never discard the automobile's velocity.

I have copied some physical date of the inertia of the planet, the embankment, Ve as infinitely larger than any conceivable Vn that is restricted to a value of motion Vn < 11.2 km/sec. Inertia, the resistance to change is so huge in Ve that one can esily assume the invariant motion of Ve with regard to the observation of any realtive motion between Ve and Vn. It is always the Vn that accelerates as a prerequisite to the observation of the Vn and Ve relative velocity.

Like SR "theory" ignores the signfcance of some "low velocity" range, I also assert, as a matter of physical law that Ve
never accelerates, never has accelerated, in a mesureable manner equivalent, in any sense, to that acceleratioon giving rise to the Vn actual motion, such as the readings of accelerometers on board Vn, a passenger train, and Ve, the planet, which will always return a zero or null result in measured Ve acceleration, and no component of relative velocity can be ascribed to Ve that is traceable to any such Vn accelerations, and . observation of relative motion of Vn and Ve will always produce the condition: Vn - Ve > 0 Vn > Ve for all time t, during which the relative velocity of Vn and Ve are observed with Vn > Ve.,

We all knlow your response to this last piece:: SR says we can ignore the constraints of physical reality if it would torpedo our Precious SR. As our PSR is theoretically perfect, any physical reality that effectiviely nullifies our PSR must be untrue and hence, such nontruth shall, by proclamation of a chorus of committed SRists, joined at the hip, proclaiming with montonic sincerity and conviction that the nullification if SR is untrue.

The NeoNewts, on the other hand, are seen gathered in a joyous assembly of glee and rousing emotional intensity, with their arms around each other, singing "Those were the days my friends, we thought they'd never end."

Geistkiesel:

Your mathematics is interesting. Since you introduced the mathematical expression, is it not your obligation to show that there is zero mass in the
P^2c^2 (did you eman (2Pc)^2?) term.
Is this request unfair?

Again, the general relation between momentum and energy is E^2 = (pc)^2 + (mc^2)^2.

For light the relation between the momentum and energy is E = pc. Replace E by pc in the first formula and you will find that for light m = 0.



Finally, what are the implications of theoretically imposed discarding of the frame velocity when measuring the relative (or direct) speed of light? To you I ask this question, no one else.

In relativity nobody discards the frame velocity. You allways use the same addition of velocity formula, which is the formula that tells you what is the velocity of something in a reference frame (call it frame A) if you know the velocity of this something in reference frame B and you know the velocity of reference frame A in reference frame B.

Let me give you an example:
Suppose an object is moving in the x-direction in reference frame B and its velocity is u. Now, suppose that I am in reference frame A that moves also in the x-direction at velocity V with respect to reference frame B. What will be the velocity u' of this object in reference frame A?

Well I do not discard the velocity of frame A wrt reference frame B, but instead I use the relativistic formula of addition of velocities:

u' = (u - V)/(1-uV/c^2).

Notice that if both u and V are very small when compared with c (this is the everyday situation), the denominator is very close to 1 and you can aproximate the transformation as u' = u - V (if you are travelling at 60 km/h and another car is travelling at 100 km/h, relative to you this car goes at 40 km/h).
If the object travels at the speed of light, i.e. its velocity is u = c, I do not discard the frame velocity, and use the same formula in order to get
u' = (c - V)/(1-cV/c^2) = c.
Oops, the velocity of light is the same in all reference frame.


I see that discarding essential data such as an observer's speed is not a universally recoginized physical practice.
If two automobiles are moving in opposite directions and they both make mesurements of the relative velocity of their combined motion, would the calculations of observers in either, or both frames, be acceptable to you as credible scentific evidence if offered as a showing of relative velocity, knowing that the author(s) specifically excluded the inclusuion of his own velocity?
No you would not consider the data credible, or scientific. Well then, at what velocity does SR kick in, where discarding the observer's motion is theoretically imposed axiomatically? Is this at all scientifically credible?

Not relevant

Notice that if both u and V are very small when compared with c (this is the everyday situation), the denominator is very close to 1 and you can aproximate the transformation as u' = u - V (if you are travelling at 60 km/h and another car is travelling at 100 km/h, relative to you this car goes at 40 km/h).
If the object travels at the speed of light, i.e. its velocity is u = c, I do not discard the frame velocity, and use the same formula in order to get
u' = (c - V)/(1-cV/c^2) = c.
Oops, the velocity of light is the same in all reference frame.

I truly appreciate your style 1100f. Without using your addition of velocity protocols I arrive at a conclusion that is similar to yours, without any of the "approximations" used for low velocity conditions.

The Va frame is taveling at, Va, the Vb frame at Vb. There is nothing requiring any structured "addition of velocity" formula to find the simple relative velocity of the cars: Va - Vb = Vab, where Vab is the measured relative velocity of the two objects.

I must understand your thinking as being based entirely on the assumption of the "relative" velocity of light being C in all reference frames. This is the fundamental basis of all Lorentz transformations, so I see your 'oops' offered as some "proof" that C is the same in all inertial frames.

Another objection iIhave is the use of the words (your words) that,

". . .the velocity of light is the same in all reference frames."

The objection is your failue to clearly state that you are claiming the "relative velocity of light is the same in all reference frames". This is not a trivial distinction as I have encounterd too many who utter the mantra without realizing the conceptual difficulty of harmonizing the "relative velocity" implications with the necessity of manufacturing time dilation and frame contraction, the discarding of simultaneity and the discarding of absolute time, space and motion.

If Va is the velocity of light, a constant, then the measure of how much faster the light travels than the Vb frame is Va - Vb = Vab < Va. This tells us the relative velocity of frame and photon. This doe not diminish the velocity of light, but it does diminish the theoretical axiom that demands that the relative velocity of light be the same in all frames.

Your "Oops" may need editing, as there is no need of any mathematical construct to tell us that the speed of light is constant. The speed of light is measured as a constant. In fact, you don't get to the Lorentz transformation (in your addition of velocities) until you have already assumed the measure of the speed of light, direct and relative, will always be C, regardless of the motion of the inertial frame. In other words, you have assumed the relative velocity of the frame is zero, then your theory and expression of that theory coincide, and this assmes the frame velocity is at rest, with respect to the photon motion.

You widen the gap from physical law when you theoretically connect the motion of light as constant with respect to the source of light, for instance, when the physical axiom is that the motion of the source of light and the emitted light are independent. To say "The velocity of light is C with respect to the motion of the source" you have imposed a restriction on the motion of the light relative to the motion of the frame, which is a clear violation, by assumption, of the independence of frame and photon. Said differently, a constant velocity of frame and photon, always, removes the independence of the motion of light.

Your mathematics is a mere reflection of assumptions, and your 'oops' is, at best, misplaced.

Thank you for your reply 1100f.

Geistkiesel.

The Va frame is taveling at, Va, the Vb frame at Vb. There is nothing requiring any structured "addition of velocity" formula to find the simple relative velocity of the cars: Va - Vb = Vab, where Vab is the measured relative velocity of the two objects.



You must prove this.

You must prove this.
OK, two automobiles are moving in a straight line observed by obervservers on the side of the road making velocity mesurements of both cars as he pass by. At the same time the observers in the automobiles are measuring the passing velocities of the automobiles as they pass each other, some tests have he cars moving in the same direction, some are moving in the oppositie direction, all measurments will be consistent from what ever frame measured.

Take a photon stream measured as moving at the speed of light C as measured friom the embankemnt, Ve, the earth, and a space ship passing is also speed measured. The data is relayed to the space ship navigfator. The navigator having observed the light stream going by then ask himself "are these data accurate?", he assumes the data from the Ve frame are accurate. The navigator then takes the values and makes the simple subtraction, C minus his own velocity as radioed to him. Both velocities are measured from a common inertial frame, what is wrong with this?

What other kind of proof do you need 1100f?

I am not sure but relativitists wil tell you that the time/distance figure is frame dependent because the time or rate of time is frame dependent. Thus simple subtractions and additions are not possible and a common inertial frame is of no value.

OK, two automobiles are moving in a straight line observed by obervservers on the side of the road making velocity mesurements of both cars as he pass by. At the same time the observers in the automobiles are measuring the passing velocities of the automobiles as they pass each other, some tests have he cars moving in the same direction, some are moving in the oppositie direction, all measurments will be consistent from what ever frame measured.

Take a photon stream measured as moving at the speed of light C as measured friom the embankemnt, Ve, the earth, and a space ship passing is also speed measured. The data is relayed to the space ship navigfator. The navigator having observed the light stream going by then ask himself "are these data accurate?", he assumes the data from the Ve frame are accurate. The navigator then takes the values and makes the simple subtraction, C minus his own velocity as radioed to him. Both velocities are measured from a common inertial frame, what is wrong with this?

What other kind of proof do you need 1100f?
This does not proof that V_AB = V_A - V_B.
What you show in fact is that if you assume that V_AB = V_A - V_B, then V_AB = V_A - V_B.

This does not proof that V_AB = V_A - V_B.
What you show in fact is that if you assume that V_AB = V_A - V_B, then V_AB = V_A - V_B.


Your very cogent reply actually made me think about the situation. Your argument has a wide range of application as d(ma)/dt = F, correct?

I wasn't assuming much. I just wanted to find out how much faster one car was moving wrt the other, or what the combined speed was if the vehicles were moving in the opposite direction.

I see how I assumed what Vab from your statement. Isn't this a very common assumption, tacit as it is? When a relativity theorist get a speeding ticket the police officer, judge and jury isn't going to be impressed by an argument that the Ve was moving at the clocked speed and he, the theorist, was stationary. If the assumption is as you observed, and I have no real objection to it, isn't this how one determines who is faster and by how much?

if photons have mass, how would you go about proving it? What kind of experiments is the claim susceptible to?
Some ideas- if photons have mass, do torches get noticeably lighter when being used?

if photons have mass, how would you go about proving it? What kind of experiments is the claim susceptible to?
Some ideas- if photons have mass, do torches get noticeably lighter when being used?
First, there would have to be a universally recognized definition of mass. I bet you have one of those don't you?

But why? I'm no physicist, but it seems to me that saying photons have mass and then saying you actually need a universally recognised definition of mass is a cop out.

So, that aside, would you settle for the "gravitational mass"?
Like here:
http://www.wordiq.com/definition/Mass

But why? I'm no physicist, but it seems to me that saying photons have mass and then saying you actually need a universally recognised definition of mass is a cop out.

So, that aside, would you settle for the "gravitational mass"?
Like here:
http://www.wordiq.com/definition/Mass
I'll buy that definition, with a slight reservation regarding the incomplete and misunderstoofd concepts of gravity. Howebewever, we'll nitpick only when necessary..

Photon have mass per the definition as the all are able to inflict a momentum component when striking material objects. The problem is, as I see it, the concept that the mass of light is zero at zero velocity, and from this the discussion isdirected ask the question is light massless througout all motion? Everytime we measure light we measure a momentum exchange, well almost always. The Mossbauer effect produces recoiless absorption of gamma radiation emitted from a frame moving a few centimeters/second with respect to the stationary target absorbing mass. If the source ios not moving the absorbtion of the gamma particle is perfect and no recoil from the absorber is observed, Move the source slightly and the absorber begins to recoil.

This says alot to me. The source of the gamma seems to be affecting the motion of the gamma and violating a postulate of light. But it seems that only the energy of the emitted gamma is affected, not the motion, which means the slight addition of velocity of the source is adoped by the moving gamma, which is consistent with my theory of photon motion I have been screaming for a while as hunt down and deprogram SR theorists.

The motion of the Ve, planet earth'e instrinsic motion is inherited or adopted by virtually all objects on the earth and that the observed relative motion of material, Vn and Ve, is an addition to the inherent motion of the Ve frame which is apparently immeasurabe directly.

First, howevr, the photons sure seem to act as if they receive and provide momentum in collisiosn with material objects and if the character of the energy of the photon is drastically altered in the collision process, so be it, the photon's mass structure has not been dnmaged, theoretically, by this alteration.

The sensitivity of the internal gamma structure in the target molecules is enlightening. It tells me that the addition of the gamma results in the radiation of a gamma at the same energy as the input particle, which is the observation of a recoil free absorption process. It seems tahthe momentum exchange is effectively zero, which show that the gamma is channelked thriough the molecule in the target with zero inertial contact, wher th eslightest perturbation fo the gamma induces a vibration that is a oerturbation of the finely balanced and tuned internal gamma structure of the molecule. This occurs even though the mass of the energetic gamma should easily induce a momentum impulse, but here momentum exchanges do not occur in a manner that is detectable.

Its is as if the input momentum perfectly matched the intenal momentum of the internal gamma structure, and while the same gamma physically that entered, is probably unrelated to the gamma leaving, and that what was not exhanged in the absorption of the gamma, is the momentum that was passed on to some mometum less aspect of the gamma particle as a baton is exchanged in a relay race. However, the structure of the incomong photon must have undergone some fancy alteration to become a momentulm less internal gamma. While hexiing gama adopted the invasion of the momentum.. It sounds like a form of pure restance less activity.

I had better stop, I am geting in over my head as this is all off the top part and I don't know if I am making any sense as I ramble on...

Final answer, yes photons have mass. Are the mass increases in high energy exchange experiments real mass or ersatz mass and not effected by gravity? I will stick with mass is mass and it all acts essentially the same. We just have to prepare ourselves to undersand the lack of rigidity in some mass forms that may have a soft wave like srructure and different that baseballs and protons.

Final answer, yes photons have mass.

Photons have NO mass.

Photons have NO mass.
a bit like asking whether a bolt of lightning has mass? or does and arc of static electricity have mass. It is certain that the effect of lightning can induce momentum. [if ever you are stuck close by lightning you woudl know what I mean] but does this force have mass....the answer would normally be no I guess.....Does giving something a negative charge change it's mass? It certainly changes the object but does it change it's mass? Mass being in part defined by it's weight....which in turn is defined by it's ambient gravitational effects.

QQ, energy has mass, as evidenced by the formula E=mc^2. It just takes a
huge amount of energy to equal a small amount of mass. Except our lowly
photon, although it carries energy, it is deemed by Special Relativity to have
NO mass.

I would consider hypothetically, a photon to be a gravitational effect that generates energy in the reflector by methods of gravitational harmonic resonance......thus our photon doesn't need to have mass. Regardless of what SR states.

Final answer, yes photons have mass.

Photons have NO mass.
what makes you so sure? special relativity theory?

If light had mass then I would have to agree but light is massless therefore has no inertia ro for that matter substance. If it had mass according to relativity it would take an infinite amount of energy to get it to travel at 'c'.

So what is our photon? I guess it could be anything you want it to be....hey?

If it has mass you would be able to see it in transit and not just in reflection.....therefore light is 2 dimensional and only discerned by effect and not substance.
If the photon has no mass how do you explain the transfer of momentun of photons striking material objects?

e=mc^2

Momentum is not soley reserved for mass.

More over, even the physics of Newtown's day had theories on photon momentum transfer which didn't include mass.

am I right in assuming that photon momentum transfer only happens if the distance of separation is reducing?

How do Solar sails work ? How will laser powered ships work ?

<iframe src="http://lighttheory.com/experiments.htm" height=500 width=700></iframe>

what makes you so sure?

Because photons can never be 'at rest.'

Hey ThinkTank, thanks for the info.

what makes you so sure?

Because photons can never be 'at rest.'

Photons can never be at rest because then they aren't photons any more. The photons have undergone a transformation of form and substance to the extent that an analysis of the altered substance gives no clue to its previous history as a photon. One can infer a history from indirect measurements such as momentum measurements.

Complete bullshit, geistkiesel.

You could try and explain the difference between mass and relativistiv mass... but that would be a waste. He could just as easily goto the library.

I was going to try and say something intelligent, but it seems others have already beaten me to it. And my challenge still stands, if photons have mass, why dont torches get lighter?

I was going to try and say something intelligent, but it seems others have already beaten me to it. And my challenge still stands, if photons have mass, why dont torches get lighter?

The torch power supply (battery) merely provides a potential to stimulate electronic motion by heating the filament in the light bulb such that acclerated electrons in the filament emit light. The mass of the photons aren't stored and released in the torch, nor the electron, as one might release a gas in a pressurized container.

And, Guthrie, there is always the question lurking in the background of the quantum mechanical nature of electromagnetic phenomena with the inherent dynamics of nonlocal force channels.

... , if photons have mass, why dont torches get lighter?

Well, U did not measure its weight properly, because that technology is yet 2b developed. How R U gonna weigh photons ? They are very small sub atomic particles

Well, U did not measure its weight properly, because that technology is yet 2b developed. How R U gonna weigh photons ? They are very small sub atomic particles
I cannot accept that photons exist in the same state when at rest and moving. It makes more sense that the photons are merely the result of stimulating the electrons to velocities that require the electrons to emit a photon when decaying back to a lower potential when the the emitted photon has left the torch. Don't ask me how the process of photon production occurs, at least not today, but ultimately it has to be a straight forward process.

U dont have to weigh photons directly, just measure the decereasing weight of the tourch.

I would expect your scales would have to be very sensitve, but eventually you would find a weight loss, well that's my opinion.

Thats right, rawthinktank, you just weigh the torch.
As for sensitive scales, methods were worked out back in the eearly 20th century to get the mass of electrons. Google for them, then try and work out how to use them on torches.

Then, Geistkiesel, that is the point at which pseudoscience deviates from "normal" science. Your opinion is that the torch will lose weight. Will you now try and perform an experiment to show this?
You'll need a very good balance, one that can go down to less than milligrams. And a sealed chamber, in which there is now way you can contaminate the torch with moisture and dirt when you switch it on and off as you would a torch in everyday use. You then need to get some new batteries, put them in the torch, weigh it, switch it on, and leave it till the batteries run down. Then, weigh it again. Repeat a few times to get an average.
[admittedly performing this is a bit tricky and will cost money, but if you do it and it shows photons have mass, then congratulations, you've just blown up a large part of conventional physics]


OK, if you think this:
"The mass of the photons aren't stored and released in the torch, nor the electron, as one might release a gas in a pressurized container."

Then, why dont you do the elemntary calculations in which you compare the energy stored in teh torches batterys with the number and mass of photons put out. You seem to have some matehmatical knowledge, so why not work it out. Or is current thermodynamics tainted as well?

That seems like a lot of trouble, guthrie, with all those batteries. Could we just measure out a specific volume, say 10 cubic centimeters, of electrolyte solution
from a charged battery and weigh that? Then we could could return it to the battery,
discharge the battery and weigh the same volume of electrolyte from the discharged
battery. Wouldn't that be simpler?

That seems like a lot of trouble, guthrie, with all those batteries. Could we just measure out a specific volume, say 10 cubic centimeters, of electrolyte solution
from a charged battery and weigh that? Then we could could return it to the battery,
discharge the battery and weigh the same volume of electrolyte from the discharged
battery. Wouldn't that be simpler?
Or, use the solution as is, insert approriate grid plates and monitor weight changes in real time?
try this for a well written outline of battery dynamics/chemistry and physics. (http://www.varta-automotive.com/eng/index2.php?p=4&s=3&content=knowhow/batterielexikon/entwicklung.html)

Yup, thats how batteries work.
The point about using the batteries is that they are sealed off. If you are measuring tiny amounts it is important to elminate outside interferences. If your weighing ot an accuracy of 1 gram in a kilogram, you have to take less effort to isolate the system than if your weighing to a milligram in a kilogram of sample.

But anyway, if as you also seem to be suggesting, the energy of the batteries is beign transformed into stuff with mass, congratulations, youve got half way to painlessly converting energy to mass and vice versa, without the need for a nuclear reactor or hydrogen bomb. Now if you can work out how to replicate it on a massive worldwide scale.....

quote:

"But anyway, if as you also seem to be suggesting, the energy of the batteries is beign transformed into stuff with mass, congratulations, youve got half way to painlessly converting energy to mass and vice versa, without the need for a nuclear reactor or hydrogen bomb. Now if you can work out how to replicate it on a massive worldwide scale....."
================================================== ======

But we would need the energy of that hydrogen bomb inside the battery to
create much mass. You've got it backwards, nuclear reactors and hydrogen
bombs convert mass into energy, not vice versa. Must be that flip-floping
of frames in relativity that has you all mixed up. Particle accelerators increase
mass by using energy.

when you heat water dos it become lighter.....thus decreasing it's weight a battery may get lighter as it charges up rather than get heavier....

when you heat water dos it become lighter.....thus decreasing it's weight a battery may get lighter as it charges up rather than get heavier....
The water heated probably doesn't move fast enough to see any"mass change'". But look at the nuclear blast. If I am not mistaken the critical mass for an uncotrolled chain reaction is around 80 kg. One separates the fuel into two equal masses, more or less. and accelerates one of the masses toward the other much like firing an artollery shell. When the mass is critical now and there is a ring of magnetic field around this critical mass that gets compressed by an explosion on the ring, compressing the field sufficient to generate a "critical Temperature" , the baby burps and here it goes. This is about it, 80 kg or so can do all that? If we here are counting photon losses for mass losses, and boiling water for mass gain, we aren't close to determining the energy in the systems we have been discussing.

Historical note: Some have quoted the co-pilot on the Enola Gay as saying, "Jesus look what we've done". The tail gunner recorded some crew conversation and he reported the co-pilot as saying: "Jesus Christ, look at that sonofabitch go!."

Hi guthrie,
I found a thought experiment called Einstein's box (http://www.phys.virginia.edu/classes/252/mass_and_energy.html) which is supposed to demonstrate that relativity predicts that the torch will lose mass.

My understanding is that the lost mass is from the lost chemical binding energy, which manifests as mass in the same way as nuclear binding energy does.

quote:

"But anyway, if as you also seem to be suggesting, the energy of the batteries is beign transformed into stuff with mass, congratulations, youve got half way to painlessly converting energy to mass and vice versa, without the need for a nuclear reactor or hydrogen bomb. Now if you can work out how to replicate it on a massive worldwide scale....."
================================================== ======

But we would need the energy of that hydrogen bomb inside the battery to
create much mass. You've got it backwards, nuclear reactors and hydrogen
bombs convert mass into energy, not vice versa. Must be that flip-floping
of frames in relativity that has you all mixed up. Particle accelerators increase
mass by using energy.
A question on your statement "particles accelerators increase mass by using energy."

Are you referring to the accelerated particle? The acceleated electron definitely goes off the 1/2mv^2 curve. I look at this as a loss of effieciency of the particle to intake the energy, suposedly increasing velocity, as the velocty increases. As the enrgy increases so does the vibrational energy of the particle which acts like it cannot take the energy in, use it to increased velociy and then take in some more. It is alsomst as of the particle hasn't enough time ot use the energy as intended before it has to stop , store what is already on board, and go get another load before the first batch can incease velocty as intended. It ends up simoply storing the unused energy intake, that doesn't get used as velocity increasing, into mass which by now is going along for the ride.

Geistkiesel

Exactly as I see it, geistkiesel. I have often 'speculated' on these pages that
the increased energy and mass of the particles was the reason the mean-life
of short-lived particles was increased, not due to 'time dilation.' Even muons
'at rest' have a very wide mean-life range, due to their ENERGY, by my thoughts. Possibly due to near light speeds of charged particles, for instance, through the magnetic field used to guide the particles in the accelerator. The
muon-in-atmosphere effect would be the same, fast moving charged particles
through Earth's magnetic field, increasing their mean-life.

Exactly as I see it, geistkiesel. I have often 'speculated' on these pages that
the increased energy and mass of the particles was the reason the mean-life
of short-lived particles was increased, not due to 'time dilation.' Even muons
'at rest' have a very wide mean-life range, due to their ENERGY, by my thoughts. Possibly due to near light speeds of charged particles, for instance, through the magnetic field used to guide the particles in the accelerator. The
muon-in-atmosphere effect would be the same, fast moving charged particles
through Earth's magnetic field, increasing their mean-life.
If we lookat he energy equation 1/2mv^2 = E and find that m =( 1/2)E/v^2 we need not necessarity look at the V^2 term as returning a number for the speed dx/dt. We can look at it physically and see hat v^2 an very well be a physical operator as oppsoed to t0o a mathematical number generator, It owuld seem reasonable to break the motion of a body into two natural functions, theordinary linear velocity and a vibrational velocit wher ehe speed of he mass in oscialltion is a measure of the mass increase of he paricle, or v^2 = f(V + F),.

Taking as simpole a model as possible we can graph the funcrion scematically as

________________________> V
___<____>_____<___>____< F

where each change of direction of the F term reflects the intrinsic vibration. Notice how on the forward cycle the F and V term add constructively, while on the reverse cycle the F terms subtracts, For low speeds the effect cancels out, or better balances out.

Now how does one constuct a mass ac -> cc inverter/filter analogous to an electronic filter/ Then one could add all the masses in the same direction that is getting wasted as just being there and hanging on for the ride, olr limiting actual speed.

The function written as f(V - F) approaching zero as F => V, but what does this mean? After all it is numerically a squared term, that of motion modulated over time the F - V could get as close to 0 as one was able with out the function blowing up as when v = c in the modern understanding. I'd like to enter one of puppies in the NASCAR tour and kick some butt, that's what I'd do. Think of all the chicks that would be hanging around.

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:
|L_________________M__________________R_|
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:
|----->| vt | vt |||
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.
|vt|<----ct------|-----------> | ||
0 1 2
|<----------|-|-|---------> | ||
|----------> | | <---------|

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.

This is totally missing the point. First you are assuming that light can be accerelated (How else can your photons arrive at the L and R reflectors at different times). Second you do not specify who is measuring the times of arrival of the photons - that makes all the difference and is the whole point of SR. And by the way, what on earth can "a little bit of Dx" mean? If you meant the derivative dx then it is itself the smallest "bit" imaginable - and when you say that "this little bit of Dx/Dt" is all that tells you there is motion, well yes. Velocity is defined as ds/dt, there's nothing else. I suspect you are confusing yourself with an over-elaborate thought experiment. SR is based on only two axioms - constancy of light velocity, and preservation of the laws of nature in all inertial frames. The rest follows as a matter of bare mathematical logic. You dont have to accept the axioms, but it sounds to me as though you have not read the original article. Try it - it isn't hard.

This is totally missing the point. First you are assuming that light can be accerelated (How else can your photons arrive at the L and R reflectors at different times). Second you do not specify who is measuring the times of arrival of the photons - that makes all the difference and is the whole point of SR. And by the way, what on earth can "a little bit of Dx" mean? If you meant the derivative dx then it is itself the smallest "bit" imaginable - and when you say that "this little bit of Dx/Dt" is all that tells you there is motion, well yes. Velocity is defined as ds/dt, there's nothing else. I suspect you are confusing yourself with an over-elaborate thought experiment. SR is based on only two axioms - constancy of light velocity, and preservation of the laws of nature in all inertial frames. The rest follows as a matter of bare mathematical logic. You dont have to accept the axioms, but it sounds to me as though you have not read the original article. Try it - it isn't hard.

Quarkhead, an interesting post, it seems you’ve been digging.
In order for you to understand what I have been doing read he following and then see if you are able to defeat the analytical results with any hing iother than your assesment iof psychiological states of mind. Try a serious physical analysis using laws of physics that are unambiguously proved from experimental results. I am not being confrintational here, in the slightest, but this seems to be the modus operandi of SR theorists protecting their motherlode.
Your question of ‘acceleration’ I don’t quite understand. The left moving light gets to the L detector before the right moving light arrives at the Right detector.

L |____|ß--------------|------|----------à____|____||R
0 1 0 1
Vt ct 0 1 ct 2vt
The sketch shows the fame at t = 0 and t = 1 after the photons emitted from the midpoint of Land R had traveled a distance ct with respect to the midpoint. L has moved toward the oncoming photons while R has moved away from the photon, hence the L detector sees the photon before the R detector.


Now looking at what the right moving photon sees is à_________*_________|||||
Which is identical to the above Where the “*” marks off the distance vt and the “||||” the little dx/dt that the photon must cross when moving through the 2vt distance, remember the fame is moving with respect to the embankment, which we are not concerned with and with respect to the original midpoint defined by the location of the source of the photons, which is an invariant location, for all purposes – theory doesn’t move it, nor do ‘ground zero’ nuclear blasts.

In the little sketch, the photon covers a distance vt + vt + vt’ in the time t’ it takes the frame to move tat little tiny distance (dx/dt)(t’) = vt’ or ct’ = 2vt + vt’, where after the algebra, t’ = t(2v)/(c –v).

If t’ not 0 then motion if t’ > 0 then motion.

The same situation above is seen by the left moving light after reflection and returning to the same position as the original midpoint now moved. Here the left photon arrives at the physical midpoint simultaneous with the now reflected right photon. The clock at the moving frame physical midpoint will record a time t’ as a difference in the process compared to where the frame is stationary on the embankment, or E frame f reference.

In a later “model” I had computers replace the mirrors and record the times of all arrivals of the photons at all detectors including the midpoint. In comparing the database of L and R computers the L detector records a photon arrival before the R computer recorded an arrival. The two photons spend the same transmission time of flight.

Let us assume that the times we discuss here are all with respect to the moving frame which synchronizes the computer clocks and midpoint clock by striking mechanical switches on the E frame that physically toggles all electronic devices simultaneously, or any other synchronization process that would satisfy you.

As the photon arrivals at the L and R computer/detector/reflector are different L, before R, and as the photons arrive simultaneously at the midpoint t’ less than the nonmoving experimental arrangements the moving frame observer sees the L => R direction inferred from the delta t of the L and R computer data.

If you assume, because I will not, that there is time dilation, this does not explain away the difference in arrival of the photons at L and R, which cannot be measured until the photons arrive back at the midpoint. In other words an observer at L cannot inform an observer at R that L got a photon before the R observer can compare and see that the R didn’t get a photon. Information travels at the speed of light, a constant. Frame contraction is useless to explain the different arrival times at L and R, as the contraction is necessarily symmetric.

What is invariant is the midpoint of the original emission of the photons located by its lonesome self. This is easily confirmed where here is an E frame for verification, and this being the case let us just remove all external references when the photos are emitted (notice I said that the E frame only verifies the invariant location of the point of emission of the photons which originally is co-located with the physical midpoint of the frame).

We perform the experiment 10^6 times then, we expose the trick by showing that the E frame was a holographic illusion and that the all 10^6 experiments were conducted in free space.

Now you may have some SR formulae you use to explain the above, so why don’t you compare some arbitrary velocity ‘v’ and total length of the frame D, which I didn’t use. All measurements are photon reference with the photon's point of emission. I do not mean the speed of light or hat photon is the reference point I mean the point of origin of the photons, which is not a moving object, not subject to the motion of the source (use an E frame source of light, the result is the same naturally), nor of any force, as that emission point is more an real coordinate frame, reality, as opposed to the abstract and mathematically contrived substitute, including Galilean coordinate frames of reference which ought to be discarded, immediately.

Caveat emptor: Do you realy believe that it is impossible to measure unaccelerated motion in umiformly moving systems then
scroll down to the appropriate frame (http://ourworld.cs.com/sandgeist/myhomepage/homepage.html) This might put a chink in your armor (very doubtful), if you can understand the process, that is. Most SRists who have looked at it answer with scorn, which leaves me believing they don't understand physics, whether there are errors or not.

Thank you Quarkhead for your interest in the post.

Geistkiesel

I have a suggestion: we should close this thread and continue all aspects of critics of SRT in one thraed, let's say "Mathematical Challenge to the Consistancy Claim of SR". It will be much more productive...
Why not a "Physical Challenge", does the word "Physics" frighten you Yuriy?

Why don't you just start a thread on your own and quit asuming you have some kind of special knowledge or power to assume control over this forum? If you are so smart and knowledge ale form your thread in words that will truly make your theses hold? You run from ideas that are, "sehr fremde, nicht var?"

You always want to close something that is making you nervous Yuriy, why is that? Anger?

Geistkiesel

Look on the entrance page of our Forum and you will (I hope so...) recognize why...
But the worse thing is that in all these threads we see the same people with the same arguments that by necessity require the same line of logic. Why we do not collect all these issues in one thread? Why we have to see your 'invention" in numerous threads, if nothing new is appearing in it principal?... And read what people say about our Forum because of that.
And do not be afraid about me: the last time I was scared many, many years ago... And notice: I never said anything about your features; by one simple reason – I do not care about that…

Look on the entrance page of our Forum and you will (I hope so...) recognize why...
But the worse thing is that in all these threads we see the same people with the same arguments that by necessity require the same line of logic. Why we do not collect all these issues in one thread? Why we have to see your 'invention" in numerous threads, if nothing new is appearing in it principal?... And read what people say about our Forum because of that.
And do not be afraid about me: the last time I was scared many, many years ago... And notice: I never said anything about your features; by one simple reason – I do not care about that…

Why don't you just accept the fact that there are some out here that don't by into your belief system. Your tired theories and BS stink Yuriy. Yuriy for all your intrusions you ahve not taen one, not a single point I have made in many posts and found a fault that you can defeat with application of the laws of physics.

How would you react to the suggestion that the tired old SR theories be excluded from this forum, or otherwise obstruct the flow of SR information and discussion? Go fuck yourself Yuriy.

Why don't you just accept the fact that there are some out here that don't by into your belief system. Your tired theories and BS stink Yuriy.The problem here is that it doesn't matter if you buy into it or not. Unless you can show it is flawed, your opinion doesn't matter. It's about predicting what happens... and relativity does that job well.

The problem here is that it doesn't matter if you buy into it or not. Unless you can show it is flawed, your opinion doesn't matter. It's about predicting what happens... and relativity does that job well.

I've showed SR fatally flawed to my understanding. This is amplified by the fact that only on rare occasions has anyone attacked what ever theses I was discussing within the four corners of the document. I must add that this modus operandi of the SR industry is not just seen by myself. It is painfully obvious. Most SR obsjections are either in the form of personal attacks or talking in terms that add nothing to the discussion such as how many experiments prove this or that, as opposed to bringing the substantive issue s of those experiments to the forum table drectly.

Your one liners are probably a relief to some, for their brevity if nothing else. However, Persol, the subject of the post to which you are referring is dangerously clsoet scuttling SR, and knowing that SRists will not admit to such gross personal mismanagament of their individual intellects that the problem will be with us for some time. If you are looking for an answer quieting the SR disturbances you will be waiting a very long time, unless that is, the SR industry can form physical arguments using other than their precious
formulae.

In my model that I diagramed above can you see where I said anything indicating that the moving frame was accelerated? This is what QuantumQuark assumed by one of his questions, as I understood him. He still hasn't answeredn my post in reply to his post. Oh well.
Geistkiesel

Geistkiesel

In my model that I diagramed above can you see where I said anything indicating that the moving frame was accelerated? This is what QuantumQuark assumed by one of his questions, as I understood him. He still hasn't answeredn my post in reply to his post. Oh well.
Geistkiesel

If youaer referring to me Geist I would apologise but am unable to find where I stated such......maybe you can point it out for me...?

If youaer referring to me Geist I would apologise but am unable to find where I stated such......maybe you can point it out for me...?
Only Geistkiesel could confuse Quarkhead with Quantum Quark. Just between you and me QQ, I thinks there's a conspiracy con-fuzh-un, me 'an U. .
All is well. The guards were alert at their posts and intercepted an intruding interloper, playing like he was a ghost, Ha, Ha..
Geistkiesel

Quarkhead, an interesting post, it seems you’ve been digging.
In order for you to understand what I have been doing read he following and then see if you are able to defeat the analytical results with any hing iother than your assesment iof psychiological states of mind. Try a serious physical analysis using laws of physics that are unambiguously proved from experimental results. I am not being confrintational here, in the slightest, but this seems to be the modus operandi of SR theorists protecting their motherlode.
Your question of ‘acceleration’ I don’t quite understand. The left moving light gets to the L detector before the right moving light arrives at the Right detector.

L |____|ß--------------|------|----------à____|____||R
0 1 0 1
Vt ct 0 1 ct 2vt
The sketch shows the fame at t = 0 and t = 1 after the photons emitted from the midpoint of Land R had traveled a distance ct with respect to the midpoint. L has moved toward the oncoming photons while R has moved away from the photon, hence the L detector sees the photon before the R detector.


Now looking at what the right moving photon sees is à_________*_________|||||
Which is identical to the above Where the “*” marks off the distance vt and the “||||” the little dx/dt that the photon must cross when moving through the 2vt distance, remember the fame is moving with respect to the embankment, which we are not concerned with and with respect to the original midpoint defined by the location of the source of the photons, which is an invariant location, for all purposes – theory doesn’t move it, nor do ‘ground zero’ nuclear blasts.

In the little sketch, the photon covers a distance vt + vt + vt’ in the time t’ it takes the frame to move tat little tiny distance (dx/dt)(t’) = vt’ or ct’ = 2vt + vt’, where after the algebra, t’ = t(2v)/(c –v).

If t’ not 0 then motion if t’ > 0 then motion.

The same situation above is seen by the left moving light after reflection and returning to the same position as the original midpoint now moved. Here the left photon arrives at the physical midpoint simultaneous with the now reflected right photon. The clock at the moving frame physical midpoint will record a time t’ as a difference in the process compared to where the frame is stationary on the embankment, or E frame f reference.

In a later “model” I had computers replace the mirrors and record the times of all arrivals of the photons at all detectors including the midpoint. In comparing the database of L and R computers the L detector records a photon arrival before the R computer recorded an arrival. The two photons spend the same transmission time of flight.

Let us assume that the times we discuss here are all with respect to the moving frame which synchronizes the computer clocks and midpoint clock by striking mechanical switches on the E frame that physically toggles all electronic devices simultaneously, or any other synchronization process that would satisfy you.

As the photon arrivals at the L and R computer/detector/reflector are different L, before R, and as the photons arrive simultaneously at the midpoint t’ less than the nonmoving experimental arrangements the moving frame observer sees the L => R direction inferred from the delta t of the L and R computer data.

If you assume, because I will not, that there is time dilation, this does not explain away the difference in arrival of the photons at L and R, which cannot be measured until the photons arrive back at the midpoint. In other words an observer at L cannot inform an observer at R that L got a photon before the R observer can compare and see that the R didn’t get a photon. Information travels at the speed of light, a constant. Frame contraction is useless to explain the different arrival times at L and R, as the contraction is necessarily symmetric.

What is invariant is the midpoint of the original emission of the photons located by its lonesome self. This is easily confirmed where here is an E frame for verification, and this being the case let us just remove all external references when the photos are emitted (notice I said that the E frame only verifies the invariant location of the point of emission of the photons which originally is co-located with the physical midpoint of the frame).

We perform the experiment 10^6 times then, we expose the trick by showing that the E frame was a holographic illusion and that the all 10^6 experiments were conducted in free space.

Now you may have some SR formulae you use to explain the above, so why don’t you compare some arbitrary velocity ‘v’ and total length of the frame D, which I didn’t use. All measurements are photon reference with the photon's point of emission. I do not mean the speed of light or hat photon is the reference point I mean the point of origin of the photons, which is not a moving object, not subject to the motion of the source (use an E frame source of light, the result is the same naturally), nor of any force, as that emission point is more an real coordinate frame, reality, as opposed to the abstract and mathematically contrived substitute, including Galilean coordinate frames of reference which ought to be discarded, immediately.

Caveat emptor: Do you realy believe that it is impossible to measure unaccelerated motion in umiformly moving systems then
scroll down to the appropriate frame (http://ourworld.cs.com/sandgeist/myhomepage/homepage.html) This might put a chink in your armor (very doubtful), if you can understand the process, that is. Most SRists who have looked at it answer with scorn, which leaves me believing they don't understand physics, whether there are errors or not.

Thank you Quarkhead for your interest in the post.

GeistkieselQuarkhead, an interesting post, it seems you've been digging. In order for you to understand what I have been doing read he following and then see if you are able to defeat the analytical results with any hing iother than your assesment iof psychiological states of mind. Try a serious physical analysis using laws of physics that are unambiguously proved from experimental results. I am not being confrintational here, in the slightest, but this seems to be the modus operandi of SR theorists