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	Look at Einstein's equation for x: x = x' + vt' (1 - v²/c²)^1/2 You can see that to calculate relative x he must be given x'. What is x'? It is local x. x measured by the train. Einstein's givens are exactly the same as mine. My question remains, though: what is v in his equation? It can't be relative v. Einstein has not derived relative v yet. It must be local v (as measured from the train). But if this is the case, it should be labelled v', as in my equations. To put it all very simply, Einstein does presume an underlying grid-- like my marked-off tunnel-- and what determines this is his given, v. As I have shown, this v is not the velocity in S. That is what we are seeking, the relative velocity. Therefore, his given v must be the velocity in S', and should have been labelled v'. The reason this v' contains a presumption of the underlying grid is that v' entails an x'. If you are given a v' in this situation, then it is understood that you are given x'. x' is the grid. v' and x' are dependent on the grid. One may ask at this point whether the grid is part of S or S'. Or is it a third, independent, S? It seems that we are measuring both S and S' against it. And it seems that the grid would be a part of S, since the traintracks are attached to the embankment, as it were. The tracks do not move. The tracks are, in fact, connected to the embankment by rigid rods. But the underlying grid is not strictly equivalent to S. S is an outcome of the measurement. S is created by the action of the embankment seeing the train. But the grid is not an outcome of the problem; it is a postulate of the problem. It is a precondition. The grid (my tunnel, Einstein's traintracks) is neither S nor S'; but it generates them both. To be technical, a grid (or a co-ordinate system) applies only when there are no velocities or accelerations. That is why Einstein specifies a rigid reference body, a rigid co-ordinate system. Moving objects cannot be connected to grids by rigid rods. A grid is a first-degree postulate. Meaning that it can generate x's, but not v's or a's. But neither S nor S' in this problem are first-degree postulates or grids. You can think of S and S' as second-degree co-ordinate systems-- velocity grids. They are generated grids. They cannot be measured directly; they must be calculated. Let S₀be the first-degree grid, the rigid co-ordinate system of displacement. Displacement is a primary, or first-degree, measurement. It is independent. It may be measured directly, without calculation, and without looking at any other quantity (specifically "time"). Velocity is a secondary, dependent, variable, as we know. Therefore, S and S' must be generated by S₀. S' is the velocity grid of an object as measured against S₀by the moving object itself. S is the velocity grid of an object as measured from a distance, by an observer who is not part of the object. I believe this way of looking at it clears up much confusion. One might say it is the key to understanding the problem as introdced by Einstein. But others might complain, "Why have S₀at all? It seems like it is the same as S. All this first-degree and second-degree stuff is just folderol. Needless complication." You can see how important it is if you ask which system, S or S', the grid belongs to. Remember that x' from above? The one we seemed to transfer willynilly from S to S'? Which co-ordinate system did it really belong to? At first it seemed to be the system of the blinker, since the blinker measured itself against it. But then it seemed to be a part of the system of the eye, since the eye and the tunnel (like the embankment and the traintrack) seemed to be connected to eachother. It seemed like the tunnel was just an extension of the system S. The answer is that the x' doesn't really belong to either S or S'. It belongs to S₀. This becomes clearer once we see how our two x-variables behave in their corresponding systems. The variable x that we used in our equations truly is the x in S. S is defined as how S' looks from a distance. S' is moving with regard to S. So S is how a moving grid is measured by a non-moving observer. x is the apparent displacement, due to velocity, of the whole system S' relative to S. But x' is not really the variable x' in S'. It is not the x-variable that belongs to S'. Not in the way that x₀is the x-variable that belongs to S₀. The x-variable in S' can be measured in S' only if it is not moving. If it is moving, then it automatically becomes another system, as Einstein rightly told us. An S", say. In other words, the x-variable in a system is used to measure non-moving things in its own system. It is a measurement of displacement or length, not a sub-measurement of velocity. x' in all our equations has not been an x of this sort. x' is a measurement by S' of its own displacement-- that is, the displacement of all of S'. x' is an external number to S', as I have said. It is a given. It is not measured by S' within S'. It is a received value. We should label it x₀, to be precise. I will now re-run the equations for you with the proper subscripts, to clarify the preceeding paragraphs. But I have tracked the problem as I have, leaving the S₀system out of it until now, for a reason. I believe it was necessary, so that you see the necessity for the labelling-- and so that you could come to see the complications of the problem just as I did, in the same order. You must first comprehend that Einstein's given v implies the existence of the tunnel, or traintracks. Only then can you see how this sub-system-- or primary system-- S₀influences the whole problem. Einstein did not recognize that his given v carried with it so much baggage, and this has been what has kept Special Relativity under a cloud for almost a century. So, to run the equations again: We have only one moving object (the blinker), whose velocity we want to measure from a distance. To do this properly we need three co-ordinate systems. We must have S₀, S, and S'. S₀is the base system or sub-system or non-velocity system that generates the two relative velocity systems S and S'. This S₀is equivalent to the non-velocity system of the eye, but should not be confused with S, since S is the how a moving object looks to the eye. S₀is how a non-moving object looks to the eye. Now, we are given two things. 1) The co-ordinate system S₀, whose measurements we know. 2) The rate of a clock at rest relative to S₀, which we will call t₀(even though, strictly, S₀has no time-- it is a displacement grid only). t₀= the initial period of both clocks. This is the period measured when the two clocks are side by side at the beginning-- WHEN THEY ARE NOT MOVING. x' = distance blinker has gone relative to tunnel marks, according to its own visual measurements. v '= velocity blinker is going, by it's own calculation. t = period that the eye sees blinks from blinker. This gives us the relative period. v = velocity eye calculates blinker to be going, based on visual evidence. This is the relative velocity. If you are with the blinker, then you will measure your own velocity like this v' = x'/t' But, x' = x₀since the blinker is at x₀^N when it records x'^N[read x prime to the N power]] So, in the equation v' = x'/t' we may use the displacement data from the S₀system-- which we have been given. v' = x₀/t' Likewise, t' = t₀, since t' remains a local measurement whether the blinker is moving or not. The blinker has no velocity, relative to itself; therefore its own clock remains a constant, to itself, whether at rest or in motion. Therefore t = t₀+ (change in x₀/c) v = x₀/t x = v/t₀ But, t₀does not = t and x₀does not = x Now your question may be, "Isn't there another way to measure velocity? You have given two methods, two operations, for determining velocity, and they both seem to require outside information. You have shown that relative velocity is dependent on local velocity (v is dependent on v'), and that local velocity is dependent upon an external grid defined as unmoving (v' is dependent on x₀). Isn't there a more straightforward way? What about the velocity in S₀? Can't we just divide some x₀ by some t₀(or differentiate something) and get a non-dependent number?" No, we can't. There are no velocities in S₀. That is the whole point of Einstein's two systems. Remember, the train is moving relative to the embankment. S is the embankment, S' is the train. As soon as you have a velocity, you must have a second system, or sub-system. Einstein himself created S' to explain velocity, and he was corect to do so. Of course you can divide x₀by t₀, if you really want to. But if you do you are back to Newton and the Galilean idea of velocity. For I have shown that x' is really x₀. And t' was defined as local time, either in S or S'. So x₀/t₀= x'/t'. If you, as the observer, measure the blinker from a negligible distance, you will get the same value for its velocity as the blinker gets for itself. This is the common idea of velocity, and is the reason why we don't usually differentiate between your measurement of your velocity and my measurement of your velocity. But remember that you always see with light, even from the shortest distances. An observed velocity will always be a relative velocity, to some degree. ~~~~~~~~ From this explanation, I think you can begin to see that the greatest problem with Einstein's transformation equations is that there is no equation for deriving the velocity of the train relative to the embankment. This is the central question of his thought problem, and yet he never answers it. It has gone unanswered and unnoticed for a century. Einstein's confusion on page 18 (Rel.) buries the given velocity (my v') so that Einstein and every other reader has forgotten it exists. Then he conflates the velocity of the train with the velocity of the man on the train, using them interchangably. Finally, he derives a relative velocity for the man on the train, never noticing that the train has a relative velocity of its own. For a century we have had no equation for the train's relative velocity. And no one has missed it. Physicists now routinely use the Lorentz transformation for v as if it were a transformation for one degree of relativity. But it is not. It is a transformation for two degrees of relativity: the man to the train and the train to the eye. Let us look at the current Lorentz transformation for velocity. W = v' + v 1 + v'v/c² This was obtained by differentiating the Lorentz equation for x with respect to time. But notice there are three variables here: W is the velocity of the man as measured from the embankment v' is the velocity of the man relative to the train, and v is the velocity of the train relative to the embankment, measured by the train In order to calculate W, Einstein must be given v' and v. That is a lot of information. And that information implies even more information (such as x' and x). If the man and train are sending information (lightrays) to the measurer on the embankment, which allows him to calculate variables such as velocity, then the measurer can easily calculate-- using his givens-- what the man and train are measuring for themselves. But subsequent scientists have acted as if Einstein knew nothing about the local situation of the man and the train. As if he was getting all his information from "visual" data. As if he was deriving his equations almost literally out of thin air. He was not. His transformation equations depend upon a set of givens, just like any other transformation equations. And his givens have turned out to include what might be called "absolute" information, as I have shown. Without this "absolute" information, no equations would be possible. Without a traintrack already known to both the train and the embankment, the measuring eye on the embankment could never have derived anything at all. His knowledge would be as limited as the knowledge of the man watching the pulse clock pass him by at night. Relative Velocity of an Approaching Object Now let us take the case where the blinker is traveling at a constant velocity toward the eye. Relativity tells us that t slows down there, too. But forgive me for not accepting that at face value. Let us first calculate when the blinks appear to arrive, in order to get t, as in the first experiment. At the first blink we see T₁ = T₁' + x'/c. Light still has x' to go, so we assume the first blink will be late relative to T₁'. At this point, you may start to think maybe Einstein was right. But be patient. For at the next interval, x' is 1km shorter. That is, the blinker is closer by 1km at the next blink. @ blink #2, T₂ = T₂' + x₂'/c where x₂' = x₁' - 1km @ blink #3, T₃ = T₃' + x₃'/c where x₃' = x₁' - 2km So, let's put in some numbers, and see what is happening. Assume the blinker starts 101km away and reaches v' instantaneously. At 101km, T' = 0 So, x₁' (x' @ T' = 1) = 100km, x₂' = 99km, etc. T₁ = 1 + (100km//300,000km/s) = 1.000333...s T₂ = 2.000330s T3 = 3.000326...s T₄ = 4.000323...s T₅ = 5.000320s What is the period here? t = T₅ - T₄ = 5.000320 - 4.000323... = .999996...s Exactly what we should have expected. If t(d) = period of departing blinker and t(a) = period of approaching blinker t(d) = 1/t(a) And, t(d) = t' + x'/c, then 1/ t(a) = t' + x'/c t(a) = 1/(t' + x'/c) = 1/[t'(1 + v'/c)] v = x'/t = x/t' v = x't'(1 + v'/c) v(a) = v' + v'²/c or, v' = [(c² + 4cv)^1/2 - c]/2 This is what I expected, from the Doppler effect. But notice that it is not the same as what Einstein predicted, and what Relativity now tells us. Time appears to speed up with objects that have a velocity approaching us. Around 1700, the Danish astronomer Ole Roemer measured the period of Io (Jupiter's moon) and proved the above assertion. When Io is moving toward the Earth, the period appears shorter. This is known and has been accepted all along, even while Relativity has tried to tell us that all moving clocks slow down. Einstein apparently did not realize that the period of Io is a clock. Scientists have never resolved these two accepted facts. If you cannot see Io as a clock, consider the binary pulsar PSR 1913+16. It revolves around its mate much like Io moves around Jupiter. It was discovered to be binary precisely because its pulses speeded up and slowed down, as in an orbit. It is a clock. It speeds up when it is moving toward us. This is admitted by everyone. This contradicts the current interpretation of Special Relativity. Relative Velocity measured at an Angle to the Line of Sight Now let us ask about velocities that are at an angle to the line of sight. This will be somewhat trickier than it seems, for this reason: Notice that the apparent change in the period of a moving object is dependent on the change in x'. Therefore, if x' does not change, then the object's period will not appear to change, and it's clocks will not appear to slow or speed up. But the line of equal distance from a stationary observer is a circle around that observer. An eye or a telescope will turn with an object moving at an angle, to allow the lightrays to continue to enter the eye directly. The angle of the object, then, must be measured relative to this turning eye. You will see what I mean as we get into the experiment. Say our blinker has an initial angle to our eye of 45^o at T' = 1 and a velocity of 10,000km/s. What is it's apparent period and apparent velocity? We desperately need an illustration here. @ T' = 1 (blink 1) let x₁' = 100,000km (so that we won't have the tiny fractions) What is x₂' @ T' = 2 ? y_n = v't_n' so, y₁ = 10,000km Obviously, angle L = 135^o so, x₂'²= x₁'² + y₁² - 2y₁ x₁cos135^o x₂'= 107304.3036km x₃' = 115014.8996km x₄' = 123055.4375km x₅' = 131365.3465km And, to find the apparent t, we use the equations we already have. T₁ = 1 + 100,000/300,000 = 1.333333s T₂ = 2.35768 T₃ = 3.383383 T₄ = 4.410185 T₅ = 5.4378845 And, apparent period t @ T₂ = T₂ - T₁ = 1.024347 @T₃ = 1.025703 @T₄ = 1.026802 @T₅ = 1.0276995 The important thing is, it is clear that the period will appear to be getting slower as the object moves away. The apparent period is not a constant in this experiment. It starts out a bit slower than t' and then continues to get even slower. But we expect it to approach a limit at t = 1.03333. Because, at infinity, it will be moving directly away. And then it will be equivalent to our first experiment. The blinker appears to get progressively slower, approaching a limit at t = t' + x'/c. It's apparent velocity depends on it's distance away. Its angle to the line of sight decreases as it departs. For an object approaching at the same trajectory, the opposite applies. At infinity it has an apparent period of t = t' + x'/c. That period decreases until the object hits the tangent of its trajectory (see illustration). In our current problem, the blinker will hit the circle at sin45o = r/100,000km where r is the radius of the circle so, r = 70711km. T= T' + .236 at the tangent. So, with some more math, we could figure out the minimum apparent period. Obviously, it is > 1, and < 1.33. Notice that beyond point D, the blinker becomes a departing object again. ~~~~~~~~ From these thought experiments, I think you can see that Special Relativity is incomplete. It states that moving clocks slow down. It does not take into account trajectory or whether the object is moving nearer or farther away. With regard to approaching objects, the Lorentz equations are flatly wrong. With regard to objects on an angled trajectory, the Lorentz equations may occasionally be a good approximation, depending on the angle at the time of measurement. Einstein himself was never clear on the implications of his theory for objects approaching us. In conversations with Karl Popper in the 40's, for instance, Popper asked him about the twin paradox. On the question of whether the time dilation coming and going would resolve, Einstein admitted he did not know. In this particular conversation he doubted the truth of the twin paradox, but never presented any equations for or against it. My theory puts the twin paradox to rest, I hope. Also notice that this New Relativity implies that objects in simple orbit do not experience time dilation, since their distance from the observer does not change. I do not have space to initiate a full discussion of this here, but those who would point to data from synchrotrons should be aware that only an observer at the center of the circle, receiving data directly (radially), would apply to my statement here. ~~~~~~~~ Now let us go back to my equation for velocity. Notice how close it is to the Lorentz transformation. my equation v = v' 1 + (v'/c) Einstein's v = v' + v" 1 + (v'v"/c²) In fact, if you think of the denominator as 1 + (v'/c)(v"/c), and you get rid of v" in the numerator and denominator, it is the same equation. I consider this strong evidence in favor of my claim that I have proceeded much like Einstein, making the same assumptions and accepting the same givens-- including the given of x'. I have just done so more explicitly But I claim that Einstein arrived at his equation by a rather circuitous route. It is obvious that if you take his equation for x and his equation for t and combine them without differentiation, like this v = x/t v = x' + v't' (1 - v ²/c²)^1/2 t' + v'x'/c² (1 - v²/c²)^1/2 v = x' + v't' = 2v't' = 2v' t' + v'x'/c² t'(1 + v'²/c²) 1 + v'²/c² you do not get the same equation. Close, but not the same. All the square roots get cancelled out, but it is still the wrong equation. But by differentiating, Einstein also fortuitously gets rid of all the square roots. The tracks are covered. And he gets an equation that now seems to work. But it only works by misuse. The process of differentiation transformed the equation into an equation for two degrees of relativity, as I have shown. That is why it has three velocity variables. But it is now routinely used for one degree of relativity. Please notice how directly and cleanly I got to my equations. My equation for relative t is simple and straightforward. Likewise my equation for relative x. And my relative v is simply x/t'. I do not need the Pythagorean theorem, or any of the ridiculous illustrations and concepts that explain it. I do not need gamma, which has proved to be an ad hoc invention. And I do not need calculus to solve an algebraic problem. Furthermore, Einstein's velocity equation is not correct for two degrees of relativity either. If it is so nearly correct for one degree of relativity, I think you can see that it will not work for two degrees. To prove this, let us make quick work of the addition of relative velocities. The Addition of Velocities I am not going to derive the equations for all trajectories. I hope you can see that that would be very complex-- much more complex than Einstein has admitted. For it depends on the trajectory of A to B and B to C. There are many possible combinations, and one equation cannot possibly cover them all. Most linear trajectories, however, will be covered by combining the three different trajectories I have provided for a single relative velocity. For now, let us ask about the situation where both velocities are receding from the observer in the same line. The question is, how do you transform the Galilean equation x/t = x'/t' + x"/t" into a relativistic equation? Notice that, logically, you must have five sets of variables: 1) The man's velocity measured by the man. 2) The man's velocity as seen from the train. 3) The train's velocity measured by the train. 4) The train's velocity as seen from the embankment. 5) Only then can you ask about the man's velocity as seen from the embankment. Let us say that you ignore all local events, as Relativity tries to do now (actually, it simply confuses local measurements with observed measurements, not even realizing the difference). If you are given the relative velocities to start with, then you can throw out 1) and 3) above. But you still have three sets of variables and three clocks, none of which are equivalent. So, what if we are given the two relative velocities, 2) and 4) above? Let v of A rel B (man to train) = v" let v of B rel C (train to embankment = v' what is v of A rel C (man to embankment)? = v What one is tempted to do is just start juggling equations, which is what everyone has done up to now. But let us stop and ask what is happening. What are we really trying to find? We are already given the relative velocities, so we do not need the equations we have discovered up to now to get them. What we need to do is visualize the problem in concrete terms. Let us start with another illustration. This always seems to help. If we know how A is observed from B, will that tell us anything about how A should be observed from C? Yes, but only indirectly. Indirectly, because remember we are dealing with observation by the use of light rays. In the observation of A from C, the light rays will travel directly from A to C. They will not necessarily pass through B. B has its own light rays from A that it is dealing with. But we should only be concerned with the light rays coming to us. That is, visual observations are made directly, and indirect evidence is dangerous in relativity. As we saw with Michelson/Morley, it can get us into trouble. We must deal only with our own light rays, the ones entering directly into our eyes. The relativity equations apply only to these rays. This is not so clear when you are dealing with relative velocities all in the same line. In this case, the light rays do pass through B. But this will not always be the case, obviously. Knowledge of A relative to B can give us A relative to A. With that knowledge we can calculate A relative to C. Like this: We are given v". Let v''' = the velocity of A measured by A. then, v''' = v''//1 - (v''/c) And we can calculate the velocity of B measured by B in the same way. If v'''' = B measured by B, then v'''' = v'//1 - (v'/c) The velocity of A to C measured by A, if ABC is a straight line, would be v = v'''' + v''' 1 + [(v'''' + v''')/c] = [v'//1 - (v'/c)] + [v''//1 - (v''/c)] 1 + {[v'//1 - (v'/c)] + [v''//1 - (v''/c)]/c} If v' = v" = .4c then, v = .57c The Galilean transformation for this problem would have given us .8c. The Einsteinian transformation would have given us .69c. ~~~~~~~~ We are now in a position to critique another equation of Einstein's. That is the equation W = c - v. On p. 18, Relativity, he defines the variables like this. c is the velocity of light relative to the embankment. v is the velocity of the train W is the velocity of the light relative to the train. He then attempts to show that this equation is incompatible with the constancy of the speed of light. He says you cannot subtract v from c, because then W would be smaller than c. He says, "If a ray of light be sent along the embankment, we see that the tip of the ray will be transmitted with a velocity c relative to the embankment." But this is simply not true. We would see no such thing. The constancy of the speed of light requires that we measure every light ray as going c (that is, every ray that comes to us.) And that the observer on the train do likewise. It says nothing about imagining light rays that we cannot see. A light ray moving along the embankment is not part of our possible data: we make a mistake if we try to plug imagined numbers into our transformation equations. The theory of relativity cannot require that we imagine every possible light ray as going c relative to every other object. This would require stopping all the objects in the universe-- except the photons. Relativity only requires that we see ourselves as stopped with regard to light. And that we calculate that every other object also sees itself as stopped. Einstein here makes the very same mistake that Lorentz made in "visualizing" the interferometer problem. He tries to see the light from both systems at the same time, and in so doing he mixes his variables. For please notice that W in this situation is not in fact the velocity of the light relative to the train. It is the velocity of light relative to the train as imagined from the embankment. It is the embankment trying to see through the eyes of the train. But an observer on the train would not use this equation in obtaining a velocity of light relative to the train. This is because the train has no velocity relative to itself. It would not use the variable v at all. The train would measure the velocity of light directly. The truth is that the embankment is free to imagine W as being less than c, if it wants to. It is perfectly allowed for an observer to calculate that an observed object has a velocity relative to light. It is done all the time. If nothing could be seen to move relative to light, nothing could be seen to move, period. Conclusion What does all this tell us about the nature of light? And what does a correction of Special Relativity imply about the theory of relativity as a whole? We will not see what it does to General Relativity, specifically, in this paper. But for now, we should have noticed that the speed of light is a local measurement. It is not itself a relative measurement. In this sense, Einstein really did "ride his ray of light." We know the local velocity of light, thanks to Special Relativity. When Michelson measured the speed of light from Mt. Wilson to Mt. Baldy, he sent a light ray over and back (with a mirror). So he was coterminous with the ray at the beginning and the end. There was no distance between the observer and the observed events. Michelson had to measure light this way, using a mirror and a single point for start and finish: not because it was more expedient, but because there would be no way for him to know when the light left Mt. Wilson if he was at Mt. Baldy waiting for it, or vice versa. You might say, "He could have had a cohort at Mt. Wilson, writing it down. Or this cohort might have signalled him." But then he would have had to know the difference between t and t': as in my thought problem, this would require knowing c. And any signal would have been a reductio ad absurdum. What would the cohort have signalled with-- a light ray? This means that Einstein's postulate that the speed of light was an absolute turns out to be true, in the only possible way it could be. The speed of light is never an observed event, therefore it will never vary from different points of view. It never takes place at a distance. It is always coterminous, to every local observer. When you see a light ray, it is always right upon you! And that is why, when you see with light, your background always appears to be stopped. You measure the speed of light with regard to yourself, and you cannot have any velocity with regard to yourself. Your system of coordinates is also stopped with regard to you (that is what makes it yours, of course), so you will always measure light the same way. And you will always measure light the same as everyone else. They also see themselves and their coordinate systems as stopped. Their measurement of light has to be a local measurement, just like Michelson's, and every local measurement is made against a stopped background. Remember this last point when thinking about the M/M interferometer. What was being attempted was a non-local measurement of light. Michelson/Morley were trying to "see" light from outside their own coordinate system. They only failed to see themselves moving with regard to themselves. This null set should not have been quite so shocking to the world. And finally, if you have been paying close attention up to now, you will have noticed something else remarkable. We have seen that my equations are nearly identical to Einstein's. I have followed his overall conception closely: we have time dilation and length dilation and equations that treat them similarly. Length contracts as time dilates. I have thrown out the Pythagorean component as untenable, but this has not affected the basic content of the equations. However, my thought problem adds a twist to the whole conception of time. I started out by making t' the given, rather than t (Einstein did this, too; but not so obviously as I did). Normally, in observing an object, we would not be given the object's own period. We would observe the period. This observed period is t. Then with c and x' we could derive the rest. This is what is done in scientific observation. But the example of the blinker has shown us more clearly that t implies a t', and that this t' applies not just to the eye, but to the blinker as well. This should have become crystal clear when we started asking whose clock was actually ticking t? The answer was, neither one. The eye saw the blinker ticking t. The blinker would have seen the eye ticking t. But each would see themselves ticking t'. Einstein's own thought problem-- which I have simply made more transparent here-- implies that in order to measure an observed velocity as dilated, one must assume that it is locally non-dilated. Look again at Einstein's equation for t: t = t' + v'x'/c² (1 - v²/c²)^1/2 What is t' here? According to Einstein's own illustration (p. 32, Rel.) t' is the time in S'. S' is the co-ordinate system of the train. That is, t' is the train's local time. This is just as it was in my thought problem. But this is not how Special Relativity has come to be interpreted. Once all the equations are solved, Einstein and everyone since has applied t (not t') to the train. The train's clock is seen to be dilated. "It is going slow." So t now belongs to the train. Before the calculation, t' was defined as the time of the train. Afterwards t is defined as the time of the train. And then t' is forgotten (or given to the embankment). And if someone clever notices this and says, "Yes, but doesn't the train only appear to be going slow?", the modern scientists say "Don't be a classicist, we only know what we observe!" And I say, "We only know what we observe and what we were given in the first place." In order to calculate the relative slowing of an observed clock, you must assume that clock is locally equivalent to your clock. What determines this equivalence? Or, to put it another way, what makes that assumption true? The speed of light itself! If the speed of light is a constant, as Einstein assumes in Special Relativity, then all local clocks will also be constant-- they will have the same period. In both my equations and Einstein's, c works as a local clock setter. The very form of the equation determines this. The reason both Einstein and I could transfer that x' into our relative equations is that the constancy of light allows us to. c is the bridge from one co-ordinate system to the other. By Einstein's definition, light travels the same x in every local system. Look at the equation c = x/t : if c is a constant, and x is a constant, then t must be a constant. ~~~~~~~~ Now you may ask, "If you have just proven that time is a constant in all local systems, how can you say that you agree with Einstein, or that you admire him? Isn't your paper a direct contradiction of Relativity?" No, it isn't. What this paper shows is that Relativity is a fact at the same time that t is a constant in all local systems. This paper is not a contradiction of Relativity, it is a re-interpretation of Relativity. No one before Einstein had ever theorized that observed data were relative data, and no one had attempted to derive equations that allowed an observer to calculate the degree of relativity. These transformation equations are very valuable, and they will be even more valuable now that they are corrected and correctly interpreted. For it is now clear that Relativity allows us to calculate local conditions from observed conditions. Up to now, it was thought that there was no direct link between your local conditions and mine. Relativity was interpreted to mean that there was only observation. "Reality" was thought to be permanently hidden, or even non-existent. But this interpretation had no basis in Relativity. When scientists used Relativity to confirm the old saw that "I cannot see through your eyes," they were forgetting that the transformation equations, read in reverse, allowed one to do just that. That is, if I can calculate x from x', then I can also calculate x' from x. x is how I see the distance. x' is how you see the distance. I can see through your eyes. ~~~~~~~~ I predict a final complaint. Some will say, "The Lorentz equations are not even used to calculate the speed of satellites and such things. We use General Relativity and Gaussian fields and tensor calculus and other tricks way beyond that Special Relativity hubbub." My answer to that is that Einstein saw Special Relativity as the limiting case for General Relativity. The equations you are using, whatever they now are, are subtle corrections of Einstein's and Grossmann's Riemann-Christoffel tensor equations, which themselves took the Lorentz equations as a starting point. Any correction in these equations of Special Relativity will imply a corresponding correction in General Relativity. Einstein's mass-increase equation is the first thing that will go by the boards. The equation E = ymc no longer pertains, since I have demonstrated that y (gamma) is a phantom. Also, Minkowski's spacetime equations rely on the Lorentz equations. Minkowski's spacetime interval = [x² - (tc)²]^1/2. Clearly this equation is derived using gamma, and must now be thrown out. As must several of Einstein's g_mn equations (not to mention Feynman's Diagrams, etc.). In my paper on gravity I will show precisely how a corrected Special Relativity affects a corrected General Relativity. Inferences First, the twin paradox relies on the assumption that all moving clocks slow down. In the second part of my thought experiment, I proved that all approaching clocks actually appear to speed up, and that the rate of increase is inversely proportional to the rate of a receding clock. That is, t(receding) = 1/t(approaching). This contradicts the twin paradox. You may ask, what of the Hafele/Keating experiment in 1971 with the atomic clocks? This experiment has been used to verify the twin paradox. But this is perhaps the most ridiculous experiment in history. It certainly proves nothing about the twin paradox with regard to Special Relativity. The scientists made absolutely no effort to limit the variables. The experiment takes place in a spinning gravitational field, with large electromagnetic variables. The clocks could be affected by any number of things, including the earth's plasma field, the sun's various fields, the moon's fields, bombardment by cosmic rays in the atmosphere, and on and on. But the most telling thing is that the airplanes carried the atomic clocks all the way around the earth. They therefore returned to the place of origin (in one sense). But this is not the same as traveling away from a point, turning around, and coming back. For one thing, it could be argued that, due to the spinning gravitational field, the point they returned to was not the point they left from, even though it was the same airport. The twin paradox is claimed to be a logical outcome, not of General Relativity, but of Special Relativity. If all moving clocks appear to slow down, regardless of trajectory (as Special Relativity now claims), then the twin paradox would follow, regardless of any additional "paradoxes" of General Relativity. The scientists therefore should have tried to minimize the affects of gravity and acceleration. And they should have avoided traveling all the way around the earth at all costs. That makes the equations so much more difficult. A spherical gravitational field with spin and magnetic and plasma fields, intersecting at least two other major gravitational fields (sun and moon)-- and then circumnavigating that field. I have made it clear that neither Einstein nor modern physicists fully understood simple translational motion. How are they to explain a difference of 59 billionths of a second in a situation so monumentally complex? Second, from Einstein's quote above-- about the necessity of a co-ordinate system-- and from the example I mentioned before (about the train clock passing at night) we can tie Relativity to another important theory of the 20th century. I said that a pulse clock on a night train with an unknown local period and an unknown local velocity could not plug into our equations, to give us any more knowledge. Connect this fact to Michelson's historical method of measuring the speed of light. Everyone knows that he sent a beam of light from Mt. Baldy to Mt. Wilson and timed its journey. But imagine if Michelson were not given the distance. What if he had to calculate the distance from Mt. Wilson to Mt. Baldy at the same time he was measuring the speed of light? How would he measure the distance? Send a laser over and back? You have to know c for that. Besides, that is what he is already doing. He has two unknowns and one observation. He cannot measure x and v at the same time. Sound familiar? The Heisenberg Uncertainty Principle holds true at the macro level as well. It is not a function of Quantum Mechanics, or of statistics. Most of all it is not a philosophical truth: the HUP does not imply that v and x do not exist at the same time. It is a fundamental truth of all measurement by observation. The only reason that the measurement of atomic particles is more indeterminate than the measurement of things on our own scale is that we can walk from Mt. Wilson to Mt. Baldy, obtain a local measurement of x', and use it in our equation. We cannot do this with atomic particles. We have no local knowledge of them. Even if we did not affect these particles with our instruments, we would still have no exact knowledge of them. We assume that the distance from Mt. Wilson to Mt. Baldy does not change spontaneously as soon as our back is turned-- we assume it remains constant from one T to the next. If we stopped making this assumption, for whatever reason, then our knowledge of reality on our own scale would also become indeterminate and probabilistic. Remember, a determination of velocity in an unknown field-- whether atomic or human scale-- always requires two observations. First, it requires a determination of x. Then it requires an observation of how much x per how much t. As I showed above with Michelson's determination of c, these two quantities cannot be gotten from the same observation. In our own world, we have no trouble combining the two observations. We assume continuity because we can see continuity. Every time we return to Mt. Baldy, it is in the same place, the same distance from old Mt. Wilson. But if we want to be difficult, we can always revert to a philosophy where Mt. Wilson disappears every time we turn our head or go into Pasadena for dinner. If we do this, the position of Mt. Wilson immediately acquires a probabilistic fuzziness. As Hume showed in the 18th century, nothing is really given. The odds are very low, based on past observations, that Mt. Wilson did a back flip when no one was looking; but, strictly, those are just odds. Knowledge is another thing entirely. Let me be very clear that I am not suggesting we stop making assumptions about Mt. Wilson. I am not proposing the adoption of a Humean philosophy or a Bohrian quantum philosophy in regards to observing mountains. That would get us precisely nowhere. I do think we should be consistent, though. I think we should allow ourselves to make the same basic assumptions about atoms that we make about mountains. Namely, that if they send us data, they exist. And do not stop existing in between data. Third, notice that my simplified equations confirm our everyday experiences, especially of the Doppler Effect. I believe Relativity at this level, the primary level, is simply the Doppler Effect on clocks, since clocks could be considered to be waves. It is especially clear in this problem, where the clock is simply a pulse with a given frequency. A pulse with a given frequency is the definition of a wave. A Prediction Now let us compare our new equations with Einstein's equations, in the space satellite problem. Let us say that we are given that a satellite is traveling 12 km/s in a direct line away from us. That velocity is the velocity of the satellite by its own instruments. Let us say the satellite has been gone for a year, earth-time. Let us also say that the satellite is far enough away from any gravitational fields that General Relativity does not pertain. How far away would we expect it to be? By my equation x = x' 1 + (v'/c) But first we must calculate x'. x' = v't' = (12km/s)(1 yr)(31,536,000s/yr) = 378,432,000km. The satellite measures itself to have traveled that far in one year. From the equation, we get x = 378,416,863.3km. That's how far we would "see" it to have gone in a year. That is because when we got a signal from the satellite saying "I have gone 378,432,000km" more than a year would have passed for us. It would be a year plus whatever time it took for the signal to travel that far. By the same token, the signal we receive at the 1 year mark, earthtime, would not be the year-end signal for the satellite. That signal we received from the satellite on day 365 was sent out sometime earlier-- when the satellite was at 378,416,863.3km. Also notice that the velocity of the satellite would appear slow, by the equation v = v' 1 + (v'/c) We would calculate the satellite to be going 11.99952km/s, from visual evidence. What would we have found if we had used Einstein's transformation equations? It is difficult to know what equation to use here. We can't use the velocity equation, since that has too many variables. We only have one given velocity. Let us use the equation for t If t' = 1s, then t = 1 (1 - v²/c²)^1/2 = 1.0000000008s And, conversely, if t = 1s, then t' = .99999999919s In one year earthtime, the satellite clock will have ticked .99999999919(31,536,000) = 31,535,999.974 x = vt' = 12km/s(31,535,999.974s) = 378,431,999.697km My equations predict a much greater apparent slowing of the satellite than do the equations of Einstein. Some may say, "Yes, your slowing is about 45,000 times as much as Einstein's. That is hardly what I would called a fractional correction. The Jet Propulsion Lab reported in Newsweek that the numbers were only off by 'one ten-billionth of the effect of gravity on earth.'" My answer to that is that I have no idea what specific equations the Jet Proplusion Lab is using to calculate the velocity of the satellites. I would assume they are not using a simple time dilation equation like I just did. I would assume they are using General Relativity equations, which factor in the gravity of all the objects in the solar system. The Lorentz equations-- and therefore Special Relativity-- are only a small part of all the math involved. Whether my correction, plugged into existing GR equations, will give the Jet Propulsion Lab correct numbers, is something I cannot say. But I will go so far as to predict that there are other problems with the mathematical methodology at the Jet Propulsion Lab-- problems that a simple fix to Special Relativity will not address. You can see the sort of major problems that have existed unknown within the rather simple mathematics of Special Relativity for a century. I would be monumentally surprised if the more difficult math of General Relativity is flawless. More than anything, that "one ten-billionth" claim seems to me to be little more than a tall flag announcing to all the sheer hubris of modern science. The JPL claims in these reports to have obsessed for twenty years over a number in the 10th position after the decimal point, and yet we can see from the mistakes addressed in this paper that the scientists and mathematicians of the twentieth century have been criminally unclear on the whole concept from the beginning. That Einstein made a few mistakes I can understand; but that they have been allowed to stand uncorrected for so long, under the noses of so many "geniuses," I cannot comprehend. The twin paradox is taught as fact to this day. As are all the other paradoxes and absurdities that have levitated not out of failures of theory, but out of failures to manipulate simple algebraic equations. I believe it was Niels Bohr who once said that only six people really understood Relativity. Now it is apparent that he overstated that number-- by six. I don't believe for a second that JPL is actually within one ten-billionth of the truth of the matter. If they have equations that are almost working, it is sheer accident. Heuristic multiple sleight-of-hand. I have shown that it is impossible to even apply Einstein's velocity equation to a satellite problem-- in which there is only one velocity. So if the Lorentz transformation for velocity has carried over into General Relativity, as part of the JPL's calculations, it is being misused on this problem. Also consider this: to be that accurate, JPL must know the masses of all the planets and their moons and the sun to the tenth decimal point. Not only that, but they must also estimate the total mass of the asteroid belt to the tenth decimal point. Then they must assume that there are no other unknowns. Modern science doesn't even know what gravity is, and yet they publicly congratulate themselves for measuring it to ten decimal points. "So what precisely are you predicting in this section?" you may ask. I am predicting that my correction to the transformation equation for velocity will force the JPL, and others, to recalibrate the complex tensor calculus they are using to calculate forces, and therefore velocities. In my paper on General Relativity, I will do a general recalibration myself, but I cannot make numerical predictions at this time without being privy to the numbers that go into this specific problem. For example, I admit to an ignorance concerning the mass of Jupiter to 10 decimal points.* By the way, it appears to me that measuring the mass of Jupiter from a distance requires the very equations I have just critiqued. All information received from Jupiter arrives on electromagnetic waves, which waves are affected by Relativity, of course. If faulty equations are yielding wrong velocities for satellites, they must also be yielding wrong masses for objects in the solar system. Therefore the velocity calculations would be doubly compromised. Appendix A The Michelson/Morley Interferometer: Let us begin with these two illustrations. They are from a recent college textbook, and may be considered typical. The first is an illustration of two boats traveling in a stream. Notice that we have, 1) two boats, 2) the stream and 3) the shore. One boat goes up and back the same distance. The other goes across the stream and back the same distance as the first. But from the shore, boat one is seen to make different movements than boat two. And both look different from the shore than they would if you were floating along with the stream. The second illustration is of the M/M interferometer. The two illustrations are meant to be analogous. But notice that in this illustration, 1) the light rays takes the place of the boats, 2) the earth (with the instrument) is the stream. 3) But what, in this analogy, is the shore? Obviously the eye should be on the shore. But in the illustration, the eye is part of the instrument. It moves along with the interferometer and the earth. So how could it possibly see the light as the "ether" sees it? It should see the light rays in the same way that the shore sees the boats! But it can't. The interferometer is a flawed instrument. But let us look further at the problem, just to be sure. "In its journey from M_S to M₂ we expect the light to travel with speed c + v, just as a boat traveling downstream acquires the speed of the river." This is what the book says. But this is not the case. If the eye were on the "shore"--that is, if the eye were off the earth and unconnected to the interferometer--then we might expect the light ray to appear to be traveling c + v, because the Mirror M₂ is moving away from the eye at v. But if the eye is a constant distance from M₂, as in the illustration, then M₂ has no velocity relative to the eye. Let me put it another way. Michelson/Morley assumed the sun was the "fixed point". They thought the ether might be the coordinate system of the sun. Therefore, using the analogy of the stream and the boats again, the velocity of the stream should be like the velocity of the earth in orbit. And the sun is like the shore. So, for the light ray to travel c + v, it must do so relative to the sun. But the eye on the interferometer is connected by a rigid support to M₂: The scientist is on the earth, moving with the interferometer. He therefore does not "see" M₂ as moving. Rather, he should be expected to see the light rays stay at velocity c before and after they pass or are reflected by the central mirror. You might say, yes, but the scientist knows that M₂ is moving. The question though, in this experiment, is not what the scientist knows, but what he sees. The book is trying to have it both ways. It is trying to see from the perspective of the shore while it is still in the river. This is not possible. But even if the eye were on the shore, even if the interferometer had been designed to send information to an observer attached to the aether, to the sun's coordinates, even then the experiment should have expected a null set. This is why. Forget the boats, forget light rays. Let us refine this analogy, this thought problem, even further. Let us say you have two men on a platform. The platform is simply a square piece of wood, 10 meters by 10 meters, say. The men are standing in one corner. Let us say they are standing on the same spot (you could say, to be more precise, that they both must push buttons that send light signals to an observer, and that these light signals are exactly equivalent: exactly the same distance from the observer.) Let us also say that the platform is moving at a constant velocity directly away from the observer. Like this: Let us go onto the platform with the men for a moment. From there, the two men move at right angles to eachother. One walks 10 meters to the edge of the platform and back. The other walks the same distance to the other edge and back. At the far end of the platform, in each direction, is a button, for each man. This button also sends a light signal to the observer, to indicate that the man is there. The men walk at the same speed. They do everything the same. What they see is that they touch the far buttons at the same time, and that they return at the same time and push the final buttons at the same time. The question now, of course, is what does the observer see? In the Galilean coordinate systems, where light has an infinite speed, the observer would be expected to see just what the men see. By Einstein's equations, the observer would not. How would they be expected to differ? They would be expected to differ at all points except these: They would both see the men together at the beginning and at the end. Both light signals travel the same distance from the end to the observer; the men are side by side. If they are there together, not even Relativity can separate them. It is true that by Einstein's equations, we would expect the total time for both walks to appear longer for the observer than for the men. Let t₀ = the time when the men depart, when they together push the signal buttons to indicate the beginning. Let t_E = the time when they return and push the same buttons. The two light signals from t_E have farther to go than the signals from t₀, so the elapsed time between the signals will be greater. And it is true that the observer would not see the men reach their far edges at the same time. They would be different total distances away from the observer, and the light signals would arrived slightly staggered. All t's between t₀ and t_E would be different for the observer than for the men. But at t_E, the men themselves and the observer would find them together. This is where we go back to the interferometer. If the two light rays reach the central mirror at the same time, they must return to it at the same time, no matter where they are observed from. No matter what they may have appeared to do from the central mirror to the outside mirrors, they must return at the same time. This is true even according to Einstein. Einstein said that any reference frame moving at constant velocity acted like a reference frame that was not moving at all. He agreed with Galileo and Newton on this. A reference frame moving at constant velocity felt no forces, and could make a valid argument that it was not moving at all. That is what Relativity means. So, imagine you are one of the men. You do not think your platform is moving. You feel no forces. When you walk, you are not heading slightly upstream or downstream, like the boat crossing the river, in order to hold a straight line. You are not leaning or making any adjustments. You are walking just like a person walks on an airplane, normally. Two people walking at right angles on an airplane will not get out of sync, no matter how fast the plane is going. They will appear to, from the earth, if they are going really fast; but even then they will be seen to return to the same point if they left from the same point. You may say, "How can you be so sure? Einstein showed that simultaneity was not an absolute. Just because these men see themselves as simultaneous does not necessarily mean the eye will." I am not saying that all things that appear simultaneous in one field will look simultaneous in another. I agree with Einstein's argument on simultaneity, for the most part. But in this particular problem, the eye must see them together at the end. In Einstein's example (to disprove simultaneity), the eye is moving toward one event and away from another. This causes the appearance of non-simultaneity. But the eye in this problem is not. The two events at t_E are not only simultaneous, they are also co-terminous. That is, they take place at the same spot. They simply cannot be separated by relativity. Now, you may say, "All this makes sense, but the previous explanation of the interferometer seemed to make sense, too. They can't both be right. Why, specifically, is the accepted explanation wrong? It seems like boat one in the first illustration, or man one in your illustration, would be travelling a zigzag path that would be longer than the up and back path of boat two or man two. Wouldn't it have to take longer to take a longer path?" No, because the times are different. According to Relativity, either mine or Einstein's, every moving object has its own time. The two men are moving in different ways relative to the eye. Therefore, from the point of view of the eye, each man has his own time. There are actually three T's, in this case. The T of man one, the T of man two, and the T of the eye. The men's times are different from the eye's, and also different from eachother. The apparent distance of man one times his apparent time is equal to the apparent distance of man two times his apparent distance, so that they return to the same spot at the same time. If this is so, you may ask why the same cannot be said for the men in their own co-ordinate system. They are moving relative to eachother-- then they should have separate clocks, and they therefore should have different velocities. It is true that they have separate clocks, clocks that do not necessarily* flash or tick equally. But we have taken the case where they do tick equally and asked what it will look like from a third point, that of the eye. We have given them equal distances and equal velocities. Therefore, in this particular case, their clocks are equivalent, relative to eachother. We have set up the situation so that this is true. It is true simply because it is given. But relative to the eye, they are not equivalent, not to eachother and not to the eye. All previous attempts to explain this situation-- that of Michelson, Lorentz, Einstein, and everybody else in the 20th century-- have agreed that the two men will have a different co-ordinate system from that of the eye. But no one has seen that from the point of view of the eye, the two men will have different co-ordinate systems from eachother. This is crucial. And this is why they end up back at the starting point at the same time. This is why Michelson's light rays end up back at the mirror at the same time, without a "fringe effect." ~~~~~~~~ Now that I have shown you how the experiment should be read, I will go back and show precisely where a specific scientist goes wrong in a specific text. This always seems to help in explaining a new theory: showing not only what is right with it, but what is wrong with the predecessor's. Let us pick on Richard Feynman this time-- a Nobel Prize winner and supposedly one of the geniuses of the 20th century. In his book Six Not-so-easy Pieces he diagrammes and explains the M/M interferometer just like everyone else. Here is his illustration. He says, "If the apparatus is at rest in the ether, the times should be precisely equal [from B to C and B to E], but if it is moving toward the right with velocity u, there should be a difference in the times." Then he jumps right into the math. To show you why he is already wrong, let me go at him in a completely different direction from those I have already taken to explain this problem. Let's look at the illustration and consider the various points of view. Specifically, let's look at it from the view of Relativity. Let us use the theory that the interferometer made possible (made necessary) to critique it. That is what science used to do, before the 20th century. It checked its findings against its assumptions. Think of it analytically. The interferometer is the postulate. We accept it as the given. Now we build a theory from its interpretation, that theory being Relativity. Surely we should go back and apply our new theory to the postulate. We should check that our theory applies to the postulate, without contradiction. Relativity is a point-of-view theory. It states, first and foremost, that measurement is an outcome of position. You must pick a position and then measure from there. All positions are different, and none of them are equivalent. Data is not directly transferable. You need a transformation equation to go from one measuring position to the next. Now, how does that apply to our illustration of the interferometer? It applies because we already have two points of view, even though we only have one illustration. What are those points of view? Well, there is the point of view of the interferometer-- or, to be more precise, the point of view of the scientist collecting data from the device. Yes, of course, you say. But what is the other point of view? The other point of view is the point of view of your own eyeballs, as you look at the illustration. Not only is your point-of-view at a right angle to the point-of-view of the device; more importantly your point-of-view is at rest, but the device is moving with regard to you. We have a train and an embankment here, where we only thought we had an illustration! For Feynman says, "Let us calculate the time required for the light to go to E and back. Let us say that the time for light to go from plate B to the mirror E is t₁ and the time for the return is t₂. While the light is on its way from B to the mirror [E], the apparatus move a distance ut₁ , so the light must traverse a distance L + ut₁ at the speed c." Well, from our eyeballs' point of view, looking at the illustration, yes. But from the point-of-view of the device, inside the illustration, no. We, looking at the illustration, conceptually see the device moving, and therefore we see the extra distance ut₁. But the device does not see itself moving. It has no velocity relative to itself. No object can calculate itself as moving, in its own co-ordinate system. Any observer attached to the device or moving along with the device will see the distance as L. Period. Feynman makes the same mistake regarding the distance from B to C. You and I see the distance as having a slope-- it will be the hypotenuse of a right triangle with L as the base. But the scientist would see no such thing. He would measure L as L, always. Since the whole theory is a theory of measurement and operation, one cannot blow by the operation of measurement without asking how lengths are seen and measured. Feynman's interpretation of the M/M interferometer assumes, like every interpretation before him, that the scientist collecting data from the device and the scientist looking at the illustration are the same person, or interchangable people. But in this case, the "theoretical scientist" and the "practical scientist" happen to be in different co-ordinate systems. They beg for a transformation equation. This is one safe that Feynman failed to crack. ~~~~~~~~ I now expect that some will say, "Well, if the M/M illustration is wrong, simply move the data-collecting scientist off the earth and run the experiment. Then you will get meaningful data. You won't because you can never get away from the lightray you are measuring. You cannot measure or see a lightray that is not in your co-ordinate system. You cannot make a measurement of light from a distance. You can set up the most elaborate device in the world, but you will still find the ray of light has ended up in your back pocket. At the end of the experiment, in order to collect any data, the light will have to come back to you. In which case it will arrive moving directly toward you. And you will not see it until it impinges on your device, at which time it is part of your co-ordinate system. ~~~~~~~~ You may also be interested to know that Einstein stated that there was no absolute test of simultaneity. He said that no observer could claim preferential status. I think he was wrong in this. Using the Doppler effect, an observer could derive that he was neither red-shifted nor blue-shifted in relation to the incoming data. This observer, unmoving in relation to both events, might easily say that his claim of simultaneity was infinitely more meaningful than anyone else's. The Doppler effect is already used to calculate relative velocities (of Quasars, for instance). It would therefore not be difficult to calculate who was in a position to determine simultaneity, and therefore an absolute simultaneity for given events. At any rate, M/M expected to find the light rays returning to their central mirror out of phase. That is, one ray was expected to be a little late, causing a fringe effect to be observed by the eye. As I have shown, this was a false expectation. As you can see, the entire problem was undermined by a poor conceptualization. This prompted the use of faulty equations to explain it (by Lorentz), which were then borrowed by Einstein to express Relativity. Appendix B This is the currently accepted derivation of the length equation from the x equation. Let L = the length of a rod in S from x₁ to x₂. L' = the length of the rod in S', from x₁' to x₂' So, L = x₂ - x₁ = y[(x₂' + vt₂') - (x₁' + vt₁')] Since an observer in S' measures x₁' and x₂' at the same time, t₁' = t₂' and the above equation therefore reduces to L = y(x₂' - x₁') And, by substitution L = yL' The trick performed here is in the statement "since an observer in S' measures the length at the same time... t₁' and t₂' are the same." But t₁' and t₂' have nothing to do with the time when L' is measured by the observer in S'. These times are fixed. They are fixed to certain points, and are dependent upon the velocity v. That is why the t' is siting next to the v in the equation. It describes a specific x. The x's are not equal, so how can the t's be equal? We know that the velocity is constant, so if the t's are equal, the x's must be equal too. And that implies that the rod has no length in S'. To be precise, t₁' and t₂' are the times when the end points of L' are seen by S' to co-oincide with external markers, if S' is measuring its own velocity. If S' is not in the process of measuring a velocity, then time will not come into the equations in S' at all. In that case, it is not that t₁' and t₂' are equal; it is that t₁' and t₂' do not exist. Time is a preparatory measurement for velocity. A measurement of length has nothing to do with time. Therefore, if you are speaking of t's in S', you must be talking about the measurement in S' of a velocity. The velocity being measured is the velocity of S' relative to S, as measured by S' (as in my blinker example). This is the only way you can have any transformation equation with v's in it. If you get rid of the t's in S', you thereby get rid of the whole idea of relative velocity and transformation equations. And if you get rid of the t's in S' you also get rid of the t's in S. Which leaves you with the equation x = x', which can yield no information, since it is meaningless. This derivation, which can be found in many current textbooks, is nothing more than number juggling. The mathematicians responsible for it needed certain information at the end, and so they coerced the equations to get what they wanted. It is very bad math. Appendix C In this appendix I critique Einstein's original paper of 1905-- probably the most famous physics paper in the 20th century. I have copied the 2 and 1/2 pages of equations that lead up to gamma here, so that you can follow along, line for line. The first part of the article is just an introduction to the problem, which I will not repeat here. It is perhaps better to approach the problem fresh, anyway. It is much simpler than history has made it. In short, Einstein has one important postulate to add to Newton and Galileo. It is this: "Any ray of light moves in the stationary system of co-ordinates with the determined velocity c, whether the ray be admitted by a stationary or moving body." This is just a restatement of the constancy of c, but it is a precise wording of it, and it will come up very soon in his math. That is why I quote it in full. It is the second of two "basic principles" of relativity. The first principle is that systems or bodies in "uniform translatory motion" relative to eachother have the same laws. That is, neither one is primary: either may be taken as "at rest." This is not so much a postulate as it is an hypothesis (as Einstein admits in this paper) but since I agree with the intent of both principles, it does not matter here. I will touch on one point before moving on. Principle 2, quoted above, already contains a confusion of terms. A moving body requires a second co-ordinate system-- that is why Einstein invents multiple systems to begin with: to explain the motion. But the second part of principle 2 ignores this fact. He says, "whether the ray be admitted by a stationary or moving body." But, obviously, if the ray is admitted by a moving body, it is not in "the stationary system of co-ordinates." It is in the moving system of co-ordinates. A better statement of principle 2 would have been that light is measured to travel c, from any and all co-ordinate systems. Very soon after this Einstein introduces his two co-ordinate systems, which in this paper he calls K and k, where K is the stationary system. Then he gives us the variables, x, y, z, and t in K; and xi, eta, zeta and tau in k. In short, given x and t, he wants to find a transformation term that will express xi and tau [ xi and tau were replaced by x' and t' in later algebraic derivations, such as the one in his book Relativity, 1916]. The last thing we are given is a velocity v. This, Einstein tells us, is the velocity of the entire system k in "the direction of the increasing x of K" But, I ask, is this the velocity of k as measured from k, or as measured from K? Einstein does not tell us, though it is absolutely crucial that we know. K and k each have their own clocks and measuring rods, which are not equivalent (by the rules of Relativity). They will therefore each measure velocity differently. In fact, they will measure the velocity of k differently. But Einstein does not assign v to either system. This is Einstein's greatest error in the whole derivation of gamma. Since v is unprimed and un-Greek, you may assume it is k as measured from K. But if this is true, then the velocity of k as measured by k must have a value too. Notice that Einstein never once, in this or any other derivation, creates a variable for that v, nor does he discover an equation which yields it. [The velocity we get in later equations-- the velocity of the current v tranformation-- is the velocity of an object moving within k. It is not the velocity of k itself.] Then we get the equation x' = x - vt. Einstein says, "If we place x' = x - vt, it is clear that a point at rest in the system k must have a system of values x', y, z independent of time. We first define tau as a function of x', y, z and t. To do this we have to express in equations that tau is nothing else than the summary of the data of clocks at rest in system k, which have been synchronized. . ." So, he states outright that x', y, and z are now the set of variables "at rest in the system k." But he has already assigned y and z to K. A point at rest relative to k is moving relative to K, since k is moving relative to K. So "at rest in the system k" is not equivalent to "at rest in K". This being true, it also means that xi, eta, and zeta have just been bumped up into a third system-- the system that is moving in k-- which he has no label for. But x' is a variable with two allegiances: he says it is now in k, but by the equation x' = x - vt it is still connected to K. Unless he means for x and v to also be "at rest in k", in which case we might as well do away with K altogether-- it has become superfluous. t is also a dual citizen. Einstein has it variously in K and k. When it is with x', it is in k. When it is with x, it is in K. Tau is in the undefined third co-ordinate system, with xi, eta, and zeta. Notice especially that Einstein has created an x' but no t'. He has three x variables but not three t variables. He then says, "From the origin of system k let a ray be emitted at the time tau₀ along the x-axis to x', and at the time tau₁ be reflected thence to the origin of the co-ordinates, arriving there at the time tau₂." So the ray takes the place of the man in the train. Einstein has a movement within a movement here: k is moving in K, and the lightray is moving in k. This provides two degrees of relativity, which Einstein is trying to calculate at once. But he does not seem to realize that he is doing this. He doesn't have enough variables to do it successfully, so his variables keep sliding from K to k. He needs three sets of variables and three co-ordinate systems. The other mistake is to use a light ray for his object moving in k. Principle 2 states that light is measured c, whether from a moving or stationary system. This makes light a special case. It is not like a man on a train. c, unlike v or v', does not change with position of measurement. This is Einstein's second greatest error in this paper. Next he prepares for a bit of calculus-- that is why he is talking of functions. He doesn't need calculus here, but it is the only way he can get the reader more confused. He gives us this strange equation: 1/2[tau(0,0,0,t) + tau(0,0,0, t + x'/(c - v) + x'/(c + v))] = tau(x', 0,0, t + x'/(c - v)) This is a preparatory step to differentiation, but we don't need to go there since this equation is faulty. It is supposed to be an expansion of 1/2(tau₀ + tau₂) = tau₁ But the corresponding time at tau₁ in either k or K is not t + x'/(c - v) . First of all, this violates Einstein's own Principle 2, which I quoted in full above. The speed of light must be measured the same from anywhere. But Einstein himself is subtracting v from it! Before giving us a partial equation for tau, supposedly from all his differentiation, he tells us that "light is always propagated along these axes [Y and Z], when viewed from the stationary system, with the velocity (c² - v²)^1/2." All that, as an aside, almost as a footnote. He gives us no equations to show why light should move like that along the Y and Z axes. Where did that square root come from? One can only suppose he borrowed it from somewhere, most likely Lorentz. But it just stands there, naked, with nothing leading into it and nothing leading out. Obviously it is supposed to be the old Pythagorean way of comparing the two lightrays in the interferometer, that I have already critiqued. But we don't have two light rays here. He has given us one lightray going up and back, in the x-direction of both K and k. So his equation for Y and Z is false-- he has given us v as the velocity in the x-direction. We have no velocity in the Y or Z direction, not in K or k. What is more astonishing is the fact that he again violates Principle 2. He says, "when viewed from the stationary system." But Principle 2 reads, "Any ray of light moves in the stationary system of co-ordinates with the determined velocity c, whether the ray be admitted by a stationary or moving body." k is that moving body. Finally, Einstein gives us this equation: tau = a[t - vx'/(c² - v²)] where "a is a function of v at present unknown." Then he gives us this xi = c(tau) This is the equation that becomes x' = ct' in the book Relativity. I have already shown that it is false [in the body of this paper], since time dilating implies a larger tau, not a smaller. But by substituting this false equation into the previous one we get, xi = ac[t - vx'/(c² - v²)] Then Einstein says that t = x'/(c - v), which I have just shown is wrong. But he substitutes it into the above equation. After simplifying he gets, xi = ac²x'/(c² - v²) then he substitutes x - vt for x', which yields, xi = ac²(x - vt)/(c² - v²) Then, as you can see, Einstein gives us some more nonsensical equations for eta and zeta. Being in the y and z directions, they can have nothing to do with v, which Einstein told us at the top of p. 43 was "in the direction of the increasing x of K." But we are basically finished. c²/(c² - v²) is our transformation term. Einstein calls it beta here, but it has come to be known as gamma. He never had a beta variable in any equation, but it suddenly appears at the end as the thing we were looking for all along. The variable "a" has been lost to the dustbin in all the rewritings of these equations, with no explanation that I have ever seen. Einstein's final error is a simple mathematical one. He assumes that c²/(c² - v²) = 1/(1 - v ² /c²)^1/2 When in fact it is simply 1/(1 - v²/c²) ---There is no square root. Either he reduced the equation wrong, or he has some hidden math steps here he is not sharing with us. It may be that he thinks that those last y and z equations will affect his transformation term beta somehow. But in his rewriting of the math in later books, he does not assume that y and z affect calculations in the x-direction. Nor does anyone who parrots this derivation of gamma. But if you think about it, the only way to get a Pythagorean square root into a linear problem in one direction, is to mess it up in just this way-- to assume, for some reason, that an x-velocity is affected by y and z. One final point. Einstein later changes the math of Special Relativity. It has been said that he did this to simplfy it for general audiences, specifically for the book Relativity. Those who say this also point to the fact that he changed the math to make it even more difficult after he completed General Relativity. For, at that time, he imported the tensor calculus used in General Relativity to express Special Relativity. In his lectures at Princeton he used this very difficult math. He had two derivations of Special Relativity working after 1916: one a revamped algebraic derivation (in the appendix of Relativity) that is very similar to the one I showed in this paper, the other a tensor calculus derivation. But both changes mask the basic errors of the original paper-- they do not fix them. In the algebraic solution, he gets rid of the Greek letters. Unfortunately he replaces xi with x'. But they are far from equivalent. x' is the x-variable in a non-moving k, as he states above. xi is the x-variable moving within k. It is the x-variable that belongs to the lightray, which is moving. Nor does he fix the equation x' = ct'. It is false whether the x-variable is xi or x'. The tensor calculus derivation also takes all the givens of the paper of 1905 as still given. It fixes none of them. All it does is make the contradictions harder to see. There is absolutely no reason to use tensor calculus to solve the problem of Special Relativity, as Einstein presents it and as it stands to this day. It does not even call for regular calculus, as Einstein proved in the appendix of his book. A problem should be solved with the simplest math that will solve it-- especially a problem of applied math. This keeps the concepts and assumptions as near to the surface as possible, where they may ride out the light of day. Now let us look at Einstein's derivation of the equation for the addition of velocities. In his 1905 paper he did not differentiate his x equation in order to find his relative velocity equation, like they do now in textbooks. He simply combined his equations algebraicly, like this: From earlier in the paper Einstein found: y = gamma = 1 /(1 - v²/c²)^1/2 tau = y(t - vx/c²) xi = y(x - vt) Now he says, if a point is moving in k, let xi = w(tau) where w "is a constant". Notice two things. One, Einstein now has a point moving in k instead of a light ray. Two, he does not define this new velocity variable at all, beyond saying it is a constant. This is really extraordinary. One must look at the sentences two or three times, to be sure that he intends w to be the velocity variable. He does not fully define w, because he can't. If he states outright that it is a velocity variable, then he must assign it to one of his co-ordinate systems. Very soon you will see why that would be fatal to him. By substitution, he gets, w(tau) = y(x - vt) wy(t - vx/c²) = y(x - vt) wt - wvx/c² = x - vt x + wvx/c² = wt + vt x(1 + wv/c²) = (w + v)t x = ( w + v )t 1 + wv/c² Now, watch this last step very closely. He reduces the above equation to: V = w + v 1 + wv/c² This is the current value for V. This equation stands to this day. But to reduce like he did he must assume that V = x/t We have been given that w = xi/tau So what does v equal? v is what x over what t? V = x/t w = xi/tau v = ?/? You may say, well maybe the x is x'. Maybe, but what is the t? He has no third t-variable anywhere, in this paper. And in later derivations, when he does have a t', he has no tau. He never has three t variables. What we need to solve for an addition of velocities, amazingly enough, is four t-variables. t₀ = the time of K from K t = the time of the point as seen from K t' = the time of k as seen from K t'' = the time of k as seen from k tau = the time of the point as seen from k. tau' = the time of the point as seen from the point but t" = tau' = t₀ We need five x-variables x = displacement of the point, as measured by K x' = displacement of k, as measured by K x" = displacement of k as measured by k. xi = displacement of the point, as measured by k xi' = displacement of the point as measured by the point Einstein says that v is the velocity of k relative to K. w is the velocity of a point relative to k. V is the velocity of that point relative to K. but to solve we also need, w' = velocity of the point measured by the point v' = velocity of k as measured by k w' = xi'/t ₀ v' = x"/t₀ tau = t₀ + xi'/c w = xi'/(t₀ + xi'/c) = w'/(1 + w'/c) w' = w/(1 - w/c) v = x"/t' t' = t₀ + x"/c v = x"/(t₀ + x"/c) = v'/(1 + v'/c) v' = v/(1 - v/c) t = t₀ + (xi' + x"/c) V = (xi' + x")/t = (xi' + x")/t₀ + (xi' + x"/c) {Eq. 5} V = v' + w' 1 + [(v' + w')/c] V = v + w 1 - (v/c) 1 - (w/c) v + w 1 + 1 - (v/c) 1 - (w/c) c = v + w - (2vw/c) 1 - v/c - w/c + vw/c² + [v + w - (2vw/c)]/c = v + w - (2vw/c) 1 - vw/c² Now, you may say, why not use "equation 5"? It looks very much like Einstein's equation, except that we are adding the velocities in the denominator rather than multiplying them. At most speeds this would only be a small correction to Einstein and would seem to imply that his math was not that far off. We can't use that equation for one very important reason. The velocity variables don't match Einstein's. Mine are prime, his were not. Mine are the local velocities of k and the point. His are actually undefined, so it is impossible to say what he meant for them to stand for. But I assume he intended them to be the measured velocities of the point and k, from k and K, respectively. I assume this because he never provides us with an equation to discover the velocity of k relative to K, or the point relative to k. He doesn't derive these equations because he thought those were his givens. But now that it turns out that his final equation suggests that his givens were local velocities, measured by the moving systems themselves. And this means that the velocity of k relative to K and the velocity of the point relative to k are still unknowns. Einstein has derived no equations for first-degree relativity! The other reason not to use equation 5 is that in most real situations we will not be given the local velocities. In using the relativity equations on quanta, for instance, the givens are not local velocities. We would be given relative, or measured-from-a-distance, numbers to begin with, and would need an equation to determine the addition of these numbers. The famous experiment of Fizeau (explained by Einstein) is another example. We are given the speed of the liquid. But this is our determination of the speed of the liquid, not the liquid's. The given is not a local measurement of the system. Please notice that my new equation for the addition of velocities gives us numbers that are very close to Einstein's in most situations. It differs from his in having another easily comprehensible term in the numerator and a minus sign instead of a plus sign in the denominator. But it may be used with confindence, since it has been derived from a thoroughly analyzed situation, as above, from five different co-ordinate systems. The true transformation equation for velocity of one degree of relativity is the one I used above, in my derivation of V. v = v' 1 + v'/c