Look at Einstein's equation for x: ~~~~~~~~ ~~~~~~~~
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. Part V: Relative Velocity Part VI: Relative Velocity ~~~~~~~~ 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. ~~~~~~~~ Part VII: The Addition of Velocities ~~~~~~~~ Part VIII: 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. ~~~~~~~~ ~~~~~~~~ Part IX: Inferences Part X: A Prediction
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.
[The following section is rather dense, and may be read as a footnote, or skipped with impunity. It analyses in depth the concept of the co-ordinate sytem, and transference of variables from one to another, using very precise notation. It will probably only confuse a first reader, and may be best saved for a second or third reading, once the rest begins to sink in.]
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 S0 be 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 S0 by 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 S0 at 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 x0 is 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 S0 system 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—S0 influences 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'. S0 is the base system or sub-system or non-velocity system that generates the two relative velocity systems S and S'. This S0 is 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. S0 is 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, S0 has 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' = x0 since the blinker is at x0 N when it records x'N[read x prime at the Nth measurement]]
So, in the equation v' = x'/t' we may use the displacement data from the S0 system—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, t0 does not = t and x0 does 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 x0 by 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.]
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.
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)
The general equation is therefore
t = t' - (change in x')/c
because change in x is negative.
t = t' - x'/c
v = x'/t
v = x'/(t' - x'/c) = v'/(1 - v'/c)
v' = v/(1 + v/c)
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.
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 45o 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 ?
yn = v'tn' so, y1 = 10,000km
Obviously, angle L = 135o
so, x2'2= x1'2 + y12 - 2y1 x1cos135o
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.
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. Notice that using Einstein's algebraic derivation above would yield only two velocity variables, v and v'. But the calculus derivation yields three. Three velocity variables should imply two degrees of relativity, but Einstein doesn't realize this. And an equation, derived by differentiation, that expresses two degrees of relativity shouldn't be equivalent in output to the same equation, derived albegraicly, that expresses only one degree of relativity.
Please notice how directly and cleanly I got to my equations. My equation for t is simple and straightforward. Likewise my equation for x. And my 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, derived by a false visualization. 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.
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}
= v' + v" - (2v'v"/c)
1 - (v'v"/c2 )
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.
To see a different, more extended, derivation of this last equation, see the end of a
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.
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 are 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 have to be corrected. The equation m = γmo no longer pertains, since I have demonstrated that γ is not true. Also, Minkowski's spacetime equations rely on the Lorentz equations. As does Einstein's gμν equation (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.
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.
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? First of all, it is difficult to see how to apply his equations to this problem. We only have one velocity, so we cannot correctly use the Lorentz transformation equation for velocity. But the JPL is applying it somehow.
V = v + w
1 + vw/c2
The only other velocity I can imagine the scientists at JPL using is the speed of the earth in orbit. So let's assume that v = the speed of the satellite, undefined as to position of measurement; and let w = the accepted speed of the earth in orbit, again undefined. I say undefined because current relativity makes no distinction between local velocity and measured-from-a-distance velocity. That is to say, JPL does not distinquish in its working equations between its measurement of the satellite's velocity, and the satellite's measurement of its own velocity. JPL uses the satellite's numbers for the given velocity. But Einstein's equations do not work that way, as I have shown. He had no conception of local velocity in his theory. He took his given v to be the velocity of a moving system from the point of view of the stationary system. So proceeding like the JPL does is wrong. I am going to follow their (likely) procedure, however, simply to discover how their discrepancies may arise. Of course, my equations here only take into account the Special Relativity part of the field equations. This is a correction only of the relative linear velocity: the other tensors will be affected in different ways that I cannot discuss here.
Let us again take
v = 12,000 m/s and let
w = 30,000 m/s
V = v + w = 41,999.99983 m/s
1 + vw/c2
This is the velocity that JPL is expecting, from its own equations.
By my equation, above, I would expect that JPL's instruments would be receiving this data:
v' = 12,000 m/s
v = v'
1 + (v'/c)
= 11,999.52 m/s
v + w = 41,999.52 m/s
My equations predict a slightly greater apparent slowing of the satellite than do the equations of Einstein.
Some may say, "Yes, your slowing is about 2,800 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 first answer to that is that my slowing is 2,800 times as much only if you compare one change to another. If you compare final numbers, my slowing is only .00114% more than theirs. That is .0000114, without the percentage.
My second 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. The fact is, they can't know any body's gravity to ten decimal points, since G is not known 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 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.
Appendices A and C are now published in separate chapters.
Click here to go to Appendix A: the Michelson/Morley Interferometer.
Click here to go to Appendix C: A Line-by-line critique of Einstein´s 1905 paper "On the Electrodynamics of Moving Bodies". 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 = γ[(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 = γ(x2' - x1')
And, by substitution
L = γL'
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.
Addendum 5/03. Equally bad is the equation finessing that has gone on to get t = γt' from Einstein's t-transformation:
t = γ(t' + vx'/c2)
This is what one graduate level textbook says—"The [period] between two events that occur at the same place (x'1 = x'2) in S' is measured to be T'2 - T'1", which gives us this equation:
T2 - T1 = γ(T'2 + v x'2/c2 - T'1 - v x'1/c2)
= γ(T'2 - T'1 )
t = γt'
Absurd, because the idea of two events occuring at the same place in S' contradicts the entire thought problem of Einstein. In that case there is no motion in S'. If there is no change in x' there is no change in t'. You cannot have two events that occur at the same place and have motion, by definition.