The Discovery of First-Degree Relativity
and the Refutation of Gamma

by Miles Mathis First written June 2001 as a compression of my longer original paper of November 2000. One or the other was submitted in 2001-2002 to PRL, ADP, CERN, Nature, and JPL, where it was either refused or ignored. For example, it was submitted November 2001 to Annalen der Physik and refused January 2002. I have the letter from Ulrich Eckern in my files. The original paper was submitted to PRL even before that, since it was refused in the summer of 2001.

Introduction to the Problem

Special Relativity is widely considered one of the most famous physical theories in history, as well as one of the most perfect. Quantum mechanics, or QED, the only other theory that is as famous, has been corrected a countless number of times in the 20th century. In that time, Special Relativity has not been corrected once. Einstein's derivations of 1905 stand to this day. Gamma and the addition-of-velocity equation have never been corrected. They are considered uncorrectable. They underlie the final equations of four-dimensional space (Minkowski) and the field equations of General Relativity. The tensor calculus takes them as given.
However in recent decades there have been a number of discrepancies found in the use of the equations on data from both accelerators and space satellites. Physical Review Letters published several papers on the satellite problem (the so-called Pioneer Anomaly) of the Jet Propulsion Lab a few years ago. It was never solved to everyone's satisfaction, using the mechanical analyses offered. I became convinced at that time, and remain convinced, that the problem is in the basic equations. So I have returned to the original derivations of nearly a century ago. I have concentrated on the algebraic derivations, ignoring the tensor calculus that was imported into the problem later. The tensor calculus is a math designed to handle a large number of variables, using matrices and other time-saving devices. It is not a good math to use for simple conceptual problems, concerning only a few linear variables. The tensor calculus unnecessarily makes a very dense theory even more difficult. It is supremely difficult, for instance, to properly analyze the basic conceptions of the theory, which are spatial and temporal, when you don't have a time variable, labeled as such. Tensor calculus may be a necessity in the field equations of General Relativity, but in Special Relativity it is more math than the job requires. This is especially true in the initial derivations, where the velocity has no angle to the x-axis, and there are no other mathematical complications.
In its inception, the math of Special Relativity was algebra. Except for one step, Einstein's 1905 derivations1 of both gamma (γ) and the addition-of-velocity equation were algebraic. Even this one step of calculus was unnecessary, as Einstein proved in the appendix to his book Relativity, where he did without it. This is not surprising, since in both places the problem concerns linear uniform motion.
After studying Einstein's various algebraic derivations closely for several years, I am now in a position to prove that his final equations, though close enough for much prediction, are not correct. He ignores one very important step, and this step completely compromises the math. Nor was this step uncovered in later emendations. All current derivations yield equations for two degrees of relativity. First-degree relativity is ignored. This paper is my announcement of the discovery of First-Degree Relativity. I rush to add that in correcting Special Relativity, I am not proposing a return to classical mechanics. Nor am I questioning the basis for time dilation. This paper is in no way a refutation of Relativity, as a whole, nor a dismissal of the need for transformation equations. I simply offer subtle corrections to the existing mathematics. My transformation equations match Einstein's, in form and theory, but they provide small differences at high speeds and distances. This solves the Pioneer Anomaly, as well as several other outstanding problems.

Part One
The Primary Error

Relativity is caused by motion. An object in motion relative to a second object no longer shares the co-ordinate system of that object. We must therefore create two systems to explain them. Specifically, the length and time variables will differ, and at least one transformation equation will be required to go from one to the other. The transformation equation(s) must include the speed of light, since the finite speed of light is what makes them necessary in the first place. If c were infinite, then all space would be one co-ordinate system, as with Galileo. This is Einstein's set-up, which I fully accept. It implies that clocks and measuring rods will not match up across systems. The result is length contraction and time dilation, which I also do not question.
In the 1905 paper, Einstein gives us two systems, K and k. K is at rest, k is moving. Then we are given a constant velocity, v�that is k relative to K (v is linear, along the x-axis). We are also given x and t in K, and ξ and τ in k. We seek the transform between them. This is all we are given involving K relative to k, but the first mistake has already been made.
What is wrong is that Einstein failed to assign the given v to either K or k. If the clocks and measuring rods in K are different from k, then K and k will measure velocity differently. That is, they must get different numbers for the velocity of k. But Einstein did not notice this. He did not notice that v, as given, is already a relative velocity. No one else has noticed this in 99 years. In this problem, we should have the velocity of k relative to K, measured from K; and the velocity of k relative to K, measured from k. One motion, two different numbers.
This is what I call First-Degree Relativity. We have a v and a v' now, and we need a transformation equation from one to the other. How can we get this?
First, let me clarify the situation. I know that some will say here that k has no velocity measured from k. Velocity is a relative term, one that requires a background against which to measure. This is true. k has no velocity relative to k. But k does have a velocity relative to K, and k can easily measure that velocity itself.

v = velocity of k rel K as measured from K.
v' = velocity of k rel K as measured from k.

But before I show you the math for achieving a transform from v to v', we must visit Einstein's equations, to see precisely what went wrong there. I suspect that few will have the fortitude to wade through my new derivations until they are convinced beyond a doubt that the current equations of Einstein are faulty. Whether or not you know of the various anomalies that have arisen in experiments, I can show you right now that Einstein's equations cannot possibly work, due to simple mathematical errors.
First of all, notice that Einstein has no transformation equation to go from v to v'. None at all. The v transform of Special Relativity is

V = v + w
1 + (vw/c2)

This is a compound velocity, as is easily seen by counting the number of velocity variables. In order to calculate V, you must be given v and w. According to Einstein's famous thought experiment with the train, v is the same as our v above. It is the velocity of the a train going by on a platform. We are on the platform. In this case, w is the velocity of a man walking away from us in the train. V is then the velocity of the walking man relative to us on the platform. But Einstein never derives an equation to go from v to v'. This is because v' is not equivalent to any of the variables above. As you can see,

V = velocity of man measured by platform
v = velocity of train measured by platform
w = velocity of man measured by (man or train?)
v' = velocity of train measured by train

Einstein has no v', so he cannot possibly calculate it. Special Relativity, as it now stands, does not recognize the existence of v'. This compromises all its equations. Put very simply, Einstein has too few coordinate systems. In the velocity transformation equation above, he has three velocity variables. But he tries to calculate these from only two coordinate systems. In order to do this he is forced to transfer variables across systems, and he does this illegally, as I will show.

[Subsection added 10/2009] Professional physicists have ignored the above analysis, telling me that SR is known to be symmetrical, by Einstein's first postulate: "The laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good." That is Einstein's wording of it from the 1905 paper. These physicists tell me that the starting velocity is symmetrical, which is why we have no v'. If I measure your velocity as v, you will measure mine as v. That is true. Under that physical transform, velocity is symmetrical. But you must see that the equations of Special Relativity are not written for that specific transform. SR is not transforming your measurement of me to my measurement of you. In fact, if it were, we would need no transforms at all. If SR were really symmetrical in this way, we would need no t' or x' either. The actual transformational equations of SR are transforming local measurements into measurements at a distance, and there is no symmetry between those operations of measurement.

Yes, Einstein's postulate 1 is correct, if it is read correctly. The mechanical laws are valid in all frames, and you can take any frame you like as being at rest. In this sense, the laws are symmetrical. But if the operation of measurement is not symmetrical between these systems, then the transforms cannot be symmetrical either. That is, the laws are always symmetrical, but the �equations of mechanics� are symmetrical only when the physical operations between the systems are analogous. In transforming length and time in one system to another, the operations are not analogous. In Einstein's transformations, we choose quite freely to measure from one system or the other. Once we do, the symmetry is broken, because the measured system is not measuring us back. The transform is between �local numbers� and �measured-at-a-distance numbers.� There is no symmetry between those numbers.

This means that although Einstein's postulate 1 is still correct and valid, the velocity is not symmetrical across the transforms. Yes, there is a symmetry between �you measure my velocity� and �I measure your velocity�, but you are not measuring my velocity in the transforms. You are also not measuring my time or length or distance traveled. All the measuring is going in one direction only. For this reason, we do have two velocities from the start, v and v', and they are not the same number.

Part Two
Another Central Error

But let us start at the beginning. Let us start with an illustration. This illustration is typical of current textbook illustrations. It is the attempt to diagram Einstein's train thought problem from the book Relativity, and here the artist has tried to graph x', x, and vt.  [To make things a bit easier, for the book Relativity Einstein exchanged the Greek variables in the 1905 paper for primed variables. I will do the same in the equations that follow.] The man is at point P. We, the observers, are understood to be watching from the embankment.

In the first part of the problem, the man is not moving inside the train: x' is simply the distance of the man from the origin at t0. We are transforming the distance, not the velocity. We transform the velocity variable later, when the man also moves inside the train. Einstein makes this very clear in the paper of 1905, and it is obvious regardless. There is only one velocity variable given. If the man were moving relative to the train, we would have to be given that velocity as well, as you can see.

To go with this thought experiment, Einstein gives us this equation (p.33, Rel.),

x' = x - vt

In his original paper of 1905 ["On the Thermodynamics of Moving Bodies"], he gives the same equation. But neither there nor in the book Relativity does he say where this equation comes from. In the 1905 paper, the equation is completely mysterious; but in Relativity he gives us a clue. Einstein says, "If in the place of law of the transmission of light we had taken as our basis the tacit assumptions of the older mechanics as to the absolute nature of times and lengths, then instead of the above we should have obtained the following equations:

x' = x - vt
y' = y
z' = z
t' = t

"This system of equations is often termed the 'Galilei transformation.' The Galilei transformation can be obtained from the Lorentz transformation by substituting an infinitely large value for the velocity of light c in the latter transformation." Einstein also called these equations an expression of "Newton's Principle of Relativity." Before Special Relativity, we transformed from one coordinate to another using these equations, he implies.
But this is not the case. There is no such thing as a Galilei transformation equation. For Galileo and Newton, no transformation was necessary for a linear problem like this. x in S' would equal x is S. Besides, the whole universe was a single co-ordinate system for Galileo, and the train would not have been given a system of its own. There is no possibility of a prime variable of this sort in a Galilei system. The only time that two Galilei systems would have two x variables is in a case where the two systems have different origins. The equation in that case would be something like x' = x - a, where a is the distance between the two origins. In Einstein's thought problem the origins are overlapping at t0. This is one of the givens. In the paper of 1905 he states outright that the origins are equal at t0. He is not trying to calculate the distance from one system to another, he is trying to export a distance in one system to another system.

This being so, the equation x' = x - vt cannot be applicable to the problem. For you can see that the true list of Galilei transformations are these.

x' = x
y' = y
z' = z
t' = t

The velocity has absolutely nothing to do with a Galilei transformation. If c is infinite, then all measurers will measure equal times, distances and velocities. x' = x, v' = v. This is because there is no difference between what I see and what the train sees. Light brings me exactly the same information that it brings the train, at exactly the same time. There can be no transformation equation: not a fancy Lorentz transformation, but also not a simple transformation like x' = x - vt .
Einstein has used the wrong first equation. Let us look at what the equation is telling us, in a specific example. Let's say at t0 the back end of the caboose is at the origin of the moving system, S'. Let's also say that x' is the distance to the front of the same caboose, as measured from inside the caboose. The whole train then leaves us at the station and travels a distance given by the term vt. The equation x = x' + vt is telling us that we, back at the station, will measure the length of the caboose as "how long the caboose is, measured from the caboose" + "the distance it has gone". As if we will add the length of the traintracks to the length of the caboose! Do you see now how utterly absurd this is? It assumes that we can't see, with our own eyes, that the back of the caboose has also traveled vt, and must therefore be subtracted from x' + vt. What we are looking for in this problem is simply "how long the caboose looks to us." This equation tells us nothing about that at all, neither classically nor relativistically. It is the wrong equation. Classically, the correct equation is just x' = x. Einstein imported an equation that Galileo would have used to find the total distance from the origin to the front of the caboose after time t, and applied it to find the length of the caboose as seen from the origin. An absolutely momentous blunder.
If light has an infinite speed, then the embankment will see both ends of any rod at the same time. The embankment will also see the back of the train and any other point on the train at the same time. So Einstein's given equation cannot be a Galilei transformation, in any sense. This mistake has never been corrected (I just saw the very same equation used by Richard Feynman to prove Special Relativity in Six Not-so-Easy Pieces�92 years after Einstein). Physics textbooks still use Einstein's series of conceptual steps to prove the equations of Special Relativity.
This is the current derivation in physics textbooks:
Assume                 x' = x - vt
Assume that the transformation from Galilean equations to Relativistic equations will be linear.  Then

Step 1:                         x' = γ(x - vt)  where γ is the transformation term we seek.
and x = γ(x' + vt')*
Now, says Einstein (following Lorentz), light travels in these coordinate systems (S and S') in this way:
Step 2:                                x = ct        and        x' = ct'

Substituting the first equations into these equations gives us:

Step 3:                               ct = γ(ct' + vt') = γ(c + v)t'     and
ct' = γ(ct - vt) = γ(c - v)t

If we substitute t' from the second equation into the first, we find that

Step 4:                              ct = γ(c + v)γ(c - v)(t/c) = γ2(c2 - v2)(t/c)

Cancel out the t on each side and solve for γ:

Step 5:                                 γ =         1/(1 - v2/c2)1/2

This is the famous transformation term gamma.  The math leading up to it is correct, but it doesn't matter, since the assumption was incorrect. You cannot start from a false equation and derive a true equation.

*Einstein used the equation x = γ(x' + vt)�without the t primed. But current physics textbooks have changed the notation in order to make gamma derivable with an internally consistent series of steps, as above. The math is correct; the postulates and givens are not.

Part Three
The Third Error

This leads us to the third major problem. Everyone knows that Einstein used the Lorentz equations to find that time appeared to slow down and x appeared to get shorter. Length contraction and time dilation. But let's look for a moment at the two light equations above. The light equations Lorentz and Einstein both used:
x = ct
x' = ct'
If these are true,
then   c = x/t      from the first of these equations
and     x' = xt'/t     by substitution
so           x'/x = t'/t

This means that in these equations the apparent change in x is proportional to the apparent change in t.

But when time slows down (in any system, or by any means of measurement), the period gets larger.  Time slowing down implies a larger  t, not a smaller t.
That is, t should appear to get larger as x appears to get smaller. Einstein even states this outright, in the book Relativity. He says (Ch.XII, p. 37) "As judged from K, the clock is moving with the velocity v ; as judged from this reference body, the time which elapses between two strokes of the clock is not one second but [γ] seconds, i.e. a somewhat larger time. As a consequence, the clock goes more slowly than when at rest." Again, he says "a somewhat larger time." Physicists have focused on the sentence after that, up to now. But time is not defined by the rate of the clock, not even by Einstein. Or stated more precisely, time is not measured that way. Relativity is primarily a theory of measurement, and so what is required is an operational definition of time. Not what time is as an abstract concept, but what time is as a measured quantity. Time is the length of the period, as Einstein flatly states here. A second is not a stroke of the clock. A second is the gap between strokes. Time is not the strokes of the clock, it is the time between strokes of the clock. As he says, a moving clock is seen by a stationary observer to have a period γt, which is larger than t. A dilated clock ticks slower because its period is longer.
[To consider this question further, see the links at the end of the paper�a discussion of this definition with several scientists and mathematicians, and a paper on the operational definition of time.]
As further proof of this very important concept, I refer you to the The Meaning of Relativity again. In Chapter 2 he provides this equation (eq. 22a):
Σ Δxv'2 - c2Δt'2 = 0
Conceptually this is obviously analogous to the equation x' = ct'. Einstein is just varying his math a bit, dressing it up. The thing to notice here is the delta t. He has now made it clear that he is referring to changes in time, not instants in time. In these equations, the variable refers to the time period, not the instant in time. Which is precisely my point. When time slows down the period gets larger.
This being true, x and t must be in inverse proportion!

So, we should find that
x/x' = t'/t     t = t'x'/x    x = x't'/t
or   xt = x't'
And, if x' = ct'
then     c = xt/t'/t'
and      x  = ct' 2/t

Only if   t = t'     does    t2/t' = t'
Einstein states that t does not equal t'

therefore  x does not equal ct
Even the light equations were wrong!

Part Four
New Transformation Equations

We are finally ready to derive new transformation equations. Going in, we know two things. 1) The current equations are mathematically flawed. 2) They are not far off, since they have been verified by many experiments.
We have jettisoned two of the most important equations, including the first and central one, so it is difficult to see how to start. You can see why no one has wanted to work on this problem for a century. It goes beyond a subtle tinkering. Since the end result of the transformation equations has always been the ability to derive a relative velocity from a local velocity (or other known quantities), we should ask, what do these terms mean?  What is a local velocity and what is a relative velocity?  It turns out that these definitions are strictly practical.  That is, these velocities are determined by how we measure them.  Historically we have always measured velocity by one of two methods:
1) We measure our own velocity by using a clock and by measuring our change in x relative to a known background.  As an example, if we were driving in a car (but did not have a built-in speedometer) we would have to make use of mile markers.  We would take note of the markers as we passed them; and then, using our on-board clock, we would calculate  the velocity.  Please notice that in this case we see the markers from a negligible distance.  The speed of light does not affect our calculation, because we are at mile marker x when we see mile marker x.
2) We measure the velocity of an object at some distance.  This measurement is arrived at in a completely different way than the first one.  Usually we are given x, as in the first problem.  We know  x  because we have already marked it off, or we have it as an accepted number from previous experiments.  But  t  is different.  We use our own clock, it is true.  But, because the object is at a distance, and because light has a finite speed, we do not see the object at the same time that the object sees itself.
To make this clearer, imagine that the object is a blinking light.  In this case, there are actually two events.  The object blinking, and our receipt of the blink.  These two events take place x distance apart, and the gap in time is the time it takes for light to travel x.
Let us make up our own thought problem to illustrate this.

Thought problem one:

Apparatus:
1) A blinker that blinks at a rate of one blink per second.
2) A tunnel marked off with lines, like a ruler, to indicate distance.
3) An eye, with a clock that ticks at a rate of one per second, at the beginning of the tunnel. Experiment:
The blinker and the eye begin at rest, next to eachother.  Their blinks and ticks are exactly synchronous.  The blinker then takes off and goes through the tunnel at a constant velocity.  It measures its own velocity based on the number of marks it passes for each blink.  It reads the marks from a negligible distance.  That is, it reads the marks as it passes them .
The eye also measures the velocity of the blinker.  It measures the velocity of the blinker relative to its own clock.  It measures by seeing the blinks, which are blinks of visible light.  The eye is given x'.  It has walked off the distance in a previous experiment (or you may want assume the eye is the one who painted the lines on the tunnel).
The blinker is set on a course directly away from the eye.  Assume that it reaches v' instantaneously.

Question:
Will the eye and the blinker measure the same velocity?
If not, how can the velocity measured by the eye be known given the velocity as measured by the blinker itself (and vice versa)?

Answer:
Let t' = the period of each clock, from its own vicinity.  This is the period measured when the two clocks are side by side at the beginning.  Notice that the blinker is a clock.  Each blink is a tick of the clock.
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.
Let  t = period that the eye sees blinks from blinker.  This gives us the apparent period.
v = velocity eye calculates blinker to be going, based on visual evidence.
This is the apparent velocity.

If you are with the blinker, then you will measure your own velocity like this

v' = x'/t'

Let us say that your first blink is at the 1km mark.  Your second at the 2km mark, and so on.
Obviously, your v' = 1km/s

What then is v, the velocity of the blinker as measured by the eye?

To discover this, we must first find  T1.  That is, when does the eye receive the first blink, according to its clock?

t = period
T = time

Well, @ T1' = 1s,
x' = 1km, so the light must travel back to the eye 1km.  It takes the light 1km/c  to do this.  So we would expect the eye to receive blink #1 at

T1 = T1' + (x'/c) = 1.000003s
And
@ receipt of second blink, T2 = 2.000006.
@ receipt of third blink, T3 = 3.00001.
and so on.

So, for a simple blinker, the general equation would be

Tn = T n ' + (xn' /c)

t  =  T2   -  T1
t  =  t' + (Δx'/c)

A blinker with a period of 1s and a local velocity of 1km/s will appear to have period of 1.000003s.
This period will be stable.

Now let us calculate the apparent velocity.

v = x'/t
= x'/[t' + (x'/c)]
= .999996km/s

You may say, "Wait, why did you use x' in that equation?  And why did you assume x' = 1km  when you said that the light must go 1km to get back to the eye, in the time equation?  You can't assume these things!  Relativity tells us that the clock will slow down and that x will shrink.    x should be less than x'."
I am not assuming x' is the distance to use in the equation for apparent velocity.  I am given it.  The velocity of an observed object is either the given distance divided by the apparent time or the apparent distance divided by the given time.  These are the only possible calculations for an observed velocity.
In the present case, v = x'/t     or   v = x/t'     but not     v = x/t
The same goes for the light ray traveling back to the eye, in the time equation.  x' is simply a given here, just as c is a given.  Without them, any equations-- mine or Einstein's-- would be useless.
If I was not given x' (or v' and t', which is the same thing), there is no way I could know it or calculate it.  And there is no way I could calculate v.
Think of it this way:  A train passes at night.  We don't know the velocity, and we can't see the mile markers.  All we can see is a pulse clock on the train.  Can we know its velocity relative to us?   No.  The Lorentz transformations, as used up to now, can tell us nothing.  We must be given a local velocity v', or we must know x'.  The apparent velocity of the pulse clock is determined by its period and its speed.  That is, it could be ticking slowly and going slowly, or ticking faster and going faster: in both cases it would look the same.
It is true, though, that x will look shorter to the observer, as Einstein said.  But this x is not x'.  Nor is it the x used in the apparent velocity equation, as I have shown.  That x is given as x'.  What we are seeking for x  here is the apparent distance.
It is calculated like this:

apparent x = (apparent v)  X   t'

If you are still unclear on why I used  t'  instead of  t, think of it this way.  What we want is to multiply the apparent velocity v by the time on our clock, right?   We want to know what x is at T1 , and T2 , and so on, on our own clock.  That is what it means to measure by your own clock.  If you know a runner's speed, and want to calculate how far he runs in a time interval, you would not check where he was as your watch ticked 1.000003, would you?  You calculate using your standard time interval, your own second hand.
You may say, "But you have defined  t  as the time for the eye, and  t'  the time for the blinker.  Now you want to switch."  No.  I never defined  t  as the time for the eye.   I calculated  t  to be the apparent period of the blinker, as measured by the eye.   This does not mean that the eye's clock is ticking every 1.000003 seconds.  It means, of course, that the blinker's clock looks like it is ticking every 1.000003 seconds, from the eye.  But the eye's clock is ticking at a normal interval, for the eye; just as the blinker's clock is ticking at a normal interval, for the blinker.  This normal interval�the rate a clock goes as seen from its own vicinity�I have defined as t'.
Notice that if the eye's clock had a period of t, then it would not see the blinker's clock as slow.  It see's the blinker's clock as having a period of t, right?  If the eye's clock also had a period of t, there would be no difference.  The blinker's clock is slow, relative to the eye's clock, which therefore is not slow.  Very simple.

So, @ T' = 1,
v  = .999996km/s.

And     x = .999996km/s = .999996km.
1s

This is just what we would expect.
t has apparently slowed down.  And  x has apparently shrunk.  That much is consistent with Einstein, at least.
But you can see that we have had to be very careful about our t's and x's and v's.  You cannot just substitute an x or a t into an equation because it looks similar to another x or t.    You must think about what is really happening.

So, to sum up:
The blinker's period will appear to slow down, but the period will remain stable (it will not continue to slow down further the farther away it gets).
Therefore, the velocity will also appear to be slow.  If the blinker sends you a message telling you that its v' is 1km/s, then it will have appeared to slow down relative to that.
If the blinker has a length along the x-axis, then the blinker will be calculated to appear shorter, because there is an apparent contraction along the x-axis.  If you measured the blinker when it was at rest next to you, then your calculation will be short relative to that.

We have found that  x = vt'
and  v = x'/t
so,   x/t'  = x'/t
and  xt = x't'   just as Relativity predicted.

x and t are inversely proportional.  As t appears to get larger, x appears to get smaller

By substituting quantities we can now easily derive the direct transformation equations, and calculate v  from v'  or  x  from  x' and v':

v =  x'/t  =  x/t'
t = t'  +  (x'/c)
= t' + (v't'/c)
= t' (1 + v'/c)
v = x'/[t'(1 + v'/c)]
x' = v't'

v =      v'
1  + (v'/c)
v' =     v
1 - (v/c)
x  =        x'
1 + (v'/c)
Or, to restate that last equation: x' = x/(1 - v/c). In this equation, you can see that the transformation term 1/(1 - v/c) can also be written c/(c - v). This is interesting because in the paper of 1905 Einstein found the transformation term to be c 2 /(c2 - v2). He improperly reduced it to 1/√(1 - (v2/c2), the current value for gamma. c2/(c2 - v2) correctly reduces to 1/[1 - (v2/c2)]. There is no square root. This has also never been corrected, or even commented on3. However, it does not matter, since I have now proved that both values for gamma are wrong. The transformation term for one degree of relativity is simply 1/(1 - v/c). This term applies when we are transforming the numbers on the train to the numbers on the platform. It does not apply to a man walking inside the train. It applies to the train itself.
You may think that the transformation term for two degrees may be gamma with or without the square root, since c2/(c2 - v2) is a sort of doubling of my term. But I will now go on to derive the velocity transform for the man walking inside the train�the velocity transform for two degrees of relativity�and in doing so I will show that gamma does not apply there either, not with the square root nor without it.

Part Five
Second-Degree Relativity

First I will show why Einstein's proof does not work. In his 1905 paper he did not differentiate his ξ 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:

γ = gamma = 1 /√(1 - v2/c2)
τ = γ(t - vx/c2)
ξ = γ(x - vt)
Now he says, if a point is moving in k,
let ξ = w(τ) 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 w at all, beyond saying it is a constant.
By substitution, he gets,
w(τ) = γ(x - vt)
wγ(t - vx/c2) = γ(x - vt)           gamma cancels out
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 were given that w = ξ/τ
So what does v equal? v is what x over what t?

V = x/t
w = ξ/τ
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 τ. 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
τ = the time of the point as seen from k.
τ' = the time of the point as seen from the point
but t" = τ' = 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.
ξ = displacement of the point, as measured by k
ξ' = 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' = ξ'/t0
v' = x"/t0
τ = t0 + ξ'/c
w = ξ'/(t0 + ξ'/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 + (ξ' + x")/c)
V = (ξ' + x")/t = (ξ' + x")/[t0 + (ξ' + 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/c 2 + [v + w - (2vw/c)]/c

V =        v + w - (2vw/c)
1 - vw/c2

Just to be sure that gamma does not apply to the transformation of two degrees for t, let us find τ' in terms of v, like Einstein did.
t = t0 + (ξ' + x"/c)
w' = ξ'/t0 and v' = x"/t0
t = t0 + (w't0+ v't0)/c
w' = w/(1 - w/c) and v' = v/(1 - v/c)
t = τ' + wτ'/c(1 - w/c) + vτ'/c(1 - v/c)
t = τ'{1 + [w/c(1 - w/c)] + v/c(1 - v/c)}
t/τ' = (1 - w/c)(1 - v/c) + (1 - v/c)w/c + (1 - w/c)v/c
(1 - w/c)(1 - v/c) (1 - w/c)(1 - v/c) (1 - w/c)(1 - v/c)
=    1 - w/c - v/c + wv/c2 + w/c - wv/c2 + v/c - wv/c2
1 - w/c - v/c + wv/c2
=       1 - wv/c2
1 - w/c - v/c + wv/c2
t/τ' =        c2 - wv        =        1 - wv/c2
(c - w)(c - v)              (1 - w/c)(1 - v/c)

Similar, but not gamma. Not surprising, since gamma only has one velocity variable. But in Einstein's derivation of gamma, regarding x and t, he already had two velocities. His set-up for the addition of velocity section is exactly the same as his set-up for x and t, in the first section. The only difference is he had a light ray moving�as his second velocity�in the first part, and a point in the second part. But in both sections he is seeking equations for two degrees of relativity.
So what if we substitute the speed of light for w in the last equation above? Does it then resolve to gamma?
t/τ' = (c2 - cv)/ c2 - c2 - cv + cv

No, it resolves to infinity, just like Einstein's t-transformation. What does Einstein's addition of velocity equation resolve to if w is replaced by c?
V =        w + v        =        c + v        =        c(c + v)        =        c
1 + wv/c 2              1 + v/c                  c + v
V resolves to c, in that case. The velocity of light is c whether it is measured from k or K. That is Principle 2 again. But then that means that Einstein's adding and subtracting of v from it in the tau expansion was pointless. My final equation for V also resolves to c if w is c, but I did not get there like he did.

Now, you may say, why not use "equation 5" above? 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. 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 have no local knowledge of quanta. 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 confidence, since it has been derived from a thoroughly analyzed situation, as above, from five different co-ordinate systems. My first-degree equation for velocity also gives us a fraction more slowing at the speed of a space satellite, which answers the Jet Propulsion Lab's decades-old problem.

1"On the Electrodynamics of Moving Bodies", Annalen der Physik, 17, 1905.
2"On the Electrodynamics of Moving Bodies", Annalen der Physik, 17, 1905, p. 8.
3Relativity, Ch.XII, last page.
4 Historical Note: Max Born used gamma without the square root, perhaps for this reason. But this does not address the other substitution errors I have shown.

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