Some physicists may now be shaking their heads, saying to themselves, "no, no, no."  They will tell me that Einstein's first equation describes a completely different situation than the one I have plotted and critiqued above.   They will say that the equation  x' = x - vt   breaks down in a wholly different way.
     The equation may be thought of in this way, they will say:

Since, in general, x = vt,    x' = x - vt   may be thought of as

       (some x) = (some x) - (some x)

It corresponds to the equation in the book  that Einstein offers:

       w = c - v

where w  is the velocity of a light ray relative to the train,
         c is the speed of light as measured from the embankment,
   and v is the speed of the train

In fact, Einstein implies an analogy between the two equations.  Therefore we may think of the first "some x" as taking the place of w. 

Therefore x'  is the displacement of the man relative to the train,
                vt is the displacement of the train relative to the embankment,
           so  x must be the displacement of the man relative to the embankment.

I say, this makes perfect sense except for one very important thing.  The notation of the variables is imprecise and confusing.  x  and  vt  appear (because of the fact that they are both unprimed) to be in the same co-ordinate system.  But they are not.  A much better notation would be the following:

             x" =  x - v't'

This tells us that we have three co-ordinate systems—the system of the embankment, the train, and the man.  And this notation stands to remind us that the given velocity is v' : the local velocity of the train.  It is the velocity of the train as measured from the train, not as measured from the embankment.  Einstein never differentiates between the two.  He never gives us an equation to find the velocity of the train as measured from the embankment—which would be simply v.  The v  he derives in Chapter XIII (on Fizeau) is the velocity of a moving man on the train, but it is for two degrees of relativity.  He gives us no equation (and we still have no equation) for a single degree of relativity—the relative velocity of the train. 
     You may say, simply set the velocity of the man to zero and run the equation.  This will give the velocity of the train.  But it won't, for several reasons.  One, because the current Lorentz equation for velocity resolves to unity if you plug in zero for one of the given v's.  It tells you that your relative velocity is equal to your given velocity—the local velocity of the train.   This is no surprise, since Einstein never differentiated between the two.   This becomes crystal clear if you set x" to zero in the last equation above.  The equation then becomes  x = v't'.     This tells us nothing.  It also does not give us an equation that can be manipulated by substitution in the way Einstein manipulated his equation.   What I mean is, x = v't'  cannot yield the term gamma. 

~~~~~~~~

[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.

~~~~~~~~

So far I have only done a critique of the algebra of Special Relativity.  But the math more commonly used in Special Relativity is calculus.  It has taken this form:

Let us say the man at point P in the illustration above is moving.  The velocity of the man as seen from the embankment is therefore

W  =  dx/dt  =  d[γ(x' + vt')]/dt'      where γ is gamma

Differentiation yields the equation

W  =     v' + v    
         1 + vv'/c²

But the form of this differentiation assumes that  W =  v' + v
   where v' =  the velocity of the man relative to the train, and
             v =  the velocity of the train 

If v' = 0, then the equation resolves to W = v.   v is a given quantity, so the equation yields no information.

Einstein's equation for velocity tells us how fast the man appears to us to be moving, if the man is moving in the train.  But if the man is not moving with regard to the train, the equation tells us nothing about the apparent velocity of the both the train and the man relative to the embankment.   No one has seemed to notice that the train has a relative velocity of its own.   Or, if you take the given v as the velocity of the train as seen from the embankment, then no one has noticed that the train will have a local velocity that is different from this observed velocity.  
       Look again at the beginning of this calculus problem as I have stated it here.  I have stated it as Einstein and the current textbooks state it:  "Let us say that the man... is moving."   Notice that there is no distinction in this sentence between 1) the man moving because he is moving with regard to the train, or 2) the man moving simply because he is seated in the train, and the train is moving. 
       By differentiating an equation of this form, Einstein has arrived at a velocity that is in fact relative to two degrees.  That is, the man relative to the train, and the train relative to the embankment.  The current transformation equations do not derive a value for the relative velocity of the train.  Einstein and all the physicists of the 20th century have not even noticed that this value is necessary—that it is, in fact, the value we were seeking in the first place.  Nor have they noticed that physics has ended up conflating, or substituting, one value for the other.  This confusion of terms has never even been noticed, much less resolved.

Part III: Lorentz, Michelson, and Pythagoras

In glossing the calculus of Special Relativity in the section above, I said that Einstein's equation for velocity gives us a number as long as the man is moving with regard to the train.  What I did not say is that it gives the wrong number for that as well.  It is wrong not only for the substitution and conceptual mistakes I have already outlined, but also for the following reason.
   
The main feature of the Lorentz equations is the term γ.  As I said above, Lorentz and Einstein calculated gamma to be

                                                  γ = 1/√[1 - (v²/c²)]

Where did they get this?  Lorentz arrrived at gamma first, and his thinking was not precisely the thinking of the substitution equations I have listed the steps for above.   It is obvious from its form that gamma comes from applying the Pythagorean theorem to something.  But what?  Lorentz intially came up with his equations to answer the findings of the Michelson-Morley interferometer experiment.  This was before Einstein proposed the theory of Special Relativity.  I break down the interferometer experiment in chapter 3, but a simpified illustration here will show where the Pythagorean theorem comes from.  This illustration was modified from a current college physics textbook.



We have already seen two algebraic errors by Einstein in the invention and derivation of the Lorentz equations.  The third—Lorentz' use of the Pythagorean theorem in deriving his original equations—arises from the problem illustrated above.
       What we find in the illustration is a spaceship with a light projector inside. The projector emits a ray of light and it travels across the spaceship to a screen on the far side. The spaceship at the top is the S' system, and it illustrates the path of the lightray as seen from inside the spaceship (A). Below is illustrated how the path of the lightray would look from the outside (B), to a stationary observer on the earth. This is the S system, obviously.
       We are told that the observer on earth would observe the process in B as it is illustrated.   But notice that in B the lightray is moving sideways to the observer on earth.  I hope is is obvious that our observer cannot see this lightray.  No one can see a tangential lightray!  We only see lightrays that come into our eyes.  Every lightray we see is coming directly toward us.  We have no knowledge of lightrays moving away from us or moving tangentially or even just missing us.  To have information about the situation on this spaceship, we must be sent a signal from the ship directly toward us.  In this illustration, the editors of the book are performing equations on imaginary light paths.  Not observed paths, but abstractions.  This is a grievous conceptual error.
      To state it another way, in the A part of the illustration, measurements are being made by observation.  In the B part, measurements are being made by the imagination.  In A the local observer is collecting real data.  If you had a person or device with the projector to collect the light rays as they returned from the screen, you could use this data to do calculations. But in B the observer is not basing his equations on collected data.  He is not even collecting any data.  There are no lightrays coming to him, and he can have no direct data concerning the motion of the given light, or indeed the motion of the spaceship itself.  In reality he would not be seeing anything.  The spaceship would pass him by, unknown.  He is making assumptions.  He is assuming that if  he could see the same lightray A sees, it would be traveling in this manner.  But this is not observation, much less measurement.  It is poor illustrating, poor thinking, and very poor applied math.
    
The Michelson-Morley interferometer was invented to to test the situation described above.  You can see how the Pythagorean theorem would be used to calculate the distance light travelled in B given the distances D and L.  D and L are the sides of the triangle and the path of the lightray in B is the hypotenuse.  The Lorentz equations applied to the interferometer work in exactly the same way.   The Lorentz transformations take us mathematically from A to B.
     This would be fine if the light ray appeared from the earth to travel that path or that distance.  But, as I said, the little man does not observe that hypotenuse.  It absolutely cannot be part of his data!
     One of the outcomes of Einstein's relativity is that all events are local.  That is, all measurements (of time, distance, etc.) are good only for the measurer.  Another measurer in another place will get different measurements.  And yet, by applying the Pythagorean theorem to this situation, the inventors of the visualization are attempting a non-local measurement.  They are taking information obtained in a local field [specifically, the distance D, obtained by local measurement in A] and transferring it into a non-local reference field [the field in B].  This is not allowed, by the very theory they are trying to prove.  In this way, the argument is circular.  In order to prove that all events are local, and that time and distance are relative, they assume that quantities can be transferred from one system to another, and that D and L in A are the same as D and L in B.  But quantities like D are transferable only if t and x are equivalent in both fields.  Besides, D is a local measurement of the co-ordinate system A, while L is an observed distance in B, and yet they are treated exactly the same.  No transformation equations are done on either one before they are plugged into the same right triangle! 
       I say "they," but it is not just the authors of textbooks who use these illustrations or the artists who create them who are to blame.  Lorentz and Einstein do the same thing.  Every illustration or conceptual analysis of this problem I have ever seen makes this same error.  The Lorentz equations came from precisely this sort of diagram or visualization, and the Michelson-Morley experiment accepts it as a given.  It is the very reason that the Lorentz equations have the form they do.  If the equivalent of this diagram had not been the accepted view at the time of the Michelson-Morley experiment, the Lorentz equations would not have had the form of the Pythagorean theorem.  As I will show, the true equations for simple time dilation have no Pythagorean component at all.  [The equation for an object moving at an angle to an observer will use plane triangle trigonometry, but not the Pythagorean theorem].*  

*To read more about Michelson-Morley, or to see a diagram of the interferometer—and see its equivalence to the diagram above—see chapter 3.

The analysis above also applies to the most current visualization of time dilation, which is the “light clock”. This visualization is now included in most up-to-date explanations, since it appears to give the shortest road to the first equation for time. You can see that the diagram below of the light clock is the same in most respects to the last diagram above. It is just a close-up view of the projector inside the spaceship, except that now the projector is made into a clock. We again use the Pythagorean theorem to find the third leg of a right triangle, and this equation yields the dilated time.



But the diagram again provides a false visualization. A light clock works by emitting a light ray. This ray reflects from a mirror opposite the clock and returns. One round-trip of the light is a tick of the clock. The diagram is meant to be a visualization of what a distant observer would see. This distant observer sees the clock moving at velocity v. The diagram must be from the point of view of a distant observer, since a local observer would not see the clock moving. The clock itself could not possibly see itself moving, since it has no velocity relative to itself. But the creator of the diagram assumes the distant observer would see light moving as it is diagrammed. In fact, he would not. A distant observer would not see the light at all. The light is moving perpendicular to his line of sight, and is invisible by definition. The vector ct/2 is imaginary. It cannot be anyone’s data. It is an assumption, and it is a false assumption. The light does not travel in that way.
       Light always travels in the local field. In the final analysis, this is simply because the local field is real and the observed field is not. A field observed from a distance is a transform field, or a mathematical field, or an optical-illusion field. It is only apparent. The vectors within this apparent field exist only in the math. They do not exist physically. Therefore there is no way that light can physically travel along these vectors that are diagrammed. Light travels straight over and straight back, and neither the light nor the clock are concerned with how that looks to us from a distance.
       It is true that the time will be dilated for the distant observer, but not for the mathematical reason diagrammed here. The time will be dilated due to differences in the signals the clock sends to us directly. To know what is going on with this light clock, the light clock must send us information in some way. Most of the time this information will arrive on light rays, whether it is visual or not. It is these light rays that actually come to us that we must do the transforms with.
       In the light clock diagram, the creator of the diagram has shown us three separate positions of the clock. If we assume that the third position is farther away from us—as distant observers—than the first position, then it must take light longer to reach us from that third position. This, by itself, will create time dilation, since the period of the clock will be stretched out.
       The transforms of Relativity must be done on light rays that are part of our data. We cannot do transforms on imaginary light rays. Special Relativity applies to real data, to real observations, and cannot be applied to non-verifiable assumptions or imaginary vectors.
       As further proof of this, let us analyze the diagram with a bit more rigor. Notice that the clock in the first position needs to send the light ray off at an angle. How does the clock know at which angle to emit the light? You will say that we are using light that is emitted in all directions, so that some of that light will reach the clock at the third position. But what if we imagined a light clock that did not emit light in all directions. What if we assume a light clock that sends out a laser? Does the creator of this diagram mean to imply that if our clock sends out a laserbeam straight across to the far mirror, it will come back and miss the clock, since the clock will have moved on? Unlikely.
       You may answer that the laserbeam emitted from the clock will act like a ball bounced by a boy on a bicycle. If a boy bounces a ball while riding a moving bicycle, the ball comes back to his hand. This is because the ball has a forward motion as well as a bouncing motion. If you answer this or assume this, you have broken Einstein’s Postulate 2. To paraphrase: “The speed of the light is independent of the motion of the source or the receiver.” The laser is emitted at speed c at one and only one angle. It cannot also have a forward motion. This forward motion would be a forward velocity, which would take its total velocity above c.
       You may now answer, “OK, so the diagram has problems. But you yourself admit that the laserbeam will end up at the third position. How else can you diagram it?” The answer is: you cannot diagram it properly. It is impossible to diagram three separate events on one piece of paper. The false light vector is caused by trying to compress three events into one diagram. You end up getting a pretty good idea of how the sequence of events might unfold relative to a distant observer, but you get a very skewed idea of the way that light is traveling. This is because the light traveling is not an “event” to the distance observer. As I keep saying, that vector is invisible: it is not a possible part of his data. That vector exists only in a diagram like this, which is a compressed and false diagram. The real event is not taking place at a distance. The event of light traveling across to the mirror and back is a strictly local event. Light can be seen only where it is. It cannot be seen where it is not. Light cannot be seen at a distance. Therefore you cannot transform the light itself. You can only use the data given you by light to transform other things.
      Einstein himself admitted this, and warned us of it. He told us that light is a special case. It is not an object like any other object. When we use light in Relativity, we must take special care not to make false assumptions. This is what Postulate 2 means: "Take care! Light is not just another object. It is emitted at velocity c from all objects, no matter how they are moving themselves. You may not add your velocity to the velocity of light, not if you are travelling in the same line as the light, and not if you are travelling at an angle to the emission."
       Therefore, the question, “where is the light right now?” is always a local question. If the light is in your eyes, you can say, “it is here.” Otherwise, you can say nothing. You cannot propose to diagram light from a distance, since your diagram is supposed to be a diagram of your data. You cannot diagram light from a distance, since there is no such data. You know only the information that light brings you, and when you receive this information the light is right in your eyes. Light in other fields is imaginary data, and if you try to diagram it you are sure to get into trouble. As a matter of data or of operation or of experience or of science, there is no such beast as “light in another field.” Light is always in the local field. That is what Einstein’s Second Postulate means.

Part IV: 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

                                        T_n = T _n ' + (x_n' /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.

Notice here the difference between my equation and Einstein's. From his paper of 1905 we find t  =  t' + Δx'/(c - v)
Einstein has subtracted a velocity from c, which is disallowed by his own Postulate Two. The correct equation is mine, which simply drops the v variable from the equation. We not need a given velocity to start with, as I have shown here; but if we are given one, it must be clear which system it belongs to. If it belongs to my blinker or Einstein's train, it should be labeled v'. If it belongs to my observer or Einstein's platform, then it is already a relative velocity. You can't discover it without a relative transform, as I am presently proving (x'/c is a transform).


Now let us calculate the apparent (or relative) 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)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.  Look above. I defined both initial times as t'. The local time is t', for both the blinker and the observer. 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)         
                 

Now that we have our new equations, I predict this complaint: "You are assuming that t and x are absolute, before you even start.  Your marked off tunnel is an absolute system of coordinates,  and your t's even coincide.  They may appear to be .000003s apart, but they are really the same.   Relativity does not rely on these assumptions.  It transcends them."
          My answer is that I have done precisely what Einstein did.   I started with given quantities and derived unknowns from them by discovering the proper equations.  You are calling my givens "absolutes," but the terminology is meaningless.  They are not absolutes, they are accepted values.  Einstein does not derive his relative velocities from thin air.   In his transformation equations, you must have certain information to start with.  I claim that my givens are no more absolute than his.  My givens are exactly the same as his.  If my elucidation of his process makes you think that these givens are absolutes, then I can force you to admit that Einstein's givens are also absolutes.
       
In fact I will do so now.  Einstein says (p.18, Rel.) that the train has a given velocity v.  The implication is that this velocity is a local velocity.  It must be the velocity of the train, as measured from the train.  He cannot mean the velocity of the train as measured from the embankment.  For if he meant as measured from the embankment, then we would already be given a relative velocity, and we would not need fancy transformation equations to find it. 
         Einstein then shows, correctly, that the train's clock will look slow to the eye on the embankment and that the train's measuring rods will look short.  And he presents the Lorentz equations for t and x.    But then he never uses his relative x and t, that he derives by these equations, to calculate a relative v.  Isn't this strange?  He never concludes that there is a relative v that is different than the v given in the problem.  He derives two t's and two x's, but never derives the second v, the relative v.
          Later, he derives the Lorentz transformation for velocity, using his t and x equations (Ch.8, on Fizeau).  But this is for the addition of velocities.  It is for the situation in which the man on the train is moving relative to the train and the train is moving relative to the embankment (eye). You have two relative velocities and you want to find the third. This equation gives us a velocity of two degrees of relativity. The velocity that this equation yields is the velocity of the man as seen by the embankment. But this is not the same as the velocity of the train.

If Einstein had derived this relative velocity for the train as seen from the embankment, then it would have been clear that something was wrong. He would then have to admit that the given v (in equation x' = x - vt) was the velocity of the train as measured by the train. And if he admitted this, then he would have to admit that this velocity was measured relative to the tracks. And if he did that, then he would have to defend himself against the charge that he had already assumed an absolute co-ordinate system, just like I am having to do. But in Einstein's explanation, it never comes up. He never defines his terms, and no one has ever asked him to. No one, til now, has ever asked what the given v is, precisely, and how it is determined.
         The given v must be the v of the train relative to the tracks, as measured from the train. Einstein must be given this velocity before he ever starts: relative v is dependent on v'. So Einstein must have his railroad track already marked off before he can calculate his relative x and t.
          He even admits this. At the top of the same page (p. 18) he says, "Of course we must refer the process of the propagation of light (and indeed every other process) to a rigid reference body (co-ordinate system)." [his parentheses]
          His problem implies the existence of a pre-existing system, like my tunnel. His pre-existing system is his traintrack. But this system remains hidden throughout the problem. Regardless, this system—whether his or mine—is not an "absolute." It is not an absolute in the sense of contradicting the relativity of measurement. It is a given, a postulate that allows for the calculation of unknowns.

You may be interested to know that Feynman also admitted the existence of what you are calling the "absolute" field. He called it the "proper" field, and the time the "proper time." This is just the local field, and Feynman admitted that all local fields were equivalent. They have to be in order to do any sort of calculations, of any kind.

You may say, "Yes, but there are two co-ordinate systems (S and S').  Distance in one will not be the same distance in the other.  x does not equal x'.  You cannot just transfer x' into your equation—as you did when the blink was traveling from the blinker to the eye—as if you already knew the distance."
         Einstein did.

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