THE SPEED OF LIGHT, AND THE SPEED OF BODIES.

This analysis of the structure of reality draws the conclusion that all matter in Space-Time is constantly on the move,
hence the Constant Motion Theory ( CMT ).



    Is MOTION possible ?

    Is CONSTANT MOTION the one and only origin of INERTIA ?

    Is INERTIA the resistance towards a change in direction of that CONSTANT MOTION ?


Observational studies of a complex reality will not lead you to an understanding of what reality is. To understand reality, the best route to take is to THINK about what existence is, and to THINK about what is required to create existence. In other words, start with the basics and understand that the complexities are simply the outcome of these basics in action. Starting from the basics does not mean starting with the basics that are already accepted. To eliminate any previous errors made by man, one must start from point zero. This also eliminates any error of assumption which may have led one to think that the basics are already exposed, when in actual fact the true basics lie below that which is presently accepted as the " BASICS ".   In other words, simpler truths which are closer to the foundation by being the actual foundation itself, may not have yet been exposed despite the majority thinking otherwise.

Unfortunately, what is accepted as today's basics, are considered to be simple. Because of this simplicity, the understanding of these simplicities can not be in error, so says the majority. This then ensures that Physics will never be complete, if these basics ARE in error. Therefore, those who do not review the basics, are those who oppose the possible completeness.

The information below will make it clear that Relativity is a necessity needed to make Variables possible, and that the Quantum units, Time Units, Space Units, Energy Units, and Most importantly Event Units, are all required to make this Relativity possible in the first place. Without units, one can not relate one to another. Without units, an Event can not be an event, for an event is a unit of occurrence. However, it is NOT being said that there is no existence of an Absolute that is also the foundation of Relative functions.


Motion exists, therefore for starters we must understand what is required to make motion possible. If "Motion" is examined closely, it appears as though it is ABSOLUTELY IMPOSSIBLE. Logic says that it would be impossible to push an object that is at rest in space, into motion!   For instance, if force is applied to an object that is at rest in space, it would first have to disappear at its original rest position to make the process of moving from such a position possible. Then after doing so, it must somehow magically reappear at a new position. This practice must then continue if the motion is to continue !

People in general, simply accept without question, and therefore have not examined this event closely at all. As the result of this lack of effort, they skip the facts and then proceed onward just thinking that the object has simply moved " FROM " the original rest position, ignoring the fact that it actually had to disappear " AT " this original rest position to begin the process of motion in the first place. Yet this is a simple fact that can not be denied. Even if one says that it disappears from  its original rest position, this still means that it had to disappear at its original rest position. Therefore logic dictates that the Universe must have been designed in such a manner that this problem is somehow overcome! To do so, there must be a replacement for the ABSOLUTE MOTION that we think exists, a replacement for what we perceive as variable motion, both replaced with something more extreme but also something that has deceived us so far ! 

REST POINT "A" TO REST POINT "B"    

Fig. #1 

To help clarify this point, imagine a hypothetical flexible object which could stretch then contract to reach a new position. In this case, the object will still be maintaining its presence, its existence, all during this entire transition of movement from point "A" to point "B".

Now imagine a solid object. It can not stretch, and so once pushed into motion it would have to disappear at its original position of rest, and then it must somehow reappear at the new position, leaving its method of existing, non-existing, then existing again, UNCLEAR.

The second simple observation is the fact that MOTION contains Two variables. 1) Distance and  2) Speed. Variables range from zero to infinity, or in this case of course the variables range from greater than zero to infinity, since zero itself would obviously mean no motion at all. Now if we were to set both of theseVariables  to infinity, the outcome would be as follows.


1) Traveling across an infinite distance  --> To continue to travel without end.

2) Traveling at an infinite speed  --> To travel across any distance without the passage of any time.

Note: If it takes time to get from point "A" to point "B", this means that you can still move faster and use less time to get from point "A" to point "B", therefore you are still traveling at a finite speed, not at an infinite speed. Only if no time at all occurs while moving across any distance, is the speed actually an infinite speed.

This combination of infinities produces a paradox because, 

1 ) + 2 )  =  To continue to travel without end, and all done without the passage of any time at all.
Or a more amusing version,

TO GO ON FOREVER IN NO TIME AT ALL ???

How many of you out there noticed these peculiar flaws ?

If not, it simply means you have followed the pattern of monkey see monkey believe, and have questioned nothing !
But at least this allows you to absorb knowledge, flawed or not, like a sponge.

The hypothetical flexible object, once stretched, has been able to exist at more than one place at the same time. Therefore the solid object somehow must also be able to exist at more than one place at the same time, to make the start of motion possible.    But How??   We will get into the details of that later on.

To go on forever in no time at all, implies that at least two clocks (1) & (2), are involved. The motion across space must continue on forever (2), but while doing so, time must come to a stop (1). Therefore, at some finite speed, the ticking of time from the moving objects point of view (1), will have to come to a complete stop while the traveling itself still continues onward at this same finite speed, continuing onward across space, forever (2).

But How??

It is as if the universe was run by a computer that can only do a limited amount of work. Although responsible for an individual objects sense of both TIME and MOTION, the computer is forced to reduce the objects awareness of the passage of time, if the object is in motion. The computers abilities are clearly Finite. Therefore, both TIME and MOTION are encapsulated within a FINITE of some kind.

The next step is a TIME / MOTION analysis.


TIME / MOTION analysis

Question 1 - Is an object that is at rest in Space at complete rest ?   Answer, - No.
  It is in motion through TIME.
 
Question 2 - Is the motion through Time at an infinite speed  ?   Answer, - No.
  As just stated on the previous page, both time and motion are
  encapsulated within a finite of some kind, therefore limiting the
  speed of motion through both SPACE and TIME.

Since objects are still in motion even if at rest in space or time, it is simple and logical to say that all objects are constantly in motion in a four dimensional reality, meaning a TIME - SPACE reality.


TIME - SPACE CONSTANT MOTION VECTORS

FIG. #2

By changing the direction of travel within TIME-SPACE as shown in Fig. 2 above, one can ---

        .    1)   Be at rest in  SPACE and move through TIME.
        .    2)   Move through both  SPACE  and  TIME.                  ( With time having slowed down. )
        .    3)   Be at rest in  TIME  and move through  SPACE.       ( To go on forever, in no time at all. )
        .    4)   Travel at an infinite Speed ( crossing distances in no time at all ), but at a finite Velocity.
        .    5)   Conclude -  All   INFINITIES   reside within a   FINITE   of some kind.

All atomic clocks in a common environment, tick at a common speed. From this it is logical to conclude that they all share the same (c)onstant motion across the four dimensions of Time-Space, since all these clocks tick at the exact same speed, and all the clocks are at rest in space relative to each other. This specific cosmological (c)onstant then determines the maximum speed at which any object can move through space or time. To clarify the term 'SPEED' when dealing with motion through both TIME and SPACE, it is best to picture one's self viewing this motion from a higher 5th dimensional point of view, looking down at the 4 dimensions of TIME-SPACE, seeing all objects moving at the same speed but in different possible directions.   Using this principle of having a (c)onstant motion within TIME-SPACE, it makes it possible to create infinities within a finite system, it allows the creation of variables. However, these variables ranging from zero to infinity, only exist RELATIVISTICALLY, due to man not being able to look across the depth of the dimension of Time, thus taking time out of the visual observation equation. The end result leaves that which appears to be a change of the magnitude of the objects motion, when in fact it is only a change of spatial velocity alone, and not a change in the overall magnitude of the object's motion across Time- Space !

With motion understood, the next step is an OBJECT / MOTION analysis


OBJECT / MOTION analysis

In FIG. #3 diagram below, we start with an object which has a 1.000 foot length. When at rest in space and moving through time, the length of the object extends entirely across space. However, if we change the direction of travel, a fascinating change occurs. If for example the object is traveling at 2/3 of the maximum possible speed of motion across space, the object now only extends 0.745" across space. So from our point of view, the observers point of view, the object has shrunk from a 1.000" length to 0.745" instead. 


Time Space Motion Vectors
FIG. # 3   

The object itself, however, has still retained the very same length. The object has now extended partially across the dimension of time as well. The faster it moves through space, the more it extends across time and the less it extends across space. If its constant motion in Time-Space was now directed across a Spatial dimension only, the objects spatial length will have been reduced right down to ZERO !  

The most important simple observation this time, is that not only do we exist in a four dimensional reality, but all objects within this reality are also four dimensional, but of a much smaller scale.   I say are four dimensional because due to the big bang that started this universe, everything is in spatial motion to some degree. Nothing is truly at complete spatial rest.

This extension across the dimension of time, which occurs if in motion across space, implies that the object will exist at more than just one Time at the same place in Space, even though the object is in motion across Space. This object is  not a flexible object that is stretching across space as in the example given in FIG. #1, so what exactly does it mean for it to be four dimensional ?

In this case, Logic says that it is ABSOLUTELY IMPOSSIBLE for a solid object to exist at more than just one time in one place in space while it is constantly moving away from its previous position. Therefore this Universe must have been designed in such a manner that this problem too is overcome !   Once again logic says that this phenomena can therefore only exist RELATIVISTICALLY, and not in the ABSOLUTE sense. Clearly, much more analysis is still required, but before we do that, let us do a geometric analysis of what we understand so far.

The next step is a GEOMETRIC   analysis.

GEOMETRIC Analysis

  L
  L'
  b

  =  the objects constant length.
  =  the objects length that extends across Space.
  =  the objects length that extends across Time.

    

c
v
t
 t'

  =  the magnitude of the objects (c)onstant motion across Space-Time.
  =  velocity. The percentage of the (c)onstant motion that is across Space only.
  =  the magnitude of motion across Time, if at rest in Space.
  =  the magnitude of motion across Time, if also in motion across Space.

Looking at the drawing above, you can see an objects length ( L ) if the objects (c)onstant motion is across Time only. However, it also shows the object if it were in motion across both Time & Space, with the Spatial velocity of ( v ). In this case, only a percentage of the ( L ) length of the object extends across Space, and it is noted as ( L' ). The remainder of ( L ) now extends across Time, and it is noted as ( b ). Within this drawing I have created two right angle triangles ( Red  ). The triangles are identical with the exception of scale. Disregarding the scale, the ratios between the side lengths of each triangle, are identical. As shown in the drawing above, we can then take advantage of the Pythagorean theorem, which allows us to create the basic equations, and perform substitutions in the production of the equations, thanks to the triangles being identical with exception of scale.

The two equations shown in the boxes at the bottom of the above drawing, make it possible to calculate the Spatial length of the object based upon its velocity of motion across Space, and also calculate the reduction of size of a time period measured while in motion at this velocity. But there is still more that has changed !

STANDARD REFERENCE UNITS
 
GEOMETRIC Analysis

On the drawing above, I have set up STANDARD REFERENCE UNITS, with each being related to " c ".

With ( c ) being the constant motion measure of all mass particles within Time-Space, this provides us with a standard reference measure of Time, Space, and Objects. Since the ( c ) constant equals 299,792.458 km per second, a one second Time Unit is equivalent in length to a 299,792.458 km Space Unit length, and equivalent to a 299,792.458 km Object Unit length.
 
In the above drawing, the Object Unit is also shown in motion in the direction across Time-Space such that it's spatial velocity is 260,000 km per second ( 259,627.8845 km per sec.). At this velocity, the Object Unit is rotated in Time-Space, and this results in the Object Unit extending across 0.8660254 sec. of the 1 second Time Unit. At the rear end of the Object Unit, there is therefore a time offset of ( + 0.4330127 sec. ), and at the front end there is a time offset of ( - 0.4330127 sec. ).

The Object Units extension of 0.8660254 sec. across the 1 second Time Unit,
is determined by v / c .

The object unit is at a 90 degree angle relative to the ( c ) vector, and thus the object unit extends across a percentage of the Time Unit that is equal to the percentage that the ( v ) vector extends across the Space Unit, whose length is also equivalent to the ( c ) vector. Hence the Object Units extension of 0.8660254 sec. across the 1 second Time Unit, is determined by v / c.

Let us picture an Object Unit being a Train which has a length of 75,000 km ( 74,948.1145 km ) when at rest in space. Rather than being the same in length as the 299,792.458 km length of the reference Object Unit shown in the STANDARD REFERENCE UNITS diagram, there is a difference, and so this difference must be taken into account if we wish to determine how far the Train extends across the dimension of Time. We can determine the difference in scale by division. ( 74,948.1145 km / 299,792.458 km = 0.25 )

Therefore, the extension of the Train across time, while the Train is in motion across space at the velocity of 259,627.8845 km per sec., equals ( 74,948.1145 km / 299,792.458 km * v / c = 0.21650635 sec. ). The time offsets are therefore ( - 0.21650635 / 2 = - 0.108253175 sec. ) at the front of the Train, and ( + 0.21650635 / 2 = + 0.108253175 sec. ) at the rear of the Train. The total of these time offsets is of course 0.21650635 sec., and so if clocks were positioned at both ends of the Train, then the clock at the back end of the Train will be ahead of the clock at the front by 0.21650635 of a second.
 
You may think that it does not make sense for one end of the Train to be ahead in time and the other to be behind, but you have to understand that the " Present Time " is still being shared by all. What this means is just as I have stated before, that all objects travel with the (c) motion in Time-Space. Therefore, at any time, all objects are at an equal distance in Time-Space from where they were positioned just moments beforehand in that Time-Space. This is what they all have in common. This is how they are all grouped together. This is what the " Present Time " actually is.

Note : Since the " 299,792.458 km " Object Unit reference length is also the distance light travels in 1 second ( c ), the equation ( 74,948.1145 km / 299,792.458 km * v / c = 0.21650635 ) can also be expressed as ...

x / c * v / c = 0.21650635
 
or
x * v / c2 = 0.21650635
or
v / c2 * x = 0.21650635
 
The ( x ) variable in this case, is the 74,948.1145 km length of the Train.

While the Train is being accelerated to 259,627.8845 km per sec., the Train is being rotated across Time-Space, and each clock onboard is slowing down. Once the Train is traveling at the velocity of 259,627.8845 km per sec. , and since t' = t * SQRT( 1- v2 / c2 ), then t' = t * 0.5 , and so the clocks are ticking at half speed.
 
If we were to send light from one end of the train to the other, and the train was at rest on the tracks, it would be observed by both those within the train, and by those at the trackside, that the light takes 0.25 sec. to complete this trip. But this agreement is about to change now that the train is in motion across the tracks at 259,627.8845 km per second. If we send light from the rear end on the train to the front end, then to an external observer standing beside the tracks, it would appear to take 0.9330127 of a second for the light to complete the trip, since the train is moving in the same direction as the light, thus the difference in speed is determined by ( c - v ). Also taken into account is the fact that the external observer sees the length of the train as ( 74,948.1145 km / 2 ) since the train now extends patially across the dimension of Time and thus extends less across the dimension of Space.

t = ( c - v )  /  ( L / SQRT( 1- v2 / c2 )
 
t = ( c - v )  /  ( 74,948.1145 km / 2 )
 
t = ( 299,792.458 km per sec. - 259,627.8845 km per sec. )  /  37,474.057 km
 
t = 40,164.5735 km per sec.  /  37,474.057 km = 0.9330127 sec.

If we were to send light in the opposite direction, to the external observer it would appear to take only 0.0669873 of a second for the light to complete the trip from end to end of the train ( 74,948.1145 km / 2 ), since the train is moving in the opposite direction than the direction that the light is moving.
 
Therefore if onboard the train, and with t' = t * 0.5 taken into account, meaning the clocks onboard the train are ticking at half speed, the measuring of a 0.9330127 sec. light path time period becomes ( 0.9330127 * 0.5 = 0.46650635 sec. = t'1 ) and the measuring of a 0.0669873 sec. light path time period becomes ( 0.0669873 * 0.5 = 0.03349365 sec. = t'2 ) .
 
With all this taken into account, and if it is agreed by those monitoring each clock at each end of the Train that the light will be released at 0.00 time, the person at the front of the Train where the light is received will be unaware that the other persons clock is ahead of his by ( + 0.21650635 sec. ) and he will think that the light was released at the 0.00 time relative to his own clock, even though it was not. By the time his clock registers 0.00, the light will already have traveled for 0.21650635 of a second. This will therefore subtract 0.21650635 sec. from the measuring of the ( t'1 ) 0.466506635 sec. time period. ( 0.466506635 - 0.21650635 = 0.25 sec. ).
 
Since v = x / t , then 74,948.1145 km / .25 sec. = 299,792.458 km per second. Therefore it appears to the observer on the Train, who still measures his Train to be 74,948.1145 km in length since his measurement instruments have also contracted in spatial length, as though the light has still traveled from one end of the Train to the other at the expected speed of light.

t'  =  0.25 sec.  =  t'1 - v / c2 x

If instead the light was released from the front end of the Train at 0.00, we would simply add the 0.21650635 time offset to the 0.03349365 sec. time period, since the clock at the rear end of the Train is already ahead by 0.21650635 of a second at the very same time that the light is released from the front end. ( 0.03349365 + 0.21650635 = 0.25 sec. ). Since v = x / t, then 74,948.1145 km / .25 sec. = 299,792.458 km per second. Therefore, in this case as well, it also would appear as though the light has traveled from one end of the Train to the other at the expected speed of light.

t'  =  0.25 sec.  =  t'2 + v / c2 x

Reviewing the example of the external observers observation of light traveling from the rear of the train to the front in 0.9330127 of a second, then what is the equation needed to Transform this view of this time period, to the view experienced while onboard the train. Let's start by using what we have uncovered so far and then reverse the process. We must take the ( t' = 0.25 sec. ) measure that was measured onboard the Train, and add this to the total time offset of ( v / c2 * x ) that was also experienced onboard the Train, and then reverse the fact the clocks onboard the Train are running at half speed by dividing ( 0.25 + v / c2 * x ) by the SQRT( 1- v2 / c2 ) .

( x = 74,948.1145 km ,   v = 259,627.8845 km per sec. )
 
0.9330127 sec. = ( 0.25 sec. + v / c2 * x ) / SQRT( 1- v2 / c2 )

This then gives us the equation such that we can calculate the 0.9330127 of a second time period that will be seen from the stationary trackside point of view, and done so by taking the time period of 0.25 of a second, as seen by those onboard the train, and then applying it to the equation. The reversal of the equation produces the final equation that allows the track side observers to determine what time period will be experienced by those onboard the Train.

( x = c * 0.9330127 sec. = 279,710.171 km ,   v = 259,627.8845 km per sec. )
 
0.25 sec. = ( 0.9330127 sec. - v / c2 * x ) / SQRT( 1- v2 / c2 )
 

Hence the Lorentz Time Transformation equation...
t' = ( t - v / c2 * x ) / SQRT( 1- v2 / c2 )


 
Train

Looking at the drawing above , we have the Train and the time offsets noted at the ends of the Train. Ahead of the Train just further down the track, there is a track loop. Now if the train goes on to the loop, the engine will eventually meet up with the caboose. This will place all four clocks at about the same location.

THE SAGNAC EFFECT

Now for the interesting part. Since the clocks onboard the Train are no longer displaced across Time nor Space from each other, then the clocks also no longer have Time offsets relative to each other. Granted these two clocks will still be running at half speed since t' = t * 0.5 when traveling at 259,627.8845 km per sec., but the light will no longer be measured as though it takes the same amount of time to go from one end of the train to the other, that it does if the light is sent in the opposite direction. ( To send light from end to end of the Train while it is on a loop, let's say that we have a fibre optic cable onboard the Train and it extends from end to end of the train, and that the light is sent across the cable. For the nit pickers, we could say that it is a hollow cable that is lined with a mirror surface, and that the hollow tube is sealed and that all gasses have been removed from the tube, hence it is a vacuum inside. )
 
In this loop example, and with t' = t * 0.5 taken into account, the measuring of the 0.9330127 sec. forward light path time period becomes ( 0.9330127 * 0.5 = 0.46650635 sec. ) and the measuring of the 0.0669873 sec. reverse light path time period becomes ( 0.0669873 * 0.5 = 0.03349365 sec. ).
 
But despite this, once the Train is back on the straight track, it will once again appear as though the light travels in either direction from end to end of the Train at the expected speed of light. ( See http://en.wikipedia.org/wiki/Sagnac_effect for details about the confirmed Sagnac effect. ). In the Train loop example, the train becomes the rotating platform.
 
So far, Relativity has not come into view, despite the fact that equations have popped up that are identical to the Lorentz-Fitzgerald contraction equation, the Time Dilation equation, and one of the Lorentz Transformation Equations.
 
Let's go one step further and proceed with the equation that will transform the object length from one point of view, to the other.
 
From the stationary observers point of view, the light traveled from the rear of the Train to the front in 0.9330127 of a second. The distance that the light travels in that time period is determined by ( 0.9330127 sec. * c = 279,710.171 km = x ). OK, the next step is to subtract the distance traveled by the train in that same time period while traveling at 259,627.8845 km per second such that all we have left is the actual spatial length of the Train itself.

( x - vt )
= ( 279,710.171 km - ( 259,627.8845 km * 0.9330127 ) )
= ( 279,710.171 km - 242,236.114 km )
= ( 37,474.057 km )

Now we have the actual contracted length of the train, but to those onboard the Train, they are unaware of the contraction due to their measurement instruments having also contracted in spatial length, and so they still think that the length is still 74,948.1145 km. We must therefore reverse the contraction by adding / SQRT( 1- v2 / c2 ) to the equation.

x' = ( x - vt ) / SQRT( 1- v2 / c2 )
x' = ( 37,474.057 ) / SQRT( 1- v2 / c2 )
x' = ( 74,948.1145 km )

The ( y' ) and ( z' ) axis are not effected,
due to the spatial motion being along the x axis,
thus the rotation across the dimension if time is from the x axis,
and so we now have the complete Lorentz Transformation Equations.

 

      x'  =  ( x - vt ) / SQRT( 1- v2 / c2 )

      y'  =  y

      z'  =  z

      t'   =  ( t - v / c2 * x ) / SQRT( 1- v2 / c2 )

 

OK let's put the equations to the test by reviewing the Train example one more time.

Train

Looking at the Train's original rest length of 74,948.1145 km, which is 1/4 the length of which light will travel in 1 second, if we sent light from one end to the other and then back again to the starting point, it would require 0.5 of a second. Based on the calculations done so far, we have also discovered that to those onboard the Train, it would still seem to take 0.5 of a second ( 0.25 sec. + 0.25 sec. = 0.5 sec. ) for light to complete the round trip even though in this case the Train was no longer at rest, but was in motion across the train tracks at a velocity of 259,627.8845 km per second, while noting of course that the onboard clocks would be ticking at half speed ( t' = t * 0.5 sec. ). Therefore if this event was monitored from a stationary point of observation on the train tracks instead, it would appear to take 1 second ( 0.9330127 sec. + 0.0669873 sec. = 1 sec. ) rather than 0.5 of a second.

Train

Looking at diagram A) above, we see two paths of light. One is outside of the Train and is being measured relative to the tracks, the other is the light inside the Train, as the Train is on the move at 259,627.8845 km per second across the tracks, moving in the same direction as the light. Let's us say that at the rear end of the Train there is a light bulb, and that the electrical conductors from that bulb then hang from that rear end of the Train. On the train tracks there are additional conductors that the Train conductors will come in contact with. These conductors lead to the light bulb at the side of the train tracks , and to a power supply. Therefore, once the rear end of the train reaches this point, both light bulbs will briefly light up. Now let's assume at the front of the train on the inside there is sign that says " ARRIVED ". Let us also assume that there is another sign that says " ARRIVED ", that is positioned 279,710.171 km further down the tracks from the location of the side track light bulb, and that this sign will light up when the flash of light reaches it, just as the sign inside the Train will also light up once the flash of light inside the Train reaches it.

If we have someone at the front of Train waiting to see the sign light up, when it happens he will also notice a sign outside of the train light up at the same time.   WHY ?   Well the measuring of the speed of light is the same whether the light is being measured inside the Train, or outside of the Train, and if you recall, both light bulbs were turned on at the same time. Therefore, from the Trains frame of reference, the only difference is that in one case the light is external, and the other is internal. This then ties our two observers, stationary and non-stationary, and our two frames of reference, both together. Neither is detached from the other in any magical way.

t'  =  ( t - v / c2 * x ) / SQRT( 1- v2 / c2 )
0.25 sec. =  ( 0.9330127 sec. - v / c2  *  279,710.171 ) / SQRT( 1- v2 / c2 )

x'  =  ( x - vt ) / SQRT( 1- v2 / c2 )
74,948.1145 km =  ( 37,474.057 km ) / SQRT( 1- v2 / c2 )

Well the Lorentz transformation equations agree with the real events, and no floating or detached inexplicable Relativity has appeared as of yet.

Train

Looking at diagram B) above, we now have the light traveling in the opposite direction. In this case we have the light bulb and electrical conductors at the front of the Train and the sign is at the rear end of the Train. Here the bulbs will light up when the front end of the train makes contact with the track conductors. The external sign is located 20,082.287 km further back down the track from the position of the track side light bulb.

t'  =  ( t + v / c2 * x ) / SQRT( 1- v2 / c2 )
0.25 sec. =  ( 0.0669873 sec. + v / c2  *  20,082.287 km ) / SQRT( 1- v2 / c2 )

x'  =  ( x + vt ) / SQRT( 1- v2 / c2 )
74,948.1145 km =  ( 37,474.057 km ) / SQRT( 1- v2 / c2 )

This time the ( - ) sign in the equations has been to changed to a ( + ) sign. It the Lorentz Transformation equations were not adjusted to compensate for the change in direction of the light, then the internal time period measured ( 0.25 sec. ), would not be correctly related to the actual time period measured on the tracks ( 0.0669873 sec. ). In the minds of the Relativists, this is seen as a case of  - v, a negative velocity.

RELATIVE POINT OF VIEW

RELATIVE POINT OF VIEW
 

In the above diagram I have taken the Standard Reference Units Diagram and have rotated it such that we are looking at the surroundings from the Object Units point of view as it is  in motion  across space at it's spatial velocity is 260,000 km per second ( 259,627.8845 km per sec.). If the Object Unit were a space ship that was 300,000 km ( 299,792.458 km ) in length, and one was positioned at one end of the space ship, as one then looks toward the opposite end of the space ship, one would also be looking across a distance of the dimension of time, and while doing so would be unaware of this. Therefore, if you were to look at another space ship that was 300,000 km ( 299,792.458 km ) in length that was  at rest  in space, then this other space ship will appear to be half the length of its actual spatial length. In other words, due to the orientation of the first space ship relative to the dimensions of Time-Space, the perception of the depth of space has been altered. With this being the case, to each observer in either of the two space ships, it will appear as though the other space ship has become contracted in length to half of the rest length. Also, due to the orientation of the first space ship relative to the dimensions of Time-Space, the perception of the depth of time has been altered. With this being the case, to each observer in either of the two space ships, it will appear as though events that occur on the other space ship, occur at a rate that is half the speed that would be expected if both ships were at rest in space relative to each other, meaning both being oriented in the same direction relative to the dimensions of Time-Space.

So at this point we , I mean I, have determined the simple reasoning as to why the speed of light is always measured as c no matter what frame of reference it is measured from, with just one exception. The exception is why is it that light always travels across Space at the c velocity no matter what the velocity is of the body that emits that light.

Well, we are beginning to see that matter is four dimensional. If one accepts the 4 dimensional structure of matter, then it is also understood that a spinning particle can and will have its axis extending across both Space and Time, and that this axis will continue to extend further into the dimension of time the faster that the particle moves across space. This also effects the release of a Photon. If the axis becomes tilted a specific amount across time due to the specific velocity across space, the spatial angular velocity is reduced in one direction, and increased in the other. This means that if a Photon is released, its velocity is ( c + v ) in one direction, or ( c - v ) if released in the other direction. Therefore, no matter what speed a particle is moving across Space, if it releases a Photon in either the direction of which the particle is moving across Space, or in the other direction, that Photon will still end up traveling across Space at the ( c ) velocity.

Meason

As shown in the diagram above, if a Meson decays, it does not matter what its velocity is across Space, the photons when released will still travel across Space at the one and only speed of light. Concerning the moment of decay of a Meson, it is to be noted that if the Meson is at rest in Space, then its entire constant motion will be across the dimension of time only. Thus, when the Meson decays and splits into two Photons, not only are the Photons now thrown across space at the ( c ) magnitude of motion, the but it is to be noted that these Photons already have an equal degree of motion across the dimension of Time. Therefore the total magnitude of motion for a Photon across Space-Time, is ( c * SQRT(2) ). Therefore Photons are not at a standstill in time, which is just as well because if they were actually frozen in Time, then their properties would also be frozen in Time, and therefore these Photons could not interact with any other particles forever onwards !
 
If time slowed down for all objects that surround you, and if there is a baseball passing by you, it would be moving very slowly, and if you tried to change its direction of travel, it would take a large amount of energy to do so. The fact that the baseball is in a slower time frame of reference, this creates a simulation as though its mass has increased. If time was at a stand still for the baseball, then even an infinite amount of force applied to it could not change its present position in space. Therefore, if the baseball was moving across space at a high velocity, it too would be within a slower time frame of reference, and this too would create the simulation of there being an increase in its mass.
 
And so, since a photon is still in full motion across the dimension of time even though it is in motion across space at 300,000 km per sec., it will not simulate there being an increase in its mass !  In fact it will appear as though the photon has no mass at all.
 
And so, overall the speed of light will always be the speed of light, and it will always be measured to be the speed of light no matter what the velocity of the body which emitted the light, nor the velocity of the moving platform that is used to measure the speed of the light.
 
The other point to be noted is that since all matter is in Spatial motion to some degree, the axis of all spinning particles extends partially across Time and therefore such matter is rotating back and forth across time. This resolves our problem of particles having to be at more than one place at one time to make motion possible. As a particle is on the move, as the result of the axis extending across time, it also reaches backward and forward across time. This then simulates the concept of actually being located at more than one place at one time, but it only occurs Relativistically, and not Holistically.

ADDING VELOCITIES

Let's get back to the Train again. At this time, we are aware that if onboard the Train, it still appears as though light takes 0.25 of a second to go in either direction from end to end of the Train even if the Train's velocity is 259,627.8845 km per second. At this velocity the clocks readings are offset from each other by 0.21650635 of a second. If the Train is then brought to a halt on the tracks, once again the clocks are synchronized now that the train no longer extends across the dimension of Time. Therefore the clocks can be set to meet this " at rest " synchronized condition even while the train is moving at a high velocity. This is done by sending light from one clock to the other, while also knowing the standard 74,948.1145 km length of the Train. Once the clocks are set by knowing how long it should take for light to travel from end to end of a train of such a length, if the train is returned to the at rest condition on the tracks, sure enough, the clocks are synchonized as judged by the trackside standards.

With that taken into account, let's imagine that we shoot a high speed bullet while onboard the Train. Relative to the train, let's say that it appears as though the speed of the bullet is 259,627.8845 km per second ( u ), and that the bullet is travelling in the same direction as the train. All of this is going on while the stationary observers whom are standing near the train tracks, see that the Train itself is moving at a velocity of 259,627.8845 km per second. Thus in turn these become our two velocities that are being added together.

With our Train length taken into account, that which is known to be a length of 74,948.1145 km ( L ) by those onboard the Train, and with the Train moving down the tracks at ( v ), which in this case is 259,627.8845 km per second, and assuming we have synchronized the clocks onboard the Train previously, we then start by having the bullet leave the rear end of the Train at 0.00 time according to the clock which also is located at the rear end of the Train. When the bullet reaches the front end of the Train, the time read at the clock located at the front of the Train will indicate ( L/u )  =  ( 74,948.1145 km / 259,627.8845 km per sec. ), and so the speed of the bullet is measured to be the expected 259,627.8845 km per second by those onboard the Train.
 
Meanwhile, from the track side stationary point of view, when the bullet reaches the front of the Train, the clock at the front end reads ( L/u ) seconds, but at the same time the clock at the rear of the Train says ( L/u + L * v/c2 ) due to the Time offset between one clock and the other, which occurs since the Train now also extends across Time and not just across Space. Also from the track side point of view, the clocks onboard the Train are running a half speed, and so the time it takes for the bullet to go from end to end of the train from the track side stationary observers point of view, is determined by the following equation below, which therefore includes   / SQRT( 1- v2 / c2 )  as part of the equation to thus take into account the slower speed at which the clocks onboard the Train are ticking.

Calc

The next step is to determine how far the bullet had traveled as seen from the track side point of view in the ( Calc ) time period.  We must take into account the distance that the Train traveled at its velocity of 259,627.8845 km per second in that ( Calc ) time period, and also take into account the length of the Train itself. This is determined by ( Calc ) plus the length of the Train, not forgetting that the Train is seen in its spatially contracted length from the track side point of view, thus   L * SQRT( 1- v2 / c2 )  must be included as part of the equation.  The total length ( Calc ) is therefore determined by the following equation.

Calc

Now that we have the time period and the distance, the next step is Calc / Calc.

Calc
Calc

Now that we have the math out of the way, let's see if this is all making sense. In the above example, ( Calc ) turns out to be 1.0103629712 seconds. Onboard the Train however, clocks are running at half speed, and so this time period becomes 0.50518149 of a second. Now if you recall the time offset between the clocks, the clock at the front of the train is behind the clock at the rear by 0.21650635 of a second, and so this must be subtracted from the measuring of the 0.5051814856 sec, time period. ( 0.50518149 sec. - 0.21650635 sec. = 0.28867514 sec. ). Onboard the Train it appears as though the bullet is moving from end to end of the Train at a velocity of 259,627.88445 km per second, in combination with the velocity of the Train, and that the train still appears to be 74,948.1145 km in length.
 
Therefore the time required to complete this trip is determined by ( 74,948.1145 km / 259,627.88445 = 0.28867514 sec. ).  And so it is a match. Everything is in agreement even though those onboard the Train think the bullet is moving at 259,627.88445 km per second within the train, yet those at the track side can see that it is moving at 296,717.5822 km per second, which is only 37,089.69775 km per second faster than the Train is moving, and the bullet is therefore actually only moving at 37,089.69775 km per second relative to the Train.

Now since velocity addition occurs in this manner of simply redirecting an objects constant motion such that it becomes closer to being directed entirely across space alone, and there is no evidence of this being the case since no change is noticed by those onboard the train, there is no clear cut way to determine if you ( and the train ) are truly at rest in Space, or not. All that can be done is determine whether or not you are moving faster across Space than another, or slower. If your clocks are ticking at a slower rate in comparison to those whom you are passing by, or are passing by you, then you are in motion across space at a higher velocity. If it is the opposite, then you are moving across Space at a slower velocity. Even this can clock comparison can only be determined by comparing their measurements of the time period taken for the completion of a specific event that has been closely measured by both of these observers.

And so we can have a frame of reference that is at rest in space, and we can have movement across this frame with the maximum speed possible being the speed of light. This applies to either direction of movement. If the frame of reference is in motion across space, then we have the velocity addition and velocity subtraction in the manner as has been described above. Thus we move closer to the speed of light, or we move away from the speed of light. If the frame of reference slows down to a stop and then reverses direction of movement across space, the this velocity addition and subtraction may begin again, but we are adding or subtracting in the opposite directions than previously, but relative to space, the maximum speed possible is still the speed of light.

C Dome

So it all comes down to what level you are at, relative to someone else, on the " c " Dome.

Now let's go one step further. Let's say we have a very long space ship that is also 74,948.1145 km long just like our Train. Now let's make it clear that we know that the distance from Earth to the moon is approximately 350,000 km. Let's now set our space ships velocity to be our popular 259,627.88445 km per second. If the space ship passes by the Earth and moon, while moving along a path line that is parallel to a line drawn between the earth and the moon, the font end of the space ship will cross the distance equal to the distance between the earth and the moon in ( 350,000 km / 259,627.88445 km = 1.34808 sec. ). However, if you recall, the clocks tick at half speed when moving at a velocity of 259,627.88445 km per second, and so to those onboard the space ship will measure only a time period of ( 1.34808 sec. / 2 = 0.67404 sec. ). Therefore from the point of view of those onboard the space ship, they will have crossed the 350,000 km distance in 0.67404 of a second. The outcome of becomes ( 350,000 km / 0.67404 sec. = 525,191 km per sec. ), and so to those onboard the space ship, it appears as though they are traveling at 525,191 km per second relative to the Earth and moon.

However, despite this being the case, based on all that has been revealed so far, if those onboard the space ship were to measure the speed of light inside or outside of the space ship, they would still measure the speed of light to be standard 299,792.458 km per second.

 

COPYRIGHT 1992 - 2007 K. SEAN PROUDLER