Chapter 1
Euclidean relativity.
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Euclidean relativity is a view of time and space that matches  what we perceive as  time and space.  It is very easy to understand but  strictly speaking it is not correct.   The correct viewpoint is called The Theory of Special Relativity which is  difficult (perhaps even impossible)   to understand.   While  the following is simplistic,   it is important in understanding The Theory of Special Relativity which follows.

The units chosen are feet for distance abbreviated ft.  and nanoseconds for time abbreviated by ns.    The speed of light is assumed to be 1 foot per nanosecond.    These units are  chosen for simplicity as  by choosing them the speed of light is 1 or one ft. per ns.    There is a small error here but it is less than 2%.

We will start by considering the situation of a high speed train passing a platform at a train station.    Thus we have two frames of reference.   The platform frame where distance and time are measured by X and T and the moving train frame where distance and time are represented by X' and T'.   The velocity of the train is represented by V.    The velocity of light is C and in this case C = 1 .  D is the distance traveled where D = V * T .

X’           0                        100                     200                   300                     400                      500
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
[__________]     [__________]     [__________]     [__________]     [__________]     [__________]                     fig 1-1
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo
---------------------------------------------------------------------------------------------------------------------------------------------
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                           !
X         0                        100                     200                   300                     400                      500

Above we see the train passing the platform in figure 1-1.    The train consists  of a  engine  4 passenger cars and a caboose.   Each car is 100 feet long.  The floors of the cars  are inscribed with feet marks starting with 0 at the center of the caboose up to 500 at the center of the engine.  The units in the train frame are X' and T'

The platform platform  is 500 feet long with markers every 100 feet.    X is used to denote the distance measured on the platform frame.   and T is used to represent time in the platform frame.

The velocity of the train is 0.6 times the speed of light  or V = 0.6  ft / nanosecond.

The train is shown at the instant when the zero on the caboose is opposite the zero on the platform.

X'                                           0                        100                     200                   300                     400                      500
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
Fig 1 - 2                                                                   [__________]     [__________]     [__________]     [__________]     [__________]     [__________]
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo

/---------------------------------------------------------------------------------------------------------------------------------------------/
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                           !
X              0                        100                     200                   300                     400                      500

The picture above in figure 1 -2  shows what happens after  500 ns have elapsed.    The train is shown at a new position relative to the platform.   As we can see the caboose of the train has progressed to the 300 feet mark.       This is predicted from D = V * T  or 0.6 ns * 500     = 300 ft.    The prediction that D = V * T  is true both  in Euclidean  Relativity and Special Relativity.
As we can see the caboose is opposite the 300  ft mark and the 100 foot marker on the train is opposite the 400 foot platform mark and the 200 foot train mark is opposite the 500 foot platform mark.

Rather than drawing this last pictures we could simply compute the answers we are looking for.
Recall we already defined X and T as distance and time in the platform frame and X' and T' as distance and time in the train frame.    The equations we use are:

X' =  X - V*T                     eq  (1 - 1)

T' = T                     eq  (1 - 2)

We should note that the first equation ( 1-1)  allow us to find both X'   as a function of two variables X and T.    The second equation  ( 1 - 2 )   seems trivial at this point.

Problem  1 - 1
Using  the above equations find X'  and T' when  X = 300 ft and T = 500ns.
X' = X - VT   =  300 - (500 * 0.6) = 0  or X' = 0.
T' = T or T' = 500 ns

Problem  1 - 2
Find X' and T' when   X = 400 ft.  and T = 500ns.
X' = X - V*T   400 -( 0.6 * 500 ) = 100
T' = T  = 500 ns

Problem  1 - 3
Find X' and T' when  X = 500 ft. and T = 500ns.
X' = X - VT  = 500 - ( 0.6 * 500) = 200
T' = T = 500ns

All of the above seem simple but this is the type of problems that are generally solved with The Special Theory of Relativity.       When we apply the Lorenz equations to the problems above  we  will correct the wrong answers given above.

Chapter 2
The  Lorenz equations
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The Special Theory of Relativity is used to solve problems like those in the preceding chapter.  It does this by replacing the equations in Chapter 1  with a new set of equations  called the Lorenz equations.    We will leave out the derivation of the Lorenz equations as this can be found in other sources.

The Lorenz equations that replace equations (1 - 1) and (1 -2)  are:
X' = G( X - V*T)             eq.    (2-1 )

T' = G ( T - X*V )                eq.  ( 2-2)

where

G =  1 / square root  ( 1 - V^2/) *    eq.   (2 - 3 )

Equation  ( 1-1) is replaced by equation  (2-1).    Comparing these two equations we see that eq. 1 -1  is similar to 2-1 with the difference being a factor G..
The second of these equations is where things get really strange.    If we compare equation  1-2 to 2-2  Not only does equation 2-2 have  the factor G in it but it has a  additional factor X*V.

G is a function of the velocity.   If V = 0.6 then G =  1/ square root ( 1 - 0.36) = 1/ square root (0.64)  =  1.25.

And now we will add clocks to the platform  picture  and to the train picture used above.     We will set all the clocks on the platform to 0 ns.     Special Relativity does allow us to set all clocks on a moving frame of reference like the train  to the same time.    To set the train clocks we will use the Lorenz equations to compute the time to set.

T' = 0

0
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
[__________]     [__________]     [__________]     [__________]     [__________]     [__________]
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo
/---------------------------------------------------------------------------------------------------------------------------------------------         fig 2 -1
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                           !
0                        100                     200                   300                     400                      500
T = 0 ns .         T = 0 ns.          T = 0 ns.               T = 0ns.               T = 0 ns.            T = 0 ns.

X' = G( X - V*T)                 (2-1 )

T' = G ( T - X*V )                  ( 2-2)

G  is known to = 1.25

V = 0.6

X = 0  T = 0  find T'   T' =   1.25 ( 0 - .6 * 0 )  =   0

X = 100  T = 0  find T'    T' =   1.25 ( 0 - .6 * 100 )  =  - 75

X = 200  T = 0  find T'   T' =   1.25 ( 0 - .6 * 300 )  =  - 150

X = 300  T = 0  find T'    T' =   1.25 ( 0 - .6 * 300 )  =  - 225

X = 400  T = 0  find T'    T' =   1.25 ( 0 - .6 * 400 )  =  -  300

X = 500  T = 0  find T'     T' =   1.25 ( 0 - .6 * 500 )  =  -  375

And also we will need to find the positions of the centerline markers on the train cars.  Again using the Lorenz equations and solving for X'  we get

X'  = G ( X - V*T )

X = 0     X' = 1.25 ( 0 - 0.6 * 0 )  = 0

X =  100  X' = 1.25 (   100 - 0.6 *  0 )   =  125

X =  200  X' = 1.25 (   200 - 0.6 *  0 )   =  250

X =  300  X' = 1.25 (   300 - 0.6 *  0 )   =  375

X =  400  X' = 1.25 (   400 - 0.6 *  0 )   =  500
centerline
X =  500  X' = 1.25 (   500 - 0.6 *  0 )   =  625

This is what we get

T' = 0                      T' = - 75        T' = -150           T' = -225           T' = -300             T'- -375

0                         125                   250                   375                     500                      625
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
[__________]     [__________]     [__________]     [__________]     [__________]     [__________]
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo
/---------------------------------------------------------------------------------------------------------------------------------------------              fig 2-2
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                           !
0                        100                     200                   300                     400                      500
T = 0 ns .         T = 0 ns.          T = 0 ns.               T = 0ns.               T = 0 ns.            T = 0 ns.

This is where Special Relativity comes into play.    In the platform frame both distance and time behave just as we expect with no surprises but when we look across to the frame of a moving train passing the platform we see a entirely different scenario.    We see distances changed so they no longer match those on the platform and when we look at the time on the moving train we see times that do not agree with the platform time.    Also they do not agree with each other.

From here lets wait 500 ns and then redraw the platform and train.    After 500 ns the caboose will have traveled to the 300 foot mark.  Now we will compute the distances and times seen from the platform as the train is passing the platform.

Using  the Lorenz equations  to  compute  T'  and X'

X' = G( X - V*T)                 (2-1 )

T' = G ( T - X*V )                  ( 2-2)

G  is known to = 1.25

V = 0.6
X =  300   T = 500    find X' and T'      X' =  1.25 ( 300 - 0.6 * 500)  =  0
T' =  1.25 (  500 - 300 * 0.6)   =  400

X =  400   T = 500    find X' and T'      X' =  1.25 ( 400 - 0.6 * 500)    =   125
T' = 1.25 * ( 500 -  400 * 0.6 )    =  325

X =  500   T = 500    find X' and T'      X' =  1.25 ( 500 -  500 * 0.6 )   =   250
T' = 1.25 ( 500  -  500 * 0.6 )   =   250

And adding them to the picture we get,

ns.                 T'=   400                  T'=   325                  T'=   250

ft.                     0                        125                      250
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
[__________]     [__________]     [__________]     [__________]     [__________]     [__________]
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo

/---------------------------------------------------------------------------------------------------------------------------------------------                    fig 2- 3
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                            !
X              0                        100                     200                   300                     400                      500

T           T = 500 ns .      T = 500 ns.          T = 500 ns.        T = 500ns.             T = 500 ns.        T =500 ns.

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CHAPTER 3     The concept of the interval.         The next question is,  is there a pattern to the numbers  above.    It turns out there is.    It is called a interval.    Lets look again at the numbers we obtained  from the two problems above.  Lets start with   fig  2-3.

ns.                 T'=   400                  T'=   325                  T'=   250

ft.                     0                        125                      250
!                            !                          !                        !                          !                           !
____
___HH_____     ___________     ___________     ___________     ___________     [        ]___V__
[__________]     [__________]     [__________]     [__________]     [__________]     [__________]
oo         oo         oo          oo         oo          oo         oo          oo          oo         oo          oo        oo

/---------------------------------------------------------------------------------------------------------------------------------------------                    fig 2- 3
/                                                                                                                                                                         /
/                                                                                                                                                                        /
!                            !                          !                        !                          !                            !
X              0                        100                     200                   300                     400                      500

T           T = 500 ns .      T = 500 ns.          T = 500 ns.        T = 500ns.             T = 500 ns.        T =500 ns.

Now lets define an Interval.      To do this we need to define a start X and T and  a finish X and T.    This is the  start and end of the interval  in  the platform frame.    This interval is given by  the square root of  ( X^2 - T^2).    The start and end in the platform frame  is the caboose first crossing the X = 0 mark on the platform and then the caboose crossing  the X = 300 mark in the platform frame.

We will then  determine the interval for the same start and finish events in the train frame.    This  is given by the square root of( X'^2 - T'^2).

The really interesting thing is that these intervals are always equal.   In fact any moving train passing these two defined events will have a X and a T.    While the individual X and T may differ the interval given by  Square RT ( T^2 - X^2 )   are all identical.      Checking the intervals in this way serves as a good way to check our answers after using the Lorenz equations.

The interval in the platform frame is,
Start                Finish                  X^2         T^2                                  T'^2- x^2         sq. rt (  T^2 - X^2  )
X = 0                 X = 300            90000
T = 0                 T = 500                     250000                             160000                    400
The interval in the train frame is

Start                                  Finish                                      X'^2                  T'^2                                     T'^2 - X'^2             sqrt ( T'^2 - x'^2 )

X' = 0                                  X' = 0                                       0

T' = 0                                   T' = 400                                                          160000                                   160000                   400

Proper time.

The proper time of an event is used when the frame looked at has  a delta X of zero.    Thus in the problem above the moving train with X' equal to zero at the start of the event and the X' = 0 at the end of the interval then 400 ns is the proper time of this event.     If we switch to the platform frame  X distance is 300 feet and the time found is 500 ns.   This is not proper time.

While up to this point space and time     electric fields and magnetic fields

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Chapter 4

Richard Haskill has a PDF available  that describes a graphic way of solving Lorenz Equations.    It is well worth looking at and may be found at,

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Chapter 5

One of the interesting results of special relativity is the fact that given a current carrying conductor by applying the Lorenz equations we can deduce the existence of the magnetic field.     That is the existent of the magnetic field is entirely due to special relativity.     For more on this subject visit Chapter 13 section 6 of the Feynman lectures in Physics.

http://www.feynmanlectures.caltech.edu/II_13.html

This approach was also taken by Purcell It is aviable here courtesy of Dan Schroeder.

http://physics.weber.edu/schroeder/mrr/MRRhandout.pdf

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