Special Relativity for the HP-41

Overview
 

 1°)  The Lorentz Transformation
 2°)  Composition of Speeds & Relative Velocity
 3°)  Doppler Effect & Aberration
 4°)  Elastic Collision

  4-a)  2 Particles with the same masses
  4-b)  2 Particles with different masses
  4-c)  Compton Effect

-See also "Einstein's Twin Paradox for the HP-41"
 

1°)  The Lorentz Transformation

                           y'
        y                  |       z'
        |                   |      /
        |       z          |    /
        |      /            |  /
        |    /              |/-------------------------- x'                ( Ox ) // ( O'x' )
        |  /               O'                                                          ( Oy ) // ( O'y' )
        |/----------------------------------- x                        ( Oz ) // ( O'z' )
      O

-We assume that O' has a constant velocity ß = ( ß_x , ß_y , ß_z ) with respect to ( Ox , Oy , Oz ).
 and that the event ( 0 , 0 , 0 , 0 ) has the same coordinates in both reference-frames.

-The following program performs the conversion  ( x , y , z , ct )  < >  ( x' , y' , z' , ct' )  for a given event
 

Formulae:     With  r(x,y,z)  and  r'(x',y',z')  ,  g = ( 1 - b² ) ^-1/2  ,   ß = || ß || = ( ß_x² +  ß_y² +ß_z² ) ^1/2

   r  =  r' + ß.g [ g/(g+1) ß.r' + c t' ]
  c t = g ( c t' + ß.r' )
                                                                           where  ß.r and ß.r' are the dot products
   r'  =  r + ß.g [ g/(g+1) ß.r - c t ]
  c t' = g ( c t' - ß.r )
 

Data Registers:              R00 = Dilatation Factor = g = ( 1 - ß² ) ^-1/2      ( Registers R01 thru R03 are to be initialized before executing "LRTF" )

                                      •  R01 = ß_x         R04 = x    ( or x' if SF 00 )             R08 = the invariant    c² t² - x² - y² - z² = c² t'² - x'² - y'² - z'²
                                      •  R02 = ß_y         R05 = y    ( or y' if SF 00 )
                                      •  R03 = ß_z          R06 = z    ( or z' if SF 00 )
                                                                  R07 = ct  ( or ct' if SF 00 )

Flags:   F00        CF 00 for the transformation  ( r , ct )   >>>  ( r' , ct' )
                            SF 00 for the transformation  ( r' , ct' )  >>>  ( r , ct )
Subroutines: /
 
 

 
   ( 98 bytes / SIZE 009 )
 
 

 
Example1:    Velocity  ß = ( 0.4 ; 0.5 ; 0.6 )   Event = ( x , y , z , c.t ) = ( 1 , 2 , 3 , 4 )

  0.4  STO 01
  0.5  STO 02
  0.6  STO 06

   CF 00

   4   ENTER^
   3   ENTER^
   2   ENTER^
   1   XEQ "LRTF"  >>>>   x =  -0.5324
                               RDN   y  =   0.0846
                               RDN   z  =   0.7015
                               RDN  c.t =   1.6681

-We also have the dilatation factor g = R00 = 2.0851  and the invariant   c² t² - x² - y² - z² = R08 = 2
 

Example2:    With the same  ß = ( 0.4 ; 0.5 ; 0.6 )   Event = ( x' , y' , z' , c.t' ) = ( 1 , 2 , 3 , 4 )

    SF 00

   4   ENTER^
   3   ENTER^
   2   ENTER^
   1   XEQ "LRTF"  >>>>   x =  6.1401
                               RDN   y  =  8.4251
                               RDN   z  = 10.7102
                               RDN  c.t = 15.0130
 

Notes:

-If R08 > 0  it is a time-like event.
-If R08 < 0  it is a space-like event.
 

2°)  Composition of Speeds & Relative Velocity
 

-We assume that the units are chosen so that  c = speed of light = 1

                           y'
        y                  |       z'
        |                   |      /
        |       z          |    /          . P = Particle
        |      /            |  /
        |    /              |/-------------------------- x'
        |  /               O'
        |/----------------------------------- x
      O

- O' has a constant velocity u = ( u_x , u_y , u_z ) with respect to ( Ox , Oy , Oz )

Problem1:   The speed of a particle P is v = ( v_x , v_y , v_z ) with respect to ( Ox' , Oy' , Oz' ). Calculate its speed w with respect to ( Ox , Oy , Oz )

Answer:      w = ( w_x , w_y , w_z )  with

   w_x = [  v_x + g.u_x ( g u.v /(g+1) + 1 ] / [ g ( 1 + u.v ) ]
   w_y = [ v_y + g.u_y ( g u.v /(g+1) + 1 ] / [ g ( 1 + u.v ) ]       u.v = dot product     g = ( 1 - u_x² - u_y² - u_z² ) ^-1/2
   w_z = [  v_z + g.u_z ( g u.v /(g+1) + 1 ] / [ g ( 1 + u.v ) ]

Problem2:   The speed of this particle P is v = ( v_x , v_y , v_z ) with respect to ( Ox , Oy , Oz ). Calculate its relative velocity w with respect to ( Ox' , Oy' , Oz' )

Answer:      w = ( w_x , w_y , w_z )  with

   w_x = [  v_x + g.u_x ( g u.v /(g+1) - 1 ] / [ g ( 1 - u.v ) ]
   w_y = [ v_y + g.u_y ( g u.v /(g+1) - 1 ] / [ g ( 1 - u.v ) ]
   w_z = [  v_z + g.u_z ( g u.v /(g+1) - 1 ] / [ g ( 1 - u.v ) ]

-Since these formulae only differ in a few changes of sign, we can use a unique program and a flag.
 

Data Registers:              R00 = Dilatation Factor = g = ( 1 - u² ) ^-1/2        ( Registers R01 thru R06 are to be initialized before executing "SPEED" )

                                      •  R01 = u_x     •   R04 = v_x         R07 = w_x
                                      •  R02 = u_y    •   R05 = v_y         R08 = w_y
                                      •  R03 = u_z     •   R06 = v_z          R09 = w_z

Flag:   F01        CF 01  for the composition of speeds
                          SF 01  for the relative velocity
Subroutines: /
 
 

 
   ( 86 bytes / SIZE 010 )
 
 

  where  w is the norm of the 3-vector w = ( w_x , w_y , w_z )

-if CF 01  w = the composition of  u  and v
-if SF 01  w = the relative speed  ( with respect to the 2nd reference-frame )

Example:    u = ( 0.4 , 0.5 , 0.6 )   v =  ( 0.27 , 0.37 , 0.47 )

    0.4  STO 01       0.27  STO 04
    0.5  STO 02       0.37  STO 05
    0.6  STO 03       0.47  STO 06

 CF 01  XEQ "SPEED"  >>>>   w_x =  0.434880406           ( execution time = 2.5 seconds )
                                      RDN    w_y =  0.553496668
                                      RDN    w_z =  0.672112930
                                      RDN    w  =  0.973249875

 SF 01  XEQ "SPEED"  >>>>    w_x =  -0.270737319
                                      RDN    w_y =  -0.301747643
                                      RDN    w_z =  -0.332757967
                                      RDN    w  =   0.524478981

Notes:

-The composition of speeds is not commutative in the general case.
-However, the norms  w  are identical
-There will be a DATA ERROR line 12 or 13 if  u = || u ||  is equal to or greater than 1
-However, the program also works if  v = || v || = 1 ( a photon )
  and if  v = || v ||  is greater than 1. In this case, the particle P is a ( hypothetic ) tachyon !

-The formulas are much simpler  if  u_y = u_z = v_y =  v_z = 0
 then  w_y = w_z = 0  and for the composition of speeds:
 w_x = ( u_x + v_x ) / ( 1 + u_x v_x )

-If you only want to compute the magnitude  w , the routine hereafter is shorter than "SPEED"
-It uses the formulae:

  w² = [ ( u + v )² - ( u x v )² ] / ( 1 + u.v )²    for the composition of speeds           where   u.v  is the dot product
  w² = [ ( u - v )² - ( u x v )² ] / ( 1 - u.v )²     for the relative velocities                     and   u x v  is the cross-product
 

Data Registers:              R00 = w                             ( Registers R01 thru R06 are to be initialized before executing "MAGW" )

                                      •  R01 = u_x     •   R04 = v_x
                                      •  R02 = u_y    •   R05 = v_y
                                      •  R03 = u_z     •   R06 = v_z

Flag:   F01        CF 01  for the composition of speeds
                          SF 01  for the relative speed

Subroutines:  "DOT" & "CROSS" ( cf "Dot-product & Cross-product for the HP-41" )
 
 

 
   ( 75 bytes / SIZE 007 )
 
 

  where  w is the norm of the 3-vector w = ( w_x , w_y , w_z )

-if CF 01  w = the composition of  u and v
-if SF 01  w = the relative speed  ( with respect to the 2nd reference-frame )

Example:    u = ( 0.4 , 0.5 , 0.6 )   v =  ( 0.27 , 0.37 , 0.47 )

    0.4  STO 01       0.27  STO 04
    0.5  STO 02       0.37  STO 05
    0.6  STO 03       0.47  STO 06

 CF 01  XEQ "MAGW"  >>>>   w  =  0.973249875

 SF 01  XEQ "MAGW"  >>>>    w  =  0.524478980
 

Remark:

-If we know the magnitudes  u & v  and the angle  µ = ( u , v )
 we can calculate the magnitude w and the angle µ' = ( u , w ) with the following program:
 
 

 
   ( 51 bytes / SIZE 003 )
 
 

-if  CF 01  w = the composition of  u and v
-if  SF 01  w = the relative speed

Example:      u = 0.7    v = 0.8    µ = ( u , v ) = 41°

    CF 01

    0.7   ENTER^
    0.8   ENTER^
    41    XEQ "U+V"   >>>>   w = 0.953563802              ---Execution time = 2.3s---
                                  X<>Y  µ' = 16.03924045°

    SF 01

    0.7   ENTER^
    0.8   ENTER^
    41    XEQ "U+V"   >>>>   w = 0.670241444
                                  X<>Y  µ' = 104.3994224°

-If you execute "U+V" after clearing F01 with

            0.7          ENTER^
   0.670241444   ENTER^
   104.3994224      R/S       it yields    0.8            ( with small roundoff errors )
                                           X<>Y     41°

Note:

-The formulas are:

     w   = ( u² + v² + 2 u v cos µ - u² v² sin² µ )^1/2 / ( 1 + u v cos µ )        simply replace µ by -µ for the relative velocity
  tan µ = [ ( 1 - u² )^1/2 v sin µ ] / ( u + v cos µ )
 
 

3°)  Doppler Effect & Aberration
 

-The frequency and the direction of a photon are changed by the relative motion of the source and the observer.
-The following program performs these conversions.

    S = Source                                                        S = Source

     S\\---> ß                                                      S/ /---> ß
        \ \                                           y                 /  /
         \  \                                         |               /   /
     c_S \   \ c₀                                   |         c_S/    / c                                              ß  = velocity of the source , we assume ß // (Ox)
          .     .                                     |        .      .                                                   c_S = velocity of a photon,  seen in the source reference-frame
          . _µS   .  _µ0                             |     . _µS   . _µ0                                                c₀  = velocity of the photon,  seen in the observer reference-frame
   ----------------------------------|---------------------------------- x
                                                     O = Observer                                                  µ_S = the angle ( ß , -c_S ) = ( Ox , -c_S )
                                                                                                                             µ₀  = the angle ( ß , -c₀ ) = ( Ox , -c₀ ) is measured by the observer
 

                     f_S = frequency in the source reference-frame                                     Of course  || c_S || = || c₀ || = 1
                     f₀  = frequency measured by the observer
 

Formulae:

  f₀  =  f_S ( 1 - ß² ) ^-1/2  ( 1 - ß cos µ_S )                cos µ₀ = ( cos µ_S - ß ) ( 1 - ß cos µ_S ) ^-1
  f_S =   f₀ ( 1 - ß² ) ^-1/2  ( 1 + ß cos µ₀ )               cos µ_S = ( cos µ₀ + ß ) ( 1 + ß cos µ₀ ) ^-1

-We assume that µ_S and µ₀  are between 0 and 180°  and  ß  is non-negative
-The source is moving away if µ₀ < 90°
 

Data Registers:  /
Flag:     F00         CF 00 for the conversion  ( f₀ , µ₀ )  >>>  ( f_S , µ_S )
                              SF 00 for the conversion  ( f_S , µ_S )  >>>  ( f₀ , µ₀ )
Subroutines: /
 
 

 
( 38 bytes / SIZE 000 )
 
 

 
Example1:      ß = 0.7     µ₀ = 41°     f₀ = 10

   CF 00

   0.7  ENTER^
   41  ENTER^
   10  XEQ "DAB"   yields     f_S = 21.4004   X<>Y   µ_S =  17.8522°

-Likewise,   ß = 0.7     µ₀ = 150°     f₀ = 10   give   f_S = 5.5141  and   µ_S =  114.9367°

Example2:      ß = 0.7     µ_S = 41°     f_S = 10

   SF 00

   0.7  ENTER^
   41  ENTER^
   10  XEQ "DAB"  gives in 1.8 second   f₀ = 6.6052   X<>Y   µ₀ =  83.3397°

-Likewise,   ß = 0.7     µ_S = 150°     f_S = 10   produce   f₀ = 22.4915  and   µ₀ =  167.1555°

Notes:

-The frequencies  f₀ & f_S are equal  iff  ( ß = 0  or  ß cos² µ₀ + 2 cos µ₀ + ß = 0 )
                                                     i-e  ( ß = 0  or  ß cos² µ_S - 2 cos µ_S + ß = 0 )

   for  ß = 0.7  the solution is  µ₀ = 114.1023165°  i-e  µ_S =  65.89768346°

-"DAB" works in all angular modes.
-However, the program listed above should not be used  if  µ  is very near 0° or 180°  because ACOS doesn't give enough accurate results in that cases.
-For example, with  ß = 0.7 and  µ₀ = 0.01°  it yields  µ_S = 0.003969568048°  whereas the correct value is µ_S = 0.004200840261°

-The following variant uses R-P instead of ACOS , it employs the formulae:

   Tan µ₀ = Sin µ_S ( cos µ_S - ß ) ^-1 ( 1 - ß² ) ^1/2
   Tan µ_S = Sin µ₀ ( cos µ₀ + ß ) ^-1 ( 1 - ß² ) ^1/2
 
 

 
   ( 53 bytes / SIZE 000 )
 
 

 
-Examples 1 and 2 produce the same results, but this time,  ß = 0.7 and  µ₀ = 0.01°  yield  µ_S = 0.004200840259°
-Moreover, this program also works if  -180° < µ < 0°

-6 bytes may be saved if you replace lines 06 to 11 by  ASIN  COS  but the execution time becomes 3.4 seconds instead of 2.5 seconds.
 

4°)  Elastic Collision
 

     a) 2 Particles with the same masses
 

                                                          ß₁
                                                       *                                           One particle is initially at rest in O
                    ß                              *  µ                                          The incoming particle has a speed  ß   ( ß < 1 = speed of light )
                   * * * * * * * * * * *
                                               O   *                                             After the shock, the speed of the incoming particle has become ß₁
                                                          *                                        and the speed of the other particle - initially at rest - is ß₂
                                                               *  ß₂                              µ is the angle between the vectors   ß₁ ,  ß₂
 

Formulae:

            cos µ = ß₁.ß₂ [ 1 + ( 1 - ß₁² )^1/2 ] ^-1 [ 1 + ( 1 - ß₂² )^1/2 ] ^-1

            ß²/( 1 - ß² ) = ß₁²/( 1 - ß₁² ) + 2.ß₁.ß₂ cos µ  ( 1 - ß₁² ) ^-1/2 ( 1 - ß₂² ) ^-1/2   + ß₂²/( 1 - ß₂² )

-We also have:       ( 1 - ß² ) ^-1/2 = ( 1 - ß₁² ) ^-1/2 + ( 1 - ß₂² ) ^-1/2 - 1   but the results would be less accurate for  ß << 1

-All these relations come from the conservation of the energy-momentum.
 

Data Registers: /
Flags: /
Subroutines: /
 
 

 
   ( 56 bytes / SIZE 000 )
 
 

 
Example:    ß₁ = 0.9    ß₂ = 0.8

   0.9   ENTER^
   0.8   XEQ "COLL"  >>>>     ß = 0.941238075       ( execution time = 2.3 seconds )
                                  X<>Y     µ = 71°73608140       ( in DEG mode )
 

-This program works in all angular modes.
-Synthetic register M may be replaced by R00 or any other data register.
-When ß is very small, µ ~ 90° , this is the classical result !

-If you know ß₁ & ß₂ , you may wish to calculate ß & µ
-You'll have to solve the following equation:

   x² sin² µ - x [ 1 + ( 1 - ß² ) ^-1/2 ] sin² µ + [ 2 + ( 1 - ß² ) ^-1/2 ] cos² µ + ( 1 - ß² ) ^-1/2  = 0    where  x = ( 1 - ß_i² ) ^-1/2     i = 1 , 2
 

Data Registers: /
Flags: /
Subroutines: /
 
 

 
   ( 53 bytes / SIZE 000 )
 
 

Example:    µ = 60°    ß = 0.986

-in DEG mode

   60        ENTER^
   0.986   XEQ "COLL2"  >>>>     ß₁ = 0.977634203       ( execution time = 2.3 seconds )
                                        X<>Y     ß₂ = 0.895051724

-This program works in all angular modes.
 

     b) 2 Particles with different masses
 
 

                                                          ß₁
                                                       *                                            One particle ( mass m₂ ) is initially at rest in O
                    ß                              *  µ₁                                          The incoming particle ( mass m₁ )  has a speed  ß   ( ß < 1 = speed of light )
                   * * * * * * * * * * *----------------------------
                                               O   *  µ₂                                        After the shock, the speed of the incoming particle has become ß₁
                                                          *                                        and the speed of the other particle - initially at rest - is ß₂
                                                               *  ß₂                              µ = µ₁ + µ₂   is the angle between the vectors   ß₁ , ß₂
 
 

Formulae:             m.ß ( 1 - ß² ) ^-1/2 = m.ß₁ cos µ₁ ( 1 - ß₁² ) ^-1/2 + ß₂ cos µ₂ ( 1 - ß₂² ) ^-1/2                                   where   m = m₁/m₂

                        m² ß²/( 1 - ß² ) + m² ß₁²/( 1 - ß₁² ) - 2.m².ß.ß₁ cos µ₁ ( 1 - ß² ) ^-1/2 ( 1 - ß₁² ) ^-1/2   =  ß₂²/( 1 - ß₂² )

                        m ( 1 - ß² ) ^-1/2 = m ( 1 - ß₁² ) ^-1/2 + ( 1 - ß₂² ) ^-1/2 - 1                  which is re-written:

                       ( 1 - ß² ) ^-1/2  = 1 + eps    with  eps = ß₁² [ 1 - ß₁² + ( 1 - ß₁² ) ^-1/2 ] ^-1  +  (1/m) ß₂² [ 1 - ß₂² + ( 1 - ß₂² ) ^-1/2 ] ^-1

   whence        ß = [ 2.eps + (eps)² ]^1/2  ( 1 + eps ) ^-1     and    ß²/( 1 - ß² ) = 2.eps + (eps)²
 

-Thus, the formulas are also accurate for small speeds  ( ß << 1 )
 

Data Registers:   R00 thru R03: temp
Flags: /
Subroutines: /
 
 

 
   ( 82 bytes / SIZE 004 )
 
 

 
Example:       m = m₁/m₂ = 2  ,  ß₁ = 0.9  ,  ß₂ = 0.8

-In DEG mode:

    2     ENTER^
   0.9   ENTER^
   0.8   XEQ "COLL3"  >>>>    ß =  0.924743364   ( in 3.4 seconds )
                                     RDN    µ₁ = 14°30751818
                                     RDN    µ₂ = 49°94058793   whence   µ = µ₁ + µ₂ = 64°24810611

-This routine works in all angular modes.
-Of course, it also works if m = 1
-For instance, with  m = m₁/m₂ = 1  ,  ß₁ = 0.9  ,  ß₂ = 0.8   it yields:

  ß =  0.941238075     µ₁ = 27°02226934    µ₂ = 44°71381201

     whence  µ = µ₁ + µ₂ = 71°73608135  which we had already found above with "COLL"     ( except in the last decimal )

-Knowing  m , ß and µ = µ₁ + µ₂  allows to compute  ß₁ &  ß₂
-This can be done by solving the cubic equation:

    x³ sin² µ - x² ( u + 2/m ) sin² µ + x ( 1/m² + 2.u/m + cos² µ ) - [ u/m² + ( u + 2/m ) cos² µ ] = 0     where  x = ( 1 - ß₁² ) ^-1/2  &  u =   ( 1 - ß² ) ^-1/2

then            ( 1 - ß₂² ) ^-1/2 =  1 -  m ( 1 - ß² ) ^-1/2 - m ( 1 - ß₁² ) ^-1/2

-So, the results may be inaccurate with small velocities ( ß << 1 )
 

Data Registers:   R00 thru R04: temp
Flag:   F01
Subroutine:   "P3"  ( cf "Polynomials for the HP-41" )
 
 

 
   ( 83 bytes / SIZE 005 )
 
 

 
Example:       m = m₁/m₂ = 2  ,  µ = 77°  ,  ß = 0.986

-In DEG mode:

    2      ENTER^
   77     ENTER^
 0.986  XEQ "COLL4"  >>>>   ß₁ =  0.985702487    ( in 6.4 seconds )
                                      RDN   ß₂ =  0.457448346

-This program gives one solution ( ß₁ , ß₂ )  and, in the above example, there is indeed a unique solution.
-But in other examples, there are 3 solutions - corresponding to the 3 solutions of the cubic ( flag F01 remains clear in this case )

-For instance,  with  m = 1.2  ,  µ = 67°  ,  ß = 0.931 ,  the 3 solutions are:

        ß₁ =  0.808532649     ß₂ =  0.895634440            If you want to calculate all the solutions,
        ß₁ =  0.768103288     ß₂ =  0.910120305            change the listing after line 41:
        ß₁ =  0.487300873     ß₂ =  0.939242706            at this step, the 3 x-solutions ( if F01 is clear ) are in registers X  Y  Z

-When µ > 90° , this program may give a wrong solution: for example the solution corresponding to µ = 83° instead of µ = 97°
-So, it's safer to check with "COLL3" if the results are exact or not!
 ( the cubic equation only involves sin² µ & cos² µ , therefore it cannot distinguish  µ from 180° - µ )
 

     c) Compton Effect
 
 

                                                         h.f '
                                                       *                                            One particle ( usually an electron ) is initially at rest in O
                    h.f                            *  µ₁                                          The incoming photon has an energy  E = h.f = h.c/L  ( f = frequency , L = wave-length ,
                   * * * * * * * * * * *----------------------------                                                                              h = Planck's constant , c = speed of light )
                                               O   *  µ₂                                        After the shock, the energy of the photon has become  E = h.f ' = h.c/L'
                                                          *                                        and the speed of the other particle - initially at rest - is ß
                                                               *  ß

Formulae:        L' - L = 2(h/mc) sin² µ₁/2

              ( 1 - ß² ) ^-1/2  = 1 + eps    with    eps = (h/mc).( L' - L )/(L.L')

             whence        ß = [ 2.eps + (eps)² ]^1/2  ( 1 + eps ) ^-1     and    ß/( 1 - ß² ) ^-1/2 = [ 2.eps + (eps)² ] ^1/2

               ß.cos µ₂ ( 1 - ß² ) ^-1/2 = ( L' - L ).[ 1 + h/(mcL) ]/L'

-With these formulas, the results are also accurate if  ß << 1.
 

Data Registers:             R00 = L' - L         R02 = L'      R04 = ß/( 1 - ß² ) ^-1/2
                                        R01 = L                R03 = µ₁      R05 = h/(mc)
Flags: /
Subroutines: /
 
 

 
   ( 70 bytes / SIZE 006 )
 
 

   where the mass unit is the mass of one electron = m_e = 9.109534 10 ^-31 kg
    and the wave-length L is expressed in Angstroem ( 1 Å = 10 ^-10 meter )

Example:       m = 1.2  ,  µ₁ = 79°  ,  L = 0.123 Å

-In DEG mode:

      1.2    ENTER^
      79     ENTER^
    0.123  XEQ "COMPT"  >>>>   L' =  0.139361229 Å
                                          RDN    µ₂ = 46°17378618
                                          RDN    ß  =  0.193671439

-Line 09, the constant   2426309 E-8 = h/(m_ec)  has been computed with  h = 6.626176 10 ^-34 J.s ,  m_e = 9.109534 10 ^-31 kg ,  c = 299792458 m/s
 and then, multiplied by E10 in order to express the wave-lengths in Angstroems.

-If  L' is very close to L , you can replace line 48 by RCL 00 = L' - L    ( L' - L  is always positive )
 

References:

[1]  Robin M. Green - "Spherical Astronomy" - Cambridge University Press - ISBN 0-521-31779-7
[2]  Stamatia Mavridès - "L'Univers relativiste" - Masson  ISBN 2-225-36080-7  ( in French )
[3]  Landau & Lifshitz - "Classical Theory of Fields" - Pergamon Press