MASSÖKNING                           MASS INCREASE

År 1901 upptäcktes genom  expe-      In  year 1901 it was discovered
riment med katodstrålar (elekt-      by cathode ray experiments  (e-
roner),  att dessa partiklar ej      lectrons), that these particles
avböjde i ett elektromagnetiskt      not inclined in an electro mag-
fält exakt så mycket som klass-      netic field in exactly this way
isk  Newtonsk teori  föreskrev.      as was awaited from pure Newto-
Detta   fenomen  kan i  princip      nian  theory.  Principally this
tolkas på  två  olika  sätt  1)      phenomenon can  be  interpreted
elektronernas  massa  ökar  med      in  two  different ways 1)  the
hastigheten eller  2) då parti-      electron mass   increases  with
keln börjar närma sig den  has-      velocity  (a real effect) or 2)
tighet som det    accelererande      when a particle begin  approac-
fältet  har, avtager verkan  av      hing the velocity of the  acce-
den accelererande  kraften.  E-      lerating  field,  the accelera-
xakt vilket av dessa alternativ      ting force diminishes. Which of
som är det riktiga går  ej utan      these  alternatives  being true
vidare att avgöra. Det kan ock-      cannot  without further  notice
så tänkas, att båda  effekterna      be  decided.  Both  effect  may
kan ha reell betydelse.              have importance.



Newtons force formula   F = d/dt(mv) ;

Deriving this formula   F = dm/dt . v + m .dv/dt

Multiplying both sides with a small distance, ds, give moving energy
(kinetic energy of the particle) :

                        F.ds = dm/dt . v.ds + m . dv/dt . ds

But  F.ds = dE (a small amount of kinetic energy) and

ds/dt is the same as velocity of the particle, givning :


                        dE = dm.v.ds/dt +m.dv.ds/dt


                                 2
                        dE = dm.v + m.v.dv


                                         2
Hence :                 dE/dv = dm/dv . v   + m.v


If we assume m increase its electromagnetic mass in accord with

            2   2 -1/2
m = mo.c.( c - v )        ( by the "Lorenz factor"), the derivative

dm/dv will be :

                                    2  2 -1/2
                       d/dv (mo.c.(c -v )     )  =


                                     2  2 -3/2
                       mo.c.(-1/2).(c -v )    .(-2v) =

                                2  2 -3/2
                       mo.c.v.(c -v )

Inserting this result into the formula above give :

                                 3
                                v                    v
            dE/dv = mo.c . ------------  + mo.c. --------
                              2  2 3/2              2  2 1/2
                            (c -v )               (c -v )


Integrating  0 --> v gives :

                   /      3                    /
                  !      v                    !    v
         E = mo.c ! ___________ dv   +  mo.c. ! ___________ dv
                  !    2  2 3/2               !    2  2 1/2
                  !  (c -v )                  !  (c -v )
                 /                            /

                            3                       2
                           x             2  2      a
Allmänt, in general : I ------- dx  =  (a -x ) + -------
                           2  2 3/2                2  2 1/2
                         (a -c )                ( a -x )


                          x              2  2 1/2
                      I ----- dx   = - (a -x )
                          2  2 1/2
                        (a -x )

         v
          !              2
          !   2  2      c              2  2 1/2
E = mo.c. ! (c -v ) + ----------- -  (c -v )
          !             2  2 1/2
          !           (c -v )
          !
         0

                               2
           !     2  2 1/2     c              2  2 1/2
   = mo.c. ! . (c -v ) +  -----------  - (c -v )   -c + c -c  =
           !                 2  2 1/2
                           (c -v )
                 2
           !    c
   = mo.c. ! --------    - c       =
           !    2  2 1/2
              (c -v )


                  !
   =   mo.c       !   2        2  2 1/2
     --------     !  c  - c. (c -v )         =
       2  2 1/2   !
     (c -v )      !


              4    2   2  2 1/2
         !   c  - c .(c -v )
   = m . !  -----------------           =
         !    2        2  2 1/2
             c  + c. (c -v )


             4   4    2  2
         !  c  -c  + c .v
   = m . ! ----------------------
         !   2         2  2 1/2
            c . (1 + (c -v )   )

=======================================================================
               2
              v                    2
   E = m. ----------------   = mo.v /(L.(1+L))
                      2  2                    2  2
          1 + SQRT(1-v /c )    if L = SQRT(1-v /c ), the Lorents factor

=======================================================================



                      2
Om/if  v= c,  E = m.c

                         2
                    m . v                  2
Om/if v<<c    E = -----------    = 1/2. m.v

                  1 + SQRT(1-0)


                           2
Vi ser att resultatet E=m.c endast är sannt om massan hos
den accelererande partikeln ökar som funktion av hastigheten
enligt sambandet m=mo/L. I annat fall är den totala energin
          2
E=1/2.mo.c .