eq. (14.26) and eq. (14.27)2017/ 12/ 07/ 11시

담당교수
Two  comments  to  add.

1.  To  derive  eq.  (14.26),  it  is  much  better  to  use  the  first  part  of  eq.  (14.25)  including  E  rather  than  p.
Then  there  is  no  beta^2  in  the  time  component  square  and  eq.  (14.26)  is  obtained.
I  strongly  recommend  you  to  compute  the  result  by  yourself.

2.  The  explanation  of  eq.  (14.27)  is  missing.
For  linear  acceleration,  |dvec{p}/dtau|=dp/dtau  and  dtau=dt/gamma.
Thus,  the  second  part  of  eq.  (14.15)  gives  dp/dt^2  after  (1-beta^2)  gamma^2  is  eliminated.

1-beta^2  from  two  terms  and  gamma^2  from  d/dtau  to  d/dt.

Thus  eq.  (14.27)  is  perfect  relativistic  expression  valid  only  for  the  linear  acceleration.