🚀 go-pugleaf

 
I will once more derive below  E = Mc^2  from Lorentz equation via
High School algebra while  mentioning and emphasizing my explanatory
insight.
I will go step by step. 
 
  Assuming that we believe in the truthfulness of the Lorentz formula:
 
 (1)                M = Mo (1 - (v^2/c^2))^(-1/2)
  
 
 I   apply to (1)  the HIGH-SCHOOL formula for the  binomial expansion
of a  power of a binomial,  as given  below:
 
  (1 - k)^(-1/2)  =  1 + 1/2 k  +  3/8  k^2  + 5/16  k^3 + ...  with |k| <1 
 
 replacing in the above  series   k  by (v^2/c^2)  and noting that  |v/c| < 1,
 we  have  the following series expansion of (1) 
 
   M = Mo (1  + 1/2 (v^2/c^2)  + 3/8 (v^4/c^4) + 5/16  v^6/c^6 + .... )

Multiplying both sides by  c^2,  I obtain  

(2)    Mc^2 = Mo c^2 + 1/2  Mo v^2 + 3/8 Mo v^4/c^2 + 5/16 Mo v^6/c^4 +....)  

From (2)  it readily follows that

(3) Mc^2 - Mo c^2 =  1/2  Mo v^2 + 3/8 Mo v^4/c^2 + 5/16 Mo v^6/c^4 +....)

NOW THE WHOLE SUBTLETY OF NY REASONING IS IN THE OBSERVATION THAT  

     Mc^2  - Mo c^2   =  1/2 Mo v^2  +  .......

says that due to acquiring velocity  v  the total Energy of the
system has acquired the "kinetic type"  energy which is  equal to
the difference of  Mc^2  and  Mo c^2.  But then what could be more
insightful, more intelligent, more reasonable  than to consider   
Mc^2  as expressing the total energy of the system at velocity v and 
to consider Mo c^2  as expressing the total energy of the system at 
rest and hence to consider  M c^2  as the general expression of the total
energy of the system  and denote it by  E. What could be more
insightful?.  Thus,

(4)        E = Mc^2  =  c^2  Mo (1 - (v^2/c^2))^(-1/2) 
 
whose series expanded form is (2), i.e.,

(2)    Mc^2 = Mo c^2 + 1/2  Mo v^2 + 3/8 Mo v^4/c^2 + 5/16 Mo v^6/c^4 +....)

The first term of the series expansion (2) provides

(5)   Mo c^2   as a first approximation of   Mc^2 = E

and the sum of the first two terms in the series expansion (2) provides

(6)   Mo c^2 + 1/2  Mo v^2  as  a better second approximation of Mv^2 = E

        
I repeat  that assuming Lorentz equation   M = Mo (1 - (v^2/c^2))^(-1/2)
the derivation of  E = Mc^2   (with my abovementioned insight)  amounts
to multiplying both sides of   M = Mo (1 - (v^2/c^2))^(-1/2)  by  c^2
and absolutely nothing else.
 You don't need even to use the series expansion of the square root. 
It played a decisive role for me  to  call  Mc^2  the system's total energy  
at velocity v  since  Mc^2 - Mo c^2, as (3) shows  indicates the acquired
"kinetic type" energy of the system due to the system's acquired velocity
v. Also,  (2)  served us to provide the two approximations (5) and 
(6)  of  E = Mc^2.
   It seems to me because of my derivation, in devising the "epoch making 
miraculous formula " E = Mc^2  Lorentz has played the most decisive role. 
-- 

--------------------------------------------------------------------------
      
         ABIAN TIME-MASS EQUIVALENCE FORMULA  T = A m^2  in Abian units 
       ALTER EARTH'S ORBIT AND TILT - STOP GLOBAL DISASTERS  AND EPIDEMICS
       ALTER THE SOLAR SYSTEM.  REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT  
                     TO CREATE A BORN AGAIN EARTH (1990)