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  Roaring Lovely April 16, 2016 at 6:06pm

  Finaly, let me use electricity nature of matter to show you the excellent way of understanding E=mc^2. Actually it is an OBVIOUS consequence of MAXWELL'S THEORY and physicists knew it long before Einstein did. But my derivation might have something unique.
  
  MOTIVATION
  
  Given that an electrically charged particle such as an electron have size, what does it keep its various parts from ripelling each other? Think about this considering that similar charges ripell each other. The model is that an electron is a pack of infinitely many infinitesmall electrically charged subparticles (or subtrons) unless a strong force keep them together, the subtrons will endlessly repell each other, approaching nothingness. This repulsion releases the stored energy. So how much is this energy?
  
  The best force that can hold the electron against its own electrostatic self repulsion (hence destruction) is MAGNETISM. To do that, electron must spin. Spinning charge can be seen as an electric current. This give rise to z-pinch, holding the particle together in a toroidal shape. The z-pinch force is given by:
  
  F=l^2/2pu
  F=force,
  I=electric current (due to spinning)
  p=pie
  u=magnetic permeability
  
  These parts also repell each other via columb's force given by:
  
  F=q^2/4per^2
  
  q=electric charge
  e=electric permitivity
  
  Now current, I=dq/dt (dq is amount of charge passing by during small change it time, dt). If the spinning is steady, then dq/dt=q/t. So I^2=(q/t)^2 the z pinching force, F=ma, now becomes
  
  ma=(q/t)^2/2pu
  
  but r=(at^2)/2 i.e if the charge acelerates till it touch eact other, let it to have take time t. Put this together and we have
  
  m=q^2/4pru
  
  But to ballance things up, the collumb force must quicly take back the charg to where it was some distance r away. To do that it must thus store energy of E=Fr=q^2/4pre (from columb's force). Then use mass for q^2/4pru and you get E=m/ue. From Maxwell's theory, we know that 1/ue=c^2
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  Roaring Lovely April 16, 2016 at 4:54pm

  Only if you patiently follow the calculation can you truely appreciate what I am saying: Einstein confused INERTIA with MASS. You see that E=mc^2 follows from Lorentz transform for mass which in turn follow from length contraction.
  
  The reasoning is that if length is contracted as seen from one frame, then things appears to accelerate slowlier as though they were heavier. Einstein says this mean mass is bigger in that frame. But this is wrong! Doing so makes the equation f=ma true by DEFINITION. Specifically, Einstein is defining mass as m=f/a, a definition for INERTIA. Rather mass as QUANTITY OF MATTER isn't expressible via an equation.
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  Roaring Lovely April 16, 2016 at 4:35pm

  To deduce E=mc^2, you can reason this way:
  
  First I must teach you simple calculus. Consider the equation:
  
  E=m'c^2{1-(v/c)^2}^(1/2) that is to say m'c^2 multiplied by the squar root of the value {1-(v/c)^2}
  
  then {E/m'c^2}^2=1-(v/c)^2
  
  now if velocity, v changes by tiny amount, write it dv, energy changes by some tiny amount, write it dE, so we have
  
  {(E dE)/m'c^2}^2=1-{(v dv)/c}^2
  
  expand things to get:
  
  (E^2 2EdE dE^2)/(m'c^2)^2=1-(v2 2vdv dv^2)/c^2
  
  since dE is soo small, dE^2 is even far smaller that we can neglect it. Same applies to dv^2. So we remain with:
  
  (E/m'c^2)^2 2EdE/(m'c^2)^2=1-(v/c)^2 2vdv/c^2
  
  but (E/m'c^2)^2=1-(v/c)^2 so this cancels themselves out and we remain with
  
  EdE=(m'c^2)^2vdv/c^2=(m'c)^2vdv
  
  But E=m'c^2{1-(v/c)^2}^(1/2)
  
  so dE=m'vdv/{1-(v/c)^2}^(1/2)
  
  but we can understand change in energy, dE as a constant force f times small change in distance. That is
  
  dE =fdx
  
  but f=ma=mdv/dt (newton's second law of motion) so
  
  dE=mdxdv/dt=mvdv (dx/dt=v)
  
  subtitute this in the above equation and we simply get
  
  m=m'/{1-(v/c)^2}^(1/2)
  
  which is the lorentz transform for mass. So reasoning backwards, we say that given the Lorentz transform for mass, it follows logically that
  
  E=m'c^2{1-(v/c)^2}
  
  This is indead the correct form of E'=m'c^2 as energy too have its own Lorentz transform as
  
  E=E'{1-(v/c)2}^(1/2)
  
  My derivation appears very long sinse I did not assume that you know INTEGRAL CALCULUS.