From: PointedEars@web.de   
      
   [Fixed quotations]   
      
   Stefan Ram wrote:   
   > "Paul.B.Andersen"  wrote or quoted:   
   > > 1n + U-235 → Ba-141 + Kr-92 + 3n   
   > > [...]   
   > > Are you claiming that the mass 3.916659E-25 kg   
   > > is not converted to 2.78E-11 J kinetic energy?   
      
   The _rest energy_ corresponding to this mass is converted _partially_ to   
   kinetic energy.  As always, one form of energy is, or several forms of   
   energy are, converted to one or more other forms of energy.   
      
   That is the/my main problem with the popular-scientific statement "mass is   
   converted to energy".   
      
   >   When you start in the rest system of "1n + U-235",   
      
   The English term is "rest _frame_" instead.   
      
   >   its momentum is p=0.   
      
   Its _total_ linear momentum which is a *vector*.   
      
   >   As the momentum is conserved, the momentum of "Ba-141 + Kr-92 + 3n"   
   >   then also is p=0, so its kinetic Energy still is zero.   
      
   [Your utter misuse of punctuation aside: ]   
      
   You are making the blunder there of calculating the total kinetic energy of   
   a system based on the vector sum of the linear momenta of its components,   
   not the sum of their separate kinetic energies.   
      
   The former approach is incorrect because the kinetic energy is supposed to   
   be a measure of motion.  In the center-of-momentum (COM) frame, previously   
   there was no motion, but now there is; so the kinetic energy of the system   
   *must* change.  (Another way to see this is relativistically: A system with   
   moving parts must have a larger total energy than one with parts at relative   
   rest; but the mass of the parts has not changed because it is   
   Lorentz-invariant, and we cannot presume any forces between the parts, so   
   the kinetic energy of the system must have changed.)   
      
   Therefore, the correct formula is (approximately)   
      
     T = T_1 + T_2 = 1/2 m_1 {V_1}^2 + 1/2 m_2 {V_2}^2   
                   = {P_1}^2/(2 m_1) + {P_2}^2/(2 m_2)   
      
   or (exact)   
      
     T = T_1 + T_2 = m_1 c^2 [gamma(V_1) - 1] + m_2 c^2 [gamma(V_2) - 1],   
      
   where the V's are the velocities, and the P's are the linear momenta; NOT   
      
     T = P^2/[2 (m_1 + m_2)] = (P_1 + P_2)^2/[2 (m_1 + m_2)],   
      
   or something like that.   
      
   [This is a consequence of the energy--momentum relation   
   E^2 = m^2 c^4 + p^2 c^2: For m != we have p = gamma m v,   
   and thus E = gamma m c^2.]   
      
   Therefore, even though the total linear momentum is conserved, and equal to   
   the zero vector in the COM frame by definition, the total kinetic energy is   
   NOT (unless the process is an ideal elastic collision, but that is not what   
   we are discussing here).   
      
   This can also be proven: Approximately,   
      
     dT/dt = m_1 V_1 A_1 + m_2 v_2 A_2 = P_1 A_1 + P_2 A_2   
      
   or (exact)   
      
     dT/dt = m_1 gamma^3(V_1) V_1 A_1 + m_2 gamma^3(V_2) V_2 A_2   
           = gamma^2(V_1) P_1 A_1 + gamma^2(V_2) P_2 A_2   
      
   where the A's are the (3-)accelerations.  That is, the total kinetic energy   
   is only conserved as long as there is no acceleration, which is what one   
   should expect.  (Notice that this is a scalar product.)   
      
   But there has to be an acceleration for the linear momenta P_1 and P_2 to   
   change from zero to non-zero; so the total kinetic energy is NOT conserved   
   from before to after that happens.   
      
   --   
   PointedEars   
      
   Twitter: @PointedEars2   
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