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   Message 178,575 of 178,769   
   Stefan Ram to Paul.B.Andersen   
   Mass and Energy   
   01 Jan 26 16:19:52   
   
   XPost: sci.physics.relativity   
   From: ram@zedat.fu-berlin.de   
      
   "Paul.B.Andersen"  wrote or quoted:   
   :that mass could be converted to energy as   
   :predicted by Einstein's E = mc².   
      
     I can spot three mistakes here right off the bat.   
      
          First, the phrase "mass could be converted to energy" is based on   
     a misconception, since mass already /is/ a form of energy - there's no   
     "conversion" happening. It's like saying, "water can be turned into a   
     liquid" - no, water /is/ a liquid.   
      
          Second, "E=mc^2" only applies to systems at rest. In general,   
     it's "E^2=(mc^2)^2+(pc)^2", where "p" is the system's momentum.   
     (That basically says mass is the magnitude of the four-momentum.)   
      
          Third, mass is conserved in nuclear fission. This point is often   
     explained incorrectly, even in otherwise solid textbooks like Grif-   
     fiths.   
      
          I've laid out the full argument elsewhere, but here's the (suffi-   
     cient) short version:   
      
          Take a nucleus that's at rest, so its total momentum p=0.   
      
          After fission, because momentum is conserved, the total momentum   
     of all the fragments is still zero. That means the relation "E^2=(mc^2   
     )^2+(pc)^2" simplifies to "E=mc^2" both before and after the split.   
      
          Since energy is also conserved, the total energy E of the nucleus   
     and its fragments - including their kinetic energy - stays the same,   
     call it E. From "E=mc^2", it immediately follows: if E stays constant,   
     m does too.   
      
   E_"before" = E_"after" - due to conservation of energy        (0)   
      
   E_"before" = m_"before" c^2 - as you said yourself            (1)   
      
   E_"after" = m_"after" c^2 - the same as (1), just later       (2)   
      
   m_"after" = m_"before" - substituting (1) and (2) into (0)    (3)   
                            and dividing by c^2   
      
     Only when you change what you call the "system", and treat the split   
     products separately, do you get systems with nonzero momentum and   
     find a mass defect and non-zero kinetic energies.   
      
   --- SoupGate-DOS v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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