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Is the Einstein equation E=Mc^2 correct?

  1. Apr 29, 2013 #1
    I wonder if Einstein equation E=mc^2 was used when the A bombes were designed and tested.
    How was the output estimated? Was all the matter thransformed into energy, as the equation indicates?
    Were there any restrictions to how this equation ( or nny other) was used?
    Thanks,
    Michael
     
  2. jcsd
  3. Apr 29, 2013 #2
    from the wiki on mass energy equivalence:

    "the "Gadget"-style bomb used in the Trinity test and the bombing of Nagasaki had an explosive yield equivalent to 21 kt of TNT. About 1 kg of the approximately 6.15 kg of plutonium in each of these bombs fissioned into lighter elements totaling almost exactly one gram less, after cooling. The electromagnetic radiation and kinetic energy (thermal and blast energy) released in this explosion carried the missing one gram of mass.[31]"

    http://en.wikipedia.org/wiki/Mass–energy_equivalence
     
  4. Apr 29, 2013 #3
    I believe so, but that only applies when stationary. When the mass is not stationary I believe I'm right in saying E2=(mc2)2 + (pc)2. So yes theoretically all the mass would have turned to energy.
     
  5. Apr 29, 2013 #4

    phinds

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    What makes you think so? IF all the mass turned into energy, then the formula would apply to all the mass. Do you really think all the mass turned into energy? Had it done so, I think it likely that instead of incinerating a large part of one city, it would have incinerated Japan.
     
  6. Apr 29, 2013 #5
    I worked out it should produce about 4.68x1018 J. That's obviously a theoretical value but how much it really produced I don't know. Anyone know what that kind of energy would do?
     
  7. Apr 29, 2013 #6

    Thanks for answering so fast.
    I don't know much about Gadget style A bombs, but my problem is this: the BRAVO bomb (March 1954) was calculated an yield of 6 Mt(between 4 to 8 Mt). The actual yield was 15 Mt an error/deviation from the original of +150%. How come?
     
  8. Apr 29, 2013 #7


    Thanks for the reply.
    I have a question related to your term of the eqn: (pc)^2. What is the meaning of p.
    I assume c is speed of light in "vacuum"
    Now some bombs were delivered by airplane. The speed of an airplane is negligible when compared with c.
    Therefore the second term is almost zero.
    But my real problem is that the BRAVO bomb (March 1954) was calculated an yield of 6 Mt(between 4 to 8 Mt). The actual yield was 15 Mt an error/deviation from the original of +150%. How is this possible?

    Thank you much.
     
  9. Apr 29, 2013 #8

    Nugatory

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    Staff: Mentor

    here p is the momentum, which is of course zero for a mass that is at rest.

    There's a pretty decent explanation in wikipedia: http://en.wikipedia.org/wiki/Castle_Bravo#Cause_of_high_yield
     
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