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Meaning of E=mc^2

  1. Jan 18, 2005 #1
    I have a question about the true meaning of E=mc^2. For starters, a certain amount of mass (m) can be converted into an amount of energy (mc^2), or vice versa. For example, two antiparticles can annihilate and leave only energy behind. So it is mass-energy that is a conserved quantity, not each one alone.

    That being said, it is my impression that mass is still one thing, and energy is another, albeit that they can be converted into each other. Mass has ties to inertia and gravity that energy does not. For example, consider a system, something like a uranium nucleus, that spontaneously decays. It is said that "the mass of the nucleus before the fission is greater than the sum of the constituent parts". This basically accounts for the potential energy (that manifests itself as kinetic energy upon fission) as extra rest mass.

    But to quote the rest mass of the uranium nucleus as including this potential energy does not make sense to me. Will the nucleus in fact behave as though it has this extra mass, for example if I apply a force to it (hypothetically) and measure the acceleration?
     
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  3. Jan 18, 2005 #2

    chroot

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    Energy and mass are actually the same thing even gravitationally; energy feels (and exerts) a gravitational force just like mass. An empty box weighs less than a box full of light; photons follow curves when moving near large masses. The factor of c^2 is very large, however, so it takes an incredible amount of energy to come close to the gravitational attraction of mass. For example, the energy released by the atomic bomb dropped on Hiroshima in WWII was equivalent to about one gram of mass. This disparity means that we don't think much about energy as exerting gravitational force or having inertia, but, indeed, it's just the same as mass.

    If you're interested, both mass and energy (all forms -- pressures, radiation, and so on) are treated quite equally as components of a mathematical object called the "stress-energy tensor" in General Relativity, Einstein's theory of gravitation. All forms of mass and energy contribute to the curvature of spacetime, which is what most people regard as the "gravitational field."

    - Warren
     
    Last edited: Jan 18, 2005
  4. Jan 18, 2005 #3
    Wow! Thank you for a great answer. That clears up a lot of misunderstanding I had. :smile:
     
  5. Jan 18, 2005 #4

    jtbell

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    Note that this "extra" mass is actually a "deficiency" because the mass of a uranium nucleus is less than the sum of the masses of the neutrons and protons that it's made of. The potential energy of the bound system is less than the potential energy of a bunch of completely separated protons and neutrons. We have to do work (a lot of it!) on the protons and neutrons in order to separate them completely.

    And we in fact measure that mass by applying a force and measuring the acceleration. In a mass spectrometer, we send a beam of uranium (or whatever) ions through a perpendicular, uniform magnetic field and measure the radius of curvature of the beam produced by the magnetic force. In most second-year "modern physics" textbooks you'll find a table of atomic masses with a precision of six significant figures or so. That's where those numbers come from.
     
  6. Jan 19, 2005 #5
    Isn't that backwards...? I thought a uranium nucleus should have a mass greater than the sum of its parts, since it is heavier than iron and thus "over the hump" on the curve of binding energy? No?
     
  7. Jan 19, 2005 #6

    chroot

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    jtbell,

    I do believe you have it backwards, also.

    - Warren
     
  8. Jan 19, 2005 #7

    jtbell

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    That depends on which set of "parts" you're talking about.

    I see now that you were thinking of a typical fission reaction in which a uranium nucleus produces two smaller nuclei and a couple of neutrons to continue the chain reaction. In this case, the sum of the masses of the products is less than the mass of the uranium nucleus, and the mass difference shows up as kinetic energy of the products.

    I was talking about taking apart the uranium nucleus completely, into individual neutrons and protons. There's no contradiction because in the fission reaction, the product nuclei are more tightly bound than the uranium nucleus, so the potential energy per nucleon is less in the products than in uranium.

    Sorry about the confusion... :frown:

    Note that I'm using "less" to mean "lower", not "smaller in magnitude." The potential energy of a nucleus is always negative, because in these situations we customarily define the potential energy to be zero when the constituent protons and neutrons are completely separated, and effectively not interacting with each other at all. So "less" here means "more negative."
     
    Last edited: Jan 19, 2005
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