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To: gondramB
gondramB said: "I don't believe this is true."

Let's try a thought experiment.

Imagine that you have a helium atom whose nucleus consists of two protons and two electrons. These four particles are held together by the strong nuclear force.

Imagine now that one could somehow separate this nucleus into two identical nuclei each containing one proton and one electron. The act of separating these tightly bound particles would require enormous forces operating over a small distance. Once the separation has been accomplished sufficiently to result in two deuterium nuclei, one should then weigh each particle.

What Einstein's theory predicts is that the "binding energy" which is added to the system will be measurable as an increase in mass of the two resulting particles such that the sum of the two particles is greater than the initial mass of the helium atom.

When we run this process in reverse, the binding energy is available to do "work" and we normally see this energy as the fireball and radiation of a fusion bomb.

Now consider what happens when you stretch a metal spring. You are causing forces to act upon a collection of atoms. Acting under the influence of that force, the atoms change position. As they change position, "work" is being done on the atoms as they move through a distance experiencing the force which is holding them together. The energies holding atoms in a solid together are typically much lower than nuclear binding energies because the forces holding atoms together in a spring are electromagnetic forces.

Nevertheless, the atoms in the spring become less tightly bound and the energy stored could theoretically be measured as an increase in mass of the system. The reason we don't measure such mass changes is because it is many, many orders of magnitude smaller than the mass changes caused by nuclear forces.

Similarly to the nuclear case above, if we now release the metal spring it can be made to do "work", that is the energy can leave the system and be observed in objects outside the spring. We could run a clock with the released spring, for example. As the spring relaxes, it exerts a force on the mechanism of the clock, causing motion. The atoms of the spring will move under the action of the electromagnetic force and the atoms will become more tightly bound than in the extended state of the spring. The mass of the entire spring will then be less by the amount predicted by Einstein's equation. But far too little for our instrumentation to measure.

24 posted on 10/01/2005 8:52:49 PM PDT by William Tell
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To: William Tell
I said: "Imagine that you have a helium atom whose nucleus consists of two protons and two electrons. "

That should have said "two protons and two neutrons" and the separation will create two deuterium nuclei each consisting of a proton and a neutron.

26 posted on 10/01/2005 9:01:52 PM PDT by William Tell
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To: William Tell

“What Einstein's theory predicts is that the 'binding energy' which is added to the system will be measurable as an increase in mass of the two resulting particles such that the sum of the two particles is greater than the initial mass of the helium atom.”

Just to be sure I understand—isn’t the sum of the two particles less than the initial mass of the helium atom, due to the loss of the binding energy?


40 posted on 10/01/2005 10:03:36 PM PDT by reasonisfaith
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To: William Tell
A HE4 nucleus is composed of two protons and two neutrons, not two electrons, they are in Bohr Radius orbit in a n=1, l=1, m=1 and s=+1/2 and -1/2, making a 3 pair boson arrangement. To wit, 1F+1F=1B or 1/2+1/2=1 in spin dynamics. This is why the HE4 atom has the highest ionization potential(13.6ev)and is the most stable of all elements. Understanding this, do you know why HE3 is different from HE4 as a super fluid? Also, can you interpret t=dKE?
47 posted on 10/01/2005 11:46:31 PM PDT by timer
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