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The mass of a nucleus if less than the mass of the protons and nucleus. The difference is knows as binding energy of the nucleus. This nuclear binding energy is derived from the strong nuclear force.

Now my question is that does the strong nuclear force provide this energy? How does the mass decrease if the force provides the energy?

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sorry, my mistake. – Qyuubi Feb 12 '11 at 12:29
To clarify, I don't understand how a force is converted into energy. – Qyuubi Feb 12 '11 at 12:31
And do you understand how potential is converted into energy? Take a rock, lift it and release. Its (gravitational) potential energy will be converted into its kinetic energy. If you understand that all forms of energy are equal and that mass is also an energy ($E = mc^2$) then what's the problem? – Marek Feb 12 '11 at 14:47
Right, Marku, +1. ;-) – Luboš Motl Feb 12 '11 at 15:41
up vote 1 down vote accepted

The question is clear enough to answer. The question does involve mass defect. The best place to start is with ordinary problems. We can consider a nonrelativistic problem of a particle in a potential well, $V(r)~=~-r^n$ for some power of $n$. For mathematical reasons $n~=~\pm 2$ have nice properties with closed form solutions. This is input into a Schrodinger equation $$ i\hbar\frac{\partial}{\partial t}\psi(r,t)~=~-\frac{\hbar^2\nabla^2}{2m}\psi(r,t)~+~V\psi(r,t) $$ So if we consider a stationary phase $\psi(r,t)~=~e^{-i\omega t}\psi(r)$ and the frequency determines energy $E~=~\hbar\omega$ $$ \hbar\omega~=~\frac{p^2}{2m}~+~V $$ where the potential energy is negative. If the potential energy were “turned off” so the particles are free then energy is larger. The energy of system is then lower, and if this energy $E~=~mc^2$ is some appreciable fraction of the mass-energy of the system $E’~=~Mc^2$, say $m/M~~\simeq~.01$ to $.1$ the system is not highly relativistic but the mass equivalence is measurable.

For atomic physics the energy levels of electrons are on the order of electron volts, while the mass of electrons are $.51$Mev. So the mass defect is pretty small. For nuclear physics of nucleons and mesons the energy levels are on the order of $10Mev$ while the collection of nucleons has mass-energy in multiples of $1Gev$. The energy levels are determined by $\pi^0,~\pi^\pm$ mesons, which are the intermediary gauge bosons between the nucleons ${P,~N}$. This in fact forms a doublet which has energy level splitting due to the electric charge difference. The above model may be made more exact if the mesons are considered to be similar to a photon, with a gauge potential $$ {\vec A}(k)~=~{\vec n}(a_ke^{-ikr}~+~a^\dagger_ke^{ikr}) $$ which interacts with a dipole formed from the nucleon doublet ${\vec{\cal P}}~=~p{vec\sigma}$ in an interaction Hamiltonian $$ H_{int}~=~-{\vec{\cal P}}\cdot{\vec A}(k). $$ We are only considering interactions with one momentum or wave number. Now expand that out and keep terms $a(k)\sigma^+$ and $a^\dagger\sigma^-$, which is the rotating wave approximation in atomic interactions with photons. This makes the above Schrodinger equation and potential more exact. This interaction Hamiltonian will then reproduce the mass-defect.

This may be further improved of course. The nucleon doublet is SU(2), and the meson potential may be extended from this naïve U(1) approximation to SU(2) as well. The momentum operator may be made covariant with respect of the gauge potential and the theory refined further. In fact the Yang-Mills theory was derived to understand this isopin theory of nuclear physics as understood in the 1950s.

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I suppose I am starting all over again, or something? – Lawrence B. Crowell Feb 12 '11 at 13:45
@Lawrence: you are probably just not logged in? – Marek Feb 12 '11 at 15:35
I never registered, and I suspect I got timed out or something. To be honest about 7 years ago I decided to limit such activities, for I think these things can easily be used to generate huge data bases about us, or be accessed by cyber-criminals, spamers and other nefarious characters. If I get cycled back again in another 28 days that might suggest I get timed out. I honestly am not terribly concerned about getting a high reputation score. – Lawrence B. Crowell Feb 13 '11 at 1:46
I merged your new user account with the old one. (I suppose I should have asked first... sorry, if it's a problem) Although it isn't actively forbidden to have multiple accounts, it is discouraged. Anyway, from a privacy standpoint, I don't see why it'd be that big of a deal to make a registered account. The only piece of information you'd be providing that you haven't already would be an Open ID, which is really just an arbitrary URL. – David Z Feb 13 '11 at 6:44

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