# E=mc^2 vs energy in strong interaction

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1. Jan 20, 2015

### Leonardo Ochoa

Hi everyone,

I have a simple and foolish question.

I want to compare the energy of a given mass (obviously e=mc2); lets say the energy of a hydrogen atom, with the energy that binds together the fundamental particles of that atom (strong interaction). I know that e=mc2 holds always true, and that the energy in strong interaction is undrainable, but do total energy of strong interaction (in a particular case) could be more or less than the energy you get when transforming mass to energy?

I know I'm confused and possibly both are the same, but appreciate an explanation.

Last edited: Jan 20, 2015
2. Jan 20, 2015

### mathman

The strong interaction holds three quarks together to form a proton. What mass are you converting to energy?

3. Jan 20, 2015

### Leonardo Ochoa

hydrogen atom, 1 proton...

4. Jan 20, 2015

### RGevo

Maybe I don't understand the question. But I'll try to answer.

An ionised hydrogen atom is a proton. This is held together by the strong force and quantitatively speaking this is a 3 quark bound state.

Although the idea of individual quark masses is a bit misleading (they always come in bound States), the effective mass of the up and down quarks is ~5 MeV. While the proton is 1 GeV. You could attribute this difference to the strong force.

Alternatively, do you mean the energy taken to expel a valence electron compared to the macroscopic strong force of a few bound nuclei? For example the energy required to fission a He nuclei out of a larger atomic mass nuclei?

5. Jan 21, 2015

### ChrisVer

atom= bound states of charged nuclei with charged electrons... charged= electrically charged and the energy of interaction comes from the electromagnetic interactions...
nuclei= bound states of protons and neutrons ... mainly by strong interactions for large distances/low energies (where the mesonic effective field theories hold)
protons= bound states of quarks and gluons...

And in general the relation $E=mc^2$ does not hold when you have interactions too, at least not in the same way - because you have to take into account the energy from the interactions... In QFT framework, the mass of such a system should get corrections from calculating further Feynman Diagrams.