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Most recent answer: 05/05/2010
Why are protons less massive than neutrons?
I tried to work out how much energy it would take to shove an electron charge into a region of diameter d, and I got: U = alpha hbar c / diameter = 197 eV nm / (137 * diameter). I hope that is right. Taking the proton size to be about 1 femtometer, I get U ~ 1.44 Mev ... that is way more than the mass difference between a proton and neutron! And it suggests the proton should be heavier! Don't the quarks have the same color charge in the proton vs. neutron, so they look the "same" according to the nuclear force? So I'd think the only difference could be the electric force. Did I calculate wrong? Either way, I don't understand how the proton can be less massive. Please help satisfy some curiousity here. Thanks!
- John Marsh (age 20)
Springfield, IL, USA
Your question puzzled physicists for many years. The answer seems to be that not only is there a difference in the electrostatic energy, but there is a difference in the mass contribution of the internal constituents. According to the quark hypothesis the proton is composed (primarily) of two up quarks and one down quark. The neutron is composed (primarily) of two down quarks and one up quark. It is difficult to ascertain exactly the masses of the bound quarks but it seems that the up quark does have a heavier effective mass. Dick Plano has some more to say about it:
(published on 04/19/2010)
Follow-Up #1: dimensional potential energy
I had never seen that "U = alpha hbar c / diameter" equation before. So I had to work it out to convince myself it is that simple (well, I was off by a factor of 2, but I probably made a little mistake). Is there some reason that this result should be "obvious", say in the order of alpha and then using dimensional analysis to get the rest? It seems "pretty" enough, that it seems this equation should be clearer to me. Any help in gaining physical intuition here would be much appreciated!
- Jack Swanson (age 19)
Chicago, IL, USA
Sure, let's use CGS, in which the potential energy is (dimensionally) just charge2
/distance. alpha is defined as q2
/hc (I'm not worried about factors of 2π here.), where q is the electron charge. So the expression you give is just q2
/diameter, the potential energy.
The alpha/hbar c factor is a red herring since this is strictly a classical problem. The correct classical answer is U = e2
3/5a where a is the radius, not the diameter. See http://farside.ph.utexas.edu/teaching/em/lectures/node56.html for a derivation.
(published on 05/05/2010)
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