Proton Radius: Is it a Sphere or Shell?

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a proton is given a radius...i think its ~10e-15
since a proton is uud is this "radius" like [1]a electron orbital density cloud that is said to be in atoms...that is to say that its a bounding volume(ie in computer geometry) or [2]an energy barrier on the uud of this radius size? Or [3]there is a shell that surrounds the uud?
 
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I think it is just an approximate limit beyond which the strong nuclear force becomes significant. In other words, get closer than about 1 femtometre and you're into the range at which the quarks bind together. Beyond that, the strength of the strong force drops off rapidly.
 
There's another really interesting aspect:get closer than 0.1 fermi and the strong force drops again...

Daniel.
 
The quarks in a proton have the same type of wave function as the electrons iin an atom. There is no definite boundary like the surface of the Earth.
The "charge radius" of the proton is defined as the square root of the expectation value <\rho r_2>=<\sum q_i r_i^2>, summed over the 3 quarks.
This rms charge radius is about 0.8 fm. The wave function is approximately spherical, corresponding to L=0 for the orbitals.
 
dextercioby said:
There's another really interesting aspect:get closer than 0.1 fermi and the strong force drops again...

Daniel.
...and becomes repulsive, no? Like in the iron-core-rebound which is responsible for some types of supernovae?
 
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I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
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