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| − | ==Origin of the scattering lengths==
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| − | The following description is adapted from [http://www.ncnr.nist.gov/programs/sans/pdf/polymer_tut.pdf Boualem Hammouda's (NCNR) SANS tutorial].
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| − | ===Neutron energy===
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| − | Consider first the energies of neutrons used in scattering experiments (recall the neutron mass is 1.67×10<sup>−27</sup> kg). A thermal neutron (~100°C) would have energy of:
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| − | :<math>
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| − | \mathrm{KE} = \frac{1}{2} m v^2 = \frac{3}{2} kT = 7 \times 10^{-21} \, \mathrm{J} = 48 \,\mathrm{meV}
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| − | </math>
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| − | The velocity of such neutrons is ~3000 m/s, and the momentum is <math>p=mv=5\times10^{-24} \, \mathrm{kgm/s}</math>. Finally, the deBroglie wavelength would be:
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| − | :<math>
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| − | \lambda = \frac{h}{p} = 1.3 \, \AA
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| − | </math>
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| − | A cold neutron (~18 K) would have energy of 4×10<sup>−22</sup> J = 2 meV, velocity of ~660 m/s, and wavelength of 6 Å.
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| − | ===Potential well===
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| − | Consider a neutron of energy <math>E_i</math> interacting with a nucleus, which exhibits an attractive square well of depth <math>-V_0</math> and width <math>2R</math>; where <math>V_0 \gg E_i</math>. The [http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation Schrödinger equation] is:
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| − | :<math>
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| − | \left[ - \frac{\hbar^2}{2 m}\nabla^2 + V(r) \right] \psi(r) = E \psi(r)
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| − | </math>
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| − | Outside of the square-well (<math>|r|>R</math>), <math>V(r)=0</math>, and so the equation is solved as simply:
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| − | :<math>
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| − | \psi_{s,\mathrm{out}} = \frac{\sin(kr)}{kr} - b \frac{e^{ikr}}{r}
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| − | </math>
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| − | where <math>k=\sqrt{2mE_i} /\hbar</math>. Inside the square-well (<math>|r|<R</math>), the potential is <math>V(r)=-V_0</math>, and the solution becomes:
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| − | :<math>
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| − | \psi_{s,\mathrm{in}} = A \frac{\sin(qr)}{qr}
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| − | </math>
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| − | where <math>q=\sqrt{2m(E_i+V_0)} /\hbar</math>. The two solutions are subject to a continuity boundary condition at <math>|r|=R</math>:
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| − | :<math>\begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} (r=R) & = \psi_{s,\mathrm{in}} (r=R) \\
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| − | \frac{\mathrm{d}}{\mathrm{d}r} \psi_{s,\mathrm{out}} (r=R) & = \frac{\mathrm{d}}{\mathrm{d}r} \psi_{s,\mathrm{in}} (r=R)
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| − | \end{alignat}
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| − | </math>
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| − | Note that <math>kR = \sqrt{2 m E_i} R 2\pi/h \ll 1</math>; because of the small neutron mass and energy (see above), as well as the small size of a nucleus (femtometers). Therefore:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} & = \frac{\sin(kR)}{kR} - b \frac{e^{ikR}}{r} \\
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| − | & \approx 1 - b/r
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| − | \end{alignat}
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| − | </math>
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| − | and so:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} (r=R) & = \psi_{s,\mathrm{in}} (r=R) \\
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| − | 1 - \frac{b}{R} & = A \frac{\sin(qR)}{qR} \\
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| − | R - b & = A \frac{\sin(qR)}{q}
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| − | \end{alignat}
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| − | </math>
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| − | And from equating the derivatives:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | 1 & = A \cos(qR) \\
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| − | A & = \frac{1}{ \cos(qR) }
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| − | \end{alignat}
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| − | </math>
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| − | Combining the two results yields:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | R - b & = \left( \frac{1}{\cos(qR)} \right) \frac{\sin(qR)}{q} \\
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| − | \frac{R}{R} - \frac{b}{R} & = \frac{ \tan(qR) }{q} \frac{1}{R} \\
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| − | \frac{b}{R} & = 1 - \frac{ \tan(qR) }{qR}
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| − | \end{alignat}
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| − | </math>
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| − | This final equation gives a first-order estimate for the scattering length, ''b'', given the radius of the nucleus (R ~ 10<sup>−15</sup> m) and the depth of the potential well (''V''<sub>0</sub> ~ MeV).
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| − |
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| − | TBD: Graph
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| − |
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| − | The extreme variation of ''b/R'' as a function of ''qR'' means that with each nucleon added, the scattering length jumps to a very different value. This also demonstrates why the scattering length can be negative (indicative of negative phase shift). This model should only be taken qualitatively, of course (e.g. we have neglected absorption, as well as the detailed form of the nuclear potential).
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