Orion1
Jan27-04, 09:25 AM
Strong Nuclear Quantum Chromodynamics: (SQCD)
\alpha_s \left( E \right) = \frac{12 \pi}{ \left(33 - 2N_f \right) ln \left[ \frac{E^2}{\Lambda^2} \right]}
Strong Nuclear Quantum Electrodynamics: (SQED)
\alpha_s \left( q^2 \right) = \frac{ \alpha_s ( q_0^2) }{1 - \left( \frac{\alpha_s (q_0^2) }{3 \pi} \right) ln \left[ \frac{q^2}{m^2} \right] }
\alpha_s \left( q^2 \right) = \alpha_s \left( E \right)
\alpha_s \left( q^2,E \right) = \frac{ \alpha_s ( q_0^2) }{1 - \left( \frac{\alpha_s (q_0^2) }{3 \pi} \right) ln \left[ \frac{q^2}{m^2} \right] } = \frac{12 \pi}{ \left(33 - 2N_f \right) ln \left[ \frac{E^2}{\Lambda^2} \right]}
SQCD, SQED Equasion Normalization:
\alpha_s (q_0^2)(33 - 2N_f) ln \left[ \frac{E^2}{\Lambda^2} \right] = 12 \pi \left( 1 - \frac{\alpha_s (q_0^2)}{3 \pi} ln \left[ \frac{q^2}{m^2} \right] \right)
\Lambda = 0.2 Gev - experimentally determined value
Nf - quark flavour number (16 max.)
q0 - strong nuclear charge
Effectivity?
\alpha_s \left( E \right) = \frac{12 \pi}{ \left(33 - 2N_f \right) ln \left[ \frac{E^2}{\Lambda^2} \right]}
Strong Nuclear Quantum Electrodynamics: (SQED)
\alpha_s \left( q^2 \right) = \frac{ \alpha_s ( q_0^2) }{1 - \left( \frac{\alpha_s (q_0^2) }{3 \pi} \right) ln \left[ \frac{q^2}{m^2} \right] }
\alpha_s \left( q^2 \right) = \alpha_s \left( E \right)
\alpha_s \left( q^2,E \right) = \frac{ \alpha_s ( q_0^2) }{1 - \left( \frac{\alpha_s (q_0^2) }{3 \pi} \right) ln \left[ \frac{q^2}{m^2} \right] } = \frac{12 \pi}{ \left(33 - 2N_f \right) ln \left[ \frac{E^2}{\Lambda^2} \right]}
SQCD, SQED Equasion Normalization:
\alpha_s (q_0^2)(33 - 2N_f) ln \left[ \frac{E^2}{\Lambda^2} \right] = 12 \pi \left( 1 - \frac{\alpha_s (q_0^2)}{3 \pi} ln \left[ \frac{q^2}{m^2} \right] \right)
\Lambda = 0.2 Gev - experimentally determined value
Nf - quark flavour number (16 max.)
q0 - strong nuclear charge
Effectivity?