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James Van Allen journal, 1951?-December 1954
Page 81
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fit a Maxwellian distribution and get an effective temperature. Maxwellian distribution in relativistic region would be interesting to work out. Might suggest form of high energy tail. Might also be of interest to see if heavy and light nuclei are in thermal equilibrium (equipartition of energy) ------------------------------------------------------------------------ ----> Room temperature (288 degrees K), (K (pi)) ~((1/40) ev). At 1 Bev, ((10 exponent 9)/(1/40)) = ((40 x (10 exponent 9))(288 exponent [1? 0?])) ~ (10 exponent 13) degrees absolute. ------------------------------------------------------------- {*KE*} E = (()/()) - ((M sub 0)(C exponent 2)) = {*TE*} E - ((M sub 0)(c exponent 2)) {*KE*} E is kinetic energy {*TE*} E is total energy (K, {*KE*} E, + rest) Proton, ((m sub 0)(C exponent 2)~= 1 Bev) beta = (square root of (1 - ((((M sub 0)(C exponent 2))/({*TE*} E)) exponent 2))) [data chart] ------------------------------------------------------------------------------------------------------------------------------- { ((((M sub 0) (C exponent 2)) / ({*TE*} E)) exponent 2) } { 0.91 } { 0.83 } { 0.70 } { 0.45 } { 0.25 } { 0.11 } { 0.028 } ------------------------------------------------------------------------------------------------------------------------------- { {*KE*} E } { 0.05 } { 0.10 } { 0.20 } { 0.50 } { 1.0 } { 2.0 } { 5.0 } Bev ------------------------------------------------------------------------------------------------------------------------------- { {*TE*} E } { 1.05 } { 1.10 } { 1.20 } { 1.50 } { 2.0 } { 3.0 } { 6.0 } ------------------------------------------------------------------------------------------------------------------------------- { beta } { 0.30 } { 0.41 } { 0.55 } { 0.74 } { 0.87 } { 0.94 } { 0.98 } -------------------------------------------------------------------------------------------------------------------------------
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fit a Maxwellian distribution and get an effective temperature. Maxwellian distribution in relativistic region would be interesting to work out. Might suggest form of high energy tail. Might also be of interest to see if heavy and light nuclei are in thermal equilibrium (equipartition of energy) ------------------------------------------------------------------------ ----> Room temperature (288 degrees K), (K (pi)) ~((1/40) ev). At 1 Bev, ((10 exponent 9)/(1/40)) = ((40 x (10 exponent 9))(288 exponent [1? 0?])) ~ (10 exponent 13) degrees absolute. ------------------------------------------------------------- {*KE*} E = (()/()) - ((M sub 0)(C exponent 2)) = {*TE*} E - ((M sub 0)(c exponent 2)) {*KE*} E is kinetic energy {*TE*} E is total energy (K, {*KE*} E, + rest) Proton, ((m sub 0)(C exponent 2)~= 1 Bev) beta = (square root of (1 - ((((M sub 0)(C exponent 2))/({*TE*} E)) exponent 2))) [data chart] ------------------------------------------------------------------------------------------------------------------------------- { ((((M sub 0) (C exponent 2)) / ({*TE*} E)) exponent 2) } { 0.91 } { 0.83 } { 0.70 } { 0.45 } { 0.25 } { 0.11 } { 0.028 } ------------------------------------------------------------------------------------------------------------------------------- { {*KE*} E } { 0.05 } { 0.10 } { 0.20 } { 0.50 } { 1.0 } { 2.0 } { 5.0 } Bev ------------------------------------------------------------------------------------------------------------------------------- { {*TE*} E } { 1.05 } { 1.10 } { 1.20 } { 1.50 } { 2.0 } { 3.0 } { 6.0 } ------------------------------------------------------------------------------------------------------------------------------- { beta } { 0.30 } { 0.41 } { 0.55 } { 0.74 } { 0.87 } { 0.94 } { 0.98 } -------------------------------------------------------------------------------------------------------------------------------
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