p45_029 - 29 From = 1 K/mc2(see Eq 38-49 and v = c = c v =c...

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29. From γ =1+ K/mc 2 (see Eq. 38-49) and v = βc = c p 1 γ 2 (see Eq. 38-8), we get v = c s 1 µ 1+ K mc 2 2 . Therefore, for the Σ 0 particle, v =(2 . 9979 × 10 8 m / s) s 1 µ 1+ 1000 MeV 1385 MeV 2 =2 . 4406 × 10 8 m / s , and for Σ 0 , v 0 =(2 . 9979 × 10 8 m / s) s 1 µ 1+ 1000 MeV 1192
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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