p38_073 - 73. The strategy is to find the speed from E =...

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Unformatted text preview: 73. The strategy is to find the speed from E = 1533 MeV and mc2 = 0.511 MeV (see Table 38-3) and from that find the time. From the energy relation (Eq. 38-45), we obtain v=c 1− mc2 E 2 = 0.99999994c ≈ c so that we conclude it took the electron 26 y to reach us. In order to transform to its own “clock” it’s useful to compute γ directly from Eq. 38-45: γ= E = 3000 mc2 though if one is careful one can also get this result from γ = 1/ 1 − (v/c)2 . Then, Eq. 38-7 leads to ∆ t0 = 26 y = 0.0087 y γ so that the electron “concludes” the distance he traveled is 0.0087 light-years (stated differently, the Earth, which is rushing towards him at very nearly the speed of light, seemed to start its journey from a distance of 0.0087 light-years away). ...
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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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