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In a reference frame Ka photon of frequency omega falls normally on a mirror approaching it with relativistic velocity V. Find the momentum inparted to the mirror during the reflection of the photon (a) in the reference frame fixed to the mirroe, (b) in the frame K. |
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Answer» Solution :(a) In the reference frame fixed to the MIRROR, the frequency of the photons is, by the Doppler shift formula `overset_(OMEGA) = omega sqrt((1+beta)/(1-beta)) (=omegasqrt(1-beta^(2))/(1-beta))` In this frame momentum imparted to the mirror is `(2cancelh overset_(omega))/(c) = (2cancelh omega)/(c) sqrt((1+beta)/(1-beta))`, (b) In the `K` frame, the incident particle CARRIES a momentum of `cancelh omega//c`and returns with momentum `(cancelh omega)/(c)(1+beta)/(1-beta)` The momentum inparted to the mirror, then, has the magnitude `(cancelh omega)/(c)[(1+beta)/(1-beta)+1] = (2cancelh omega)/(c)(1)/(1-beta)` Hence `beta = (V)/(c)`. |
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