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In a reference frame `K`a 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`. |
Answer» (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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