Physics · Special Relativity
An atom emits light of angular frequency ω0 in its own rest frame. The atom moves with speed v along the line joining it to a stationary observer. Which option correctly gives the observed angular frequency ω when the motion is directly away from or towards the observer?
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1Away: ω = ω0√(c + vc - v), Towards: ω = ω0√(c - vc + v)
- 2Away: ω = ω0(c - vc + v), Towards: ω = ω0(c + vc - v)
- 3Away: ω = ω0√(c - vc + v), Towards: ω = ω0√(c + vc - v)✓
- 4Away: ω = γ(1 + β)ω0, Towards: ω = γ(1 - β)ω0
Correct answer
Away: ω = ω0√(c - vc + v), Towards: ω = ω0√(c + vc - v)
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