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Relativity
The Relativistic Doppler shift
Consider a source of very short-duration electromagnetic pulses (e.g., a radar antenna) at rest at the origin of reference frame S. An observer moves in the x-direction at velocity v as seen in frame S. The observer is at rest in frame S’. Suppose a first pulse is sent at t = 0 when the observer is at position x = xo Further suppose the (n + 1)th pulse is sent out at t = nτ where τ is the period.
- Carefully draw the Minkowski diagram (space-time diagram) for frame S identifying the two events representing the arrival of the first and (n + 1)th pulse at the observer.
- Find the time interval between those two pulses in the observer’s frame S’.
- And therefore find the frequency ν' in the observer’s frame in terms of the source frequency ν and the observer’s velocity factor β. This is the relativistic Doppler shift.
S’(2) is moving right with a speed v. S(1) is standing. At (X0,0) in S the first light is emitted. These events are the red circles in the S and S’ axes. At (X0 + ντ,τ) in S' the second light is emitted. These events are the blue circles in S and S’. The interval is Δt' = τ' in S' (the brown segment). In S the interval is Δt = τ. Thus τ is the time between two consecutive emissions of light.
From the diagram we can immediately deduce that 
The speed v of S’ moving is
To find tB, suppose one has a mirror at τ' that reflects back the light emitted at tA = τ towards S. This light arrives in S at tC = K * τ' (from symmetry considerations). Therefore 
Therefore the speed of S’ is 
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