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**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 = x*_{o} Further suppose the (*n* + 1)* ^{th}* pulse is sent out at

- Carefully draw the Minkowski diagram (space-time diagram) for frame S identifying the two events representing the arrival of the first and (
*n*+ 1)pulse at the observer.^{th } - 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 (*X*0 + *ντ,τ)* in* S'* the second light is emitted. These events are the blue circles in S and S’. The interval is Δ*t**' *= *τ' *in* S' *(the brown segment). In S the interval is Δ*t* = *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 *t _{B}*, suppose one has a mirror at

Therefore the speed of S’ is

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