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Relativistic Doppler Shift With Equations

I have already looked at the non-relativistic Doppler effect on Ask Will Online. This lead us onto the equation λ / λ = v / c when v << c (velocity of the object is very much less than the speed of light) with the maximum velocity being around 0.1c. This is because after 0.1c, due to time dilation  time starts to significantly slow down. 

As we already know, t = γτ. At low speeds, γ will equal extremely close to 1. This means at low speeds, the time of the observed time with the moving object will be the same as a wristwatch time of someone stationary. The equation λ / λ = v / c works up to around 0.1c because it is an exponential graph. This means that towards 0 to around 0.1c, the time dialtion factor will still be roughly 1. However, after roughly 0.1c, the gradient starts to increase and increase until the graph gets to an isotope. Therefore, we need a new equation for when the speed of which we are trying to measure Doppler shift is more than 0.1c.


We will take ‘k’ to be the relativistic Doppler shift.
  • k = λ / λ…
  • This also equals √1+v/c / 1-v/c…
  • Which is γ (1+v/c) where γ = 1 / √1-v²/c².
If v = 0, k will equal 1 which means the change in wavelength will 0. But, because γ is roughly 1..


  • Time dilation is only a theory. If you wanted to include this theory when calculating the Doppler shift factor, you can use √1+v/c / 1-v/c or  γ (1+v/c) where γ = 1 / √1-v²/c².
  • Not taking time dilation into account, λ / λ = v / c.
  • The time dilation, from the graph, does not move significantly above 1 until speeds very close to the speed of light.

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