Doppler effect

The Doppler effect or Doppler shift is the change in frequency of a Wave due to its motion with respect to an observer. The frequency is increased if the wave is moving towards the observer or reduced otherwise. Being a relative effect, its description depends on whether we are using Galilean relativity or Einstein relativity. It is a consequences of the invariance of a phase across frames of reference.

Classical Doppler effect

Consider two frames of reference, SS and S′S', and a wave of angular frequency ω\omega and ω′\omega' depending on which frame you are measuring it in. The frame SS is attached to the observer, whereas S′S' is the source frame, so ω\omega will be the observed frequency and ω′\omega' will be the "emission" frequency.

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The modified angular frequency is

ω′=ω(1−n^⋅v^μ)\boxed{\omega'=\omega\left( 1- \frac{\hat{\mathbf{n}}\cdot \hat{\mathbf{v}}}{\mu} \right)}

Say the source is standing still and the source is moving. If the source is moving away, then n^⋅v^<0\hat{\mathbf{n}}\cdot \hat{\mathbf{v}}<0 and ω<ω′\omega<\omega'. This means that the observed frequency is lower: in case of light, it will be redder, and in case of sound, it will be lower pitched. Similarly, if the source is moving towards the observer, then n^⋅v^>0\hat{\mathbf{n}}\cdot \hat{\mathbf{v}}>0 and ω>ω′\omega>\omega'. Light becomes bluer and sound becomes higher pitched.

Say now the source is standing still and the observer is the one moving. If the observer is moving away, then n^⋅v^>0\hat{\mathbf{n}}\cdot \hat{\mathbf{v}}>0 and ω>ω′\omega>\omega'. Light becomes bluer and sounds go higher. As you may guess, if the observer is moving towards the source, then n^⋅v^<0\hat{\mathbf{n}}\cdot \hat{\mathbf{v}}<0 and ω<ω′\omega<\omega'. Light becomes redder and sounds go deeper.

In general, if the distance between source and observer is getting smaller, the perceived frequency increases, otherwise it decreases.

Relativistic Doppler effect

The relativistic version of the effect is essentially the same as the classical one, with the key difference that we employ Lorentz transformations instead of Galilean ones. It is still a consequence of phase invariance.

Consider a wave of phase cωt−ck⋅x=cω′t′−ck′⋅x^′c\omega t-c\mathbf{k}\cdot \mathbf{x}=c\omega't'-c\mathbf{k}'\cdot \hat{\mathbf{x}}'. The Lorentz transformation from (x0,x1,x2,x3)(x_{0},x_{1},x_{2},x_{3}) to (x0′,x1′,x2′,x3′)(x_{0}',x_{1}',x_{2}',x_{3}') says

{x0′=γ(x0−βx1)x1′=γ(x1−βx0)x2′=x2x3′=x3\begin{cases} x_{0}'=\gamma(x_{0}-\beta x_{1}) \\ x_{1}'=\gamma(x_{1}-\beta x_{0}) \\ x_{2}'=x_{2} \\ x_{3}'=x_{3} \end{cases}

With the scalar products expanded, the phase reads

ωx0−c(k1x1+k2x2+k3x3)=ω′x0′−c(k1′x1′+k2′x2′+k3′x3′)\omega x_{0}-c(k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3})=\omega'x_{0}'-c(k_{1}'x_{1}'+k_{2}'x_{2}'+k_{3}'x_{3}')

Applying the reverse transformation (from x′\mathbf{x}' to x\mathbf{x}) yields

ωγ(x0′+βx1′)−ck1γ(x1′+βx0′)−ck2x2′−ck3x3′=ω′x0′−c(k1′x1′+k2′x2′+k3′x3′)\omega \gamma(x_{0}'+\beta x_{1}')-ck_{1}\gamma(x_{1}'+\beta x_{0}')-ck_{2}x_{2}'-ck_{3}x_{3}'=\omega'x_{0}'-c(k_{1}'x_{1}'+k_{2}'x_{2}'+k_{3}'x_{3}')

We'll do some algebra to extract the wavenumbers:

(ωγ−cγβk1)x0′+(ωγβ−ck1γ)x1′−ck2x2′−ck3x3′=ω′x0′−c(k1′x1′+k2′x2′+k3′x3′)(\omega \gamma-c\gamma \beta k_{1})x_{0}'+(\omega \gamma \beta-ck_{1}\gamma)x_{1}'-ck_{2}x_{2}'-ck_{3}x_{3}'=\omega'x_{0}'-c(k_{1}'x_{1}'+k_{2}'x_{2}'+k_{3}'x_{3}')

Divide through by cc and compare term by term

ωγc−γβk1=ω′c,ωγβc−k1γ=−k1′,k2=k2′,k3=k3′\frac{\omega \gamma}{c}-\gamma \beta k_{1}=\frac{\omega'}{c},\quad \frac{\omega \gamma \beta}{c}-k_{1}\gamma=-k_{1}',\quad k_{2}=k_{2}',\quad k_{3}=k_{3}'

Further comparing this with the Lorentz transformation

k0′=γ(k0−βk1),k1′=γ(k1−βk0),k2′=k2,k3′=k3k_{0}'=\gamma(k_{0}-\beta k_{1}),\quad k_{1}'=\gamma(k_{1}-\beta k_{0}),\quad k_{2}'=k_{2},\quad k_{3}'=k_{3}

we can define the wave four-vector as

kμ=(ωc,k)k^{\mu}=\left( \frac{\omega}{c},\mathbf{k} \right)

???

k0=ωc∣k∣,k0′=ω′c=∣k′∣k_{0}=\frac{\omega}{c}\lvert \mathbf{k} \rvert ,\quad k_{0}'=\frac{\omega'}{c}=\lvert \mathbf{k}' \rvert

From this we can write

ω′c=γ(ωc−βkx)=γ(ωc−βkcos⁡θ)\frac{\omega'}{c}=\gamma\left( \frac{\omega}{c}-\beta k_{x} \right)=\gamma\left( \frac{\omega}{c}-\beta k\cos \theta \right)

where θ\theta is the angle between the velocity of the wave and the line between the wave and the observer. Using k=ω/ck=\omega/c and extracting ω′\omega' gives use the relativistic Doppler effect:

ω′=γω(1−βcos⁡θ)\boxed{\omega'=\gamma \omega(1-\beta \cos \theta)}

The equation reads a little different, but the gist of it is the same. ω′\omega' is ω\omega rescaled by some factor, in this case γ(1−βcos⁡θ)\gamma(1-\beta \cos \theta). This factor is still determined by relative velocity and whether the motion is towards or away the observer.

Let's see what happens if θ=0\theta=0. In this case, k\mathbf{k} is perpendicular to v\mathbf{v}. Let's call νs\nu_{s} the "proper frequency" of the wave, the frequency as measured in the source's frame of reference. The observed frequency νobs\nu_{obs} will be given by

νobs=γνs(1∓β)=νs1∓β1−β2=νs(1∓β)2(1−β)(1+β)=νs1∓β1±β\nu_{obs}=\gamma \nu_{s}(1\mp\beta)=\nu_{s} \frac{1\mp \beta}{\sqrt{ 1-\beta ^{2} }}=\nu_{s}\sqrt{ \frac{(1\mp \beta)^{2}}{(1-\beta)(1+\beta)} }=\nu_{s}\sqrt{ \frac{1\mp \beta}{1\pm \beta} }

The sign depends on whether the source and observer are getting closer (++) or farther (−-). If they are getting closer, νobs>νs\nu_{obs}>\nu_{s}, whereas if they are getting farther, νobs<νs\nu_{obs}<\nu_{s}, as we expect. In relativity and most often in astrophysics, these are called Redshift and blueshift respectively.