# Deriving relativistic Doppler shift in terms of wavelength? [closed]

Consider a star moving with velocity $$v$$ at an angle $$\theta$$ with respect to its line of sight to Earth. Show that the relativistic Doppler shift is

$$\lambda_{obs} = \frac{1 - \frac{v}{c} cos(\theta)}{\sqrt{1 - \frac{v^2}{c^2}}} \lambda_{em}$$

in which $$c$$ is the speed of light, $$\lambda_{obs}$$ is the observed wavelength, and $$\lambda_{em}$$ is the emitted wavelength.

Can someone show me to derive this equation? So far, I have been using a reference frame $$S'$$ for a certain angle $$\theta'$$ in which the $$y' = ct'\sin(\theta')$$ and $$x' = ct'\cos(\theta')$$. I used the Lorentz transformation to find that

$$x = \frac{x' + vt'}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{ct'\cos(\theta')}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{ct'(\cos(\theta') + \frac{v}{c})}{\sqrt{1 - \frac{v^2}{c^2}}}$$

I am not sure what do from here. Also, what happens for velocities that are much smaller than c? How can I use this equation to write how at $$v\ll c$$ the equation reduces to the usual expression for a Doppler shift such that

$$\lambda_{obs} = (1 + \frac{v_r}{c})\lambda_{em}$$

in which $$v_r$$ is the radial velocity?

• The wording of the first two sentences make this sound like homework. I've added the homework-and-exercises tag. – user4552 Sep 10 '17 at 19:35

Hint :

Suppose two light pulses $p'_1$ and $p'_2$ are emitted successively from the Star towards the Earth at time moments $t'_1$ and $t'_2$, apart by an infinitesimal time interval $\mathrm{d}t'=t'_2\!-\!t'_1$. Time $t'$ is the time in the rest frame $\mathrm{S'}$ of the Star.

These two events happen in the rest frame $\mathrm{S}$ of the Earth at time moments $t_1$ and $t_2$, apart by the dilated infinitesimal time interval $\mathrm{d}t=t_2\!-\!t_1=\gamma\left(v\right)\mathrm{d}t'$. Time $t$ is the time in the rest frame $\mathrm{S}$ of the Earth.

Now, let the two light pulses arrive to Earth at Earth time moments $\hat{t}_{\!1}$ and $\hat{t}_{\!2}$, apart by an infinitesimal time interval $\mathrm{d}\hat{t}=\hat{t}_{\!2}\!-\!\hat{t}_{\!1}$. If the Star would be at rest relatively to Earth or its motion would be transverse (no radial motion : $v_\mathrm{r}=0$) then $\mathrm{d}\hat{t}=\mathrm{d}t$. But because of the radial motion of the Star relatively to Earth the 2nd pulse, which emitted later, has to run a larger distance than the 1rst pulse if the Star is moving away or has to run a smaller distance than the 1rst pulse if the Star is approaching. In the first case $\mathrm{d}\hat{t}>\mathrm{d}t$. In the second case, that shown in the Figure, $\mathrm{d}\hat{t}<\mathrm{d}t$.

So, if you could estimate the time interval $\mathrm{d}\hat{t}$ then you would solve the problem since the time intervals are inversely proportional to frequencies that is proportional to wavelengths : $$\dfrac{\mathrm{d}\hat{t}}{\mathrm{d}t'}=\dfrac{\nu'}{\nu}=\dfrac{\lambda}{\lambda'}=\dfrac{\lambda \text{(observed)}}{\lambda'\text{(emitted)}}$$

$===================================================$

Solution 1 (related to the Hint)

As shown in Figure-02 above

$$\mathrm{d}t=t_2\!-\!t_1=\gamma(v)\left(t'_{\!2}\!-\!t'_{\!1}\right) =\gamma(v)\mathrm{d}t' \tag{1.01}$$

$$\mathrm{dr}\approx \mathrm{r_2}-\mathrm{r_1}=-v_\mathrm{r}\,\mathrm{d}t=-v \cos\theta\, \gamma(v)\,\mathrm{d}t' \tag{1.02}$$

$$\mathrm{d}\hat{t}=\hat{t}_2\!-\!\hat{t}_1=\left(t_2\!+\!\dfrac{ \mathrm{r}_2}{c}\right)\!-\!\left( t_1\!+\!\dfrac{ \mathrm{r}_1}{c} \right)=\mathrm{d}t\!+\!\dfrac{\mathrm{dr}}{c} =\gamma(v)\mathrm{d}t'\!-\!\dfrac{v \cos\theta\, \gamma(v)\,\mathrm{d}t' }{c} \Longrightarrow \nonumber$$

$$\dfrac{\mathrm{d}\hat{t}}{\mathrm{d}t' } = \dfrac{1\!-\!\dfrac{v \cos\theta}{c}}{ \sqrt{1\!-\!\dfrac{v^2}{c^2}}} \stackrel{\left(\beta=\tfrac{v}{c}\right)}{=\!=\!=}\dfrac{1\!-\!\beta \cos\theta}{\sqrt{1\!-\!\beta^2}}=\dfrac{\nu' \text{(emitted)}}{\nu\:\text{(observed)}} =\dfrac{\lambda\:\text{(observed)}}{\lambda' \text{(emitted)}} \tag{1.03}$$ QED.

$===================================================$

Solution 2

For a plane wave the angular frequency 4-vector $$\boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \tag{2.01}$$ is transformed between frames under the Lorentz transformation. This is proved in the link for a more general configuration of two frames (see the Figure in the end of the link). In (2.01) $$\omega= 2\pi\nu \tag{2.02}$$ is the angular frequency and $\:\nu\:$ the frequency. Also $$\mathbf{k}= \dfrac{ 2\pi}{\lambda} \;\mathbf{m} , \qquad \Vert \mathbf{m}\Vert =1 \tag{2.03}$$ is the wave 3-vector and $\:\lambda\:$ the wavelength. The plane wave $^\prime$$^\primepropagates^\prime$$^\prime$ with velocity vector $$\mathbf{w}= \dfrac{ \omega}{\Vert \mathbf{k}\Vert } \;\mathbf{m}=\lambda\nu\;\mathbf{m} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert^{2}}\mathbf{k}, \qquad \Vert \mathbf{w}\Vert \equiv \mathrm{w} = \dfrac{ \omega}{\Vert \mathbf{k}\Vert }=\lambda\nu \tag{2.04}$$ From the Lorentz equation (A-14b) in the link we have $$\omega^{\boldsymbol{\prime}} =\gamma\left(\omega\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}c\mathbf{k}}{c}\right) \tag{2.05}$$ For a light wave $\: \mathbf{k}=(2\pi\nu/c)\mathbf{m}\:$ so $$\nu^{\boldsymbol{\prime}} =\gamma\left(1\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}\mathbf{m}}{c}\right)\nu \tag{2.06}$$ In above equation $\:\mathbf{v}=\!-\boldsymbol{v}\:$ is the velocity vector of the Earth relatively to the Star , the vector $\:\boldsymbol{v}\:$ shown in Figures-01,-02 and $\:\mathbf{m}\:$ the unit vector parallel to its radial component $\:\boldsymbol{v}_{\mathrm{r}}\:$ $$\mathbf{m}=\dfrac{\boldsymbol{v}_{\mathrm{r}}}{\Vert\boldsymbol{v}_{\mathrm{r}}\Vert} \tag{2.07}$$ so that finally $$\dfrac{\nu' \text{(emitted)}}{\nu\:\text{(observed)}}=\gamma\left(1\!-\!\dfrac{ v \cos \theta}{c}\right)= \dfrac{1\!-\!\dfrac{v \cos\theta}{c}}{ \sqrt{1\!-\!\dfrac{v^2}{c^2}}} \tag{2.08}$$