3
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ The wording of the first two sentences make this sound like homework. I've added the homework-and-exercises tag. $\endgroup$ – user4552 Sep 10 '17 at 19:35
3
$\begingroup$

enter image description here

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 : \begin{equation} \dfrac{\mathrm{d}\hat{t}}{\mathrm{d}t'}=\dfrac{\nu'}{\nu}=\dfrac{\lambda}{\lambda'}=\dfrac{\lambda \text{(observed)}}{\lambda'\text{(emitted)}} \end{equation}


enter image description here

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

Solution 1 (related to the Hint)

As shown in Figure-02 above

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

\begin{equation} \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} \end{equation}

\begin{equation} \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 \end{equation}

\begin{equation} \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} \end{equation} QED.

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

Solution 2

Link : My answer in About de Broglie relations

For a plane wave the angular frequency 4-vector \begin{equation} \boldsymbol{\Omega} \equiv \left(\omega,c\mathbf{k} \right) \tag{2.01} \end{equation} 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) \begin{equation} \omega= 2\pi\nu \tag{2.02} \end{equation} is the angular frequency and $\:\nu\:$ the frequency. Also \begin{equation} \mathbf{k}= \dfrac{ 2\pi}{\lambda} \;\mathbf{m} , \qquad \Vert \mathbf{m}\Vert =1 \tag{2.03} \end{equation} is the wave 3-vector and $\:\lambda\:$ the wavelength. The plane wave $^\prime$$^\prime$propagates$^\prime$$^\prime$ with velocity vector \begin{equation} \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} \end{equation} From the Lorentz equation (A-14b) in the link we have \begin{equation} \omega^{\boldsymbol{\prime}} =\gamma\left(\omega\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}c\mathbf{k}}{c}\right) \tag{2.05} \end{equation} For a light wave $\: \mathbf{k}=(2\pi\nu/c)\mathbf{m}\:$ so \begin{equation} \nu^{\boldsymbol{\prime}} =\gamma\left(1\!+\!\dfrac{ \mathbf{v}\boldsymbol{\cdot}\mathbf{m}}{c}\right)\nu \tag{2.06} \end{equation} 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 $\:\boldsymbol{v}_{\mathrm{r}}\:$ \begin{equation} \mathbf{m}=\dfrac{\boldsymbol{v}_{\mathrm{r}}}{\Vert\boldsymbol{v}_{\mathrm{r}}\Vert} \tag{2.07} \end{equation} so that finally \begin{equation} \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} \end{equation}

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.