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I am after to model the Doppler shift of an LEO satellite. At [1] (pp 183) I have found the following formula:

$$\frac{f_r}{f_s}=\frac{1-\frac{u}{c}\cos\theta}{\sqrt{1-\frac{u^2}{c^2}}}$$

$f_s$ stable frequency transmitted from the satellite

$f_r$ received (Doppler shifted) frequency at the observation point.

$u$ satellite velocity

$c$ velocity of light

$\theta$ angle between the velocity vector of the satellite and the observer-satellite line of sight

After some further research, I see that falls in one of the theories of estimating the Doppler effect.

Can I please ask how to derive the nominator of the above formula? I have found several references but am not yet convinced.

[1]Seeber, G. (2003). Satellite Geodesy.New York, Walter de Gruyter

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In the following replace $\:^{\prime}$$^{\prime}$Star$\:^{\prime}$$^{\prime}$ by $\:^{\prime}$$^{\prime}$satellite$\:^{\prime}$$^{\prime}$.

Solution 1 enter image description here

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 we could estimate the time interval $\mathrm{d}\hat{t}$ then we 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)}} \tag{01} \end{equation}


enter image description here

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{A} \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{B} \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{C} \end{equation} QED.

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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}

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Strangely enough, it does not seem the transverse Doppler effect has been demonstrated on Physics SE. This is a trivial consequence of the Lorentz transformation of 4-wave vector. Denoting by $\omega$ the pulsation and by $(k_x, k_y, k_z)$ the wave-vector, $(\omega, k_x, k_y, k_z)$ is a 4-vector and therefore the transformation between the frame $S$ where the satellite is at rest and the frame $R$ where the receiver is at rest is

$$\begin{align} \omega_R &= \gamma(\omega_S + u\,k_{S,x})\\ k_{R,x} &= \gamma(k_{S,x} + u\, \omega_S)\\ k_{R,y}&=k_{S,y}\\ k_{R,z}&=k_{S,z}\\ \gamma&=(1-u^2)^{-\frac{1}{2}}\\ \end{align} $$

where the $x$-axis is oriented along the speed of $S$ wrt to $R$. I wrote all of them but we are going to use the first one only. Oh, and I used units where c=1. The next step is to use the relation between the pulsation and the length of the wave vector and introduce an angle for the direction of propagation

$$k_{S,x} = \omega_S\cos\phi,$$

where $\phi$ is the angle between the wave vector and the direction $\overrightarrow{SR}$. Then

$$\frac{\omega_R}{\omega_S}=\gamma(1+u\cos\phi)$$

This is your formula since $\phi = \pi - \theta$ as you consider the angle between $\overrightarrow{RS}$ and the wave vector instead. After re-introducing $c$ of course…

A great reference for undergraduate on that and much more is the classic

Wolfgang Rindler, Introduction to Special Relativity, 2nd ed. Oxford University Press, 1991.

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  • $\begingroup$ Thanks very much for your reply,Is it possible to point me on some references for background reading on that? $\endgroup$ – Giwrgos Rizeakos Jun 13 '17 at 14:32
  • $\begingroup$ @Luc J. Bourhis. Let’s observer is on the Earth. According to the task in this case $cos \theta = 0$. This will be purely Transverse Doppler Effect. Right? Does (that formula and) your calculation mean, that frequency will be higher on the reception, or $\sqrt {1-v^2/c^2}$ times blueshifted? Does that mean, that observer on the Earth will see, that a clock on the moving satellite is ticking faster, than his own? Kind regards, $\endgroup$ – Albert Nov 1 '17 at 20:25

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