# LOG#035. Doppler effect and SR (I).

The Doppler effect is a very important phenomenon both in classical wave motion and relativistic physics. For instance, nowadays it is used to the detect exoplanets and it has lots of applications in Astrophysics and Cosmology.

Firstly, we remember the main definitions we are going to need here today.

$\omega =\dfrac{2\pi}{T}=2\pi\nu=2\pi f$

Sometimes, we will be using the symbol $f$ for the frequency $\nu$. We also have:

$\mathbf{n}=\dfrac{\mathbf{k}}{\vert \mathbf{k}\vert}$ $k=\vert \mathbf{k}\vert=\dfrac{2\pi}{\lambda}=\dfrac{\omega}{c}$

A plane (sometimes electromagnetic) wave is defined by the oscillation:

$A(\mathbf{r},t)=A_0\exp\left(iKX\right)$

where

$KX=\mathbb{K}\cdot\mathbb{X}=\mbox{PHASE}=\mathbf{k}\cdot \mathbf{r}-\omega t$

If $K^2=0$, then the wave number is lightlike (null or isotropic) and then $(K^0)^2=\dfrac{\omega^2}{c^2}$

Using the Lorentz transformations for the wave number spacetime vector, we get:

$K'^0=\gamma \left(K^0-\beta k^1\right)$

i.e., if the angle with the direction of motion is $\alpha$ so $k^1=\dfrac{\omega}{c}\cos\alpha$

$\dfrac{\omega '}{c}=\gamma \left(\dfrac{\omega}{c}-\beta\left(\dfrac{\omega}{c}\cos \alpha\right)\right)$

we deduce that

$\boxed{\omega=\omega_0\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}}$

or for the normalized frequency shift

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}-1}$

This is the usual formula for the relativistic Doppler effect when we define $\omega'=\omega_0$, and thus, the angular frequency (also the frecuency itself, since there is only a factor 2 times the number pi of difference) changes with the motion of the source. When the velocity is “low”, i.e., $\beta<<1$, we obtain the classical Doppler shift formula:

$\omega \approx (1+\beta\cos\alpha)\omega_0$

We then calculate the normalized frequency shift from $\Delta \omega=\omega-\omega_0$

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{V\cos\alpha}{c}}$

The classical Doppler shift states that when the source approaches the receiver ($\cos\alpha>0$), then the frequency increases, and when the source moves away from the receiver ($\cos\alpha<0$), the frequency decreases. Interestingly, in the relativistic case, we also get a transversal Doppler shift which is absent in classical physics. That is, in the relativistic Doppler shift, for $\alpha=\pi/2$, we obtain

$\boxed{\omega=\dfrac{\omega_0}{\gamma_V}=\omega_0\sqrt{1-\beta^2}}$

and the difference in frequency would become

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\sqrt{1-\beta^2}-1}$

There is an alternative deduction of these formulae. The time that an electromagnetic wave uses to run a distance equal to the wavelength, in a certain inertial frame S’ moving with relative speed V to another inertial frame S at rest, is equal to:

$t=\dfrac{\lambda}{c-V}=\dfrac{c}{(c-V)f_s}=\dfrac{1}{(1-\beta_V)f_s}$

where $f_s$ is the frequency of the source. Due to the time dilation of special relativity

$t=t_0\gamma$ and thus

$f_0=\dfrac{1}{t_0}=\gamma(1-\beta_V)f_s=\sqrt{\dfrac{1-\beta_V}{1+\beta_V}}f_s$

so we get

$\boxed{\dfrac{f_s}{f_o}=\dfrac{f_{source}}{f_{obs}}=\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}$

The redshift (or Doppler displacement) is generally defined as:

$\boxed{z=-D=\dfrac{\lambda_o-\lambda_s}{\lambda_s}=\dfrac{f_s-f_o}{f_o}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}-1}$

When $\beta\rightarrow 0$, i.e., $V\rightarrow 0,V<, we get the classical result

$z\approx\beta=\dfrac{v}{c}$

The generalization for a general non-parallel motion of the source/observer is given by

$\boxed{f_0=\dfrac{f_s}{\gamma\left(1+\dfrac{v_s\cos\theta_o}{c}\right)}}$

If we use the stellar aberration formula:

$\boxed{\cos\theta_o=\dfrac{\cos\theta_s-\dfrac{v_s}{c}}{1-\dfrac{v_s}{c}\cos\theta_s}}$

the last equation can be recasted in terms of $\theta_s$ instead of $\theta_o$ as follows:

$\boxed{f_0=\gamma\left(1-\dfrac{v_s}{c}\cos\theta_s\right)}$

and then

$\boxed{z=\dfrac{f_s-f_0}{f_0}=-D=\dfrac{\Delta f}{f_0}=\dfrac{1}{\gamma\left(1-\dfrac{v_s\cos\theta_s}{c}\right)}-1}$

Again, for a transversal motion, $\theta_o=\pi/2$, we get a transversal Doppler effect:

$\boxed{f_0=\dfrac{f_s}{\gamma}=\sqrt{1-\beta^2}f_s} \leftrightarrow \boxed{z_T=-D=\dfrac{f_o-f_s}{f_s}=\sqrt{1-\beta_s^2}-1}$

Remark: Remember that the Doppler shift formulae are only valid if the relative motion (of both source and observer/receiver) is slower than the speed of the (electromagnetic) wave, i.e., if $v\leq c$.

In the final part of this entry, we are going to derive the most general formula for Doppler effect, given an arbitrary motion of source and observer, in both classical and relativistic Physics. Recall that the Doppler shift in Classical Physics for an arbitrary observer is given by a nice equation:

$\boxed{f'=f\left[\dfrac{v-v_o\cos\theta_o}{v-v_s\cos\theta_s}\right]}$

Here, $v$ is the velocity of the (electromagnetic) wave in certain medium, $v_o$ is the velocity of the observer in certain direction forming an angle $\theta_o$ with the “line of sight”, while $v_s$ is the velocity of the source forming an angle $\theta_s$ with the line of sight/observation. If we write

$V_o=-v_o\cos\theta_o$ $V_s=-v_s\cos\theta_s$ $V_{so}=V_s-V_o$

we can rewrite this last Doppler formula in the following way ( for $v=c$):

$f'=\left(\dfrac{c+V_o}{c+V_s}\right)f$ or with $f'=f_o$ and $f=f_s$

$\boxed{f_o=f_s\left(\dfrac{c+V_s}{c+V_o}\right)=\left(1+\dfrac{V_s-V_o}{c+V_o}\right)f_s=\left(1+\dfrac{V_{so}}{c+V_o}\right)f_s}$

and

$\boxed{z=-D=\dfrac{f_0-f_s}{f_0}=\left(\dfrac{c+V_0}{c+V_s}\right)-1=\dfrac{V_o-V_s}{c+V_s}=\dfrac{-V_{so}}{c+V_s}}$

The most general Doppler shift formula, in the relativistic case, reads:

$\boxed{\dfrac{f_o}{f_s}=\dfrac{1-\dfrac{\vert\vert \mathbf{v}_o\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{co}}{1-\dfrac{\vert\vert \mathbf{v}_s\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{cs}}\sqrt{\dfrac{1-\dfrac{v_s^2}{c^2}}{1-\dfrac{v_o^2}{c^2}}}}$

or equivalently

$\boxed{\dfrac{f_o}{f_s}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}}$

and where $\mathbf{v}_s,\mathbf{v}_o$ are the velocities of the source and the observer at the time of emission and reception, respectively, $\beta_s,\beta_o$ are the corresponding beta boost parameters, $\mathbf{c}$ is the “light” or “wave” velocity vector and we have defined the angles $\theta_{cs},\theta_{co}$ to be the angles formed at the time of emission and the time of reception/observation between the source velocity and the “wave” velocity, respectively, between the source and the wave and the observer and the wave. Two simple cases of this formula:

1st. Parallel motion with $\mathbf{c}\parallel\mathbf{v}_s\rightarrow \theta_{cs}=0\textdegree\rightarrow \cos\theta_{cs}=1\rightarrow f_o>f_s$.

2nd. Antiparallel motion with $\mathbf{c}$ going in the contrary sense than that of $\mathbf{v}_s$. Then, $\theta_{cs}=180\textdegree\rightarrow \cos\theta_{cs}=-1\rightarrow f_o.

The deduction of this general Doppler shift formula can be sketched in a simple fashion. For a signal using some propagating wave, we deduce:

$\vert \mathbf{r}_o-\mathbf{r}_s\vert ^2=\vert\mathbf{C}\vert^2\left(t_o-t_s\right)^2$

with

$\mathbf{C}=\dfrac{\mathbf{r}_o-\mathbf{r}_s}{t_o-t_s}$

Differentiating with respect to $t_o$ carefully, it provides

$\mathbf{C}\cdot\left[\mathbf{v}_o-\mathbf{v}_s\dfrac{dt_s}{dt_o}\right]=\vert \mathbf{C}\vert^2\left(1-\dfrac{dt_o}{dt_s}\right)$

Solving for $\dfrac{dt_o}{dt_s}$ we get

$\dfrac{dt_o}{dt_s}=\dfrac{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_o }{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_s}=\dfrac{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_o\right)}{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_s\right)}=\dfrac{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_o}{\vert\mathbf{C}\vert^2}}{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_s}{\vert\mathbf{C}\vert^2}}$

Using the known formula $\vert \mathbf{r}\cdot\mathbf{s}\vert=\vert\mathbf{r}\vert\vert\mathbf{s}\vert\cos\theta_{rs}$, we obtain in a simple way:

$\dfrac{dt_o}{dt_s}=\dfrac{1-\dfrac{\vert\mathbf{v}_o\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_o}}{1-\dfrac{\vert\mathbf{v}_s\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_s}}=\dfrac{\vert\mathbf{C}-\mathbf{v}_o\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}}{\vert\mathbf{C}-\mathbf{v}_s\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}}$

Finally, using the fact that $\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}=\cos\theta_{\mathbf{C},\mathbf{v}_s}$, the similar result $\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}=\cos\theta_{\mathbf{C},\mathbf{v}_o}$, and that the proper time induces an extra gamma factor due to time dilation

$dt=d\tau\gamma$

and we calculate for the frequency:

$\dfrac{n_o}{n_s}=\dfrac{\nu_o}{\nu_s}=\dfrac{f_o}{f_s}=(\mbox{PREFACTOR})\dfrac{\gamma_s dt_s}{\gamma_odt_o}$

where the PREFACTOR denotes the previously calculated ratio between differential times. Finally, elementary algebra let us derive the expression:

$\dfrac{f_o}{f_s}=\dfrac{d\tau_s}{d\tau_o}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}$

Q.E.D.

An interesting remark about the Doppler effect in relativity: the Doppler effect allows us to “derive” the Planck’s relation for quanta of light. Suppose that a photon in the S’-frame has an energy $E'$ and momentum $(p'_x,0,0)=(-E',0,0)$ is being emitted along the negative x’-axis toward the origin of the S-frame. The inverse Lorentz transformation provides:

$E=\gamma (E'+vp'_x)=\gamma (E'-\beta E')=\dfrac{1-\beta}{\sqrt{1-\beta^2}}E'$, i.e.,

$E=\sqrt{\dfrac{1-\beta}{1+\beta}}E'$

By the other hand, by the relativistic Doppler effect we have seen that the frequency f’ in the S’-frame is transformed into the frequency f in the S-frame if we use the following equation:

$f=f'\sqrt{\dfrac{1-\beta}{1+\beta}}$

If we divide the last two equations we get:

$\dfrac{E}{f}=\dfrac{E'}{f'}=constant \equiv h$

Then, if we write $h=6.63\cdot 10^{-34}J\cdot s$, and $E=hf$ and $E'=hf'$.

AN ALTERNATIVE HEURISTIC DEDUCTION OF THE RELATIVISTIC DOPPLER EFFECT

In certain rest frame S, there is an observer receiving light beams/signals. The moving frame is the S’-frame and it is the emitter of light. The source of light approaches at velocity V, and it sends pulses with frequency $f_0=\nu_0$. What is the frequency that the observer at S observes? Due to time dilation, the observer at S observes a longer period

$T=T_0\dfrac{1}{\sqrt{1-\beta_V^2}}$

with

$\beta_V=\dfrac{V}{c}$

The distance between two consecutive light beams seen by the observer at S will be:

$\lambda=cT-VT=(c-V)T=(c-V)\dfrac{T_0}{\sqrt{1-\beta_V^2}}$

Therefore, the observer frequency in the S-frame is:

$f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V)}=\nu_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}$

i.e.

$\boxed{f=f_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}$

If the source approaches the observer, then $f>f_0$. If the source moves away from the observer, then $f. In the case, the velocity of the source forms certain angle $\alpha$ in the direction of observation, the same argument produces:

$f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V\cos\alpha)}=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}$

that is

$\boxed{f=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}}$

In the case of transversal Doppler effect, we get $\alpha=90\textdegree=\pi/2$, and so:

$\boxed{f=f_0\sqrt{1-\beta^2}}$

Q.E.D.

Final remark (I): If $\theta=\alpha=\dfrac{\pi}{2}$ or $\theta=\alpha=\dfrac{3\pi}{2}$ AND $v<, therefore we have that there is no Doppler effect at all.

Final remark (II): If $v\approx c$, there is no Doppler effect in certain observation directions. Those directions can be deduced from the above relativistic Doppler effect formula with the condition $f=f_0$ and solving for $\theta$. This gives the next angular direction in which Doppler effect can not be detected

$\boxed{\cos\theta=\dfrac{\sqrt{1-\beta^2}-1}{\beta}}$