# LOG#045. Fake superluminality.

Before becoming apparent superluminal readers, we are going to remember and review some elementary notation and concepts from the relativistic Doppler effect and the starlight aberration we have already studied in this blog.

Let us consider and imagine the next gedankenexperiment/thought experiment. Some moving object emits pulses of light during some time interval, denoted by $\Delta \tau_e$ in its own frame. Its distance from us is very large, say

$D>>c\Delta \tau_e$

Question: Does it (light) arrive at time $t=D/c$? Suppose the object moves forming certain angle $\theta$ according to the following picture

Time dilation means that a second pulse would be experiment a time delay $\Delta t_e=\gamma \Delta \tau_e$, later of course from the previous pulse, and at that time the object would have travelled a distance $\Delta x=v\Delta t_e\cos\theta$ away from the source, so it would take it an additional time $\Delta x/c$ to arrive at its destination. The reception time between pulses would be:

$\Delta t_r=\Delta t_e+\beta \Delta t_e\cos\theta=\gamma (1+\beta \cos\theta)\Delta \tau_e$

i.e.

$\boxed{\Delta t_r=(1+\beta\cos\theta)\gamma \Delta \tau_e}$

In the range $0<\theta<\pi$, the time interval separation measured from both pulses in the rest frame on Earth will be longer than in the rest frame of the moving object. This analysis remains valid even if the 2 events are not light beams/pulses but succesive packets or “maxima” of electromagnetic waves ( electromagnetic radiation).

Astronomers define the dimensionless redshift

$\boxed{(1+z)\equiv \dfrac{\Delta t_r}{\Delta \tau_e}=\gamma (1+\beta \cos\theta)}$

where, as it is common in special relativity, $\beta=v/c$, $\gamma^2=\dfrac{1}{1-\beta^2}$

The 3 interesting limits of the above expression are:

1st. Receding emitter case. The moving object moves away from the receiver. Then, we have $\theta=0$ supposing a completely radial motion in the line of sight, and then a literal “redshift” ( lower frequencies than the proper frequencies)

$(1+z)=\sqrt{\dfrac{1+\beta}{1-\beta}}$

2nd. Approaching emitter case. The moving object approaches and goes closer to the observer. Then, we get $\theta=\pi$, or motion inward the radial direction, and then a “blueshift” ( higher frequencies than those of the proper frequencies)

$(1+z)=\sqrt{\dfrac{1-\beta}{1+\beta}}$

3rd. Tangential or transversal motion of the source. This is also called second-order redshift. It has been observed in extremely precise velocity measurements of pulsars in our Galaxy.

$(1+z)=\gamma$

Furthermore, these redshifts have all been observed in different astrophysical observations and, in addition, they have to be taken into account for tracking the position via GPS, geolocating satellites and/or following their relative positions with respect to time or calculating their revolution periods around our planet.

Remark: Quantum Mechanics and Special Relativity would be mutually inconsistent IF we did not find the same formual for the ratios between energy and frequencies at different reference frames.

EXAMPLE: The emission line of the oxygen (II) [O(II)] is, in its rest frame, $\lambda_0=3727\AA$. It is observed in a distant galaxy to be at $\lambda=9500\AA$. What is the redshift z and the recession velocity of this galaxy?

Solution.  From the definition of wavelength in electromagnetism $cT=\lambda$, adn $c\tau=\lambda_0$. Then,

$(1+z)=\dfrac{T}{\tau}=\dfrac{\lambda}{\lambda_0}=\dfrac{9500}{3727}=2.55$, and thus $z=1.55$

From the radial velocity hypothesis, we get

$(1+z)=\sqrt{\dfrac{1+\beta}{1-\beta}}$ or

$\beta=\dfrac{(1+z)^2-1}{(1+z)^2+1}=0.73$

and thus $\beta=0.73$ or $v=0.73c$
Note that this result follows from the hypothesis of the expansion of the Universe, and it holds in the relativistic theory of gravity, General Relativity, and it should also holds in extensions of it, even in Quantum Gravity somehow!

Remember: Stellar aberration causes taht the positions on the sky of the celestial objects are changing as the Earth moves around the Sun. As the Earth’s velocity is about $v_E\approx 30km/s$, and then $\beta_E\approx 10^{-4}$, it implies an angular separation about $\Delta \theta\approx 10^{-4}rad$. Anyway, it is worth mentioning that the astronomer Bradley observed this starlight aberration in 1729! A moving observer observes that light from stars are at different positions with respect to a rest observer, and that the new position does not depend on the distance to the star. Thus, as the relative velocity increases, stars are “displaced” further and further towards the direction of observation.

Now, we are going to the main subject of the post. I decided to review this two important effects because it is useful to remember then and to understand that they are measured and they are real effects. They are not mere artifacts of the special theory of relativity masking some unknown reality. They are the reality in the sense they are measured. Alternative theories trying to understand these effects exist but they are more complicated and they remember me those people trying to defend the geocentric model of the Universe with those weird metaphenomenon known as epicycles in order to defend what can not be defended from the experimental viewpoint.

In order to make our discussion visual and phenomenological, I am going to consider a practical example. Certain radio-galaxy, denoted by 3C 273 moves with a velocity

$\omega=0.8 miliarc sec/yr=4\cdot 10^{-9}\dfrac{rad}{yr}$

Note that $1 miliarc sec=\left(\dfrac{10^{-3}}{3600}\right)^{\textdegree}$

Knowing the rate expansion of the universe and the redshift of the radiogalaxy, its distance is calculated to be about $2.6\cdot 10^9 lyr$. To obtain the relative tangential velocity, we simply multiply the angular velocity by the distance, i.e. $v_{r\perp}=\omega D$.

From the above data, we get that the apparent tangential radial velocity of our radiogalaxy would be about $v_{r\perp}\approx 10c$. Indeed, this observation is not isolated. There are even jets of matter flowin from some stars at apparent superluminal velocities. Of course this is an apparent issue for SR. How can we explain it? How is it possible in the SR framework to obtain a superluminal velocity? It shows that there is no contradiction with SR. The (fake and apparent) superluminal effect CAN BE EXPLAINED naturally in the SR framework in a very elegant way. Look at the following picture:

It shows:

-A moving object with velocity $v=\vert \mathbf{v}\vert$ with respect to Earth, approaching to Earth.

-There is some angle $\theta$ in the direction of observation. And as it moves towards Earth, with our conventions, $lates \theta\approx\pi=180\textdegree$

-The moving object emits flashes of light at two different points, A and B, separated by some time interval $\Delta t_e$ in the Earth reference frame.

-The distance between those two points A and B, is very small compared with the distance object-Earth, i.e., $d(A,B)<< D$.

Question: What is the time separation $\Delta t_r$ between the receptions of the pulses at the Earth surface?

The solution is very cool and intelligent. We get

A: time interval $\Delta t_e=t_A=\dfrac{D}{c}$

B: time interval $t_B=t_A+\dfrac{v\Delta t_e\cos\theta}{c}$

Note that $\cos\theta<0$!

From this equations, we get a combined equation for the time separation of pulses on Earth

$\boxed{\Delta t_r=\Delta t_e (1+\beta \cos\theta)}$

The tangential separation is defined to be

$\Delta Y=Y_B-Y_A=v\Delta t_e\sin\theta$

so, the apparent velocity of the source, seen from the Earth frame, is showed to be:

$\boxed{v_a=\dfrac{\Delta Y}{\Delta t_r}=\dfrac{\beta\sin\theta}{1+\beta\cos\theta}c}$

Remark (I): $v_a>>c$ IFF $\beta\approx 1$ AND $\cos\theta\approx -1$!

Remark (II): There are some other sources of fake superluminality in special relativity or general relativity (the relativist theory of gravity). One example is that the phase velocity and the group velocity can indeed exceed the speed of light, since from the equation $v_{ph}v_{g}=c^2$, it is obvious that whenever that one of those two velocities (group or phase velocity) are lower than the speed of light at vacuum, the another has to be exceeding the speed of light. That is not observable but it has an important rôle in the de Broglie wave-particle portrait of the atom. Other important example of apparent and fake superluminal motion is caused by gravitational (micro)lensing in General Relativity. Due to the effect of intense gravitational fields ( i.e., big concentrations of mass-energy), light beams from slow-movinh objects can be magnified to make them, apparently, superluminal. In this sense, gravity acts in an analogue way of a lens, i.e., as it there were a refraction index and modifying the propagation of the light emitted by the sources.

Remark (III): In spite of the appearance, I am not opposed to the idea of superluminal entities, if they don’t break established knowledge that we do know it works. Tachyons have problems not completely solved and many physicists think (by good reasons) they are “unphysical”.  However, my own experience working with theories beyond special/general relativity and allowing superluminal stuff (again, we should be careful with what we mean with superluminality and with “velocity” in general) has showed me that if superluminal objects do exist, they have observable consequences. And as it has been showed here, not every apparent superluminal motion is superluminal!Indeed, it can be handled in the SR framework. So, be aware of crackpots claiming that there are superluminal jets of matter out there, that neutrinos are effectively superluminal entities ( again, an observation refuted by OPERA, MINOS and ICARUS and in complete disagreement with the theory of neutrino oscillations and the real mass that neutrino do have!) or even when they say there are superluminal protons and particles in the LHC or passing through the atmosphere without any effect that should be vissible with current technology. It is simply not true, as every good astronomer, astrophysicist or theoretical physicist do know! Superluminality, if it exists, it is a very subtle thing and it has observable consequences that we have not observed until now. As far as I know, there is no (accepted) observation of any superluminal particle, as every physicist do know. I have discussed the issue of neutrino time of flight here before:

https://thespectrumofriemannium.wordpress.com/2012/06/08/

Final challenge: With the date given above, what would the minimal value of $\beta$ be in order to account for the observed motion and apparent (fake) superluminal velocity of the radiogalaxy 3C 273?

# 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

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