I Headlight Effect of Moving Light Source: Formula

[Hiero]

Suppose we have a light source emitting photons isotropically at a constant rate. Call the rest frame of the source S’ and call the photon-flux-density in this frame $\sigma ‘$. (I’m treating it as a scalar instead of a vector because I’m assuming the light should travel radially in either frame.)

Let the S’ frame move with velocity $\beta$ in the S frame. Then how will the the photon density look in the S frame? By symmetry it can only depend on the angle $\theta$ with the direction of motion.

Using the aberration formulas and assuming the number of photons (per unit time) in a cone between $\theta$ and $\theta + d\theta$ is the same as the number between $\theta’$ and $\theta’ + d\theta’$ ... then I get the following result:

$$\frac{\sigma}{\sigma ‘} = \frac{1-\beta ^2}{(1-\beta \cos\theta)^2}\approx 1+2\beta \cos\theta$$

I can’t find anything similar online. Is this formula an accurate quantification of the so called “headlight effect”? I can show the steps of my derivation upon request.

Thanks.

 

[jartsa]

Using the aberration formulas and assuming the number of photons (per unit time) in a cone between θ\theta and θ+dθ\theta + d\theta is the same as the number between θ′\theta’ and θ′+dθ′\theta’ + d\theta’ ... then I get the following result:

So you did not take into account time dilation and Doppler shift?

Doppler shift causes the space between consecutive photons to be shorter or longer, time dilation causes the time between emission of two consecutive photons to be longer.

 

[vanhees71]

What we need is the energy-momentum tensor of the electromagnetic field. We consider it in a large distance from the source. Then everywhere this tensor must be given by
$$T^{\mu \nu}=\epsilon(t,r) u^{\mu} u^{\nu},$$
where $u^{\mu}$ is the "four-velocity of a photon". It's everywhere in radial spatial direction,
$$u^{\mu}=\begin{pmatrix}1 \\ \vec{e}_r \end{pmatrix},$$
with
$$\vec{e}_r=\begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix}.$$
Note that $\epsilon(t,r)$ is a Lorentz scalar quantity, namely the energy density $T^{00}$ in the rest frame of the source.

To get the function $\epsilon(t,r)$ we have to fulfill local energy-conservation, i.e.,
$$\partial_{\mu} T^{\mu 0}=\partial_{\mu} [\epsilon u^{\mu}]=\dot{\epsilon} - \vec{\nabla} (\epsilon \vec{e}_r)=0.$$
Using the formula for the divergence in spherical coordinates, you get
$$\dot{\epsilon}=\frac{1}{r^2} \partial_r (r^2 \epsilon).$$
The general solution obviously is
$$\epsilon(t,r)=\frac{f(t-r)}{r^2}.$$
Now we have to express $T^{\mu \nu}$ in the frame, where the source moves with velocity $\beta$ in $z$ direction. This is most easily achieved by writing everything in covariant form, but this is very easy since
$$T^{\prime \mu \nu}=\epsilon(x) u^{\prime \mu} u^{\prime \mu}.$$
The energy density in the moving frame thus is
$$\epsilon'(x')=\epsilon(x) (u^{\prime 0})^2,$$
but
$$u^{\prime 0}=\gamma (1+\beta \cos \vartheta) = \frac{1+\beta \cos \vartheta}{\sqrt{1-\beta^2}}$$
and thus indeed
$$\epsilon'=\frac{(1+\beta \cos \vartheta)^2}{1-\beta^2} \epsilon,$$
but now you must consider that $\cos \vartheta$ is the polar angle in the restframe of the source. But you want this in terms of the polar angle in the frame, where the source is moving.

To this end we first write everything in covariant form in the restframe. There the four-velocity of the source is $V=(1,0,0,0)$ and the covariant expression for the spatial distance from the origin in the restframe of the source thus is
$$R=r=\sqrt{(V\cdot x)^2-x \cdot x}$$
and
$$u=\frac{1}{R} [x+V (R-V\cdot x)].$$
Finally
$$\epsilon=\frac{f(u \cdot x)}{R}.$$
Now we can simply rewrite everything in the frame where the source moves with velocity $\beta$ along the 3-axis:
$$V'=\gamma(1,0,0,\beta), \quad \gamma=\frac{1}{\sqrt{1-\beta^2}}.$$
Then
$$R=\sqrt{(V' \cdot x')^2-x' \cdot x'}=\sqrt{\gamma^2(t'-\beta r' \cos \vartheta'))^2-t^{\prime 2} + r^{\prime 2}}.$$
Further
$$u' \cdot x'=x' \cdot V'-R=\gamma(t'-\beta r' \cos \vartheta'))-R$$
and again
$$T^{\prime \mu \nu}=\frac{1}{R^2} f(u' \cdot x') u^{\prime \mu} u^{\prime \mu}.$$
Thus you finally get
$$\frac{\epsilon'}{\epsilon}=\frac{T^{\prime 00}}{T^{00}}=\frac{(u^{\prime 0})^2}{(u^0)^2}=(u^{\prime 0})^2.$$
Now
$$u^{\prime 0}=\frac{1}{R} [t'(1-\gamma^2) + \gamma R + \gamma^2 \beta r' \cos \vartheta'].$$
BTW you can easily check that in both frames energy and momentum are conserved,
$$\partial_{\mu} T^{\mu \nu}=\partial_{\mu}' T^{\prime \mu \nu}=0.$$

It's also clear that this calculation takes into account all kinematical effects, i.e., Doppler shift of the frequency/wave length and aberration.

 

[kent davidge]

@vanhees71 you are awesome at this , I hope one day I will be like you