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IOAA 2022, Теория — Задача 13. Relativistic Beaming

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International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 50 т.

Условие

Consider an isotropic light source of frequency $f_R$ in a frame which is fixed to the source (i.e. rest frame). In this rest frame, consider a light ray emitted from the source that makes an angle $\theta_R$ with the $X$ -axis. The light source is moving along positive $X$ direction with (relativistic) speed $v$ as measured in the lab frame.

(a) Find an expression for the frequency $f_L$ of this ray in the lab frame, and the cosine of the angle that this ray makes with the $X$ -axis in the lab frame.

Hint: In relativistic mechanics, energy $E$ and momentum $p$ of a particle between rest and lab frame are related in the following way:

\begin{aligned} \frac{E_L}{c} &= \gamma\left(\frac{E_R}{c} + p_{x_R}\frac{v}{c}\right) \\ p_{x_L} &= \gamma\left(p_{x_R} + \frac{E_R v}{c^2}\right) \\ p_{y_L} &= p_{y_R} \\ p_{z_L} &= p_{z_R} \end{aligned}

where:

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

[11 т.]

(b) For the following cases:

i) $\theta_R = 0^{\circ}$

ii) $\theta_R = \cos^{-1}(-v/c)$

iii) $\theta_R = 90^{\circ}$

iv) $\theta_L = 180^{\circ}$

draw direction vectors of the beam in $XY$ plane of the rest frame as well as separately in $X'Y'$ plane of the lab frame. [4 т.]

In accretion disks around black holes, the charged particles are orbiting at relativistic speeds and in their rest frames may be considered as isotropic point sources of light. Consider such a particle $K$ in a circular orbit of radius $r$ and angular speed $\omega$ around a central object located at $O$ (see figure).

A circle centred at O; axis OY horizontal to the right towards P and axis OX vertically down; particle K on the circle in the lower right with an arrow tangent to the circle showing its motion and a horizontal line from K towards a second P on the right.

Let us assume that our lab frame is fixed to an observer located at a point $P$ on the $OY$ axis, which is stationary with respect to $O$ . $OP = R \gg r$ . Let $t_{L0} = t_{R0} = 0$ correspond to the moment when K is seen crossing the OX axis. As $K$ is moving with relativistic speed, the duration $\Delta t_R$ measured by an observer in the rest frame of the source $K$ is related to the duration measured in the lab frame $\Delta t_L$ at $P$ by the expression $\Delta t_L = \gamma\Delta t_R$ .

(c) Derive an expression for $f_L$ as a function of $t_L$ ( $t_L > R/c$ )? [7 т.]

Let us consider a fraction of the light from the source that is emitted in an infinitesimal solid angle $\Delta\Omega_R = -\Delta(\cos\theta_R) \cdot \Delta\phi$ in the direction making an angle $\theta_R$ with respect to the $X$ axis in the rest frame, as it is shown on the figure below.

A hemisphere with the source (yellow dot) at the centre of its base; axes X up, Y and Z in the base plane; a thin ring of angular width Δθ at polar angle θ from X and a narrow lune of width Δφ meeting it in a small shaded element; the azimuth φ marked along the base.

(d) Show that, as measured in the lab frame

\Delta\Omega_L = \frac{\Delta\Omega_R}{\gamma^2\left(1 + \frac{v}{c}\cos\theta_R\right)^2}

[10 т.]

(e) If the intrinsic luminosity of the light source is $L$ , what is the energy flux $F_L$ observed by the observer at point $P$ at the moment $t_L$ ( $t_L > R/c$ )?

Hint: In the rest frame of the source, you may assume $N_R$ number of photons are directed within the solid angle $\Delta\Omega_R$ during the time interval $\Delta t_R$ . [15 т.]

(f) Charged particles in the relativistic beam shot from the supermassive black hole at the centre of the galaxy M87 have speeds up to $0.95c$ . What would be the maximum and minimum amplification factor for the energy flux for a relativistic beam from M87? [3 т.]

Отговори

Решение

(a) Energy of a photon:

E = hf

while the momentum:

\vec{p} = \frac{hf}{c}\vec{n} \qquad \mathbf{2.0}

where $\vec{n}$ is an unit vector in the direction of the motion of the photon. Substituting this into energy-momentum transformation law we have:

\begin{aligned} \frac{hf_L}{c} &= \gamma\left(\frac{hf_R}{c} + p_{x_R}\frac{v}{c}\right) \\ p_{x_L} &= \gamma\left(p_{x_R} + \frac{hf_R v}{c^2}\right) \\ p_{y_L} &= p_{y_R} \\ p_{z_L} &= p_{z_R} && \mathbf{2.0} \end{aligned}

in our case:

\begin{array}{lll} p_{x_L} = |p_L|\cos\theta_L, & p_{y_L} = |p_L|\sin\theta_L, & p_{z_L} = 0 \\ p_{x_R} = |p_R|\cos\theta_R, & p_{y_R} = |p_R|\sin\theta_R, & p_{z_R} = 0 \end{array} \qquad \mathbf{3.0}

where

|p| = \frac{hf}{c}

Thus,

\begin{aligned} f_L &= \gamma\left(f_R + \frac{v}{h}|p_R|\cos\theta_R\right) \\ &= \gamma\left(f_R + \frac{v}{h}\frac{hf_R}{c}\cos\theta_R\right) \\ \therefore f_L &= \gamma f_R\left(1 + \frac{v}{c}\cos\theta_R\right) && \mathbf{2.0} \\ cos(\theta_L) &= \frac{p_{x_L}}{|p_L|} = \frac{cp_{x_L}}{hf_L} \\ &= \frac{c\gamma\left(p_{x_R} + \frac{hf_R v}{c^2}\right)}{h\gamma f_R\left(1 + \frac{v}{c}\cos\theta_R\right)} \\ &= \frac{c\left(\frac{hf_R}{c}\cos\theta_R + \frac{hf_R v}{c^2}\right)}{hf_R\left(1 + \frac{v}{c}\cos\theta_R\right)} \\ \therefore \cos\theta_L &= \frac{\cos\theta_R + \frac{v}{c}}{1 + \frac{v}{c}\cos\theta_R} && \mathbf{2.0} \end{aligned}

(b) Now just by plugging in the values of cosine we see that in case (i) (orange arrow) $\cos\theta_L = 1$ , So the photon keeps moving in the same direction, in case (ii) (Green arrow) $\cos\theta_L = 0$ , in case (iii) (Yellow arrow) $\cos\theta_L = v/c$ ,(iv) (Grey arrow) $\cos\theta_L = -1$ .

Two sets of axes titled 'Lab Frame' and 'Rest Frame', each with Y up and X to the right, with coloured direction arrows: in the lab frame an orange arrow along +X, two yellow arrows close to +X, green arrows along +Y and −Y and a grey arrow along −X; in the rest frame an orange arrow along +X, yellow arrows along +Y and −Y, and green and grey arrows pointing near −X.

Marking scheme,
0.5 pt for each answer in the rest frame and lab frame.

KP = OP - OK\sin\theta = R - r\sin\theta \qquad \mathbf{1.0}

Circle centred at O with the X axis pointing down and a line from O to the right towards P; the particle K (yellow dot) on the circle at the lower right, with OK making the angle θ with the X axis; from K an orange arrow labelled Y inclined at θ to a horizontal line from K towards a second P.

where $\theta = \omega t$ .

Let $t_L = T$ be the time at which photon reaches $P$ after leaving $K$ at a time $t$ (in lab frame) then:

\begin{aligned} c(T - t) &= KP = R - r\sin\theta \approx R \\ \therefore t &= t_L - \frac{R}{c} && \mathbf{1.0} \end{aligned}

This photon makes angle $\theta = \omega t$ to the direction of its motion in lab frame. Now,

\begin{aligned} \cos\theta_L &= \frac{\cos\theta_R + \frac{v}{c}}{1 + \frac{v}{c}\cos\theta_R} \\ \cos\theta_L + \frac{v}{c}\cos\theta_R\cos\theta_L &= \cos\theta_R + \frac{v}{c} \\ \cos\theta_R(1 - \frac{v}{c}\cos\theta_L) &= \cos\theta_L - \frac{v}{c} \\ \therefore \cos\theta_R &= \frac{\cos\theta_L - \frac{v}{c}}{1 - \frac{v}{c}\cos\theta_L} && \mathbf{2.0} \\ f_L &= \gamma f_R\left(1 + \frac{v}{c}\cos\theta_R\right) \\ &= \gamma f_R\left(1 + \frac{\frac{v}{c}\left(\cos\theta_L - \frac{v}{c}\right)}{1 - \frac{v}{c}\cos\theta_L}\right) \\ &= \gamma f_R\left(\frac{1 - \frac{v}{c}\cos\theta_L + \frac{v}{c}\cos\theta_L - \left(\frac{v}{c}\right)^2}{1 - \frac{v}{c}\cos\theta_L}\right) \\ &= \frac{f_R}{\gamma\left(1 - \frac{v}{c}\cos\theta_L\right)} = \frac{f_R}{\gamma\left(1 - \frac{v}{c}\cos(\omega t)\right)} \\ \therefore f_L &= \frac{f_R}{\gamma\left(1 - \frac{v}{c}\cos\left[\omega\left(t_L - \frac{R}{c}\right)\right]\right)} && \mathbf{3.0} \end{aligned}

(d)

\begin{aligned} \Delta\Omega_R &= -\Delta(\cos\theta_R)\cdot\Delta\phi \\ &= [\cos\theta_R - \cos(\theta_R + \Delta\theta_R)]\cdot\Delta\phi \\ &= [\cos\theta_R - \cos\theta_R\cos(\Delta\theta_R) + \sin\theta_R\sin(\Delta\theta_R)]\cdot\Delta\phi \\ &= [\cos\theta_R - \cos\theta_R\cdot(1) + \sin\theta_R(\Delta\theta_R)]\cdot\Delta\phi \\ \Delta\Omega_R &= \sin\theta_R(\Delta\theta_R)(\Delta\phi) \\ \therefore \Delta\Omega_L &= \sin\theta_L(\Delta\theta_L)(\Delta\phi) && \mathbf{3.0} \\ \sin^2\theta_L &= 1 - \cos^2\theta_L \\ &= 1 - \left(\frac{\cos\theta_R + \frac{v}{c}}{1 + \frac{v}{c}\cos\theta_R}\right)^2 \\ &= \frac{1 + 2\frac{v}{c}\cos\theta_R + \left(\frac{v}{c}\right)^2\cos^2\theta_R - \cos^2\theta_R - 2\frac{v}{c}\cos\theta_R - \left(\frac{v}{c}\right)^2}{\left(1 + \frac{v}{c}\cos\theta_R\right)^2} \\ &= \frac{\left(1 - \left(\frac{v}{c}\right)^2\right)\left(1 - \cos^2\theta_R\right)}{\left(1 + \frac{v}{c}\cos\theta_R\right)^2} = \frac{\sin^2\theta_R}{\gamma^2\left(1 + \frac{v}{c}\cos\theta_R\right)^2} \\ \sin\theta_L &= \frac{\sin\theta_R}{\gamma\left(1 + \frac{v}{c}\cos\theta_R\right)} && \mathbf{3.0} \end{aligned}

\begin{aligned} \text{Also, } \Delta\theta_L \approx \sin(\Delta\theta_L) &= \frac{\sin(\Delta\theta_R)}{\gamma\left(1 + \frac{v}{c}\cos\theta_R\right)} \\ \Delta\theta_L &\approx \frac{\Delta\theta_R}{\gamma\left(1 + \frac{v}{c}\cos\theta_R\right)} && \mathbf{2.0} \\ \frac{\Delta\Omega_L}{\Delta\Omega_R} &= \frac{\sin\theta_L(\Delta\theta_L)(\Delta\phi)}{\sin\theta_R(\Delta\theta_R)(\Delta\phi)} \\ &= \frac{\sin\theta_R(\Delta\theta_R)}{\gamma^2\left(1 + \frac{v}{c}\cos\theta_R\right)^2}\cdot\frac{1}{\sin\theta_R(\Delta\theta_R)} \\ \therefore \Delta\Omega_L &= \frac{\Delta\Omega_R}{\gamma^2\left(1 + \frac{v}{c}\cos\theta_R\right)^2} && \mathbf{2.0} \end{aligned}

(e) Let $N(\theta_L)\Delta\Omega_L\Delta T$ be the number of photons arriving in the vicinity of $P$ within the element of solid angle $\Delta\Omega_L$ and time interval $T, T + \Delta T$ (it is important that $\Delta T \neq \Delta t$ where $\Delta t$ is the emission time of the same photons (in lab frame)). These being photons emitted by $K$ into the solid angle $\Delta\Omega$ in the time interval $t_{0R}, t_{0R} + \Delta t_{0R}$ , so that:

N(\theta_L)\Delta\Omega_L\Delta T = N\Delta\Omega_R\Delta t_{0R} \qquad \mathbf{2.0}

From an earlier part:

\begin{aligned} cT &= ct + R - r\sin\theta_L = ct + R - r\sin(\omega t) \\ \therefore c\Delta T &= c\Delta t - r[\sin(\omega t)\cos(\omega\Delta t) + \sin(\omega\Delta t)\cos(\omega t) - sin(\omega t)] \\ c\Delta T &= c\Delta t - r\omega\cos(\omega t)\Delta t && \mathbf{2.0} \\ \frac{\Delta T}{\Delta t_{0R}} &= \frac{\Delta t}{\Delta t_{0R}}\frac{\Delta T}{\Delta t} = \gamma\left(1 - \frac{v}{c}\cos(\omega t)\right) \\ \frac{\Delta t_{0R}}{\Delta T} &= \frac{1}{\gamma\left(1 - \frac{v}{c}\cos(\omega t)\right)} && \mathbf{3.0} \\ \frac{N(t_L)}{N_R} &= \frac{\Delta\Omega_R\Delta t_{0R}}{\Delta\Omega_L\Delta T} \\ &= \frac{\gamma^2\left(1 + \frac{v}{c}\cos\theta_R\right)^2}{\gamma\left(1 - \frac{v}{c}\cos(\omega t)\right)} \\ &= \frac{\gamma\left(1 + \left(\frac{v}{c}\right)\frac{\cos\theta_L - \frac{v}{c}}{1 - \frac{v}{c}\cos\theta_L}\right)^2}{\left(1 - \frac{v}{c}\cos\theta_L\right)} \\ &= \frac{\gamma}{\gamma^4\left(1 - \frac{v}{c}\cos\theta_L\right)^3} \\ N(t_L) &= \frac{N_R}{\gamma^3\left(1 - \frac{v}{c}\cos(t_L - R/c)\right)^3} && \mathbf{3.0} \end{aligned}

In terms of energy fluxes

\begin{aligned} F_0 &= \frac{hf_R N_R}{R^2} = \frac{L}{4\pi R^2} \\ F(t_L) &= \frac{hf_L N(t_L)}{R^2} && \mathbf{2.0} \\ \frac{F(t_L)}{F_0} &= \frac{hf_L N(t_L)}{hf_R N_R} \\ &= \frac{1}{\gamma^4\left(1 - \frac{v}{c}\cos\left[\omega\left(t - \frac{R}{c}\right)\right]\right)^4} \\ F(t_L) &= \frac{L}{4\pi R^2\gamma^4\left(1 - \frac{v}{c}\cos\left[\omega\left(t - \frac{R}{c}\right)\right]\right)^4} && \mathbf{3.0} \end{aligned}

The radiation is strongly beamed in the direction of motion of the source so that a remote observer in or near the orbital plane of the source sees strongly pulsed radiation.

(f) The amplification (for a given $v$ ) is highest when $\cos\theta_L = 1$ . 1.0

\begin{aligned} F(t_L) &= \frac{F_0}{\left[\gamma\left(1 - \frac{v}{c}\right)\right]^4} \\ A_{\mathrm{max}} &= \frac{1}{[0.0975 \times 0.05]^4} \\ A_{\mathrm{max}} &\approx 1.8 \times 10^9 && \mathbf{1.0} \\ \text{Similarly, } A_{\mathrm{min}} &= \frac{1}{[0.0975 \times 1.95]^4} \\ A_{\mathrm{min}} &\approx 770 && \mathbf{1.0} \end{aligned}

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf