IOAA 2015, Теория, кратки задачи, английски — Задача 8

1

During 700 µs, the radio wave travels about r = c t = $3\times10^{10}\times700\times10^{-6} = 2.1\times10^{7}$ cm. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. (r = $2.1\times10^{7}$ cm, R = 2.3 kpc)

[10]

Flux density observed at 1660 MHz is

$S_{1660MHz} = 0.35$ kJy

$= 0.35\times10^{3}\times10^{-23}$ ergs s$^{-1}$ cm$^{-2}$ Hz$^{-1}$

$= 3.5\times10^{-21}$ ergs s$^{-1}$ cm$^{-2}$ Hz$^{-1}$

[10]

2

The solid angle subtended by this source of radiation is

$\Omega = \pi\,(r/R)^2 = 3.14\,(2.1\times10^{7}/2.3\times10^{3}\times3.086\times10^{18})^2$

$= 2.75\times10^{-29}$ sr

[30]

3

The flux density is related to the total brightness by the relation:

$S_{1660MHz} = B_{1660MHz}\,\Omega$

[20]

while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:

$B_{1660MHz} = 2kT_b\nu^2/c^2$

[20]

where $T_b$ is the brightness temperature. Then we have:

$T_b = S_{1660MHz}c^2/2k\nu^2\Omega$

$T_b = [3.5\times10^{-21}\times(3\times10^{10})^2]/[2\times(1.38\times10^{-16})\times(1660\times10^{6})^2\times(2.75\times10^{-29})]$

$= 1.5\times10^{26}$ K

[10]