2015IOAA理论短问题第8题-亮温度

英文题目

A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.

中文翻译

一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度

英文解答

part1

• (10%)

During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. 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. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)

• (10%)

   Flux density observed at 1660 MHz is

$$ \begin{aligned} \mathrm{S}_{1660 \mathrm{MHz}} &=0.35 \mathrm{kJy} \\ &=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\ &=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1} \end{aligned} $$

part2

• (30%)

   The solid angle subtended by this source of radiation is

$$ \begin{aligned} \Omega=\pi(r / R)^{2} &=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\ &=2.75 \times 10^{-29} \mathrm{sr} \end{aligned} $$

part3

• (20%)

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

$$ \mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega $$

• (20%)

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

$$ \mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2} $$

• (10%)

   where Tb is the brightness temperature. Then we have:

$$ \begin{aligned} \mathrm{T}_{\mathrm{b}} &=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\ \mathrm{T}_{\mathrm{b}} &=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\ &=1.5 \times 10^{26} \mathrm{~K} \end{aligned} $$