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

标准答案（英文）

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:

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

• (10%)

   where Tb is the brightness temperature. Then we have:

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

标准答案（中文）

part1

• (10%)

在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）

• (10%)

   在1660Mhz频率处，流量密度为

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

part2

• (30%)

   在观测者处，源所张立体角为

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

part3

• (20%)

   流量密度与总光度有以下关系

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

• (20%)

   在该频率下，总光度可以用瑞利-金斯公式近似得到：

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

• (10%)

   其中Tb为亮温度。于是我们有

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