2017年IOAA理论第4题-21厘米HI星系巡天

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IOAA赛场上通常是使用英文回答问题，所以建议各位用英文回答本题

英文原题

(T4) 21-cm HI galaxy survey [10 marks]

A radio telescope is equipped with a receiver which can observe in a frequency range from 1.32 to 1.52 GHz. Its detection limit is 0.5 mJy per beam for a 1-minute integration time. In a galaxy survey, the luminosity of the HI spectral line of a typical target galaxy is 10²⁸ W with a linewidth of 1 MHz. For a large beam, the HI emitting region from a far-away galaxy can be approximated as a point source. The HI spin-flip spectral line has a rest-frame frequency of 1.42 GHz.

What is the highest redshift, z , of a typical HI galaxy that can be detected by a survey carried out with this radio telescope, using 1-minute integration time? You may assume in your calculation that the redshift is small and the non-relativistic approximation can be used. Note that 1 Jy = 10^-26 W m^-2 Hz^-1. [10]

中文翻译

21厘米HI星系巡天

一台射电望远镜的观测频率为1.32到1.52GHz。它在1分钟的积分时间内的探测极限为每波束0.5mJy。在一次星系巡天中，典型目标星系的HI谱线光度为10²⁸W，谱线宽度为1MHz。对于一个大的波束来说，来自遥远星系的HI辐射区域可以近似为点源。HI自旋翻转谱线具有1.42GHz的静止参考系频率。

使用这台射电望远镜进行星系巡天，可以探测到的典型HI星系的最高红移z是多少？在你的计算中，可以采用低红移情况下的非相对论近似。注意1Jy=10^-26Wm^-2Hz^-1。

（《天文爱好者》上给出的翻译是“星系寻天”，但是通常使用的翻译是“星系巡天”。对翻译有修改，特此标注。）

官方解答

英文原文

Case 1: If we only consider the frequency range where HI spectral line can be detected by the receiver of this radio telescope,

$$z\approx\frac{f_0-f}{f_0}$$

[1 Mark]

The lowest frequency, 1.32 GHz, can be used to observe the highest redshifted HI, at

$$z=\frac{1.42-1.32}{1.42}=0.0704</center><div style="float:right">[1 Mark]<div> '''Alternative solution''': Above is the cosmological redshift formula in frequency conventionally used in radio astronomy. However, if the cosmological redshift in optical astronomy is used, student may use <center>$$z\approx\frac{f_0-f}{f}$$</center> =\frac{1.42-1.32}{1.32}=0.0758 <div style="float:right">and the student should be awarded full mark for this part.</div> '''Case 2''': Use the information given to calculate the redshift limit set by the flux limit due to furthest distance that typical HI galaxy can be detected by the telescope <center>for low z, non-relativistic redshift $$d=\frac{v_r}{H_0}=\frac{cz}{H_0}$$</center><div style="float:right">[1 Mark]</div> The HI spectral-line flux density for a galaxy with HI luminosity L and line-width $$Δf$$ at distance $$d$$ is <center>$$\frac{L}{4πd^2}*\frac{1}{Δf}(Wm^-2Hz^-1}

[2 Mark]

Setting this to the detection limit and solve correctly, and taking special care of converting various units

$$S=\frac{LH_0^2}{4πΔfc^2z^2}≥S_{lim}=0.5*10^26</center><div style="float:right"> [1 Mark]</div> Therefore, <center>$$z≤\frac{H_0}{c}√{\frac{L}{4πΔfS_{lim}}}

[1 Mark]

Correctly substitute numerical values,

$$z≤\frac{67.8(km^-1Mpc^-1)}{2.99*10^5(km s^-1) 3.08*10^22 (m Mpc^-1 )}√{\frac{10^28W}{4π*10^6(Hz)*0.5*10^-29(Wm^-2Hz^-1)}</center><div style="float:right"> [1 Mark]</div> <center>$$z≤0.0929$$</center><div style="float:right">[1 Mark]</div> '''Conclusion''': For typical HI galaxy, the highest redshift is limited by the receiver’s lowest frequency limit and we get $$z_{max}$$=0.0704 (or 0.0758 for alternative solution given above) <div style="float:right">[1 Mark]</div> $$The last part is marked only if student calculate both instrumental factors.$$ $$If students use more precise cosmological effects, they will be marked accordingly.$$