2017年IOAA理论第4题-21厘米HI星系巡天

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

$$z=\frac{1.42-1.32}{1.42}=0.0704$$

[1 Mark]

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

$$z\approx\frac{f_0-f}{f}=\frac{1.42-1.32}{1.32}=0.0758$$

and the student should be awarded full mark for this part.

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

for low z, non-relativistic redshift $$d=\frac{v_r}{H_0}=\frac{cz}{H_0}$$

[1 Mark]

The HI spectral-line flux density for a galaxy with HI luminosity L and line-width $$Δf$$ at distance $$d$$ is

$$\frac{L}{4πd^2}*\frac{1}{Δf(Wm^{-2}Hz^{-1})}$$

[2 Mark]

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

$$S=\frac{LH_0^2}{4πΔfc^2z^2}≥S_{lim}=0.5*10^{26}$$

[1 Mark]

Therefore,

$$z≤\frac{H_0}{c}√{\frac{L}{4πΔfS_{lim}}}$$

[1 Mark]

Correctly substitute numerical values,

$$z≤\frac{67.8(km^{-1}Mpc^{-1})}{2.99*10^5(km s^{-1}) 3.08*10^{22} (m Mpc^{-1})}√{\frac{10^{28}W}{4π*10^6(Hz)*0.5*10^{-29}(Wm^{-2}Hz^{-1})}}$$

[1 Mark]

$$z≤0.0929$$

[1 Mark]

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)

[1 Mark]

The last part is marked only if student calculate both instrumental factors.

If students use more precise cosmological effects, they will be marked accordingly.

中文翻译

情况1 : 如果我们只考虑HI 谱线可以被这台射电望远镜观测到的谱线频率，

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

[1 分]

最低的观测频率, 1.32 GHz, 可以被用于计算 HI的最大红移量, 为

$$z=\frac{1.42-1.32}{1.42}=0.0704$$

[1 分]

另一个可行的解: 以上是常用于射电天文学的宇宙学红移公式。但是，如果宇宙学红移被用于可见光天文学，学生可以使用

$$z\approx\frac{f_0-f}{f}=\frac{1.42-1.32}{1.32}=0.0758$$

然后学生应该得到这部分的全部分数。

情况2: 用已有信息去计算因为望远镜的探测极限而能观测到的最遥远的典型HI星系的红移。

在较低红移的情况下, 非相对论红移 $$d=\frac{v_r}{H_0}=\frac{cz}{H_0}$$

[1 分]

一个光度为 L ，谱线宽度为 $$Δf$$ ，距离为 $$d$$ 的星系的HI自旋翻转谱线流量密度为，

$$\frac{L}{4πd^2}*\frac{1}{Δf(Wm^{-2}Hz^{-1})}$$

[2 分]

代入到探测极限，计算正确，并且特别注意几个不同的单位，

$$S=\frac{LH_0^2}{4πΔfc^2z^2}≥S_{lim}=0.5*10^{26}$$

[1 分]

因此,

$$z≤\frac{H_0}{c}√{\frac{L}{4πΔfS_{lim}}}$$

[1 分]

代入数值计算,

$$z≤\frac{67.8(km^{-1}Mpc^{-1})}{2.99*10^5(km s^{-1}) 3.08*10^{22} (m Mpc^{-1})}√{\frac{10^{28}W}{4π*10^6(Hz)*0.5*10^{-29}(Wm^{-2}Hz^{-1})}}$$

[1 分]

$$z≤0.0929$$

[1 分]

结论: 对于一个典型HI 星系，它可以被观测到的最大红移量由射电望远镜的最低频率决定，即为 $$z_{max}$$=0.0704 (或者是另一个可行的解给出的 0.0758 )

[1 分]

结论部分仅当学生考虑到两个影响因素时才会评分。

如果学生使用了更精细的宇宙学效应，将会根据他的解答来给分。