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IOAA 2021, Теория — Задача 4. ALMA - Calculating photons (10 points).

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Всички материали от IOAA 2021

Theory — Q4-1, English (Official) — ALMA - Calculating photons · 10 т.

Standard Table of Constants — values used in this problem

Speed of light in vacuum: $c=2.998\times10^8\ \mathrm{m\,s^{-1}}$ . Planck constant: $h=6.626\times10^{-34}\ \mathrm{J\,s}$ . Boltzmann constant: $k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$ .

Standard Table of Constants — values used in this problem

1 jansky: $1\ Jy=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$ .

Standard Table of Constants — values used in this problem

Rayleigh-Jeans law is given by $B_\nu=\frac{2k_BT}{c^2}\nu^2$ which is the power emitted per unit emitting area, per steradian, per unit frequency.

Условие

ALMA is a radio observatory with a revolutionary design. It consists of 66 high-precision antennas, operating in the wavelength range from $0.32\ \mathrm{mm}$ to $8.60\ \mathrm{mm}$ . The principal array has fifty antennas of $12\ \mathrm{m}$ diameter each that can work together as a single telescope in the so-called interferometric mode. There is also another array of four $12\ \mathrm{m}$ antennas, and twelve smaller antennas of $7\ \mathrm{m}$ diameter each.

Imagine that a single $12\ \mathrm{m}$ antenna is being calibrated, pointing to a source with a known incident flux of $1 \times 10^{-20}\ \mathrm{W/m^2}$

4.1 Assuming that all the flux arrives at the shortest wavelength of ALMA sensitivity, determine the average number of photons that would reach the detector every second. [2 т.]

4.2 Compare it to the average number of photons that would have reached the detector, if all the flux arrived at the longest wavelength of operation. [2 т.]

4.3 What is the angular resolution (in arcsec) of a single $12\ \mathrm{m}$ antenna, operating at $74.9\ \mathrm{GHz}$ ? [2 т.]

4.4 Imagine the principal array operating at $74.9\ \mathrm{GHz}$ in the interferometric mode. Assuming for simplicity that the spatial resolution is solely given by the longest baseline (largest distance between any pair of antennas), which turns to be $D_{\max} = 16\ \mathrm{km}$ , what would be the angular resolution (in arcsec) in this case? Treat this case as a single slit aperture instead of a circular one. [2 т.]

4.5 For a radio antenna, the term SEFD refers to 'System Equivalent Flux Density', which is a characteristic energy flux density of the antenna, depending on its temperature and size. We also note that for energy estimation of radio photons, Rayleigh-Jeans approximation is valid. Assuming a system temperature of $691\ K$ , what would be the SEFD of the full ALMA observatory in Jansky if all the 66 antennas could work together? [2 т.]

Отговори

Решение

4.1 To get the number of photons per second we have to multiply the Flux by the area of the dish, to know the incoming energy per second, and divide this by the energy of a photon.

For $\lambda_1=0.32\ mm=3.2\times10^{-4}m$ :

Energy of a photon: $E=\frac{hc}{\lambda}=6.2076\times10^{-22}Jules$ [1 point]

Area of the disk: $A=\pi R^2=113\ m^2$

Number of photons per second:

n_1=\frac{flux\times A}{E}\approx1820\ photon/s

[1 point]

4.2 Same calculation, just changing the energy of the individual photons:

For $\lambda_2=8.6\ mm=8.6\times10^{-3}m$

Energy of a photon: $E=\frac{hc}{\lambda}=2.31\times10^{-23}Jules$ [1 point]

Number of photons per second:

n_2=\frac{flux\times A}{E}\approx48900\ photon/s

[1 point]

4.3 Spatial resolution of a single telescope is given by:

\theta=1.22\frac{\lambda}{D}

where D represents the diameter of the dish. This value will be given in radians, so it must be converted to arcsec afterwards.

For a frequency of 74.9 GHz, the corresponding wavelength is:

\lambda=\frac{c}{f}=4\ mm

[1 point]

And the spatial resolution:

\theta=1.22\frac{4mm}{12\ m}\approx83.9\ arcsec

[1 point]

4.4 For an array the correct expression is:

\theta=\frac{\lambda}{B}\qquad\text{or}\qquad\theta=\frac{\lambda}{2B}

[1 point]

being B the longest baseline in the array.

So in this case:

\theta=\frac{4\ mm}{16\ km}\approx0.0516\ arcsec\qquad\text{or}\qquad0.0258\ arcsec

[1 point]

4.5 The SEFD is a characteristic flux of a system, found by dividing the characteristic energy associated to the so-called temperature of the antenna by its effective area:

SEFD=\frac{2kT_{sys}}{A_e}

[1 point]

As no additional information is given about the effective area, the actual physical area of the array should be used:

A=54\left(\pi\times6^2\right)+12\left(\pi\times3.5^2\right)\approx6569\ m^2

[0.5 point]

Substituting the Boltzmann constant, the given temperature of the antenna, and converting the answer to Jansky we get:

SEFD\approx290.5\ Jy

[0.5 point]

NOTES:

Question	We indicate the answer must be:	Tolerance
4.a	Approximated to the nearest integer	+/- 10 photons
4.b	Approximated to the nearest integer	+/- 10 photons
4.c	2 digit of precision	[83.0 , 85.0]
4.d	2 digits of precision	0.05 exact
4.e	2 digit of precision	[289.0 , 291.0]

Official marking-note discrepancy

The worked solution gives both 0.0516 arcsec and 0.0258 arcsec for 4.4, while its tolerance table states “0.05 exact”. Both source statements are retained.

Оригинал в Архива: TQ-4-Q.pdf · официални решения: TQ4.pdf · допълнителни данни: TQ-4-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf