APhO 2024, Теория — Задача 3. The First Discovered Quasar: Unveiling the Mysteries of the Astrophysical Source 3C 273

Part A: Moon's Apparent Motion Against the Background Stars (1,8 т.)

The Moon takes 27.3 days to complete 1 orbit of the Earth with respect to the stars, a period known as a sidereal month. Through a telescope, the moon’s motion is readily apparent, but to the eye, it requires careful observation over several hours to notice the Moon’s changing position. Remembering that 3600 arcseconds = 60 arcminutes = 1 degree then, through how many degrees, arcminutes, and arcseconds does the Moon move against the background of stars in

A.1 one hour? One minute? One second? [1,3 т.]

A.2 How long does it take the Moon to move a distance equal to its own diameter on the sky? (Note: the angular diameter of the moon as viewed from Earth is 30 arcminutes) [0,5 т.]

Part B: Using Lunar Occultations to Precisely Determine Radio Source Positions: The Case of 3C 273 (1,8 т.)

In the late 1950s and early 1960s, radio position measurements with this precision were not available. While occultations had been used previously, Hazard was specifically interested in their ability to determine radio positions and structure with arcsecond precision. Such measurements were made in the optical reference frame thereby allowing reliable optical identifications. Hazard had noted that the strong Class II radio source 3C 273 would be hidden by the Moon (also known as occultation) several times during 1962 and 1963.

The 1962 August 5 occultation was undertaken at 410 and 136 MHz. Both disappearance and reappearance were observed and the disappearance record revealed the presence of two components, A and B, in the source. The disappearance on the right and the reappearance records on the left, at 410 MHz are plotted in Fig. 1.

B.1 What is causing the oscillation in the observed intensity?

A. The diffraction pattern of the telescope

B. Instabilities in the mount of the telescope

C. The features of the surface of the Moon

D. Earth's atmospheric conditions [0,6 т.]

B.2 Why there is no extra bump for the B component in the left plot for the reappearance of the source? How does it help to understand the structure of the source?

A. The two components have been orbiting around the Moon.

B. Our line-of-sight has changed with respect to the two components.

C. Both components are aligned with the limb of the Moon when they reappear.

D. The time scales of the physics behind the luminosity of the two components were long enough to make one of them fade. [0,6 т.]

B.3 Based on your answer in A.1 to A.4, and the data in Fig. 1, estimate the approximate apparent (projected) angular separation between the two components of 3C 273. [0,6 т.]

Part C: The Breakthrough Discovery of 3C 273's True Nature (1,8 т.)

In 1962, the year of these observations, Maarten Schmidt was working on the programme of optical identification and spectroscopy of the optical objects identified with radio sources. Whiteoak mentions as an afterthought that the ‘current Caltech thinking’ is that the potential 3C 273 identification is with a star and a strange jet. Given that no other bright star had been proposed as a radio source identification, he assumed that the bright magnitude 13 ‘star’ (magnitude accounts as a way to measure how bright a star is; the brighter the star, the smaller the number) was merely a confusing foreground very bright star. To obtain a spectrum of the faint jet, which he saw as by far the most likely identification, it was inevitable that the bright confusing star some arcseconds away would spill over into any spectrum of the jet he would obtain. To offset this, Maarten Schmidt had decided to first obtain a spectrum of this bright star. On the night of December 29, he managed to obtain a spectrum of the bright ‘star’ which showed some faint emission lines (Fig. 2), but with no obvious explanation in terms of any expected stellar lines. Only when Schmidt decided to compare the strange spectrum with the Balmer lines of hydrogen, things became clear:

C.1 Compare the wavelength ratios of the lines shown in Fig. 2. Based on these ratios, what is the redshift $z = \frac{\lambda - \lambda_0}{\lambda_0}$ of this source? [0,6 т.]

C.2 Consider that this redshift is arising due to relativistic gravitational effects from the mass of the object. In this case, we can estimate the gravitational redshift as $z = \frac{1}{\sqrt{1 - \frac{2GM}{c^2 r}}} - 1$. Prove that, whether you put 3C 273 at any distance (e.g, edge of the Milky Way, $r \sim$ 100 kpc $\sim 3 \times 10^{21}$ m; edge of the Solar System, $r \sim$ 100 AU $\sim 1.5 \times 10^{13}$ m), the mass of 3C 273 would be so large it would disrupt the entire galaxy / solar system. [0,6 т.]

C.3 Hubble's law states that the farther the galaxies, the faster they are moving away from Earth. The Hubble's constant is the ratio between the speed and the distance of those galaxies, and equal to H=75 km/s/Mpc. With this idea in mind, we can consider instead that the calculated redshift is due to the cosmological expansion, and we can approximate that $z \sim v/c$. In that case, what is the distance of this object? Compare this with the size of the Milky Way. [0,6 т.]

Part D: The Intrinsic Luminosity of the Radio Source 3C 273 (1,8 т.)

The flux of the source 3C 273 has been measure across the radio spectrum and has been found to follow the equation $F_\nu \sim 25000\nu^{(-0.3)}\ \mathrm{Jy}$, where $\nu$ is the observed frequency in Hertz and a Jansky is defined as $1\ \mathrm{Jy} = 10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$. Of course, since the source is radiating over a sphere of radius the distance from us, the intrinsic luminosity will be given by $L_\nu = 4\pi d^2 F_\nu$. With this,

D.1 What is the luminosity of the source per unit frequency?

(help: $1\ \mathrm{pc} = 3.09 \times 10^{16}\ \mathrm{m}$) [0,6 т.]

D.2 What is the total luminosity in the radio band (ie., from $10^7$ Hz to $10^{11}$ Hz)? [0,6 т.]

D.3 How does the luminosity of 3C 273 compare with the luminosity of the Sun ( $L_{sun} = 3.82 \times 10^{26}$ W) and the Milky Way ($L_{MW} = 1.5 \times 10^{10} L_{sun}$) in the same frequency range? [0,6 т.]

Part E: The Power Source of 3C 273 (1,4 т.)

We have seen that the brightness of 3C 273 is too large for a single star, even for an entire galaxy. Therefore, there should be another mechanism, other than stars, producing all this power.

E.1 Check that the annihilation of matter with antimatter (e.g., $\mathrm{e}^+ + \mathrm{e}^- \to 2\gamma$) would not explain this, since it would produce very energetic X-rays rays rather than optical and radio emission. [0,7 т.]

E.2 Something that was immediately proposed was the accretion of matter into a supermassive black hole. Let’s check that potential energy can give all this energy: Consider that the black hole accretes one solar mass ($2 \times 10^{30}$ kg) towards its Schwarzschild radius ($R_s = \frac{2GM}{c^2}$) every year. What is the power that can be obtained for the accretion? Would this be enough to explain the observations discussed in D1 to D3? [0,7 т.]

Part F: Modern Observations and the Nature of 3C 273's Components (1,4 т.)

Modern images from various telescopes (see e.g., Fig. 3) have found that the two components A and B that were measured with the lunar occultation actually refer to a compact core, which hosts the black hole, and a jet that extends the distance that you calculated before. This jet is thought to be produced via acceleration of the accreted particles via the strong black hole magnetic field, in a similar way particles from the Solar wind hit the magnetic field of Earth to produce the auroras near the poles, but at much larger scale.

The energy density required for the production of such jet can be written in terms of the magnetic fields as the sum of the particle energy density $U_e \sim B^{(-3/2)}$ and the magnetic energy density $U_B \sim B^2$.

F.1 Show that the minimum energy density is given by $\frac{U_e}{U_B} = \frac{4}{3}$ [0,7 т.]

F.2 If we assume that the total power of the 3C 273 jet is given by half of the energy released by 3C 273 in one second calculated in D.1 and D.2, and the jet has a volume of $10^{45}\ \mathrm{m^3}$, estimate the magnetic field of the jet.

(hint: use that the magnetic permeability is $4\pi \times 10^{-7}\ \mathrm{T\,m\,A^{-1}}$) [0,7 т.]