Black hole calculations, the "Event Horizon Telescope"

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SUMMARY

Astronomers are set to image the "shadow" of supermassive black holes using the Event Horizon Telescope, leveraging three key factors: the large event horizons of these black holes, the ability of radio wavelengths to penetrate interstellar dust, and the construction of a telescope array the size of Earth. Calculations reveal the Schwarzschild radii for the Milky Way's black hole (4x106 MSun) as 1.18x109 km and for M87 (3.5x109 MSun) as 1.0325x1010 km. The expected angular diameters can be determined using the formula for angular resolution, which requires careful unit conversion and consideration of the distances involved.

PREREQUISITES
  • Understanding of Schwarzschild radius calculations
  • Familiarity with angular diameter and resolution formulas
  • Knowledge of radio wavelengths and their properties
  • Basic unit conversion skills for astronomical measurements
NEXT STEPS
  • Study the Schwarzschild radius formula and its applications in astrophysics
  • Learn about the principles of radio astronomy and the Event Horizon Telescope
  • Explore angular resolution calculations in detail, including practical examples
  • Investigate the impact of interstellar dust on different wavelengths of light
USEFUL FOR

Astronomy students, astrophysicists, and researchers interested in black hole imaging and radio astronomy techniques will benefit from this discussion.

Ellie Snyder
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Homework Statement


Soon astronomers will be imaging the “shadow” of light from the event horizons of black holes. Since black holes are very small, this achievement seems impossible. Nevertheless, it is possible for three reasons:

1. Supermassive black holes have large event horizons.
2. Radio wavelengths of light are not affected by interstellar dust and can explore regions close to the black holes.
3. By building a radio telescope effectively as large as the Earth, astronomers can overcome the relatively poor resolution associated with long wavelengths and achieve high angular resolution.

a. Calculate the Schwarzschild radii (in kilometers) of the supermassive black holes associated with our Milky Way galaxy (4x10^8 M_Sun) and the giant elliptical galaxy M87 (3.5x10^9 M_Sun).

b. The distances of these black holes are 8.4 kpc and 16.4 mega-parsecs, respectively. What are the expected angular diameters of the innermost circular orbits, assuming a non-rotating black hole?

c. Show that radio astronomers can resolve (1.22*(wavelength/D)) these objects at =0.7 mm (c=wavelength*f; f=450 GHz) if they combine separate radio telescopes over the diameter of the Earth (12,742 km).

Homework Equations


a. R_Sch=(2GM)/c^2=(2.95 km/M_Sun)*M
b. Diam.=2arctan(d/2D)
c. Angular resolution=1.22*(wavelength/D)

The Attempt at a Solution


a. Using the eqn with 2.95/solar mass I calculated Schwarzschild Radii of 1.18*10^9 km for our galaxy's black hole, and 1.0325*10^10 km for M87.
b. Would I plug in the calculated radii for the variable d and the given distances for D to calculate the angular diameter?
c. Just convert the 0.7 mm and 12,742 km into similar units, then plug them into the angular resolution eqn?
 
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Ellie Snyder said:

The Attempt at a Solution


a. Using the eqn with 2.95/solar mass I calculated Schwarzschild Radii of 1.18*10^9 km for our galaxy's black hole, and 1.0325*10^10 km for M87.
Our Milky Way BH is only 4x10^6 Msun, not 4x10^8.
b. Would I plug in the calculated radii for the variable d and the given distances for D to calculate the angular diameter?
Yes.
c. Just convert the 0.7 mm and 12,742 km into similar units, then plug them into the angular resolution eqn?
Yes. Make sure an keep track of your units and whether you are working in radians or degrees/minutes/seconds.
 

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