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## Imaging the Galaxy’s supermassive blackhole – part 1

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

## The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity $v_{esc}$ when that velocity is equal to the speed of light $c$. That is

$v_{esc} = c = \sqrt{ \frac{2GM }{ R } }$

where $M$ is the mass of the blackhole, $G$ is the universal gravitational constant, and $R$ is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius $R_{s}$). So, we can write

$R_{s} = \frac{ 2GM }{ c^{2} }$

Putting in a mass of 4.5 million solar masses, we find

$R_{s} = 1.33 \times 10^{10} \text{ metres}$

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
$\theta = 6.17 \times 10^{-9} \text{ degrees}$ (remember to double $R_{s}$ to get the diameter of the event horizon). This is the same as

$\boxed{ \theta = 22.22 \text{ micro arc seconds} }$

Converting this to radians, we get

$\theta ( \text{in radians}) = 1.08 \times 10^{-10}$

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

$\theta ( \text{in radians}) \approx 4 \times 10^{-10}$

## The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

$\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }$

where $D$ is the diameter of the telescope and $\lambda$ is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of $50 \times 10^{-4} \text{ radians }$ at various wavelengths.

For visible light, assuming $\lambda = 550 \text{ nanometres}$

$D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }$

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

$D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }$

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

$D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }$

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!