An overview of active galaxies
An overview of active galaxies

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An overview of active galaxies

3.5 Example 1

(a) A gravitationally bound uniform density sphere, of radius r, is composed of a large number of subelements, with total mass M. Use the virial theorem,

The gravitational energy is given bywhere G is Newton's constant of gravitation, and that the kinetic energy is

Here we have used the angle brackets to indicate a mean, typical, or expected value of the quantity enclosed. The right-hand side of the expression for EKE comes from summing up all the individual m2 contributions, giving the total mass multiplied by the typical value of 2. The in the equation indicates the velocity dispersion, i.e. how fast the subelements of the gravitationally bound system are moving with respect to the centre of gravity.

Show that the typical velocity, , of a subelement is related to M according to:

(b) Astronomers observing a Seyfert galaxy obtain some spectra and high-quality images, in which the light from a point source was spread over only a fraction of an arc second. Despite the excellent spatial resolution of the images, the nucleus of the galaxy is unresolved, and hence its size is known to be r < 50 pc. The emission lines have full width at zero intensity of 6000 km s−1. Estimate the mass of the central black hole.

Solution

(a) Substituting the above expressions into Equation 5 we obtain:Making 2 the subject of the equation, therefore

Since measurements of 2 in extragalactic astrophysics are often somewhat imprecise, we can ignore the difference between 3/5 and 1, and hence we obtainas required.

(b) We can apply this approximate version of the virial theorem to the observations of the Seyfert galaxy. From the information given, we need to substitute values of r and to make the estimate. We are told r is the ‘size’ of the nucleus, which you could think of as its diameter, however in estimates of this type astronomers are often vague about distinctions between quantities such as radius and diameter. Such distinctions are unnecessary whenever the calculation includes unavoidable uncertainties much greater than a factor of 2.

Clearly we can use the upper limit on r straight from the information supplied, but we should work in consistent units. Since Peterson adopts the cgs system, we will generally adopt it in this course, and will use this example to illustrate how (just as in SI) systematic use of the cgs system will give you answers in the appropriate cgs unit. The full width of the emission lines gives us an indication of the velocity dispersion, and the value of the full width is 6000 km s−1. This presumably means that some material is moving away from us at 3000 km s−1, while other material is moving towards us at this speed. Therefore a value of = 3000 km s−1 seems an appropriate estimate. (Note: Since is going to be squared to give our mass estimate, we keep the factor of 2 here.)

We now have values for both the unknowns, but we must convert them into cgs units. Recalling that ‘cgs’ stands for ‘centimetre-grams-seconds’, the cgs unit of length is clearly the centimetre (cm); similarly, the cgs unit of velocity is centimetres per second (cm s−1). Therefore we need to convert from parsecs (pc) to centimetres (cm).

1 pc = 3.086 × 1016 m, and 100 cm = 1 m, therefore 1 pc = 3.086 × 1018 cm. Since none of the measurements we are given have more than 1 significant figure, and we are asked only for an estimate, it is probably safe in this instance to use 1 pc = 3 × 1018 cm. In other calculations you may need to use higher precision.

Using this conversion factor, r < 50 pc becomes r < 50 × 3 × 1018 cm, i.e. r < 1.5 × 1019 cm. Similarly we should express in cgs units by converting km s−1 into cm s−1. Since 1 km = 1 × 103 m, and 100 cm = 1 m, 1 km = 1 × 105 cm, = 3000 km s−1 becomes = 3000 × 105 cm s−1, i.e. = 3 × 108 cm s−1.

Substituting in these values, therefore,

The symbol ‘ ’ is used to indicate ‘less than approximately’. The ‘less than’ is because r < 1.5 × 1019 cm, and the ‘approximately’ because we have used estimates. Cancelling units and multiplying out numbers we arrive at the estimate:

     M 2 × 1043g.

This has provided the mass estimate we wanted, but this is not exactly the answer Peterson would have given. The example above uses pure cgs units, but in fact if you return to Peterson, you will see that velocities are generally measured in km s−1, and masses in M. These units are generally used by astronomers, irrespective of whether they are working in the SI or cgs system, because they are the most convenient (and easily visualised) choice. Hence we will convert the mass above from grams into solar masses. Noting that 1M ≈ 2 × 1030 kg ≈ 2 × 1033 g, we obtainwhich reduces toTherefore we conclude that the mass of the central black hole is less than but comparable to the mass of 10 billion stars like the Sun. Hence the label supermassive seems to be justified.

Question 2

What assumption is implied in the use of the virial argument in Example 1? Do you think the assumption is justified?

Answer

The virial theorem is derived from the equation of hydrostatic equilibrium: therefore the assumption behind the use of this equation is that a state of equilibrium under self-gravity holds.

If the Seyfert galaxy was far from equilibrium, it would be unlikely that the emission would last for as long as 108 years. So, while it is clear that some material must be falling under the gravitational influence of the supermassive black hole, it is probably safe to assume a near-equilibrium configuration for our first estimate of the mass of the central black hole.

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