Integrating over the Radio Luminosity Function

I assume that there is energy equipartition between fields and radiating particles in the source (although see Leahy 1989). I further assume that all of the energy E in particles (there is also an energy E in the field) is radiated to give a source luminosity L for a lifetime . These three quantities are related by

L = E / . (2)

If the radio luminosity function (the number of sources per unit volume per unit luminosity interval) is denoted by (L), then the birthrate function is

d (L) dt = (L) / (L) = L (L) / E. (3)

The rate of energy creation in radio sources is the source birthrate multiplied by the energy of each source, or

d^2 / dL dt = E d(L) / dt = L(L), (4)

where is the energy density in magnetic fields. If the universe were not expanding, this would also equal the energy density in radiation, but the radiation energy density is decreased because the radio waves are redshifted away. The rhs of equation (4) is equal (to within a factor ln 10) to the logarithmic radio luminosity function.

To calculate the total energy input into radio sources we must then integrate equation (4) over both luminosity and time, so that

= L (L) dt dL = (1/ln 10}(d N(L) / d log L) dt dL. (5)

The radio luminosity function at 2.7 GHz has been given by Dunlop & Peacock (1990). For the steep spectrum sources, which are those of interest here, they gave a luminosity function of the form

dN(L) / d log L = / [(L/)^+(L/)^] (6)

where

= 10^-6.91 Mpc^-3, = 0.69, = 2.17, (7)

and

(z) = 10^24.89 10^(1.26z-0.26z^2) W Hz^-1 sr^-1. (8)

We then get

= (1 / ln 10) / [(L/)^+(L/)^] dt dL. (9)

In a simple = 1 cosmology, the time t is related to the age of the universe t_0 and redshift z as

t = t_0/(1+z)^3/2, (10)

so that, integrating over all sources more luminous than and up to a maximum redshift ,

With the substitution y = L/, this is

Strictly, is a function of z (because it depends on ). However, very little error in evaluating the integral over y in Eq. (12) is made if the lower limit is replaced by = 0, so I will make this assumption, which allows the integrals over y and z to be separated into two factors,

() = _^ d y / (y^+y^) (13)

and

() = _0^ [(z) / (0)] (3/2)d z/(1+z)^5/2. (14)

I show the values of the integrals in Figs 1 and 2. As can be seen quite clearly, the choice of and makes very little difference to the final result, so I take limiting values which are = 3.47 and = 7.87.

Fig. 1. The values of the integrals () [(a), left] and () [(b), right].

The accumulated energy density is then

= 11.86 t_0 (0). (15)

We aren't quite yet finished, as is in W Hz^-1 sr^-1 at 2.7 GHz, and we require the bolometric luminosity, which involves integrating over frequency and solid angle. If the radio spectrum is a power law of spectral index n (so that I() ^-n), then

= 4 I() d = 4 I()(/)^-n d/. (16)

Integrating over a range _min to _max gives

= 4I() [(_max/)^(1-n) - (_min/)^(1-n)] /(1-n). (17)

For the sources in their sample Dunlop & Peacock find n = 0.85, which has the advantage that the total luminosity is not very sensitive to the cutoff frequencies chosen. With _max a few times and _min << , we get

4I()/(1-n). (18)

The bolometric luminosity corresponding to (0) is

= 410^24.89/(1-n) = 1.75x10^36 W. (19)

The total energy density (equation 15) is then

= 1.07x10^48 J Mpc^-3. (20)

• On to section 5