Integrating over the Radio Luminosity Function

To calculate the magnetic energy available for deposition in the intracluster medium, we must calculate the total energy input into magnetic fields in radio sources. Rather than attempt to estimate the energy content of a large number of radio sources and integrate over the radio source population, it proves much easier to integrate over the much better known radio luminosity function, and to relate the observed luminosity to the source's energy content.

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 tau. These three quantities are related by

L = E / tau. (2)

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

d Phi(L) dt = Phi(L) / tau(L) = L Phi(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 rho_E / dL dt = E dPhi(L) / dt = LPhi(L), (4)

where rho_E 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

rho_E =  double integral L Phi(L) dt dL =  double integral (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 = rho_0 / [(L/L_c)^alpha+(L/L_c)^beta] (6)


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


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

We then get

rho_E =  double integral (1 / ln 10) / [(L/)^alpha+(L/L_c)^beta] dt dL. (9)

In a simple Omega = 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 L_min and up to a maximum redshift z_max,

With the substitution y = L/L_c, this is

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

E_1(y_min) = integral_y_min^infinity d y / (y^alpha+y^beta) (13)


E_2(z_max) = _0^z_max [L_c(z) / L_c(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 z_max makes very little difference to the final result, so I take limiting values which are E_1 = 3.47 and E_2 = 7.87.

Fig. 1. The values of the integrals E_1(y_min) [(a), left] and E_2(z_max) [(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(nu) nu^-n), then

L_bol = 4pi I(nu) dnu = 4pi nu_0I(nu_0)(nu/nu_0)^-n dnu/nu_0. (16)

Integrating over a range nu_min to nu_max gives

L_bol = 4I(nu_0) [(nu_max/nu_0)^(1-n) - (nu_min/nu_0)^(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 nu_max a few times nu_0 and nu_min << nu_0, we get

L_bol approx 4pinu_0I(nu_0)/(1-n). (18)

The bolometric luminosity corresponding to (0) is

L_bol = 4pi10^24.89/(1-n) = 1.75x10^36 W. (19)

The total energy density (equation 15) is then

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

Peter Tribble,