of 1 MHz. By the time that power has reached the observer, the power is spread over the surface of a sphere with area or about , so its flux density is . More generally, for sources at cosmological distances, a k-correction must be made for the spectral index α of the source, and a relativistic correction must be made for the fact that the frequency scale in the emitted rest frame is different from that in the observer's rest frame. So the full expression for radio luminosity, assuming isotropic emission, is where Lν is the luminosity in , Sobs is the observed flux density in , DL is the luminosity distance in metres, z is the redshift, α is the spectral index (in the sense , and in radio astronomy, assuming thermal emission the spectral index is typically equal to 2.) For example, consider a 1 Jy signal from a radio source at a redshift of 1, at a frequency of 1.4 GHz. Ned Wright's cosmology calculator calculates a luminosity distance for a redshift of 1 to be 6701 Mpc = 2×1026 m giving a radio luminosity of . To calculate the total radio power, this luminosity must be integrated over the bandwidth of the emission. A common assumption is to set the bandwidth to the observing frequency, which effectively assumes the power radiated has uniform intensity from zero frequency up to the observing frequency. In the case above, the total power is . This is sometimes expressed in terms of the total (i.e. integrated over all wavelengths) luminosity of the Sun which is , giving a radio power of . Luminosity formulae The Stefan–Boltzmann equation applied to a black body gives the value for luminosity for a black body, an idealized object which is perfectly opaque and non-reflecting: where A is the surface area, T is the temperature (in kelvins) and is the Stefan–Boltzmann constant, with a value of Imagine a point source of light of luminosity that radiates equally in all directions. A hollow sphere centered on the point would have its entire interior surface illuminated. As the radius increases, the surface area will also increase, and the constant luminosity has more surface area to illuminate, leading to a decrease in observed brightness. where is the area of the illuminated surface. is the flux density of the illuminated surface. The surface area of a sphere with radius