Spectral flux density

The flux from an unresolved (point-like) object may be considered to be a parallel beam of light or a plane wave originating at "infinity". It impinges on the telescope with a given amount of energy deposited per second, per square meter, and per unit frequency interval at frequency v (e.g., in the interval v - 0.5 Hz to v + 0.5 Hz). This is known as the spectral flux density,

but it could be more properly called the spectral energy flux density to distinguish it from a photon flux density (photons m-2 Hz-1). In fact it is often called simply the "flux density". Here we reserve that term for another related use.

The definition serves to standardize the values of spectral flux density reported in the literature from telescopes with different areas and frequency bands (band-widths). The standard quantity is simply the observed power divided by the area of the telescope and by the bandwidth in Hz. Thus observations with different equipment of the same source should yield approximately the same value for S.

The motivation for this normalization is similar to that for reporting an automobile speed as distance per unit time, e.g., 100 km/h, rather than in terms of the actual elapsed time of the journey, 20 km/12 min. The standard reference time makes it much easier to compare the average speeds from journeys of different durations.

The actual energy received by the telescope per second (power P in watts), in a narrow frequency band Av, is the product of the spectral flux density (averaged over the band Av), the effective area Aeff of the telescope, and the bandwidth Av,

For example, a relatively bright celestial radio source might yield a spectral flux density S(v) at the earth of

S(v) = 1.0 x 10-26 Wm-2Hz-1 = 1.0 Jy (jansky) (8.3)

at frequency v = 100 MHz. This particular spectral flux density is known as 1.0 jansky; Carl Jansky was the discoverer of radio radiation from the (MW) Galaxy.

Now, suppose that this signal is detected with a perfect antenna of diameter 16 m (Aeff ~ 200 m2) that is tuned to v = 100 MHz with a narrow bandwidth Av = 104 Hz. This means that the detection system accepts radiation in the frequency band 104 Hz wide at v = 100 MHz, i.e., 99 995 000 Hz to 100 005 000 Hz. Further assume that S(v) is constant, or nearly so, across the band Av so that S(v) ~ S(v )av. The actual power received by the antenna would then be, substituting into (2),

This is a very small amount of energy; it would take 0.5 x 1020 s to accumulate the energy of one joule, enough to light a 1 W bulb for 1 s. This time corresponds to about 1013 years or ~1000 times the age of the universe! The detection of such small (and even smaller) amounts of power requires high-sensitivity detectors. Do not be misled by such tiny power levels; they are associated with tremendous total power outputs from the source, as we demonstrate below.

The calculation above was made under the assumption that the spectral flux density S(v) is constant over the bandwidth of frequencies accepted by the antenna. For a bandwidth narrow compared to the measured frequency, this is often a reasonable assumption. However, it is possible that much of the energy lies in a spectral line, an enhancement of flux in a very narrow, e.g. 10 Hz, frequency band. In this case, the spectral flux density could be close to zero for a large part of the 104 Hz band. A proper calculation of the power received by the antenna for a variable S(v) requires that the product S(v) dv A be integrated (summed) over the frequency band,

P v2

J vi where Aeff is the effective area of the telescope, taking into account inefficiencies that dissipate some of the incident energy, and where the integration is over the bandwidth of detected frequencies. In practical cases, the effective area is a function of the frequency Aeff(v), so it too would go inside the integral.

The spectral flux density is properly a vector S, where S = |S|, because the flux at any point in space must have a direction; it is a "flow". The direction of the flux may be determined by immersing a test surface of fixed size into the flow and then rotating the surface to various orientations. When the flux through the surface reaches its maximum value, the surface normal lies along the flow lines. This is analogous to the vector current flux density J (A m-2) in electromagnetic theory. Do not confuse the quantity S (W m-2 Hz-1) used in this text with that used in electromagnetic theory for the Poynting vector (W/m2).

Flux density

The total power flowing across unit area is called the flux density F (W/m2). It is obtained by integrating the spectral flux density S(v) over the frequency band of interest, p v2

J vi

Again, this quantity is properly a vector since it is a flow that has direction; a surface immersed in the flow can be oriented to give the direction of flow as above. For our purposes, we usually use the scalar quantity F = | F"|. In electromagnetic theory, this vector is the quantity known as the Poynting vector; see (7.13).

The flux F can be written in terms of the average over frequency of the spectral flux density S by multiplying and dividing the right side of (6) by Av = v2 - v1 and recalling the definition of an average, Sav = /S dv/Av. Thus, F = Sav Av.


The luminosity L of a source is, in its usual meaning, the total power output (W) summed over all frequencies. In practice, one usually must specify the band of frequencies (of radiation) that are being measured, e.g., the visual V band, the entire optical band, or the 1-10 keV x-ray band. Normal stars like the sun emit only a tiny fraction of their power in the radio and x ray, and these emissions were long unknown, so these bands were ignored in traditional astronomy. The luminosity over the entire optical band including the spillover into the adjacent infrared and ultraviolet bands is called the bolometric luminosity; it is this which is usually given the symbol L where L = L boi.

If the distance r to a source is known, an estimate of the luminosity in a specified band Av is obtained by multiplying the flux density F(W/m2 in the band Av) by the area of a sphere centered on the source with its surface passing through the earth (antenna) as shown in Fig. 1,

= 4nr 2 Sav Av where the (unconventional) subscript Av reminds one that the luminosity is restricted to the chosen band Av.

An assumption implicitly adopted in (7) is that the emission from the source is isotropic, that is, the energy is radiated equally into all directions. Only in this case does our antenna get its expected share of the total radiation that leads to the factor 4nr2 in (7). The assumption of isotropy is a common one because an antenna on

Photons of fn

Antenna area A

Photons of fn

Antenna area A

Sky Constellations Map
Figure 8.1. A point-like source at distance r from an antenna radiates equally in all directions. The luminosity of the source in a given frequency band is the flux density (W/m2) detected in that band multiplied by 4nr2.

the earth can sample only one emission direction, and isotropy is often the most reasonable guess. Some objects, e.g., pulsars, active galactic nuclei and gamma-ray bursts, emit beams of radiation that are demonstrably non-isotropic; in these cases, (7) is clearly incorrect.

Consider the luminosity of a hypothetical radio source radiating isotropically with a constant spectral flux density at the earth of S = 1.0 Jy over the 50-150 MHz band. If it is at the center of the Galaxy, at a distance of ~25 000 LY (2.4 x 1020 m), its luminosity from 50 to 150 MHz (Av = 108 Hz) would be

This is a lot of watts, equivalent to almost 10 billion trillion 100-watt light bulbs! It is about 1/600 of the sun's luminosity of 4 x 1026 W. The quasar 3C273 at a distance of 2.1 MLY with spectral flux densities ranging from 100 to 50 Jy over the range 100 to 1000 MHz, has a radio luminosity of ~1036 W. Inclusion of optical, x-ray and gamma-ray radiation raises this to more than 1038 W, or ~1012 L Q (solar luminosities). This would be in error, probably an overestimate, if the radiation is beamed.

Over a wide bandwidth, such as a factor of two change in frequency, the spectral flux density S(v) is likely to change substantially; in this case one must integrate over frequency to obtain the luminosity,

L = 1.0 x 10-26 x 108 x 4n x (2.4 x 1020)2 = 7 x 1023 W (8.8)

+ L = 4nrFAv = 4nr2 S(v) dv (W; isotropic emission) (8.9)

J v1

One can often estimate roughly the luminosity without integrating by substituting for the integral the product of the bandwidth Av and a typical or average value of the spectral flux density Sav in that band. If the functional form of S(v) is simple, formal integration is quite straightforward.


Some astronomical sources emit occasional isolated bursts of radiation which might last for 10-100 seconds. It is convenient to define a quantity that gives the flux of energy integrated over the duration of the burst. This is called the fluence S (J m-2), or more precisely, the energy fluence, which is defined as the time integral of the flux density, ft2 2 S = & dt (J m-2; fluence) (8.10)

Again this quantity is properly a vector S = \S |. A spectral fluence (J m-2 Hz-1) could also be defined if desired.

Note that the quantities above are all derived from an integrations of S(v) over one or more of the variables: area, frequency, and time.

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  • Monica Lind
    How to change flux into power astronomy?
    3 months ago

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