What is radio luminosity?
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by
raynorris
scientist
Jean Tate asked me how you calculate radio luminosity. Here goes.
First, there are two useful pages http://en.wikipedia.org/wiki/Luminosity explains what luminosity is, and http://www.astro.soton.ac.uk/~td/flux_convert.html gives you a handy calculatorWe measure observed flux densities (i.e. received signal strength) of radio sources in Jansky, and 1 Jy =10^{-26} Wm^{-2)Hz^{-1} where A^{b} means A to the power of b. So that's (10 to the power of -26) Watts per square metre per Hertz. I've included he brackets for clarity but we often omit them.
So suppose you had a 10W transmitter at a distance of 1 million metres, radiating over a bandwidth of 1 MHz. By the time that power has reached you, it is spread over the surface of a sphere with area 4piradius^2 or about 10^13 m^2 (where I'm approximating 4*pi=10)
So its flux density is 10 (W) / 10^6 (bandwidth in Hz) /(10^13) Wm^-2Hz^-1 =10^-18 Wm^-2Hz^-1 = 10^8 Jy
Radio luminosity is measured in W/Hz , not W, so that the bandwidth doesn't need to be specified.
Now lets look at a real source. Suppose we receive a 1 Jy signal from a radio source at a redshift of 1, at a frequency of 1.4 GHz
First, you've got to convert that redshift to a distance in metres. The easiest way to do that is to use Ned Wrights cosmology calculator on
http://www.astro.ucla.edu/~wright/CosmoCalc.html
Put in a redshift of 1, click on general, and it tells you the luminosity distance is 6701 Mpc = 6701310^22 m = 2*10^26 mSo in this case, the radio luminosity = (10^-26) * 4 pi(210^26)^2 = 510^27 W/Hz. This means its a very strong double-lobed radio source, since anything above 10^24 W/Hz is an FRII double-lobed source.
Incidentally, you may want to calculate the total radio power. For this, you need to integrate over the bandwidth of the emission.
What bandwidth is the power being radiated over? Actually we generally don't know, and a common assumption is to set the bandwidth to the observing frequency (this is often called setting the fractional bandwidth to 1), in which case you're basically assuming the power radiated is uniform from zero frequency up to the observing frequency. In practice, we know the flux of AGN at lower frequencies often peaks at a few hundred MHz, and then decrease below that, so assuming uniform probably isn't a bad approximation. In the case above, we'd calculate the total power to be 510^27 * 1.410^9 = 710^36 W. You can compare this to the total (i.e. integrated over all wavelengths) luminosity of the Sun which is 3.810^26W, so the radio power of our source can also be written as 2*10^10 LoPosted
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raynorris
scientist
I've now inserted an abbreviated version of the above into the wikipedia article on luminosity on http://en.wikipedia.org/wiki/Luminosity
Posted
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JeanTate
in response to raynorris's comment.
Thank you very much! 😄
If I may, I'll try an example. Suppose the FIRST integrated flux density of a source is 110 mJy, and it is (securely, robustly) associated with a galaxy whose spectroscopic redshift is reliably observed to be 0.055. As it's FIRST, the observed frequency is 1.4 GHz. Assuming H0 is 71 km/s/Mpc, ΩM is 0.23, and ΩΛ is 0.73 (i.e. Ωtotal=1), CosmoCalc gives the luminosity distance as 243 Mpc.
Putting those values into the online calculator, I get 7.8x1023 W/Hz as the source's luminosity density.
Is that right?
Right at the top of calculator's webpage is this: "Note: No allowance is made for K-corrections or redshift-bandwidth effects"
From what I've read, radio lobes often have spectral indices of ~(-)0.7*, so if we assume the FIRST source has a spectral index of 0.7, wouldn't the estimate for L1.4GHz have to be reduced by a factor of (something like) (1+z)(1+0.7)** ?
*the sign depends on how they're defined; physically, less power/unit frequency as frequency increases
**or is it (1+z)(1-0.7)??Posted
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JeanTate
in response to raynorris's comment.
The central part of that section now (i.e. just a minute or so ago) reads (some formatting may be lost):
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 observers rest frame. So the full expression for radio luminosity, assuming isotropic emission, is
Lν = (Sobs * 4 * π * (DL)2 / (1+z)(1+α)
where Lν is the luminosity in W/Hz, Sobs is the observed flux density in Wm-2Hz-1, DL is the luminosity distance in metres, z is the redshift, α is the spectral index (in the sense I ∝ να, and is typically -0.7).
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 = 21026 m giving a radio luminosity of (10-26) * 4 π * (21026)2 / (1+1)(1.7) = 31027 W/Hz.
Checking, I find:
- CosmoCalc gives a DL of 6701 Mpc at z=1 if H0=69.6 (km/s/Mpc), ΩM = 0.286, and Ωvac = 0.714
- 6701 Mpc = 2.07*1026 m
- (10-26) * 4 π * (21026)2 / (1+1)(1.7) = 1.55*1027 W/Hz (not 3)
- to get 3*1027 W/Hz, the exponent of (1+1) needs to be ~0.74; i.e. α =~-0.34
I also plugged the numbers into the radio luminosity calculator (link given in the Wikipedia article, and the OP); I got:
- DL = 6706 Mpc
- Lν = 5.38*1027 W/Hz
- multiplying Lν by 1/(1+z)(1+α) and solving for α for 3, α =~-0.34
What am I missing? 😦
Posted
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raynorris
scientist
Hi Jean
I'm glad you spotted my deliberate mistake - I wondered if you would. 😃
That'll teach me not to do these things on the fly! Yes I had the wrong sign for alpha. I'm correcting the wikipedia version now.
Thanks
RayPosted
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raynorris
scientist
And BTW for anyone else following this thread go to the wikipedia version on https://en.wikipedia.org/wiki/Luminosity#Radio_Luminosity, rather than the version above, in which I now include the full relativistic corrections
Posted