1
$\begingroup$

I have some radio observation data in jansky/beam from NVSS at 1.4GHZ, and I want to convert these values into luminosity values (erg/sec).

I have the distance data and I made a formula to convert it:

$$L = \frac{4\pi d^2 F}{A_{\text{eff}} \lambda^2}$$

Is it correct? I ask because my luminosity calculations look like they are wrong. But i think that the denominator was wrong. So I am stuck to convert Jasnky/beam to erg/s.

How to proceed?

$\endgroup$

1 Answer 1

2
$\begingroup$

The luminosity of an object in general is

$$L=4\pi\cdot d^2\cdot F$$

where $d$ is the distance and $F$ the observed flux (power per unit area).

Now Jansky is a frequency specific flux defined as

$$1 Jansky = 10^{-26} W/m^2/Hz = 10^{-23} erg/cm^2/sec/Hz$$

So from this you get

$$L[erg/sec/Hz] = 4\pi\cdot d^2[cm^2]\cdot 10^{-23}F[Jansky]$$

or if you prefer the distance $d$ in parsec

$$L[erg/sec/Hz] = 4\pi\cdot d^2[pc^2]\cdot 9.5\cdot 10^{13}F[Jansky]$$

Note that in this case you can set $Jansky/beam = Jansky$ because the object must be fully contained in the field of view of the antenna beam for being able to determine its luminosity. Be aware though that this is merely a frequency specific luminosity as given by the frequency and frequency bandwidth you are observing on. For the total luminosity you would have to integrate over all frequencies.

Edit:

Since it seems from your comment that you are relating this to extragalactic objects with possibly substantial redshifts, you have to modify the above formula to account for the redshift. In this case the distance in the flux/luminosity formula is not the usual distance anymore but the 'luminosity distance'

$$d_L = d\cdot (1+z)$$

with $z$ the redshift factor

$$z=\frac{\nu_0}{\nu}-1$$

where $\nu_0$ is the frequency when the light was emitted and $\nu$ when it was observed

So the luminosity/flux formula above is then

$$L_{\nu_0} [erg/sec/Hz] = 4\pi\cdot d^2[pc^2]\cdot (1+z)^2\cdot 9.5\cdot 10^{13}F_{\nu} [Jansky]$$

so without the redshift factor you would be underestimating the luminosity in this case as energy is effectively reduced by the redshift.

Note that the frequencies for the emitted luminosity and the observed flux are now not identical anymore. Whilst on the left hand side you have $\nu_0$, you have $\nu = \nu_0/(1+z)$ on the right hand side.

Of course, if the distance values you have are already 'luminosity distances' rather than geometric distances (which may well be the case), then you don't need the correction for the redshift anymore and the earlier formula (without the (1+z) factor) applies. You still have to bear in mind though that your frequency will have been redshifted on its way to the observer i.e. the original frequency will have been higher.

$\endgroup$
4
  • $\begingroup$ well. I convert my distance data from megaparsecs to meters. So i took the formula as J=10^-26W/m^2/Hz. Then i put J value in luminosity formula and multiplied with the Hz value to get Luminosity as erg/sec. I guess it's correct right? $\endgroup$
    – Ege Tunç
    May 30, 2023 at 13:47
  • $\begingroup$ @EgeTunç In meters the factor in front of F in the formula above would be $10^{-19}$. I am not sure though what you mean by 'multiplied with the Hz' value. The luminosity on the left hand side of the formula is frequency specific as the flux on the right hand side is frequency specific if its unit is Jansky. It seems you are approaching this the wrong way around:: you should first be clear what exactly you understand under 'luminosity' and then try to connect this to the observed flux data. $\endgroup$
    – Thomas
    May 30, 2023 at 17:44
  • $\begingroup$ So You say I misunderstood the part of multiplying the value by the hertz in which the observation was made. As I understand, i can calculate luminosity as second formula that you wrote for a specific hertz. $\endgroup$
    – Ege Tunç
    May 31, 2023 at 10:10
  • $\begingroup$ @EgeTunç Please see my extended answer. $\endgroup$
    – Thomas
    May 31, 2023 at 21:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .