<p>The luminosity of an object in general is</p>
<p><span class="math-container">$$L=4\pi\cdot d^2\cdot F$$</span></p>
<p>where <span class="math-container">$d$</span> is the distance and <span class="math-container">$F$</span> the observed flux (power per unit area).</p>
<p>Now Jansky is a frequency specific flux defined as</p>
<p><span class="math-container">$$1 Jansky = 10^{-26} W/m^2/Hz = 10^{-23} erg/cm^2/sec/Hz$$</span></p>
<p>So from this you get</p>
<p><span class="math-container">$$L[erg/sec/Hz] = 4\pi\cdot d^2[cm^2]\cdot 10^{-23}F[Jansky]$$</span></p>
<p>or if you prefer the distance <span class="math-container">$d$</span> in parsec</p>
<p><span class="math-container">$$L[erg/sec/Hz] = 4\pi\cdot d^2[pc^2]\cdot 9.5\cdot 10^{13}F[Jansky]$$</span></p>
<p>Note that in this case you can set <span class="math-container">$Jansky/beam = Jansky$</span> 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.</p>
<p><strong>Edit:</strong></p>
<p>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'</p>
<p><span class="math-container">$$d_L = d\cdot (1+z)$$</span></p>
<p>with <span class="math-container">$z$</span> the redshift factor</p>
<p><span class="math-container">$$z=\frac{\nu_0}{\nu}-1$$</span></p>
<p>where <span class="math-container">$\nu_0$</span> is the frequency when the light was emitted and <span class="math-container">$\nu$</span> when it was observed</p>
<p>So the luminosity/flux formula above is then</p>
<p><span class="math-container">$$L_{\nu_0} [erg/sec/Hz] = 4\pi\cdot d^2[pc^2]\cdot (1+z)^2\cdot 9.5\cdot 10^{13}F_{\nu} [Jansky]$$</span></p>
<p>so without the redshift factor you would be underestimating the luminosity in this case as energy is effectively reduced by the redshift.</p>
<p>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 <span class="math-container">$\nu_0$</span>, you have <span class="math-container">$\nu = \nu_0/(1+z)$</span> on the right hand side.</p>
<p>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.</p>

CC BY-SA 4.0
2023-05-30T13:47:02.733Z
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?
Ege Tunç
CC BY-SA 4.0
2023-05-30T17:44:37.080Z
@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.
Thomas
CC BY-SA 4.0
2023-05-31T10:10:20.697Z
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.
Ege Tunç
CC BY-SA 4.0
2023-05-31T21:50:41.630Z
@EgeTunç Please see my extended answer.
Thomas
CC BY-SA 4.0
2023-05-29T16:04:26.297Z
2023-06-01T07:28:47.907Z
2023-06-01T07:28:47.907Z
Thomas
<p>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).</p>
<p>I have the distance data and I made a formula to convert it:</p>
<p><span class="math-container">$$L = \frac{4\pi d^2 F}{A_{\text{eff}} \lambda^2}$$</span></p>
<p>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.</p>
<p>How to proceed?</p>

<radio-astronomy><luminosity><radio><flux>
Help converting Jansky/beam to erg s^-1
Ege Tunç
# Help converting Jansky/beam to erg s^-1

(Asked by: username_0 on 2023-05-29)

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?

**Tags:** radio-astronomy, luminosity, radio, flux

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## 1 answers

### Answer by username_1 on 2023-05-29. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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.  
> **username_0**: 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.  
> **username_1**: @EgeTunç Please see my extended answer.  

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