<p>This <span class="math-container">$f_\nu$</span> is giving you the scaling. The units would come from the normalisation factors that you don't write here.</p>
<p>Try to think of the Planck <a href="https://en.wikipedia.org/wiki/Planck%27s_law#The_law" rel="nofollow noreferrer">spectrum</a> for example - the frequency scaling has nothing to do with the units of <em>specific intensity</em>.</p>
<p>Hope this helps!</p>

CC BY-SA 4.0
2024-11-02T01:50:27.587Z
Makes sense. If I normalize $f_\nu^{disk}$ dividing by the integral over the frequency range it ends with ${\rm Hz}^{-1}$ units. Thanks!
Matías Liempi
CC BY-SA 4.0
2024-10-31T19:04:35.267Z
2024-10-31T19:05:01.447Z
2024-10-31T19:05:01.447Z
MN3
<p>I have estimated the bolometric luminosity of an accreting BH (from a model) as <span class="math-container">$$L_{bol}= \eta \dot{M}_{BH}c^2,$$</span>
where <span class="math-container">$\eta$</span> is the radiative eficciency, <span class="math-container">$\dot{M}_{BH}$</span> the accretion rate of the BH and <span class="math-container">$c$</span> the speed of light. This is in units of <span class="math-container">$L_\odot$</span>.</p>
<p>I want to build a model to compute the luminosity in a given filter, so the logic step is to obtain <span class="math-container">$L_{\nu}$</span>. My initial guess is to asume that the emission comes from the accretion disk and the jet (with efficiency <span class="math-container">$\eta_{jet}$</span>) as  <span class="math-container">$L_{\nu}=L_{\nu}^{disk}+L_{\nu}^{jet}$</span>. My question starts here, if the accretion disk is modeled as a thin disk, I can approximate <span class="math-container">$$L_{\nu}^{disk}=\frac{\eta}{\eta+\eta_{jet}}L_{bol}f_{\nu}^{disk},$$</span>
where <span class="math-container">$f_{\nu}^{disk}$</span> is ormalized spectral profile of the disk component. Assuming a simple model <span class="math-container">$f_{\nu}^{disk}\propto \nu^{1/3}$</span>. My question is here, what are the units of last equation? would it be <span class="math-container">$[L_\odot\, {\rm Hz^{1/3}}]$</span>?.
Because then I want to compute the luminosity in a given filter as
<span class="math-container">$$L_{filter}= \frac{\int L_{\nu} R_{\nu}(\nu)d\nu}{\int R_{\nu}(\nu)d\nu}$$</span>, where <span class="math-container">$R_\nu (\nu)$</span> is the response function of the filter, and the units should be in units of  <span class="math-container">$4.4659\times 10^{13}~[{\rm W/Hz}]$</span> which is the zero point of the AB magnitude system. It is trivial to convert <span class="math-container">$L_\odot$</span> to <span class="math-container">$W$</span>, but my question is related to the <span class="math-container">${\rm Hz}$</span> units, which I really want to understand.
Many thanks!</p>

<luminosity><modeling><accretion-discs><smbh>
Bolometric luminosity in solarluminosity units to Luminosity per Hz modelling
Matías Liempi
# Bolometric luminosity in solarluminosity units to Luminosity per Hz modelling

(Asked by: username_0 on 2024-10-27)

I have estimated the bolometric luminosity of an accreting BH (from a model) as $$L\_{bol}= \eta \dot{M}\_{BH}c^2,$$
where $\eta$ is the radiative eficciency, $\dot{M}\_{BH}$ the accretion rate of the BH and $c$ the speed of light. This is in units of $L\_\odot$.

I want to build a model to compute the luminosity in a given filter, so the logic step is to obtain $L\_{\nu}$. My initial guess is to asume that the emission comes from the accretion disk and the jet (with efficiency $\eta\_{jet}$) as $L\_{\nu}=L\_{\nu}^{disk}+L\_{\nu}^{jet}$. My question starts here, if the accretion disk is modeled as a thin disk, I can approximate $$L\_{\nu}^{disk}=\frac{\eta}{\eta+\eta\_{jet}}L\_{bol}f\_{\nu}^{disk},$$
where $f\_{\nu}^{disk}$ is ormalized spectral profile of the disk component. Assuming a simple model $f\_{\nu}^{disk}\propto \nu^{1/3}$. My question is here, what are the units of last equation? would it be $[L\_\odot\, {\rm Hz^{1/3}}]$?.
Because then I want to compute the luminosity in a given filter as
$$L\_{filter}= \frac{\int L\_{\nu} R\_{\nu}(\nu)d\nu}{\int R\_{\nu}(\nu)d\nu}$$, where $R\_\nu (\nu)$ is the response function of the filter, and the units should be in units of $4.4659\times 10^{13}~[{\rm W/Hz}]$ which is the zero point of the AB magnitude system. It is trivial to convert $L\_\odot$ to $W$, but my question is related to the ${\rm Hz}$ units, which I really want to understand.
Many thanks!

**Tags:** luminosity, modeling, accretion-discs, smbh


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## 1 answers

### Answer by username_1 on 2024-10-31. Score: 0 | Accepted answer

This $f\_\nu$ is giving you the scaling. The units would come from the normalisation factors that you don't write here.

Try to think of the Planck [spectrum](https://en.wikipedia.org/wiki/Planck%27s_law#The_law) for example - the frequency scaling has nothing to do with the units of *specific intensity*.

Hope this helps!

> **Comments:**
> **username_0**: Makes sense. If I normalize $f_\nu^{disk}$ dividing by the integral over the frequency range it ends with ${\rm Hz}^{-1}$ units. Thanks!  

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