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Above is the model I fit my data too in Xspec. Xspec is a piece of software where you can model spectra from observations with their set models. It handles things like absorption, reddening, etc. to produce spectral energy distributions. This is a blackbody spectrum with normalization proportional to the surface area.

kT is measured in keV.

norm = K = $R_{km}^2/D_{10}^2$, where $R_{km}$ is the source radius in km and $D_{10}$ is the distance to the source in units of 10 kpc.

I understand that in this form it is the energy flux. But I can't understand where this $1.0344\times10^{-3}$ comes from.

I know that $$B_{\nu}=\frac{2h\nu^{3}}{c^{2}}\frac{1}{\exp(\frac{h\nu}{kT})-1}.$$

I'm modelling radiation from a star for example, and therefore:

$$F=\int I\cos\theta \,\mathrm d\Omega = B_{\nu} \int_{0}^{2\pi}\,\mathrm d\phi\int_{0}^{\theta_{c}}\sin\theta \cos\theta \,\mathrm d\theta,$$

such that $\theta_{c}=\sin^{-1}R/r$, where $R$ is the radius of the star and $r$ is the distance to from the observer to the star.

This gives $$F=\pi B_{\nu}\left(\frac{R}{r}\right)^{2}$$

Therefore, my specific flux for a blackbody is:

$$F_{\nu}=\frac{2\pi h\nu^{3}}{c^{2}}\frac{1}{\exp(\frac{h\nu}{kT})-1}\frac{R^{2}}{r},$$ where $R$ is the radius of the star and $r$ is the distance to from the observer to the star.

Is my calculation of the specific flux wrong? Or am I just not seeing an obvious unit conversion somewhere in the first $A(E)$ equation.

Also, I think the $\mathrm dE$ is included in $A(E)$ here, so try to ignore that. I'm mostly confused where the $1.0344\times10^{-3}$ comes from.

  • 1
    $\begingroup$ What makes you think it might be wrong? $\endgroup$
    – JMLCarter
    Mar 30, 2017 at 15:36
  • $\begingroup$ What is the origin of the fitting equation in your first image? I would have expected $A(E)$ to have an $E^3$ in the numerator, since $E=h\nu$ and your $F_\nu \propto A(E)/dE$ correctly has a $\nu^3$ in its numerator. $\endgroup$
    – rob
    Mar 31, 2017 at 1:32
  • $\begingroup$ Why has a power of $r$ been lost on the last line? $\endgroup$
    – ProfRob
    Jan 19, 2019 at 10:54

3 Answers 3


I recently ran into the same situation while comparing the xspec models bbody and bbodyrad. I'm guessing that you're taking some of these formulae from Rybicki and Lightman, since you're using their notations. I will refer to these equations with RL#__, for Rybicki and Lightman copyright 2004 version.

Just under RL#1.38, the text states: $I_{\nu{}}=B_{\nu{}}$, since the Planck function $B_{\nu{}}$ is the thermal radiation in an optically thick medium. Back in $\S{}\S{}$1.3, the properties of $I_{\nu{}}$ are discussed, and we see that the units are erg$~$s$^{-1}$$~$cm$^{-2}$$~$ster$^{-1}$$~$Hz$^{-1}$, and that to get to flux we must integrate over solid angles, which you did and R&L do in RL#1.13:$~F=\pi{}B\left(\frac{R}{D}\right)^{2}$.

But we should be careful, as this is still dependent on $\nu{}$! If we want to get the units right, the formula should really look like: $\mathrm dF_{\nu{}}=\pi{}B_{\nu{}}\left(\frac{R}{D}\right)^{2}\,\mathrm d\nu{}$ with units on both sides erg$~$s$^{-1}$$~$cm$^{-2}$. Now we're talking.

First convert the xspec model to our $\nu{}$-dependent formulation.

\begin{align} A(E)&=K\times{}1.0344\times{}10^{-3}\frac{E^{2}\,\mathrm dE}{e^{E/kT}-1}\\ &= \left(\frac{R_{km}}{D_{10}}\right)^{2}\times{}1.0344\times{}10^{-3}\frac{E^{2}\,\mathrm dE}{e^{E/kT}-1} \end{align}

$$A(\nu{})=\left(\frac{R_{km}}{D_{10}}\right)^{2}\times{}1.0344\times{}10^{-3}\frac{(h\nu{})^{2}h\,\mathrm d\nu{}}{e^{h\nu{}/kT}-1}$$ where we sub. $E=h\nu{}$

In cgs, the conversion to radius in cm and distance in cm is $\frac{R_{km}}{D_{10}} = \frac{3.086\times{}10^{22}}{10^{5}}\times{}\frac{R}{D}$. Now

\begin{align} A(\nu{})&=\left(\frac{R_{km}}{D_{10}}\right)^{2}\times{}1.0344\times{}10^{-3}h^{3}\frac{\nu{}^{2}\,\mathrm d\nu{}}{e^{h\nu{}/kT}-1}\\ &=2.9\times{}10^{-47}\left(\frac{R}{D}\right)^{2}\times{}\frac{\nu{}^{2}\,\mathrm d\nu{}}{e^{h\nu{}/kT}-1} \end{align}

Comparing this to the R&L version:

$$\mathrm dF_{\nu{}}=\left(\frac{2\pi{}h}{c^{2}}\right)\left(\frac{R}{D}\right)^{2}\frac{\nu{}^{3}\,\mathrm d\nu{}}{e^{h\nu{}/kT}-1}= 4.6\times{}10^{-47}\left(\frac{R}{D}\right)^{2}\frac{\nu{}^{3}\,\mathrm d\nu{}}{e^{h\nu{}/kT}-1}$$

We get a similar coefficient, however the qualitative form is off by a power of $\nu{}$.


I think the XSpec gives you A(E) which returns photon flux (photons/cm^2/s), not the energy flux. If you manually want to calculate flux using any of the components, you have to add another "E" to the function before integration. Therefore integrating A(E)EdE should give you flux in units of ergs/cm^2/s. Hope it makes sense. You can test this on a simple power law spectrum and calculate the unabsorbed flux manually and the results should match. Hope this helps.

  • $\begingroup$ The question is where the factor of $1.0344\times10^{-3}$ comes from. Can you explain that? $\endgroup$ Jun 26, 2018 at 2:21

As you mentioned, we have $$F_{\nu}d\nu = \frac{2\pi h\nu^3}{c^2} \frac{1}{exp(h\nu/kT)-1} (\frac{R}{D})^2 d\nu$$ Here $F_{\nu}$ is the energy flux, in unit of erg/cm$^2$/s/Hz. Then this can be converted to $$F_{E}dE = \frac{2\pi E^2}{h^3 c^2} \frac{1}{exp(E/kT)-1} (\frac{R}{D})^2 dE$$ since $E = h\nu$ and $F_{E}$ is photon flux, of which the unit is photons/cm$^2$/s/keV.

The norm K in Xspec is in unit of $(\frac{R_{km}}{D_{10 kpc}})^2$, hence, $$(R/D)^2 = K (\frac{R_{km}}{D_{10 kpc}})^2 = \frac{K}{9.5214\times10^{34}}$$ The constants $h$ and $c$ are $h =4.1356677\times10^{-18}$keV s and $c=2.998\times10^{10}$ cm, respectively.

Okay, now we can calculate this factor. $$F_{E}dE = \frac{2\pi}{9.5214\times10^{34}h^3 c^2} \frac{KE^2dE}{exp(E/kT)-1} \\ = 1.038\times10^{-3} \frac{KE^2dE}{exp(E/kT)-1} $$

Well, it's close to that of $1.0344\times10^{-3}$ .


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