<p>As I understand it, spectral flux is defined as</p>
<p><span class="math-container">$ F_{\nu} = \int_\Omega I_{\nu}(\theta, \phi)\cos{\theta} d\Omega $</span></p>
<p>where <span class="math-container">$d\Omega$</span> is the solid angle element over which the integral is performed. Here <span class="math-container">$I_{\nu}$</span> is the specific intensity. The <span class="math-container">$\nu$</span> subscript denotes frequency dependence.</p>
<p>According to A. Choudhuri &quot;Astrophysics for Physicists&quot; p. 25, blackbody sources are isotropic emitters therefore we can drop the angular dependence in the spectral intensity and <span class="math-container">$I_{\nu}(\theta, \phi)$</span> becomes your Planck formula <span class="math-container">$B_{\nu}(T)$</span>.</p>
<p>Hence</p>
<p><span class="math-container">$ F_{\nu} = \pi B_{\nu}(T)$</span></p>

CC BY-SA 4.0
2020-07-01T18:55:13.380Z
would that be the same if I used wavelength dependent? So that both F and B are dependent on the wavelength
Melvin
CC BY-SA 4.0
2020-07-03T13:35:04.590Z
@Melvin, yes, the flux law should be the same, if we just assume the specific intensity I was wavelength dependent, but we can't simply substitute the relationship (nu = c / lambda) in the Planck formula, as it has a different form depending on whether the dependence is frequency or wavelength. Explained here I think https://en.wikipedia.org/wiki/Planck%27s_law#Different_forms
sunra
CC BY-SA 4.0
2020-07-01T11:10:18.703Z
2020-07-01T15:47:51.067Z
2020-07-01T15:47:51.067Z
sunra
<p>I'm currently working on a fun project in my free time where I'm trying to calculate the temperature, among other things, of stars based on their spectra. Since I have essentially no prior experience in astronomy, except for quite a strong interest, this may sound like quite a dumb question, but I would appreciate an answer. The data provided from SDSS (Sloan Digital Sky Survey) that I'm using plots the spectral flux density, F<sub>λ</sub>, over wavelength, λ. Now my problem is, I'm trying to do a curve fit in Python using Planck's law, (I'm aware that it is also possible to use the proportions between red and blue light, but I first want to see how close I can get using this method) but Planck's law (as below) gives spectral radiance. This means that while the data I'm using is in different units than the numbers I get from Planck's law. Therefore, (as you can see below) they are on greatly different scales. So my question is, how do I fix this? How can I use Planck's law to be able to fit the data? Is there any relationship between units that I have not been able to find? I want to clarify that I have tried my best to find the answer on my own, using Wikipedia and other sources, but my attempts have, obviously, not been rather successful.</p>
<p><a href="https://i.sstatic.net/ogmQE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ogmQE.png" alt="enter image description here" /></a></p>
<p>The data that I'm using (The y-axis is 10^-17 erg/s/cm^2/Å):</p>
<p><a href="https://i.sstatic.net/HZaPe.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HZaPe.png" alt="enter image description here" /></a></p>
<p>The graphs I have got using Planck's law:</p>
<p><a href="https://i.sstatic.net/QswzD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QswzD.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2020-06-29T23:20:51.243Z
As far as I can tell, you have not correctly calculated $B_{\lambda}$. Units appear to be a possible issue.
ProfRob
CC BY-SA 4.0
2020-06-29T23:21:43.253Z
@RobJeffries why do you say that? :)
Melvin
CC BY-SA 4.0
2020-06-29T23:23:11.863Z
Because a 6000 K blackbody doesn't peak in the ultraviolet. Could it be that the x-axis is in nm, but you have not realised the x-axis of the data is in Angstroms?
ProfRob
CC BY-SA 4.0
2020-06-29T23:24:34.870Z
@RobJeffries oh yeah. I’ll have a look at that tomorrow, thanks. My issue with the spectral radiance Vs. spectral flux density still remains though :(
Melvin
CC BY-SA 4.0
2020-06-29T23:28:33.127Z
The y-axis scaling is irrelevant here. $\pi B_{\lambda}$ is the specific flux in W m$^{-2}$ per wavelength interval at the surface of the star. The observed flux is of course measured at the Earth. The normalisation is a free parameter that depends on the size of the star and how far away it is.
ProfRob
CC BY-SA 4.0
2020-06-29T23:31:09.640Z
@RobJeffries I don’t quite understand since I’m not familiar with all the terms :/ Would you mind clarifying? :)
Melvin
CC BY-SA 4.0
2020-06-29T23:32:33.673Z
The difference between the two quantities, for a star, is just a factor of pi, i.e. $F_\lambda = \pi B_\lambda$ (which comes from integrating the emitted light from the star’s surface over all solid angles). Since you’ll need an arbitrary scaling factor anyway to get the quantities to agree (to account for the surface area of the star, and for the star’s distance from Earth), this factor of pi can just get folded into your scaling.
Eric Jensen
CC BY-SA 4.0
2020-06-29T23:33:59.430Z
@ELNJ oh ok, thanks
Melvin
CC BY-SA 4.0
2020-06-30T14:20:46.047Z
related - not a solution, but an interesting bit of "news" about stars and spectral radiance curves  https://aasnova.org/2018/10/31/perfect-blackbodies-in-the-sky/
Carl Witthoft
CC BY-SA 4.0
2020-07-02T06:24:17.080Z
It really would help if the axes of the figures had a name and units. Anyhow, as was said, the energy scaling is arbitrary and can be chosen to fit
planetmaker
<star><spectra><units>
How to use spectral flux density data while using Planck's law giving spectral radiance?
Melvin
# How to use spectral flux density data while using Planck's law giving spectral radiance?

(Asked by: username_0 on 2020-06-29)

I'm currently working on a fun project in my free time where I'm trying to calculate the temperature, among other things, of stars based on their spectra. Since I have essentially no prior experience in astronomy, except for quite a strong interest, this may sound like quite a dumb question, but I would appreciate an answer. The data provided from SDSS (Sloan Digital Sky Survey) that I'm using plots the spectral flux density, Fλ, over wavelength, λ. Now my problem is, I'm trying to do a curve fit in Python using Planck's law, (I'm aware that it is also possible to use the proportions between red and blue light, but I first want to see how close I can get using this method) but Planck's law (as below) gives spectral radiance. This means that while the data I'm using is in different units than the numbers I get from Planck's law. Therefore, (as you can see below) they are on greatly different scales. So my question is, how do I fix this? How can I use Planck's law to be able to fit the data? Is there any relationship between units that I have not been able to find? I want to clarify that I have tried my best to find the answer on my own, using Wikipedia and other sources, but my attempts have, obviously, not been rather successful.

[![enter image description here](https://i.sstatic.net/ogmQE.png)](https://i.sstatic.net/ogmQE.png)

The data that I'm using (The y-axis is 10^-17 erg/s/cm^2/Å):

[![enter image description here](https://i.sstatic.net/HZaPe.png)](https://i.sstatic.net/HZaPe.png)

The graphs I have got using Planck's law:

[![enter image description here](https://i.sstatic.net/QswzD.png)](https://i.sstatic.net/QswzD.png)

**Tags:** star, spectra, units
> **Comments:**
> **username_1**: As far as I can tell, you have not correctly calculated $B_{\lambda}$. Units appear to be a possible issue.  
> **username_0**: @RobJeffries why do you say that? :)  
> **username_1**: Because a 6000 K blackbody doesn't peak in the ultraviolet. Could it be that the x-axis is in nm, but you have not realised the x-axis of the data is in Angstroms?  
> **username_0**: @RobJeffries oh yeah. I’ll have a look at that tomorrow, thanks. My issue with the spectral radiance Vs. spectral flux density still remains though :(  
> **username_1**: The y-axis scaling is irrelevant here. $\pi B_{\lambda}$ is the specific flux in W m$^{-2}$ per wavelength interval at the surface of the star. The observed flux is of course measured at the Earth. The normalisation is a free parameter that depends on the size of the star and how far away it is.  
> **username_0**: @RobJeffries I don’t quite understand since I’m not familiar with all the terms :/ Would you mind clarifying? :)  
> **username_2**: The difference between the two quantities, for a star, is just a factor of pi, i.e. $F_\lambda = \pi B_\lambda$ (which comes from integrating the emitted light from the star’s surface over all solid angles). Since you’ll need an arbitrary scaling factor anyway to get the quantities to agree (to account for the surface area of the star, and for the star’s distance from Earth), this factor of pi can just get folded into your scaling.  
> **username_0**: @ELNJ oh ok, thanks  
> **username_3**: related - not a solution, but an interesting bit of "news" about stars and spectral radiance curves  https://aasnova.org/2018/10/31/perfect-blackbodies-in-the-sky/  
> **username_4**: It really would help if the axes of the figures had a name and units. Anyhow, as was said, the energy scaling is arbitrary and can be chosen to fit  

---

## 1 answers

### Answer by username_5 on 2020-07-01. Score: 1 | Accepted answer

As I understand it, spectral flux is defined as

$ F\_{\nu} = \int\_\Omega I\_{\nu}(\theta, \phi)\cos{\theta} d\Omega $

where $d\Omega$ is the solid angle element over which the integral is performed. Here $I\_{\nu}$ is the specific intensity. The $\nu$ subscript denotes frequency dependence.

According to A. Choudhuri "Astrophysics for Physicists" p. 25, blackbody sources are isotropic emitters therefore we can drop the angular dependence in the spectral intensity and $I\_{\nu}(\theta, \phi)$ becomes your Planck formula $B\_{\nu}(T)$.

Hence

$ F\_{\nu} = \pi B\_{\nu}(T)$

> **Comments:**
> **username_0**: would that be the same if I used wavelength dependent? So that both F and B are dependent on the wavelength  
> **username_5**: @Melvin, yes, the flux law should be the same, if we just assume the specific intensity I was wavelength dependent, but we can't simply substitute the relationship (nu = c / lambda) in the Planck formula, as it has a different form depending on whether the dependence is frequency or wavelength. Explained here I think https://en.wikipedia.org/wiki/Planck%27s_law#Different_forms  

---

