<p>There can be large deviations from a Planck distribution, the <a href="https://en.wikipedia.org/wiki/Balmer_jump" rel="nofollow noreferrer">Balmer jump</a>, for example. On a finer scale, there are the <a href="https://en.wikipedia.org/wiki/Fraunhofer_lines" rel="nofollow noreferrer">Fraunhofer lines</a>.</p>

CC BY-SA 4.0
2022-11-07T13:32:28.620Z
2022-11-07T13:32:28.620Z
John Doty
<p>Radiant intensity depends on both the the (effective) temperature and emitting area of the star. If the spectrum can be represented as a blackbody, then the radiant intensity is proportional to <span class="math-container">$R^2 T^4$</span>, where <span class="math-container">$R$</span> is the radius of the (assumed spherical) star and <span class="math-container">$T$</span> is the temperature of the blackbody.</p>
<p>The <em>shape</em> of the spectrum of the star is the same as the Planck function, which depends only on temperature, but it is the <em>area under the spectrum</em> that is the radiant intensity and that depends on both radius and temperature.</p>
<p>In your example, we can therefore say
<span class="math-container">$R_A^2 T_A^4 = S$</span> and <span class="math-container">$R_B^2T_B^4=4S$</span>. However, the degeneracy can be broken by noting that the peak of the Planck spectrum of A is at about half the wavelength of the peak for B, which tells you what the ratio of <span class="math-container">$T_B/T_A$</span> is.</p>
<p>NB: Stellar spectra are not exactly blackbodies, but they can be assigned an &quot;effective&quot; temperature <span class="math-container">$T_{\rm eff}$</span>, which is the temperature of a blackbody with radius <span class="math-container">$R$</span> that would give the same radiant intensity as observed.</p>

CC BY-SA 4.0
2022-11-06T08:21:49.120Z
2022-11-06T08:21:49.120Z
ProfRob
<p>Figure shows the intensity of the radiant energy emitted from stars A and B over a unit time according to the wavelength. The area between the graph and the horizontal axis is S and 4S, respectively. <a href="https://i.sstatic.net/7Xrsb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7Xrsb.jpg" alt="enter image description here" /></a></p>
<p>Is the distribution of radiation energy emitted per unit time by wavelength only affected by temperature or by star size and temperature?  In other words, does the intensity of each wavelength of the radiant energy per unit time not take the form of a Planck Curve, but a form of multiplying the area of the planck function?</p>

<spectroscopy><stellar-astrophysics><spectral-type><spectrometry><electromagnetic-spectrum>
Does the luminosity of a star have the form of a Planck curve?
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# Does the luminosity of a star have the form of a Planck curve?

(Asked by: username_0 on 2022-11-06)

Figure shows the intensity of the radiant energy emitted from stars A and B over a unit time according to the wavelength. The area between the graph and the horizontal axis is S and 4S, respectively. [![enter image description here](https://i.sstatic.net/7Xrsb.jpg)](https://i.sstatic.net/7Xrsb.jpg)

Is the distribution of radiation energy emitted per unit time by wavelength only affected by temperature or by star size and temperature? In other words, does the intensity of each wavelength of the radiant energy per unit time not take the form of a Planck Curve, but a form of multiplying the area of the planck function?

**Tags:** spectroscopy, stellar-astrophysics, spectral-type, spectrometry, electromagnetic-spectrum

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## 2 answers

### Answer by username_2 on 2022-11-06. Score: 10 | Accepted answer

Radiant intensity depends on both the the (effective) temperature and emitting area of the star. If the spectrum can be represented as a blackbody, then the radiant intensity is proportional to $R^2 T^4$, where $R$ is the radius of the (assumed spherical) star and $T$ is the temperature of the blackbody.

The *shape* of the spectrum of the star is the same as the Planck function, which depends only on temperature, but it is the *area under the spectrum* that is the radiant intensity and that depends on both radius and temperature.

In your example, we can therefore say
$R\_A^2 T\_A^4 = S$ and $R\_B^2T\_B^4=4S$. However, the degeneracy can be broken by noting that the peak of the Planck spectrum of A is at about half the wavelength of the peak for B, which tells you what the ratio of $T\_B/T\_A$ is.

NB: Stellar spectra are not exactly blackbodies, but they can be assigned an "effective" temperature $T\_{\rm eff}$, which is the temperature of a blackbody with radius $R$ that would give the same radiant intensity as observed.


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### Answer by username_1 on 2022-11-07. Score: 1

There can be large deviations from a Planck distribution, the [Balmer jump](https://en.wikipedia.org/wiki/Balmer_jump), for example. On a finer scale, there are the [Fraunhofer lines](https://en.wikipedia.org/wiki/Fraunhofer_lines).


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