I'm Having trouble replicating the Black Body model for sun shown on this plot

enter image description here

To my understanding I should only use Planck's formula:

$$I(\lambda) = \frac{2\pi hc^2}{\lambda^5}\dfrac{1}{exp\left(\frac{hc}{k_bT\lambda}\right)}\cdot 10^{-9}$$

Being $\lambda$ in [m] and having the last factor, $10^{-9}$, added for the units presented. What i get is

enter image description here

Can anyone see why?

  • $\begingroup$ That looks pretty close to me. Why do you think there is a problem? $\endgroup$ – John Rennie Mar 6 '16 at 6:34
  • $\begingroup$ The vertical axis scale. The numbers are way off. $\endgroup$ – Jaime Paredes Mar 6 '16 at 6:34
  • $\begingroup$ Ah, OK, I didn't look at the vertical scale. The absolute power at the Earth depends on the Earth-Sun distance. If you were on Pluto the curve would still have the same shape but obviously the vertical scale would be completely different. You need to correct for the Earth-Sun distance in some way that I'll have to think about. $\endgroup$ – John Rennie Mar 6 '16 at 6:37

The equation for the black body spectrum is:

$$ B(\lambda,T) = \frac{2\pi hc^2}{\lambda^5}\frac{1}{exp\left(\frac{hc}{k_bT\lambda}\right)-1} $$

where $B$ is the spectral radiance and has units of watts per unit area of the emitter per unit wavelength. So if we multiply by $10^{-9}$ as you have this gives us the power per square metre of the Sun's surface per nm. When I graph this I get:

Black body curve

which is broadly the same as your curve. As Benjamin says there's an error in your equation though as it happens that error doesn't make too much difference.

To get the total power per nm we have to multiply by the surface area of the Sun, $4\pi r^2$, to get:

$$ P_\text{total} = B\,4\pi r^2 $$

where $r$ is the radius of the Sun. Then divide by the area of the sphere with the radius of the Earth-Sun distance, $4\pi R^2$, to get:

$$ P_\text{Earth} = B\,\frac{r^2}{R^2} $$

The factor $r^2/R^2 \approx 2.17 \times 10^{-5}$, and including this in the calculation gives:

Spectrum at Earth

And this is pretty close to the curve you cite.

  • $\begingroup$ Note that $B_{\lambda}$ usually refers to the specific intensity of blackbody radiation. There's a factor of $\pi$ difference. In fact I'll go further, your formula is not spectral radiance, because that is expressed per unit solid angle. $\endgroup$ – Rob Jeffries Aug 11 '19 at 7:42

The way you have expressed the power density or irradiance is not correct. There is a (-1) in the denominator which is subtracted from the exponential term. Look at the following link to correctly express your result: black body radiation curves


What is plotted on the y-axis of the graph is not specific intensity, but the spectral irradiance at the Earth.

The flux from the surface of the Sun is $\pi B_{\lambda}$, where $B_{\lambda}$ is the specific intensity of a blackbody. This appears to be what you have defined as $I$ (with the correction of -1 to the denominator).

The total luminosity as a function of wavelength, $L=4\pi R^2 I$, where $R$ is the solar radius. Finally, we divide by the surface area of a sphere at the Earth to get the spectral irradiance $f = (R^2/D^2) I$, where $D$ is the Earth-Sun distance.

Thus you curve should be multiplied by $2.16\times 10^{-5}$.

That seems to work (please check you have also used temperature in kelvin).


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