Colors and temperature

In my previous few posts I have made several references to stars' colors and their temperature, and the link between the two properties. I used the color index of each star in the database to determine the temperature, and the temperature was used to determine both the radius of the star and the color in which the star would be displayed in the Hertzsprung-Russel diagram. In this post I will elaborate a bit on how the temperature of a body relates to its color.

The starting assumption is that stars are so-called black bodies. In physics, the term "black body" does no imply that an object would actually be black when viewed; rather, it signifies that the object is able to fully absorb light of any wavelength (black bodies should not be confused with black holes, which are an entirely different concept). Black bodies emit electromagnetic radiation depending on their temperature, i.e., they would glow visibly when sufficiently heated. This radiation is what we refer to as black body radiation . Real objects are rarely black bodies in this sense, as they reflect, transmit, absorb or scatter different wavelengths differently. Still, treating an object like a black body is usually good as an initial approximation.

The total power radiated per surface area and solid angle (or the spectral radiance) by a black body of temperature $T$ is given by Planck's law: \begin{align} B_{\lambda}(\lambda, T) &= \frac{2 h c^2}{\lambda^5}\frac{1}{e^{\frac{h c}{\lambda k_B T}} - 1}\,, \end{align} where $h = 6.62601\cdot 10^{-34}\, \mathrm{J\, s}$ is the Planck constant, $c = 2.99792\cdot 10^8\,\mathrm{m/s}$ is the speed of light, and $k_B = 1.38065\cdot 10^{-23}\,\mathrm{J/K}$ is the Boltzmann constant.

The plot below shows the spectral radiance as a function of wavelength for six different temperatures, including the temperature of the Sun, $T = 5778\,\mathrm{K}$. The area under each curve has the approximate color of the light emitted by a black body at that temperature.

The color corresponding to each temperature in the range from $0\,\mathrm{K}$ to $16000\,\mathrm{K}$ is shown here, with the temperature of the Sun marked by the vertical dashed line:

The light emitted by the Sun thus has a whitish color similar to the one found under the curve corresponding to $T=5778\,\mathrm{K}$, in the upper figure, or at the vertical line in the lower figure. When viewed from Earth the apparent color of light from the Sun is modified by scattering in the atmosphere, where predominantly short wavelengths are scattered. The light received directly from the Sun therefore has a somehwat reduced blue component, thus appearing more yellow than the light emitted from the Sun. The scattering of blue light is also the reason why the clear daytime sky appears blue.

The total power radiated per unit area can at a given temperature be found by integrating the radiance over the solid angle and the wavelength, \begin{align} j^* &= \iint B_{\lambda}(\lambda, T) \cos(\theta) \,\mathrm{d}\Omega \, \mathrm{d}\lambda\, = \sigma\cdot T^4\,. \end{align}

This result, $j^* = \sigma\cdot T^4$, is the Stefan-Boltzmann law which I used here for determining the radius of a star from its luminosity and temperature. The law states that the power radiated is proportional to the temperature to the fourth power, i.e., the power emitted increases dramatically when the temperature is increased. In the fist plot above the power per unit area is proportional to the area under the curve, and the growth of this area with increasing temperature is a reflection of the Stefan-Boltzmann law.

Planck's law can also be used to determine the wavelength at which the spectral irradiance is largest, i.e., the most prominent wavelength in the spectrum. To do this, find the maximum of the spectral radiance by setting $\frac{\partial}{\partial \lambda} B_{\lambda}(\lambda, T) = 0$. The derivative with respect to $\lambda$ is \begin{align} \frac{\partial}{\partial \lambda} B_{\lambda}(\lambda, T) &= \frac{2 h c^2}{\lambda^6} \left(\frac{-5}{e^{\frac{h c}{\lambda k_B T}} - 1} + \frac{h c}{\lambda k_B T} \cdot \frac{ e^{\frac{h c}{\lambda k_B T}}}{\left(e^{\frac{h c}{\lambda k_B T}} - 1\right)^2}\right)\,. \end{align}

Putting the paranthesised terms equal to zero and letting $x \equiv \frac{h c}{\lambda k_B T}$ now gives \begin{align} \frac{-5}{e^x - 1} + x \cdot \frac{e^x}{\left(e^x - 1\right)^2} &= 0\,, \end{align} or, multiplying with $(e^x-1)^2$ and simplifying, \begin{align} \left(x - 5\right)e^{x} + 5 &=0\,. \end{align}

The solutions can be found by numerical means to be $x=0$ and $x \approx 4.96511$, with only the latter having physical relevance. $\lambda_{\text{max}}$ can now be obtained from the definition of $x$, giving Wien's displacement law: \begin{align} \lambda_{\text{max}} &= \frac{h c}{4.96511\cdot k_B T} \approx \frac{2.89778\cdot 10^{-3}\,\mathrm{m\, K}}{T}\,. \end{align}

Wien's displacement law gives the location of the peak of each curve in the first plot above. The inversly proportional relationshp between $\lambda_{\text{max}}$ and $T$ can be seen as the peak shifting further to the left with increasing temperature.

In a later post I will discuss how to convert the Planck spectrum to rgb colors, i.e., how to display the approximate color of an object from its temperature.