• What determines the colour of an object?
• What is a black-body in the physical sense?
• How can the generation of thermal radiation be explained on an atomic level?
• How are black bodies realized in practice?
• How can the different annealing colours of objects be explained?
• How can the emitted wavelength spectrum be described mathematically?
• What does the Stefan Boltzmann Law state?
• What does Wien’s displacement law state?

## Blackbody

When white light hits an opaque object, a certain part of that light is always absorbed and the rest is reflected. Depending on which wavelengths are absorbed, the reflected light consists of certain wavelengths. These reflected wavelengths determine the perceived color of the object.

A green leaf absorbs almost all wavelengths except the green wavelength range between 500 nm and 550 nm. Thus, the reflected light contains exactly the wavelength range that gives us the green color impression.

If, on the other hand, an object absorbs all incident wavelengths of visible light, then obviously no visible radiation is reflected. The object appears black to us under normal conditions. A black object therefore does not emit any visible radiation. An object that absorbs all incident radiation is therefore also called a blackbody (or black body). A perfect black body, however, only exists in the ideal conception. In reality, every object will always reflect a part of the incident radiation. That’s why even black objects often show slight shading. For this reason, the charcoal shown in the figure below also shows slight shading on the left hand side.

A blackbody is an idealized object that absorbs all incident electromagnetic radiation!

One could now jump to the conclusion that a blackbody does not emit any electromagnetic radiation. From an energetic point of view, however, this conclusion is quickly disproved. Electromagnetic radiation is always connected with radiant energy, depending on the wavelength. If this energy is absorbed by an object, then the absorbed energy cannot simply disappear. The absorbed radiant energy becomes noticeable in a temperature increase of the object (increase of the internal energy)!

As long as a black body is irradiated and absorbs energy, its temperature would have to rise steadily. Experience shows, however, that objects placed in the sun and thus irradiated do not heat up permanently. It does not matter whether it is an ideal or a real black body. As long as the radiation is only partially absorbed, the temperature should rise steadily due to the permanently absorbed radiant energy.

In reality, however, one notices that the temperature of irradiated objects does not rise permanently. At some point a thermal equilibrium will be established. This also applies to black painted objects, which can be regarded as ideal black bodies in very good approximation. Objects that absorb radiant energy must therefore also somehow release energy again. Otherwise the forming of a thermal equilibrium could not be explained.

One could now think that an object placed in the sun is cooled by the surrounding air by convection or by thermal conduction with the ground. These heat transfers will certainly take place and contribute to the thermal equilibrium, but there must be another mechanism. This becomes apparent when one looks at an object in a vacuum. Thermal convection and conduction require particles that transport the thermal energy. In a vacuum, however, there are no particles, so that these heat transfers cannot take place. And yet one will notice that even irradiated bodies in a vacuum sooner or later come into a thermal equilibrium.

An everyday example is the earth. The earth is irradiated by the sun in the vacuum of space and absorbs part of the solar radiation. Obviously, there is no permanent heating of the earth. The earth is in a stable thermal equilibrium, which leads to an average surface temperature of about 15 °C. This thermal equilibrium in a vacuum can only be explained by the fact that the earth itself emits energy through electromagnetic radiation.

This radiation emitted by a body in thermal equilibrium is also referred to as thermal radiation. Thermal radiation occurs with all objects, not only those in vacuum! With very strongly heated objects as for example a glowing bar of metal, this thermal radiation can not only be felt very clearly, but can also be seen by the glowing reddish color!

In incandescent light bulbs, this thermal radiation is technically used to emit visible light. In fact, most of the thermal radiation in this case is emitted in the infrared wavelength range. In this wavelength range we cannot see the radiation, but we can still perceive it. We perceive the infrared radiation through our skin in the form of heat.

The emitted wavelength spectrum depends strongly on the temperature of the body. At low current and therefore low temperature, the filament is slightly reddish. At high current and therefore high temperature, the filament glows intensively yellowish. This spectral distribution of the wavelength as a function of temperature is discussed in more detail in the section on blackbody radiation.

Thermal radiation is the radiation emitted by a body in thermal equilibrium due to its temperature!

Note: Thermal radiation in the narrower sense is often referred to only as invisible radiation in the infrared range. In the broadest sense (and also in this article), the term thermal radiation refers to the entire emitted wavelength range of a body due to its temperature, i.e. the entire energetic radiation!

The generation of the thermal radiation can be explained by the oscillation of the atoms. In principle, every accelerated movement of charged particles leads to the generation of electromagnetic waves, i.e. radiation. The higher the temperature, the stronger and faster the particles vibrate and the more radiation is emitted (higher intensity). Only at absolute zero are there no atomic motions and the body does not emit any thermal radiation.

Thermal radiation comes from oscillations of the atoms!

## Remark on the term blackbody

As already explained, an ideal black body must also emit thermal radiation. Otherwise it could not be explained why a complete absorption of radiant energy leads to a thermal equilibrium at some point! Even if by definition a blackbody absorbs all incident radiation, it still emits radiation. This is not reflected radiation, but radiation which the body emits “from inside” due to the oscillation of the atoms. See also the article Stefan-Boltzmann law.

Note that the definition of a blackbody is simply that all incident radiation is absorbed and not that it cannot emit any radiation itself! It can do this very well, but under ambient conditions this emitted radiation lies in the non-visible wavelength range (infrared radiation). The object therefore initially appears black to the eye, which is why it is called a black-body.

Only if the temperature is increased very strongly and the body starts to glow, radiation in the visible wavelength range is emitted. Even if the body now has a color depending on the temperature (“annealing colors”), this object is still called a blackbody by definition, since all incident radiation is still absorbed. So a blackbody does not necessarily have to be black! An impressive example is the sun. In fact, the sun is an almost perfect black body. Only due to the enormous temperature of 5778 K it appears incandescent in the sky.

The definition of a blackbody is not that it appears black under all circumstances, but that it absorbs all incident radiation! Depending on the temperature, even a supposedly black object emits visible radiation and “glows” in different colors!

If an object is not heated too much, the thermal radiation emitted is in the infrared wavelength range. This infrared radiation is invisible to the human eye, which is why a blackbody actually appears black! With a thermal imaging camera, however, this thermal radiation can be made “visible”. One can then draw conclusions about the temperature from this.

The emitted thermal radiation, however, changes its wavelength range with temperature. At sufficiently high temperatures, the emitted wavelength spectrum shifts into the visible range and even beyond (i.e. in the ultraviolet range). Even a supposedly “black” body then begins to “glow” in color. This glow is not reflected radiation but self-emitted radiation from “inside”!

This is where the blackbody comes into play. In order to examine the emitted radiation spectrum of an object, it is necessary that only the thermal radiation emitted by the body itself is actually measured and not the reflected radiation by other radiation sources. Therefore, the body to be examined must not reflect any radiation but must absorb all incident radiation. To examine the wavelength spectrum of thermal radiation (also known as spectral distribution), one needs a blackbody!

A blackbody is particularly suitable for the investigation of the spectral distribution of thermal radiation, since the radiation does not contain any reflecting parts of interference sources!

The emitted thermal radiation of a black body is called blackbody radiation!

## Realization of blackbodies in practice (cavity radiation)

In practice, a blackbody can be realized relatively easily in a very good approximation. For this purpose only a hole is drilled into a hollow object. If radiation enters the cavity through the hole, it is reflected several times by the walls. With each reflection a certain part of the radiation is always absorbed. After several reflections, the radiation is almost completely absorbed. Consequently, no more radiation exits through the hole (at least not the incident radiation). All incident radiation is “swallowed” by the cavity and the hole therefore appears black. The cavity can therefore be regarded as a blackbody in a very good approximation.

The figure above shows an upside-down flowerpot with a hole and an empty beverage can. Due to the absorbed incident radiation, the interiors appear black through the openings. Even with the beverage can made of highly reflective aluminium, almost no visible radiation gets to the outside! Even reflective materials such as metals can thus become a black hollow body in very good approximation. For this reason, blackbody radiation is often referred to as cavity radiation.

Note that the term black body actually refers only to the cavity or hole and not to the object itself! Because the body itself usually has a color, which is caused by reflections of the incident light. So the outside of the object is by no means a black body!

Even if no visible light is emitted from the cavity at first and therefore appears black, radiation in terms of thermal radiation is emitted due to the oscillation of the atoms. With a thermal imaging camera, this initially invisible blackbody radiation can be made visible in the form of infrared radiation (white areas in the illustration below).

## Spectral distribution of blackbody radiation (Planck spectrum)

With cavity radiation, the emitted wavelength spectrum of different materials can now be examined at different temperatures. Due to the thermal resistance even at relatively high temperatures, it makes sense to use metals. Holes are then drilled into the hollow metal blocks. Through the holes the blackbody radiation can finally pass to the outside and the spectral distribution of the blackbody radiation can be determined with a detector. The temperature can be changed relatively easily by heating the metal blocks and the resulting effects on the spectral distribution can be examined.

The animation below shows the principle of cavity radiation, in which radiation enters through a hole, but no longer exits it due to absorption processes.

The evaluation of the experiment takes place with a diagram in which the so-called spectral intensity is plotted over the wavelength. In simplified terms, such a diagram indicates which wavelength is radiated with which intensity (radiant power per unit area). Usually a double logarithmic division is used in order to be able to cover a large wavelength range at the different temperatures.

In the physical sense, the term intensity refers to the radiant power per unit area (surface power density), i.e. the radiant energy emitted per unit area and per unit time. In this case, the surface of the cavities is to be understood as the surface, since the radiation emanates from this surface and exits through the hole. The intensity plotted in the diagram is related to the wavelength itself, since the intensity can always only be measured within a wavelength range and cannot be determined for an exact wavelength. This intensity per wavelength is therefore also referred to as spectral intensity. In such a diagram, the area under the curve corresponds to the intensity with which the radiation is perceived in the corresponding wavelength range. The area under the entire curve thus corresponds to the total radiated intensity.

The experiment shows that the emitted wavelength spectrum depends only on the temperature and not on the material. The cavity radiation of an aluminium block thus has the same spectral distribution as a steel or brass block.

The wavelength spectrum of blackbody radiation depends only on the temperature and not on the material!

A more detailed examination of the wavelength spectrum shows that the maximum spectral intensity shifts with increasing temperature to ever shorter wavelengths. At relatively low temperatures the maximum lies in the infrared range and the radiation is not visible to our eyes. However, with increasing temperature the spectrum shifts into the visible range. The hole or blackbody now begins to glow due to the visibly emitted radiation.

From approx. 1000 K it is a slightly reddish glow, as known from glowing metal. The figure below shows the glow of a heating element of a baking oven. The temperature is about 800 °C.

At 2000 K, the proportion of the yellow wavelength range in the spectrum has increased. The body tends to radiate yellowish. However, since the emitted intensity is relatively high in the entire visible range, all color receptors in our eyes are “overexposed”. In this case, all receptors are excited almost identically and the yellowish radiation usually appears white to our eye (see also article Color vision).

The white color of the electric arc during welding with temperatures of over 3000 K can also be explained by the overexposure of the color receptors in our eyes. In fact, the emitted spectrum is rather a yellowish wavelength spectrum. This yellowish color can only be perceived as such if the intensity is greatly reduced, e.g. if it is observed indirectly as scattered light on a white sheet of paper.

In addition, the spectral distribution shows that at such high temperatures of over 3000 K more ultraviolet radiation (UV radiation) is emitted. This explains the protective equipment required for welding, which not only protects the eyes but also the arms. Without covered arms one would get a “sunburn” by the UV radiation otherwise!

At even higher temperatures of about 6000 K, almost all visible wavelengths with the same intensity are present in the radiated spectrum. This radiation is therefore a white “color”. This also explains the white solar radiation, since the sun as an almost perfect blackbody has a surface temperature of 5778 K! The sun also radiates UV radiation to a not inconsiderable extent. Fortunately, however, a large part of this UV radiation is absorbed by the earth’s atmosphere.

Significantly higher temperatures than our sun have so-called blue giants. Some of these astronomical objects have 50 times the mass of the sun. The surface temperatures can reach several 10,000 Kelvin. At these temperatures the blue wavelength range is more present in the radiated spectrum than the reddish parts. The light of such blue giants therefore appears bluish, which is the reason why such stars are called blue giants.

## Mathematical description

The emitted wavelength spectrum of a blackbody could not be explained for a long time. Until then, it was always assumed that energy would be distributed continuously. It was only by introducing discrete energy levels that the physicist Max Planck succeeded in describing blackbody radiation mathematically. Although he did not know how to interpret the introduction of discrete energy levels physically at first, he laid the foundation for quantum mechanics.

#### Wavelength form

Planck could derive the following formula for the distribution of the spectral intensity $$I_s$$ as a function of wavelength $$\lambda$$. This formula is also known as Planck’s law.

\begin{align}
\label{planck}
&\boxed{I_s(\lambda) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\lambda k_B T}\right)-1} } ~~~\text{Planck’s Law} \\[5px]
\end{align}

#### Frequency form

Since wavelength $$\lambda$$ and frequency $$f$$ are related by the speed of propagation $$c$$ ($$\lambda = \frac{c}{f}$$), Planck’s law of radiation can also be expressed as a function of frequency. However, the wavelength $$\lambda$$ must not simply be replaced by the expression $$\frac{c}{f}$$. This has to do with the fact that the spectral intensity is a quantity related to the wavelength. Therefore, one also has to convert the wavelength intervals $$\text{d}\lambda$$ into corresponding frequency intervals $$\text{d}f$$!

Only the radiated intensities are really comparable, but not the spectral intensities. The emitted intensity $$\text{d}I(\lambda)$$ in a wavelength range between $$\lambda$$ and $$\lambda + \text{d}\lambda$$ is calculated by the product of the spectral intensity $$\text{d}I(\lambda)$$ and the wavelength interval $$\text{d}\lambda$$ (“area under the graph”):

\begin{align}
\label{c}
&\text{d} I(\lambda) =I(\lambda) \cdot \text{d}\lambda= \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\lambda k_B T}\right)-1} \cdot \text{d}\lambda \\[5px]
\end{align}

The wavelength interval $$\text{d}\lambda$$ must therefore be assigned a corresponding frequency interval $$\text{d}f$$. This is done by deriving the function $$f=\frac{c}{\lambda}$$ with respect to the variable $$\lambda$$:

\begin{align}
\label{a}
&\boxed{f = \frac{c}{\lambda}} ~~~\text{or}~~~\boxed{\color{red}{\lambda=\frac{c}{f}}}\\[5px]
&\frac{\text{d}f}{\text{d}\lambda} = \frac{c}{-\lambda^2} \\[5px]
\label{z}
&\text{d}\lambda = – \frac{\lambda^2}{c} \cdot \text{d}f \\[5px]
\end{align}

The negative sign in equation (\ref{z}) merely expresses that an increase of frequency by $$\text{d}f>0$$ results in a decrease of wavelength by $$\text{d}\lambda<0$$. At this point, however, only the magnitudes of the intervals are relevant, so that the negative sign can be omitted. Taking equation (\ref{a}) into account, one obtains:

\begin{align}
&\text{d}\lambda = \frac{\lambda^2}{c} \cdot \text{d}f = \frac{\left( \color{red}{\frac{c}{f}}\right)^2}{c} \cdot \text{d}f = \frac{c}{f^2} \cdot \text{d}f \\[5px]
\label{b}
&\boxed{\color{blue}{\text{d}\lambda = \frac{c}{f^2} \cdot \text{d}f}} \\[5px]
\end{align}

If now the equations (\ref{a}) and (\ref{b}) are used in equation (\ref{c}), then the spectral intensity $$I_s(f)$$ as a function of frequency is obtained:

\begin{align}
\text{d} I(f) &= \frac{2\pi h c^2}{\color{red}{\left(\frac{c}{f}\right)}^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\color{red}{\frac{c}{f}} k_B T}\right)-1} \cdot \color{blue}{\frac{c}{f^2} \cdot \text{d}f} \\[5px]
&= \frac{2\pi h c^2 f^5}{c^5} \cdot \frac{1}{\exp\left(\dfrac{h f}{ k_B T}\right)-1} \cdot \color{blue}{\frac{c}{f^2} \cdot \text{d}f} \\[5px]
&= \underbrace{\frac{2\pi h f^3}{c^2} \cdot \frac{1}{\exp\left(\dfrac{h f}{ k_B T}\right)-1}}_{I_s(f)} \cdot \text{d}f \\[5px]
\label{freq}
&\boxed{I_s(f) = \frac{2\pi h f^3}{c^2} \cdot \frac{1}{\exp\left(\dfrac{h f}{k_B T}\right)-1} } \\[5px]
\end{align}

For the conversion from the wavelength form to the frequency form of the spectral distribution the following relationship applies:

\begin{align}
&\boxed{I_s(f) =\frac{\lambda^2}{c} \cdot I_s(\lambda)} \\[5px]
\end{align}

Note, that when it comes to spectral intensity, it is no longer possible to simply convert the wavelength to a frequency by the formula $$f=\frac{c}{\lambda}$$. This is because wavelength and frequency behave reciprocally. Both quantities are therefore based on different interval widths! This will also play a role in the section Wien’s displacement law.

### Stefan-Boltzmann law

As already mentioned, the radiated intensity results from the area under the spectral intensity distribution. Planck’s law must therefore be integrated over the entire wavelength range or frequency range. The integration is to be carried out using the frequency form (\ref{freq}):

\begin{align}
&I= \int_{0}^{\infty} I_s(f) ~ \text{d}f\\[5px]
&I= \int_{0}^{\infty} \frac{2\pi h f^3}{c^2} \cdot \frac{1}{\exp\left(\dfrac{h f}{k_B T}\right)-1} ~ \text{d}f\\[5px]
\label{her}
&I= \frac{2\pi h}{c^2} \cdot \int_{0}^{\infty} \frac{f^3}{\exp\left(\dfrac{hf }{k_B T}\right)-1} ~ \text{d}f\\[5px]
\end{align}

This integral can be solved by replacing the argument $$\frac{hf}{k_BT}$$ of the exponential function by $$x$$ (integration by substitution). Thus, the following relationships apply between the variable $$x$$ and the variable $$f$$:

\begin{align}
&x := \frac{hf}{k_B T} ~~~\Rightarrow \boxed{\color{red}{f = \frac{x k_B T}{h}}} \\[5px]
&\frac{\text{d}x}{\text{d}f} = \frac{h}{k_B T} ~~~\Rightarrow \boxed{\color{blue}{\text{d}f = \frac{k_B T}{h}\text{d}x} } \\[5px]
\end{align}

If these relations are used in equation (\ref{her}), then one obtains:

\begin{align}
&I= \frac{2\pi h}{c^2} \cdot \int_{0}^{\infty} \frac{\left(\color{red}{\frac{x k_B T}{h} }\right)^3}{\exp\left(\color{red}{x} \right)-1} ~ \color{blue}{\frac{k_B T}{h}\text{d}x }\\[5px]
&I= \frac{2\pi h}{c^2} \cdot \int_{0}^{\infty} \frac{k_B^3 T^3}{h^3} \frac{x^3}{\exp\left(x \right)-1} ~ \frac{k_B T}{h}\text{d}x\\[5px]
&I= \frac{2\pi h}{c^2} \cdot \frac{k_B^3 T^3}{h^3} \cdot \frac{k_B T}{h} \cdot \int_{0}^{\infty} \frac{x^3}{\exp\left(x \right)-1} ~\text{d}x\\[5px]
&I= \frac{2\pi k_B^4}{h^3 c^2} T^4 \int_{0}^{\infty} \frac{x^3}{\exp\left(x \right)-1} ~\text{d}x\\[5px]
\end{align}

The integral $$\int_{0}^{\infty} \frac{x^3}{\exp\left(x \right)-1} ~\text{d}x$$ cannot be solved so easily in a conventional way. But a look at the mathematics formula collection shows that the result is $$\frac{\pi^4}{15}$$. Thus the intensity of the blackbody radiation can be calculated as follows:

\begin{align}
&I= \frac{2\pi k_B^4}{h^3 c^2} T^4 \cdot \frac{\pi^4}{15}\\[5px]
&I= \underbrace{\frac{2\pi^5 k_B^4}{15 h^3 c^2}}_{\sigma} \cdot T^4 \\[5px]
&\boxed{I= \sigma \cdot T^4}~~~~~\text{and}~~~~~\boxed{\sigma = \frac{2\pi^5 k_B^4}{15 h^3 c^2}}= 5,670 \cdot 10^{-8} \frac{\text{W}}{\text{m²K}^4} \\[5px]
\end{align}

The constant quantities can be combined to a new constant, the so-called Stefan-Boltzmann constant $$\sigma$$ (not to be confused with the Boltzmann constant $$k_B$$!). The radiated intensity of a black body is therefore only dependent on the temperature. It increases with the fourth power of the temperature. This is also called Stefan-Boltzmann law.

The Stefan-Boltzmann law states that the intensity of the blackbody radiation in thermal equilibrium is proportional to the fourth power of the temperature!

The intensity $$I$$ can now be used to determine the radiant power $$\Phi) of a blackbody (also called radiant flux), i.e. its radiant energy emitted per unit time. For this the intensity \(I$$ (as surface power density) has to be multiplied by the surface area $$A$$ of the blackbody:

\begin{align}
&\boxed{\Phi(T,A) = \sigma \cdot A \cdot T^4} \\[5px]
\end{align}

More information on the Stefan Boltzmann Law and its derivation from thermodynamics can be found in the main article Stefan-Boltzmann law.

### Real bodies: The emissivity

In practice, real objects do not radiate with the intensity of a blackbody, but have a lower radiant power. This is expressed by the unitless emissivity $$\varepsilon<1$$. The emissivity represents the radiant power of a real body compared to an ideal black body:

\begin{align}
&\boxed{I_{real}=\varepsilon \cdot \sigma \cdot T^4}\\[5px]
&\boxed{\Phi_{real} = \varepsilon \cdot \sigma \cdot A \cdot T^4} \\[5px]
\end{align}

For non-metallic surfaces, the emissivity is in many cases above 0.9. Many objects can therefore be regarded as black bodies in very good approximation with regard to the emitted radiation. This makes it relatively easy to determine the temperature of real objects with the help of a thermal imaging camera or a pyrometer, since surface properties have a rather minor influence (unless the surfaces are extremely reflective).

### Wien’s displacement law

The spectral distribution as a function of temperature is now to be examined more closely. It turns out that the maximum of the curve shifts with increasing temperature to ever shorter wavelengths. The dependence of this wavelength $$\lambda_{max}$$ on the temperature is given by the following equation. This equation is also known as Wien’s displacement law.

\begin{align}
&\boxed{\lambda_{max}=\frac{2897,8 \text{ µm K}}{T}}~~~\text{Wien’s displacement law} \\[5px]
\end{align}

The Wien’s displacement law can be obtained by determining the maxima of Planck’s law. For this purpose, the function (\ref{planck}) must be derived with respects to the wavelength $$\lambda$$. By using the product rule and setting the derivative equal to zero, one gets:

\begin{align}
\label{abl}
&\frac{\text{d}I_s(\lambda)}{\text{d}\lambda} \overset{!}{=} 0 ~~~~~\text{mit}~~~~~ I_s(\lambda) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\lambda k_B T}\right)-1} \\[5px]
\end{align}

\begin{align}
\frac{\text{d}I_s(\lambda)}{\text{d}\lambda} &= 2 \pi c^2 \left(\frac{hc}{k_B T \lambda^7} \cdot \frac{\exp\left( \frac{hc}{k_B T \lambda}\right)}{\left[\exp\left( \frac{hc}{k_B T \lambda} \right) -1\right]^2} – \frac{1}{\lambda^6} \cdot \frac{5}{\exp\left(\frac{hc}{k_B T \lambda}\right)-1} \right) \\[5px]
&= \underbrace{\frac{2 \pi c^2}{\lambda^6 \cdot \left[ \exp\left( \frac{hc}{k_B T \lambda} \right) -1 \right] }}_{>0} \underbrace{\left(\frac{hc}{k_B T \lambda} \cdot \frac{\exp\left( \frac{hc}{k_B T \lambda}\right)}{\exp\left( \frac{hc}{k_B T \lambda} \right) -1} – 5 \right)}_{=0} =0
\end{align}

This equation will only be zero if the term in the round bracket becomes zero:

\begin{align}
&\frac{hc}{k_B T \lambda} \cdot \frac{\exp\left( \frac{hc}{k_B T \lambda} \right)}{\exp\left( \frac{hc}{k_B T \lambda} \right) -1} – 5 = 0 \\[5px]
\end{align}

With the substitution given below, this equation can be simplified:

\begin{align}
\label{max}
&\boxed{x := \frac{hc}{\lambda k_B T}} ~~~\Rightarrow~~~ x \cdot \frac{\exp\left(x\right)}{\exp\left(x\right) -1} – 5 = 0 \\[5px]
\end{align}

This equation can only be solved numerically, e.g. with the Newton’s method. The result will be x = 4,9651. With this result the wavelength $$\lambda_{max}$$ can be determined as a function of the temperature by rearranging the equation (\ref{max}):

\begin{align}
& \lambda_{max}= \frac{hc}{x k_B T} = \frac{\tfrac{hc}{x k_B}}{T}= \frac{0,0028978 \text{ m K}}{T}= \frac{2897,8 \text{ µm K}}{T} \\[5px]
\end{align}

\begin{align}
&\boxed{\lambda_{max}=\frac{2897,8 \text{ µm K}}{T}}\\[5px]
\end{align}

The maximum of the spectral intensity can also be determined for the frequency form $$I_s(f)$$. For this the function $$I_s(f)$$ must be derived with respect to frequency $$f$$ and setting the derivative equal to zero:

\begin{align}
&\frac{\text{d}I_s(f)}{\text{d}\lambda} \overset{!}{=} 0 ~~~\Rightarrow~~~ \boxed{f_{max} = 5,879 \cdot 10^{10} \tfrac{\text{Hz}}{\text{K}} \cdot T}
\end{align}

Remark

Note that Wien’s displacement law indicates the wavelength $$\lambda_{max}$$ at which the spectral intensity has a maximum. This maximum is not to be equated with the maximum of the intensity itself or with the maximum of the radiant power! This leads for example to the fact that although the general relationship $$f=\frac{c}{\lambda}$$ applies, it does not apply in this particular case $$f_{max}=\frac{c}{\lambda_{max}}$$!

This has to do with the fact that the spectral intensity is a quantity related to the wavelength. One measures the radiant power in a certain wavelength interval $$\text{d}\lambda$$ and refers to it the radiant power. A comparison of different radiant powers is therefore only possible if the same wavelength intervals are always considered. Since the frequency is not proportional to the wavelength but reciprocally proportional, equidistant wavelength intervals do not also mean equidistant frequency intervals!

A simple example is the wavelength range between 1 and 10 µm, which is divided into intervals of 1 µm each. This results in the following equidistant series:

\begin{align}
&1-2-3-4-5-6-7-8-9-10 \\[5px]
\end{align}

The reciprocal values of this, which in the figurative sense have the meaning of the frequency as a reciprocal value of the wavelength, no longer result in an equidistant series:

\begin{align}
&\frac{1}{1}-\frac{1}{2} -\frac{1}{3} -\frac{1}{4} -\frac{1}{5} -\frac{1}{6} -\frac{1}{7} -\frac{1}{8} -\frac{1}{9} -\frac{1}{10} \\[5px]
&1-0,5-0,333-0,25-0,25-0,2-0,167-0,143-0,125-0,111-0,1 \\[5px]
\end{align}

So one cannot compare equidistant wavelength intervals with equidistant frequency intervals. Therefore a different frequency $$f_{max}$$ than one could expect from the formula $$f_{max}=\frac{c}{\lambda_{max}}$$ is obtained when using the frequency form of the spectral intensity.