Cooking on Mount Everest

With increasing altitude above sea level, the air pressure decreases more and more (see also the article on barometric formula). This shows the phenomenon that water begins to boil at significantly lower temperatures than one is used to at lower altitudes. At sea level at a pressure of 1.013 bar, water begins to boil at a temperature of 100 °C.

However, on Mount Everest at an altitude of 8849 m, the air pressure is only around 0.325 bar. Due to this significantly reduced pressure, the water already begins to boil at a temperature of around 71°C. However, since the temperature does not rise any further during boiling, the cooking of foods such as potatoes or pasta thus takes significantly longer (see also the article Why does the temperature remain constant during a change of state?).

Explanation with the particle model

The fact that the boiling point depends on the ambient pressure applies not only to water, but ultimately to all liquids. In particular, it is true that the boiling point decreases with decreasing pressure. This phenomenon can be explained qualitatively with the particle model of matter.

During boiling, the liquid vaporizes and becomes gaseous. In this vaporization process, energy is absorbed by the liquid and added to the molecules, allowing them to break free from the molecular binding forces of the liquid and enter the gas phase. At an ambient air pressure of 1 bar, vaporization of water takes place at a temperature of 100 °C.

However, if the ambient air pressure is increased, the air molecules collide more strongly with the surface of the liquid. In the process, the air molecules push the liquid molecules back into the liquid, so to speak. It thus becomes more difficult for the molecules in the liquid to pass into the gas phase. The water molecules consequently require greater energy and thus a higher temperature in order to escape the liquid phase. For this reason, with increased ambient air pressure, a higher boiling temperature is required to vaporize a liquid or bring it to a boil.

The boiling temperature of a liquid increases with increasing ambient pressure!

Increase of boiling temperature at elevated ambient pressure (pressure cooker)

Under high ambient pressure, water consequently also boils at higher temperatures. This is used, for example, in so-called pressure cookers to heat the water to over 100 °C. A pressure cooker seals the pot of water gas-tight. During vaporization, water normally expands 1700 times. However, since this is not possible with a sealed pot, the pressure consequently increases. A pressure relief valve usually limits the pressure to a maximum of 2 bar. The boiling temperature rises to around 120 °C at this increased pressure. As a result, food prepared in the pot is no longer cooked at just 100 °C, but at 120 °C!

Decrease of boiling temperature at reduced ambient pressure

If an increase in the ambient pressure leads to an increase in the boiling temperature, then in the opposite case this means that a decrease in the ambient pressure results in a decrease in the boiling temperature. And this is exactly what explains why water on Mount Everest boils at already 71 °C due to the lower pressure of only 0.325 bar. Preparing food that normally requires a temperature of 100 °C in water is therefore not so easy at high altitudes. At this point, one would have to use the pressure cooker already explained to obtain increased pressure and raise the boiling temperature.

The following experiment provides an impressive demonstration of the decrease in boiling temperature with decreasing pressure. For this purpose, a glass with water is placed under a vacuum chamber. A thermometer is placed in the glass to observe the temperature. The thermometer indicates a temperature of 20 °C. Now the vacuum pump is switched on and thus the pressure is reduced step by step. Below a pressure of about 0.023 bar, one then observes small bubbles rising in the water. This is the typical phenomenon when water boils, with the temperature still at 20 °C. And indeed, at a pressure of 0.023 bar, the water already begins to boil at 20 °C.

Constant temperature during vaporization and melting

When water is heated with an immersion heater, one first observes a rise in temperature. But during vaporization, the temperature does not increase any further. The temperature remains constant at 100 °C (boiling point), and this despite the fact that heat is obviously still being supplied by the immersion heater.

A similar behavior can be observed when ice melts. To demonstrate this, place ice cubes from a refrigerator in a bowl and heat them with a heat lamp, for example. The emitted heat causes the temperature of the ice cubes to rise at first. However, if the ice starts to melt at a temperature of 0 °C (melting point), the temperature of the water-ice mixture does not increase any further. The temperature remains constant at 0 °C, even though heat is obviously being supplied by the heat lamp. Only when all the ice has completely liquefied does the temperature increase again.

Not only when water melts or vaporizes do the temperatures remain constant, but also in the reverse cases,when gaseous water condenses or liquid water solidifies. This phenomenon of constant temperature can generally be observed when the state of matter of a substance changes (also called phase transition or phase change). This is not only true for water, but can be observed for all pure substances.

The question arises as to why the temperature does not change despite the transfer of heat energy during a phase change. And is this also true for mixtures of substances?

Cause of temperature increase when heat is transferred

If energy is transferred to a substance as heat, this causes the molecules to move more violently. In solids, for example, the vibration of the atoms increases as a result. In liquids and gases, the transferred heat increases the kinetic energy and thus the speed of the molecules. Since the temperature of a substance is a measure of the kinetic energy of the molecules, this explains the generally observable increase in temperature when heat is supplied to a substance (see also the article Temperature and particle motion).

Since, on the other hand, the temperature remains constant in the case of a phase transition, the energy supplied can obviously no longer benefit the kinetic energy of the molecules. Using the example of the vaporization of a liquid, the atomic processes that take place are explained in more detail below.

Atomic processes during vaporization

In the liquid state, the individual molecules are bound together by intermolecular forces (Van der Waals forces). These forces ensure that the molecules in the liquid do not distribute freely throughout the space, as is the case with gases, but form a coherent substance. The intermolecular binding forces can be thought of as rubber bands that hold the molecules of the liquid together.

If the liquid is now heated, the binding forces are loosened up by the stronger particle movements. In a figurative sense, this would correspond to an overstretching of the rubber bands due to the increasing movement (increasing distance). At some point, the motion of the molecules will be so strong that rubber bands will wear out and thus lose elasticity. In this state, the boiling point of the liquid is reached and the molecules are hardly elastically connected with each other.

At this boiling point, the kinetic energies of the individual molecules are greater than the binding energies between the molecules. The motion of the molecules is, so to speak, stronger than the bond between the molecules. In the figurative sense, this would correspond to the point where the molecules have enough energy to break the rubber bands that normally hold them together. Those molecules that have broken free of the bonds can now move freely and are no longer bound to the liquid – they have become gaseous. Note that in general, intermolecular binding forces also act in the gaseous state, but these are significantly lower compared to the binding forces in the liquid or solid state!

The heat energy supplied during vaporization therefore does not benefit the increase in kinetic energy and thus the increase in temperature, because the heat energy is used to break the molecules loose from the intermolecular binding forces (change in internal energy). For this reason, the temperature remains constant during vaporization until the change of state is complete. Only then can the kinetic energy and therefore the temperature be further increased.

During a phase transition the supplied energy is not used to increase the kinetic energy of the molecules, but to change the binding energies (increase in internal energy)!

The amount of heat required to completely vaporize a liquid is called the heat of vaporization. More information specifically on this can be found in the article Specific heat of vaporization and condensation (latent heat).

Atomic processes during condensation

When a gaseous substance condenses, it emits the previously absorbed heat of vaporization (in this case called heat of condensation). This process can also be illustrated with rubber bands. While the molecules in the gaseous phase can move relatively free, the molecules in the liquid state are held together by stronger intermolecular forces. The process of condensation thus corresponds to the “capture” of the molecules with the help of rubber bands. Thereby, the flying molecules hit the network of already captured molecules of the liquid phase with full force.

On impact, part of the kinetic energy of the molecules is transferred to the molecules in the liquid. However, in order to prevent molecules that have already been captured by the binding forces from being kicked out of the liquid phase again, energy must be removed from the molecules upon impact. This corresponds to the dissipation of the heat of condensation so that the condensed substance remains permanently liquid and the molecules in it cannot break away again from the liquid phase. Thus, although heat (of condensation) is dissipated, there is no decrease in temperature because of the simultaneous internal release of energy due to the impact processes during condensation.

Atomic processes during melting and solidification

It is not only during the transition from the liquid to the gaseous phase (or vice versa) that the binding energies between the molecules change abruptly. Also during the transition from the solid to the liquid state, a sudden change of the binding energy occurs. While the molecules in the solid state are firmly bound to a specific location due to the great binding forces, the molecules in the liquid state can move relatively freely due to the weak binding forces.

Therefore, energy is also required to break the molecules free from the strong binding forces during melting. This is provided by the heat input during melting. This heat input does not lead to a further increase in temperature until all intermolecular bonds have been broken and the substance has melted. Only then can the supplied heat be used to increase the kinetic energy – the temperature of the liquid rises.

The amount of heat required to completely melt a substance is called the heat of fusion. More information specifically on this can be found in the article Specific heat of fusion and heat of solidification (latent heat).

In the reverse case, i.e. during solidification, the previously supplied heat of fusion must be dissipated (in this case called heat of solidification) in order to completely solidify the liquid substance. Here, too, the temperature remains constant until the liquid has completely solidified.

Changes in the state of matter at non-constant pressure

In the article Why does water boil faster at high altitudes? it has already been explained in detail that the boiling temperature changes with the ambient pressure. Such pressure dependence occurs not only in vaporization or condensation, but generally in any kind of phase transition. Therefore, melting temperatures or solidification temperatures are also pressure-dependent. Thus, the temperature remains constant during a change of state only if the pressure remains constant at the same time.

If, for example, water were to be brought to the boil in a so-called pressure cooker, the temperature would no longer remain constant during vaporization. A pressure cooker seals the pot of water gas-tight. Compared to liquid water, however, gaseous water occupies a much larger space. In a pressure cooker, however, gaseous water cannot expand. The pressure therefore increases continuously as the water vaporizes (a relief valve usually limits the pressure to a maximum of 2 bar). With the continuous increase in pressure, the boiling temperature also rises permanently during vaporization. Consequently, the temperature does not remain constant in this case.

In the case of phase transitions of pure substances, the temperature remains constant only if the pressure is kept constant at the same time (isobaric process)!

Phase of transition of mixtures of substances

While in the case of phase transitions of pure substances the temperature remains constant, in the case of mixtures of substances there is usually only a slowing down of the temperature change. In this case, only part of the transferred heat is used to change the binding energies, while the other part simultaneously causes a change in temperature. It is therefore by no means the case that the temperature of all substances remains constant during phase transitions.

In the case of mixtures of substances, the temperature generally no longer remains constant during phase transitions, but the temperature change merely slows down in the process!

Adiabatic mixing

If two bodies with different initial temperatures are brought into contact with each other, the temperatures will become more and more equal. Eventually, thermodynamic equilibrium will be reached. The temperatures have then completely equalized and a common final temperature has been established, which is also referred to as the mixing temperature.

One can observe such an equalization of temperatures, for example, when pouring hot water into a cold glass. While the glass is heated by the hot water, the water cools down on the relatively cold glass. After some time, the different initial temperatures have equalized and the glass has the same temperature as the water inside it. The final temperature lies between these two initial temperatures.

Depending on how much water is poured into the glass, the final temperature is shifted towards higher or lower values. With a larger amount of water, it can be assumed that higher final temperatures result, since more hot water is present and causes greater heating of the glass. The equalization of temperatures can be explained using the particle model. This is discussed in detail in the article Heat and thermodynamic equilibrium.

In the following, it will be shown how the final temperature can be determined when two bodies are in thermal contact with each other. It is assumed that heat is only transferred between the two considered bodies. Heat transfer to the surroundings is thus neglected. Such a thermal mixing process neglecting an unwanted heat transfer to the surroundings is also called an adiabatic mixing (the term adiabatic system is explained in more detail in the article Thermodynamic systems).

Derivation of the formula for calculating the final temperature

The basic relationship between transferred heat Q and temperature change ΔT of a body is given by the heat capacity C of the considered object

\begin{align}
\label{q}
\boxed{Q = C \cdot \Delta T}~\text{,} \\[5px]
\end{align}

where the heat capacity C for a homogeneous body can be determined from the specific heat capacity c of the substance and its the mass m:

\begin{align}
\label{c}
&\boxed{C = c \cdot m} \\[5px]
\end{align}

In the following, we will consider the example of hot water and a colder glass. If the hot water is poured into the cool glass, then heat is transferred from the water to the glass. This causes the glass to heat up due to absorbed heat. At the same time, the water cools due to the heat given off. The amount of heat emitted by the water (Q_w) is equal to the amount of heat absorbed by the glass (Q_g):

\begin{align}
\label{e}
Q_\text{g} = Q_\text{w} \\[5px]
\end{align}

Note that heat transfer to the surroundings has been neglected and, for reasons of energy conservation, the heat emitted by the water must therefore be completely absorbed by the glass.

The heat Q_w emitted by the water causes the temperature of the water to decrease by a certain amount ΔT_w according to equation (\ref{q}). For a given initial temperature T_w, the heat given off can be determined as follows when a common final temperature T_f has been established in thermal equilibrium:

\begin{align}
Q_\text{w} = C_\text{w} \cdot \underbrace{\left( T_\text{w}-T_\text{f}\right)}_{\Delta T_\text{w}>0} \\[5px]
\end{align}

In the analogous way, the absorbed heat by the glass Q_g can be determined on the basis of the initial temperature T_g and the final temperature T_f:

\begin{align}
Q_\text{g} = C_\text{g} \cdot \underbrace{\left( T_\text{f}-T_\text{g}\right)}_{\Delta T_\text{g}>0} \\[5px]
\end{align}

Note that the initial temperature of the water is greater than the final temperature, while the initial temperature of the glass is lower than the final temperature (T_w>T_f>T_g). The temperature differences in the two upper equations were therefore chosen to give positive values for the amounts of heat in each case. In this way, the amounts of heat can now be equated according to equation (\ref{e}) and solved for the final temperature:

\begin{align}
Q_\text{g} &= Q_\text{w} \\[5px]
C_\text{g} \cdot (T_\text{f} – T_\text{g})&= C_\text{w} \cdot (T_\text{w} – T_\text{f})  \\[5px]
C_\text{g} \cdot T_\text{f} – C_\text{g} \cdot T_\text{g} &= C_\text{w} \cdot T_\text{w} – C_\text{w} \cdot T_\text{f}\\[5px]
C_\text{g} \cdot T_\text{f} +C_\text{w} \cdot T_\text{f}  &= C_\text{w} \cdot T_\text{w} + C_\text{g} \cdot T_\text{g} \\[5px]
T_\text{f} \cdot (C_\text{w} +C_\text{g})   &= C_\text{w} \cdot T_\text{w} + C_\text{g} \cdot T_\text{g} \\[5px]
T_\text{f}  &= \frac{C_\text{w} \cdot T_\text{w} + C_\text{g} \cdot T_\text{g}}{C_\text{w} +C_\text{g}} \\[5px]
\end{align}

\begin{align}
\label{tm}
\boxed{T_\text{f}  = \frac{C_\text{w} \cdot T_\text{w} + C_\text{g} \cdot T_\text{g}}{C_\text{w} +C_\text{g}}} \\[5px]
\end{align}

For homogeneous substances, equation (\ref{c}) can be used in equation (\ref{tm}) to determine the final temperature on the basis of the specific heat capacities c and the masses m of the substances:

\begin{align}
\boxed{T_\text{f}  = \frac{c_\text{w} \cdot m_\text{w} \cdot T_\text{w} + c_\text{g} \cdot m_\text{g} \cdot T_\text{g}}{c_\text{w} \cdot m_\text{w} + c_\text{g} \cdot m_\text{g}}} \\[5px]
\end{align}

Numerical example

In the following, the final temperature is to be determined on the basis of concrete values. We assume a glass with a mass of m_g = 100 g. The specific heat capacity of glass can be assumed to be c_g = 0.72 kJ/(kg⋅K). The initial temperature of the glass is room temperature with T_g = 293 K (20 °C). Now 200 ml of water with a mass of m_w = 200 g is poured into the glass. The initial temperature of the water is T_w = 333 K (60 °C). The specific heat capacity of water can be assumed to be c_w = 4.2 kJ/(kg⋅K).

In fact, it makes no difference at this point whether the temperatures are used in the unit degrees Celsius or in the unit Kelvin in the upper formula for calculating the final temperature. If the temperatures are used in the unit degrees Celsius, the final temperature also results in the unit degrees Celsius. If, on the other hand, the temperatures are used in the unit Kelvin, the final temperature also results in the unit Kelvin.

We use the temperatures in the unit degree Celsius and obtain in this way a final temperature of T_f = 56.8 °C:

\begin{align}
\underline{T_\text{f}}  = \frac{4.2 \tfrac{\text{kJ}}{\text{kg K}} \cdot 100 \text{ g} \cdot 60 \text{ °C} + 0.72 \tfrac{\text{kJ}}{\text{kg K}} \cdot 200 \text{ g} \cdot 20 \text{ °C}}{4.2 \tfrac{\text{kJ}}{\text{kg K}} \cdot 200 \text{ g} + 0.72 \tfrac{\text{kJ}}{\text{kg K}} \cdot 100 \text{ g}} = \underline{56.8 \text{ °C}}\\[5px]
\end{align}

The water obviously cools down by only 3.2 °C, while the glass heats up by 36.8 °C. This is due to the significantly greater heat capacity of the water, which is caused on the one hand by the greater mass and on the other hand by the significantly higher specific heat capacity. If only 20 ml of water were added, a final temperature of 41.5 °C would result. In this case, the heat capacities of water and glass are approximately equal and the temperature of the water decreases by the same amount as the temperature of the glass increases (for more information see section Special cases of Richmann’s law of mixtures).

Extending to any two bodies in thermal contact

The formula for calculating the final temperature was derived using the example of water and glass. However, this formula can be transferred to any two substances. It does not matter whether it is a solid and a liquid or two solids that are brought into thermal contact. This relationship also applies to mixtures of two liquids!

Therefore, in general, the final temperature T_f in thermodynamic equilibrium for any two bodies (1) and (2) with different initial temperatures T₁ and T₂ and different heat capacities C₁ and C₂ can be calculated using the following formula:

\begin{align}
\label{rr}
&\boxed{T_\text{f}  = \frac{C_{1} \cdot T_{1} + C_{2} \cdot T_{2}}{C_{1} +C_{2}}} \\[5px]
\end{align}

For bodies consisting of homogeneous substances, the heat capacities can be determined from the specific heat capacities c and their masses m:

\begin{align}
&\boxed{T_\text{f}  = \frac{c_1 \cdot m_1 \cdot T_{1} + c_2 \cdot m_2 \cdot T_{2}}{c_1 \cdot m_1 +c_2 \cdot m_2}} \\[5px]
\end{align}

These two equations are also known as Richmann’s law of mixtures! For the use of this formula it does not matter which of the two bodies (1 or 2) is the warmer and which is the colder. It should also be mentioned again that the temperatures do not necessarily have to be used in the unit Kelvin, but can also be used in the unit degrees Celsius.

The validity of Richmann’s law must be limited for the case of phase transitions that occur during the equalization of temperatures. This would be the case, for example, when ice cubes are poured into a warm beverage. In these cases, (latent) amounts of heat due to the change in the state of matter must also be taken into account (heat of fusion).

Richmann’s law of mixtures describes the resulting final temperature when two bodies with different initial temperatures are brought into thermal contact, provided that no phase transition occurs and it is assumed that heat is transferred only between these two bodies!

It can be seen from a closer look at equation (\ref{rr}) that the final temperature results from the weighted arithmetic mean of the initial temperatures, where the weighting is done by the heat capacities! Depending on the heat capacity, the final temperature therefore shifts to higher or lower values.

Note that in practice, when two bodies are brought into thermal contact, the heat energy transferred from the hotter body does not completely benefit the cooler body. A certain part of the heat energy emitted is also transferred to the surroundings. In the case of mixing two liquids in a vessel, a part of the heat is also transferred to the vessel. The final temperature of two bodies will therefore be lower than the theoretically calculated final temperatures due to these (unwanted) heat losses.

Special cases of Richmann’s law of mixtures

When mixing two identical substances (for example when pouring hot water into cold water), the specific heat capacities are identical (c₁=c₂=c), so that the final temperature T_f is independent of these specific heat capacities. That means, no matter which substances are brought into thermal contact with each other, as long as they are identical the final temperature will be the same in all cases!

\begin{align}
\require{cancel}
&T_\text{f} = \frac{\bcancel{c} \cdot m_1 \cdot T_{1} +\bcancel{c} \cdot m_2 \cdot T_{2}}{\bcancel{c} \cdot m_1 + \bcancel{c} \cdot m_2} \\[5px]
&\boxed{T_\text{f} = \frac{m_1 \cdot T_{1} + m_2 \cdot T_{2}}{m_1 + m_2}} ~~~\text{only applies for identical substances}\\[5px]
\end{align}

If, furthermore, the masses of the two substances are equal (m₁=m₂=m), the final temperature T_f results from the arithmetic mean of the initial temperatures (average temperature):

\begin{align}
&T_\text{E} = \frac{m \cdot T_{1} + m \cdot T_{2}}{m + m} \\[5px]
&\boxed{T_\text{f} = \frac{T_{1} + T_{2}}{2}} ~~~\text{only applies to identical substances with the same mass} \\[5px]
\end{align}

The negative thermal expansion of water (density anomaly) has an existential importance for life on earth. More precisely: for life under water. Due to the density anomaly, layers of water with different temperatures form in still waters. This thermal stratification is due to the different density of water, which is caused by the different temperatures. Depending on the temperature, heavy water sinks to the bottom of a lake while lighter water rises to the top. The transitions between the temperatures of the individual layers are of course smooth. Since water has the highest density at 4 °C and is therefore the heaviest, it will sink to the bottom of the water (bottom layer).

In summer, the warmer water layers will settle above the 4 °C cold ground layer due to the lower density. As a result of the decreasing density with increasing temperature, the water temperature will steadily increase towards the water surface. While the water surface of deep waters can be relatively warm in summer, depending on the intensity of the sun, the water at the bottom will generally not heat up above 4 °C.

This is especially the case in very deep or still waters, where there are hardly any currents that lead to a mixing of the layers. When the temperature of the bottom layer rises above 4°C in shallow water or during longer periods of heat, the layer of water with the lowest temperature accumulates at the bottom and warmer layers of water lie above it.

In autumn the water will cool down gradually. The temperature of the relatively warm surface layer will drop accordingly. Thus, the temperature difference between bottom layer and surface layer will become smaller and smaller. The temperatures of the different water layers will equalize more and more. Sooner or later, a uniform temperature of 4 °C will be reached throughout the entire water.

This will also be the case if the water has warmed up to 5°C in summer, for example. In this case, a uniform temperature of 5 °C will first form when the water cools down. Further cooling may result in a 4 °C layer on the surface for a short time. However, due to its greater density, this water layer will then sink to the ground. The warmer layers are thus forced to the surface and also cool down. Finally, even in such a case, sooner or later a uniform temperature of 4 °C will be reached throughout the entire water.

If the cold water of 4 °C continues to cool down further in winter, the colder layers will no longer sink to the ground. Because due to the thermal negative expansion, those cooler layers are lighter (lower density). In the temperature range between 4 °C and 0 °C applies: The colder the water, the lighter it will be. This is the actual density anomaly of water, since it no longer contracts but expands as it cools. As a result, the warmer and therefore heavier layers of water will no longer accumulate on the water surface in winter as in summer, but rather the colder and therefore lighter layers of water will be at the surface! Finally, the solidification temperature of 0 °C is reached first on the water surface. For this reason, ice always begins to form first on the surface of a lake.

If the water is deep enough, the cold will usually not penetrate to the deep layers and thus the water will not freeze completely (note that water has only a very low thermal conductivity!). In the deeper layers, the water usually remains liquid at around 4 °C. The fact that a body of water freezes from above due to the density anomaly and thus usually does not freeze completely, ensures the survival of the animals in it.

Without the density anomaly of water, the cold water layers would sink to the ground in winter. The warmer layers would be displaced and rise to the surface. This would lead to a rapid cooling of the water until it freezes completely. The fish in such waters would not survive.

Introduction

In this article, the Stefan-Boltzmann Law is to be derived with the laws of thermodynamics. In order to be able to do this, fundamental relationships must first be clarified.

Relationship between energy density and pressure

Kinetic theory of gases with classical particles

With the help of the kinetic theory of gases, the following relationship between the pressure of a gas p and the speed v of the particles contained therein was derived in the article Pressure and temperature:

\begin{align}
\label{p}
&\boxed{p = \frac{1}{3}\frac{N}{V}m\cdot \overline{v^2}} \\[5px]
\end{align}

Here N denotes the number of gas particles in a volume V and m refers to the mass of one particle (not the entire gas mass!). The speed refers to the root mean square speed.

In a classical view of ideal gases, in which the particle speeds are far below the speed of light, the mean kinetic energy of the particles can be described in very good approximation with ½⋅m⋅v² (Note that this formula only applies to non-relativistic mechanics!) The following relationship between the pressure and the mean energy of a particle becomes obvious:

\begin{align}
&p = \frac{2}{3} \cdot \frac{N}{V} \cdot \overbrace{\frac{1}{2} m \overline{v^2}}^{\overline{W_{kin}}} = \frac{2}{3} \cdot \frac{N}{V} \cdot \overline{W_{kin}} = \frac{2}{3} \cdot \frac{\overbrace{N \cdot \overline{W_{kin}}}^{\text{internal energy } U}}{V} = \frac{2}{3}\cdot \overbrace{\frac{U}{V}}^{\text{energy densitiy } u_v} = \frac{2}{3}\cdot u_v \\[5px]
&\boxed{p = \frac{2}{3} u_v} ~~~~~\text{only valid for a classical gas}
\end{align}

In the above derivation it was used that the product of the number of particles N and the mean kinetic energy of a particle corresponds to the total energy contained in the gas, i.e. the so-called internal energy U. The quotient of internal energy U and volume V can therefore be interpreted as (volumetric) energy density u_v (to avoid misunderstandings with the specific internal energy u as a mass-related quantity, a “v” is added to the energy density in the index).

Due to the fact that classical mechanics was used for the derivation under the assumption of the validity of W_kin=½⋅m⋅v², the obtained relationship between pressure and energy density applies only to classical ideal gases.

Kinetic theory of gases with relativistic particles (photons)

Due to the wave-particle duality, electromagnetic radiation can also be imagined as a beam of photons, as has already been done for the derivation of the radiation pressure. In this way, the radiation of a blackbody can now be described mathematically. In the article Blackbody radiation the realization of such a blackbody has already been explained in detail. For this purpose, a small hole is drilled into an object leading into a cavity.

All incident radiation is absorbed by the cavity, which also emits radiation itself. In thermal equilibrium, the emitted and absorbed radiation energy is the same, so that in principle a gas of photons is formed in the cavity. Instead of massive particles, in this case we are dealing with massless, relativistic photons. Here, too, a pressure p can be found in the cavity volume V due to the moving photons, the so-called radiation pressure.

Equation (\ref{p}) can now also be used, it only has to be interpreted against the background of theory of relativity. For photons, the speed of the particles v corresponds to the speed of light c and the particle mass m must be considered as the relativistic mass m_rel. At this point it is not necessary to differentiate between the mean value of the speed squares \(\overline{c^2}\) and the speed square c², since in this case there is no classical speed distribution but all photons are travelling with the same speed of light. Both values are therefore identical.

\begin{align}
\label{pp}
&p = \frac{1}{3}\frac{N}{V}\cdot m_{rel} \cdot c^2 \\[5px]
\end{align}

Note that although photons do not have a rest mass, they do have a relativistic mass! For example, Einstein showed with his famous equation on energy-mass equivalence that every thing with energy can be assigned a corresponding (relativistic) mass m_rel:

\begin{align}
\label{e}
&\boxed{E =mc^2} ~~~~~\text{or}~~~~~ \boxed{W=m_{rel}\cdot c^2} \\[5px]
\end{align}

According to quantum mechanics, the energy of a photon results from the product of frequency f and Planck constant h, so that the relativistic mass can be determined from this:

\begin{align}
&W = h\cdot f \\[5px]
\end{align}

The product of the (relativistic) mass m (=m_rel) and the square of the speed of light c² contained in equation (\ref{pp}) corresponds to the energy of a photon W according to equation (\ref{e}). Thus the following relationship applies between the radiation pressure p of the photon gas and its energy density u_v:

\begin{align}
&p = \frac{1}{3}\frac{N}{V}\cdot \overbrace{ m_{rel}\cdot c^2}^{W}=\frac{1}{3}\frac{N}{V}\cdot W =\frac{1}{3}\frac{\overbrace{N\cdot W}^{\text{“internal” energy } U}}{V} =\frac{1}{3} \cdot \overbrace{\frac{U}{V}}^{\text{energy density }u_v} = \frac{1}{3} u_v \\[5px]
\label{h}
&\boxed{p = \frac{1}{3} u_v } ~~~~~\text{only valid for a photon gas (homogeneous radiation)}
\end{align}

The product of photon energy and number of photons corresponds to the total (internal) energy contained in the photon gas, i.e. the radiant energy (photon gas energy). The quotient of radiant energy and cavity volume can in turn be understood as the energy density of the photon gas or the energy density of the radiation.

The relationship between pressure and energy density of a classical gas thus differs by a factor of 2 from that of a relativistic photon gas.

Derivation from radiation pressure

The relationship between radiation pressure and energy density for a homogeneous photon gas can also be derived from the radiation pressure of a directed beam. In the article Radiation pressure it was shown that the radiation pressure p_beam, which a directed beam exerts on an object at complete reflection, corresponds to twice the value of the energy density:

\begin{align}
\label{s}
&\boxed{p_{beam} =2 \cdot u_v} ~~~~~\text{only valid for complete reflection of directed beams} \\[5px]
\end{align}

This equation does not contradict equation (\ref{h})! Because equation (\ref{s}) is only valid for directed radiation, i.e. when all photons move in the same direction and hit a boundary surface. In the present case of cavity radiation, however, it is a matter of radiation that is permanently reflected, absorbed and emitted inside, i.e. the photons all move in completely random directions; it is actually a photon gas.

In order to make a completely random photon motion out of a directed photon beam, all photons contained in it would suddenly have to move in different directions. Assuming a homogeneous statistical distribution, one sixth of the photons would move upwards, one sixth downwards, one sixth to the left and one sixth to the right, and one sixth to the rear, and finally one sixth of the photons would move forwards towards the boundary surface under consideration, on which a pressure would be exerted.

At the same energy density (photon density), the pressure of a homogeneous photon gas is thus one sixth lower than that of a directed photon beam, since only one sixth of the photons effectively collide with the boundary surface. Thus, for a homogeneous photon gas, one obtains the same relationship:

\begin{align}
& p = \frac{1}{6} \cdot p_{beam} = \frac{1}{6} \cdot 2 \cdot u_v= \frac{1}{3} \cdot u_v \\[5px]
\end{align}

One could now argue at this point that the cavity radiation is not based on a complete reflection and thus equation (\ref{s}) is not valid at all. Rather, a blackbody or a cavity radiation is a complete absorption of the radiation. This is correct, but in thermal equilibrium exactly the same number of photons are absorbed as emitted. This is where the thermodynamic equilibrium comes into being.

Kinematically speaking, a reflection is nothing more than an absorption (“impact”) with subsequent emission (“rebound”). Only because absorption and emission do not take place at identical times in the case of cavity radiation, they are essentially identical to reflection, in which both processes in principle take place immediately one after the other. For this reason, a completely reflective behavior must be assumed for cavity radiation in thermodynamic equilibrium!

Relationship between a change in volume and the change in photon gas energy

As the experimental study of blackbody radiation has already shown, the energy of the radiation emitted depends only on the temperature. The energy density is therefore only a function of the temperature. At constant temperature and thus constant energy density u_v, an increase of the cavity volume by dV means a corresponding increase of the photon gas energy dU:

\begin{align}
& U(T,V) = u_v(T) \cdot V \\[5px]
& \frac{\text{d}U(T,V)}{\text{d}V} =\frac{\text{d}(u_v(T) \cdot V )}{\text{d}V} = \overbrace{\underbrace{\frac{\partial{u_v(T)}}{\partial{V}}}_{=0} \cdot V +\underbrace{\frac{\partial{V} }{\partial{V}}}_{=1} \cdot u_v(T)}^{\text{product rule}} = u_v(T) = u_v \\[5px]
\label{t}
&\boxed{\left(\frac{\partial{U}}{\partial{V}}\right)_T = u_v = 3p} \\[5px]
\end{align}

Note that the (partial) derivation of the energy density with respect to the volume equals to zero, since the energy density is not a function of the volume. With respect to the variable V, this is the derivation of a constant. In the last step it was used that the energy density u_v corresponds to three times the value of the pressure p [see equation (\ref{h})].

The fact that the photon gas energy increases at a constant energy density to the same extent as the volume also becomes clear, because a constant energy density also means a constant photon density. If one increases the volume of the cavity, then with constant photon density there are now more photons in it (the cavity then emits more photons until the thermal equilibrium is restored). Since each photon is associated with a certain energy, the total energy contained in the volume has also increased.

First law of thermodynamics

Since the photon gas is considered analogous to a classical ideal gas, the first law of thermodynamics can also be applied at this point. In differential form, it is represented as follows (in which S denotes the entropy):

\begin{align}
&\boxed{\text{d}U = T \cdot \text{d}S – p \cdot \text{d}V } ~~~~~\text{First law of thermodynamics} \\[5px]
\end{align}

If this equation is divided by dV, the following relationship is obtained at constant temperature:

\begin{align}
\label{ss}
& \left(\frac{\partial{U}}{\partial{V}}\right)_T = T \left(\frac{\partial{S}}{\partial{V}}\right)_T – p \\[5px]
\end{align}

Without going deeper into the derivation of the so-called Maxwell relations of thermodynamics, these provide another important relationship at this point: A change in entropy and a change in volume at constant temperature are in the same relationship as a change in pressure and a change in temperature at constant volume. Mathematically this is expressed as follows:

\begin{align}
&\boxed{\left(\frac{\partial{S}}{\partial{V}}\right)_T = \left(\frac{\partial{p}}{\partial{T}}\right)_V} ~~~~~\text{Maxwell relation} \\[5px]
\end{align}

This Maxwell relation can now be used in equation (\ref{ss}):

\begin{align}
\label{x}
& \left(\frac{\partial{U}}{\partial{V}}\right)_T = T \left(\frac{\partial{p}}{\partial{T}}\right)_V – p \\[5px]
\end{align}

According to equation (\ref{t}) the term (∂U/∂V)_T can be replaced by the energy density u_v. In addition, the pressure p according to the equation (\ref{h}) corresponds to just one third of the energy density:

\begin{align}
u_v &= T \left(\frac{\partial{\frac{1}{3}u_v}}{\partial{T}}\right)_V – \frac{1}{3} u_v \\[5px]
u_v &= \frac{1}{3} T \left(\frac{\partial{u_v}}{\partial{T}}\right)_V – \frac{1}{3} u_v \\[5px]
3 u_v &= T \left(\frac{\partial{u_v}}{\partial{T}}\right)_V – u_v \\[5px]
\end{align}

Since the energy density is only a function of temperature, the partial derivative with respect to the temperature can be expressed as an ordinary derivative. After separation of variables (Fourier method) we obtain:

\begin{align}
3 u_v &= T \frac{\text{d}u_v}{\text{d}T} – u_v \\[5px]
4 u_v &= T \frac{\text{d}u_v}{\text{d}T} \\[5px]
\frac{\text{d}u_v}{u_v} &=4 \frac{\text{d}T}{T} \\[5px]
\end{align}

Both sides of the equation can now be integrated, whereby constants of integration must be taken into account when calculating the primitive function, which are summarized in the constant a (in this case, a generally stands for a constant, not for a special value!):

\begin{align}
\int \frac{\text{d}u_v}{u_v} &= \int 4 \frac{\text{d}T}{T} \\[5px]
\ln{u_v} &= 4 \cdot \ln{T} + a\\[5px]
e^{\ln{u_v}} &= e^{4 \cdot \ln{T}+a} \\[5px]
e^{\ln{u_v}} &= e^{4 \cdot \ln{T}} \cdot e^{a} \\[5px]
e^{\ln{u_v}} &= \left(e^{\ln{T}}\right)^4 \cdot a \\[5px]
&\underline{u_v = a \cdot T^4} \\[5px]
\end{align}

The energy density u_v is thus proportional to the fourth power of the absolute temperature T. A volume V filled with photons (e.g. the volume of the considered cavity), which are emitted by a blackbody at the temperature T, contains the following total energy U:

\begin{align}
&U = u_v \cdot V = a \cdot T^4 \cdot V \\[5px]
\label{st}
&\boxed{U = a \cdot T^4 \cdot V} \\[5px]
\end{align}

The constant a could now be determined experimentally. Usually, however, it is not the energy of a photon gas that is of interest but the power with which the body emits the photons.

Radiant power

In the following, a blackbody with a surface area A is considered, which is in thermodynamic equilibrium at the temperature T. This black body emits photons that move away from the surface at the speed of light c (in the case of a hollow body into the cavity and then through the hole into the environment, or in the case of a solid body directly into the environment). Within the infinitesimal time dt these photons cover a certain distance ds:

\begin{align}
& \text{d}s = c \cdot \text{d}t \\[5px]
\end{align}

Since the photons are emitted from the surface area A, they occupy a certain shell volume around the blackbody within the time dt. Due to the infinitesimal distance, the volume dV of this emitted photon shell results from the product of base area A and height ds:

\begin{align}
& \text{d}V = A \cdot \text{d}s = A \cdot c \cdot \text{d}t \\[5px]
\end{align}

At a given temperature T the energy dU contained in the emitted photon volume dV is given by equation (\ref{st}).

\begin{align}
&\text{d}U = a \cdot T^4 \cdot \text{d}V= a \cdot T^4 \cdot A \cdot c \cdot \text{d}t \\[5px]
\end{align}

Within the time dt the blackbody radiates the energy dU. From this the radiant power Φ can be determined as the emitted radiant energy per unit time:

\begin{align}
\require{cancel}
&\Phi =\frac{\text{d}U}{\text{d}t} = \frac{a \cdot T^4 \cdot A \cdot c \cdot \bcancel{\text{d}t} }{\bcancel{\text{d}t}} =\underbrace{a\cdot c}_{\sigma} \cdot A \cdot T^4 \\[5px]
\end{align}

The constant a and the constant speed of light c can be combined to a new constant, the so-called Stefan-Boltzmann constant σ (not to be confused with the Boltzmann constant k_B!). Thus the radiant power of a black body in thermodynamic equilibrium at a given temperature T results from the following formula:

\begin{align}
\label{bb}
&\boxed{\Phi = \sigma \cdot A \cdot T^4} ~~~~~ \sigma = 5,670 \cdot 10^{-8} \frac{\text{W}}{\text{m²K}^4} \\[5px]
\end{align}

This law was experimentally derived by the physicist Josef Stefan and later mathematically derived by Ludwig Boltzmann. This law is therefore called the Stefan-Boltzmann Law.

If the radiant power Φ at this point is related to the surface area A of the blackbody, then the intensity I is obtained:

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

The Stefan-Boltzmann constant σ, which was initially determined empirically by experiments, is a fundamental physical constant that could actually only be derived from other fundamental constants by quantum mechanics:

\begin{align}
&\boxed{\sigma = \frac{2 \pi^5k_B^4}{15h^3c^2} } \\[5px]
\end{align}

As already explained at the beginning of this article, Blackbodies only exist in the ideal. In reality, therefore, bodies do not radiate with the intensity of a black body but with lower power. This is expressed by the emissivity ε<1:

\begin{align}
&I=\varepsilon \cdot \sigma \cdot T^4\\[5px]
\end{align}

Introduction

Planck’s law of radiation describes the radiation emitted by black bodies. However, there are different forms of representation, which will be discussed in more detail in the following. These differences are mainly due to the different quantities that are considered: e.g. intensity per wavelength or the intensity per solid angle or the energy density. Furthermore, these quantities can be considered either as a function of the wavelength or the frequency of the radiation.

Distribution of spectral intensity as a function of wavelength

The figure below shows the spectral intensity of the emitted radiation of a black body as a function of wavelength for different temperatures. The mathematical description of these curves has already been discussed in more detail in the article on 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{spectral intensity (wavelength form)} \\[5px]
\end{align}

In this form of representation, the term intensity means a surface power density. The intensity thus indicates the radiant power of the black body, which it emits per unit area. Using the illustrative example of a light bulb, this would mean that for the calculation of its intensity, the radiant power is divided by the surface area of the light bulb. With a radiant power of 50 watts and a surface area of 100 cm², this would result in a radiation intensity of 0.5 W/cm² (“0.5 watts per square centimeter”).

Such an indication of the intensity does not, however, allow any statement as to whether the radiation emitted contains short wavelengths or long wavelengths. For our light bulb this would mean that no statement can yet be made on the basis of the intensity whether the light bulb emits its power more in the short-wave range and thus appears bluish or shines more strongly in the long-wave range and thus shines reddish.

In order to be able to examine the distribution of the wavelengths, the entire wavelength spectrum would first have to be divided into many small intervals. These wavelength intervals would then have to be examined separately from each other for their respective power. With the help of filters, such wavelength intervals could be separated from other wavelengths. The intensity with which the wavelength interval under consideration is present in the emitted radiation would then be determined by dividing the measured power by the size of the surface of the radiating body.

In this case, however, the measured power or intensity still depends decisively on the size of the chosen wavelength interval. To put it simply, this means: If you choose a wavelength interval twice as large, then twice as many wavelengths pass the filter and therefore twice as much power is measured (at least if the intervals are chosen very small). Therefore the measured power or intensity is related to the chosen wavelength interval! In this way one gets a constant intensity independent of the width of the wavelength interval. Such an intensity per unit wavelength interval ist finally called spectral intensity.

This spectral intensity is finally shown as a function of wavelength in the diagram. In this form of representation, the area under the curve corresponds to the radiated intensity in the wavelength range under consideration.

To determine the total intensity I with which the black body radiates, the spectral intensity I(λ) must therefore be integrated with respect to the entire wavelength spectrum from zero to infinity (total area under the curve):

\begin{align}
&\boxed{I=\int \limits_{\lambda=0}^\infty I_s(\lambda) ~\text{d}\lambda } ~~~\text{radiation intensity of a black body} \\[5px]
\end{align}

The intensity refers to the power per unit area. The total radiant power P of the black body is finally obtained by integrating the intensity over the entire surface A of the black body:

\begin{align}
&\boxed{P=\int\limits_{(A)} I ~\text{d}A = \int\limits_{(A)} \int \limits_{\lambda=0}^\infty I_s(\lambda) ~\text{d}\lambda ~\text{d}A} ~~~\text{radiant power of a black body} \\[5px]
\end{align}

If the black body emits radiation that is evenly distributed over its surface (isotropic black body), the intensity is equal at every point on the surface. In this case, the total radiated power is simply the product of intensity I and surface area of the black body A:

\begin{align}
&\boxed{P=I \cdot A} ~~~\text{radiant power of an isotropic black body} \\[5px]
\end{align}

The solution of the integrals shown above finally leads to Stefan-Boltzmann law, which describes the radiant power of a body as a function of its temperature. In the linked article you will find more information especially on this topic.

Distribution of spectral intensity as a function of frequency

Since wavelength λ and frequency f are related by the speed of propagation c (λ=c/f), Planck’s law of radiation can also be expressed as a function of frequency. However, the wavelength λ must not simply be replaced by the expression 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 dλ into corresponding frequency intervals df!

Only the radiated intensities are really comparable, but not the spectral intensities. The emitted intensity dI(λ) in a wavelength range between λ and λ+dλ is calculated by the product of the spectral intensity dI(λ) and the wavelength interval dλ (“area under the graph”):

\begin{align}
\label{c}
&\text{d} I(\lambda) =I_s(\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 dλ must therefore be assigned a corresponding frequency interval df. This is done by deriving the function f=c/λ with respect to the variable λ:

\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 df>0 results in a decrease of wavelength by dλ<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=c/λ. 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.

Distribution of the specific spectral intensity (spectral flux)

To experimentally determine the emitted spectral intensity of a surface element of a black body, the emitted radiation within a considered wavelength interval would have to be measured with a detector. In the article Blackbody radiation the realization of a blackbody by means of an opening leading into a cavity has already been described in detail. In this case, the surface element to be examined would correspond to the area of the opening, which completely absorbs incident radiation and thereby emits blackbody radiation itself (also called cavity radiation). The radiation propagates spherically into the half-space (see figure below).

Influence on the measured radiant power: The ” direction of view”

For the measurement of the emitted radiant power, however, it is not sufficient to place the detector somewhere in a fixed position where only a part of the radiation is detected. One has to detect the radiation around the whole half space to actually measure the total radiant power. However, the radiant power of the surface element (here: area of the opening) is not the same in all directions! Thus, one cannot assume that the power measured with the detector at one place is the same at another place.

Everyday experience already shows that the radiant power of a (flat) surface element is dependent on direction. Imagine a large glowing metal plate. If you look at this plate frontally, you will perceive a relatively strong thermal radiation (this thermal radiation is part of blackbody radiation!) The radiant power is therefore relatively high when looking at the plate from the front. If you look at the glowing plate from the side, however, the thermal radiation appears less intense. This means that the radiant power is lower. This can be explained by the fact that the plate appears much smaller when viewed from the side and thus has a less effective radiating surface. Obviously, the area appearing in the viewing direction, the so-called projected area, has an influence on the perceived or measured radiant power.

The projected area can be clearly illustrated by imagining the detector as a “flashlight”. The shadow that the illuminated surface element then creates on a screen placed behind it corresponds to the projected area.

Influence on the measured radiant power: The solid angle

On the one hand, the direction in which the the detector is pointing is therefore relevant for the measured radiation power and on the other hand, of course, the area the detector occupies (its “field of view”, so to speak). The latter is expressed by the so-called solid angle. A solid angle Ω is defined by the ratio of an area on a sphere A to the square of the radius of the sphere r²:

\begin{align}
\label{r}
&\boxed{\Omega =\frac{A}{r^2}} ~~~\text{solid angle} \\[5px]
\end{align}

The solid angle can be understood analogously to the radian. In two-dimensional space, the radian is the ratio of the length of an arc to the radius of the circle. In three-dimensional space, the solid angle is the ratio of a segment of the sphere to the square of the radius of the sphere.

If a segment includes the entire sphere surface with an area of 4π⋅r², then a solid angle of 4π is obtained. The angle occupies the entire space of the sphere, so to speak. If the area is only half the entire sphere surface, the solid angle is 2π. For such an solid angle of 2π, the detector would just cover the entire half-space of radiation and detect the entire radiant power of the surface element. The detector would also have to be hemispherical.

Note: As long as a considered segment is part of a spherical surface, a solid angle can be defined for any arbitrarily shaped surface according to the equation (\ref{r}). The outline of the surface does not necessarily have to form a circle as is often suggested in order to specify a solid angle! In other words: Our detector does not need a circular sensor, it can also be rectangular or otherwise arbitrarily shaped. It only needs to be part of a spherical surface. For large radii and small surfaces, however, the curvature of the surface is so small that flat surfaces can be used in very good approximation.

Spectral intensity per solid angle (spectral flux)

If one examines the emitted radiant power of the opening of the cavity acting as a blackbody with a detector, then one observes two things:

1. the measured radiant power dP is proportional to the area projected in the direction the detector is pointing dA_p,
2. the measured radiant power dP is proportional to the solid angle dΩ the detector covers (as long as small solid angles are considered).

Point (1) has already been explained: the larger the radiating surface facing the detector, the greater the power. Point (2) can also be clearly understood. If two very small detectors are placed directly next to each other, they both register the same power. If one takes both detectors together in thought, which then take up twice the solid angle, then one obviously receives twice the power. Strictly speaking, this applies only as long as the two solid angles are very small and the surface segments are close together, so that they do not differ in the direction of pointing. Otherwise, according to point (1), one of the detectors would measure a slightly different radiant power.

Furthermore, the measured radiant power is of course dependent on the size of the wavelength interval dλ, which the filter of the detector allows to pass. A wavelength interval twice as large means that twice as much radiation passes through the filter. The measured power is therefore twice as large. This proportionality between power and wavelength interval only applies to very small wavelength intervals. For a given wavelength range between λ and λ+dλ, the above relationships can be mathematically represented as follows:

\begin{align}
& \text{d}P \sim \text{d}\Omega \cdot \text{d}A_p \cdot \text{d}\lambda \\[5px]
\label{p}
&\frac{ \text{d}P}{\text{d}\Omega \cdot \text{d}A_p \cdot \text{d}\lambda} =\text{konstant} = B_s \\[5px]
\end{align}

Thus, if the radiant power of a black body is no longer related only to a wavelength interval and to a surface element as with the spectral intensity I_s, but next to the wavelength interval dλ to the solid angle dΩ and to the projected area dA_p, then one also obtains a constant quantity. This quantity is called specific spectral intensity or spectral flux B_s. The spectral flux thus represents the radiant power emitted per unit wavelength interval and per unit solid angle and per unit projected area.

Spectral flux is the radiant power of a projected surface element per unit of solid angle and per unit of wavelength interval!

Applied to our example this means: No matter at what distance and with what orientation the detector faces the surface element of the black body, the same value B_s is measured for a certain wavelength interval. Thus, the radiation of a blackbody can also be characterized by exactly this quantity B_s. In practice, this quantity has the advantage that one does not need the entire emitted radiation of a surface element to characterize the radiation, as is the case with the spectral intensity. With the spectral flux, only a part of the radiation within a certain solid angle needs to be examined.

The following formula applies to the spectral flux B_s depending on the wavelength or frequency:

\begin{align}
\label{bs}
&\boxed{B_s(\lambda) = \frac{2 h c^2}{\lambda^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\lambda k_B T}\right)-1} } ~~~\text{wavelength form} \\[5px]
\end{align}

\begin{align}
&\boxed{B_s(f) = \frac{2 h f^3}{c^2} \cdot \frac{1}{\exp\left(\dfrac{h f}{k_B T}\right)-1} } ~~~\text{frequency form} \\[5px]
\end{align}

Relationship between spectral intensity and spectral flux

If one looks at the formulas (\ref{planck}) and (\ref{bs}), they obviously differ only by the factor π. How this relationship comes about will be explained in the following. First of all, it should again be pointed out that the spectral flux B_s indicates the radiation intensity per unit solid angle, whereas the spectral intensity I_s refers to the total radiation intensity. Since an (infinitesimal) surface element radiates spherically into half-space, the spectral flux B_s must be added up over the entire half-space, so to speak, to calculate the spectral intensity I_s. For the spherical half-space the solid angle is 2π. Therefore, shouldn’t both quantities differ by 2π instead of only by π?

When summing-up over the half-space, it must be noted that the spectral flux does not refer to the full area of the surface element, but only to the projected area. However, both areas are only identical for the special case when the surface element is viewed from the front. Otherwise the projected area is always smaller. On average over all solid angles, the projected area is half as small as the area itself. Therefore the result is not 2π, but only π. In the following, this will also be shown mathematically.

Mathematically, the summation corresponds to the integration of the spectral flux over the radiation hemisphere:

\begin{align}
\text{d}P &= \text{d}P \\[5px]
I_s \cdot \text{d}A \cdot \text{d}\lambda &= \int\limits_{(A)} B_s \cdot \text{d}A_p \cdot \text{d}\Omega \cdot \text{d}\lambda \\[5px]
\label{l}
I_s \cdot \text{d}A &= \int\limits_{(A)} B_s \cdot \text{d}A_p \cdot \text{d}\Omega \\[5px]
\end{align}

First of all, a solid angle dΩ must be assigned a direction so that the projected area dA_p can be determined. For this purpose a segment of a surface dσ required for the solid angle to be determined is best described by spherical coordinates. A single point on the surface of a sphere is fully defined in by specifying two angles. The angle φ is the angle in the x-y plane and the angle θ is the angle to the z axis.

If one now allows both angles to vary within a certain range dφ and dθ, the result is a spherical segment dσ. The area of this segment can be determined as follows:

\begin{align}
& \text{d}\sigma = r^2 \cdot \sin(\theta) ~ \text{d}\theta \cdot \text{d}\varphi \ \\[5px]
\end{align}

With the definition of a solid angle dΩ as the ratio of the area dσ to the square of the radius r², the solid angle is determined as follows:

\begin{align}
\label{x}
& \text{d}\Omega = \frac{\text{d}\sigma}{r^2}= \sin(\theta) ~ \text{d}\theta \cdot \text{d}\varphi \ \\[5px]
\end{align}

The area dA_p projected in the direction of the position vector results from the cosine of the angle θ:

\begin{align}
\label{y}
& \text{d}A_p = \text{d}A \cdot \cos(\theta) \ \\[5px]
\end{align}

If the equations (\ref{x}) and (\ref{y}) are used in equation (\ref{l}), then the following relationship results:

\begin{align}
\require{cancel}
I_s \cdot \text{d}A &= \int \limits_{(A)} B_s \cdot \text{d}A_p \cdot \text{d}\Omega \\[5px]
I_s \cdot \bcancel{\text{d}A} &= \int \limits_{(A)} B_s \cdot \bcancel{\text{d}A} \cdot \underbrace{\cos(\theta) \cdot \sin(\theta)}_{=\tfrac{1}{2}\cdot \sin(2\theta)} ~ \text{d}\theta \cdot \text{d}\varphi \\[5px]
I_s &= B_s \cdot \tfrac{1}{2} \int \limits_{(A)} \sin(2\theta) ~ \text{d}\theta \cdot \text{d}\varphi \\[5px]
\end{align}

Note that, as explained above, B_s is constant everywhere on the surface of the sphere, and therefore is a constant quantity when integrating. Only the integration over the hemisphere has to be done, i.e. within the limits of φ=0…2π and θ=0…π/2:

\begin{align}
I_s &= B_s \cdot \tfrac{1}{2} \int \limits_{\varphi=0}^{2\pi} \int \limits_{\theta=0}^{\tfrac{\pi}{2}} \sin(2\theta) ~ \text{d}\theta \cdot \text{d}\varphi \\[5px]
&= B_s \cdot \tfrac{1}{2} \int \limits_{\varphi=0}^{2\pi} \underbrace{\left[ -\frac{1}{2} \cos(2\theta) \right]_{\theta=0}^{\tfrac{\pi}{2}}}_{=1} \text{d}\varphi \\[5px]
&= B_s \cdot \tfrac{1}{2} \cdot \int \limits_{\varphi=0}^{2\pi} \text{d}\varphi \\[5px]
&= B_s\cdot \tfrac{1}{2} \cdot \left[\varphi \right]_{\varphi=0}^{2\pi} \\[5px]
&= B_s\cdot \tfrac{1}{2} \cdot 2\pi \\[5px]
&= B_s \cdot \pi \\[5px]
\end{align}

Spectral intensity I_s and spectral flux B_s differ only by the factor π. This applies to both the wavelength form and the frequency form:

\begin{align}
\label{gg}
\boxed{I_s = B_s \cdot \pi} \\[5px]
\end{align}

Distribution of the spectral energy density of cavity radiation

The spectral flux makes it possible to draw conclusions about the so-called spectral energy density u_s of cavity radiation. Spectral energy density means the radiant energy contained in the volume of the cavity per unit wavelength interval.

For this we again look at a hollow object with a tiny hole. All radiation coming in through the opening is absorbed by the inner walls with every reflection until after a few reflections all radiation is absorbed. The hole is by definition a blackbody that absorbs all incident radiation. However, depending on the temperature of the cavity, the inner walls themselves emit radiation. In thermodynamic equilibrium, the walls absorb as much radiant energy as they emit, so that neither further heating nor cooling of the cavity or body occurs.

Inside the cavity thus a kind of photon gas is formed (the Derivation of the Stefan-Boltzmann law is based exactly on this idea of photon gas). The radiant energy or energy density present in such a cavity can be determined from the blackbody radiation emitted through the opening.

For this we consider a surface element dA at the opening to the cavity. The surface element is thus part of the photon gas. As usual with gases, the photons in the photon gas move in different directions. Half of the radiation thus escapes into free space, where it can be detected. The other half of the photons moves back into the cavity.

At any solid angle dΩ radiation is emitted from the projected area dA_p. The photons moving at the speed of light c cover the distance dl within the time dt:

\begin{align}
&\text{d}l = c \cdot \text{d}t \\[5px]
\end{align}

Within the time dt the photons thus occupy the following volume dV:

\begin{align}
&\text{d}V = \text{d}A_p \cdot \text{d}l = \text{d}A_p \cdot c \cdot \text{d}t \\[5px]
\end{align}

The radiant energy dU in the wavelength interval dλ, which is contained in the emitted radiation, can be determined by the spectral flux B_s and the time interval dt [see equation (\ref{p})].

\begin{align}
\require{cancel}
&\text{d}U = \overbrace{B_s \cdot \text{d}A_p \cdot \text{d}\lambda \cdot \text{d}\Omega}^{\text{d}P} \cdot \text{d}t \\[5px]
\end{align}

For the energy density u as the ratio of energy dU and volume dV the following formula applies:

\begin{align}
\require{cancel}
&u = \frac{\text{d}U}{\text{d}V} = \frac{B_s \cdot \cancel{\text{d}A_p} \cdot \text{d}\lambda \cdot \text{d}\Omega \cdot \cancel{\text{d}t}}{ \cancel{\text{d}A_p} \cdot c \cdot \cancel{\text{d}t} } \\[5px]
&u = \frac{\text{d}U}{\text{d}V} = \frac{B_s \cdot \text{d}\lambda \cdot \text{d}\Omega}{c} \\[5px]
\end{align}

The energy density in this form still depends on the wavelength interval and the solid angle. Therefore, the energy density is usually expressed as spectral energy density u_s, i.e. as energy density per unit wavelength interval. Furthermore, in the case of a volume element, it makes no sense to relate the energy density to a solid angle. After all, the entire volume element is considered with its energy density. Therefore the energy density must be related to the “entire volume”, i.e. to a full sphere with a solid angle of dΩ=4π:

\begin{align}
&u_s = \frac{u}{\text{d}\lambda} = \frac{B_s}{c} \cdot \text{d}\Omega = \frac{B_s}{c} \cdot 4\pi \\[5px]
&\boxed{u_s = \frac{4\pi}{c} \cdot B_s} \\[5px]
\end{align}

Finally, the following relationship exists between the spectral energy density and the spectral intensity [see equation (\ref{gg})]:

\begin{align}
&\boxed{u_s = \frac{4}{c} \cdot I_s} \\[5px]
\end{align}

This relationship between the spectral energy density and the (specific) spectral intensity applies not only to the wavelength form but also to the frequency form. In this case the spectral energy density then means the energy density per unit frequency interval.

\begin{align}
&\boxed{u_s(\lambda) = \frac{8 \pi h c}{\lambda^5} \cdot \frac{1}{\exp\left(\dfrac{h c}{\lambda k_B T}\right)-1} } ~~~\text{wavelength form} \\[5px]
\end{align}

\begin{align}
&\boxed{u_s(f) = \frac{8 \pi h f^3}{c^3} \cdot \frac{1}{\exp\left(\dfrac{h f}{k_B T}\right)-1} } ~~~\text{frequency form} \\[5px]
\end{align}

The total energy density inside the cavity of a black body would finally be obtained by integrating these equations over the entire wavelength or frequency range from zero to infinity. This total energy density is only dependent on temperature and is spatially constant throughout the cavity. If the energy density would differ in two points, then more energy would be contained in one volume element than in the other. The volume element with the higher energy density would radiate “more” than the other. However, the photon gas or the entire cavity would then not be in thermodynamic equilibrium.

Furthermore, the energy density is not dependent on the volume of the cavity. The spectral distribution of the energy density is solely determined by the temperature. So whether the cavity is small or large is irrelevant for the (spectral) energy density contained in it!

The energy density inside the cavity of a hollow black body depends only on the temperature!

Blackbody radiation

The emitted wavelength spectrum of a blackbody as shown in the figure below 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.

Planck could derive the following formula for the distribution of the spectral intensity I_s as a function of wavelength λ. 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 (wavelength form)} \\[5px]
\end{align}

Intensity means the radiant power of the black body emitted per unit area (surface power density). If, as in this case, the intensity is related to the wavelength interval within which the power is emitted, this is called the spectral intensity. If the spectral intensity is plotted over the wavelength, then in such a diagram the area under the curve corresponds to the emitted intensity in the wavelength range under consideration.

There is a clear relationship between the wavelength λ of a radiation and its frequency f. This relationship results from the speed of propagation of the radiation, which in this case corresponds to the speed of light c (c=λ⋅f). Therefore the spectral distribution of the intensity I_s can also be expressed as a function of frequency:

\begin{align}
\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} } ~~~\text{Planck’s law (frequency form) } \\[5px]
\end{align}

In the article Different forms of Planck’s law, the derivation of this frequency form from the wavelength form and other forms of Planck’s law are discussed in detail.

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 h⋅f/(k_B⋅T) 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 ∫₀^∞ x³/(exp(x)-1) dx cannot be solved so easily in a conventional way. But a look at the mathematics formula collection shows that the result is π⁴/15. 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 σ (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 Φ 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 ε<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 λ_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 λ. 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 λ_max can be determined as a function of the temperature by solving equation (\ref{max}) for λ_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 λ_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=c/λ applies, it does not apply in this particular case f_max=c/λ_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 dλ 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]
&1-0,5-0,333-0,25-0,2-0,167-0,143-0,125-0,111-0,1 \\[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=c/λ_max is obtained when using the frequency form of the spectral intensity.

Introduction

In the article Blackbody radiation it has already been explained in detail why every object emits radiation above absolute zero. This radiation is also called thermal radiation. Thermal radiation is caused by the motions of atoms that emit electromagnetic waves, i.e. radiation.

Thermal radiation can not only be proven by the fact that it is able to heat other objects, as one could derive from the term thermal radiation. At sufficiently high temperatures, the radiated wavelength spectrum shifts into the visible range and can thus be observed directly by the human eye. The reddish annealing of a heated metal rod during forging is, for example, the result of such a visible thermal radiation.

The visible glow of the filament of a light bulb at over 3000 °C is also a typical example which can be traced back to the phenomenon of thermal radiation. More than 90 % of the energy is radiated in the non-visible infrared range and can therefore only be perceived as heat. The remaining part, however, is in the visible wavelength spectrum and can be directly observed as a yellowish glow (a small part is also radiated as ultraviolet light).

Radiant power

As a heat lamp or a glowing steel block clearly shows, emitting radiation is obviously associated with emitting energy. How much radiant energy ΔQ an object emits per unit time Δt, i.e. how high its radiant power Φ is, depends mainly on the temperature T, but also on the area of the surface A as well as on the radiative property of the body (the so-called emissivity ε

\begin{align}
&\boxed{\Phi = \frac{\Delta Q}{\Delta t}}=\Phi(T,A,\varepsilon) \\[5px]
\end{align}

Note that the radiant power (also called radiant flux) does not only refer to the thermal energy in the infrared spectrum or to the radiated energy in the visible wavelength range, but to the energy related to the entire wavelength spectrum, i.e. to the entire energetic radiation!

If, for example, an incandescent light bulb is operated at low current, the temperature of the filament is correspondingly low. The light bulb does not only glow less but it also does not heat up as much. Overall, the radiant power is relatively low at low temperatures. With a large current, on the other hand, the filament heats up strongly and the temperature is correspondingly high. It then not only glows intensely yellow but also radiates infrared radiation to a high degree, which is clearly noticeable as heat. The higher the temperature of a body, the higher the radiant power is!

Note: The fact that the filament is weakly reddish at low temperature and bright yellow at high temperature is due to the wavelength spectrum emitted, which shifts to the yellowish range with increasing temperature (more information on this in the article Blackbody radiation).

In addition to the temperature, the area of the surface of the radiating body also influences the radiant flux. The larger the surface, the more atoms can vibrate and emit radiation. Note that radiation emitted by the atoms inside the body is directly reabsorbed by the surrounding atoms. Thus only the atoms on the surface are relevant for the radiation of the electromagnetic waves. If the surface is twice as large, the radiant power should therefore be twice as high.

More detailed studies by the physicists Josef Stefan and Ludwig Boltzmann at the end of the 19th century showed that the radiant power is actually directly proportional to the surface area of the emitting object. The influence of temperature on radiant power, on the other hand, is far greater. It increases with the fourth power of the absolute temperature. A doubling of the temperature from e.g. 1000 K to 2000 K thus increases the radiant power by a factor of 16! For an ideal thermal radiator, a so-called blackbody, the radiant power Φ_ideal results as a function of temperature T and surface area A as follows:

\begin{align}
&\Phi_{ideal} \sim A \cdot T^4 \\[5px]
&\boxed{\Phi_{ideal} = \sigma \cdot A \cdot T^4} ~~~~~ \sigma = 5,670 \cdot 10^{-8} \frac{\text{W}}{\text{m²K}^4} \\[5px]
\end{align}

The Stefan-Boltzmann law states that the radiant power of an object in thermal equilibrium is proportional to the fourth power of temperature and directly proportional to its surface area!

The proportionality factor σ is called the Stefan-Boltzmann constant and is a universal constant, i.e. it does not depend on the material of the radiating object as long as the body absorbs all incident radiation and can therefore be regarded as a blackbody.

A black-body is an ideal thermal radiator that absorbs all incident radiation and therefore radiates at maximum power!

Why a black body is not only a perfect absorber of radiation but also a perfect emitter of radiation will be explained later.

If the radiant power Φ of the black body is related to its surface area A, then one also speaks of the so-called intensity I (surface power density). The intensity indicates the strength of the radiant power per unit area. The intensity of black-body radiation depends only on temperature:

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

The Stefan-Boltzmann law can be derived from Planck’s law or from a thermodynamic approach. You can read more about this in the linked articles.

Kirchhoff’s law of thermal radiation

In the following a blackbody is considered, which is irradiated by a heat lamp. By definition, the blackbody will absorb all incident radiation. The absorbed energy leads to an increase in temperature and the blackbody begins to emit more and more radiation. Finally, over time, a thermal equilibrium will be reached in which the temperature no longer rises. In thermodynamic equilibrium, the radiant energy emitted within a certain time (emitted radiant power Φ_e) must therefore be the same as the absorbed radiant energy (absorbed radiant power Φ_a):

\begin{align}
\label{kirch}
&\boxed{\Phi_a \overset{!}{=} \Phi_e} \\[5px]
\end{align}

This radiation equilibrium between emitted radiation and absorbed radiation basically applies to every body in thermal equilibrium, including non-ideal black bodies that do not radiate at maximum power. After all, a constant temperature and thus a thermal equilibrium will be reached for every object after a certain time, in which emission and absorption must take place to the same extent. This law is also called Kirchhoff’s law of radiation.

Kirchhoff’s law of radiation states that in thermal equilibrium of a radiating body, emission and absorption take place to the same extent!

In the same way as the emitted radiant power increases with the fourth power of the temperature, the absorbed radiant power must also increase with the fourth power of the temperature. Otherwise there would be no thermal equilibrium. The law with which an ideal blackbody emits radiation according to the Stefan-Boltzmann law must therefore also apply to the absorbed radiation in thermal equilibrium!

\begin{align}
&\boxed{\Phi_{a,ideal} = \Phi_{e,ideal} = \sigma \cdot A \cdot T^4} \\[5px]
\end{align}

Real bodies

This situation can now be transferred to real objects, which are not perfect black bodies. With real objects a certain part of the radiation is always reflected and not completely absorbed like with black bodies. The absorbed radiant power of a real body will therefore be a factor α<1 less than that of an ideal black body. This factor α, which describes the absorbed part of the incident radiation and thus indicates the absorbed radiant power in comparison to an ideal black body, is also referred to as absorptivity α.

\begin{align}
&\boxed{\Phi_{a,real} = \alpha \cdot \Phi_{a,ideal} } ~~~~~\alpha<1 \\[5px]
&\Phi_{a,real} = \alpha \cdot \sigma \cdot A \cdot T^4 \\[5px]
\end{align}

Absorptivity α is the absorbed radiation portion of an incident radiation, i.e. the absorbed radiant energy of a real body compared to an ideal blackbody!

According to Kirchhoff’s law of radiation, the emitted radiant power in thermal equilibrium is just as high as the absorbed radiant power. A real body that has a lower absorption power than an ideal black body will therefore also have a lower emission power to the same extent!

The emission power of a real body in thermal equilibrium is as much lower as the absorption power!

However, this also means that a body that absorbs to the maximum also emits to the maximum. A blackbody is therefore not only an ideal absorber of radiation but also a perfect emitter of radiation!

A blackbody is an ideal thermal radiator with the maximum possible radiant power!

In purely formal terms, the emission power of a real body can also be expressed with a factor which then indicates the ratio of the emitted radiation of the real body compared to an ideal black body! This factor is referred to as emissivity ε.

\begin{align}
&\boxed{\Phi_{e,real} = \varepsilon \cdot \Phi_{e,ideal} } ~~~~~\varepsilon<1 \\[5px]
&\Phi_{e,real} = \varepsilon \cdot \sigma \cdot A \cdot T^4 \\[5px]
\end{align}

An emissivity of e.g. ε=0.9 means that the considered object has 90 % of the radiant power of an ideal thermal radiator (perfect black body). Many non-metallic objects have an emissivity of more than 90 % and can therefore be regarded as blackbodies in very good approximation with respect to their emitted radiation. The emissivity for an ideal blackbody is ε=1.

The emissivity represents the ratio of the radiation actually emitted by a real body to that of an ideal thermal radiator, a perfect blackbody!

Although the absorptivity and the emissivity are formally differentiated, in thermal equilibrium this distinction is obsolete, since both quantities have the same value according to Kirchhoff’s law of radiation (\ref{kirch}):

\begin{align}
\require{cancel}
&\Phi_{e,real} \overset{!}{=} \Phi_{e,real} \\[5px]
&\alpha \cdot \bcancel{\sigma \cdot A \cdot T^4} = \varepsilon \cdot \bcancel{\sigma \cdot A \cdot T^4} \\[5px]
&\boxed{\alpha = \varepsilon} \\[5px]
\end{align}

In thermal equilibrium, the absorptivity is equal to the emissivity!

Remark

Real bodies which have a constantly lower (spectral) emission power in the entire radiated wavelength spectrum compared to an ideal black body are also called gray bodies. Gray bodies therefore have an absorptivity or emissivity that is not dependent on the wavelength.

In some cases, however, a body will also absorb or emit radiation to different degrees depending on the wavelength. The absorptivity or emissivity then depends on the wavelength. Such bodies are called selective absorbers or selective radiators.

Thermal imaging cameras are based on the same principle as the pyrometer. These cameras capture the radiation spectrum of an object, which then allows conclusions to be drawn about the temperature (see article Black-body radiation). In comparison to an infrared thermometer, however, the thermal imaging camera does not only detect the temperature at a single point but in a wide optical range and displays it.

Since the infrared radiation emitted by objects is not visible to the human eye, the radiation is converted into a spectrum visible to us and displayed. Since the image displayed by the camera does not correspond to the radiation actually emitted, one speaks of false colors in this context.

The figure below shows the thermal image of a driven car taken by a thermal imaging camera. The hot spots can be seen in white and the cold spots in dark blue.

Thermal imaging cameras are always suitable when the distribution of temperature is to be determined. They are used, for example, by the fire brigade to quickly identify potential sources of heat or fire.

The thermometers presented in the previous articles must directly touch the object from which the temperature is to be determined. This can be a disadvantage in many cases, for example with very hot materials such as metal melts or corrosive liquids. In such cases non-contact measuring methods are used.

The so-called pyrometer (IR thermometer) detects the thermal radiation (infrared radiation) from objects, which each body emits due to its temperature (see article Black-body radiation). The radiation spectrum recorded by the measuring device thus allows conclusions to be drawn about the temperature of the objects. With the help of such a radiation measuring device, the object to be measured no longer has to be touched directly in order to measure the temperature.

Pyrometers (infrared thermometers) use the heat radiation of objects invisible to the human eye to determine the temperature!

Infrared thermometers are often equipped with a laser. However, this laser has only a targeting function. In this way, it is easier to determine the exact location from which the temperature is measured.