Introduction

As explained in the article on solidification conditions, the presence of nuclei is a basic requirement for liquids to solidify. The forming of a nucleus is called nucleation. In principle, the nucleation can take place in two different ways. Either as heterogeneous nucleation or homogeneous nucleation.

Heterogeneous nucleation

If foreign particles serve as nuclei (i.e. particles of a different substance than the ones of the melt itself), one also speaks of heterogeneous nuclei. Heterogeneous nuclei can be caused by impurities in the melt. But the vessel walls in which the melt is held also serve as heterogeneous nuclei with their foreign particles.

Such foreign particles are the most common reason in everyday life why undercooled liquids are generally not observed and one could therefore prematurely conclude that only the condition of falling below the solidification temperature (melting point) would be necessary for a solidification process.

The formation of heterogeneous nuclei preferentially occur at places where the foreign particles are located (e.g. on the vessel wall). The probability of nucleation is therefore not homogeneously distributed over the melt but concentrates on the contaminated areas. This is the reason why it is called “heterogeneous” nucleation.

Nucleation caused by foreign substances is referred to as heterogeneous nucleation!

Homogeneous nucleation

Not only foreign particles but also the own particles of the substance can serve as nuclei. In this context one also speaks of homogeneous nucleation.

Homogeneous nuclei can be formed, for example, by the atoms arranging themselves randomly in the melt in the form of an unit cell or lattice structure. Due to the large number of particles in a melt, the formation of such homogeneous nuclei is not as unlikely as it may sound at first. Homogeneous nuclei can also be caused by unmelted residues, which is, however, rather unlikely in molten metals.

The probability of a random arrangement of atoms in the form of a lattice structure is homogeneously distributed over the entire melt. So there are no preferred places where the particles come together. This is the reason why it is called “homogeneous” nucleation.

The nucleation caused by own particles is called homogeneous nucleation!

Influencing the nucleation

With the knowledge of the necessary prerequisites for the solidification process, the formation of microstructures can now be specifically controlled.

Thus, with increasing undercooling (supercooling), the probability that the melt will form heterogeneous nuclei increases to a certain extent. After all, strong supercooling means greater inertia of the particles and thus the probability that self-formed nuclei will dissolve decreases.

A large undercooling usually leads to an increased homogeneous nucleation. The melt then begins to solidify at many nuclei simultaneously. This creates a very fine-grained microstructure with many grain boundaries, which is characterised by very good strength and toughness.

The stronger the supercooling, the more nuclei form in the melt and the finer the resulting grains!

Another possibility to achieve a fine-grained structure is the targeted admixture of foreign particles, which then serve as heterogeneous nuclei. This process is called seeding.

For the seeding of nodular cast iron, for example, a compound of iron and silicon is often used. However, since the added foreign particles tend to dissolve in the melt after some time, the seeding process should take place immediately before solidification or during the casting of the metal.

A fine-grained structure can also be achieved by seeding the melt with specifical foreign particles!

The formation of the microstructure can basically be divided into two phases: nucleation and nucleus growth. Both the phase of nucleation and the phase of nucleus growth can be specifically influenced in order to shape the structure according to the properties desired. Due to their complexity, these phases are dealt with in more detail in separate articles.

Polycrystal

A metallic material generally does not have a uniform lattice orientation (see crystallographic defects). Exceptions to this are single crystals, which, however, can only be produced at great technical expense (more on this later).

In general, metals have microscopic areas, in which the orientation of the lattice structure change from area to area (the term lattice orientation must not be confused with the term lattice structure!). These small zones of constant lattice orientation are called grains or crystallites. The different grains are separated by relatively structureless regions, which are only a few atomic distances wide. These regions are called grain boundaries.

Grains (crystallites) are small areas within a crystalline material which have a uniform lattice orientation. Individual grains are separated by grain boundaries!

The light micrograph above shows iron in an almost carbon-free state. One can see the individual grains (white areas) with their grain boundaries and silicate inclusions (dark spots).

One calls such a microscopically visible structure in the interior of a material with its grains, grain boundaries, precipitates, etc. also microstructure.

Microstructures are the microscopic structure of a substance!

In a micrograph, microstructures can be seen. For this, however, the sample must be specially prepared in most cases. In particular by polishing and etching (for example with alcoholic nitric acid). Since the different lattice orientation of the grains is accompanied by a different reflection behavior, they are clearly distinguishable under a microscope. Such a crystal, which is characterized by many small crystallites (grains), is also called polycrystal.

In cases such as polycrystalline silicon solar cells or galvanized sheet, the individual grains are visible even to the naked eye. The picture below shows a galvanized component made of steel and a polycrystalline silicon solar cell.

It is often useful to draw a micrograph only schematically to make the essential structure of the material visible. The figure below shows the schematic microstructure of a polycrystalline material.

Solidification process

A grainy microstructure as seen in the upper section is typical for metals. The reason for this lies in the solidification process. Liquid metals usually do not start to solidify in a single point. Rather, the solidification process starts at many points of the melt simultaneously. Such solidification points are generally referred to as nuclei. The formation of the crystal structure is also called crystallization.

The solidification process in crystalline materials is also referred to as crystallization!

Each nucleus initiated the formation of the lattice structure. The spatial orientation of the lattice is different from nucleus to nucleus, depending on how the nucleus is aligned in the melt. Each nucleus will later become a grain with a uniform lattice orientation. Where the grains collide after crystallization, the grain boundaries arise.

Grain boundaries are among the two-dimensional crystallographic defects, since they interfere with the uniform lattice orientation. However, this does not have to be a disadvantage, as one could prematurely draw from the term “defect”. Grain boundaries even contribute in particular to increase the strength (grain boundary hardening), since the deformation mechanism is blocked at the grain boundaries. Fine-grained metals with many grain boundaries are thus characterized by a relatively high strength combined with good toughness.

Fine-grained microstructures usually have good toughness and high strength at the same time!

This raises the question of how one can influence the structure of a metal so that many grains are formed in the structure. To clarify this question, the process of solidification needs to be considered more closely. Therefore, in a separate article, the conditions of solidification are dealt with in more detail.

Monocrystal (single crystal)

In contrast to a polycrystal, however, a monocrystal consists of only one single grain. At the atomic level, such a  single crystal  is characterized by a uniform lattice orientation. This means in particular that the microstructure has no grain boundaries.

A single crystal consists of only one single grain with a certain spatial lattice orientation!

Single crystals can be specially grown out of the melt by certain methods. The solidification process is not initiated simultaneously as usual in many places at the same time but only triggered at a single point (on a so-called seed crystal). The melt then solidifies with a uniform lattice orientation.

Such a principle is used, for example, as a so-called Czochralski process in chip manufacture. There, single crystals are needed so that the electrical conductivity is not adversely affected by grain boundaries or other lattice orientations.

But also as a construction material single crystals are used in special cases. For example, as high-strength turbine blades (nickel superalloy) in high-temperature range. Especially in high-temperature applications, a polycrystalline material has the disadvantage that the grain boundaries start to shift above the recrystallization temperature (called grain boundary sliding). The material “softens”, so to speak. Consequently, to prevent this “softening”, a high temperature material must not have grain boundaries, especially at high load.

Introduction

In the article homogeneous nucleation or heterogeneous nucleation, the formation of nuclei was explained in more detail. If such a stable nucleus emerges from the undercooled melt, then further atoms will accumulate from the melt (by releasing crystallisation heat). The nucleus now begins to grow and initiates the stage of nucleus growth or crystal growth.

The particles are aggregated by diffusion processes through the melt, where they hit the surface of the solidified crystal. The atoms diffuse there at energetically favourable locations. The direction of heat dissipation at the solidification front has a decisive influence on the shape of the growing crystal.

Depending on whether the heat of crystallisation is dissipated via the solidified crystal or the adjacent melt, a distinction is made between polygonal crystal growth and dendritic crystal growth. These two types of crystal growth are described in more detail below.

Polygonal crystal growth

If, for example, a nucleus forms on the surface of a mold wall, the heat of crystallisation released at the growth front is often dissipated via the solidified crystal and then via the mold wall.

This is due to the fact that the melt generally has a higher temperature than the mold wall and heat dissipation finally always takes place in the direction of decreasing temperature (positive temperature gradient in the direction of the melt).

This is also the case when the melt is relatively slightly supercooled. The crystal then grows in an area with a higher temperature than itself. A possible (premature) branching of the crystal would then melt again. Therefore, the solidification front remains relatively flat. This is also referred to as a stable growth front.

Depending on how strong the heat dissipation via the crystal is, rather roundish crystals (globulites) or elongated crystals grow (columnar crystals).

In polygonal (stable) crystal growth, the heat of crystallization is dissipated via the crystal itself.

Excursus: Microstructure of a casting

Polygonal crystal growth also explains the typical three-zone microstructure of a solidified casting block (primary microstructure). Such a casted block is also called ingot. In the vicinity of the mold wall, a very fine-grained structure with roundish grains (globulites) forms due to the strong isotropic undercooling caused by the cool mold wall (zone I).

Subsequently, a zone with elongated grains (columnar crystals) is obtained due to the strongly directed, anisotropic heat dissipation towards the mold wall (transcrystalline zone II). The relatively low (isotropic) cooling or undercooling inside the casting block produces a very coarse-grained microstructure (zone III).

By subsequent heat treatment, this heterogeneous structure can be transformed into a homogeneous structure with the desired properties (secondary microstructure).

Dendritic crystal growth

In the case of dendritic crystal growth, heat is not dissipated across the crystal as is the case with polygonal crystal growth, but over the melt.

This will be the case when the melt is severely supercooled and the crystal forms freely in the melt. This results in a negative temperature gradient in the direction of the melt. The crystal now grows in an area with a lower temperature than itself.

Any (advance) branching that may form will continue to grow very quickly, as the cooler melt crystallises rapidly at the branching. The crystal forms branches that are very reminiscent of a fir-tree like structure – these are called dendrites.

In dendritic (unstable) crystal growth, the heat of crystallization is dissipated via the surrounding melt.

The micrograph below shows copper with an increased oxygen content. During solidification, the bluish shimmering copper (I) oxide Cu₂O (also called copper oxide) formed, which shows dendritic growth. Due to the microstructure cut, only the dendrite branches cut in the plane can be seen. The dendrites are embedded in the eutectic matrix with finely distributed copper and copper oxide.

With dendritic crystal growth, there is a fundamental danger that no melt can flow into the gaps between the branches . Small microscopic cavities are then formed, which are called microshrinkage.

Excursus: Snowflake

A dendritic crystal growth also shows water when it crystallises into a snowflake on cold winter days. The hexagonal lattice structure of the ice leads to the typical six main branches of the snowflake, from which several smaller branches branch off.

Introduction

The nucleation considered in the article homogeneous nucleation referred to the own particles in the melt. The greatest threshold for the homogeneous nucleation to take place is the surface energy that needs to be applied. This results in small nuclei not being able to grow and therefore dissolving again.

A reduction in surface energy therefore means that less activation energy (free energy barrier) is required for nucleation. Thus even small nuclei are stable and do not dissolve immediately. The thermal fluctuations would then be able to form more growth-capable nuclei.

At this point, impurities, foreign particles or the vessel walls can be very helpful (generally referred to as substrate), as they apply part of the surface energy required for nucleation. This is called heterogeneous nucleation, which requires less activation energy than homogeneous nucleation.

Foreign particles can apply part of the activation energy, so that even smaller nuclei can grow!

Since a melt usually always contains such foreign particles, heterogeneous nucleation is therefore much more likely than homogeneous nucleation.

Wetting

The laws for heterogeneous nucleation are derived in the same way as for homogeneous nucleation. The following example shows a smooth vessel wall (mold) towards the melt on which a nucleus forms. Due to the surface tensions, the solidified nucleus has the shape of a cut-off sphere.

The angle at the contact point between the nucleus surface and the vessel wall is referred to as the contact angle \(\Theta\). Depending on the interaction between nucleus, melt and wall, the wall is very strongly wetted by the nucleus (small contact angle) or hardly wetted (large contact angle). This has a corresponding effect on the shape of the spherical cap.

Decisive for the contact angle \(\Theta\) are the specific surface energies or surface tensions \(\gamma\) between the interfaces nucleus-melt (\(\gamma_{nm}\)), nucleus-wall (\(\gamma_{nw}\)) and melt-wall (\(\gamma_{mw}\)). From an equilibrium analysis, the following relationship results (Young relation):

\begin{align}
\label{benetzungswinkel}
\boxed{\cos\Theta = \frac{\gamma_{mw}-\gamma_{nw}}{\gamma_{nm}}} ~~~\text{Young relation} \\[5px]
\end{align}

Using the contact angle \(\Theta\), the volume \(V\) of the spherical cap at a given nucleus radius \(r\) can be determined by the following equation:

\begin{align}
&V= \tfrac{1}{3} h^2 \pi \left(3r – h \right) ~~~\text{with}~~~ h=r\left(1-\cos\Theta \right) ~~~\text{follows:} \\[5px]
\label{kuppelvolumen}
&\boxed{V= \tfrac{1}{3} \pi \left(2-3\cos\Theta+\cos^3\Theta \right)r^3} \\[5px]
\end{align}

The contact angle \(\Theta\) can also be used to determine the surface of the spherical cap \(A_c\), which forms the interface between the nucleus and the melt:

\begin{align}
&A_c= 2\pi r h ~~~\text{with}~~~ h=r\left(1-\cos\Theta \right) ~~~\text{follows:} \\[5px]
\label{mantelflaeche}
&\boxed{A_c= 2 \pi \left(1-\cos\Theta \right) r^2} \\[5px]
\end{align}

The base surface \(A_w\) of the spherical segment that forms towards the wall also depends on the contact angle \(\Theta\):

\begin{align}
&A_w= \pi a^2 ~~~\text{with}~~~ a=r~\sin\Theta ~~~\text{follows:} \\[5px]
\label{basisflaeche}
&\boxed{A_w= \pi \sin^2\Theta \cdot r^2} \\[5px]
\end{align}

Gibbs energy of a nucleus

In the case of heterogeneous nucleation, the change in Gibbs energy \(\Delta G_n\) of a forming nucleus consists of three parts:

\begin{align}
\label{0}
\boxed{\Delta G_n = \Delta G_v + \Delta G_{s,nm} + \Delta G_{s,nw}}
\end{align}

• Reduction of volume energy \(\Delta G_v\) due to phase change
• Surface energy to be applied \(\Delta G_{s,nm}\) at the interface between nucleus and melt
• Change of surface energy \(\Delta G_{s,nw}\) during the phase change between nucleus and wall

These terms will be discussed in more detail below.

Reduction of volume energy due to phase change

The change in volume energy \(\Delta G_v\) is given by the volume \(V\) of the nucleus in combination with the density \(\rho\) and the specific Gibbs energy change \(\Delta g_v\) (Note that this part leads to a reduction of the energy and therefore bears a negative sign in contrast to the surface energy to be applied):

\begin{align}
&\Delta G_v = – m \cdot \Delta g_v ~~~~~\text{with}~~~~m = V \cdot \rho ~~~~ \text{follows:} \\[5px]
\label{321}
&\underline{\Delta G_v = – V \cdot \rho \cdot \Delta g_v } \\[5px]
\end{align}

The specific change in Gibbs energy \(\Delta g_v\) can be expressed by the relationship to the specific heat of solidification \(q_s\), solidification temperature \(T_s\) and undercooling \(\Delta T\) (derivation see here):

\begin{align}
\label{schmelz}
&\underline{\Delta g_v = q_s \cdot \frac{\Delta T}{T_s}} \\[5px]
\end{align}

If the equations (\ref{kuppelvolumen}) and (\ref{schmelz}) are entered in equation (\ref{321}), then the volume energy \(\Delta G_v\) to be applied results on the basis of a phase change of a nucleus as follows:

\begin{align}
\label{2}
&\boxed{\Delta G_v = – \frac{\pi~\rho~q_s}{3~T_s} \Delta T \left(2-3\cos\Theta+\cos^3\Theta \right) \cdot r^3} \\[5px]
\end{align}

Surface energy to be applied at the interface between nucleus and melt

The surface energy \(\Delta G_{s,nm}\) to be applied to produce the nucleus surface towards the melt results via the spherical cap \(A_c\) and the corresponding specific surface energy \(\gamma_{nm}\):

\begin{align}
&\underline{\Delta G_{s,nm} = \gamma_{nm} \cdot A_c} \\[5px]
\end{align}

The surface energy \(\Delta G_{s,nm}\) to produce the nucleus surface towards the melt is given by means of equation (\ref{mantelflaeche}) as follows:

\begin{align}
\label{3}
&\boxed{\Delta G_{s,nm} = 2 \pi~\gamma_{nm}~\left(1-\cos\Theta \right) r^2} \\[5px]
\end{align}

Change of the surface energy at the interface to the wall

In the molten state, the surface energy \(E_{mw}\) is obtained at the contact surface between the not yet formed nucleus (= melt) and the wall as follows:

\begin{align}
E_{mw}=\gamma_{mw} \cdot A_w \\[5px]
\end{align}

Due to the formation of the nucleus, the surface energy \(E_{mw}\) between the former melt / wall changes to a different surface energy \(E_{nw}\) between nucleus / wall:

\begin{align}
E_{nw}=\gamma_{nw} \cdot A_w \\[5px]
\end{align}

As a result, the surface energy at the contact area with the wall – and with it Gibbs’ energy of the nucleus – has changed as follows:

\begin{align}
\label{d}
&\underline{\Delta G_{s,nw} = (\gamma_{nw}-\gamma_{mw}) \cdot A_w} \\[5px]
\end{align}

The energetic advantage of heterogeneous nucleation can already be seen at this point. The surface energy of the nucleus towards the wall does not have to be applied from zero but only has to change (\(\gamma_{mw}>0\))!
While the contact surface \(A_w\) is determined by the equation (\ref{basisflaeche}), the difference of surface tensions can be replaced by the equation (\ref{benetzungswinkel}):

\begin{align}
\label{young}
&\underline{\gamma_{nw}-\gamma_{mw} = -\gamma_{nm}\cdot \cos(\Theta)} \\[5px]
\end{align}

If equations (\ref{basisflaeche}) and (\ref{young}) are inserted into equation (\ref{d}), then change of a surface energy at a contact area to a wall \(\Delta G_{s,nw}\) can be calculated as follows:

\begin{align}
\underline{\Delta G_{s,nw} = -\gamma_{nm} \cdot \cos\Theta \cdot \pi \sin^2\Theta \cdot r^2} \\[5px]
\end{align}

Furthermore, the geometric expression \( \cos(\Theta) \cdot \sin^2(\Theta) \) can be replaced by the term \( \cos(\Theta) – \cos^3(\Theta) \). This finally leads to:

\begin{align}
\label{1}
&\boxed{\Delta G_{s,nw} = -\pi~\gamma_{nm}~ \left(\cos\Theta – \cos^3\Theta \right) \cdot r^2} \\[5px]
\end{align}

Total change of Gibbs energy of the germ

The change in Gibbs energy \(\Delta G_n\) of the solidifying nucleus can be expressed by inserting the equations (\ref{2}), (\ref{3}) and (\ref{1}) into equation (\ref{0}) and then summarizing as follows:

\begin{align}
\Delta G_n = &- \frac{\pi~\rho~q_s}{3~T_s} ~\Delta T \left(2-3\cos\Theta+\cos^3\Theta \right) \cdot r^3 \\[5px]
&+ 2 \pi~\gamma_{nm}~\left(1-\cos\Theta \right) r^2 \\[5px]
&- \pi~\gamma_{nm}~ \left(\cos\Theta – \cos^3\Theta \right) \cdot r^2 \\[5px]
\Delta G_n = &- \frac{\pi~\rho~q_s}{3~T_s} ~\Delta T \left(2-3\cos\Theta+\cos^3\Theta \right) \cdot r^3 \\[5px]
&+ \pi~\gamma_{nm}~\left( 2-3\cos\Theta+\cos^3\Theta \right) ~r^2 \\[5px]
\end{align}

By further summaries, the change in Gibbs energy of a wettening nucleus assumed to be spherical with the radius \(r\) at the given undercooling \(\Delta T\) can be represented as follows:

\begin{align}
\label{heterogene}
\boxed{\Delta G_n(\Delta T, r) =\tfrac{1}{4}\left(2-3\cos\Theta+\cos^3\Theta \right) ~ \left[-\tfrac{4~\pi~\rho~q_s}{3~T_s}~\Delta T~r^3 + 4\pi~\gamma_{nm}~r^2 \right]} \\[5px]
\end{align}

This equation is discussed in more detail in the following section.

Comparison of heterogeneous and homogeneous nucleation

A comparison with the energy change of a nucleus during homogeneous nucleation is interesting at this point:

\begin{align}
\label{keim}
&\boxed{\Delta G_{hom}(\Delta T, r) = -\tfrac{4 \pi \rho ~ q_s}{3 ~ T_S} \cdot \Delta T \cdot r^3 + 4 \pi ~ \gamma \cdot r^2} \\[5px]
\end{align}

With the same radius and same undercooling, the energy change by heterogeneous nucleation differs by the geometric factor located outside the square brackets of equation (\ref{heterogene}) (wetting factor).

The term inside the square brackets corresponds precisely to the change in energy during homogeneous nucleation. From an energetic point of view, the heterogeneous nucleation on a smooth wall is thus linked to the homogeneous nucleation as follows:

\begin{align}
\boxed{\Delta G_{het} =f(\Theta) \cdot \Delta G_{hom}} ~~~ \text{with} ~~~ \boxed{f(\Theta)=\tfrac{1}{4}\left(2-3\cos\Theta+\cos^3\Theta \right)} \le 1 \\[5px]
\end{align}

The critical nucleation radius is basically the same for heterogeneous nucleation as for homogeneous nucleation, only the activation energies differ (under otherwise identical conditions). Note that the heterogeneous nucleation has the same (fictitious) nucleus radius, but ultimately only part of the sphere volume actually forms the actual nucleus.

The critical nucleus radius can in principle be obtained in the same way as for homogeneous nucleation by deriving and zeroing the equation (\ref{heterogene}):

\begin{align}
&\frac{d (\Delta G_n)}{dr} = 0 \\[5px]
& \tfrac{1}{4}\left(2-3\cos\Theta+\cos^3\Theta \right) \cdot \left[-\tfrac{4 \pi \rho ~ q_s}{T_s} \cdot \Delta T \cdot r^2 + 8 \pi \gamma_{nm} \cdot r \right] = 0 \\[5px]
& \frac{4 \pi \rho ~ q_s}{T_s} \cdot \Delta T \cdot r = 8 \pi \gamma_{nm} \\[5px]
& \boxed{r_{c,het} = \frac{2 \gamma_{nm} ~ T_s}{q_s~ \rho ~\Delta T}}=r_{c,hom}
\end{align}

Insertion of the critical radius \(r_{c,het}\) in equation (\ref{heterogene}) provides the activation energy of heterogeneous nucleation \(\Delta G_{c,het}\) (free energy barrier), with the already described correlation to homogeneous nucleation:

\begin{align}
&\boxed{\Delta G_{c,het} = \tfrac{1}{4}\left(2-3\cos\Theta+\cos^3\Theta \right) \cdot \left[\frac{16 ~\pi ~ \gamma_{nm}^3 ~ T_s^2}{3 ~ q_s^2 ~ \rho^2 ~ \Delta T^2}\right] } \\[5px]
\end{align}

or

\begin{align}
&\boxed{\Delta G_{c,het} = f(\Theta) \cdot \Delta G_{c,hom}} ~~~ \text{with} ~~~ \boxed{f(\Theta)=\tfrac{1}{4}\left(2-3\cos\Theta+\cos^3\Theta \right)} \le 1 \\[5px]
\end{align}

The term, which is purely dependent on the contact angle, is always less than 1, so that it becomes apparent at this point that the heterogeneous nucleation requires less activation energy than the homogeneous nucleation.

For a contact angle of \(\Theta\)=90° the nucleus is a hemisphere and the wetting factor has the value \(f(\Theta)\)=0.5. In this case, the nucleus requires only half the activation energy. After all it has only half as much volume as a spherical nucleus during homogeneous nucleation.

With larger contact angles, the wetting factor increases more and more. At an angle of \(\Theta\)=180°, the wetting factor finally reaches its maximum of 1. the spherical cap has become a complete sphere that no longer wet the wall. As a special case, the homogeneous nucleation is then obtained!

Conversely, very small wetting factors are obtained for small contact angles, i.e. for very strong wetting. As a result, the activation energy required for nucleation is greatly reduced. With a smaller contact angle and the same critical radius, the volume of the nucleus also decreases considerably. It will then no longer be necessary to provide as many atoms for nucleation.

At a contact angle of \(\Theta\)=20°, for example, the required number of atoms drops from 300,000 atoms to only 800 atoms (under otherwise identical conditions) compared to the example already presented for homogeneous nucleation.

The activation energy decreases by the same factor from the original 9.5⋅10^-16 J to 0.025⋅10^-16 J. Thus it becomes clear once again that heterogeneous nucleation is much more likely than homogeneous nucleation.

The heterogeneous nucleation works particularly well if the introduced foreign particles (seed crystals) or the vessel walls can be wetted very well.

When introducing seed crystals into the melt, make sure that the seed particles have a larger radius than the critical radius of nucleation. At the same time, however, there is a risk that the particles could dissolve in the hot melt before nucleation could have begun. For this reason, the seed crystals are often introduced directly into the undercooled melt during pouring.

Gibbs energy

In order to understand the processes of nucleation, an energetic analysis is indispensable. In this context the so-called Gibbs energy plays a central role. Gibbs energy is often referred to as Gibbs free energy, Gibbs function or free enthalpy.

The Gibbs energy balances the enthalpy \(H\) (and with it the internal energy \(U\) as well as pressure \(p\) and volume \(V\)) and the entropy \(S\) at a given temperature \(T\) of a substance:

\begin{equation}
\label{gibbs_energie}
\boxed{G=H – T \cdot S } ~~~ \text{bzw.} ~~~ \boxed{G= U + pV – S \cdot T}
\end{equation}

The Gibbs energy is the decisive measure of whether a chemical or thermodynamic reaction at constant temperature and constant pressure (which is usually the case in a solidification process!) takes place voluntarily with the release of energy or must be forced with the supply of energy.

If the Gibbs energy increases during a reaction (\(\Delta G>0\)), this requires a external energy input (endergonic reaction). The energetically higher final state means that the reaction does not take place voluntarily.

However, if the Gibbs energy decreases with a reaction (\(\Delta G<0\)), then the reaction will take place voluntarily with a release of energy. The energetically more favourable state will occur accordingly (exergonic reaction).

If the Gibbs energy does not change during a reaction (\(\Delta G=0\)), then the reaction will take place to the same extent in one direction as in the other (thermodynamic equilibrium).

The direction of thermodynamic processes can be assessed on the basis of the Gibbs energy:
\begin{align}
\label{unfreiwillig}
&\Delta G > 0 \text{ : Reaction is not voluntary!} \\[5px]
\label{freiwillig}
&\Delta G < 0 \text{ : Reaction is voluntary!} \\[5px]
&\Delta G = 0 \text{ : Reaction takes place to the same extent in both directions!} \\[5px]
\notag \\[5px]
\end{align}

Gibbs energy of different states of matter

The figure below shows the schematic curve of Gibbs energy as a function of the temperature for the solid, liquid and gaseous states of a substance.

It is shown that below a certain temperature the Gibbs energy of the solid state is energetically more favourable than that of the liquid state. This intersection of the corresponding Gibbs curves indicates the solidification temperature or, conversely, the melting temperature of the material. At the solidification point, the Gibbs energy of the liquid and solid states is obviously equal.

In the analogous way, the evaporation or condensation temperature can be defined as the intersection of the Gibbs curves between the liquid and the gaseous state.

Basically, only the state of matter with the lowest Gibbs energy is in a thermodynamic equilibrium. If the temperature falls below the solidification point, the liquid state is no longer in thermodynamic equilibrium due to the higher Gibbs energy.

Rather, at this point there is a reduction in Gibbs energy at the transition from the solid to the liquid phase (\(\Delta G<0\)). In accordance with the above-mentioned law, phase transformation from liquid to solid state takes place voluntarily [see equation (\ref{freiwillig})]. So a thermodynamic system always voluntarily assumes the state of lowest Gibbs energy.

At a certain temperature, only the state of matter with the lowest Gibbs energy is thermodynamically stable (thermodynamic equilibrium)!

The difference between the Gibbs energies of two states is thus the driving force for a substances to change their state of matter when falling below or exceeding a certain temperature and thus take up the state of lowest possible Gibbs energy.

The difference in Gibbs energy is the driving force for a change of state!

This circumstance already expresses the fact that simply reaching the solidification temperature is not sufficient to cause a liquid substance to solidify. This is because a drive is only given when there is a difference in Gibbs energy and thus a fall below the solidification temperature (undercooling!).

Latent heat

The difference in Gibbs energies between the solid and liquid states of a substance at absolute zero corresponds to the heat of solidification\(Q_s\) or enthalpy of solidification of the substance.

In the same way, the condensation heat \(Q_c\) or enthalpy of condensation can be determined in the difference of the Gibbs energy between the liquid and the gaseous state at absolute zero.

The sum of both enthalpies then corresponds to the heat of desublimation \(Q_d\) or enthalpy of desublimation.

Activation energy

Even if subcooling results in principle in a drive for phase transformation (\(\Delta G_v\)), this process does not yet run automatically. An important factor that also influences the Gibbs energy of a solidifying nucleus has not yet been taken into account.

The situation can be compared to a standing car on a hill, which has just passed the highest point. Although the car is basically ready to roll down the hill on its own towards the state of equilibrium, the last energy needed to overcome the static friction and get the rolling started is missing.

In the analogous way, even after the temperature has fallen below the solidification point a “resistance” must first be overcome before the solidification can actually take place on its own. This “resistance” is mainly the surface energy to be applied, which is necessary to form the solid surface during crystallization. Even if the forming particle is to grow, surface energy must be used so that the crystal can increase its surface area. Such an energy input to trigger a process is generally referred to as activation energy.

The surface energy can be understood as resistance against phase transformation (activation energy)!

Note that the Gibbs curves shown in the upper section refer only to the energetic conditions in liquid and solid state. This does not take into account energy sales that are necessary for the “initiation” of the conversion processes, such as surface energy.

Gibbs energy of a nucleus

The change in Gibbs energy \(\Delta G_{n}\) of a considered nucleus which is currently solidifying when the solidification temperature \(T_s\) is reached can ultimately be traced back to two essential terms, which are discussed in more detail in the following sections:

\begin{align}
\label{unterkuehlung}
&\boxed{\Delta G_n = \Delta G_v + \Delta G_s}  \\[5px]
&\Delta G_v<0 \text{ für } T<T_s\\
&\Delta G_v>0 \text{ für } T>T_s\\
&\Delta G_s>0\\
\end{align}

Note: Deformation energies can be neglected when transforming from a liquid state to a solid state. Only phase transformations within in the solid state require deformation energy. In such a case a third (positive) term which considers these effects will be present in equation (\ref{unterkuehlung}).

Volume energy

The first term \(\Delta G_{v}\) in equation (\ref{unterkuehlung}) results on the basis of lower Gibbs-energy of a solid state in comparison with a liquid state. As long as the temperature is below the solidification temperature, this energy contribution is always negative (\(\Delta G_{v}<0\)). After all, the solid state below the solidification temperature has a lower Gibbs energy than the liquid state.

Put simply, this means nothing else than that the solidified volume below the solidification temperature has a “volume energy” lower by the amount \(\Delta G_{v}\) than in the liquid state.

This term would only be positive if it were assumed that a solid nucleus is formed above the solidification temperature due to statistical fluctuations, because above the solidification point, the solid state has a higher Gibbs energy than the liquid phase (more about this later).

For the term \(\Delta G_{v}\) the solidifying volume is decisive, i.e. the volume of the (short-term solidified) nucleus that is formed. If a spherical nucleus with radius \(r\) is assumed, the reduction of Gibbs energy \(\Delta G_{v}\) is proportional to the spherical volume \( V=\frac{4}{3} \pi r^3 \):

\begin{align}
\label{volumenenergie1}
&\Delta G_v = -\underbrace{\frac{4}{3} \pi r^3 \cdot \rho}_{\text{nucleus mass}} \cdot \Delta g_v  \\[5px]
\label{volumen}
& \Delta G_v \sim r^3  \\[5px]
\end{align}

In equation (\ref{volumenenergie1}) the term \(\Delta g_v\) denotes the specific Gibbs energy change (“Gibbs energy per solidified mass”), which is obviously dependent on the untercooling \(\Delta T\) of the nucleus. With given supercooling the term \(\Delta G_{v}\) is thus proportional to the third power of the considered nucleus radius.

The change in specific Gibbs energy \(\Delta g_v\) can be approximately assumed to be proportional to the undercooling \(\Delta T\), if it is related to the solidification temperature \(T_s\). The proportionality factor corresponds precisely to the specific heat of solidification \(q_s\). This consideration is obtained by using the intercept theorem (the error due to the curved lines instead of straight lines can often be neglected in practice):

\begin{align}
&\frac{\Delta g_v}{\Delta T} \approx \frac{q_s}{T_s}   \\[5px]
\label{schmelzwaerme}
&\boxed{\Delta g_v \approx q_s \cdot \frac{\Delta T}{T_s}}   \\[5px]
\end{align}

If equation (\ref{schmelzwaerme}) is combined with equation (\ref{volumenenergie1}), the change in Gibbs energy of the solidified nucleus is dependent on the nucleus radius \(r\) and undercooling \(\Delta T\):

\begin{align}
\label{volumenenergie}
\boxed{\Delta G_v = -\frac{4 \pi \rho ~ q_s}{3 ~ T_s} \cdot \Delta T \cdot r^3} ~~~~~\text{(decrease of “volume energy”)} \\[5px]
\end{align}

The undercooling of metals during homogeneous nucleation is in the order of about 25 % of the solidification temperature. For iron with a solidification temperature of \(T_s=1810 \text{ K}\) this means, for example, a undercooling of \(\Delta T = 450 \text{ K}\).

Surface energy

The second term \(\Delta G_{s}\) in equation (\ref{unterkuehlung}) refers to the surface energy  \(\Delta G_s\) which must be considered at occurrence of the solidified nucleus. This part causes the Gibbs energy to increase with the use of energy. The term \(\Delta G_s\) is therefore always positive, regardless of whether the nucleus itself forms above or below the solidification temperature due to thermal fluctuations in the melt (\(\Delta G_s>0\)).

The surface energy is obviously proportional to the surface of the spherical nucleus \( S = 4 \pi r^2 \) and thus proportional to the square of the nucleus radius \(r\):

\begin{align}
\label{a}
&\boxed{\Delta G_s = 4 \pi r^2 \cdot \gamma} ~~~~~~\text{(“surface energy” to be applied)} \\[5px]
\label{oberflaeche}
& \Delta G_s \sim r^2 \\[5px]
\end{align}

In equation (\ref{a}) the term \(\gamma\) denotes the specific surface energy (also called surface tension). Overall, the change in Gibbs energy of a nucleus assumed to be spherical with the radius \(r\) at a given undercooling \(\Delta T\) can be calculated as follows:

\begin{align}
\label{keim}
&\boxed{\Delta G_n(r, \Delta T) = -\frac{4 \pi \rho ~ q_s}{3 ~ T_S} \cdot \Delta T \cdot r^3 + 4 \pi ~ \gamma \cdot r^2}   \\[5px]
\end{align}

This equation is interpreted in more detail in the following section.

Critical nucleus radius

For temperatures above the solidification temperature the undercooling is mathematically considered negative (\(\Delta T<0\)) and thus both terms in equation (\ref{keim}) positive. In this case, the figure below shows the change in Gibbs energy \(\Delta G_{n}\) with increasing nucleus radius \(r\).

In all cases the Gibbs energy of the nucleus \(\Delta G_{n}\) will increase (positive change of Gibbs energy). As a result, there will be no thermodynamic equilibrium between the solidified nucleus and the remaining melt. It may well be that a nucleus forms, but due to the unstable state, it will quickly dissolve by releasing energy and strives for a minimum of Gibbs’ energy.

Above the solidification temperature, a formed nucleus will dissolve immediately with release of energy!

However, the diagram only becomes interesting when the temperature of the melt or the nucleus is below the solidification temperature and thus there is a positive undercooling (\(\Delta T>0\)). In this case the first term in equation (\ref{keim}) becomes negative. The diagram below shows the course of Gibbs energy change \(\Delta G_{n}\) of the nucleus as a function of its radius. The diagram now shows a maximum in the change of Gibbs energy (green dot in the figure). This curve will be interpreted in more detail below.

First, it is assumed that a solid nucleus with the radius \(r_1\) is formed, which is to the left of the maximum of the Gibbs curve (see figure below). To form such a nucleus, the Gibbs energy obviously needs to be increased. This is done by statistical fluctuations in the thermal energy of the melt, which can provide this amount of energy in the short term.

If the nucleus is then to continue to grow and thus initiate the solidification process, the nucleus must obviously become larger. According to the diagram, however, an increase in the radius is only possible with a further increase in Gibbs energy. Such an increase in Gibbs energy is not voluntary! Rather, the reduction of Gibbs energy and thus the reduction of the nucleus occurs voluntarily and is therefore more likely. The result is that the randomly formed nucleus dissolves again with a release of energy.

Now a randomly formed nucleus with the radius \(r_2\) is considered, which is to the right of the maximum of the Gibbs energy curve. This formation also occurs through thermal fluctuations in the melt, which causes the corresponding increase in Gibbs energy in the short term.

In such a case, according to the diagram, a reduction of the nucleus radius is only possible with the use of energy (increase of Gibbs energy) and will therefore not take place voluntarily. The nucleus will not dissolve on its own! Rather, the solidified germ will now voluntarily increase its radius while reducing Gibbs energy. Of course, this can only be done by adding further particles from the melt. Thus the beginning of the voluntary nucleus growth is given and the solidification process is initiated!

Only nucleus of a certain size are capable of growth and can initiate the solidification process!

This critical nucleus radius \(r_c\) results from the maximum of the Gibbs function, which can be obtained by derivation:

\begin{align}
\label{maximum}
&\frac{d (\Delta G_n)}{dr} = 0 \\[5px]
& -\frac{4 \pi \rho ~ q_s}{T_s} \cdot \Delta T \cdot r^2 + 8 \pi \gamma \cdot r = 0 \\[5px]
& \frac{4 \pi \rho ~ q_s}{T_s} \cdot \Delta T \cdot r = 8 \pi \gamma \\[5px]
\label{kritisch}
& \boxed{r_c = \frac{2 \gamma ~ T_s}{q_s~ \rho ~\Delta T}}
\end{align}

Typical values for the surface tensions of technically relevant metals are in the range between 1 and 4 J/m². If a value of \(\gamma\) = 2.5 J/m² and a solidification temperature of \(T_s = 1810 K\) is assumed for iron and a specific solidification heat of 268 kJ/kg and a density of \(\rho\) = 7870 kg/m³ is considered, then a undercooling of \(\Delta T = 450 K\) results in a critical nucleus radius \(r_c\) of approximately 9.5 nm. Approximately 300,000 atoms are then required to form such a nucleus.

Free energy barrier for nucleation

As can be seen from the diagram above for the change in Gibbs energy, energy must be used to produce the nucleus in order to increase Gibbs energy accordingly. This is achieved by thermal fluctuations in the melt, which can supply this amount of energy at local level in the short term. This amount of energy to be overcome is also called free energy barrier (activation energy for nucleation).

The amount of energy that has to be supplied for a stable formation of a nucleus is called free energy barrier (activation energy for nucleation)!

Only under the assumption that a randomly formed nucleus has a radius greater than \(r^*\) does crystallization take place directly with energy output. However, due to the large number of atoms required for this, the probability that such a large nucleus will form is low.

Undercooling \(\Delta T\) has obviously decisive influence on the critical radius according to the equation (\ref{kritisch}). The greater the supercooling \(\Delta T\) the smaller the critical radius. At the same time, the free energy barrier decreases as the critical radius decreases. As undercooling progresses, it becomes increasingly likely that nuclei capable of growth will form. Many nuclei can then form, which leads to a fine-grained structure. However, this only applies as long as the supercooling is not too great (more on this later).

Within certain limits a large supercooling leads to increased nucleation!

The free energy barrier \(\Delta G_c\) can be determined by inserting equation (\ref{kritisch}) into equation (\ref{keim}), so that the direct relationship to undercooling \(\Delta T\) becomes apparent:

\begin{align}
&\boxed{\Delta G_c = \frac{16 ~\pi ~ \gamma^3 ~ T_S^2}{3 ~ q_s^2 ~ \rho^2 ~ \Delta T^2} } \\[5px]
\end{align}

For the numerical example above, the energy barrier is in the order of about \(9.5 \cdot 10^{-16} \text{J}\).

Note that both the activation energy for nucleation and the critical nucleus radius increase with low undercooling and even become infinitely large when the solidification temperature is reached. A stable nucleation is therefore no longer possible when the solidification point is reached!

Nucleation rate

In the previous section it was shown that the critical nucleus radius and the energy barrier decrease with progressive undercooling. The latter is applied by thermal fluctuations of the melt, whereby several smaller fluctuations are more likely than single large fluctuations. This explains why the number of growth-capable nuclei increases with progressive undercooling. A strong supercooling is therefore usually desired, as this leads to a fine-grained and thus solid and tough microstructure.

However, there are also limits to the increasing number of nuclei with increasing undercooling. This is because diffusion processes in the melt and in the nucleus are necessary to form a nucleus or to allow it to grow, so that the atoms can attach to each other. With increasing undercooling, however, the diffusion processes due to the “sluggish” particles decrease. This applies to diffusion in the melt, diffusion across the phase boundary and diffusion within the nucleus. There are then no longer enough particles which could come together to form a nucleus and make it grow. As a result, fewer nuclei will form from a critical undercooling.

The diffusion-related agglomeration of the particles per time \(\dot N_d\) can be described by an Arrhenius ansatz that predicts an exponential dependence on temperature:

\begin{align}
\label{diffusion}
& \dot N_d \sim \text{exp}{\left(-\frac{Q_d}{k_B \cdot T}\right)}  \\[5px]
\end{align}

\(Q_d\) denotes the temperature-independent activation energy for diffusion and \(k_B\) the Boltzmann constant.

At the same time, it must be considered that the particles must also reach the critical radius. Analogous to the equation (\ref{diffusion}), the number of nuclei formed per time \( \dot N_c\), which exceed the critical radius, can be determined:

\begin{align}
\label{keimzahl}
&\dot N_c \sim \text{exp}{\left(-\frac{\Delta G_c(\Delta T)}{k_B \cdot T}\right)}  \\[5px]
\end{align}

\(\Delta G_c(\Delta T)\) denotes the temperature-dependent activation energy for nucleation.

The rate of nucleation \(\dot N_{n}\) ultimately depends on both influences, so that the total dependency is given by multiplying both equations:

\begin{align}
\label{keimbildungsrate}
&\boxed{\dot N_{n} \sim \text{exp}{\left(-\frac{Q_d}{k_B \cdot T}\right)} \cdot \text{exp}{\left(-\frac{\Delta G_c(\Delta T)}{k_B \cdot T}\right)}}  \\[5px]
\end{align}

Note that the energy barrier for nucleation depends on undercooling. The figure below schematically shows the influence of undercooling on the nucleation rate according to the equation (\ref{keimbildungsrate}).

With low supercooling, the activation energy for nucleation \(\Delta G_c\) is considerably greater than the energy activation for diffusion \(Q_d\). The formation of nuclei is therefore inhibited by the great energy needed for nucleation.

If the undercooling is too high, the whole thing turns into the opposite. Due to the low temperature, the activation energy for diffusion \(Q_d\) is considerably greater and now inhibits the formation of a nucleus, although the energy barrier for nucleation \(\Delta G_c\) would be very low for this.

Consequently, there is a maximum of nucleation rate with a certain undercooling. By this optimal undercooling most nuclei can obviously form, which leads to an extremely fine-grained structure.

However, the diagram of the nucleation rate also shows that excessive undercooling can even completely suppress nucleation. Such a strong supercooled melt then solidifies without crystallization. The particles are “frozen” so to speak before they could have attached themselves to a crystal structure. Such metals are then also called amorphous metals (metallic glasses).

In the article on types of nuclei it was explained that with stronger cooling and thus undercooling, more nuclei will be formed in the melt (high nuclei formation rate). As a result of the many nuclei, a fine-grained and thus a high-strength and tough structure is obtained.

However, there are also limits to such a strong cooling. If it cools down too quickly, the particles do not have time to settle down in a lattice structure. Due to the rapid temperature decrease, the particles become so quickly inert that they can only move to a limited extent. They are practically “frozen” in their molten state.

How quickly a melt must be cooled down to prevent crystallization depends on the chemical properties of the substance. With glass, for example, this condition is achieved by the already low cooling speeds in air. This is also the reason why glass does not usually have a crystalline structure, but belongs to the amorphous substances. However, if glass were to be cooled very slowly and in a controlled manner, glass would also crystallize! From a thermodynamic point of view, glass is therefore a highly supercooled liquid!

However, because the particles are much too inert in this undercooled state, glass seems “solid”. But as the temperature increases more and more, the flow behavior gradually becomes apparent, as the inertia decreases with higher temperature. As is usual for liquids, the glass then gradually begins to melt.

Glasses therefore do not have a fixed melting point (one speaks of a so-called glass transition temperature)! But even at room temperature, a flow of the glass can theoretically be observed. In practice, however, this requires periods of several thousand to millions of years. This is why glass is considered a solid for practical reasons.

In the same way that crystallization in glass can be suppressed by rapid cooling, amorphous solidification of metals can also be achieved if cooling is sufficiently strong. These metals are then called amorphous metals (also called metallic glasses).

The production of such amorphous metals requires cooling rates in the order of 1000 °C within a few milliseconds. Larger components cannot (yet) be produced in this way because the required cooling rates inside the workpiece cannot be achieved.

Amorphous metals are characterized by extremely high hardness and strength and are limited to a few special applications due to the expensive manufacturing process. Among other things, high-quality golf clubs are made of metallic glass.

Introduction

The hand warmer described in the article solidification conditions shows another phenomenon during the solidification of liquids (melts), which still needs an explanation.

Obviously, heat is released during the crystallization of the hand warmer. This ultimately means that the hand warmer can be used as a heat source on cold winter days.

In principle, this phenomenon of heating during solidification is not limited to the hand warmer. Ultimately, such a heat release takes place in every solidification process (crystallization). This effect will therefore be explained in more detail below.

Pure substances

During the cooling of a melt, the expected temperature decrease is shown in the cooling curve. The temperature reduction is due to the continuously withdrawn heat energy from the melt. This leads to a reduction in the kinetic energy of the particles.

When the solidification temperature (melting point) is reached, the kinetic energy of the particles has fallen to such an extent that the mutual attraction forces cause the particles to accumulate. With the creation of the lattice structure, however, the kinetic energy of the atoms decreases suddenly, as the particles show only a relatively small (vibrational) movement in the crystal compound. However, the difference in binding energies has not simply disappeared but has been converted into heat energy!

In principle, this process can be imagined as an impact of a particle from the melt on the already solidified lattice structure, which then attaches itself to it. The impact also causes the grid structure to vibrate, which is equivalent to a (heat) energy input. This ultimately means a local temperature increase (note, that a higher vibrational movement means a higher temperature!). Such internal heat release during solidification is called heat of solidification. In the case of crystalline substances it is also referred to as crystallization heat.

The heat of solidification released by crystalline materials is also called heat of crystallisation!

The internal crystallization heat (“temperature increase”) released during solidification ultimately compensates for the external heat dissipation (“temperature reduction”). In fact, a permanent oscillation around the solidification temperature will be observed at the microscopic level.  No heat is effectivly added to or removed from the solidifying substance during crystallization. The temperature therefore remains constant from a macroscopic point of view.

This keeping constant of the temperature is also called thermal arrest and becomes noticeable in the cooling curve as a horizontal line. Only when all particles have adhered to the lattice structure and the solidification is finished, no more heat of crystallization can be released. From this point on, only the external heat dissipation becomes effective and the temperature finally begins to fall again.

The effect that the temperature remains constant during solidification is a typical phenomenon of pure substances.

During the solidification of pure substances, the temperature usually remains constant (thermal arrest)!

Mixtures

In principle, mixtures of substances also show such heat release during solidification. However, due to the mutual chemical influence of the substances, the heat of crystallisation can generally no longer fully compensate for cooling.

As a result, the temperature decrease is no longer stopped completely but only slowed down. The cooling curve is only flatter during solidification.

During the solidification of mixtures, the temperature decrease usually slows down!

In fact, there are also some alloys that have both, a thermal arrest as well as a flattend cooling curve (so-called solid solution alloys). Such alloy types are described in more detail in a separate article.

Note

The process of external heat dissipation and internal heat release during crystallization, which normally takes place simultaneously, is carried out separately in hand warmers due to undercooling. Because if the supercooled hand warmer is activated, it will not be cooled by an external heat dissipation at all in the first moment.

Therefore, at first you only have the effect of internal heat release due to crystallization! Only when the hand warmer has become warm and firm does heat dissipation take place due to the temperature difference to the environment and it cools down slowly. In this case, internal heating and external cooling take place separately.

Introduction

The origin for the formation of a microstructure is usually the molten state. The metal is then cast and solidified. During this process the microstructure finally forms. The processes involved are explained in more detail below.

In the liquid state, the metal atoms are distributed completely randomly. Due to the high temperature, the particles have a relatively high kinetic energy. In this state, the force of attraction between the atoms is not sufficient to permanently bind them together. Due to the violent impact processes with other atoms, possibly short-term existing bonds are immediately broken up again.

Supercooling

If the melt is now gradually cooled, the kinetic energy of the particles also decreases as the temperature drops. If the temperature falls below a critical value, the mutual attraction between the particles gets the upper hand. Bonds between atoms can now become permanently effective. The solidification temperature (“melting point”) has been undercut and the microstructure will be formed.

Due to the forces of attraction that become effective, more and more particles from the melt attach to the already solidified lattice structure. This process continues until all particles have finally adhered to the metal lattice. Since the crystal structure of the metal forms during this solidification process, this is also referred to as the crystallization process.

The solidification process of a crystalline structure is also called crystallization!

The temperature of the melt must therefore be below the solidification temperature of the material so that solid bonds remain permanently stable. The difference between the local temperature of the melt and the solidification temperature is also referred to as undercooling or supercoolung. Supercooling is absolutely necessary to trigger the solidification process; simply reaching the solidification temperature is not enough.

In order to trigger a solidification process, the melt must be supercooled (undercooled)!

Although supercoolung is a necessary condition for the solidification process, it alone will not cause a melt to solidify! The second necessary prerequisite is explained in more detail in the following section.

Nuclei

Although the undercooling described in the previous chapter is a necessary condition for the solidification process, it alone is not enough to cause a melt to solidify! This may sound a little strange at first; a simple everyday example will illustrate this in the following.

With the help of a hand warmer it quickly becomes clear that a mere supercooling of a melt (liquid) alone is not sufficient to trigger a solidification process! The liquid used in hand warmers is often sodium acetate trihydrate, the solidification temperature of which is around 58 °C.

However, even at temperatures down to -20 °C, this liquid state of sodium acetate trihydrate remains. In this case, even with supercooling of up to 78 °C, the melt cannot be induced to solidify. This state is also referred to as supercooled melt or undercooled melt.

Obviously, in addition to supercooling, another necessary prerequisite must also be fulfilled for a melt to solidify. This further condition can be seen very clearly from the hand warmer. Because if a metal plate inside is pressed, the solidification process suddenly begins.

The reason for the solidification is the tiny unmelted residues of the sodium acetate trihydrate (or other impurities), which have settled in the fine grooves of the metal platelet. Only when the platelet is pressed the melt come into contact with the unmelted residues oder other impurities. These act as so-called nuclei from which solidification can begin.

Note: Until today it is not exactly clear whether exactly this principle actually triggers solidification in a hand warmer. What is certain, however, is that solidification makes nuclei absolutely necessary in order to trigger a crystallization process.

The presence of nuclei is therefore another condition to be met when a solidification process wants to be triggered.

There must be nuclei in the melt to trigger a crystallization process!

Incidentally, the formation of a grain can be vividly reproduced with the hand warmer. Starting from the nucleus, more and more particles from the melt accumulate around this nucleus and crystallization gradually progresses. Once the melt is consumed, the crystallite hits the surrounding shell, just as the grain in a metal hits its neighboring grains after solidification.

Solidification temperature

It can be concluded that even in the presence of nuclei, solidification does not occur as long as there is no supercooling (the metal plate of a hand warmer can be pressed as often as it is needed above the melting temperature – the melt remains liquid).

Conversely, the solidification process is not triggered even if no nuclei are present in the melt despite supercooling (despite severe undercooling, the hand warmer remains liquid as long as there is no nucleus in the melt).

The stated solidification temperatures are ultimately theoretical values above which any self-generated nuclei dissolve immediately. The solidification process is therefore not triggered in such cases.

Only below the solidification temperature is a nucleus capable of growth and crystallization can begin!

Transferred to the microstructure of metals, supercooling and nuclei are therefore always a prerequisite for solidification. However, nuclei do not always have to be unmelted residues of the liquid as in the case of the hand warmer. Nuclei usually develop in other ways, especially in molten metals. This is discussed in more detail in a separate article.

Excursus: Subcooled water (freezing rain)

Water can also be put into a supercooled state of well below -10 °C and still be liquid! However, this requires highly pure water without any impurities that could serve as solidifying nuclei. Of course, the vessel within which the water is stored must also be free of impurities.

However, as soon as the water comes into contact with other particles (for example when pouring onto a table) the nucleation is triggered and the water begins to solidify immediately. Even the smallest vibrations can trigger the solidification process as well.

In nature supercooled water plays a important role in so-called freezing rain. These are subcooled raindrops. These initially rain in liquid form onto the ground. On impact, however, the impurities there serve as crystallization nuclei. The raindrops immediately begins to crystallize (solidify) and thus turns to ice!