Lecture 6: Maxwell-Boltzmann Distribution (Thermal energy in gases, part 3)
0) Orientation, learning outcomes, and admin
Today's lecture builds on our understanding of thermal energy in gases. We'll construct the Maxwell-Boltzmann distributions, starting with one-dimensional velocity components and then extending to two- and three-dimensional speeds. Our approach begins with Boltzmann's statistical weight, $e^{-E/kT}$, which describes how particles populate energy levels. We'll then connect these distributions to important average quantities, such as mean square speeds and average kinetic energies, and demonstrate how they align with the equipartition theorem.
By the end of this lecture, you should be able to derive the Maxwell-Boltzmann distribution in one, two, and three dimensions, including its normalisation. You'll also be able to calculate average speeds and average kinetic energies directly from these distributions. Finally, you should be able to relate these results to the equipartition principle, which states that each quadratic degree of freedom contributes $\frac{1}{2}kT$ to the average energy.
For administrative matters, all lecture recordings are accessible via the "Replay" section on the left-hand menu of Blackboard. Tomorrow's problems class will focus on selected derivations and worked examples from this lecture, providing an opportunity to deepen your understanding of these concepts.
⚠️ Exam Alert! The lecturer explicitly stated: "I recommend you definitely come to the problems class tomorrow."
1) Recap: Boltzmann factor and extreme-temperature limits
The Boltzmann factor, $e^{-\beta E}$, where $\beta = 1/(kT)$, provides a statistical weight for a system in a state of energy $E$ at an absolute temperature $T$. This factor fundamentally describes how populations of particles are distributed across different energy levels at a given temperature. At very low temperatures, particles tend to occupy lower energy states, while at higher temperatures, they spread out across a wider range of energy levels.
Consider the high-temperature limit, as $T \to \infty$. In this scenario, $\beta \to 0$, which causes the Boltzmann factor $e^{-\beta E}$ to approach $e^0 = 1$. This means that at infinitely high temperatures, all energy levels become equally populated. This illustrates a fundamental aspect of how thermal energy influences the distribution of particles.
Side Note: It's worth briefly considering a unique situation called population inversion. If a system were to have more particles in higher energy levels than in lower ones, the Boltzmann factor $e^{-E/kT}$ would mathematically imply a negative absolute temperature. This non-equilibrium state is famously realised in lasers, where energy is directed in a way that can be considered "hotter" than infinite temperature. However, for all systems we will study in this course, we'll assume normal, positive absolute temperatures.
2) From microstate weights to a probability density
To describe the full state of a particle, including its position and momentum, we use a probability density. This probability for a particle to be found within infinitesimal ranges of position ($\text{d}x$, $\text{d}y$, $\text{d}z$) and momentum ($\text{d}p_x$, $\text{d}p_y$, $\text{d}p_z$) is given by:
$$
P[x, y, z, p_x, p_y, p_z] \, dx \, dy \, dz \, dp_x \, dp_y \, dp_z = A e^{-E / kT} \, dx \, dy \, dz \, dp_x \, dp_y \, dp_z
$$
Here, $A$ represents a normalisation constant. This constant ensures that the total probability of finding the particle somewhere in the entire phase space (all possible positions and momenta) sums to one.
A crucial aspect of this expression is its separability. If the total energy $E$ can be expressed as a sum of independent energy components (e.g., kinetic energy in the $x$ -direction, $K(p_x)$, and potential energy $V(x)$), then we can analyse each component separately. This allows us to simplify the problem by first deriving distributions for individual velocity components in one dimension before combining them to understand more complex speed distributions in two or three dimensions.
As we move through derivations involving these distributions, you'll encounter various Gaussian integrals. Please be reassured that these standard integrals will be provided in exams, and we'll work through their application in the problems class.
3) 1D velocity-component distribution: derivation and averages
We begin by deriving the probability distribution for a single velocity component, $v_x$. We start with the probability distribution for momentum in one dimension, $P[p_x]$, where the kinetic energy is $p_x^2/(2m)$. The unnormalised distribution is proportional to $e^{-p_x^2/(2mkT)}$. To normalise this, we integrate from $-\infty$ to $\infty$ and set the result equal to one. Using the standard integral $\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\pi/\alpha}$, where $\alpha = 1/(2mkT)$, we find the normalisation constant.
Next, we convert this momentum distribution to a velocity distribution. Since $p_x = mv_x$, it follows that $\text{d}p_x = m \, \text{d}v_x$. Substituting these into the normalised momentum distribution yields the one-dimensional velocity distribution:
$$
P[v_x] dv_x = \left( \frac{m}{2 \pi k T} \right)^{1/2} e^{- \frac{m v_x^2}{2 k T}} dv_x
$$
This distribution, as shown in the accompanying graph, is a symmetric Gaussian curve centred at $v_x = 0$. This symmetry immediately tells us that the average velocity, $\overline{v_x}$, is zero. To obtain a meaningful measure of the typical motion, we calculate the mean square velocity, $\overline{v_x^2}$.
To find $\overline{v_x^2}$, we integrate $v_x^2$ multiplied by the probability distribution $P[v_x]$ over all possible velocities:
$$
\overline{v_x^2} = \int_{-\infty}^{\infty} v_x^2 P[v_x] dv_x
$$
Using another standard integral (which would be provided in an exam), this integral evaluates to:
$$
\overline{v_x^2} = \frac{kT}{m}
$$
From this, we can determine the average kinetic energy for this single degree of freedom:
$$
\frac{1}{2} m \overline{v_x^2} = \frac{1}{2} m \left( \frac{kT}{m} \right) = \frac{1}{2} kT
$$
This result perfectly aligns with the equipartition theorem, which states that each quadratic degree of freedom contributes $\frac{1}{2}kT$ to the average energy. This one-dimensional distribution is conceptually useful for understanding phenomena such as Doppler broadening in spectroscopy and simple models of an isothermal atmosphere.
4) Why speed distributions differ from component Gaussians: state counting
The distribution of speeds for particles in two or three dimensions differs significantly from the simple Gaussian shape of a single velocity component. This difference arises from a crucial concept known as "state counting" or the "density of states." While a small interval $\text{d}v_x$ directly corresponds to a small interval of speed $\text{d}c$ in one dimension, this is not true in higher dimensions. Many different combinations of velocity components (e.g., $v_x, v_y, v_z$) can result in the same overall speed $c = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
Consider the geometry in velocity space. In two dimensions, all velocity vectors $(\vec{v} = v_x \hat{i} + v_y \hat{j})$ that have the same speed $c$ lie on a circle of radius $c$. The number of available "speed states" within a small speed interval from $c$ to $c+\text{d}c$ is proportional to the circumference of this circle, which is $2\pi c \, \text{d}c$.
Extending this to three dimensions, all velocity vectors with the same speed $c$ lie on the surface of a sphere of radius $c$. The number of available speed states in the interval from $c$ to $c+\text{d}c$ is now proportional to the surface area of this sphere, which is $4\pi c^2 \, \text{d}c$.
This geometric factor has a profound consequence for the shape of speed distributions. The speed distribution is formed by multiplying the exponentially decaying Boltzmann factor ($e^{-mc^2/2kT}$) by this "states factor." As a result:
- In two dimensions, the speed distribution is proportional to $c \, e^{-mc^2/2kT}$.
- In three dimensions, it's proportional to $c^2 \, e^{-mc^2/2kT}$.
Unlike the symmetric Gaussian for velocity components, these speed distributions start at zero (because there are no states with zero speed), rise to a peak as the states factor dominates, and then decay exponentially at higher speeds as the Boltzmann factor takes over.
Combining the Boltzmann factor with the two-dimensional state-counting factor ($2\pi c \, \text{d}c$) and then normalising, we arrive at the two-dimensional speed distribution:
$$
P[c] dc = c \frac{m}{kT} e^{-mc^2 / 2kT} dc
$$
To find the average kinetic energy in two dimensions, we first calculate the mean square speed, $\overline{c^2}$. This involves integrating $c^2$ multiplied by the distribution $P[c]$ over all possible speeds from $0$ to $\infty$. Using a standard integral for the $c^3$ -term, we find:
$$
\overline{c^2} = \frac{2kT}{m}
$$
Therefore, the average kinetic energy in two dimensions is $\frac{1}{2}m\overline{c^2} = \frac{1}{2}m \left( \frac{2kT}{m} \right) = kT$. This result perfectly aligns with the equipartition theorem, which predicts $\frac{1}{2}kT$ for each of the two quadratic degrees of freedom (for $v_x$ and $v_y$), summing to $kT$. The shape of this distribution, as shown in the graph, illustrates two competing effects: the linear rise at small speeds is due to the increasing number of available states, while the decay at larger speeds is governed by the Boltzmann exponential term.
6) 3D Maxwell-Boltzmann speed distribution: full form and anatomy
To arrive at the full three-dimensional Maxwell-Boltzmann speed distribution, we begin with the joint probability distribution for the three velocity components $v_x, v_y, v_z$:
$$
P[v_x v_y v_z] \, dv_x dv_y dv_z = \left( \frac{m}{2 \pi k T} \right)^{3/2} e^{\frac{-m(v_x^2 + v_y^2 + v_z^2)}{2kT}} dv_x dv_y dv_z
$$
Using the relationship $c^2 = v_x^2 + v_y^2 + v_z^2$ and incorporating the three-dimensional state-counting factor $4\pi c^2 \, \text{d}c $, we can convert this into a distribution for speed$ c$. After normalisation, the result is the standard Maxwell-Boltzmann speed distribution:
$$
P[c] \, dc = 4 \pi \left( \frac{m}{2 \pi k T} \right)^{3/2} c^2 e^{- \frac{m c^2}{2 k T}} \, dc
$$
The anatomy of this curve is crucial for understanding its characteristic shape. The $c^2$ term, originating from the increasing number of available speed states (the surface area of a sphere in velocity space), causes the probability to initially rise from zero at small speeds. Conversely, the exponential Boltzmann term, $e^{-mc^2/2kT}$, rapidly suppresses the probability at large speeds. The product of these two competing factors results in the familiar bell-shaped Maxwell-Boltzmann distribution, which peaks at an intermediate speed, representing the most probable speed for particles in the gas. It's important to remember that this is a distribution of speeds, which are always positive, unlike the velocity component distributions which are symmetric about zero.
7) What was deferred and what comes next
In this lecture, the emphasis was placed on the conceptual construction of the Maxwell-Boltzmann distributions. We completed the full derivation for the one-dimensional velocity component distribution, including its normalisation and the calculation of its average kinetic energy. For the two- and three-dimensional distributions, we focused on understanding the crucial role of the "state-counting" factor (proportional to $c$ in 2D and $c^2$ in 3D) and presented the final forms of these distributions, sketching the normalisation and integral results.
A full, worked example using these distributions and the necessary standard integrals will be covered in tomorrow's problems class. This session will provide an opportunity to practise these calculations and clarify any questions you may have regarding normalisation and the computation of average values from the distributions.
Key takeaways
- The Boltzmann factor $e^{-E/kT}$ governs how particles populate energy levels; as $T \to \infty$, all levels become equally populated. Population inversion, as seen in lasers, implies a negative absolute temperature but is not typical for the systems we study.
- Starting from $P \propto e^{-E/kT}$, the one-dimensional velocity component distribution is a Gaussian: $P[v_x] dv_x = \left( \frac{m}{2 \pi k T} \right)^{1/2} e^{- \frac{m v_x^2}{2 k T}} dv_x$. The average velocity $\overline{v_x}$ is zero, but the mean square velocity $\overline{v_x^2} = kT/m$, leading to an average kinetic energy per component of $\frac{1}{2}kT$.
- Speed distributions differ from component Gaussians due to the geometric effect of "state counting" in velocity space. In 2D, $P[c] \propto c \, e^{-mc^2/2kT} $, with an average kinetic energy of$ kT = 2 \times (\frac{1}{2}kT) $. In 3D, the full Maxwell-Boltzmann speed distribution is$ P[c] \, dc = 4 \pi \left( \frac{m}{2 \pi k T} \right)^{3/2} c^2 e^{- \frac{m c^2}{2 k T}} \, dc$.
- The characteristic shape of the Maxwell-Boltzmann speed distributions (starting at zero, rising to a peak, then decaying) is a result of the competition between a growing states factor ($c$ or $c^2$) and a decaying Boltzmann exponential term.
- Standard Gaussian integrals are essential for normalising these distributions and calculating average values. These integrals will be provided in exams and thoroughly practised in the problems class.
## Lecture 6: Maxwell-Boltzmann Distribution (Thermal energy in gases, part 3)
### 0) Orientation, learning outcomes, and admin
Today's lecture builds on our understanding of thermal energy in gases. We'll construct the Maxwell-Boltzmann distributions, starting with one-dimensional velocity components and then extending to two- and three-dimensional speeds. Our approach begins with Boltzmann's statistical weight, $e^{-E/kT}$, which describes how particles populate energy levels. We'll then connect these distributions to important average quantities, such as mean square speeds and average kinetic energies, and demonstrate how they align with the equipartition theorem.
By the end of this lecture, you should be able to derive the Maxwell-Boltzmann distribution in one, two, and three dimensions, including its normalisation. You'll also be able to calculate average speeds and average kinetic energies directly from these distributions. Finally, you should be able to relate these results to the equipartition principle, which states that each quadratic degree of freedom contributes $\frac{1}{2}kT$ to the average energy.
For administrative matters, all lecture recordings are accessible via the "Replay" section on the left-hand menu of Blackboard. Tomorrow's problems class will focus on selected derivations and worked examples from this lecture, providing an opportunity to deepen your understanding of these concepts.
> **⚠️ Exam Alert!** The lecturer explicitly stated: "I recommend you definitely come to the problems class tomorrow."
### 1) Recap: Boltzmann factor and extreme-temperature limits
The Boltzmann factor, $e^{-\beta E}$, where $\beta = 1/(kT)$, provides a statistical weight for a system in a state of energy $E$ at an absolute temperature $T$. This factor fundamentally describes how populations of particles are distributed across different energy levels at a given temperature. At very low temperatures, particles tend to occupy lower energy states, while at higher temperatures, they spread out across a wider range of energy levels.
Consider the high-temperature limit, as $T \to \infty$. In this scenario, $\beta \to 0$, which causes the Boltzmann factor $e^{-\beta E}$ to approach $e^0 = 1$. This means that at infinitely high temperatures, all energy levels become equally populated. This illustrates a fundamental aspect of how thermal energy influences the distribution of particles.
*Side Note:* It's worth briefly considering a unique situation called population inversion. If a system were to have more particles in higher energy levels than in lower ones, the Boltzmann factor $e^{-E/kT}$ would mathematically imply a negative absolute temperature. This non-equilibrium state is famously realised in lasers, where energy is directed in a way that can be considered "hotter" than infinite temperature. However, for all systems we will study in this course, we'll assume normal, positive absolute temperatures.
### 2) From microstate weights to a probability density
To describe the full state of a particle, including its position and momentum, we use a probability density. This probability for a particle to be found within infinitesimal ranges of position ($\text{d}x$, $\text{d}y$, $\text{d}z$) and momentum ($\text{d}p_x$, $\text{d}p_y$, $\text{d}p_z$) is given by:
$$ P[x, y, z, p_x, p_y, p_z] \, dx \, dy \, dz \, dp_x \, dp_y \, dp_z = A e^{-E / kT} \, dx \, dy \, dz \, dp_x \, dp_y \, dp_z $$
Here, $A$ represents a normalisation constant. This constant ensures that the total probability of finding the particle *somewhere* in the entire phase space (all possible positions and momenta) sums to one.
A crucial aspect of this expression is its separability. If the total energy $E$ can be expressed as a sum of independent energy components (e.g., kinetic energy in the $x$-direction, $K(p_x)$, and potential energy $V(x)$), then we can analyse each component separately. This allows us to simplify the problem by first deriving distributions for individual velocity components in one dimension before combining them to understand more complex speed distributions in two or three dimensions.
As we move through derivations involving these distributions, you'll encounter various Gaussian integrals. Please be reassured that these standard integrals will be provided in exams, and we'll work through their application in the problems class.
### 3) 1D velocity-component distribution: derivation and averages
We begin by deriving the probability distribution for a single velocity component, $v_x$. We start with the probability distribution for momentum in one dimension, $P[p_x]$, where the kinetic energy is $p_x^2/(2m)$. The unnormalised distribution is proportional to $e^{-p_x^2/(2mkT)}$. To normalise this, we integrate from $-\infty$ to $\infty$ and set the result equal to one. Using the standard integral $\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\pi/\alpha}$, where $\alpha = 1/(2mkT)$, we find the normalisation constant.
Next, we convert this momentum distribution to a velocity distribution. Since $p_x = mv_x$, it follows that $\text{d}p_x = m\,\text{d}v_x$. Substituting these into the normalised momentum distribution yields the one-dimensional velocity distribution:
$$ P[v_x] dv_x = \left( \frac{m}{2 \pi k T} \right)^{1/2} e^{- \frac{m v_x^2}{2 k T}} dv_x $$
This distribution, as shown in the accompanying graph, is a symmetric Gaussian curve centred at $v_x = 0$. This symmetry immediately tells us that the average velocity, $\overline{v_x}$, is zero. To obtain a meaningful measure of the typical motion, we calculate the mean square velocity, $\overline{v_x^2}$.
To find $\overline{v_x^2}$, we integrate $v_x^2$ multiplied by the probability distribution $P[v_x]$ over all possible velocities:
$$ \overline{v_x^2} = \int_{-\infty}^{\infty} v_x^2 P[v_x] dv_x $$
Using another standard integral (which would be provided in an exam), this integral evaluates to:
$$ \overline{v_x^2} = \frac{kT}{m} $$
From this, we can determine the average kinetic energy for this single degree of freedom:
$$ \frac{1}{2} m \overline{v_x^2} = \frac{1}{2} m \left( \frac{kT}{m} \right) = \frac{1}{2} kT $$
This result perfectly aligns with the equipartition theorem, which states that each quadratic degree of freedom contributes $\frac{1}{2}kT$ to the average energy. This one-dimensional distribution is conceptually useful for understanding phenomena such as Doppler broadening in spectroscopy and simple models of an isothermal atmosphere.
### 4) Why speed distributions differ from component Gaussians: state counting
The distribution of speeds for particles in two or three dimensions differs significantly from the simple Gaussian shape of a single velocity component. This difference arises from a crucial concept known as "state counting" or the "density of states." While a small interval $\text{d}v_x$ directly corresponds to a small interval of speed $\text{d}c$ in one dimension, this is not true in higher dimensions. Many different combinations of velocity components (e.g., $v_x, v_y, v_z$) can result in the same overall speed $c = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
Consider the geometry in velocity space. In two dimensions, all velocity vectors $(\vec{v} = v_x \hat{i} + v_y \hat{j})$ that have the same speed $c$ lie on a circle of radius $c$. The number of available "speed states" within a small speed interval from $c$ to $c+\text{d}c$ is proportional to the circumference of this circle, which is $2\pi c\,\text{d}c$.
Extending this to three dimensions, all velocity vectors with the same speed $c$ lie on the surface of a sphere of radius $c$. The number of available speed states in the interval from $c$ to $c+\text{d}c$ is now proportional to the surface area of this sphere, which is $4\pi c^2\,\text{d}c$.
This geometric factor has a profound consequence for the shape of speed distributions. The speed distribution is formed by multiplying the exponentially decaying Boltzmann factor ($e^{-mc^2/2kT}$) by this "states factor." As a result:
* In two dimensions, the speed distribution is proportional to $c \, e^{-mc^2/2kT}$.
* In three dimensions, it's proportional to $c^2 \, e^{-mc^2/2kT}$.
Unlike the symmetric Gaussian for velocity components, these speed distributions start at zero (because there are no states with zero speed), rise to a peak as the states factor dominates, and then decay exponentially at higher speeds as the Boltzmann factor takes over.
### 5) 2D speed distribution: form and average kinetic energy
Combining the Boltzmann factor with the two-dimensional state-counting factor ($2\pi c\,\text{d}c$) and then normalising, we arrive at the two-dimensional speed distribution:
$$ P[c] dc = c \frac{m}{kT} e^{-mc^2 / 2kT} dc $$
To find the average kinetic energy in two dimensions, we first calculate the mean square speed, $\overline{c^2}$. This involves integrating $c^2$ multiplied by the distribution $P[c]$ over all possible speeds from $0$ to $\infty$. Using a standard integral for the $c^3$-term, we find:
$$ \overline{c^2} = \frac{2kT}{m} $$
Therefore, the average kinetic energy in two dimensions is $\frac{1}{2}m\overline{c^2} = \frac{1}{2}m \left( \frac{2kT}{m} \right) = kT$. This result perfectly aligns with the equipartition theorem, which predicts $\frac{1}{2}kT$ for each of the two quadratic degrees of freedom (for $v_x$ and $v_y$), summing to $kT$. The shape of this distribution, as shown in the graph, illustrates two competing effects: the linear rise at small speeds is due to the increasing number of available states, while the decay at larger speeds is governed by the Boltzmann exponential term.
### 6) 3D Maxwell-Boltzmann speed distribution: full form and anatomy
To arrive at the full three-dimensional Maxwell-Boltzmann speed distribution, we begin with the joint probability distribution for the three velocity components $v_x, v_y, v_z$:
$$ P[v_x v_y v_z] \, dv_x dv_y dv_z = \left( \frac{m}{2 \pi k T} \right)^{3/2} e^{\frac{-m(v_x^2 + v_y^2 + v_z^2)}{2kT}} dv_x dv_y dv_z $$
Using the relationship $c^2 = v_x^2 + v_y^2 + v_z^2$ and incorporating the three-dimensional state-counting factor $4\pi c^2\,\text{d}c$, we can convert this into a distribution for speed $c$. After normalisation, the result is the standard Maxwell-Boltzmann speed distribution:
$$ P[c] \, dc = 4 \pi \left( \frac{m}{2 \pi k T} \right)^{3/2} c^2 e^{- \frac{m c^2}{2 k T}} \, dc $$
The anatomy of this curve is crucial for understanding its characteristic shape. The $c^2$ term, originating from the increasing number of available speed states (the surface area of a sphere in velocity space), causes the probability to initially rise from zero at small speeds. Conversely, the exponential Boltzmann term, $e^{-mc^2/2kT}$, rapidly suppresses the probability at large speeds. The product of these two competing factors results in the familiar bell-shaped Maxwell-Boltzmann distribution, which peaks at an intermediate speed, representing the most probable speed for particles in the gas. It's important to remember that this is a distribution of *speeds*, which are always positive, unlike the velocity component distributions which are symmetric about zero.
### 7) What was deferred and what comes next
In this lecture, the emphasis was placed on the conceptual construction of the Maxwell-Boltzmann distributions. We completed the full derivation for the one-dimensional velocity component distribution, including its normalisation and the calculation of its average kinetic energy. For the two- and three-dimensional distributions, we focused on understanding the crucial role of the "state-counting" factor (proportional to $c$ in 2D and $c^2$ in 3D) and presented the final forms of these distributions, sketching the normalisation and integral results.
A full, worked example using these distributions and the necessary standard integrals will be covered in tomorrow's problems class. This session will provide an opportunity to practise these calculations and clarify any questions you may have regarding normalisation and the computation of average values from the distributions.
## Key takeaways
* The Boltzmann factor $e^{-E/kT}$ governs how particles populate energy levels; as $T \to \infty$, all levels become equally populated. Population inversion, as seen in lasers, implies a negative absolute temperature but is not typical for the systems we study.
* Starting from $P \propto e^{-E/kT}$, the one-dimensional velocity component distribution is a Gaussian: $P[v_x] dv_x = \left( \frac{m}{2 \pi k T} \right)^{1/2} e^{- \frac{m v_x^2}{2 k T}} dv_x$. The average velocity $\overline{v_x}$ is zero, but the mean square velocity $\overline{v_x^2} = kT/m$, leading to an average kinetic energy per component of $\frac{1}{2}kT$.
* Speed distributions differ from component Gaussians due to the geometric effect of "state counting" in velocity space. In 2D, $P[c] \propto c \, e^{-mc^2/2kT}$, with an average kinetic energy of $kT = 2 \times (\frac{1}{2}kT)$. In 3D, the full Maxwell-Boltzmann speed distribution is $P[c] \, dc = 4 \pi \left( \frac{m}{2 \pi k T} \right)^{3/2} c^2 e^{- \frac{m c^2}{2 k T}} \, dc$.
* The characteristic shape of the Maxwell-Boltzmann speed distributions (starting at zero, rising to a peak, then decaying) is a result of the competition between a growing states factor ($c$ or $c^2$) and a decaying Boltzmann exponential term.
* Standard Gaussian integrals are essential for normalising these distributions and calculating average values. These integrals will be provided in exams and thoroughly practised in the problems class.