bridge derivation: Stiffness (Young’s Modulus) from Interatomic Potentials

[bhpayne]

prompt

In the Latex file below, where there are two equations with no explanatory text between them, modify the Latex file by adding explanatory text to document the mathematical transformations that relate each equation in the file to other equations in the file.

Modify the Latex file below by adding explanatory text about each step of the derivation. Document the mathematical transformations that relate each equation in the file to other equations in the file.

For the following Latex file, write out the file and insert a label for each equation. The label that is added should be descriptive of the equation and contain no spaces, only underscores.

Labels for equations should be unique within the document.

The following Latex file contains a derivation with many steps. Sometimes a previous expression is referenced implicitly. Edit the following text to manually insert references where appropriate. For example replace text like "substitute the forces" with "substitute Eq.~\ref{} to get". Be sure to reference the relevant equation labels so that Latex can compile the document.

Just write the revised Latex file content as your answer.

\documentclass{article}
\usepackage{amsmath}

\title{Derivation of Stiffness (Young's Modulus) from Interatomic Potentials}
\date{\today}
\author{Ben Payne with Gemini 2.0}

% margins of 1 inch:
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}

\usepackage[pdftex]{hyperref} % hyperlink equation and bibliographic citations


\begin{document}
\maketitle


\section{Introduction}

Stiffness, as quantified by Young's modulus ($E$), describes a material's resistance to elastic deformation under tensile or compressive stress. At the atomic level, stiffness arises from the interatomic forces that hold atoms together in a solid. When a material is stretched, the interatomic distances increase, and the restoring forces between atoms resist this change, leading to macroscopic stiffness.

Let's consider a simplified model of a solid where atoms are arranged in a regular lattice and interact through a pairwise potential.  We will focus on a one-dimensional chain of atoms for simplicity, but the principles extend to three-dimensional solids.

\subsection{Step 1: Interatomic Potential}

We assume that the interaction between two adjacent atoms can be described by a potential energy function $U(r)$, where $r$ is the distance between the atoms.  A common model potential is the Lennard-Jones potential, but for simplicity in this derivation, we can consider a general potential $U(r)$ that has a minimum at the equilibrium distance $r_0$.

The force $F(r)$ between two atoms is the negative derivative of the potential energy with respect to the distance:
\begin{equation}
F(r) = - \frac{dU(r)}{dr}
\end{equation}
At equilibrium distance $r_0$, the force is zero, i.e., $F(r_0) = 0$, and the potential energy is at a minimum.

\subsection{Step 2: Restoring Force for Small Displacements}

When we stretch or compress the material slightly, the interatomic distance deviates from the equilibrium distance $r_0$. Let's consider a small displacement $\delta r = r - r_0$. For small displacements, we can approximate the force using a Taylor expansion around $r_0$:
\begin{equation}
F(r) \approx F(r_0) + \left. \frac{dF}{dr} \right|_{r=r_0} (r - r_0) + \mathcal{O}((r-r_0)^2)
\end{equation}
Since $F(r_0) = 0$ and we consider only linear elasticity (small displacements), we can neglect higher-order terms. Also, $\frac{dF}{dr} = -\frac{d^2U}{dr^2}$. Let $k = \left. \frac{d^2U}{dr^2} \right|_{r=r_0}$ be the second derivative of the potential energy at the equilibrium distance. This $k$ represents the "spring constant" for the interatomic bond. Then, the restoring force for small displacements is approximately:
\begin{equation}
F(r) \approx -k (r - r_0) = -k \delta r
\end{equation}
Here, $k$ represents the stiffness of the interatomic bond.

\subsection{Step 3: Macroscopic Stress and Strain}

Consider a one-dimensional chain of atoms with equilibrium spacing $r_0$.  Let $A$ be the effective cross-sectional area per chain (in 3D, this would be the area per atom in a plane perpendicular to the tensile force). When we apply a macroscopic tensile force $\mathcal{F}$ to stretch the material, each interatomic bond experiences a force approximately equal to $\mathcal{F}$. The macroscopic stress $\sigma$ is the force per unit area:
\begin{equation}
\sigma = \frac{\mathcal{F}}{A}
\end{equation}
The macroscopic strain $\epsilon$ is the fractional change in length. If we consider a length $L_0 = N r_0$ consisting of $N$ atoms in equilibrium, and the length becomes $L = N r$ after stretching, the strain is:
\begin{equation}
\epsilon = \frac{L - L_0}{L_0} 
\end{equation}
\begin{equation}
\epsilon = \frac{N r - N r_0}{N r_0} 
\end{equation}
\begin{equation}
\epsilon = \frac{r - r_0}{r_0} 
\end{equation}
\begin{equation}
\epsilon =  \frac{\delta r}{r_0}
\end{equation}

\subsection{Step 4: Young's Modulus}

From Hooke's Law, the macroscopic stress is related to the macroscopic strain by Young's modulus $E$:
\begin{equation}
\sigma = E \epsilon
\end{equation}
Substituting the expressions for stress and strain from atomic considerations, we have:
\begin{equation}
\frac{\mathcal{F}}{A} = E \frac{\delta r}{r_0}
\end{equation}
Also, from Step 2, we have $\mathcal{F} \approx k \delta r$ (magnitude of restoring force equals applied force in equilibrium). Substituting this into the previous equation:
\begin{equation}
\frac{k \delta r}{A} = E \frac{\delta r}{r_0}
\end{equation}
We can cancel out $\delta r$ (for small non-zero strain), and solve for Young's modulus $E$:
\begin{equation}
E = \frac{k r_0}{A}
\end{equation}
\begin{equation}
E = \frac{r_0}{A} \left. \frac{d^2U(r)}{dr^2} \right|_{r=r_0}
\end{equation}
Thus, Young's modulus $E$ is related to the second derivative of the interatomic potential at the equilibrium distance, the equilibrium interatomic distance $r_0$, and the effective cross-sectional area per chain $A$.  For a given material and crystal structure, $r_0$ and $A$ are determined by atomic packing, and $k$ is determined by the strength and shape of the interatomic potential.

\subsection{Lennard-Jones Potential}

For the Lennard-Jones potential:
\begin{equation}
U(r) = 4 \epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\end{equation}
where $\epsilon$ is the depth of the potential well and $\sigma$ is the distance at which $U(r) = 0$. The equilibrium distance $r_0 = 2^{1/6} \sigma$. Calculating the second derivative and evaluating it at $r_0$ will give a specific expression for $k$ in terms of $\epsilon$ and $\sigma$, which can then be used to estimate Young's modulus.


\end{document}

thoughts

The user wants me to modify a Latex file.
The Latex file describes a derivation of Young's Modulus from interatomic potentials.
I need to perform the following modifications:

1. Add explanatory text between equations to describe the mathematical transformations.
2. Add labels to each equation. The labels should be descriptive and use underscores, no spaces, and be unique.
3. Add references to equations where appropriate. Replace phrases like "substitute the forces" with "substitute Eq.~\ref{eq:label} to get".

Let's go through the Latex file step by step and apply these modifications.

Step 1: Interatomic Potential
Equation: F(r) = - dU(r)/dr
Label: Force_from_Potential
Explanatory text before equation: "The force $F(r)$ between two atoms is the negative derivative of the potential energy with respect to the distance:"

Step 2: Restoring Force for Small Displacements
Equation 1: F(r) approx F(r_0) + dF/dr|_{r=r_0} (r - r_0) + O((r-r_0)^2)
Label 1: Taylor_Expansion_Force
Explanatory text before equation: "For small displacements, we can approximate the force using a Taylor expansion around $r_0$:"
Equation 2: F(r) approx -k (r - r_0) = -k delta r
Label 2: Restoring_Force_Harmonic_Approx
Explanatory text between equations: "Since $F(r_0) = 0$ and neglecting higher-order terms in Eq.~\ref{Taylor_Expansion_Force}, and letting $k = d^2U/dr^2|_{r=r_0}$, we obtain the restoring force for small displacements:"

Step 3: Macroscopic Stress and Strain
Equation 1: sigma = F/A
Label 1: Macroscopic_Stress_Definition
Explanatory text before equation: "The macroscopic stress $\sigma$ is the force per unit area:"
Equation 2: epsilon = (L - L_0)/L_0
Label 2: Strain_Definition_Length_Change
Explanatory text between equations: "The macroscopic strain $\epsilon$ is the fractional change in length:"
Equation 3: epsilon = (Nr - Nr_0)/(Nr_0)
Label 3: Strain_Length_Expanded
Explanatory text between equations: "Expressing the lengths in terms of the number of atoms $N$ and interatomic distances $r$ and $r_0$:"
Equation 4: epsilon = (r - r_0)/r_0
Label 4: Strain_Simplified_N_Cancelled
Explanatory text between equations: "Simplifying by cancelling $N$:"
Equation 5: epsilon = delta r / r_0
Label 5: Strain_in_terms_delta_r
Explanatory text between equations: "Letting $\delta r = r - r_0$, we can write the strain as:"

Step 4: Young's Modulus
Equation 1: sigma = E epsilon
Label 1: Hookes_Law
Explanatory text before equation: "From Hooke's Law, the macroscopic stress is related to the macroscopic strain by Young's modulus $E$:"
Equation 2: F/A = E delta r / r_0
Label 2: Stress_Strain_Atomic_Expressions
Explanatory text between equations: "Substituting the expressions for stress from Eq.\ref{Macroscopic_Stress_Definition} and strain from Eq.\ref{Strain_in_terms_delta_r} into Eq.\ref{Hookes_Law}, we have:"
Equation 3: (k delta r)/A = E delta r / r_0
Label 3: Force_k_delta_r_Substitution
Explanatory text between equations: "Substituting $\mathcal{F} \approx k \delta r$ from Eq.\ref{Restoring_Force_Harmonic_Approx} into Eq.\ref{Stress_Strain_Atomic_Expressions}:"
Equation 4: E = k r_0 / A
Label 4: Youngs_Modulus_k_r0_A
Explanatory text between equations: "Cancelling out $\delta r$ and solving for Young's modulus $E$:"
Equation 5: E = r_0/A d^2U(r)/dr^2|_{r=r_0}
Label 5: Youngs_Modulus_Potential_Second_Derivative
Explanatory text between equations: "Substituting $k = d^2U/dr^2|_{r=r_0}$ into Eq.\ref{Youngs_Modulus_k_r0_A}:"

Lennard-Jones Potential
Equation 1: U(r) = 4 epsilon [ (sigma/r)^12 - (sigma/r)^6 ]
Label 1: Lennard_Jones_Potential
Explanatory text before equation: "For the Lennard-Jones potential:"

Finally, I will assemble all the modified parts into a complete Latex file. I will double check the equation labels for uniqueness and correctness of references.

response

\documentclass{article}
\usepackage{amsmath}

\title{Derivation of Stiffness (Young's Modulus) from Interatomic Potentials}
\date{\today}
\author{Ben Payne with Gemini 2.0}

% margins of 1 inch:
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}

\usepackage[pdftex]{hyperref} % hyperlink equation and bibliographic citations


\begin{document}
\maketitle


\section{Introduction}

Stiffness, as quantified by Young's modulus ($E$), describes a material's resistance to elastic deformation under tensile or compressive stress. At the atomic level, stiffness arises from the interatomic forces that hold atoms together in a solid. When a material is stretched, the interatomic distances increase, and the restoring forces between atoms resist this change, leading to macroscopic stiffness.

Let's consider a simplified model of a solid where atoms are arranged in a regular lattice and interact through a pairwise potential.  We will focus on a one-dimensional chain of atoms for simplicity, but the principles extend to three-dimensional solids.

\subsection{Step 1: Interatomic Potential}

We assume that the interaction between two adjacent atoms can be described by a potential energy function $U(r)$, where $r$ is the distance between the atoms.  A common model potential is the Lennard-Jones potential, but for simplicity in this derivation, we can consider a general potential $U(r)$ that has a minimum at the equilibrium distance $r_0$.

The force $F(r)$ between two atoms is the negative derivative of the potential energy with respect to the distance:
\begin{equation}
F(r) = - \frac{dU(r)}{dr}
\label{eq:Force_from_Potential}
\end{equation}
At equilibrium distance $r_0$, the force is zero, i.e., $F(r_0) = 0$, and the potential energy is at a minimum.

\subsection{Step 2: Restoring Force for Small Displacements}

When we stretch or compress the material slightly, the interatomic distance deviates from the equilibrium distance $r_0$. Let's consider a small displacement $\delta r = r - r_0$. For small displacements, we can approximate the force using a Taylor expansion around $r_0$:
\begin{equation}
F(r) \approx F(r_0) + \left. \frac{dF}{dr} \right|_{r=r_0} (r - r_0) + \mathcal{O}((r-r_0)^2)
\label{eq:Taylor_Expansion_Force}
\end{equation}
Since $F(r_0) = 0$ and we consider only linear elasticity (small displacements), we can neglect higher-order terms in Eq.~\ref{eq:Taylor_Expansion_Force}. Also, $\frac{dF}{dr} = -\frac{d^2U}{dr^2}$. Let $k = \left. \frac{d^2U}{dr^2} \right|_{r=r_0}$ be the second derivative of the potential energy at the equilibrium distance. This $k$ represents the "spring constant" for the interatomic bond. Then, the restoring force for small displacements is approximately:
\begin{equation}
F(r) \approx -k (r - r_0) = -k \delta r
\label{eq:Restoring_Force_Harmonic_Approx}
\end{equation}
Here, $k$ represents the stiffness of the interatomic bond.

\subsection{Step 3: Macroscopic Stress and Strain}

Consider a one-dimensional chain of atoms with equilibrium spacing $r_0$.  Let $A$ be the effective cross-sectional area per chain (in 3D, this would be the area per atom in a plane perpendicular to the tensile force). When we apply a macroscopic tensile force $\mathcal{F}$ to stretch the material, each interatomic bond experiences a force approximately equal to $\mathcal{F}$. The macroscopic stress $\sigma$ is the force per unit area:
\begin{equation}
\sigma = \frac{\mathcal{F}}{A}
\label{eq:Macroscopic_Stress_Definition}
\end{equation}
The macroscopic strain $\epsilon$ is the fractional change in length. If we consider a length $L_0 = N r_0$ consisting of $N$ atoms in equilibrium, and the length becomes $L = N r$ after stretching, the strain is:
\begin{equation}
\epsilon = \frac{L - L_0}{L_0}
\label{eq:Strain_Definition_Length_Change}
\end{equation}
Expressing the lengths in Eq.~\ref{eq:Strain_Definition_Length_Change} in terms of the number of atoms $N$ and interatomic distances $r$ and $r_0$:
\begin{equation}
\epsilon = \frac{N r - N r_0}{N r_0}
\label{eq:Strain_Length_Expanded}
\end{equation}
Simplifying Eq.~\ref{eq:Strain_Length_Expanded} by cancelling $N$:
\begin{equation}
\epsilon = \frac{r - r_0}{r_0}
\label{eq:Strain_Simplified_N_Cancelled}
\end{equation}
Letting $\delta r = r - r_0$, we can write the strain from Eq.~\ref{eq:Strain_Simplified_N_Cancelled} as:
\begin{equation}
\epsilon =  \frac{\delta r}{r_0}
\label{eq:Strain_in_terms_delta_r}
\end{equation}

\subsection{Step 4: Young's Modulus}

From Hooke's Law, the macroscopic stress is related to the macroscopic strain by Young's modulus $E$:
\begin{equation}
\sigma = E \epsilon
\label{eq:Hookes_Law}
\end{equation}
Substituting the expressions for stress from Eq.~\ref{eq:Macroscopic_Stress_Definition} and strain from Eq.~\ref{eq:Strain_in_terms_delta_r} into Eq.~\ref{eq:Hookes_Law}, we have:
\begin{equation}
\frac{\mathcal{F}}{A} = E \frac{\delta r}{r_0}
\label{eq:Stress_Strain_Atomic_Expressions}
\end{equation}
Substituting $\mathcal{F} \approx k \delta r$ from Eq.~\ref{eq:Restoring_Force_Harmonic_Approx} into Eq.~\ref{eq:Stress_Strain_Atomic_Expressions}:
\begin{equation}
\frac{k \delta r}{A} = E \frac{\delta r}{r_0}
\label{eq:Force_k_delta_r_Substitution}
\end{equation}
Cancelling out $\delta r$ in Eq.~\ref{eq:Force_k_delta_r_Substitution} and solving for Young's modulus $E$:
\begin{equation}
E = \frac{k r_0}{A}
\label{eq:Youngs_Modulus_k_r0_A}
\end{equation}
Substituting $k = \left. \frac{d^2U(r)}{dr^2} \right|_{r=r_0}$ into Eq.~\ref{eq:Youngs_Modulus_k_r0_A}:
\begin{equation}
E = \frac{r_0}{A} \left. \frac{d^2U(r)}{dr^2} \right|_{r=r_0}
\label{eq:Youngs_Modulus_Potential_Second_Derivative}
\end{equation}
Thus, Young's modulus $E$ is related to the second derivative of the interatomic potential at the equilibrium distance, the equilibrium interatomic distance $r_0$, and the effective cross-sectional area per chain $A$.  For a given material and crystal structure, $r_0$ and $A$ are determined by atomic packing, and $k$ is determined by the strength and shape of the interatomic potential.

\subsection{Lennard-Jones Potential}

For the Lennard-Jones potential:
\begin{equation}
U(r) = 4 \epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\label{eq:Lennard_Jones_Potential}
\end{equation}
where $\epsilon$ is the depth of the potential well and $\sigma$ is the distance at which $U(r) = 0$. The equilibrium distance $r_0 = 2^{1/6} \sigma$. Calculating the second derivative and evaluating it at $r_0$ will give a specific expression for $k$ in terms of $\epsilon$ and $\sigma$, which can then be used to estimate Young's modulus.


\end{document}