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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 6: Matter VI: LJF Meets Young Modulus and Breaking Materials</b><br>
Thu 05 Feb 2026 13:00
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== Young’s Modulus Meets Lennard-Jones ==

On a microscopic scale, applying a linear force extends the distance between neighbouring atoms:

[[File:TM-Y1S2-lec06-01.png|thumb|center|600px]]

The mean interatomic distance changes from <nowiki>$r_0 \to r_0 + dr$</nowiki> for an infinitesimal force <nowiki>$dF$</nowiki>. Near the equilibrium point, where the material is in the elastic region, we expect the extension <nowiki>$dr \propto dF$</nowiki>, as in this region <nowiki>$\Delta L \propto F$</nowiki>.

Going back to the Lennard-Jones Force, we have a small region around the equilibrium point:

[[File:TM-Y1S2-lec06-02.png|thumb|center|600px]]

If we zoomed into this region, we would see it is approximately linear:

[[File:TM-Y1S2-lec06-03.png|thumb|center|600px]]

Hence, around <nowiki>$r_0$</nowiki>, <nowiki>$dF/dr$</nowiki> is constant. Going back to Young’s Modulus, we have:

<div class="pn-import-equation"><nowiki>$$Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\delta Y / Y}$$</nowiki></div>

Here, the force is just <nowiki>$dF$</nowiki>. The area this force is acting over (microscopically) is actually a square, as the material is a cubic lattice, <nowiki>$A = (r_0 + dr)^2$</nowiki>. Hence stress is: <div class="pn-import-equation"><nowiki>$$\text{stress} = -\frac{dF}{(r_0 + dr)^2}$$</nowiki></div>

The negative sign compensates for the fact that the LJF opposes the deformation (is negative under tension), ensuring the stress and strain have the same sign and their ratio is positive.

<div class="pn-import-equation"><nowiki>$$\text{strain} = \frac{dr}{r_0}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies Y = -\frac{\left( \frac{dF}{(r_0 + dr)^2}\right)}{\left(\frac{dr}{r_0}\right)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= -\frac{dF r_0}{dr (r_0 + dr)^2}$$</nowiki></div>

We take the quadratic to first order, assuming <nowiki>$dr \ll r_0$</nowiki>:

<div class="pn-import-equation"><nowiki>$$= - \frac{1}{r_0} \frac{dF}{dr}\biggr\rvert_{r = r_0}$$</nowiki></div>

As force is <nowiki>$- dV / dr$</nowiki> we can show that: <div class="pn-import-equation"><nowiki>$$Y = \frac{1}{r_0} \frac{d^2 V}{dr^2} \biggr\rvert_{r = r_0}$$</nowiki></div>

And using the definition: <div class="pn-import-equation"><nowiki>$$V = \epsilon \left(\frac{r_0^{12}}{r^{12}} - 2 \frac{r_0^6}{r^6}\right)$$</nowiki></div> Hence: <div class="pn-import-equation"><nowiki>$$F = - \frac{dV}{dr} = -\epsilon \left(\frac{12r_0^{12}}{r^{13}} - 12 \frac{r_0^6}{r^7}\right) = \epsilon \left(12 \frac{r_0^6}{r^7} - 12\frac{\epsilon_0^{12}}{r^{13}}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$Y = - \frac{dF}{dr} = \frac{d^2 V}{dr^2} = \epsilon \left(\frac{156r_0^{12}}{r^{14}} - \frac{84 r_0^6}{r^8}\right)_{r = r_0}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$Y = \frac{156 \epsilon}{r_0^2} - \frac{84 \epsilon}{r_0^2} = \boxed{\frac{72 \epsilon}{r_0^2}}$$</nowiki></div>

This provides a microscopic definition of the Young’s Modulus to complement our previous macroscopic definition.

== Breaking Materials ==

Materials break when the attractive force between atoms is less than the force applied to the material. This occurs when: <div class="pn-import-equation"><nowiki>$$\frac{dF}{dr} = 0 \implies F = F_\text{max}$$</nowiki></div>

If we pull continuously with a force greater than <nowiki>$F_\text{max}$</nowiki> we will pull atoms apart and break the material. The maximum definable stress is based on the maximum attractive LJF over the area of a single molecule:

<div class="pn-import-equation"><nowiki>$$\text{Breaking Stress} = \frac{F_\text{max}}{r^2_0}$$</nowiki></div>

=== Deriving <nowiki>$r$</nowiki> at <nowiki>$F_\text{max}$</nowiki> ===

<nowiki>$F_\text{max}$</nowiki> occurs where <nowiki>$\frac{dF}{dr} = 0$</nowiki>. Using the earlier result: <div class="pn-import-equation"><nowiki>$$F(r) = 12\epsilon \left(\frac{r_0^{12}}{r^{13}} - \frac{r_0^6}{r^7}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{dF}{dr} = 12\epsilon \left(-\frac{13r_0^{12}}{r^{14}} + \frac{7r_0^6}{r^{8}}\right) = 0$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{7r_0^6}{r^8} = \frac{13r_0^{12}}{r^{14}} \implies 7r^6 = 13r_0^6$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies \boxed{r = \sqrt[6]{\frac{13}{7}}\, r_0}$$</nowiki></div>

=== Finding <nowiki>$F_\text{max}$</nowiki> ===

Substituting <nowiki>$r^6 = \frac{13}{7}r_0^6$</nowiki> into <nowiki>$F(r)$</nowiki>. Simplifying each term: <div class="pn-import-equation"><nowiki>$$\frac{r_0^{12}}{r^{13}} = \frac{1}{r}\cdot\left(\frac{r_0^6}{r^6}\right)^2 = \frac{1}{r}\cdot\frac{49}{169}, \qquad \frac{r_0^{6}}{r^{7}} = \frac{1}{r}\cdot\frac{r_0^6}{r^6} = \frac{1}{r}\cdot\frac{7}{13}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$F_\text{max} = \frac{12\epsilon}{r}\left(\frac{49}{169} - \frac{91}{169}\right) = -\frac{504\epsilon}{169\,r}$$</nowiki></div>

Substituting <nowiki>$r = \left(\frac{13}{7}\right)^{1/6} r_0$</nowiki>: <div class="pn-import-equation"><nowiki>$$\implies \boxed{|F_\text{max}| = \frac{504\epsilon}{169\,r_0}\left(\frac{7}{13}\right)^{1/6} \approx \frac{2.69\,\epsilon}{r_0}}$$</nowiki></div>

=== Breaking Stress ===

Hence the breaking stress is: <div class="pn-import-equation"><nowiki>$$BS = \frac{F_\text{break}}{A} = \frac{F_\text{max}}{r_0^2}$$</nowiki></div> Macroscopically and microscopically.

== Fractures ==

Say your material has some tiny imperfection of depth <nowiki>$d$</nowiki> and width <nowiki>$2r$</nowiki>.

[[File:TM-Y1S2-lec06-04.png|thumb|center|600px]]

This causes stress to increase by a factor of: <div class="pn-import-equation"><nowiki>$$\left(1 + 2 \sqrt{\frac{d}{r}}\right)$$</nowiki></div>

The breaking force decreases with increasing <nowiki>$d$</nowiki>, which makes sense as a deeper groove will cause a much smaller cross sectional area and a larger stress.

The breaking force increases with a decreasing <nowiki>$r$</nowiki>, where grooves on atomic length scales wide causes a huge increase in the stress.

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc