From 2b08403958bbf804b7067b6dca8f0c68050bd48c Mon Sep 17 00:00:00 2001
From: Bowy La Riviere <b.m.lariviere@student.tudelft.nl>
Date: Sun, 14 Mar 2021 14:32:03 +0000
Subject: [PATCH] Apply 17 suggestion(s) to 1 file(s)

---
 src/10_xray.md | 48 +++++++++++++++++++++++-------------------------
 1 file changed, 23 insertions(+), 25 deletions(-)

diff --git a/src/10_xray.md b/src/10_xray.md
index 8e72ef0a..ea986823 100644
--- a/src/10_xray.md
+++ b/src/10_xray.md
@@ -265,16 +265,15 @@ fig.show()
 
 ```
 
-Despite us knowing that both real-space and reciprocal lattice vectors are somehow related to eachother, we do not yet know their exact relation.
-We do however know that they are subject to
+To find the reciprocal lattice vectors, we use the relation
 $$
 \mathbf{a_i}\cdot\mathbf{b_j}=2\pi\delta_{ij}.
 $$ 
-This tells us that
+The relation leads to several simple conclusions.
+One such conclusion is the orthogonality between several real-space and reciprocal lattice vectors: 
 $$
-\mathbf{a}_1\cdot\mathbf{b}_2=\mathbf{a}_2\cdot\mathbf{b}_1=0,
+\mathbf{a}_1\cdot\mathbf{b}_2=\mathbf{a}_2\cdot\mathbf{b}_1=0.
 $$
-showing us that $\mathbf{a}_1$ is perpendicular to $\mathbf{b}_2$ and $\mathbf{a}_2$ is perpendicular to $\mathbf{b}_1$.
 We also find
 $$
 \mathbf{a}_1\cdot\mathbf{b}_1=\mathbf{a}_2\cdot\mathbf{b}_2=2\pi.
@@ -284,16 +283,16 @@ $$
 \lvert \mathbf{a}_1 \rvert \lvert \mathbf{b}_1 \rvert =\frac{2\pi}{\cos(\theta_1)} \:\: \text{and} \:\: \lvert \mathbf{a}_2 \rvert \lvert \mathbf{b}_2 \rvert =\frac{2\pi}{\cos(\theta_2)},
 $$
 where $\theta_i$ is the angle between the vectors $\mathbf{a}_i$ and $\mathbf{b}_i$.
-To find the angles $\theta_1$ and $\theta_2$, we use the fact that the angle between $\mathbf{a}_1$ and $\mathbf{a}_2$ is $60^\circ$,that $\mathbf{a}_1 \perp \mathbf{b}_2$ and that $\mathbf{a}_2 \perp \mathbf{b}_1$. 
-From this we conclude that $\theta_1 = \theta_2 = 30^\circ$.
+To find the angles $\theta_1$ and $\theta_2$, we use the orthogonality relations above and the fact that the angle between $\mathbf{a}_1$ and $\mathbf{a}_2$ is $60^\circ$.
+From this we conclude that $\theta_1 = \theta_2 = 30^\circ$.```
 Because $\lvert \mathbf{a}_1 \rvert = \lvert \mathbf{a}_2 \rvert = a$, we find
 $$
 \lvert \mathbf{b}_1 \rvert = \lvert \mathbf{b}_2 \rvert = \frac{4\pi}{a\sqrt{3}}. 
 $$
-Unsurprisingly, we find that the lengths of the reciprocal lattice vectors are the same and that the reciprocal lattice has a reciprocal dependence on the lattice constant $a$.
+Unsurprisingly, we find that the lengths of the reciprocal lattice vectors are equal and have inverse dependence on the lattice constant $a$.
 With $\lvert \mathbf{b}_2 \rvert$ and $\mathbf{a}_1 \perp \mathbf{b}_2$, we easily find 
 $$
-\mathbf{b}_2 = \frac{4\pi}{a\sqrt{3}} \mathbf{\hat{y}}
+\mathbf{b}_2 = \frac{4\pi}{a\sqrt{3}} \mathbf{\hat{y}}.
 $$
 
 We follow the same procedure to find $\mathbf{b}_1$:
@@ -304,7 +303,7 @@ $$
 
 ??? Question "Is the choice of a set of reciprocal lattice vectors unique? If not, which other ones are possible?"
     No. As is the case for the real-space lattice vectors, the choice of a set of reciprocal lattice vectors is not unique.
-    There are multiple sets of reciprocal lattcie vectors that fullfil all criteria.
+    There are multiple sets of reciprocal lattice vectors that fulfill all criteria.
     The reciprocal lattice vectors that also fullfil the criteria are
     $$
     \mathbf{b}_1 = \frac{4\pi}{a\sqrt{3}} \left(-\frac{\sqrt{3}}{2} \mathbf{\hat{x}} + \frac{1}{2}\mathbf{\hat{y}} \right) \quad \text{and} \quad \mathbf{b}_2 = -\frac{4\pi}{a\sqrt{3}} \mathbf{\hat{y}}.
@@ -327,39 +326,38 @@ $$
 \mathbf{b_3}=\frac{2\pi(\mathbf{a}_1\times\mathbf{a}_2)}{ \mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a_3})}
 $$
 
-Note that the denominator $V = \mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a_3})$ is the volume of the real-space unit cell spanned by the real-space lattice vectors $\mathbf{a}_1$, $\mathbf{a}_2$ and $\mathbf{a}_3$. 
-These definitions of the reciprocal lattice vectros are cyclic: $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a_3})=\mathbf{a}_2\cdot(\mathbf{a_3}\times\mathbf{a}_1)=\mathbf{a_3}\cdot(\mathbf{a}_1\times\mathbf{a}_2)$.
+Note that the denominator $V = \mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a_3})$ is the volume of the real-space unit cell spanned by the lattice vectors $\mathbf{a}_1$, $\mathbf{a}_2$ and $\mathbf{a}_3$. 
+The definitions of the reciprocal lattice vectros are cyclic: $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a_3})=\mathbf{a}_2\cdot(\mathbf{a_3}\times\mathbf{a}_1)=\mathbf{a_3}\cdot(\mathbf{a}_1\times\mathbf{a}_2)$.
 
 ### The reciprocal lattice as a Fourier transform
-One can also think of the reciprocal lattice as being a Fourier transform of the real-space lattice. 
-For simplicity, we illustrate this for a 1D lattice (the same principles can be applied to a 3D lattice).
-We imagine the real-space lattice as a density function consisting of delta peaks:
+One can also think of the reciprocal lattice as a Fourier transform of the real-space lattice. 
+For simplicity, we illustrate this for a 1D lattice (the same principles apply to a 3D lattice).
+We model the real-space lattice as a density function consisting of delta peaks:
 
 $$
 \rho(x)=\sum_{n} \delta(x-na)
 $$
 
-We take the Fourier transform of this function:
+We take the Fourier transform of this function to find:
 
 $$
-{\mathcal F}_{k}\left[\rho(x)\right]=\int_\infty^\infty \mathrm{d}x\ \mathrm{e}^{ikx} \rho(x)=\sum_{n} \int_\infty^\infty \mathrm{d}x\ \mathrm{e}^{ikx} \delta(x-na)=\sum_{n} \mathrm{e}^{ikna}
+{\mathcal F}_{k}\left[\rho(x)\right]=\int_{-\infty}^\infty \mathrm{d}x\ \mathrm{e}^{ikx} \rho(x)=\sum_{n} \int_{-\infty}^\infty \mathrm{d}x\ \mathrm{e}^{ikx} \delta(x-na)=\sum_{n} \mathrm{e}^{ikna}
 $$
 
 This sum is non-zero only if $k=2\pi m/a$.
-Therefore, we can rewrite this as:
+If we recall the beginning of the lecture, then these points correspond to reciprocal lattice points $G$. 
+Therefore, we rewrite this into the form
 
 $$
-{\mathcal F}_{k}\left[\rho(x)\right]=\frac{2\pi}{|a|}\sum_{m} \delta\left(k-\frac{2\pi m}{a}\right)
+{\mathcal F}_{k}\left[\rho(x)\right]=\frac{2\pi}{|a|}\sum_{m} \delta\left(k-G\right).
 $$
-However, in the beginning of the lecture we studied a 1D reciprocal lattice and saw that the reciprocal lattice points are given by $G = 2\pi m/a$.
-Thus the values of $k$ lie on reciprocal lattice points and the Fourier tranform of the real-space lattice is thus a reciprocal lattice.
-In other words, Fourier transforming a real-space lattice yields a reciprocal lattice!
-The above can easily be generalized to three dimensions:
+Therefore, we see that the Fourier transform is non-zero only at reciprocal lattice points.
+In other words, Fourier transforming a real-space lattice yields a reciprocal lattice! 
+The above result generalizes directly to three dimensions:
 
 $$
-{\mathcal F}_\mathbf{k}\left[\rho(\mathbf{r})\right]=\int \mathrm{d}\mathbf{r}\ \mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}} \rho(\mathbf{r}) = \sum_\mathbf{G}\delta(\mathbf{k}-\mathbf{G}),
+{\mathcal F}_\mathbf{k}\left[\rho(\mathbf{r})\right]=\int \mathrm{d}\mathbf{r}\ \mathrm{e}^{i\mathbf{k}\cdot\mathbf{r}} \rho(\mathbf{r}) = \sum_\mathbf{G}\delta(\mathbf{k}-\mathbf{G}).
 $$
-with $\mathbf{G}$ being the reciprocal lattice.
 
 
 ### Periodicity of the reciprocal lattice
-- 
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