From b9840b68687d47e6d42eead2d182fe60c902fddb Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Wed, 12 Feb 2020 09:21:57 +0000
Subject: [PATCH] Update 2_debye_model.md - typo
---
src/2_debye_model.md | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/src/2_debye_model.md b/src/2_debye_model.md
index 3fc73951..ab2cbab0 100644
--- a/src/2_debye_model.md
+++ b/src/2_debye_model.md
@@ -129,7 +129,7 @@ Periodic boundary conditions imply that the atomic displacement $\mathbf{\delta
$$
\mathbf{\delta r}(\mathbf{r} + L\mathbf{\hat{x}}) = \mathbf{\delta r}(\mathbf{r})
$$
-To satisfy this equation, we arrive at the condition $k_x=p 2 \pi/L$]$, with $p= ..., -2, -1, 0, 1, 2$ in $\mathbb{Z}$.
+To satisfy this equation, we arrive at the condition $k_x=p 2 \pi/L]$, with $p= ..., -2, -1, 0, 1, 2$ in $\mathbb{Z}$.
We see that periodicity implies that not all the points in $k$-space are allowed.
Instead only waves for which each component $k_x, k_y, k_z$ of the $\mathbf{k}$-vector belongs to the set
@@ -140,7 +140,7 @@ In 3D the allowed $k$-vectors form a regular grid:
-There is therefore exactly one allowed ${\bf k}$ per volume $\left(\frac{2\pi}{L}\right)^3$ in reciprocal space.
+There is therefore exactly one allowed ${\bf k}$-value per volume $\left(\frac{2\pi}{L}\right)^3$ in reciprocal space.
When we consider larger and larger box sizes $L→∞$, the volume per allowed mode becomes smaller and smaller, and eventually we obtain an integral:
$$
@@ -150,7 +150,7 @@ $$
## Density of states
-Continuing with our calculation of the total energy, we get:
+Let's use this knowledge and continue our calculation of the total energy:
$$
\begin{align}
E &= \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z × 3×\left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right),\\
--
GitLab