From f9f062d4382a5f501420ffbe4f19902fe94cf480 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 31 Oct 2021 13:54:31 +0100
Subject: Expand knowledge base

---
 content/know/concept/boltzmann-relation/index.pdc | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

(limited to 'content/know/concept/boltzmann-relation')

diff --git a/content/know/concept/boltzmann-relation/index.pdc b/content/know/concept/boltzmann-relation/index.pdc
index ddaa22f..b892745 100644
--- a/content/know/concept/boltzmann-relation/index.pdc
+++ b/content/know/concept/boltzmann-relation/index.pdc
@@ -39,16 +39,16 @@ $$\begin{aligned}
     = - k_B T_e \nabla n_e
 \end{aligned}$$
 
-At equilibrium, we demand that $\vb{f}_e = \vb{f}_p$,
+At equilibrium, we demand that $\vb{f}_e = - \vb{f}_p$,
 and isolate this equation for $\nabla n_e$, yielding:
 
 $$\begin{aligned}
     k_B T_e \nabla n_e
-    = q_e n_e \nabla \phi
+    = - q_e n_e \nabla \phi
     \quad \implies \quad
     \nabla n_e
-    = \frac{q_e \nabla \phi}{k_B T_e} n_e
-    = \nabla \bigg( \frac{q_e \phi}{k_B T_e} \bigg) n_e
+    = - \frac{q_e \nabla \phi}{k_B T_e} n_e
+    = - \nabla \bigg( \frac{q_e \phi}{k_B T_e} \bigg) n_e
 \end{aligned}$$
 
 This equation is straightforward to integrate,
-- 
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