From d60dc1dc00693e2f0a142a215a34baba8f11b835 Mon Sep 17 00:00:00 2001
From: Jack Halford <jack@0x5.be>
Date: Tue, 4 Jun 2024 15:25:34 +0200
Subject: [PATCH] 13 done

---
 1/13-the-reduced-mass.md | 57 ++++++++++++++++++++--------------------
 1/index.md               |  2 +-
 2 files changed, 30 insertions(+), 29 deletions(-)

diff --git a/1/13-the-reduced-mass.md b/1/13-the-reduced-mass.md
index a374871..42fa046 100644
--- a/1/13-the-reduced-mass.md
+++ b/1/13-the-reduced-mass.md
@@ -2,35 +2,36 @@
 title: 13. The reduced mass
 ---
 
-A complete general solution can be obtained for an extremely important
-problem, that of the motion of a system consisting of two interacting particles
-(the two-body problem).
-As a first step towards the solution of this problem, we shall show how it
-can be considerably simplified by separating the motion of the system into
-the motion of the centre of mass and that of the particles relative to the centre
-of mass.
+A complete general solution can be obtained for an extremely important problem, that of the motion of a system consisting of two interacting particles (the *two-body problem*).
+
+As a first step towards the solution of this problem, we shall show how it can be considerably simplified by separating the motion of the system into the motion of the centre of mass and that of the particles relative to the centre of mass.
+
 The potential energy of the interaction of two particles depends only on
 the distance between them, i.e. on the magnitude of the difference in their
 radius vectors. The Lagrangian of such a system is therefore
-L =
-(13.1)
-Let r III r1-r2 - be the relative position vector, and let the origin be at the
-centre of mass, i.e. M1r1+M2r2 = 0. These two equations give
-= m2I/(m1+m2),
-r2
--M1I/(m1+m2).
-(13.2)
-Substitution in (13.1) gives
-L
-(13.3)
+
+```load
+1/13.1
+```
+
+Let $\v{r} \equiv \v{r}_1-\v{r}_2$ be the relative position vector, and let the origin be at the centre of mass, i.e. $m_1\v{r}_1+m_2\v{r}_2 = 0$. These two equations give
+
+```load
+1/13.2
+```
+
+Substitution in `1/13.1` gives
+
+```load
+1/13.3
+```
+
 where
-m=mym
-(13.4)
-is called the reduced mass. The function (13.3) is formally identical with the
-Lagrangian of a particle of mass m moving in an external field U(r) which is
-symmetrical about a fixed origin.
-Thus the problem of the motion of two interacting particles is equivalent
-to that of the motion of one particle in a given external field U(r). From the
-solution r = r(t) of this problem, the paths r1 = r1(t) and r2 = r2(t) of the
-two particles separately, relative to their common centre of mass, are obtained
-by means of formulae (13.2).
+
+```load
+1/13.4
+```
+
+is called the *reduced mass*. The function `1/13.3` is formally identical with the Lagrangian of a particle of mass $m$ moving in an external field $U(\v{r})$ which is symmetrical about a fixed origin.
+
+Thus the problem of the motion of two interacting particles is equivalent to that of the motion of one particle in a given external field $U(\v{r})$. From the solution $\v{r} = \v{r}(t)$ of this problem, the paths $\v{r}_1 = \v{r}_1(t)$ and $\v{r}_2 = \v{r}_2(t)$ of the two particles separately, relative to their common centre of mass, are obtained by means of formulae `1/13.2`.
diff --git a/1/index.md b/1/index.md
index e0f6847..8ef3016 100644
--- a/1/index.md
+++ b/1/index.md
@@ -21,12 +21,12 @@ II.  CONSERVATION LAWS
 III.  INTEGRATION OF THE EQUATIONS OF MOTION
 11. [Motion in one dimension](11-motion-in-one-dimension.html)
 12. [Determination of the potential energy from the period of oscillation](12-determination-of-the-potential-energy-from-the-period-of-oscillation.html)
+13. [The reduced mass](13-the-reduced-mass.html)
 
 <span style="background-color: yellow; color: white: width: 100%;">
 🚧 WORK IN PROGRESS BELOW THIS POINT 🚧
 </span>
 
-13. [The reduced mass](13-the-reduced-mass.html)
 14. [Motion in a central field](14-motion-in-a-central-field.html)
 15. [Kepler's problem]()
 IV.  COLLISION BETWEEN PARTICLES