diff --git a/1/14-motion-in-a-central-field.md b/1/14-motion-in-a-central-field.md
index 5f45e33..4f61c93 100644
--- a/1/14-motion-in-a-central-field.md
+++ b/1/14-motion-in-a-central-field.md
@@ -20,163 +20,144 @@ In the present case, the generalised momentum $p_\phi=mr^2\dot{\phi}$ is the sam
1/14.2
```
-This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The expression $\mfrac{1}{2}r\cdot r\dd{\phi}$ is the area of the sector bounded by two neighbouring radius vectors and an element of the path `1/fig8`
+This law has a simple geometrical interpretation in the plane motion of a single particle in a central field. The expression $\mfrac{1}{2}r\cdot r\dd{\phi}$ is the area of the sector bounded by two neighbouring radius vectors and an element of the path `1/fig8`. Calling this area $\dd{f}$, we can write the angular momentum of the particle as
-§14
-Motion in a central field
-31
-(Fig. 8). Calling this area df, we can write the angular momentum of the par-
-ticle as
-M = 2mf,
-(14.3)
-where the derivative f is called the sectorial velocity. Hence the conservation
-of angular momentum implies the constancy of the sectorial velocity: in equal
-times the radius vector of the particle sweeps out equal areas (Kepler's second
-law).t
-rdd
-dd
-0
-FIG. 8
-The complete solution of the problem of the motion of a particle in a central
-field is most simply obtained by starting from the laws of conservation of
-energy and angular momentum, without writing out the equations of motion
-themselves. Expressing in terms of M from (14.2) and substituting in the
-expression for the energy, we obtain
-E = =
-(14.4)
-Hence
-(14.5)
-or, integrating,
-constant.
-(14.6)
-Writing (14.2) as do = M dt/mr2, substituting dt from (14.5) and integrating,
-we find
-constant.
-(14.7)
-Formulae (14.6) and (14.7) give the general solution of the problem. The
-latter formula gives the relation between r and , i.e. the equation of the path.
-Formula (14.6) gives the distance r from the centre as an implicit function of
-time. The angle o, it should be noted, always varies monotonically with time,
-since (14.2) shows that & can never change sign.
-t The law of conservation of angular momentum for a particle moving in a central field
+```load
+1/14.3
+```
+
+where the derivative $\dot{f}$ is called the *sectorial velocity*. Hence the conservation of angular momentum implies the constancy of the sectorial velocity: in equal times the radius vector of the particle sweeps out equal areas (*Kepler's second law*)[^\dagger]
+
+[^\dagger]: The law of conservation of angular momentum for a particle moving in a central field
is sometimes called the area integral.
-32
-Integration of the Equations of Motion
-§14
-The expression (14.4) shows that the radial part of the motion can be re-
-garded as taking place in one dimension in a field where the "effective poten-
-tial energy" is
-(14.8)
-The quantity M2/2mr2 is called the centrifugal energy. The values of r for which
-U(r)
-(14.9)
-determine the limits of the motion as regards distance from the centre.
-When equation (14.9) is satisfied, the radial velocity j is zero. This does not
-mean that the particle comes to rest as in true one-dimensional motion, since
-the angular velocity o is not zero. The value j = 0 indicates a turning point
-of the path, where r(t) begins to decrease instead of increasing, or vice versa.
-If the range in which r may vary is limited only by the condition r > rmin,
-the motion is infinite: the particle comes from, and returns to, infinity.
-If the range of r has two limits rmin and rmax, the motion is finite and the
-path lies entirely within the annulus bounded by the circles r = rmax and
-r = rmin- This does not mean, however, that the path must be a closed curve.
-During the time in which r varies from rmax to rmin and back, the radius
-vector turns through an angle Ao which, according to (14.7), is given by
-Mdr/r2
-(14.10)
-The condition for the path to be closed is that this angle should be a rational
-fraction of 2n, i.e. that Ao = 2nm/n, where m and n are integers. In that case,
-after n periods, the radius vector of the particle will have made m complete
-revolutions and will occupy its original position, so that the path is closed.
-Such cases are exceptional, however, and when the form of U(r) is arbitrary
-the angle is not a rational fraction of 2nr. In general, therefore, the path
-of a particle executing a finite motion is not closed. It passes through the
-minimum and maximum distances an infinity of times, and after infinite time
-it covers the entire annulus between the two bounding circles. The path
-shown in Fig. 9 is an example.
-There are only two types of central field in which all finite motions take
-place in closed paths. They are those in which the potential energy of the
-particle varies as 1/r or as r2. The former case is discussed in §15; the latter
-is that of the space oscillator (see §23, Problem 3).
-At a turning point the square root in (14.5), and therefore the integrands
-in (14.6) and (14.7), change sign. If the angle is measured from the direc-
-tion of the radius vector to the turning point, the parts of the path on each
-side of that point differ only in the sign of for each value of r, i.e. the path
-is symmetrical about the line = 0. Starting, say, from a point where = rmax
-the particle traverses a segment of the path as far as a point with r rmin,
-§14
-Motion in a central field
-33
-then follows a symmetrically placed segment to the next point where r = rmax,
-and so on. Thus the entire path is obtained by repeating identical segments
-forwards and backwards. This applies also to infinite paths, which consist of
-two symmetrical branches extending from the turning point (r = rmin) to
-infinity.
-'max
-min
-so
-FIG. 9
-The presence of the centrifugal energy when M # 0, which becomes
-infinite as 1/22 when r -> 0, generally renders it impossible for the particle to
-reach the centre of the field, even if the field is an attractive one. A "fall" of
-the particle to the centre is possible only if the potential energy tends suffi-
-ciently rapidly to -00 as r 0. From the inequality
-1mr2 = E- U(r) - M2/2mr2
-or r2U(Y)+M2/2m < Er2, it follows that r can take values tending to zero
-only if
-(14.11)
-i.e. U(r) must tend to - 8 either as - a/r2 with a > M2/2m, or proportionally
-to - 1/rn with n > 2.
-PROBLEMS
-PROBLEM 1. Integrate the equations of motion for a spherical pendulum (a particle of mass
-m moving on the surface of a sphere of radius l in a gravitational field).
-SOLUTION. In spherical co-ordinates, with the origin at the centre of the sphere and the
-polar axis vertically downwards, the Lagrangian of the pendulum is
-1ml2(02 + 62 sin20) +mgl cos 0.
-2*
-34
-Integration of the Equations of Motion
-§14
-The co-ordinate is cyclic, and hence the generalised momentum Po, which is the same as the
-z-component of angular momentum, is conserved:
-(1)
-The energy is
-E = cos 0
-(2)
-= 0.
+
+`1/fig8`
+
+The complete solution of the problem of the motion of a particle in a central field is most simply obtained by starting from the laws of conservation of energy and angular momentum, without writing out the equations of motion themselves. Expressing in terms of $M$ from `1/14.2` and substituting in the expression for the energy, we obtain
+
+```load
+1/14.4
+```
Hence
-(3)
-where the "effective potential energy" is
-Ueff(0) = COS 0.
-For the angle o we find, using (1),
-do
-(4)
-The integrals (3) and (4) lead to elliptic integrals of the first and third kinds respectively.
-The range of 0 in which the motion takes place is that where E > Ueff, and its limits
-are given by the equation E = Uell. This is a cubic equation for cos 0, having two roots
-between -1 and +1; these define two circles of latitude on the sphere, between which the
-path lies.
-PROBLEM 2. Integrate the equations of motion for a particle moving on the surface of a
-cone (of vertical angle 2x) placed vertically and with vertex downwards in a gravitational
-field.
-SOLUTION. In spherical co-ordinates, with the origin at the vertex of the cone and the
-polar axis vertically upwards, the Lagrangian is sin2x) -mgr cos a. The co-
-ordinate is cyclic, and Mz = mr2 sin²a is again conserved. The energy is
-= a.
-By the same method as in Problem 1, we find
-==
-The condition E = Ueff(r) is (if M + 0) a cubic equation for r, having two positive roots;
-these define two horizontal circles on the cone, between which the path lies.
-PROBLEM 3. Integrate the equations of motion for a pendulum of mass M2, with a mass M1
-at the point of support which can move on a horizontal line lying in the plane in which m2
-moves (Fig. 2, §5).
-SOLUTION. In the Lagrangian derived in §5, Problem 2, the co-ordinate x is cyclic. The
-generalised momentum Px, which is the horizontal component of the total momentum of the
-system, is therefore conserved
-Px = cos = constant.
-(1)
-The system may always be taken to be at rest as a whole. Then the constant in (1) is zero
-and integration gives
-(m1+m2)x+m2) sin = constant,
-(2)
-which expresses the fact that the centre of mass of the system does not move horizontally.
+
+```load
+1/14.5
+```
+
+or, integrating,
+
+```load
+1/14.6
+```
+
+Writing `1/14.2` as $\dd{\phi} = M\dd{t}/mr^2$, substituting $\dd{t}$ from `1/14.5` and integrating, we find
+
+```load
+1/14.7
+```
+
+Formulae `1/14.6` and `1/14.7` give the general solution of the problem. The latter formula gives the relation between $r$ and $\phi$, i.e. the equation of the path. Formula `1/14.6` gives the distance $r$ from the centre as an implicit function of time. The angle $\phi$, it should be noted, always varies monotonically with time, since `1/14.2` shows that $\phi$h can never change sign.
+
+The expression `1/14.4` shows that the radial part of the motion can be regarded as taking place in one dimension in a field where the "effective potential energy" is
+
+```load
+1/14.8
+```
+The quantity $M^2/2mr^2$ is called the *centrifugal energy*. The values of $r$ for which
+
+```load
+1/14.9
+```
+
+determine the limits of the motion as regards distance from the centre. When equation `1/14.9` is satisfied, the radial velocity $\dot{\v{r}}$ is zero. This does not mean that the particle comes to rest as in true one-dimensional motion, since the angular velocity $\phi$ is not zero. The value $\dot{\v{r}} = 0$ indicates a *turning point* of the path, where $r(t)$ begins to decrease instead of increasing, or *vice versa*.
+
+If the range in which $r$ may vary is limited only by the condition $r > r_{\min}$,
+the motion is infinite: the particle comes from, and returns to, infinity.
+
+If the range of $r$ has two limits $r_{\min}$ and $r_{\max}$, the motion is finite and the path lies entirely within the annulus bounded by the circles $r = r_{\max}$ and $r = r_{\min}$. This does not mean, however, that the path must be a closed curve. During the time in which $r$ varies from $r_\max$ to $r_\min$ and back, the radius vector turns through an angle $\Delta\phi$ which, according to `1/14.7`, is given by
+
+```load
+1/14.10
+```
+
+The condition for the path to be closed is that this angle should be a rational fraction of $2\pi$, i.e. that $\Delta\phi=2\pi m/n$, where $m$ and $n$ are integers. In that case, after $n$ periods, the radius vector of the particle will have made $m$ complete revolutions and will occupy its original position, so that the path is closed.
+
+Such cases are exceptional, however, and when the form of $U(r)$ is arbitrary the angle is not a rational fraction of $2pi$. In general, therefore, the path of a particle executing a finite motion is not closed. It passes through the minimum and maximum distances an infinity of times, and after infinite time it covers the entire annulus between the two bounding circles. The path shown in `1/fig9` is an example.
+
+There are only two types of central field in which all finite motions take place in closed paths. They are those in which the potential energy of the particle varies as $1/r$ or as $r^2$. The former case is discussed in `1/15`; the latter is that of the space oscillator (see `1/23p3`).
+
+At a turning point the square root in `1/14.5`, and therefore the integrands in `14.6` and `14.7`, change sign. If the angle $\phi$ is measured from the direction of the radius vector to the turning point, the parts of the path on each side of that point differ only in the sign of $\phi$ for each value of $\v{r}$, i.e. the path is symmetrical about the line $\phi = 0$. Starting, say, from a point where $r = r_\max$ the particle traverses a segment of the path as far as a point with $r=r_\min$, then follows a symmetrically placed segment to the next point where $r = r_\max$, and so on. Thus the entire path is obtained by repeating identical segments forwards and backwards. This applies also to infinite paths, which consist of two symmetrical branches extending from the turning point ($r = r_\min$) to infinity.
+
+```fig
+9
+```
+
+The presence of the centrifugal energy when $M \neq 0$, which becomes infinite as $1/r^2$ when $r\to 0$, generally renders it impossible for the particle to reach the centre of the field, even if the field is an attractive one. A "fall" of the particle to the centre is possible only if the potential energy tends suffi-
+ciently rapidly to $-\infty as $r\to 0$. From the inequality
+
+$$
+\frac{1}{2}m\dot{\v{r}}^2=E-U(r)-M^2/2mr^2 > 0,
+$$
+
+or $r^2U(r)+M^2/2m < Er^2$, it follows that $r$ can take values tending to zero only if
+
+```load
+14.11
+```
+
+i.e. $U(r)$ must tend to $-\infty$ either as $-\alpha/r^2$ with $\alpha > M^2/2m$, or proportionally
+to $-1/r^n$ with $n > 2$.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
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diff --git a/1/15-keplers-problem.md b/1/15-keplers-problem.md
index 0e60f9b..e84e0e7 100644
--- a/1/15-keplers-problem.md
+++ b/1/15-keplers-problem.md
@@ -1,8 +1,6 @@
---
-title: 15-keplers-problem
+title: 15 Kepler's problem
---
-Kepler's problem
-35
Using (1), we find the energy in the form
E
(3)
diff --git a/1/2-the-principle-of-least-action.md b/1/2-the-principle-of-least-action.md
index e975007..6beafd4 100644
--- a/1/2-the-principle-of-least-action.md
+++ b/1/2-the-principle-of-least-action.md
@@ -2,8 +2,7 @@
title: 2. The principle of least action
---
-The most general formulation of the law governing the motion of mech-
-anical systems is the principle of least action or Hamilton's principle, according to which every mechanical system is characterised by a definite function
+The most general formulation of the law governing the motion of mechanical systems is the principle of least action or Hamilton's principle, according to which every mechanical system is characterised by a definite function
$L(q_1,q_2,q_s,...,\dot{q}_1,\dot{q}_2,\dot{q}_s,t)$, or briefly $L(q,\dot{q},t)$, and the motion of the system is such that a certain condition is satisfied.
Let the system occupy, at the instants $t_1$ and $t_2$, positions defined by two
diff --git a/1/equations/14.3.tex b/1/equations/14.3.tex
index 6f2f2d0..dbf7caf 100644
--- a/1/equations/14.3.tex
+++ b/1/equations/14.3.tex
@@ -1 +1 @@
-2m\dd{A}=M\dd{t}
+M=2m\dot{f}
diff --git a/1/equations/14.6.tex b/1/equations/14.6.tex
index 8138a2c..10b14bd 100644
--- a/1/equations/14.6.tex
+++ b/1/equations/14.6.tex
@@ -1 +1 @@
-t=\int\dd{r}\left(\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}\right)^{-1/2}+\text{constant}
+t=\int\dd{r}\bigg/\sqrt{\frac{2}{m}(E-U(r))-\frac{M^2}{m^2r^2}}+\text{constant}
diff --git a/1/index.md b/1/index.md
index 0b1897e..3449a68 100644
--- a/1/index.md
+++ b/1/index.md
@@ -22,12 +22,12 @@ 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)
+14. [Motion in a central field](14-motion-in-a-central-field.html)
🚧 WORK IN PROGRESS BELOW THIS POINT 🚧
-14. [Motion in a central field](14-motion-in-a-central-field.html)
15. [Kepler's problem](15-keplers-problem.html)
IV. COLLISION BETWEEN PARTICLES
16. [Disintegration of particles](16-disintegration-of-particles.html)
diff --git a/tools/math.tex b/tools/math.tex
index d3f3879..43e2116 100644
--- a/tools/math.tex
+++ b/tools/math.tex
@@ -12,6 +12,8 @@
\newcommand{\qop}[2][]{\hat{#2}\vphantom{#2}^{#1}}
\newcommand{\mfrac}[2]{\textstyle\frac{#1}{#2}\displaystyle}
\newcommand{\ptag}[2][]{\label{#1/#2}\tag*{(#2)}}
+\newcommand{\min}{\text{min}}
+\newcommand{\max}{\text{max}}
\DeclareMathOperator{\ch}{ch}
\DeclareMathOperator{\sh}{sh}
diff --git a/tools/pf-filter.py b/tools/pf-filter.py
index c3a87a5..dd33909 100644
--- a/tools/pf-filter.py
+++ b/tools/pf-filter.py
@@ -1,6 +1,7 @@
import panflute as pf
import subprocess
import os
+import re
def prepare(doc):
pass
@@ -14,15 +15,20 @@ def action(elem, doc):
# elem.url = '../' + elem.url[:-3] + '.html'
# return elem
+
# reference
if isinstance(elem, pf.Code):
# todo sketchy if/else, replace with regex
- if '.' in elem.text:
+ if '/' in elem.text:
+ book, text = elem.text.split('/')
+ else:
+ book = '1'
+ text = elem.text
+ if '.' in text:
# equation ref
- book, eq = elem.text.split('/')
+ eq = text
res = subprocess.run(["./tools/tex_to_html.sh", "inline", f"{book}/equations/{eq}.tex"], capture_output=True)
tooltiptext = res.stdout.decode('utf-8')
- book, eq = elem.text.split('/')
html = f'''
({eq})
@@ -32,9 +38,9 @@ def action(elem, doc):
# {tooltiptext}
elem = pf.RawInline(html, format='html')
return elem
- elif '/' in elem.text:
+ else:
# section ref
- book, pnum = elem.text.split('/')
+ _pnum = text
try:
slug = [f for f in os.listdir(book) if f.startswith(pnum)][0]
except:
@@ -45,14 +51,16 @@ def action(elem, doc):
html = f'§{title.strip('\n')}'
elem = pf.RawInline(html, format='html')
return elem
- else:
- # todo reference to whole book
- return elem
# load equation
if isinstance(elem, pf.CodeBlock):
+ if '/' in elem.text:
+ book, text = elem.text.split('/')
+ else:
+ book = '1'
+ text = elem.text
if 'load' in elem.classes:
- book, eq = elem.text.split('/')
+ eq = text
res = subprocess.run(["./tools/tex_to_html.sh", "block", f"{book}/equations/{eq}.tex"], capture_output=True)
html = res.stdout.decode('utf-8')
elem = pf.RawBlock(html, format='html')