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295
1/14-motion-in-a-central-field.md
View file

@ -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 $n > 2$.
<!-- 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. -->
<!-- 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. -->

4
1/15-keplers-problem.md
View file

@ -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)

3
1/2-the-principle-of-least-action.md
View file

@ -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

2
1/equations/14.3.tex
View file

@ -1 +1 @@
2m\dd{A}=M\dd{t}
M=2m\dot{f}

2
1/equations/14.6.tex
View file

@ -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}

2
1/index.md
View file

@ -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)
<span style="background-color: yellow; color: white: width: 100%;">
🚧 WORK IN PROGRESS BELOW THIS POINT 🚧
</span>
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)

2
tools/math.tex
View file

@ -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}

26
tools/pf-filter.py
View file

@ -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'''
<span class="tooltip" class="math inline">
({eq})
@ -32,9 +38,9 @@ def action(elem, doc):
# <span class="tooltiptext">{tooltiptext}</span>
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'<a href="{slug}">§{title.strip('\n')}</a>'
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')