s/LL1/1/g

1/10-mechanical-similarity.md

@@ -2,12 +2,12 @@
 title: 10. Mechanical similarity
 ---
 
-Multiplication of the Lagrangian by any constant clearly does not affect the equations of motion. This fact (already mentioned in `LL1/2`) makes possible, in a number of important cases, some useful inferences concerning the properties of the motion, without the necessity of actually integrating the equations.
+Multiplication of the Lagrangian by any constant clearly does not affect the equations of motion. This fact (already mentioned in `1/2`) makes possible, in a number of important cases, some useful inferences concerning the properties of the motion, without the necessity of actually integrating the equations.
 
 Such cases include those where the potential energy is a homogeneous function of the co-ordinates, i.e. satisfies the condition
 
 ```load
-LL1/10.1
+1/10.1
 ```
 
 where $\alpha$ is any constant and $k$ the degree of homogeneity of the function.
@@ -18,20 +18,20 @@ $\alpha^2/\beta^2$. The potential energy is multiplied by $\alpha^k$. If $\alpha
 A change of all the co-ordinates of the particles by the same factor corresponds to the replacement of the paths of the particles by other paths, geometrically similar but differing in size. Thus we conclude that, if the potential energy of the system is a homogeneous function of degree k in the (Cartesian) co-ordinates, the equations of motion permit a series of geometrically similar paths, and the times of the motion between corresponding points are in the ratio
 
 ```load
-LL1/10.2
+1/10.2
 ```
 
 where $l'/l$ is the ratio of linear dimensions of the two paths. Not only the times but also any mechanical quantities at corresponding points at corresponding times are in a ratio which is a power of $l'/l$. For example, the velocities, energies and angular momenta are such that
 
 ```load
-LL1/10.3
+1/10.3
 ```
 
 The following are some examples of the foregoing.
 
-As we shall see later, in small oscillations the potential energy is a quadratic function of the co-ordinates ($k=2$). From `LL1/10.2` we find that the period of such oscillations is independent of their amplitude.
+As we shall see later, in small oscillations the potential energy is a quadratic function of the co-ordinates ($k=2$). From `1/10.2` we find that the period of such oscillations is independent of their amplitude.
 
-In a uniform field of force, the potential energy is a linear function of the co-ordinates (see `LL1/5.8`), i.e. $k = 1$. From `LL1/10.2` we have $t'/t = \sqrt{l'/l}$.  Hence, for example, it follows that, in fall under gravity, the time of fall is as 2the square root of the initial altitude.
+In a uniform field of force, the potential energy is a linear function of the co-ordinates (see `1/5.8`), i.e. $k = 1$. From `1/10.2` we have $t'/t = \sqrt{l'/l}$.  Hence, for example, it follows that, in fall under gravity, the time of fall is as 2the square root of the initial altitude.
 
 In the Newtonian attraction of two masses or the Coulomb interaction of two charges, the potential energy is inversely proportional to the distance apart, i.e. it is a homogeneous function of degree $k = 1$. Then $t'/t = (l'/l)^{3/2}$, and we can state, for instance, that the square of the time of revolution in the orbit is as the cube of the size of the orbit (Kepler's third law).
 
@@ -40,7 +40,7 @@ If the potential energy is a homogeneous function of the co-ordinates and the mo
 Since the kinetic energy $T$ is a quadratic function of the velocities, we have by Euler's theorem on homogeneous functions $\sum_a \v{v}a\cdot\partial T/\partial\v{v}_a = 2T$, or, putting $\partial T/\partial \v{v}_a = \v{p}_a$, the momentum,
 
 ```load
-LL1/10.4
+1/10.4
 ```
 
 Let us average this equation with respect to time. The average value of any function of time $f(t)$ is defined as
@@ -58,27 +58,27 @@ $$
 =0
 $$
 
-Let us assume that the system executes a motion in a finite region of space and with finite velocities. Then $\sum_a\v{p}_a\cdot\v{r}_a$ is bounded, and the mean value of the first term on the right-hand side of `LL1/10.4` is zero. In the second term we replace $\v{\dot{p}}_a$ by $-\partial U/\partial \v{r}_a$ in accordance with Newton's equations `LL1/5.3`, obtaining[^1]
+Let us assume that the system executes a motion in a finite region of space and with finite velocities. Then $\sum_a\v{p}_a\cdot\v{r}_a$ is bounded, and the mean value of the first term on the right-hand side of `1/10.4` is zero. In the second term we replace $\v{\dot{p}}_a$ by $-\partial U/\partial \v{r}_a$ in accordance with Newton's equations `1/5.3`, obtaining[^1]
 
 ```load
-LL1/10.5
+1/10.5
 ```
 
 If the potential energy is a homogeneous function of degree $k$ in the radius
-vectors $\v{r}_a$, then by Euler's theorem equation `LL1/10.5` becomes the required relation:
+vectors $\v{r}_a$, then by Euler's theorem equation `1/10.5` becomes the required relation:
 
 ```load
-LL1/10.6
+1/10.6
 ```
 
-Since $\bar{T}+\bar{U}=\bar{E}=E$, the relation `LL1/10.6` can also be expressed as
+Since $\bar{T}+\bar{U}=\bar{E}=E$, the relation `1/10.6` can also be expressed as
 
 ```load
-LL1/10.7
+1/10.7
 ```
 
 which express $\bar{U}$ and $\bar{T}$ in terms of the total energy of the system.
 
-In particular, for small oscillations ($k=2$) we have $\bar{T} = \bar{U}$, i.e. the mean values of the kinetic and potential energies are equal. For a Newtonian interaction $(k = - 1)$, $2\bar{T} = - \bar{U}$, and $E=-\bar{T}$, in accordance with the fact that, in such an interaction, the motion takes place in a finite region of space only if the total energy is negative (see `LL1/15`).
+In particular, for small oscillations ($k=2$) we have $\bar{T} = \bar{U}$, i.e. the mean values of the kinetic and potential energies are equal. For a Newtonian interaction $(k = - 1)$, $2\bar{T} = - \bar{U}$, and $E=-\bar{T}$, in accordance with the fact that, in such an interaction, the motion takes place in a finite region of space only if the total energy is negative (see `1/15`).
 
-[^1]: The expression on the right of `LL1/10.5` is sometimes called the virial of the system.
+[^1]: The expression on the right of `1/10.5` is sometimes called the virial of the system.

1/11-motion-in-one-dimension.md

@@ -5,20 +5,20 @@ title: 11. Motion in one dimension
 The motion of a system having one degree of freedom is said to take place in one dimension. The most general form of the Lagrangian of such a system in fixed external conditions is
 
 ```load
-LL1/11.1
+1/11.1
 ```
 
 where $a(q)$ is some function of the generalised co-ordinate $q$. In particular,
 if $q$ is a Cartesian co-ordinate ($x$, say) then
 
 ```load
-LL1/11.2
+1/11.2
 ```
 
-The equations of motion corresponding to these Lagrangians can be integrated in a general form. It is not even necessary to write down the equation of motion; we can start from the first integral of this equation, which gives the law of conservation of energy. For the Lagrangian `LL1/11.2` (e.g.) we have $\mfrac{1}{2}m\dot{x}^2+U(x)=E$. This is a first-order differential equation, and can be inte- grated immediately. Since $\dd{x}/\dd{t} = \sqrt{2[E - U(x)]/m}$, it follows that
+The equations of motion corresponding to these Lagrangians can be integrated in a general form. It is not even necessary to write down the equation of motion; we can start from the first integral of this equation, which gives the law of conservation of energy. For the Lagrangian `1/11.2` (e.g.) we have $\mfrac{1}{2}m\dot{x}^2+U(x)=E$. This is a first-order differential equation, and can be inte- grated immediately. Since $\dd{x}/\dd{t} = \sqrt{2[E - U(x)]/m}$, it follows that
 
 ```load
-LL1/11.3
+1/11.3
 ```
 
 The two arbitrary constants in the solution of the equations of motion are here represented by the total energy E and the constant of integration.
@@ -26,17 +26,17 @@ Since the kinetic energy is essentially positive, the total energy always exceed
 The points at which the potential energy equals the total energy,
 
 ```load
-LL1/11.4
+1/11.4
 ```
 
 give the limits of the motion. They are *turning points*, since the velocity there is zero. If the region of the motion is bounded by two such points, then the motion takes place in a finite region of space, and is said to be *finite*. If the region of the motion is limited on only one side, or on neither, then the
 motion is *infinite* and the particle goes to infinity.
 
 A finite motion in one dimension is oscillatory, the particle moving repeatedly back and forth between two points (in Fig. 6, in the potential well AB between the points X1 and x2). The period T of the oscillations, i.e. the time during which the particle passes from X1 to X2 and back, is twice the time
-from X1 to X2 (because of the reversibility property, `LL1/5`) or, by `LL1/11.3`),
+from X1 to X2 (because of the reversibility property, `1/5`) or, by `1/11.3`),
 
 ```load
-LL1/11.5
+1/11.5
 ```
 
-where $x_1$ and $x_2$ are roots of equation `LL1/11.4` for the given value of $E$. This formula gives the period of the motion as a function of the total energy of the particle.
+where $x_1$ and $x_2$ are roots of equation `1/11.4` for the given value of $E$. This formula gives the period of the motion as a function of the total energy of the particle.

1/2-the-principle-of-least-action.md

@@ -10,27 +10,27 @@ Let the system occupy, at the instants $t_1$ and $t_2$, positions defined by two
 sets of values of the co-ordinates, $q^{(1)}$ and $q^{(2)}$. Then the condition is that the system moves between these positions in such a way that the integral
 
 ```load
-LL1/2.1
+1/2.1
 ```
 
-takes the least possible value[^1]. The function L is called the Lagrangian of the system concerned, and the integral `LL1/2.1` is called the action.
+takes the least possible value[^1]. The function L is called the Lagrangian of the system concerned, and the integral `1/2.1` is called the action.
 
 The fact that the Lagrangian contains only $q$ and $\dot{q}$, but not the higher
 derivatives $\ddot{q}$, $\dddot{q}$, etc., expresses the result already mentioned, that the mechanical state of the system is completely defined when the co-ordinates and velocities are given.
 
-Let us now derive the differential equations which solve the problem of minimising the integral `LL1/2.1`. For simplicity, we shall at first assume that the system has only one degree of freedom, so that only one function $q(t)$ has to
+Let us now derive the differential equations which solve the problem of minimising the integral `1/2.1`. For simplicity, we shall at first assume that the system has only one degree of freedom, so that only one function $q(t)$ has to
 be determined.
 
 Let $q = q(t)$ be the function for which $S$ is a minimum. This means that $S$ is increased when $q(t)$ is replaced by any function of the form $q(t)+sq(t)$, 
 
 ```load
-LL1/2.2
+1/2.2
 ```
 
-where $\delta q(t)$ is a function which is small everywhere in the interval of time from $t_1$ to $t_2$; $\delta q(t)$ is called a variation of the function $q(t)$. Since, for $t = t_1$ and for $t = t_2$, all the functions `LL1/2.2` must take the values $q^{(1)}$ and $q^{(2)}$ respectively, it follows that
+where $\delta q(t)$ is a function which is small everywhere in the interval of time from $t_1$ to $t_2$; $\delta q(t)$ is called a variation of the function $q(t)$. Since, for $t = t_1$ and for $t = t_2$, all the functions `1/2.2` must take the values $q^{(1)}$ and $q^{(2)}$ respectively, it follows that
 
 ```load
-LL1/2.3
+1/2.3
 ```
 
 The change in $S$ when $q$ is replaced by $q + \delta q$ is
@@ -43,7 +43,7 @@ $$
 When this difference is expanded in powers of $\delta q$ and $\delta\dot{q}$ in the integrand, the leading terms are of the first order. The necessary condition for $S$ to have a minimum[^2] is that these terms (called the first variation, or simply the variation, of the integral) should be zero. Thus the principle of least action may be written in the form
 
 ```load
-LL1/2.4
+1/2.4
 ```
 
 or, effecting the variation,
@@ -56,10 +56,10 @@ $$
 Since $\delta\dot{q} = \dd{\delta q}/\dd{t}$, we obtain, on integrating the second term by parts,
 
 ```load
-LL1/2.5
+1/2.5
 ```
 
-The conditions `LL1/2.3` show that the integrated term in `LL1/2.5` is zero. There remains an integral which must vanish for all values of $\dd{q}$. This can be so only if the integrand is zero identically. Thus we have
+The conditions `1/2.3` show that the integrated term in `1/2.5` is zero. There remains an integral which must vanish for all values of $\dd{q}$. This can be so only if the integrand is zero identically. Thus we have
 
 $$
 \frac{\dd{}}{\dd{t}}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0\qquad
@@ -71,29 +71,29 @@ functions $q_i(t)$ must be varied independently in the principle of least action
 We then evidently obtain $s$ equations of the form
 
 ```load
-LL1/2.6
+1/2.6
 ```
 
-These are the required differential equations, called in mechanics Lagrange's equations[^3]. If the Lagrangian of a given mechanical system is known, the equations `LL1/2.6` give the relations between accelerations, velocities and coordinates, i.e. they are the equations of motion of the system.
+These are the required differential equations, called in mechanics Lagrange's equations[^3]. If the Lagrangian of a given mechanical system is known, the equations `1/2.6` give the relations between accelerations, velocities and coordinates, i.e. they are the equations of motion of the system.
 
-Mathematically, the equations `LL1/2.6` constitute a set of $s$ second-order equations for $s$ unknown functions $q_i(t)$. The general solution contains $2s$ arbitrary constants. To determine these constants and thereby to define uniquely the motion of the system, it is necessary to know the initial conditions which specify the state of the system at some given instant, for example the initial values of all the co-ordinates and velocities.
+Mathematically, the equations `1/2.6` constitute a set of $s$ second-order equations for $s$ unknown functions $q_i(t)$. The general solution contains $2s$ arbitrary constants. To determine these constants and thereby to define uniquely the motion of the system, it is necessary to know the initial conditions which specify the state of the system at some given instant, for example the initial values of all the co-ordinates and velocities.
 
 $Let a mechanical system consist of two parts $A$ and $B$ which would, if closed, have Lagrangians $L_A$ and $L_B$ respectively. Then, in the limit where the distance between the parts becomes so large that the interaction between them may be neglected, the Lagrangian of the whole system tends to the value $lim L = L_A+L_B.
 
 ```load
-LL1/2.7
+1/2.7
 ```
 
 This additivity of the Lagrangian expresses the fact that the equations of motion of either of the two non-interacting parts cannot involve quantities pertaining to the other part.
 
 It is evident that the multiplication of the Lagrangian of a mechanical system by an arbitrary constant has no effect on the equations of motion.
 
-From this, it might seem, the following important property of arbitrariness can be deduced: the Lagrangians of different isolated mechanical systems may be multiplied by different arbitrary constants. The additive property, however, removes this indefiniteness, since it admits only the simultaneous multiplication of the Lagrangians of all the systems by the same constant.  This corresponds to the natural arbitrariness in the choice of the unit of measurement of the Lagrangian, a matter to which we shall return in `LL1/4`.
+From this, it might seem, the following important property of arbitrariness can be deduced: the Lagrangians of different isolated mechanical systems may be multiplied by different arbitrary constants. The additive property, however, removes this indefiniteness, since it admits only the simultaneous multiplication of the Lagrangians of all the systems by the same constant.  This corresponds to the natural arbitrariness in the choice of the unit of measurement of the Lagrangian, a matter to which we shall return in `1/4`.
 
 One further general remark should be made. Let us consider two functions $L'(q,\dot{q},t) and $L(q,\dot{q},t)$, differing by the total derivative with respect to time of some function $f(q,t)$ of co-ordinates and time:
 
 ```load
-LL1/2.8
+1/2.8
 ```
 
 The integrals (2.1) calculated from these two functions are such that

1/3-galileos-relativity-principle.md

@@ -11,7 +11,7 @@ It is found, however, that a frame of reference can always be chosen in which sp
 We can now draw some immediate inferences concerning the form of the Lagrangian of a particle, moving freely, in an inertial frame of reference.  The homogeneity of space and time implies that the Lagrangian cannot contain explicitly either the radius vector $\v{r}$ of the particle or the time $t$, i.e. $L$ must be a function of the velocity $\v{v}$ only. Since space is isotropic, the Lagrangian must also be independent of the direction of $\v{v}$, and is therefore a function only of its magnitude, i.e. of $\v{v}^2 = v^2$.
 
 ```load
-LL1/3.1
+1/3.1
 ```
 
 Since the Lagrangian is independent of $\v{r}$, we have $\partial L/\partial\v{r} = 0$, and so Lagrange's equation is[^1]
@@ -23,7 +23,7 @@ $$
 whence $\partial L/\partial \v{v} = \text{constant}$. Since $\partial L/\partial \v{v}$ is a function of the velocity only, it follows that
 
 ```load
-LL1/3.2
+1/3.2
 ```
 
 Thus we conclude that, in an inertial frame, any free motion takes place with a velocity which is constant in both magnitude and direction. This is the law of inertia.
@@ -41,19 +41,19 @@ The complete mechanical equivalence of the infinity of such frames shows also th
 The co-ordinates $\v{r}$ and $\v{r}'$ of a given point in two different frames of reference $K$ and $K'$, of which the latter moves relative to the former with velocity $\v{V}$, are related by
 
 ```load
-LL1/3.3
+1/3.3
 ```
 
 Here it is understood that time is the same in the two frames:
 
 ```load
-LL1/3.4
+1/3.4
 ```
 
 The assumption that time is absolute is one of the foundations of classical
 mechanics[^2].
 
-Formulae `LL1/3.3` and `LL1/3.4` are called a Galilean transformation. Galileo's
+Formulae `1/3.3` and `1/3.4` are called a Galilean transformation. Galileo's
 relativity principle can be formulated as asserting the invariance of the mech-
 anical equations of motion under any such transformation.
 

1/4-the-lagrangian-for-a-free-particle.md

@@ -2,7 +2,7 @@
 title: 4. The Lagrangian for a free particle
 ---
 
-Let us now go on to determine the form of the Lagrangian, and consider first of all the simplest case, that of the free motion of a particle relative to an inertial frame of reference. As we have already seen, the Lagrangian in this case can depend only on the square of the velocity. To discover the form of this dependence, we make use of Galileo's relativity principle. If an inertial frame $K$ is moving with an infinitesimal velocity $\v{\epsilon}$ relative to another inertial frame $K'$, then $\v{v}' = \v{v}+\v{\epsilon}$. Since the equations of motion must have the same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of co-ordinates and time (see the end of `LL1/2`).
+Let us now go on to determine the form of the Lagrangian, and consider first of all the simplest case, that of the free motion of a particle relative to an inertial frame of reference. As we have already seen, the Lagrangian in this case can depend only on the square of the velocity. To discover the form of this dependence, we make use of Galileo's relativity principle. If an inertial frame $K$ is moving with an infinitesimal velocity $\v{\epsilon}$ relative to another inertial frame $K'$, then $\v{v}' = \v{v}+\v{\epsilon}$. Since the equations of motion must have the same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of co-ordinates and time (see the end of `1/2`).
 
 We have $L' = L(v'^2) = L(v2+2\v{v}\cdot\v{\epsilon}+\v{\epsilon}^2)$. Expanding this expression in powers of $\v{\epsilon}^2$ and neglecting terms above the first order, we obtain
 
@@ -13,7 +13,7 @@ $$
 The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $\v{v}$. Hence $\partial L/\partial v^2$ is independent of the velocity, i.e. the Lagrangian is in this case proportional to the square of the velocity, and we write it as
 
 ```load
-LL1/4.1
+1/4.1
 ```
 
 From the fact that a Lagrangian of this form satisfies Galileo's relativity principle for an infinitesimal relative velocity, it follows at once that the Lagrangian is invariant for a finite relative velocity $\v{V}$ of the frames $K$ and $K'$.
@@ -35,13 +35,13 @@ L'=L
 +\dd{(m\v{r}\cdot\v{V}+\mfrac{1}{2}mV^2t)}/\dd{t}.
 $$
 
-The second term is a total time derivative and may be omitted. The quantity m which appears in the Lagrangian `LL1/4.1` for a freely moving particle is called the mass of the particle. The additive property of the Lagrangian shows that for a system of particles which do not interact we have[^1].
+The second term is a total time derivative and may be omitted. The quantity m which appears in the Lagrangian `1/4.1` for a freely moving particle is called the mass of the particle. The additive property of the Lagrangian shows that for a system of particles which do not interact we have[^1].
 
 ```load
-LL1/4.2
+1/4.2
 ```
 
-It should be emphasised that the above definition of mass becomes meaningful only when the additive property is taken into account. As has been mentioned in `LL1/2`, the Lagrangian can always be multiplied by any constant without affecting the equations of motion. As regards the function `LL1/4.2`, such multiplication amounts to a change in the unit of mass; the ratios of the masses of different particles remain unchanged thereby, and it is only these ratios which are physically meaningful.
+It should be emphasised that the above definition of mass becomes meaningful only when the additive property is taken into account. As has been mentioned in `1/2`, the Lagrangian can always be multiplied by any constant without affecting the equations of motion. As regards the function `1/4.2`, such multiplication amounts to a change in the unit of mass; the ratios of the masses of different particles remain unchanged thereby, and it is only these ratios which are physically meaningful.
 
 It is easy to see that the mass of a particle cannot be negative. For, according
 to the principle of least action, the integral
@@ -55,7 +55,7 @@ has a minimum for the actual motion of the particle in space from point 1 to poi
 It is useful to notice that
 
 ```load
-LL1/4.3
+1/4.3
 ```
 
 Hence, to obtain the Lagrangian, it is sufficient to find the square of the ele-
@@ -63,19 +63,19 @@ ment of arc dl in a given system of co-ordinates. In Cartesian co-ordinates,
 for example, $\dd{l^2} = \dd{x^2}+\dd{y^2}+\dd{z^2}$, and so
 
 ```load
-LL1/4.4
+1/4.4
 ```
 
 In cylindrical co-ordinates $\dd{l^2}=\dd{r^2}+r^2\dd{\phi^2}+\dd{z^2}$ whence
 
 ```load
-LL1/4.5
+1/4.5
 ```
 
 In spherical co-ordinates $\dd{l^2}=\dd{r^2}+r^2\dd{\theta^2}+r^2\sin^2\theta\dd{\phi^2}$, and
 
 ```load
-LL1/4.6
+1/4.6
 ```
 
 [^1]: We shall use the suffixes a, b, c, ... to distinguish the various particles, and i, k,l, to distinguish the co-ordinates.

1/5-the-lagrangian-for-a-system-of-particles.md

@@ -2,40 +2,40 @@
 title: 5. The Lagrangian for a system of particles
 ---
 
-Let us now consider a system of particles which interact with one another but with no other bodies. This is called a closed system. It is found that the interaction between the particles can be described by adding to the Lagrangian `LL1/4.2` for non-interacting particles a certain function of the co-ordinates, which depends on the nature of the interaction[^1]. Denoting this function by $-U$, we have
+Let us now consider a system of particles which interact with one another but with no other bodies. This is called a closed system. It is found that the interaction between the particles can be described by adding to the Lagrangian `1/4.2` for non-interacting particles a certain function of the co-ordinates, which depends on the nature of the interaction[^1]. Denoting this function by $-U$, we have
 
 [^1]: This statement is valid in classical mechanics. Relativistic mechanics is not considered in this book. (See `LL2`)
 
 ```load
-LL1/5.1
+1/5.1
 ```
 
-where $\v{r}_a$ is the radius vector of the $a$th particle. This is the general form of the Lagrangian for a closed system. The sum $T=\sum\mfrac{1}{2}m_av_a^2$ is called the kinetic energy, and U the potential energy, of the system. The significance of these names is explained in `LL1/6`.
+where $\v{r}_a$ is the radius vector of the $a$th particle. This is the general form of the Lagrangian for a closed system. The sum $T=\sum\mfrac{1}{2}m_av_a^2$ is called the kinetic energy, and U the potential energy, of the system. The significance of these names is explained in `1/6`.
 
 The fact that the potential energy depends only on the positions of the particles at a given instant shows that a change in the position of any particle instantaneously affects all the other particles. We may say that the interactions are instantaneously propagated. The necessity for interactions in classical mechanics to be of this type is closely related to the premises upon which the subject is based, namely the absolute nature of time and Galileo's relativity principle. If the propagation of interactions were not instantaneous, but took place with a finite velocity, then that velocity would be different in different frames of reference in relative motion, since the absoluteness of time necessarily implies that the ordinary law of composition of velocities is applicable to all phenomena. The laws of motion for interacting bodies would then be different in different inertial frames, a result which would contradict the relativity principle.
 
-In `LL1/3` only the homogeneity of time has been spoken of. The form of the Lagrangian `LL1/5.1` shows that time is both homogeneous and isotropic, i.e. its properties are the same in both directions. For, if $t$ is replaced by $-t$, the Lagrangian is unchanged, and therefore so are the equations of motion. In other words, if a given motion is possible in a system, then so is the reverse motion (that is, the motion in which the system passes through the same states in
+In `1/3` only the homogeneity of time has been spoken of. The form of the Lagrangian `1/5.1` shows that time is both homogeneous and isotropic, i.e. its properties are the same in both directions. For, if $t$ is replaced by $-t$, the Lagrangian is unchanged, and therefore so are the equations of motion. In other words, if a given motion is possible in a system, then so is the reverse motion (that is, the motion in which the system passes through the same states in
 the reverse order). In this sense all motions which obey the laws of classical
 mechanics are reversible.  Knowing the Lagrangian, we can derive the equations of motion:
 
 ```load
-LL1/5.2
+1/5.2
 ```
 
-Substitution of `LL1/5.1` gives
+Substitution of `1/5.1` gives
 
 ```load
-LL1/5.3
+1/5.3
 ```
 
 In this form the equations of motion are called Newton's equations and form
 the basis of the mechanics of a system of interacting particles. The vector
 
 ```load
-LL1/5.4
+1/5.4
 ```
 
-which appears on the right-hand side of equation `LL1/5.3` is called the force on the $a$th particle. Like $U$, it depends only on the co-ordinates of the particles, and not on their velocities. The equation `LL1/5.3` therefore shows that the acceleration vectors of the particles are likewise functions of their co-ordinates only.
+which appears on the right-hand side of equation `1/5.3` is called the force on the $a$th particle. Like $U$, it depends only on the co-ordinates of the particles, and not on their velocities. The equation `1/5.3` therefore shows that the acceleration vectors of the particles are likewise functions of their co-ordinates only.
 
 The potential energy is defined only to within an additive constant, which has no effect on the equations of motion. This is a particular case of the nonuniqueness of the Lagrangian discussed at the end of §2 (todo link). The most natural and most usual way of choosing this constant is such that the potential energy tends to zero as the distances between the particles tend to infinity.
 
@@ -51,7 +51,7 @@ Substituting these expressions in the function $L=\mfrac{1}{2}\sum m_a(\dot{x}_a
 we obtain the required Lagrangian in the form
 
 ```load
-LL1/5.5
+1/5.5
 ```
 
 where the $a_{ik}$ are functions of the co-ordinates only. The kinetic energy in
@@ -68,23 +68,23 @@ U(q_A,q_B)$, where the first two terms are the kinetic energies of the systems $
 For example, when a single particle moves in an external field, the general form of the Lagrangian is
 
 ```load
-LL1/5.6
+1/5.6
 ```
 
 and the equation of motion is
 
 ```load
-LL1/5.7
+1/5.7
 ```
 
 A field such that the same force F acts on a particle at any point in the field
 is said to be uniform. The potential energy in such a field is evidently
 
 ```load
-LL1/5.8
+1/5.8
 ```
 
-To conclude this section, we may make the following remarks concerning the application of Lagrange's equations to various problems. It is often necessary to deal with mechanical systems in which the interaction between different bodies (or particles) takes the form of constraints, i.e. restrictions on their relative position. In practice, such constraints are effected by means of rods, strings, hinges and so on. This introduces a new factor into the problem, in that the motion of the bodies results in friction at their points of contact, and the problem in general ceases to be one of pure mechanics `LL1/25`. In many cases, however, the friction in the system is so slight that its effect on the motion is entirely negligible. If the masses of the constraining elements of the system are also negligible, the effect of the constraints is simply to reduce the number of degrees of freedom $S$ of the system to a value less than $3N$. To determine the motion of the system, the Lagrangian `LL1/5.5` can again be used, with a set of independent generalised co-ordinates equal in number to the actual degrees of freedom.
+To conclude this section, we may make the following remarks concerning the application of Lagrange's equations to various problems. It is often necessary to deal with mechanical systems in which the interaction between different bodies (or particles) takes the form of constraints, i.e. restrictions on their relative position. In practice, such constraints are effected by means of rods, strings, hinges and so on. This introduces a new factor into the problem, in that the motion of the bodies results in friction at their points of contact, and the problem in general ceases to be one of pure mechanics `1/25`. In many cases, however, the friction in the system is so slight that its effect on the motion is entirely negligible. If the masses of the constraining elements of the system are also negligible, the effect of the constraints is simply to reduce the number of degrees of freedom $S$ of the system to a value less than $3N$. To determine the motion of the system, the Lagrangian `1/5.5` can again be used, with a set of independent generalised co-ordinates equal in number to the actual degrees of freedom.
 
 ## Problems
 
@@ -92,5 +92,5 @@ Find the Lagrangian for each of the following systems when placed in a uniform g
 
 #### PROBLEM 1. A coplanar double pendulum
 
-`LL1/fig1`
+`1/fig1`
 

1/6-energy.md

@@ -4,7 +4,7 @@ title: 6. Energy
 
 During the motion of a mechanical system, the $2s$ quantities $q_i$ and $\dot{q}_i$, $(i = 1, 2, s)$ which specify the state of the system vary with time. There exist, however, functions of these quantities whose values remain constant during the motion, and depend only on the initial conditions. Such functions are called integrals of the motion.
 
-The number of independent integrals of the motion for a closed mechanical system with S degrees of freedom is $2s-1$. This is evident from the following simple arguments. The general solution of the equations of motion contains $2s$ arbitrary constants (see the discussion following equation `LL1/2.6`). Since the equations of motion for a closed system do not involve the time explicitly, the choice of the origin of time is entirely arbitrary, and one of the arbitrary constants in the solution of the equations can always be taken as an additive constant to in the time. Eliminating $t + t_0$ from the $2s$ functions $q_i = q_i(t+t_0, C_1, C_2, C_{2s_1})$, $\dot{q}_i = \dot{q}_i(t + t_0, C_1, C_2, ..., C_{2s-1})$, we can express the $2s-1$ arbitrary constants $C_1, C_2, C_{2s_1}$ as functions of $q$ and $\dot{q}$, and these functions will be integrals of the motion.
+The number of independent integrals of the motion for a closed mechanical system with S degrees of freedom is $2s-1$. This is evident from the following simple arguments. The general solution of the equations of motion contains $2s$ arbitrary constants (see the discussion following equation `1/2.6`). Since the equations of motion for a closed system do not involve the time explicitly, the choice of the origin of time is entirely arbitrary, and one of the arbitrary constants in the solution of the equations can always be taken as an additive constant to in the time. Eliminating $t + t_0$ from the $2s$ functions $q_i = q_i(t+t_0, C_1, C_2, C_{2s_1})$, $\dot{q}_i = \dot{q}_i(t + t_0, C_1, C_2, ..., C_{2s-1})$, we can express the $2s-1$ arbitrary constants $C_1, C_2, C_{2s_1}$ as functions of $q$ and $\dot{q}$, and these functions will be integrals of the motion.
 
 Not all integrals of the motion, however, are of equal importance in mechanics. There are some whose constancy is of profound significance, deriving from the fundamental homogeneity and isotropy of space and time. The quantities represented by such integrals of the motion are said to be conserved, and have an important common property of being additive: their values for a system composed of several parts whose interaction is negligible are equal to the sums of their values for the individual parts.
 
@@ -41,14 +41,14 @@ $$
 Hence we see that the quantity
 
 ```load
-LL1/6.1
+1/6.1
 ```
 
-remains constant during the motion of a closed system, i.e. it is an integral of the motion; it is called the energy of the system. The additivity of the energy follows immediately from that of the Lagrangian, since `LL1/6.1` shows that it is a linear function of the latter.
+remains constant during the motion of a closed system, i.e. it is an integral of the motion; it is called the energy of the system. The additivity of the energy follows immediately from that of the Lagrangian, since `1/6.1` shows that it is a linear function of the latter.
 
 The law of conservation of energy is valid not only for closed systems, but also for those in a constant external field (i.e. one independent of time): the only property of the Lagrangian used in the above derivation, namely that it does not involve the time explicitly, is still valid. Mechanical systems whose energy is conserved are sometimes called conservative systems.
 
-As we have seen in `LL1/5`, the Lagrangian of a closed system (or one in a
+As we have seen in `1/5`, the Lagrangian of a closed system (or one in a
 constant field) is of the form $L = T(q, \dot{q}) - U(q)$, where $T$ is a quadratic function of the velocities. Using Euler's theorem on homogeneous functions, we have
 
 $$
@@ -57,16 +57,16 @@ $$
 = 2T
 $$
 
-Substituting this in `LL1/6.1` gives
+Substituting this in `1/6.1` gives
 
 ```load
-LL1/6.2
+1/6.2
 ```
 
 in Cartesian co-ordinates,
 
 ```load
-LL1/6.3
+1/6.3
 ```
 
 Thus the energy of the system can be written as the sum of two quite different terms: the kinetic energy, which depends on the velocities, and the potential energy, which depends only on the co-ordinates of the particles.

1/7-momentum.md

@@ -16,10 +16,10 @@ $$
 where the summation is over the particles in the system. Since $\v{\epsilon}$ is arbitrary, the condition $\delta L = 0$ is equivalent to
 
 ```load
-LL1/7.1
+1/7.1
 ```
 
-From Lagrange's equations `LL1/5.2` we therefore have
+From Lagrange's equations `1/5.2` we therefore have
 
 $$
 \sum_a \frac{\dd{}}{\dd{t}} \frac{\partial L}{\partial\v{v}_a}
@@ -30,25 +30,25 @@ $$
 Thus we conclude that, in a closed mechanical system, the vector
 
 ```load
-LL1/7.2
+1/7.2
 ```
 
 remains constant during the motion; it is called the momentum of the system.
-Differentiating the Lagrangian `LL1/5.1`, we find that the momentum is given in
+Differentiating the Lagrangian `1/5.1`, we find that the momentum is given in
 terms of the velocities of the particles by
 
 ```load
-LL1/7.3
+1/7.3
 ```
 
 The additivity of the momentum is evident. Moreover, unlike the energy, the momentum of the system is equal to the sum of its values $\v{p}_a = m_a\v{v}_a$ for the individual particles, whether or not the interaction between them can be neglected.
 
 The three components of the momentum vector are all conserved only in the absence of an external field. The individual components may be conserved even in the presence of a field, however, if the potential energy in the field does not depend on all the Cartesian co-ordinates. The mechanical properties of the system are evidently unchanged by a displacement along the axis of a co-ordinate which does not appear in the potential energy, and so the corresponding component of the momentum is conserved. For example, in a uniform field in the z-direction, the $x$ and $y$ components of momentum are conserved.
 
-The equation `LL1/7.1` has a simple physical meaning. The derivative $\partial L/\partial\v{r}_a = - \partial U\partial\v{r}_a$ is the force $\v{F}_a$ acting on the $a$th particle. Thus equation `LL1/7.1` signifies that the sum of the forces on all the particles in a closed system is zero:
+The equation `1/7.1` has a simple physical meaning. The derivative $\partial L/\partial\v{r}_a = - \partial U\partial\v{r}_a$ is the force $\v{F}_a$ acting on the $a$th particle. Thus equation `1/7.1` signifies that the sum of the forces on all the particles in a closed system is zero:
 
 ```load
-LL1/7.4
+1/7.4
 ```
 
 In particular, for a system of only two particles, $\v{F}_1+\v{F}_2 = 0$: the force exerted by the first particle on the second is equal in magnitude, and opposite in direction, to that exerted by the second particle on the first. This is the equality of action and reaction (Newton's third law).
@@ -56,20 +56,20 @@ If the motion is described by generalised co-ordinates $q_i$, the derivatives
 of the Lagrangian with respect to the generalised velocities
 
 ```load
-LL1/7.5
+1/7.5
 ```
 
 are called generalised momenta, and its derivatives with respect to the general-
 ised co-ordinates
 
 ```load
-LL1/7.6
+1/7.6
 ```
 
 are called generalised forces. In this notation, Lagrange's equations are
 
 ```load
-LL1/7.7
+1/7.7
 ```
 
 In Cartesian co-ordinates the generalised momenta are the components of the vectors $\v{p}_a$. In general, however, the $p_i$ are linear homogeneous functions of the generalised velocities $\dot{q}_i$, and do not reduce to products of mass and velocity.

1/8-centre-of-mass.md

@@ -14,36 +14,36 @@ $$
 or
 
 ```load
-LL1/8.1
+1/8.1
 ```
 
 In particular, there is always a frame of reference $K'$ in which the total
-momentum is zero. Putting $P' = 0$ in `LL1/8.1`, we find the velocity of this frame:
+momentum is zero. Putting $P' = 0$ in `1/8.1`, we find the velocity of this frame:
 
 ```load
-LL1/8.2
+1/8.2
 ```
 
-If the total momentum of a mechanical system in a given frame of reference is zero, it is said to be at rest relative to that frame. This is a natural generalisation of the term as applied to a particle. Similarly, the velocity V given by `LL1/8.2` is the velocity of the "motion as a whole" of a mechanical system whose momentum is not zero. Thus we see that the law of conservation of momentum makes possible a natural definition of rest and velocity, as applied to a mechanical system as a whole.
+If the total momentum of a mechanical system in a given frame of reference is zero, it is said to be at rest relative to that frame. This is a natural generalisation of the term as applied to a particle. Similarly, the velocity V given by `1/8.2` is the velocity of the "motion as a whole" of a mechanical system whose momentum is not zero. Thus we see that the law of conservation of momentum makes possible a natural definition of rest and velocity, as applied to a mechanical system as a whole.
 
-Formula `LL1/8.2` shows that the relation between the momentum $\v{P}$ and the velocity $V$ of the system is the same as that between the momentum and velocity of a single particle of mass $\mu = \sum m_a$, the sum of the masses of the particles in the system. This result can be regarded as expressing the additivity of mass.
+Formula `1/8.2` shows that the relation between the momentum $\v{P}$ and the velocity $V$ of the system is the same as that between the momentum and velocity of a single particle of mass $\mu = \sum m_a$, the sum of the masses of the particles in the system. This result can be regarded as expressing the additivity of mass.
 
-The right-hand side of formula `LL1/8.2` can be written as the total time derivative of the expression
+The right-hand side of formula `1/8.2` can be written as the total time derivative of the expression
 
 ```load
-LL1/8.3
+1/8.3
 ```
 
-We can say that the velocity of the system as a whole is the rate of motion in space of the point whose radius vector is `LL1/8.3`. This point is called the centre of mass of the system.
+We can say that the velocity of the system as a whole is the rate of motion in space of the point whose radius vector is `1/8.3`. This point is called the centre of mass of the system.
 
-The law of conservation of momentum for a closed system can be formulated as stating that the centre of mass of the system moves uniformly in a straight line. In this form it generalises the law of inertia derived in `LL1/3` for a single free particle, whose "centre of mass" coincides with the particle itself.
+The law of conservation of momentum for a closed system can be formulated as stating that the centre of mass of the system moves uniformly in a straight line. In this form it generalises the law of inertia derived in `1/3` for a single free particle, whose "centre of mass" coincides with the particle itself.
 
 In considering the mechanical properties of a closed system it is natural to use a frame of reference in which the centre of mass is at rest. This eliminates a uniform rectilinear motion of the system as a whole, but such motion is of no interest.
 
 The energy of a mechanical system which is at rest as a whole is usually called its internal energy $E$. This includes the kinetic energy of the relative motion of the particles in the system and the potential energy of their interaction. The total energy of a system moving as a whole with velocity $V$ can be written
 
 ```load
-LL1/8.4
+1/8.4
 ```
 
 Although this formula is fairly obvious, we may give a direct proof of it. The energies $E$ and $E'$ of a mechanical system in two frames of reference $K$ and $K'$ are related by
@@ -58,7 +58,7 @@ E
 \end{align}
 
 ```load
-LL1/8.5
+1/8.5
 ```
 
-This formula gives the law of transformation of energy from one frame to another, corresponding to formula `LL1/8.1` for momentum. If the centre of mass is at rest in $K'$, then $\v{P}' = 0$, $E' = E_i$, and we have `LL1/8.4`.
+This formula gives the law of transformation of energy from one frame to another, corresponding to formula `1/8.1` for momentum. If the centre of mass is at rest in $K'$, then $\v{P}' = 0$, $E' = E_i$, and we have `1/8.4`.

1/9-angular-momentum.md

@@ -9,13 +9,13 @@ We shall use the vector $\delta\v{\phi}$ of the infinitesimal rotation, whose ma
 Let us find, first of all, the resulting increment in the radius vector from an origin on the axis to any particle in the system undergoing rotation. The linear displacement of the end of the radius vector is related to the angle by $|\delta\v{r}|=r\sin\theta\delta\phi$ `TODO link to fig5`. The direction of $\delta\v{r}$ is perpendicular to the plane of $\v{r}$ and $\delta\v{\phi}$. Hence it is clear that
 
 ```load
-LL1/9.1
+1/9.1
 ```
 
 When the system is rotated, not only the radius vectors but also the velocities of the particles change direction, and all vectors are transformed in the same manner. The velocity increment relative to a fixed system of co-ordinates is
 
 ```load
-LL1/9.2
+1/9.2
 ```
 
 TODO [fig5]
@@ -56,7 +56,7 @@ $$
 Since $\delta\v{\phi}$ is arbitrary, it follows that $(\dd{}/\dd{t})\sum\v{r}_a\times\v{p}_a$, and we conclude that the vector
 
 ```load
-LL1/9.3
+1/9.3
 ```
 
 called the angular momentum or moment of momentum of the system, is conserved in the motion of a closed system. Like the linear momentum, it is additive, whether or not the particles in the system interact.
@@ -73,7 +73,7 @@ Since the definition of angular momentum involves the radius vectors of the part
 \end{align}
 
 ```load
-LL1/9.4
+1/9.4
 ```
 
 It is seen from this formula that the angular momentum depends on the choice of origin except when the system is at rest as a whole (i.e. $\v{P} = 0).  This indeterminacy, of course, does not affect the law of conservation of angular momentum, since momentum is also conserved in a closed system.
@@ -87,18 +87,18 @@ $$
 +\sum_a m_a\v{r}_a\times\v{V}
 $$
 
-The first sum on the right-hand side is the angular momentum $M'$ in the frame $K'$; using in the second sum the radius vector of the centre of mass `Ll1/8.3`, we obtain
+The first sum on the right-hand side is the angular momentum $M'$ in the frame $K'$; using in the second sum the radius vector of the centre of mass `1/8.3`, we obtain
 
 ```load
-LL1/9.5
+1/9.5
 ```
 
-This formula gives the law of transformation of angular momentum from one frame to another, corresponding to formula `LL1/8.1` for momentum and `LL1/8.5` for energy.
+This formula gives the law of transformation of angular momentum from one frame to another, corresponding to formula `1/8.1` for momentum and `1/8.5` for energy.
 
 If the frame $K'$ is that in which the system considered is at rest as a whole, then $\v{V}$ is the velocity of its centre of mass, $\mu\v{V}$ its total momentum $\v{P}$ relative to $K$, and
 
 ```load
-LL1/9.6
+1/9.6
 ```
 
 In other words, the angular momentum $\v{M}$ of a mechanical system consists of its "intrinsic angular momentum" in a frame in which it is at rest, and the angular momentum $\v{R}\times\v{P}$ due to its motion as a whole.
@@ -113,7 +113,7 @@ The component of angular momentum along any axis (say the z-axis) can
 be found by differentiation of the Lagrangian:
 
 ```load
-LL1/9.7
+1/9.7
 ```
 
 where the co-ordinate is the angle of rotation about the z-axis. This is evident from the above proof of the law of conservation of angular momentum, but can also be proved directly. In cylindrical co-ordinates $r, \phi, z$ we have (substituting $x_a = r_a\cos\phi_a, y_a = -r_a\sin\phi_a$
@@ -122,9 +122,9 @@ $$
 M_z=\sum_a m_a(x_a\dot{y}_a-y_a\dot{x}_a)
 $$
 ```load
-LL1/9.8
+1/9.8
 ```
 
 The Lagrangian is, in terms of these co-ordinates,
 
-and substitution of this in `LL1/9.7` gives `LL1/9.8`.
+and substitution of this in `1/9.7` gives `1/9.8`.

package.json

@@ -0,0 +1 @@
+{ "dependencies": { "vite": "^5.2.12" } }