Curves in Parametric Form. Foundations of Mechanics

The illustration below is an example of something which we customarily call a curve, but which cannot be described as the graph of a function; for there are points where several $y$-values correspond to one value of $x$.

Then how can we define a curve so as to take in this case? When we drew the curve, we started with the pencil at some point and moved it. After $t$ seconds the pencil was at a new point $(x(t),y(t))$ of the curve, and any point of the curve could be reached after some time (perhaps including travel backward in time). Now $x(t)$ is a function of $t$, namely, "the $x$-coordinate of the pencil at time $t$", and likewise $y(t)$ is a function of $t$. The fact that we draw the curve in one continuous motion is expressed by saying that $x(t)$ and $y(t)$ are continuous functions. Thus we are led to make the following definition:

The graph of a function is a special case; the curve described by the graph of $y=f(x)$ can be written in this "parametric form" by using $x$ for the variable $t$. Then the curve is $(t,f(t))$. For a given $t$, we can regard

$$X(t)=(x_1(t),x_2(t),\dots ,x_n(t))$$

either as a point in $n$-space or as a vector. Thus it is possible to perform vector operations with this quantity. In many of our references to a curve, we shall regard $t$ as time. For instance, we define the velocity at time $t$ of a point on the curve to be

X'(t) = (x_1'(t), x_2'(t), \dots, x_n'(t))

if it exists. We define the acceleration at time $t$ to be

X''(t) = (x_1''(t), x_2''(t), \dots, x_n''(t))

if it exists.

Since we can really get no adequate intuitive notion of acceleration, we may regard our definition of acceleration as not requiring justification. The definition of velocity is a limiting case of an "average velocity", as follows:

At time $t$, let the corresponding point of the curve be $X(t)$; at time $t+h$, the point is then $X(t+h)$. If we only knew these two points, the simplest type of motion to assume between them is straight-line motion at a constant rate. This means that the vector $X(t+h)-X(t)$ is traversed in time $h$ at a constant rate. This rate, together with its direction, is then given by

$$\frac{X(t+h)-X(t)}{h}=(\frac{x_1(t+h)-x_1(t)}{h},\dots ,\frac{x_n(t+h)-x_n(t)}{h})\cdot$$

The velocity at time $t$ is then the limit of this average velocity as the time $h$ over which the average is taken goes to zero, and the result is simply X'(t).

We use the velocity to give us a definition of the tangent to the curve: the tangent line to $X(t)$ at $t=t_0$ is the line through $X(t_0)$ and parallel to the velocity vector at $t_0$, or X'(t_0). Thus it is the set of points

X(t_0) + s X'(t_0),

where $s$ ranges through all numbers.

With the concepts we have now developed, we can formulate the basic notions of Newton's mechanics and derive quite easily a number of important consequences. To begin with, we have certain concepts which we shall call primary notions; they are mathematical formulations of quantities observable in nature. They are:

These notions are subject to certain fundamental laws, which are really those of Newton. The first is simple:

I. F = m X''(t), or in words:
The force acting on a particle at $X(t)$ is the product of its mass and its acceleration.

The other laws require a little more explanation. A mechanical system is a collection of particles with positions $X_1$, $X_2$, $\dots$, $X_N$, each depending on time, and having respective masses $m_1$, $m_2$, $\dots$, $m_N$. Let $F_i$ be the force acting on the $i$-th particle. Then by I, F_i=m_i X_i^{\prime\prime}. $F_i$ is composed of an external force ${\overline{F}}_i$ , coming from outside of the system, and other "internal forces" $F_{ij}$, where $F_{ij}$ is the force exerted by the $j$-th particle on the $i$-th. Thus

$$F_i={\overline{F}}_i+F_{i1}+F_{i2}+\cdots +F_{iN},$$

which we write more briefly as

$$F_i={\overline{F}}_i+\sum _{j=1}^N{F}_{ij}\cdot$$

The other laws concern the $F_{ij}$:

II. $F_{ij}=-F_{ji}$ ("action equals reaction"; in particular, $F_{ii}=0$).

III. $F_{ij}$ is parallel to $X_i-X_j$, i.e., the force exerted by one particle on another acts along the line joining them.

Let us consider for a moment the case where all the external forces ${\overline{F}}_i$ are zero (a closed system). Then

m_i X_i'' = F_i = \sum_{j=1}^N F_{ij} \cdot

If we add up the forces acting on all particles, we see that

\sum_{i=1}^N m_i X_i'' = \sum_{i=1}^N \sum_{j=1}^N F_{ij} \cdot

Now all $F_{ii}=0$, and if $i\ne j$, each of $F_{ij}$ and $F_{ji}$ occurs exactly once in the double sum on the right. Since $F_{ij}=-F_{ji}$, their sum is $0$, and so the whole double sum is zero.
Thus

\sum_{i=1}^N m_i X_i'' = 0 \cdot

Since (m_i X_i')' = m_i X_i'', we have

\left( \sum_{i=1}^N m_i X_i' \right)' = 0,

or

the total momentum \sum m_i X_i' is a constant $A$.

Then $\sum m_i{X}_i=At+B$ for some fixed vector $B$ (all our results for vector-valued functions follow from those for ordinary functions by considering the components separately). If we let $m=\sum m_i$ be the total mass and let $Y$ be the vector $\frac{1}{m}\sum m_i{X}_i=\sum \frac{m_i}{m}{X}_i$, we have

$$mY=At+B\cdot$$

The point $Y$ is called the center of mass of the system. We see that

If $X(t)$ and $Y(t)$ are vector-valued functions in 3-space, it is left as an exercise to verify that

(X \times Y)' = (X' \times Y) + (X \times Y').

In particular,

(X \times X')' = (X' \times X') + (X \times X'') = X \times X'' \cdot.

In a closed system,

m_i X_i'' = \sum_{j=1}^N F_{ij},

and so

m_i(X_i \times X_i'') = X_i \times m_i X_i'' = \sum_{j=1}^N (X_i \times F_{ij}) \cdot

Again let us sum over all particles. We shall show that

\sum_{i=1}^N m_i (X_i \times X_i'') = \sum_{i=1}^N \sum_{j=1}^N (X_i \times F_{ij}) = 0 \cdot

As when we showed

$$\sum _{i=1}^N\sum _{j=1}^N{F}_{ij}=0,$$

it will be enough to show that each pair of terms

$$X_i\times F_{ij}\text{and}{X}_j\times F_{ji}$$

cancel when $i\ne j$. Now $F_{ji}=-F_{ij},$ so the sum of these two is

$$(X_i-X_j)\times F_{ij}.$$

By III, this is $0$. Thus our assertion is proved.
But

\sum_{i=1}^N m_i (X_i \times X_i'') = \left( \sum_{i=1}^N m_i (X_i \times X_i') \right)'

is the derivative of the total angular momentum (about $0$). Hence we conclude that

(It is left as an exercise in case $0$ is replaced by any other point $P$.)

Now let us consider the case with external forces ${\overline{F}}_i$. By the fact that the sum of the internal forces is $0$, we see that

\sum_{i=1}^N m_i X_i'' = \sum_{i=1}^N \bar{F}_i,

or that

\left( \sum_{i=1}^N m_i X_i' \right)' = \sum_{i=1}^N \bar{F}_i \cdot

i.e.,

Since mY'' = \sum m_i X_i'', we see that the center of mass of the system moves like a particle of mass $m$ acted on by the total external forces.

Also,

m_i(X_i \times X_i'') = (X_i \times \bar{F}_i) + \sum_{j=1}^N (X_i \times F_{ij}) \cdot

Adding as before,

\left( \sum_{i=1}^N m_i (X_i \times X_i') \right)' = \sum_{i=1}^N m_i(X_i \times X_i'') = \sum_{i=1}^N (X_i \times \bar{F}_i),

or the derivative of the total angular momentum is the sum of the moments of the external forces.

When we have only one particle, all $F_{ij}=0$, and ${\overline{F}}_i$ is parallel to $X_i\cdot$ We assume the origin is the source of the force ${\overline{F}}_1\cdot$ We drop the indices $i$ as unnecessary. Then (m(X \times X'))' = (X \times \bar{F}) = 0. Applied to the solar system with the assumption that the only non-negligible force acting on a planet is that of the sun, we see that the derivative of the angular momentum of a planet about the sun is $0$, or that its angular momentum about the sun is a constant. This is one formulation of Kepler's law that a planet sweeps out area at a constant rate.