Review of Linear Motion

Alexei Gilchrist

1 Linear Motion

A simple review of the notation and some of the expressions for kinematics and dynamics along a line, for classical physics.

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1 Kinematics

The basic kinematic variable for an object is its position \(x\). In two or more dimensions this is a vector, e.g.

\[\begin{equation*}\vec{r} = \begin{pmatrix} x\\ y\\ z \end{pmatrix} \equiv x \hat{x}+y \hat{y}+z \hat{z},\end{equation*}\]

where \(\hat{x}\), \(\hat{y}\), \(\hat{z}\), are the unit basis vectors (the directions of the axes). Generally writing vectors as a column of the components is more useful than using symbols for the unit vectors. It’s much clearer to see what is going on, and linear transformations are represented as matrices written in the same basis.

Note that the basis, or co-ordinate system you choose is just that—a choice. There are an infinite number of possible bases and they don’t even have to be an orthogonal set, though this usually makes calculations much easier. The basis elements just have to be linearly independent and span the space (that is, you can get to any point by some linear combination of basis elements). You should choose whatever basis makes the problem at hand look the simplest. We’ll almost always choose an orthonormal basis (orthogonal and unit length basis vectors).

The length of the position vector (distance from the origin) is \(\sqrt{\vec{r}.\vec{r}} = r\), and if the basis is orthonormal it’s just the square root of the the sum of the squares of the components, \(r=\sqrt{x^2+y^2+z^2}\).

The instantaneous change in time of the position gives the velocity:

\[\begin{equation*}v = \frac{dx}{dt} \equiv \dot{x}\end{equation*}\]

Often you’ll see an over-dot on a variable to denote differentiation with respect to time as shown above. Again in two or more dimensions this is a vector, and we just differentiate each of the components

\[\begin{equation*}\vec{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \dot{\vec{r}} =\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{pmatrix} = \dot{x} \hat{x}+\dot{y} \hat{y}+\dot{z} \hat{z}.\end{equation*}\]

and \(v^2\equiv \dot{x}^2+\dot{y}^2+\dot{z}^2\) as before.

The final kinematic variable is the acceleration which is the change in the velocity with time

\[\begin{equation*}\vec{a} = \dot{\vec{v}} = \ddot{\vec{r}} = \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix}\end{equation*}\]

and the double dot indicates differentiating with respect to time twice.

2 Dynamics

The various time derivatives of position give a description of how an object moves with time, usually expressed as an equation of motion. Why an object should move in a particular way is the subject of dynamics.

The next quantity to add to our list is the mass of an object, \(m\). Essentially this captures an objects resistance to changes in velocity. This relationship is expressed neatly by Newton’s second law:

\[\begin{equation*}F = m\dot{v} = m\ddot{x}.\end{equation*}\]

If we can determine the force \(F\) by other means, then this equation will tell us how the object will move. Of course, in two or more dimensions the force will be a vector:

\[\begin{equation*}\vec{F} = m\ddot{\vec{r}} = m \begin{pmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{pmatrix},\end{equation*}\]

and it’s an under appreciated fact that in many cases the components of this equation are independent of each other (a notable exception is a charged particle moving in a magnetic field). This means that in general we can often split a problem up into three independent, and simpler problems in each dimension, or by a clever choice of co-ordinates introduce a lot of simplifications.

Rewriting Newton’s second law slightly,

\[\begin{equation*}\vec{F} = \frac{d}{dt} (m\vec{v}),\end{equation*}\]

suggests the momentum \(\vec{p} = m\vec{v}\) as a useful quantity. So we can see that a force on an object will change an objects momentum

\[\begin{equation*}\vec{F} = \dot{\vec{p}}.\end{equation*}\]

Finally we come to energy \(E\), where we distinguish between two types: kinetic \(T\) and potential \(U\). The kinetic energy is energy associated with motion:

\[\begin{equation*}T = \frac{1}{2}m \dot{x}^2 = \frac{1}{2m}p^2\end{equation*}\]

For many particles in three dimensions this becomes the energy of each independent degree of freedom

\[\begin{equation*}T = \sum_j \frac{1}{2} m_j \dot{\vec{r}}_j^2 = \sum_j \frac{1}{2} m_j (\dot{x}_j^2+\dot{y}_j^2+\dot{z}_j^2).\end{equation*}\]

For a conservative system the potential energy \(U\) determines the force:

\[\begin{equation*}F = -\frac{dU}{dx}.\end{equation*}\]

In three dimensions \(U\equiv U(x,y,z)\) and the derivatives become partial derivatives

\[\begin{equation*}\vec{F} = -\begin{pmatrix} \partial U/\partial x \\ \partial U/\partial y \\ \partial U/\partial z \end{pmatrix}\end{equation*}\]

Given this relationship between the potential energy and the force, it is tempting to plot the potential energy and imagine a small ball rolling on top of this potential energy surface (or a curve in 1-D). Certainly this analogy reproduces some of the behavior of an object moving under such a potential. For instance the direction of motion is correct—the ball will roll downhill which corresponds to a force in the opposite direction to the slope. The ball will sit still at points where the surface is flat and indeed there is no force on the object at this point as \(dU/dx=0\), and you can even tell if this equilibrium point is stable or unstable by considering what will happen if you give the ball a little nudge. That increasing slope corresponds to increasing force is also correct. However it’s possible to take this analogy too far. For instance, the velocity of the ball has no connection to the object. A vertical wall in the potential would correspond to an infinite force, but for the ball there would be no change in position (motion downwards has no meaning in the analogy). It’s a great analogy, just be mindful of it’s limitations.