Kinematics in phase space

For this short article, as a stop gap before another one, I thought I would look at how the movement of a single particle (its kinematics) look like in a plot of speed versus position. This plot is very similar to a plot of conjugate momenta (or generalized coordinates) from classical mechanics, but rather than using the momentum of the particle, we use its speed (or, you can assume a particle of mass \(1\)).

To begin, consider the case of a particle at rest, \(v = 0\). In this case, the particle is at \(x_0\) and will never move. So, the plot is just a dot:

Next, let’s consider the case of constant velocity \(v\). Here, too, the plot is simple. The particle is either moving to the right (positive velocity), or to the left (negative speed).

Now, what about movement with a constant acceleration? Here, we start from the equations of motion (or the result of integrating \(\ddot x = a\) twice):

\[v = at + v_0\]

\[x = \frac{1}{2}at^2 + v_0t+x_0\]

Extract the time from the first and replace into the second to obtain

\[x = \frac{1}{2a} v^2 + \left(x_0 - \frac{v_0^2}{2a}\right)\]

Which is the equation of a parabola with no linear term. Plotting it we have

This assumes that the acceleration, the velocity and the position vector are all in the same direction. It is left as an exercise to the reader to consider the other 3 cases.

What is more interesting is to determine how the value of \(a\) influences the shape of the parabola. Observe that because we don’t have a linear term, the extreme point of the parabola is always on the \(v=0\) axis, regardless of the initial orientation of acceleration and velocity vectors. But what if we are looking at two plots of the same starting position and same speed, but with different accelerations?

Out of these two lines, which one has higher acceleration? Why?

And, even more interesting, try determining how the curve would look like for a constant jerk.