Analytical positioning of bodies (WIP)

Introduction

In games it is often necessary to move objects to a desired position. Imagine a ghost chasing Pacman or wanting to return to the ghosts’ home. This can be done in many different ways. For example, the velocity (or applied force) can be updated frame by frame, so that it points towards the destination at all times. Another possibility is to use Bezier curves or splines, if the trajectories are known beforehand.

A primitive like the cubic Bezier curve can be very useful, except that it is not quite fit for purpose in a Special Relativity context. This post is the first step in trying to find a curve primitive that works in Special Relativity. Such a primitive would make it easier to construct complex paths that are sane from a relativistic standpoint.

In the relativistic video game I am developing, every physical object moves along a piece-wise hyperbolic trajectory. Hyperbolic motion is what you get when you uniformly accelerate a body in Special Relativity. It is the relativistic counterpart of parabolic motion in classical mechanics and is probably the simplest kind of motion after rectilinear motion. Despite this, hyperbolic motion is mathematically hard to handle. The main source of pain comes from the \(\gamma = 1/\sqrt{1 - v^2/c^2}\) factors and the square root in them. Prepare for a bit of a ride!

Problem statement

We have a point-like particle in 3D space with known position, \(\mathbf{r}_{\mathrm{I}}\), and velocity, \(\mathbf{v}_{\mathrm{I}}\), at an initial time, \(t_{\mathrm{I}}\). We want to move the particle so that it has a different position, \(\mathbf{r}_{\mathrm{F}}\), and a different velocity, \(\mathbf{v}_{\mathrm{F}}\), at a later time \(t_{\mathrm{F}} > t_{\mathrm{I}}\).

The problem is sketched in the figure below:

We want to move the particle in the “simplest possible” way… For example, we could apply just a constant force (uniform acceleration) between time \(t_{\mathrm{I}}\) and \(t_{\mathrm{F}}\). Alternatively, if applying a single constant force is not enough to solve the problem generally, we could subdivide the time interval [\(t_{\mathrm{I}}\), \(t_{\mathrm{F}}\)] in two or more intervals and apply a different constant force in each subsequent interval.

Classical case

Clearly, it is not possible to solve the problem generally by applying one single force. Take the case where the initial and final velocities are both zero but the initial and final positions are distinct. We must apply a force to move the particle away from its initial position, but a constant force would then only keep increasing the particle’s velocity, which we want to be again zero at the destination point.

Let’s try, instead, to subdivide the trajectory in two pieces. From time \(t_{\mathrm{I}}\) to \(t_{\mathrm{M}}\) a uniform acceleration \(\mathbf{a}_{\mathrm{I}}\) is applied. From time \(t_{\mathrm{M}}\) to \(t_{\mathrm{F}}\) a different uniform acceleration \(\mathbf{a}_{\mathrm{F}}\) is applied.

Before we write down any equations, we apply the following coordinate transformation:

\[\newcommand{\vini}[1]{\mathbf{#1}_{\mathrm{I}}} \newcommand{\vmid}[1]{\mathbf{#1}_{\mathrm{M}}} \newcommand{\vfin}[1]{\mathbf{#1}_{\mathrm{F}}} \newcommand{\ini}[1]{#1_{\mathrm{I}}} \newcommand{\fin}[1]{#1_{\mathrm{F}}} \newcommand{\vvr}{\mathbf{r}} \newcommand{\vvv}{\mathbf{v}} \newcommand{\vva}{\mathbf{a}} \begin{equation} \mathbf{r}' = \mathbf{r} - \frac{\vini{r} + \vfin{r}}{2} - \frac{\vini{v} + \vfin{v}}{2} \, \left(t - \frac{\ini{t} + \fin{t}}{2}\right). \label{eq:transf_r} \end{equation}\]

Deriving with respect to time, we obtain the transformation equation for velocities:

\[\begin{equation} \mathbf{v}' = \mathbf{v} - \frac{\vini{v} + \vfin{v}}{2}. \label{eq:transf_v} \end{equation}\]

In this reference frame, final position and velocity are just the opposite of the initial position and velocity, respectively. This can be seen by inserting \(\vvr = \vini{r},\, \vfin{r}\) in Eq. \(\eqref{eq:transf_r}\) and \(\vvv = \vini{v},\, \vfin{v}\) in Eq. \(\eqref{eq:transf_v}\):

\[\begin{eqnarray} \vini{r}' & = & \frac{\vini{r} - \vfin{r}}{2} - \frac{\vini{v} + \vfin{v}}{2} \, \frac{\ini{t} - \fin{t}}{2}, \nonumber\\ \vfin{r}' & = & \frac{\vfin{r} - \vini{r}}{2} - \frac{\vini{v} + \vfin{v}}{2} \, \frac{\fin{t} - \ini{t}}{2} = -\vini{r}', \nonumber\\ \vini{v}' & = & \frac{\vini{v} - \vfin{v}}{2}, \nonumber\\ \vfin{v}' & = & \frac{\vfin{v} - \vini{v}}{2} = -\vini{v}',\nonumber \end{eqnarray}\]

Let us now write down the trajectory of the particle for the two parts:

\[\begin{eqnarray} \vvr'(t) & = & -\mathbf{R} - \mathbf{V} \, (t - \ini{t}) + \frac{1}{2} \vini{a} \, (t - \ini{t})^2 \nonumber\\ \vvr'(t) & = & \mathbf{R} + \mathbf{V} \, (t - \fin{t}) + \frac{1}{2} \vfin{a} \, (t - \fin{t})^2 \nonumber \end{eqnarray}\]

Above, we renamed the quantities as follows: \(\mathbf{R} = \vfin{r}'\), \(\mathbf{V} = \vfin{v}'\).

These two trajectories have to join up at a time \(t_M\), with \(\ini{t} < t_M < \fin{t}\). In particular, positions and velocities of these two trajectories must match at time \(t_M\). The constraint on matching positions is easily written as:

\[\vini{a} \, \ini{T}^2 - \vfin{a} \, \fin{T}^2 = 4 \mathbf{R} + 2 \mathbf{V} \, (\ini{T} - \fin{T})\]

where we have introduced \(\ini{T} = t_M - \ini{t}\) and \(\fin{T} = \fin{t} - t_M\). The constraints on matching velocities at time \(t_M\) gives:

\[\vini{a} \, \ini{T} + \vfin{a} \, \fin{T} = 2\mathbf{V}\]

This is solved to:

\[\begin{eqnarray*} \vini{a} & = & \frac{2 \mathbf{V}}{T} + \frac{4 \mathbf{R}}{\ini{T} T},\\ \vfin{a} & = & \frac{2 \mathbf{V}}{T} - \frac{4 \mathbf{R}}{\fin{T} T}, \end{eqnarray*}\]

where we have introduced \(T = \ini{T} + \fin{T} = \fin{t} - \ini{t}\).

These equations represent different solutions for different choices of \(t_M\). One possible choice of \(t_M\) is:

\[\begin{equation} \bbox[lightyellow, 10px, border: 2px solid orange]{ t_M = \frac{\ini{t} + \fin{t}}{2}, } \label{eq:tm_choice_middle} \end{equation}\]

i.e. the mid-point between \(\ini{t}\) and \(\fin{t}\), which corresponds to \(\ini{T} = \fin{T} = T/2\). This is probably the simplest choice of \(t_M\) and makes the trajectory construction “time-symmetrical”. In other words, the trajectory constructed by flipping \(\mathbf{R} \rightarrow -\mathbf{R}\) goes from \(-\mathbf{R}\) to \(\mathbf{R}\), following exactly the reversed path. This would not be the case, if we chose – say – \(t_M = (\ini{t} + \fin{t})/3\). It should be noted, however, that there may be other ways of choosing \(t_M\), based on the initial conditions of the problem, that have the same time-reversal property.

Let’s conclude this section by rewriting the result in the case \(\ini{T} = \fin{T} = T/2\). We get \(\vini{a} = \frac{\mathbf{2V}}{T} + \frac{\mathbf{8R}}{T^2}\) and \(\vfin{a} = \frac{\mathbf{2V}}{T} - \frac{\mathbf{8R}}{T^2}\), or, in terms of the original positions and velocities,

\[\bbox[lightyellow, 10px, border: 2px solid orange]{ \begin{eqnarray*} \vini{a} & = & -\frac{3\vini{v} + \vfin{v}}{\fin{t} - \ini{t}} + 4 \frac{\vfin{r} - \vini{r}}{(\fin{t} - \ini{t})^2} \\ \vfin{a} & = & \frac{\vini{v} + 3\vfin{v}}{\fin{t} - \ini{t}} - 4 \frac{\vfin{r} - \vini{r}}{(\fin{t} - \ini{t})^2} \end{eqnarray*} }\]

Some observations:

• The solution we found is just a couple of parabolas stiched together. In three or more dimensions, these two parabolas are not guaranteed to be coplanar.
• A reference frame that sees \(\vmid{r}\) not moving, i.e. \(\vmid{v} = 0\), also sees all the motion taking place in two straight segments. The velocity for parabolic motion clearly is never zero, unless the parabola is degenerate.
• By construction, the two curves that compose the trajectory meet at time \(t = t_M\) and thus at position \(\vmid{r} = (\vini{r} + \vfin{r})/2 + (\vfin{v} - \vini{v})(\fin{t} - \ini{t})/8\) and velocity \(\vmid{v} = 2(\vfin{r} - \vini{r})/(\fin{t} - \ini{t}) - (\vini{v} + \vfin{v})/2\). In the special reference frame where initial position and velocity are opposite of the final ones, \(\vmid{r}' = -\mathbf{V}T/2\) and \(\vmid{v}' = 2\mathbf{R}/T\).
• Increasing \(T \rightarrow sT\), \(s > 1\) leads to a wider curve. The same can be achieved by simultaneously increasing both the initial and final velocities, \(\vini{v},\,\mathbf{v}_F \rightarrow s\vini{v},\,s\mathbf{v}_F\). The choice of \(\ini{t}\) and \(\fin{t}\) are not crucial to determine the shape of the curve. Of course, they are crucial for determining the time parametrisation of the curve.

The video below shows some of the features of the solution derived above in this page.