Alternatively, how there’s appears to be a “phase transition”-like phenomenon for a special case of 3D projectile motion

What do projectile motion and phase transitions have in common? More than meets the eye it seems!

Motivated by my last post, I was kind of curious to see what happens when we try to solve the same problem in 3D. Since the results are quite ugly and long, I won’t post them here, but just the general idea on how to approach the problem.

As a start, we again consider a projectile with velocity $v_0$, but this time thrown in some arbitrary direction $\hat{\mathbf{n}}$, and we have an additional force in the $y$ direction. A (probably difficult to setup) physical scenario this problem can model would be a particle in mutually orthogonal electric and gravitational fields. Since we’re given the absolute value of the initial velocity, and a direction, it’s simplest to work in spherical coordinates. Furthermore, the initial conditions are such that the projectile is launched from the origin. The equations of motion are of course:

$\ddot x = 0, \quad \ddot y = g_y, \quad \ddot z = - g_z$

All of the above gives us the following equations for the trajectory:

$x(t) = v_0 \cos\phi \sin\theta\, t \\ y(t) = \frac{1}{2} g_y t^2 + v_0 \sin\phi \sin\theta\, t \\ z(t) = - \frac{1}{2} g_z t^2 + v_0 \cos\theta \, t$

As the equation for the $z$ component remains identical (save for the definition of the angle, which is complementary to the one in the previous post), the time of flight actually stays the same, $T = 2 v_0 \cos\theta / g_z$. The longest trajectory is given by maximizing the following integral:

$L(\theta, \phi) = \int_0^T \mathrm{d}t\, \sqrt{\dot x^2 + \dot y^2 + \dot z^2}$

The integral above is of exactly the same type as in the previous post, albeit the solution is now much more cumbersome to write down. There’s an interesting simplification though: if we take define $|\mathbf{g}| = \sqrt{\sum_i g_i^2}$, $R_y = g_y / |\mathbf{g}|$, and $R_z = g_z / |\mathbf{g}|$, it turns out that $L$ can be factorized in the following way:

$L(\theta, \phi) = f(v_0, |\mathbf{g}|) \, h(\theta, \phi, R_y, R_z)$

Additionally, we have the constraint $R_y^2 + R_z^2 = 1$, so in reality $h$ depends only on $R_z$.
From here on, I will just focus on one particular case, namely $\phi = 0$, which has a very interesting property. Since we only care about the angular dependence of the above, I’ll write down just the rescaled distance, $h(\theta, R_z) = 2 |\mathbf{g}| L /v_0^2$, here:

$h(\theta, R_z) = \frac{1}{R_z} \bigg[-R_z^2 \cos (\theta ) \sqrt{\left(\frac{4}{R_z^2}-4\right) \cos ^2(\theta )+1}+2 \cos (\theta ) \sqrt{\left(\frac{4}{R_z^2}-4\right) \cos ^2(\theta )+1}+R_z^2 \cos (\theta )+R_z \left(R_z^2 \cos ^2(\theta )-1\right) \left(\log (1-R_z \cos (\theta ))-\log \left(\sqrt{\left(\frac{4}{R_z^2}-4\right) \cos ^2(\theta )+1}+\left(\frac{2}{R_z}-R_z\right) \cos (\theta )\right)\right)\bigg]$

This enormous equation reduces to a similar-looking one from the previous post for $R_z = 1$, though in this case for the complementary angle. What’s interesting is to see what happens when we vary $R_z$, which is limited to the interval $[0, 1]$: So, what’s so special about the above behavior? Well, there seems to be a threshold value, $R_z^\text{crit.}$, above which the largest trajectory is achieved by throwing the projectile at some nonzero angle $\theta^\textrm{max.}$, while below this value, the longest trajectory will always be achieved at an angle $\theta = 0$, that is, a completely vertical throw!
This is very reminiscent of a phase transition (see for instance the Wiki article on Landau theory for an overview), where now $R_z$ plays a role analogous to temperature, $h$ can be considered as the equivalent of the free energy function, and $\theta^\textrm{max.}$, the angle at which the trajectory is maximal, is the order parameter of the system.
To show that this is indeed very much like a second order phase transition, we can plot the angle $\theta^\textrm{max.}$ as a function of $R_z$: As we can see, while $\theta^\textrm{max.}$ is a continuous function of the “temperature” parameter $R_z$ (so it cannot be a first order transition), there is a “bump” in the graph around some value of $R_z$, at which point $\partial \theta^\textrm{max.} / \partial R_z$ diverges, indicating the existence of a second order (or continuous, since $\theta^\textrm{max.}(R_z)$ is continuous) phase transition. We can compute this critical value by solving the equation:

$\frac{\partial h}{\partial \theta} = 0$

by setting $\theta = 0$ and solving for $R_z$, which gives us $R_z^\textrm{crit.} \approx 0.903186$. I leave the calculation of the critical exponents as an exercise for the reader :)
Overall, I was quite surprised to see a phase transition-like phenomenon showing up in a in problem involving projectile motion!

As a bonus, here’s a video of the largest trajectories for various values of $R_z$; as you can see, as soon as $R_z > R_z^\textrm{crit.}$, we have a nonzero initial throwing angle (with respect to the $z$ axis):

Click on the video if it doesn't play automatically

Note that as $R_z \rightarrow 1$, we must have $R_y \rightarrow 0$, and consequently we get a smaller and smaller displacement in the $y$ direction, and for $R_z = 1$ we recover the 2D problem.