Air Resistance as an Analog Relativity Computer

A Good High School Science Fair Idea

Suppose you drop a hamster off a tall building. It will fall, initially with an acceleration of $g$, about $10 m/s^2$. But after the hamster has swept out a volume of air whose mass equals the hamster’s own, the effects of air resistance dominate and the acceleration is small.

A hamster is about as dense as water, which is about a thousand times as dense as air. So air resistance becomes important when a hamster falls 1000 times its own thickness. The thickness of a spread-eagled hamster is 5 cm, so air resistance dominates falls over 50m.

(Compare this to a sheet of paper. A sheet of paper is as dense as a hamster, but 1000 sheets of paper are only as thick as a 1000-page book, so for the sheet of paper, air resistance is important even if you fall just a few centimeters.)

50m is about 20 stories, so if we want the speed of the hamster when dropping it off a building at least that tall, we had better include air resistance.

A reasonable model for air resistance is that its force is proportional to the square of the hamster’s velocity, because if the hamster falls twice as fast, it strikes the air twice as hard, but also strikes twice as much air per second. Thus

$F_{air} = -kv^2$

for some constant $k$. To use Newton’s second law, include the force of gravity and set the force equal to $ma$.

$F_{air} + F_{grav} = m a$

$-kv^2 + gm = m a$

Let $k/m = b$ and you get

$-bv^2 + g = a = \frac{\textrm{d}v}{\textrm{d}t}$

This is a separable equation. We get

$\textrm{d}t = \frac{\textrm{d}v}{-b v^2 + g}$

which integrates to

$t+c = \frac{1}{\sqrt{bg}}\textrm{arctanh}\left(\sqrt{\frac{b}{g}} v\right)$.

Solving for $v$ yields

$v = \sqrt{\frac{g}{b}}\tanh\left(\sqrt{bg}(t+c)\right)$

If the hamster is dropped from rest, the initial condition gives $c = 0$ so

$v = \sqrt{\frac{g}{b}}\tanh\left(\sqrt{bg}t\right)$

The hamster speeds up, approaching its terminal velocity

$v_{ter} = \sqrt{\frac{g}{b}}$

To estimate the terminal velocity, go back to the heuristic that air resistance dominates when the hamster sweeps out its own mass worth of air. That happened at 50m, and in a 50m free-fall you accelerate to about 30m/s, or 70mph (the hamster should survive, I’m told).

If we measure time in units of $\sqrt{gb}$ and then choose units of length so that the velocity $\sqrt{g/b} = 1$, the hamster’s velocity during free fall is just

$v = \tanh(t)$

An Impractical High School Science Fair Idea

Suppose the hamster survives, so we put it in a rocket blasting into deep space. We stay behind on Earth. The hamster has constant acceleration $g$, as viewed from its own reference frame.

The effects of relativity will be important when $gt \approx c$, which happens after about a year. If the hamster’s trip will be longer than that, we better include the effects of relativity.

At a given moment, the hamster is going along at some speed $v$. To transform from the reference frame of Earth to the hamster, we use the Lorentz transformation

$\left( \begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x' \\ t' \end{array}\right)$

where $\theta$ is the rapidity, defined by

$\tanh\theta = v/c$. We’ll choose the proportion of units of length and time such that $c = 1$.

The hamster accelerates, so this transformation depends on time. Let’s try to find $\theta(\tau)$, the rapidity as a function of the proper time experienced by the hamster.

In a small proper time $\textrm{d}\tau$, the hamster accelerates by $g \textrm{d}\tau$. The rapidity of this transformation is given by

$\tanh\theta' = g \textrm{d}\tau$.

For small $\theta'$, $\tanh\theta' = \theta'$, so

$\theta' = g \textrm{d}\tau$.

We can measure time in units of $1/g$. Then the transformation from the hamster’s comoving frame from one moment to the next is

$\left( \begin{array}{cc} \cosh \textrm{d}\tau & \sinh \textrm{d}\tau \\ \sinh \textrm{d}\tau & \cosh \textrm{d}\tau \end{array} \right) \left(\begin{array}{c} x' \\ t' \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

To find the hamster’s new velocity in Earth’s frame, we simply compose the transformations

$\left( \begin{array}{cc} \cosh \textrm{d}\tau & \sinh \textrm{d}\tau \\ \sinh \textrm{d}\tau & \cosh \textrm{d}\tau \end{array} \right) \left( \begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

That’s an easy matrix multiplication – it’s almost a rotation matrix. We get

$\left( \begin{array}{cc} \cosh(\theta + \textrm{d}\tau)& \sinh(\theta + \textrm{d}\tau)\\ \sinh(\theta + \textrm{d}\tau) & \cosh(\theta + \textrm{d}\tau) \end{array} \right) \left(\begin{array}{c} x \\ t \end{array} \right) = \left( \begin{array}{c} x'' \\ t'' \end{array}\right)$

We see that $\theta(\tau + \textrm{d}\tau) = \theta(\tau) + \textrm{d}\tau$

Since $\theta(0) = 0$ (i.e. the hamster has 0 speed when $t = 0$), this integrates to

$\theta(\tau) = \tau$.

Returning to the velocity $v = \tanh(\theta)$ gives

$v = \tanh(\tau)$.

The velocity of the hamster dropped from the tall building, as a function of time, follows the same equation as the velocity of the hamster accelerating in the spaceship, as a function of proper time. All we have to do is substitute $c = v_{ter}$ and, choosing the correct units of length (depending on the constant in the equation for the hamster’s drag), the analogy is complete.