# Average distance of a freefalling body in random intervals

This is taken from the first example in Griffith's intro to QM:

Suppose I drop a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distances? That is to say, what is the time average of the distance traveled?

Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance must be less than h/2. Ignoring air resistance, the distance x at time t is

$$x(t) = \frac 1 2 gt^2$$

The velocity is $$dx/dt = gt$$, and the total flight time is $$T = \sqrt{2h/g}$$. The probability that the camera flashes in the interval dt is dt/T, so the probability that a given photograph shows a distance in the corresponding range dx is

$$\frac{dt}{T} = \frac{dx}{gt}\sqrt{\frac{g}{2h}} = \frac{1}{2\sqrt{hx}} dx$$.

The solution goes on from there, but there are some things that kind of seem ambiguous about the description here. I could ask a ton of questions about exactly what scenario is being described. But rather than try to ask all these questions, I feel like I might understand things better if I try to think about it in more standard probability language.

In other probabilistic analyses, we need: (1) events, (2) their probabilities, and (3) an associated value. Then we can apply the expected value formula.

So what exactly is the event that we're talking about here? I grab a random time $$t_a$$ and a moment soon after $$t_a+\varepsilon$$, and I grab a random photograph, and find the probability that the duration when the camera shutter was open lies entirely inside of $$[t_a,t_a+\varepsilon]$$?

So maybe we can model it by first fixing some small $$\varepsilon>0$$ and using a uniform pdf over the interval $$[0,T-\varepsilon]$$. There's our events and the associated probabilities.

Now what's the associated values? The length of the rock's path through this interval? If that's right then it's precisely $$\frac 1 2 g(t_a+\varepsilon)^2 - \frac 1 2 gt_a^2 = gt_a\varepsilon + \varepsilon^2/2$$.

Now we apply the expected value formula and get

$$\int_0^{T-\varepsilon}(gt_a\varepsilon + \varepsilon^2/2)\frac{1}{T-\varepsilon}dt_a$$

But at this point it looks like something went wrong. If you compute the integral and take the limit as $$\varepsilon \rightarrow 0$$ you get nonsense. And at the very least I'm not getting the answer Griffiths does.

I don't think I got my events wrong, or their probabilities, although maybe I did. More likely I think I got the associated values wrong. But if it's not the distance the rock goes through from time $$t_a$$ to $$t_a+\varepsilon$$ then what is it?

Note that I saw this but it didn't clarify anything for me: Probability Density of a freely falling body

[Edit: As I think more about my analysis, it probably has something to do with the relationship between $$t_a$$ and $$\varepsilon$$. In fact I'm realizing it doesn't make sense to integrate with respect to $$t_a$$ since that is explicitly a single point ... I think ... right? But then I'm not sure how to properly set up the integral.

Sorry, I've never been very good at modeling physical systems with infinitesimals. The approximations always confuse me.]

If the goal is just to compute the time average of distance traveled, then it needn't be so complicated. The (continuous) random variable is distance traveled, and the probability space is the interval $$[0,T]$$ with the uniform distribution. So, the average (or expected value) of distance is $$E[x(t)]=\int_0^T \frac{x(t)}{T}\mathrm dt=\frac{1}{T}\int_0^{T}\frac12 gt^2\mathrm dt\\ =\frac12 g\frac{T^2}{3}=\frac h3$$ after plugging in $$T=\sqrt{2h/g}$$.
The benefit of the books approach is in finding a density describing the distribution of the distances traveled, $$x$$, which is nice and intuitively appealing. Mathematically this is just the change of variables $$\int_0^T\frac{1}{T}\mathrm dt\stackrel{t=\sqrt{2x/g}}=\int_0^h \frac{1}{2\sqrt{hx}} \mathrm dx$$ yielding the density he claims: $$\rho(x)=\frac{1}{2\sqrt{hx}}\mathbb{1}_{[0,h]}(x)$$. Note the intuition that the rock should spend more time near the top is borne out by the shape of the density.