Probability Density of a freely falling body

Question

The following question has been taken from David J Griffith's Intro to QM. This is not my homework! :D

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 ?

Now the book gives the solution. I do not understand the question.The solution goes on to develop a probability density as a solution although I do not understand why.The probability density $\rho(x)$ has been shown below. I am aware of the mechanics behind the freely falling body. I fail to understand the question. What is the idea behind the question ? The distance the rock travels has to be h, in reality. But how does taking photos change the average distance ? The total time of travel is given by $T = \sqrt{2h/g}$. If we are to take a million snaps, then certainly a snap has to be taken after every fixed number of units of time, in order to accomodate a million snaps within the time $T$. Kindly help me to get the physical reality behind this problem. Thank you !

$\rho(x) = \frac{1}{2\sqrt{hx}}$ where $(0\leq x\leq h)$

Answer 1

I fail to understand the question. What is the idea behind the question ? The distance the rock travels has to be h, in reality. But how does taking photos change the average distance ?

This seems like an exercise in mathematical methods of the probability theory, there is not really any physical reality behind it.

What Griffiths is probably referring to by his million photo idea is calculating average value of ( distance traveled in time $t$, where $t\in [0; \sqrt{2h/g}]$), with each interval of $t$ getting weight proportional to its length. That is done most easily by integration of continuous probability. Summation of exact million of values would be problematic (and it is hard to see what for it would be good).

Answer 2

I'm sorry my answer came 7 years later... but Ján is correct. The idea of this problem is to get a grasp of the concept of probability distribution, since in the context of QM it is a key concept. I agree with the fact that the statement of the problem is not too clear.

There are probably two ways of thinking about the proposed situation to get a physical intuition or a physical viewpoint:

1. Imagine you record a video of the falling body in such a way that during the total time of the fall ($T=\sqrt{2h/g}$) the final video has a million frames. In each frame you can measure the distance from the top of the cliff that the body has fallen, which we call $x$. Then you can just average all these values to get the fallen average distance, which is $\langle x\rangle=h/3$. The probability distribution $\rho(x)=\displaystyle\frac{1}{2\sqrt{gh}}$ just tells you the fraction of those million frames that caught the body at a position $x$ in the range between $x$ and $x+\mathrm{d}x$. Check that for smaller values of $x$ this function takes higher values, which makes sense since the body moves slower there (at the beginning of the fall) and for the most part of the time it will be located at lower values of $x$. In other words, if you randomly choose one video frame it is more likely or more probable to pick one in which the body is at lower values of $x$. Recall that the average value of $x$ (also known as the expectation value of $x$ in QM) can be calculated as $$\langle x\rangle=\int_0^hx\rho(x)\mathrm{d}x$$

2. Now imagine that you repeat the experiment in exactly the same conditions a million times. I mean, suppose you drop the body from the top of the cliff a million times in exactly the same way. Then, each time you drop the body you take a single picture at a random instant of time during the fall (we can safely assume that the elapsed time during one picture snap is negligible if compared to $T$). The point here is that the instant at which you decide to take the picture should be truly random in order to cover all posible values for $x$ (it doesn't work if you decide to take the picture always at the same instant during the fall). Subsequently you measure in all those pictures the position $x$ of the body for each fall and average all obtained values. If you do this with enough randomness you will get the same average result, $\langle x\rangle=h/3$. This second viewpoint makes use of the notion of ensemble, which is a key concept in statistical mechanics (also in QM).

It is quite instructive to calculate the average of the squared values of $x$, i.e. $\langle x^2\rangle$. With it you can then obtain the standard deviation of the distribution, through $$\sigma_x=\sqrt{\langle x^2\rangle-\langle x\rangle^2}$$ where $$\langle x^2\rangle=\int_0^hx^2\rho(x)\mathrm{d}x$$ which results in $\sigma_x=2h/(3\sqrt{5})$. By comparing this value with the average $\langle x\rangle=h/3$ you can see that the dispersion around the average is quite high, $$\frac{\sigma_x}{\langle x\rangle}=\frac{\displaystyle\frac{2h}{3\sqrt{5}}}{h/3}=\frac{2}{\sqrt{5}}\approx 0.89$$

Hope somebody finds this useful!