To continue the discussion of the Planck-versus-Einstein time averages for the energies of Planck’s electromagnetic resonators, let’s start with an elementary example of a time average from the David J. Griffiths and Darrell F. Schroeter book Introduction to Quantum Mechanics, 3rd edition, 2018.
Griffiths
& Schroeter (G&S) give an example of a discrete variable average
and an example of a continuous variable average. We’re interested in the
continuous average, but the discrete data example is a good way to prepare for
the continuous case.
Discrete probability example
You
probably know the drill on this one. I’m purposely choosing this simplest example
because I want to embellish it into a more complicated example--related to the
Boltzmann-Planck complexion-counting versus the Einstein time-average counting—later.
There
are N = 14 people in a room with ages ranging from 14 to
25. Let N(j) represent the number of people who
are j years old. The given dataset is
N(14) = 1
N(15) = 1
N(16) = 3
N(22) = 2
N(24) = 2
N(25) = 5
I
won’t belabor this example too much by repeating it here. I’ll just note that
G&S use it to introduce a few statistical concepts: the most probable age,
the average (mean) age, and the median age. They then show how to calculate the
variance and standard deviation from the average.
In
making these calculations, the concept of probability isn’t even needed. The
“most probable age,” for instance, is based on looking at the data and finding
which age is represented by the largest N(j) value. In this example, it’s
N(25) = 5. And you know how to calculate the average age, right? No concept of probability required for that
either, but it’s a good place to “invent” the concept. Writing out the sum of
the weighted ages, the weight being N(j), and dividing by the number of people in the room gives the average age,
$$\frac{1(14)+1(15)+3(16)+2(22)+2(24)+5(25)}{14}
= 21,$$
from
which we can extract a notion of probability by observing that, when the
divisor 14 is distributed to each term separately, each term has a weighting
factor equal to (number of people that
age) divided by (total number of people), i.e., N(j)/N. This is what
G&S call the probability,
$$P( j ) = \frac{N(j)}{N},$$
But
we know, for instance, that there are two people in the room who are 24, so
what’s probability got to do with it? The probability that a person in the room
is 24 is unity, and the same is true for the probability of two people in the
room being 24. The probability that three or more people in the room are 24 is
zero. These numbers are known from the data set.
To
have any meaning, a probability must describe something that is uncertain or
has a random chance of having different values when we measure it.
Therefore, the randomness in this example is introduced when we ask: what is
the probability that a randomly chosen person in this room is 24? Since there
are two 24-year-olds, the probability is 2/14 or 1/7.
Something
G&S don’t discuss is how things would change if the ages of the people in
this room were given in months, or even in weeks or days. If everyone in
the room knew the exact time of day they were born, their ages could even be
calculated in hours. But with every passing hour their numerical hourly age
increases by one, so we’d need to say something like “everyone write down their
age in hours today at 3:10 p.m.”
I
mention this because we’re about to make the transition from discrete variables
to continuous variables. G&S just sort of jump to the continuous case in
one fell swoop (which I’ll get to in a minute) but I want to make a slower
transition.
So
here’s a question to think about: is it less likely for two or more people in
the room to be the same age in months compared with the same age in years? If
so, why is that the case, and how would you calculate how much less likely it
is? Now we’re getting into some nontrivial probability calculations. For
instance, we’re given that there are three 16-year-olds in the room, so how
would we calculate the probability that all three were born in the same month?
But I’m getting off the immediate subject, so I’ll just put a footnote here.*
The
point I’m making is when you ask for the probability that someone selected at
random from this small group is a certain number of months old, you’re more
likely to get zero, and you’re also more likely get a zero probability that
anyone in the room is the same age in months. Change the unit of measurement to
days or hours and these probabilities become even smaller.
This
is where G&S jump all the way into the existential void—the continuum of
numbers on the number line—when they make the transition to the continuous
case. They say the probability of picking someone whose age “is exactly 16
years, 4 hours, 27 minutes, and 3.33… seconds is zero.” They mean
exactly zero, whereas in my above discussion you can see that as the time
interval becomes smaller, the probability of someone being a certain age
measured in the smaller time intervals gets smaller.
Thus,
choosing the feminine pronoun, G&S say:
“The
only sensible thing to speak about is the probability that her age lies in
some interval—say, between 16 and 17. If the interval is
sufficiently short, this probability is proportional to the length of
the time interval. For example, the chance that her age is between 16 and
16 plus two days is presumably twice the probability that it
is between 16 and 16 plus one day. (Unless, I suppose, there
was some extraordinary baby boom 16 years ago, on exactly that day—in which
case we have simply chosen an interval too long for the rule to apply. If the
baby boom lasted six hours, we’ll take intervals of a second or less, to be on
the safe side. Technically, we’re talking about infinitesimal intervals.)
Continuous probability example
The fact that G&S mention this fluctuation in birth rates is the reason I chose to use this example. Fluctuations are the deciding factor in considering Planck’s shorter time average versus Einstein’s longer time average. Planck uses the idea of fluctuations, without saying that word, to define the entropy of a resonator:
Entropy means disorder, and I thought that one should find this disorder in the irregularity with which even in a completely stationary radiation field the vibrations of the resonator change their amplitude and phase, as long as one considers time intervals long compared to the period of one vibration, but short compared to the duration of a measurement.
He famously found in this same paper, based on the 14 Dec 1900 talk he gave, that the resonators could only "change their amplitude" (amplitude-squared actually) by very small discrete amounts \(\epsilon = h\nu\) so that nothing smaller is allowed, although this seems to be what Planck implies by the "irregularity" of a change in amplitude. To be discussed in a later post!
For now, we’ll just look at what G&S give as the continuum definition of probability, starting with ρ(x)dx = probability of a randomly chosen individual’s age being between x and x + dx.
“The proportionality factor, ρ(x), is often loosely called ‘the probability of getting x,’ but this is sloppy language; a better term is probability density. The probability that x lies between a and b (a finite interval) is given by the integral of ρ(x):
$$P(x)=\int_{b}^{a} \rho (x)dx. $$
“Suppose
someone drops 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?”
Neglecting air resistance, the distance x the rock has fallen in time t is
$$x(t) = ½ gt^2.$$
The speed variable is dx/dt, and the total time of the fall is
$$T=\sqrt \frac{2h}{g}.$$
Thus, say G&S, “the probability that a particular photograph was taken between t and t + dt is dt/T, so the probability that it shows a distance in the corresponding range x to x + dx is
$$\frac{dt}{T} =\frac{dx}{gt}\sqrt \frac{g}{2h}=\frac{1}{2\sqrt hx}dx,$$
and the probability density is
$$\rho (x)=\frac{1}{2\sqrt hx}.$$
I leave it to you to write and solve the time integral based on the above relations, and to use the total time in terms of total distance to get the average distance, \(\frac{h}{3}\). Could we have gotten this same result by using the actual distances (from the top of the cliff to the point where each photo was taken) added up and divided by the number of photos? I'll be applying these ideas to the Planck and Einstein time averages in the next post, i.e., part 3b. And then, after that, back to the future: the final particles-to-waves post! We hope.
* A question you may be contemplating is what happens to these probabilities for a much larger population, say the approximately eight billion people on Earth right now? Then you will have nonzero probabilities of people being the same age in months, even in minutes. There is an age continuum. (Okay, we'd need a statistically significant sample from this overall population to calculate anything from actual numbers. Let's assume we have that.) You can ask for the probability that two randomly selected people were born in the same minute and get a nonzero answer. But you still will get an answer of zero when you pick a person at random and ask for the probability that he or she is “exactly 16 years, 4 hours, 27 minutes, and 3.33… seconds” old. And, not forgetting the question leading to this footnote, the probability that the three 16-year-olds were born in the same month is
$$P = \frac{1}{1} \frac{1}{12}\frac{1}{12}$$
