Bus Stops and the Electron Charge

Posted on November 29, 2007
Filed Under Physics |

While I’m preparing the next part in our neural networks tutorial, let’s digress a bit and talk about something completely different.

You’re standing outside, waiting for the bus to arrive. Being a physicist, you busy yourself with building some model until it does. Assume the rate of bus arrivals is \lambda - for example, 5 buses per hour - and that buses are “memoryless” - that is, you can start your calculations and observations from the moment you’ve arrived at the bus stop without caring about what happened prior to your arrival, or what the hour was when you arrived. Since there are \lambda buses per time unit, the probability that a bus will stop in the next few seconds, dt, is proportional to \lambda dt. Let P(T) be the probability to not see a bus after time T. To derive an expression for P(T), we “chop up” the interval into N subintervals of equal length, T/N:

Exponential distribution waiting times

Assuming T/N to be sufficiently short, \lambda T/N is the probability to see a bus in that time interval and 1-\lambda T/N is the probability to not see one. The probability to not see a single bus in the interval P(T) is simply the product of individual probabilities:

P(T) \approx \left(1-\frac{\lambda T}{N} \right)^N

Taking the limit N \rightarrow \infty, and recalling that

e^x = \lim_{N \rightarrow \infty} \left(1+\frac{x}{N} \right)^N

we obtain

P(T) = e^{-\lambda T}

Puzzle: where did we make use of the “memoryless” property?
The next question we’ll ask ourselves is what is the number of buses we expect to see in a time interval \Delta T. Let’s denote the probability of seeing n buses in that time interval by P_n (\Delta T). I’m going to ask you to take it on faith for now that that this is given by:

P_n(\Delta T) = \frac{ (\lambda \Delta T)^n e^{-\lambda \Delta T}}{n!}

This probability distribution is called a poisson distributions and events that are memoryless, such as bus arrivals (well, in theory at least!), are called poisson processes. It’s convenient to remember that a memoryless process with a rate lambda will be described by a poisson process, as derived above. Poisson processes are particularly easy to analyze because there is only one constant, \lambda, which characterizes them. For example, the average number of events in the time interval \Delta T will be:

E[P_n(\Delta T)] = \lambda \Delta T

and the variance will be:

Var[P_n(\Delta T)] = \lambda \Delta T

exactly the same! This is a unique property of the poisson process.

There are plenty of natural Poisson processes in nature. Here is an ingenious example of how to use them to measure a fundamental constant of nature, the electron charge: consider the weak current of electrons in a wire. Let’s assume we have a small number of electrons, N, such that the rate of electrons passing through a particular point is \lambda, and let’s assume their flux is governed by a memoryless process, so the number of electrons we expect to see within a time interval \Delta T will be governed by the poisson distribution. The mean number will be \lambda \Delta T and the variance will equally be \lambda \Delta T. Now, when you think about it, the current we measure will be given simply by the charge per electron times the number of electrons which pass a given point in a given unit time interval:

I = e \dot{n}

where \dot{n} \equiv \frac{dn}{dt} is the number of electrons passing the point per unit time. Its mean will be:

\bar{I} = e \bar{\dot{n}}

If we assume the rate of passage of electrons is modeled by a Poisson process (since it’s electrons per unit time we’ll take \Delta T = 1) characterized by a rate \lambda (where \lambda has units of electrons/unit time), then:

\bar{I} = e \lambda

The current’s variance will be:

\sigma_I^2 \equiv \bar{I^2} - \bar{I}^2 = e^2 \left[ \bar{\dot{n}^2} - \bar{\dot{n}}^2 \right] = e^2 \lambda = e \bar{I}

where we have made use of the fact that the rate \dot{n} is a poisson process, and hence its variance is equal to its mean. Our result tells us that, if we measure the current’s variance and divide it by its mean, the result will be equal to the electron charge:

\frac{ \sigma_I^2 } { \bar{I}} = e

The following figure gives a visual representation of the mean and variance of the current:

Mean and variance of electric current

For this trick to work, though, one must be able to measure the variance due to the Poisson nature of the current. As it so happens, there are other sources of noise in a given system - such as thermal noise - which dominate the picture except for low temperatures and weak enough currents. Needless to say, this method of measuring e is only meaningful when all other sources of noise have been eliminated or sufficiently suppressed.

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