Let's let $p(\xi)$ be the probability a molecule travels at least $\xi$ between collisions, lets say $\xi=0.01$. When I think of this statement, I think it it is the probability the molecule is able to travel 0.01 OR greater since the key word is AT LEAST. The formula of determining this probability is $$p(\xi)=e^{\frac{\xi}{\Lambda}}$$

Now what if I want to determine when the probability the molecule travels GREATER THAN 0.01? I feel like there should be a different formula for this because it is only greater than 0.01 and it doesn't include 0.01 like before. I can't seem to find the formula for this and I don't know enough molecular theory to derive it.

I am also trying to determine the root mean square free path. I know how to find the mean free path but I cannot find anything online about the root mean square free path


2 Answers 2


Regarding your "at least" vs "greater than":

I think your intuition about "at least" vs "greater than" works for discreet outcomes, but not continuous ones like what you are considering. Well, at least I think your parameter squiggle is continuous? Read on if so.

Think about a six-sided die. This is a discreet case. The probability to roll at least 3 is 4/6, while the probability to roll greater than 3 is 3/6. Note here that the difference in the probabilities $4/6 - 3/6 = 1/6$ characterizes the difference between "at least" and "greater than."

Now think of a 100 sided die. The difference between corresponding probabilities will be 1/100. Smaller.

Continue this limiting process. It seems that for the continuous cases there isn't a difference between "at least" and "greater than."

(Someone who knows statistics might be able to clean up my messy language.)

  • $\begingroup$ I'm not really sure what kind of variable the $\xi$ is. It is just the distance the molecule will travel without a collision. I'm guessing it's continuous since it can be 0.01 or 0.012 or 0.013. Are you essentially saying that I should take "at least" and "greater than" to be the same? $\endgroup$ Apr 11, 2014 at 6:10
  • $\begingroup$ That's where my own logic leads me. It's possible I'm wrong. Others will chime in and agree or disagree. $\endgroup$
    – BMS
    Apr 11, 2014 at 6:11

Actually, this is a very general issue that affects all continuous measurable quantities. Whenever you make a measurement of a continuous quantity, you are choosing from all possible real numbers within the allowed range for that quantity. There are an infinite number of possible values, and thus the probability of getting a specific one of them - like $\xi = 0.01$, in your example - is zero. And it should make sense that the probability of getting $\xi > 0.01$ differs from the probability of getting $\xi \ge 0.01$ by the probability of getting $\xi = 0.01$: zero, which means the two former probabilities are the same.

$$\begin{align} P(\xi > 0.01) &= P(\xi \ge 0.01) - P(\xi = 0.01) \tag{1} \\ &= P(\xi \ge 0.01) - 0 \end{align}$$

Of course, that's not a rigorous mathematical argument, just a way to intuitively understand why the two functions might be the same.

If you want a more formal demonstration, you have to consider the probability distribution function $p(\xi)$. which is defined such that the probability of $\xi$ being between $a$ and $b$ is equal to

$$\int_a^b p(\xi)\mathrm{d}\xi$$

We use this because it doesn't make sense to talk of the measured value of $\xi$ being equal to something because you can never actually establish that it is equal, only that it is close to within some precision. In your specific case, you can't talk about $\xi$ being equal to 0.01, but you can talk about it being in some small range around 0.01. So, whereas equation (1) from above doesn't really make sense, this equation does make sense:

$$\int_{0.01+\epsilon}^\infty p(\xi)\mathrm{d}\xi = \int_{0.01-\epsilon}^\infty p(\xi)\mathrm{d}\xi - \int_{0.01-\epsilon}^{0.01+\epsilon} p(\xi)\mathrm{d}\xi$$

Here $\epsilon$ is some small value which you can think of as the precision of your measuring device. If $\lvert\xi - 0.01\rvert < \epsilon$ then your measuring device will read 0.01, otherwise it will read something else. (This is a highly simplified model of measurement error, but it works for my purposes here.)

Because $p(\xi)$ is finite around 0.01, you can in principle use a measuring device that is precise enough (i.e. small enough $\epsilon$) that the last term, $\int_{0.01-\epsilon}^{0.01+\epsilon} p(\xi)\mathrm{d}\xi$, is small enough to consider negligible, no matter what your definition of "negligible" is. When you do that, then under your definition of "negligible", you get

$$\int_{0.01+\epsilon}^\infty p(\xi)\mathrm{d}\xi = \int_{0.01-\epsilon}^\infty p(\xi)\mathrm{d}\xi$$

which is exactly the statement you're looking for, that the probability of measuring $\xi > 0.01$ is the same as the probability of measuring $\xi \ge 0.01$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.