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I am trying to determine the mathematical difference between mean free path and root-mean-square free path. For an ideal gas, the relaxation time is $$\tau=\frac{1}{\sqrt2 \pi nd^2 \bar v}$$ and the mean free path is $$\Lambda=\tau \bar v $$ so the velocities cancel. When I am calculating the RMS free path, I am assuming I would use $$\Lambda_{rms}=v_{rms}\tau_{rms}$$ and I am assuming $$\tau_{rms}=\frac{1}{\sqrt2 n \pi d^2 v_{rms}}$$

This would again cause $v_{rms}$ to cancel leaving the RMS free path as the same value as the mean free path which seems odd to me. Should I just use regular $\tau$ instead of $\tau_{rms}$?

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    $\begingroup$ $\Lambda_{rms} = \sqrt{\mathrm{E}[(v\tau{})^2]}$, where E() is the expectation value. I don't think this will work out to be the same as $v_{rms}\tau_{rms}$. $\endgroup$
    – The Photon
    Apr 13, 2014 at 23:18
  • $\begingroup$ I have no idea what to even do with that honestly. I am not a molecular physics guy. I just need to find a formula for RMS free path and there is nothing on google. $\endgroup$ Apr 13, 2014 at 23:38
  • $\begingroup$ To be honest if I knew what to do with it I would post an answer rather than a comment. I'm just saying I think you need to step back and question the assumptions you stated in the post before you can move forward. Hopefully someone else will actually know the answer and be able to give you more help. $\endgroup$
    – The Photon
    Apr 14, 2014 at 0:29

2 Answers 2

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You're running into a tricky property of statistical variables: what is true for an individual particle is not necessarily true when averaged across a distribution. In particular, you can say that the distance one particle travels between collisions, its free path length $\ell$, is equal to that one particle's speed times its free path time $t$:

$$\ell = vt$$

but that does not necessarily mean that the mean free path length will be equal to the mean (or RMS) speed times the mean free time. So one needs to be careful when trying to write relations like $\Lambda = \bar{v}\tau$.

Relating RMS and mean lengths, times, and speeds

Consider an event in which a particle traveling at a speed $v$ bounces off one particle and then off another particle a time $t$ later, traversing a distance $\ell$ between bounces. As I mentioned before, $\ell = vt$ for this one event.

Now, in a gas there are many of these events happening. The times $t$, the speeds $v$, and the lengths $\ell$ all have probability distributions: respectively,

$$\begin{align} &p_t(t) & &p_v(v) & &p_L(\ell) \end{align}$$

This means, for example, the probability that a randomly chosen event's path length is between $a$ and $b$ is

$$\int_{a}^{b}p_L(\ell)\mathrm{d}\ell$$

The mean and RMS free paths are then defined as

$$\begin{align} \Lambda &= \int_0^\infty \ell p_L(\ell)\mathrm{d}\ell & \Lambda_\text{rms} &= \sqrt{\int_0^\infty \ell^2 p_L(\ell)\mathrm{d}\ell} \end{align}$$

respectively, and similarly for mean ($\bar{v}$) and RMS ($v_\text{rms}$) speed and time ($\tau$ and $\tau_\text{rms}$).

Now, these probability distributions are uncorrelated except that we know $\ell = vt$ for each event. So we can write

$$p_L(\ell) = \int_0^\infty\int_0^\infty p_v(v) p_t(t)\delta(\ell - vt)\mathrm{d}v\,\mathrm{d}t$$

This is essentially stating that for each length $\ell$, an event with that length can occur for any speed $v$ (with probability $p_v(v)$) and any time $t$ (with probability $p_t(t)$) such that $vt = \ell$. Then you integrate over all possible $v$'s and $t$'s.

Plugging this into the definition of the mean free path,

$$\begin{align} \Lambda &= \int_0^\infty \ell p_L(\ell)\mathrm{d}\ell \\ &= \int_0^\infty \int_0^\infty\int_0^\infty \ell p_v(v) p_t(t)\delta(\ell - vt)\mathrm{d}v\,\mathrm{d}t\,\mathrm{d}\ell \\ &= \int_0^\infty \int_0^\infty vt p_v(v) p_t(t)\mathrm{d}v\,\mathrm{d}t \\ &= \int_0^\infty vp_v(v)\,\mathrm{d}v \int_0^\infty t p_t(t)\,\mathrm{d}t \\ &= \bar{v}\tau \end{align}$$

You can do the same thing for the RMS free path:

$$\begin{align} \Lambda_\text{rms}^2 &= \int_0^\infty \ell^2 p_L(\ell)\mathrm{d}\ell \\ &= \int_0^\infty \int_0^\infty\int_0^\infty \ell^2 p_v(v) p_t(t)\delta(\ell - vt)\mathrm{d}v\,\mathrm{d}t\,\mathrm{d}\ell \\ &= \int_0^\infty \int_0^\infty v^2t^2 p_v(v) p_t(t)\mathrm{d}v\,\mathrm{d}t \\ &= \int_0^\infty v^2p_v(v)\,\mathrm{d}v \int_0^\infty t^2 p_t(t)\,\mathrm{d}t \\ &= v_\text{rms}^2\tau_\text{rms}^2 \end{align}$$

or

$$\Lambda_\text{rms} = v_\text{rms}\tau_\text{rms}$$

thus confirming that the RMS free path is the product of RMS speed and RMS free time, and the mean free path is the product of mean speed and mean free time.

RMS and mean free time and path in terms of speed

The other thing to check is whether $\tau = \frac{1}{\sqrt{2}\pi nd^2\bar{v}}$ carries over to RMS quantities. First, though, let's see where that relationship comes from.

In this other answer of mine, I explain why the probability distribution for a particle to experience an interaction after a time $\Delta t$ is proportional to time for very short times:

$$P_\text{int}(\Delta t) \sim n\pi d^2\sqrt{2}\bar{v}\Delta t \equiv a\Delta t$$

(where $n = \frac{N}{V}$ is the number density of particles and $a$ is defined to be that combination of constants). Over longer times, this produces an exponential distribution:

$$p_t(t) = ae^{-at}$$

The mean free time is then

$$\tau = \int_0^\infty tp(t)\mathrm{d}t = \int_0^\infty ate^{-at}\mathrm{d}t = \frac{1}{a} = \frac{1}{n\pi d^2\sqrt{2}\bar{v}} $$

Now, doing the same thing for the RMS free time:

$$\tau_\text{rms}^2 = \int_0^\infty t^2p(t)\mathrm{d}t = \int_0^\infty at^2e^{-at}\mathrm{d}t = \frac{2}{a^2} = \frac{1}{(n\pi d^2\bar{v})^2} $$

or

$$\tau_\text{rms} = \frac{1}{n\pi d^2\bar{v}}$$

So the RMS free time actually differs from the mean free time by a factor of $\sqrt{2}$.

Incidentally, you can express this in terms of the RMS speed, by finding $\bar{v}$ as a function of $v_\text{rms}$, but doing so requires the use of the probability distribution for speed $p_v(v)$, which will be the Maxwell-Boltzmann distribution. I'll omit the derivations and simply copy the results from that Wikipedia page:

$$\begin{align} \bar{v} &= \frac{2}{\sqrt{\pi}}v_p & v_\text{rms} &= \sqrt{\frac{3}{2}}v_p \end{align}$$

so

$$\bar{v} = \sqrt{\frac{8}{3\pi}}v_\text{rms}$$

which finally yields

$$\tau_\text{rms} = \frac{\sqrt{3\pi/8}}{n\pi d^2v_\text{rms}}$$

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  • $\begingroup$ Can you explain the delta function you used in your answer? Specifically what the double integral of this delta function itself over the entire v-t plane looks like? Actually I am not familiar with these kinds of delta functions... $\endgroup$
    – user266637
    Aug 16, 2023 at 8:02
  • $\begingroup$ Conceptually, the effect of the delta function is to turn the integral into a one-dimensional integral taken along the curve (in this case, a line) where the argument of the delta function is zero. If it helps, consider a simple example $\iint f(v)\delta(u)\mathrm{d}u\mathrm{d}v$ where you can verify that the delta function turns the 2D integral over the $u$-$v$ plane into a 1D integral along the $v$-axis, and then consider the coordinate transformation $x = u + v$, $y = u - v$. If you want more detail, you may want to ask about this on Mathematics. $\endgroup$
    – David Z
    Aug 17, 2023 at 5:20
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Although this is an old post I feel compelled to write something here to clarify some issues regarding the computation of the mean free path. First, I would like to answer the original post by Greg and address his comments, some of which suggest he might benefit from a definition of statistical average. Second, I would like to point out that a molecule's speed between collisions and its time of flight are correlated.

Greg is indeed correct that the mean free path (defined precisely in David's answer) is equal to the product of $\tau$ and $\bar{v}$ but, as David rightly pointed out, one should not take this for granted. The reason is that a mean (or expected value) of some random variable, in the statistical sense, is to be computed as the first moment of the probability distribution for that variable. (The concept of moment is familiar from the definition of center of mass: you will see that what one is calculating is a first moment of a mass distribution normalized by the total mass.) The mean free path length is therefore the expected value of $\ell$ computed from the distribution of free path lengths, whatever that may be. Likewise the mean free time is the expected value of $t$ computed from the distribution of free flight times (i.e., time between successive collisions), and the mean speed is the expected value of $v$ computed from the distribution of speeds given by the well-known Maxwellian speed distribution for an ideal gas in thermal equilibrium. As a shorthand let us denote mean by angled brackets $<...>$. In general, $<\ell> = <vt> \neq <v><t>$. For both sides to be equal the distribution for $\ell$ must be a product of the distributions for $v$ and $t$. That is, $v$ and $t$ would have to be independent variables.

For the computations that Greg wishes to do the facts are as follows. It turns out that $<\ell> = <v><t>$ and this happens in spite of the fact that $v$ and $t$ are not independent! The value that he surmises, $<\ell> = (\sqrt{2}n\pi d^2)^{-1}$ is correct. As for the RMS version, $<\ell^2> \neq <v^2><t^2>$ except for special interaction potentials. For a classical gas composed of identical rigid spheres, one can show that the ratio of the two sides is $(<\ell^2>/(<v^2><t^2>))^{1/2} = 0.9269$.

Now on to the second point. As someone pointed out in the comments, it is the case that faster molecules do collide more often. In fact, the collision rate $r(v)$, which describes the probability per unit time that a given molecule with speed $v$ hits another molecule, scales asymptotically like $v$. The easiest way to see this is to pick a molecule in the tail of the Maxwellian speed distribution. For such a molecule, almost all other molecules are frozen in place. Therefore, the rate at which it scatters off of these ``fixed'' targets is proportional to its speed. One may calculate the rate for any $v$, large or small, by evaluating the mean relative speed in the equilibrium velocity distribution.

The critical assumption that one often makes for a dilute gas is a Markov property for collision events. That is, the collision rate is independent of the history of the molecule under scrutiny. Using this, one can regard $r$ as a function of $v$ only.

Let us construct a joint probability distribution $P(\vec{v},t)d^3v dt$. A specification of a tuplet of numbers $(\vec{v},t)$ means that we wish to know the probability of finding a molecule that has just undergone a collision which is traveling ballistically with velocity centered around $\vec{v}$ for a time centered around $t$ until its next collision. It must be stressed that $$ P(\vec{v},t) \neq P(\vec{v}) P(t), $$ where the latter two densities correspond to the marginal distributions (i.e., what you get if you integrate over all times or all velocities of the joint distribution, respectively). Moreover, what is the appropriate marginal distribution for velocity? It is not the Maxwell distribution $f(v)d^3v$ because one is interested in the likelihood of finding a range of velocities plucked from a randomly selected collision, not a randomly selected molecule anywhere along its flight path! The appropriate thing to do turns out to weight the Maxwell distribution by a factor of $Ar(v)$ because that is the probability (per unit time) that a molecule experiences a collision. Here $A$ is a normalization factor that ensures unit total probability. $A = 1/\int r(v) f(v)d^3v$. Once we know the probability of finding any molecule within our tolerance for velocity we need to know the probability that it flies without interruption for at least a time $t$. This conditional probability is exponential because it may be thought of as a product of a large number of independent probabilities to avoid a collision. Formally, for a time interval discretized into $n$ steps, it is $\lim_{n \to \infty}(1-r(v)t/n)^n$. So the survival probability is $e^{-r(v)t}$. Finally, the probability that it will collide with something else just after is $r(v)dt$. Hence, $$ P(\vec{v},t)d^3vdt = Ar(v)f(v)d^3v \times e^{-r(v)t} \times r(v)dt. $$ Evidently, because of the $v$-dependence of $r$, $v$ and $t$ are dependent random variables.

From the joint distribution we obtain the one for the path length $\ell$ by enforcing the constraint $\ell = vt$. Following David, $$ P(\ell) \equiv \int_0^\infty dt \int d^3v \, P(\vec{v},t) \delta(\ell-vt). $$ Write the delta function as $|v|^{-1}\delta(t-\ell/v)$ and do the time integral. We find $$ P(\ell) = A\int \frac{r(v)^2}{v} e^{-r(v)\ell/v} f(v)d^3v. $$ The expected value of $\ell$ may be evaluated from $\int_0^\infty \ell P(\ell)d\ell$ by exchanging the order of integration and recognizing that the remaining velocity integral is just the mean speed of the Maxwell distribution. Hence, $<\ell> = A<v>$. Finally, to prove that $A = <t>$ go back and find the marginal distribution $P(t)dt$ and calculate its first moment.

While having formal expressions for the joint and marginal distributions is helpful, for investigating Greg's RMS question one will also need to know the explicit formula for the collision rate. For a gas of rigid spheres, $r(v)$ is known to be $$ Ar(v) = \frac{1}{2\sqrt{2}}e^{-v^2/2} + \frac{\sqrt{\pi}}{4}(v+v^{-1}) \mathrm{erf}(v/\sqrt{2}). $$ Of course, interaction potentials other than the infinitely repulsive short-distance one assumed for rigid spheres will give different expressions for $r$.

Some of these issues were discussed in an article I wrote for Am J Phys.

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