I've learnt how to calculate the collision frequency of ideal gas molecules with the walls of container using Maxwell Boltzmann distribution given as

$$ f=\frac{1}{4} n\langle v\rangle$$

However I was wondering if we could find the same result using some basic arguments. Could anyone please help.


3 Answers 3


The formula for collision frequency per unit area, $f = \frac 14 n \overline v$ is independent of the speed distribution.

We sum the number of molecules with speed $v$ arriving at a small patch (area $A$) of wall from different angles (See Fig 1). Then we sum over $v$. enter image description here

Assuming isotropy, the fraction of molecules that are travelling towards $A$ at polar angles (angles to the normal) between $\theta$ and $\theta + d\theta$ and azimuthal angles between $\phi$ and $\phi + d\phi$ is

$$\frac{r\ d\theta\ r\sin \theta\ d\phi}{4 \pi r^2}= \frac{\sin \theta\ d\theta d\phi}{4 \pi} $$

[This is the surface area of the portion of a hemisphere centred on $A$ that subtends angles $d\theta$ and $d\phi$ at $A$, as a fraction of the total surface area of the sphere.]

So if there are $\nu_v$ molecules per unit volume with speed $v$ (or if you like, with speeds between $v\ \text{and}\ v+dv$) the number per unit volume travelling in the direction we've specified is $$\nu_v\frac{\sin \theta\ d\theta d\phi}{4 \pi} $$

Now consider a 'column' (Fig 2) at angles $\theta$ and $\phi$ with an obliquely sliced base coinciding with $A$. The cross-sectional area of the column is $A \cos \theta$. All the molecules travelling at speed $v$ in the specified direction in a length $v \Delta t$ of this column, that is in a volume $A \cos \theta\ v \Delta t$, will hit $A$ in a time $\Delta t$. So the number of molecules with speed $v$ in this column that will hit $A$ per unit time, per unit area is $$f_1(v, \theta, \phi)=\frac{v \Delta t\ A \cos \theta}{\Delta t\ A} \times \nu_v\frac{\sin \theta\ d\theta d\phi}{4 \pi}=\nu_v v \frac{\cos \theta\ \sin \theta\ d \theta\ d\phi}{4 \pi}$$

Integrating over the azimuthal angle, $\phi$, from 0 to $2 \pi$ we get $$f_2(v, \theta) =\frac12 \nu_v v \cos \theta\ \sin \theta\ d \theta$$

Integrating over the polar angle, $\theta$, between 0 and $\frac \pi2$ gives

$$f_3(v) =\ \frac 14 \nu_vv $$ So for all speeds of molecules at once $$f =\ \frac 14 \sum \nu_vv =\ \frac 14 n \overline v\ \ \ \ \ \ \ \text{in which}\ \ \ \ \ \ \ \ n=\sum \nu_v.$$

  • $\begingroup$ I don't think that there's an intuitive way to understand why the fraction is $\frac 14$ – unless you can see intuitively that the integration is going to yield that factor! $\endgroup$ Commented Aug 25, 2020 at 22:26
  • $\begingroup$ Could you please explain how you got the first term of :$\frac fA\ =\ \frac{\nu_v v A \cos \theta}{A} \times \frac{2 \pi r \sin \theta\ r d\theta}{4 \pi r^2}$ $\endgroup$
    – Kashmiri
    Commented Aug 26, 2020 at 4:24
  • $\begingroup$ You are quite right to ask about the first fraction. The general idea is to consider a cylinder at angle $\theta$ to the normal, with its end sliced off obliquely so as to meet the patch of surface, so the cross-section of the cylinder is $A \cos \theta$, and to calculate the number of molecules in that cylinder that will hit $A$ per unit time. This is still not a very good explanation, though. A diagram is needed and the answer could do with some restructuring. I'll work on it over the next day or so and edit the answer above. Watch this space! $\endgroup$ Commented Aug 26, 2020 at 7:51
  • $\begingroup$ Thank you a lot. $\endgroup$
    – Kashmiri
    Commented Aug 26, 2020 at 12:11
  • $\begingroup$ I've attempted to solve it and have got the correct expression , would you like to see it? $\endgroup$
    – Kashmiri
    Commented Aug 26, 2020 at 12:18

In order to gain an intuitive understanding of the collision flux formulae, it might be instructive and insightful to deliberately choose a somewhat wrong ansatz first1. For our gedankenexperiment we assume a cube filled with gas.

Since we want to derive a relation that gives us the 'number of collisions against a wall, per unit of time (e.g., lets say seconds), per unit area (e.g., $m^2$)', we have to demand the following physical dimension for our desideratum,

$$ \left[f \right]_{SI} = \frac{1}{s \, m^2} \quad.$$

Next we can ponder about which quantities f propably depends, that is of course the higher number density $n$ is in the cube the more likely it is for any particle to hit on of the walls. Moreover, if the particles move more rapidly, i.e. their kinetic energy being higher, the more likely it is for the particle to hit the walls more often in a given amount of time. Assuming a uniform, isotropic distribution of the particle velocities into all directions, we can use the average velocity $\langle v \rangle$. Thus, we can confidently assume,

$$ f \propto n \, \langle v \rangle$$

and it also nicely complies with our condition above (the required dimension of $f$).

Next we know that the particles can move in 3 different directions (i.e. parallel any of the three cartesian axes), thus giving us at least a proportionality factor of $1/3$. However, since we want to discuss the collision flux of one specific wall, we only can count half of the particles, as the other half would hit the opposite wall, consequently giving us $1/6$. We arrive at

$$ f = \frac{1}{6} \, n \, \langle v \rangle \quad.$$

However our argumentation is flawed and quite superficial, since we did not take the intrinsic probability distribution of the velocity into account, i.e. the famous Maxwell–Boltzmann distribution. In the correct derivation, as you know, $1/6$ is replaced by $1/4$ which is an artefact of the non-Gaussian probability distribution, which makes the average velocity $\langle v\rangle$ not equivalent with the most likely velocity $v_{max}$ of this distribution (as a pure Gaussian distribution would guarantee).

A fully, water-proof way to arrive at the correct result - without taking the unique and characteristic Maxwell-Boltzmann distribution into account - is diffcult to come by.

Personally, I guess that one part, $1/2$ is very certainly due to the fact that we only count collisions on one, specific wall, e.g. particles traveling in the positive $z$ direction. The missing $1/2$ factor is more elusive but I reckon, it has something to do that you do not want to count particles double in your calculations.

P.S. I have done some digging into a few textbooks on thermodynamics and statistical mechanics and all sources that derive the equation in question (also commonly referred to as collision flux or impinging rate of molecules on a surface, in the literature I consulted) do not specifically discuss or ponder over the, not-really-intuitive, proportionality factor $1/4$. I would be very grateful for any literature or online resources that could improve this answer (and also my definitely not perfect understanding).


  1. As it is also commonly done in other derivations, e.g. the free-fall time $\tau_{ff}$ in Astrophysics.

Let's say we've a cube of length $L$ filled with a gas having $N$ number of particles in it, with concentration of molecules as $n$ . We assume that every molecule has a velocity of $\bar{v}$. The velocity space for our gas is shown in the first figure below, which shows each particle's velocity vector, the arrows represent the velocities of our gas particles( I've drawn only a few velocity vectors but in reality the sphere would be filled by arrows pointing in every direction representing the possible velocities of the gas particles). The sphere has a radius of$|\bar{v}|$ since all the particles have a magnitude of velocity equal to $|\bar{v}|$ .
Let $\frac{N}{4 \pi|v|^{2}}$=$\sigma$.

Now let's compute the average positive $x$ component of the velocity $\bar{v}$and denote it by $\bar{V}_{x}$. Since only the particles with velocity vectors in the hemisphere of velocity space facing us have non negative $x$ component of velocities so only they contribute to the average positive $x$ component of the velocity (See the second figure below). So we sum up it's contribution and we have $$ \bar{V}_{x}=\frac{1}{N / 2} \int_{Hemi} \sigma v_{x} d s $$ where $$v_{x}=\bar{v} \sin \theta \cdot \cos \phi$$ Therefore $$\left.V_{x}=\frac{1}{N / 2} \int_{Hemi} \sigma \cdot \bar{v} \sin \theta \cos \phi\right) d s$$ which reduces to $$\bar{V}_{x}=\frac{1}{2} \bar{v}$$. Now let's go back to our cube and calculate the collision frequency(refer to last figure) The number of particles going to the right is $\frac{n \Delta V}{2}$ where $\Delta V=A \cdot \Delta l$ and all of them collide with the right wall in time $$\Delta t=\frac{\Delta L}{\bar{V}_{x}}$$ the collision frequency per unit area is therefore$$\frac{n\Delta V}{2 \Delta t \cdot A}$$ which is $$ \frac{1}{2} n \bar{V}_{x} $$ and using the value of $V_x$we have the value of collision frequency as $$ \frac{1}{4} n \bar{v}$$ 1enter image description hereenter image description hereenter image description here

  • $\begingroup$ I'm still editing it, I've 2G network from the past year. $\endgroup$
    – Kashmiri
    Commented Aug 26, 2020 at 14:51
  • $\begingroup$ You will define your $\theta$ and your $\phi$, won't you? I'm surprised to see $\sin \theta \cos \phi$, but I'll await developments. $\endgroup$ Commented Aug 26, 2020 at 14:56
  • $\begingroup$ I'm on third floor trying to upload pictures, I've got 2Kb speed. But I'm soo much thankful that you are taking time to see. Thank you $\endgroup$
    – Kashmiri
    Commented Aug 26, 2020 at 15:00
  • $\begingroup$ I'm finding it quite hard to follow, I'm afraid. $r$ and $\Delta L$ are undefined? I'm also worried about $\overline {v_x}=\frac12 \overline v$ [for $v_x > 0$, presumably]; my intuition – perhaps not good enough – is that to establish this, if it is true, would require knowledge of the distribution. $\endgroup$ Commented Aug 26, 2020 at 16:44
  • $\begingroup$ Using the Maxwell distribution I find $\overline {v_x}=\frac14 \overline v$ [for 𝑣𝑥>0] , But this needs checking. $\endgroup$ Commented Aug 26, 2020 at 23:16

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