# Is it possible to experimentally verify this collision rate formula?

## Background

Let's say I have the $$i$$'th gas molecule with velocity $$\vec v_i(t)$$ at time $$t$$. To find the net displacement $$s_i$$ we integrate with respect to $$t$$:

$$\vec s_i = \int_{0}^{t} \vec v_i(t') dt'$$

Since, the gas molecule under goes collisions. Initially the $$i$$'th gas molecule has velocity $$v_{i,1}$$ for time interval $$t_1$$ and then velocity $$v_{i,2}$$ for time interval $$t_2$$ and so on ...

$$\int_{0}^{t} \vec v_i(t') dt' = \vec v_{i,1} t_1 + \vec v_{i,2} t_2 + \dots + \vec v_{i,n} t_n$$

Now, we can write the velocity as the infinitesimal ratio of displacement and time:

$$\int_{0}^{t} \vec v_i(t') dt' = t_1 \frac{d \vec x_{i,1}}{dt} + t_2 \frac{d \vec x_{i,1}}{dt} + \dots + t_n \frac{dx_{i,n}}{dt}$$

Multiplying both sides by $$dt$$:

$$\int_{0}^{t} \Big( \int_{0}^{t''} \vec v_i(t') dt' \Big) dt'' = \int^{ \vec x_i(t)}_{\vec x_i(0)} t_1 d \vec x_{i,1} + t_2 d \vec x_{i,2} + \dots + t_n dx_{i,n}$$

Using some number theory and analysis and assuming the $$\vec v_i$$ which appear follow a particular distribution:

$$\int_{0}^{t} \Big( \int_{0}^{t''} \vec v_i(t') dt' \Big) dt'' = \langle t_c \rangle \int^{ \vec x_i(t)}_{\vec x_i(0)} d \vec x_i$$

where $$\langle t_c \rangle$$ is the average time interval for a collision. Since we have only done this for one particle we now proceed to do this over all the particles in a volume $$V$$ (and assume the collision rate in the block of volume is the same/the temperature is the same in the volume $$V$$).

$$\int_{0}^{t} \Big( \int_{0}^{t''} \sum_{i=\text{all particle in V}}m_i \vec v_i(t') dt' \Big) dt'' /M = \langle t_c \rangle \sum_{i=\text{all particle in V}} \int_{ \vec x_i(0)}^{\vec x_i(t)} m_i d \vec x_i /M$$

where $$M$$ is the total mass $$M$$ in $$V$$ and $$m_i$$ is the mass of the $$i$$'th particle. Since the center of mass is immune to collisions:

$$\int_{0}^{t} \Big( \int_{0}^{t''} \vec v_{CM}(t') dt' \Big) dt'' = \langle t_c \rangle \int_{0}^t \vec v_{CM} (t'') dt''$$

where $$\vec v_{CM}$$ is the velocity of the center of mass and we used:

$$\sum_{i=\text{all particle in V}} m_i \int_{ \vec x_i(0)}^{\vec x_i(t)} d \vec x_i /M = \sum_i m_i \int_0^t ( \vec v_{i,1} + \vec v_{i,2} + \vec v_{i,3} + \dots ) dt /M$$

## Question

Is the derivation correct? Is there some clever computational experiment way to verify this?

• This discussion has been moved to chat
– Tyberius
Jul 7, 2021 at 13:25

$$\frac{A_\text{CM}}{\Delta x_\text{CM}}=\langle t_\text{c} \rangle$$ where $$\Delta x_\text{CM}$$ is the net displacement of the center of mass traveled in some time interval and $$A_\text{CM}$$ is the mean absement or time integral of the displacement over that interval. Absement isn't a commonly used metric (I hadn't heard of it before researching this question), but you could think of it as a time-weighted displacement (or a displacement-weighted time). So in words, I think your equation would say "the average collision time is equal to the displacement-weighted time divided by the net displacement".