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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?

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

1 Answer 1

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If I'm interpreting your last equation correctly, this is more or less true by definition. My interpretation of the equation (rearranging and subbing in variables):

$$\frac{A_\text{CM}}{\Delta x_\text{CM}}=\langle t_\text{c} \rangle$$ 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".

While I think you could verify this by running an MD simulation and computing these quantities individually, I think this relationship basically just contains the same information as the typical expression for collision time (mean free time), which is just the mean free path divided by the average speed.

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  • $\begingroup$ First let me thank you for your answer. I thought about it and I think it may be worthwhile doing an experiment for this. Because it is not "this is more or less true by definition" since we assume a particular convergence. The number theoretic formula is not true for any arbitrary sequence. Thus, I also think running a simulation may not be helpful since I suspect the simulation would at some point use a random generator to model the randomness of a gas. This formula is only valid for particular random sequences. $\endgroup$ Jan 26 at 0:33

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