# Calculating the "mean values" in the thermodynamic limit

• fxdung
In summary: Only if the function is linear.In thermodynamics the mean free path is determined by the product of the mean speed and the mean time of collision. It is the distance covered between collisions.
fxdung
In thermodynamics limit, does function of many mean values(of some physical quantities) equal mean value of the function of the values?

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Only if the function is linear.

Simple counter-example: if ##x \in (-1,0,1) ## then ##<x>=0## and if then ##f(x) = x^2 ## you see that ##<f> \ne f(<x>)##

Why in Thermodynamics we calculate the mean free path(of molecules of gas) equal product of mean speed and mean time of collision? It would be wrong?

Any reference ? (So we can look at the expressions)

It is in Concepts in Thermal Physics of Blundell page 73:
8.3 The mean free path
Having derived the mean collision time, it is tempting to derive the mean free path as
λ=<v><τ>=<v>/nσv (8.15)

Don't have the book. But it looks linear ?
However, 'Tempting' begs for a context ...

Reality one substitute v=squaroot(2)*<v>(because v is relative speed) but the formula is not changed.Here we have product of two mean value.

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Ah, the 'tempting' refers to the need of relative speed, same as in hyperpyhsics link.

And the side note (eq 7.23) doesn't throw a tantrum over an error of some 8%, so not much reason to worry

But what I like to ask is why can we write mean lamda=product of mean speed and mean time?

Because the free path is the distance covered between collisions, and the time between collisions is worked out in 8.1.
Distance = ## v ## times ##t##...

Note that the derivation is an estimate: just like ##v## has a distribution, so will ##\lambda##

If v and t are independent values then the mean value of the product equals product of the mean values, But why we know that they are independent value(Why probability of product equals product of probabilities of v and t)?

They are clearly not independent ##t\propto {1\over v}##

Note that the derivation is an estimate: Main takeaway is to show that ##\lambda >> d##

In Classical Physics we calculate all mean values as calculating the values!Why?

Maybe because it's good enough ?

It seems clear for the transition from Quantum Mechanics to Classical Physics(the uncertainty is able to omit in Classical Physics).But how about Classical Statistics Mechanics?Can we omit the error deviating from the mean values?

That's so general a question I don't feel qualified to answer. @vela ? @vanhees71 ?

The idea that always ##\langle f(\vec{x},\vec{p}) \rangle=f(\langle \vec{x} \rangle,\langle \vec{p}##, \rangle) is obviously WRONG.

Take an ideal gas in the classical limit. Then the phase-space distribution function is given by the Maxwell-Boltzmann distribution,
$$\frac{\mathrm{d} N}{\mathrm{d}^3 x \mathrm{d}^3 p}=\frac{1}{(2 \pi \hbar)^3} \exp\left (-\frac{\vec{p}^2}{2m k T} \right)=f(\vec{x},\vec{p}).$$
The average momentum is obviously
$$\langle \vec{p} \rangle=\frac{1}{Z} V \int_{\mathbb{R}^3} \mathrm{d}^3 p \vec{p} f(\vec{p})=0$$
but
$$\langle \vec{p}^2 \rangle=2m \langle E \rangle=3m k T \neq \langle \vec{p} \rangle^2.$$

If all physical values >0, then in thermodynamic limit,the mean value of function of variables is equal the function of mean values of the variables?

## 1. What is the thermodynamic limit?

The thermodynamic limit refers to the theoretical concept of taking a system to an infinitely large size. In statistical mechanics, this limit is used to study the behavior of a system with a large number of particles or degrees of freedom.

## 2. Why is the thermodynamic limit important in calculating mean values?

The thermodynamic limit is important because it allows us to make predictions about the behavior of a system at the macroscopic level. By taking the limit of a large number of particles, we can eliminate the effects of fluctuations and obtain more accurate mean values.

## 3. How is the thermodynamic limit used in statistical mechanics?

In statistical mechanics, the thermodynamic limit is used to calculate the average properties of a system, such as energy, temperature, and pressure. This is done by taking the limit of a large number of particles and using statistical methods to calculate the mean values of these properties.

## 4. What is the significance of mean values in the thermodynamic limit?

The mean values calculated in the thermodynamic limit are important because they represent the average behavior of a system at the macroscopic level. They allow us to make predictions about the system's behavior and understand its thermodynamic properties.

## 5. Can the thermodynamic limit be applied to all systems?

The thermodynamic limit can be applied to most systems, but there are some exceptions. For example, systems with long-range interactions or systems that are not in equilibrium may not behave as expected in the thermodynamic limit. In these cases, other methods must be used to calculate mean values.

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