Calculating the "mean values" in the thermodynamic limit

[fxdung]

In thermodynamics limit, does function of many mean values(of some physical quantities) equal mean value of the function of the values?

 

[BvU]

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 $<f> \ne f(<x>)$

 

[fxdung]

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?

 

[BvU]

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

 

[fxdung]

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)

 

[BvU]

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

 

[fxdung]

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.

 

[BvU]

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

 

[fxdung]

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

 

[BvU]

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 $\lambda$

 

[fxdung]

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)?

 

[BvU]

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

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

 

[fxdung]

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

 

[BvU]

Maybe because it's good enough ?

 

[fxdung]

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?

 

[BvU]

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

 

[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
$$\langle \vec{p}^2 \rangle=2m \langle E \rangle=3m k T \neq \langle \vec{p} \rangle^2.$$

 

[fxdung]

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?