1
$\begingroup$

I understand that Maxwell-Boltzmann distributions arise for distributions of weakly interacting particles at equilibrium. But I'd like to know if there's a deeper reason behind why they are specifically Maxwellian.

I apologize if the question is poorly formed, it just popped into my head and I thought I'd ask about it.

$\endgroup$
2
$\begingroup$

Maxwell derived it from simple assumptions about collisions of the molecules. If the interaction is weak (decreases fast enough with distance), use of the Boltzmann-Gibbs probability distribution of states of the molecule

$$ \rho(\mathbf r,\mathbf p) = \frac{e^{-\frac{E(\mathbf r,\mathbf p)}{k_B T}}}{Z}, $$ where $$ Z = \int e^{-\frac{E(\mathbf r,\mathbf p)}{k_B T}}\,d^3\mathbf r\,d^3\mathbf p, $$

should be valid and from this one may directly derive the M-B probability distribution $f(v)$ for speeds.

$\endgroup$
0
$\begingroup$

The most lucid answer to this question is the original half-page deduction of the distribution by Maxwell himself which can be found in his collected papers or, more accessibly, in Hawking's anthology "On the Shoulders of Giants". He deduces the form of the probability law from the physical assumptions

a) the probabilities of the components of the molecular velocities in the three coordinate directions are statistically independent;

b) the probablity depends only on the speed, i.e., the absolute value of the velocity vector, and not on its direction.

Mathematically, this means that the only functions of $r$ which split as the product of functions of $x$, $y$ and $z$ are those of the form $a \exp(br^2)$. This leads to what we now call the normal or Gaussian distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.