3
$\begingroup$

In the top answer to this question (Is the Boltzmann constant really that important?) I read that the Boltzmann constant is just a dummy factor which converts energy to temperature.

But that allows me to put another numerical value in place of the Boltzmann constant but keep the dimension J/K.

I.e. what if, in $$S=c\ln W,$$ I put $c=56 \, \mathrm J/\mathrm K$ in place of $c=k\approx 1.38\cdot 10^{-23} \, \mathrm J/\mathrm K$?

On the page Thermodynamic beta, the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to Derivation of Boltzmann Distribution Law) the Boltzmann distribution.

So the Boltzmann distribution depends on the numerical value of Boltzmann constant. Then why is the Boltzmann constant just a dummy factor?

For example, the mean speed of molecules depends on

$$S=k\ln W.$$

Changing the numerical value of $k$ would make the speed totally different.

$\endgroup$
  • 1
    $\begingroup$ Guessing but the definition of temperature probably depends on the arbitrary value of the Boltzmann constant. If this dependence is as $\textbf{K} \cdot k_B = $const. (in Joules) his would make the temperature dependent terms invariant under arbitrary constant selection. $\endgroup$ – acarturk May 15 at 21:25
  • 1
    $\begingroup$ Re Changing the numerical value of $k$ would make the speed totally different: In the expression $S = k\,\ln\Omega$, the $S$ on the left hand side has nothing to do with speed. That $S$ instead represents entropy, which has units of energy/temperature. $\endgroup$ – David Hammen May 17 at 11:11
2
$\begingroup$

In thermodynamics, the temperature is actually defined in terms of the entropy:

$\frac{1}{T} = \left(\frac{\partial S}{\partial U} \right)_{V,N}$.

Therefore, if you would change the definition of the bolzmann constant to $k' = a k$, the temperature would scale accordingly with a factor $1/a$, and the combination $kT$, which appears in the Boltzmann distribution, would remain the same.

$\endgroup$

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.