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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][1], the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to [Derivation of Boltzmann Distribution Law][2]) 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. [1]: https://en.wikipedia.org/wiki/Thermodynamic_beta [2]: https://bouman.chem.georgetown.edu/S98/boltzmann/boltzmann.htm

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.

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][1], the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to [Derivation of Boltzmann Distribution Law][2]) 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. [1]: https://en.wikipedia.org/wiki/Thermodynamic_beta [2]: https://bouman.chem.georgetown.edu/S98/boltzmann/boltzmann.htm

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][1], the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to [Derivation of Boltzmann Distribution Law][2]) 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. [1]: https://en.wikipedia.org/wiki/Thermodynamic_beta [2]: https://bouman.chem.georgetown.edu/S98/boltzmann/boltzmann.htm

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 $c\approx 56 \, \mathrm J/\mathrm K$ in place of $c\approx 1.38\cdot 10^{-23} \, \mathrm J/\mathrm K$?

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?

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][1], the Boltzmann entropy using the Boltzmann constant implies thermodynamic beta, which implies (according to [Derivation of Boltzmann Distribution Law][2]) 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. [1]: https://en.wikipedia.org/wiki/Thermodynamic_beta [2]: https://bouman.chem.georgetown.edu/S98/boltzmann/boltzmann.htm