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The statistical thermodynamics definition of entropy: $S = kN \ln (W)$ troubles me a lot with the problem of dimenstions. where $S$ is entropy; $k$, the Boltzmann constant; $N$ the number of particles in the system and W the number of microstates corresponding to a given macrostate of the system.

For the equation to be dimensionally correct, $W$ must be a dimensionless number.

But if $W$ were to be a dimensionless number, then rewriting the equation in the form: $S = \ln [(W)^kN]$, we find the quantity in square brackets doesn't make sense - since a pure number is raised to a power that has dimensions! More over, since the argument of a logerthmic quantity must be a dimensionless number, the RHS will have no dimensions while LHS has dimensions leading to a paradox.

Again, statistical thermodynamics is full of equations that give elaborations of the quantity $W$. These elaborations contain logerithmic terms with arguments having dimentions (mostly, properties such as $U$ or $E$ and $V$, OR the corresponding molar quantities such as $U/N$ etc.) which not only defies normal mathematics rules but also gives rise to the perennial problem of Gibbs paradox.

Quantum mechanics, information theory etc are broughin to account for Gibbs paradox - which arises in the first place due to a confusing mathematical expression for entropy.

While there is no confusion in equilibrium (classical) thermodynamics about the fact that S is an extensive property, statistical thermodynamics results/equations lead to arguments wheteher $S$ is an intensive quantity or an extensive quantity - all because of the statistical thermodynamics definition of entropy through the equation $S = kN \ln (W)$ - that is confusing with its inherent problem of dimensions.

Any clarifications regarding the dimensional analysis of the defining equation of entropy in statistical thermodynamics/statistical mechanics is requested.

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    $\begingroup$ Notice this is not a problem in thermo specifically, but in the log identities. There is nothing wrong with say $x=x_0 ln(N)$ (for any dimensionful $x$ and dimensionless $N$), but $x=ln(N^{x_0})$ has the same problem you worry about here. $\endgroup$
    – levitopher
    Commented Jul 13, 2014 at 3:47
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    $\begingroup$ Microcanonical entropy in classical statistical physics is usually defined as $k_B\ln W$ or $k_B\ln(W/N!)$ where $W$ is phase volume corresponding to given volume and energy and $N$ is number of identical particles. There is no $N$ (number of particles) in front of the logarithm. $\endgroup$
    – Ján Lalinský
    Commented Jul 13, 2014 at 7:45
  • $\begingroup$ You are probably confused because you assume that in the expression $\ln W$, $W$ has to be dimensionless. That is not true; you can logarithmize any quantity regardless of its units. If this quantity is varied just by changing its units, due to property of logarithm $\ln (cW) = \ln W + \ln c$, the result is always dimensionless. $\endgroup$
    – Ján Lalinský
    Commented Jul 13, 2014 at 7:50
  • $\begingroup$ I don't understand why you think W has to be dimensionful. S and k have the same units, so the dimensions balance when W is dimensionless, which in fact it is, since it's just a number. $\endgroup$
    – N. Virgo
    Commented Aug 14, 2014 at 22:50
  • $\begingroup$ The first part of my question contained a mistake. Instaed of the statement: For the equation to be dimensionally correct, W must be a dimensional number, it must read: For the equation to be dimensionally correct, W must be a dimensionless number. The question therefore is:S=ln[(W)kN], we find the quantity in square brackets doesn't make sense - since a pure number is raised to a power that has dimensions! More over, since the argument of a logerthmic quantity must be a dimensionless number, the RHS will have no dimensions while LHS has dimensions leading to a paradox. $\endgroup$
    – user55356
    Commented Aug 15, 2014 at 3:36

3 Answers 3

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Strictly speaking, you're using the relation $a\ln b = \ln(b^a)$ outside of its domain of validity. When $a$ has dimensions and $b$ is dimensionless, it's perfectly valid to write $a\ln b$, but it is not equal to $\ln(b^a)$, because $b^a$ is undefined and so is its logarithm.

If you want, for notational convenience you could specify that the logarithm of a power, $\ln(b^a)$, is defined to be equal to $a\ln b$ even if $b^a$ is not itself defined. This would be consistent but fairly confusing, so you won't see it very often. In this specific case, everybody writes $k_B \ln W$, never $\ln\bigl(W^{k_B}\bigr)$, for exactly that reason.

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  • $\begingroup$ Could you please give some reference, so that I can get convinced with your answer. $\endgroup$
    – user55356
    Commented Aug 16, 2014 at 15:01
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    $\begingroup$ What do you want a reference for? $\endgroup$
    – David Z
    Commented Aug 18, 2014 at 7:00
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I'd have used $\Omega$ rather than $W$, and I'd say either way that it's dimensionless. Additonally the Boltzmann constant serves only to define the temperature unit (in the case of SI that's kelvins). So your presmise is incorrect and that's the origin of your problem. Read a textbook, a good one is Schreoeder.

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A reference that demonstrates the fact:'When a is a quantity with units (and b is dimensionless), it's perfectly valid to write alnb, but it is not equal to ln(ba), because ba is undefined and so is its logarithm.'

I want to get more information/details that asserts the information you provided through your useful answer.

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