# Why are expressions such as $\operatorname{ln}T$ used in thermodynamics where $T$ is not dimensionless?

In all thermodynamics texts that I have seen, expressions such as $\operatorname{ln}T$ and $\operatorname{ln}S$ are used, where $T$ is temperature and $S$ is entropy, and also with other thermodynamic quantities such as volume $V$ etc. But I have always thought that this is incorrect because the arguments $x$ in expessions such as $\operatorname{ln}x$ and $e^x$ ought to be dimensionless. Indeed at undergraduate level I always tried to rewrite these expressions in the form $\operatorname{ln}\frac{T}{T_0}$. So is it correct to use expressions such as $\operatorname{ln}T$ at some level?

You are absolutely right about the dimensional analysis. The use of $\ln T$ etc. is always a shorthand for $\ln \left(\frac{T}{T_0}\right)$ which is okay to use if for some reason you don't care about $T_0$, i.e. because it cancels out or you are interested in the asymptotic behaviour only.

In any expression where you have to take derivatives to get observable quantities (partition function, generating functional etc.), it's okay to leave off the scale:

$$\mathrm{d} \ln \left(\frac{T}{T_0}\right) = T^{-1} \mathrm{d}T$$

independent of $T_0$.

So: it's a lazy shorthand - the kind of thing much beloved by physicists. :)

Also, mathematically, there's nothing wrong as such with taking the logarithm of a value expressed in arbitrary units. The logarithm just turns unknown multiplicative factors into unknown additive terms. For example, $$\log(T\ {\rm K}) = \log(T) + \log(1\ {\rm K}).$$

Sure, you end up with a constant term of $\log(1\ {\rm K})$ which has no fixed value, but it's no worse in that regard than any other unknown constant. If it cancels out, great! If not, you just carry it around until you can do something to it (like, say, exponentiate it to get back a factor of $\rm K$ again).

As long as you agree that $T\ {\rm K}$ represents a positive scalar value, it has a well defined logarithm, even if its exact value is unknown. It's no less defined than $\log(2x) = \log(2) + \log(x)$, where $x$ is any unknown positive value. (Of course, you're free to work in a system where you arbitrarily insist that $1\ {\rm K}$ isn't a positive scalar and $\log(1\ {\rm K})$ isn't defined. But the point is that we can define it, in the "obvious" way, without introducing any inconsistencies.)

Ps. It's perhaps worth noting that the reason we can do that is because $\rm K$ is an ordinary multiplicative unit which obeys all the usual rules of arithmetic, such as $1\ {\rm K} + 1\ {\rm K} = 2\ {\rm K}$ and $1\ {\rm K} \cdot 1\ {\rm K} = 1\ {\rm K}^2$. We would not be able to do that for "affine units" like $\rm ^\circ C$, which already conceal an additive constant in the notation. Arguably, this is because writing $x\ {\rm ^∘C}=(273.15+x)\ {\rm K}$ is itself an abuse of notation, retained for historical reasons.

• But mathematically I think you cannot even perform this separation as $\operatorname{ln}(T K)$ does not exist in the first place. I am more inclined to agree with the previous answer - I think this is a sloppy shortcut which isn't truly valid. Jan 2 '13 at 15:08
• Sure you can. As long as you agree that $T\ {\rm K}$ represents a positive scalar value, it has a well defined logarithm, even if its exact value is unknown. It's no less defined than $\log(2x)=\log(2)+\log(x)$, where $x$ is any unknown positive value. (Of course, you're free to work in a system where you arbitrarily insist that $1\ \rm K$ isn't a positive scalar and $\log(1\ {\rm K})$ isn't defined. But the point is that we can define it, in the "obvious" way, without introducing any inconsistencies.) Jan 2 '13 at 17:15
• The key is the lack of inconsistencies. We don't know what ln(1K) means physically just like we don't know what sqrt(-1) means. Both are valid to use mathematically however, the final answers will not have them. However, exp(1K) is NOT valid as it generates inconsistencies. Oct 25 '17 at 5:28