I have troubles understanding a seemingly simple integral in a physical context. Take a look at $\int_{V_1}^{V_2} \frac{\mathrm{d}V}{V}$ which appears in isothermal expansions (V being the volume of a gas). Now of course the result is $\ln{V_2}-\ln{V_1}$ or using log laws $\ln{\frac{V_2}{V_1}}$.

The first expression would require you to evaluate the logarithm of a unit of volume which to my unterstanding is impossible. The second expression makes sense, though. How can you account for this apparent discrepancy?

  • $\begingroup$ Related: physics.stackexchange.com/q/109995/2451 and links therein. $\endgroup$ – Qmechanic Feb 26 '15 at 21:35
  • 2
    $\begingroup$ Note that the integrand is dimensionless and so must be the limits. That is to say, the limits are normalized (dimensionless) volumes. $\endgroup$ – Alfred Centauri Feb 26 '15 at 21:40
  • $\begingroup$ @AlfredCentauri Um, shouldn't the limits of integration match the units of the differential, not the whole integrand? $\endgroup$ – user10851 Feb 26 '15 at 22:05
  • $\begingroup$ @ChrisWhite, for $f(V) = \ln(V)$, $V$ is dimensionless and $df(V) = \frac{1}{V}dV$ which is dimensionless. $\endgroup$ – Alfred Centauri Feb 26 '15 at 22:52
  • $\begingroup$ Basically you can't do $\ln V_1 - \ln V_2$ if $V_i$ have units, but you can do $\ln (V_1 / v) - \ln (V_2 / v)$ where $v$ is an arbitrary unitful constant, so in practice you can always just cross out the units as long as they are both the same unit. $\endgroup$ – CR Drost Feb 27 '15 at 0:03

How can you account for this apparent discrepancy?

There is no discrepancy in the value of the difference of the logs $$ \log(V_1)-\log(V_2) $$ and the value of the log of the ratio $$ \log{V_1/V_2}\;. $$ And neither of these depends on the choice of units. This is because, for both the difference and the ratio, the units cancel. For example, suppose I represent the "units" as a number "U" such that $$ V=\alpha U\;, $$ where $\alpha$ is unitless.

Then $$ \log(V_1)-\log(V_2)=\log(\alpha_1 U)-\log(\alpha_2 U)=\log(\alpha_1)+\log(U)-\log(\alpha_2)-\log(U) $$ $$ =\log(\alpha_1)-\log(\alpha_2)\;. $$

And, similarly, $$ \log(V_1/V_2)=\log(\frac{\alpha_1 U}{\alpha_2 U})=\log(\frac{\alpha_1}{\alpha_2})\;. $$

Not much else to say but that... Clearly, trying to interpret $\log(V_1)$ on its own doesn't make sense since changing the unit changes the numerical value... but the interpretation and sensibleness of the difference of the logs certainly makes sense and is independent of the choice of units.

  • $\begingroup$ Thanks for the quick reply! I do understand that the ratio $\frac{V_1}{V_2}$ is dimensionless. What I don't quite understand is your rearrangement of the difference $\ln{V_1}-\ln{V_2}$. Why are you "allowed" to apply logarithmic identities to an expression including physical units? Does $\ln(U)$ even mean anything (where U is a unit as in your example)? $\endgroup$ – df42 Feb 26 '15 at 21:57
  • $\begingroup$ @df42 well, think about it like this: does $U$ even mean anything? $\endgroup$ – David Z Feb 26 '15 at 22:36
  • 1
    $\begingroup$ Nice answer. +1. This explanation works for subtraction only. Is there a similar way to show that units are "eliminated" for addition? $\endgroup$ – Steeven Feb 26 '15 at 23:21
  • 2
    $\begingroup$ No, because they're not. $\endgroup$ – hft Feb 26 '15 at 23:47
  • 1
    $\begingroup$ Note at the very start the units of the terms in the integral cancel $$dV/V$$ $\endgroup$ – docscience Feb 27 '15 at 3:55

If you introduce any unit $v$ in terms of which you express the numerical values of $V_1,V_2$, then you see


So if $V_1$ and $V_2$ were the integral bounds of a single integral sign, it's only important that make sure you express the numerical value of $V_1,V_2$, in the difference of the two log-expressions, as multiples of the same arbitrary scale.

Writing $\ln\left(V_2\right)-\ln\left(V_1\right)$ implies you fixed a unit for them and it wasn't explicitly emphasised. People don't get in trouble here because the result is really independent of the unit, as it being equal to $\ln\left(\frac{V_2}{V_1}\right)$ demonstrates.


In mathematics when we put a number x in ln x, we assume x belong to a field, so x must be a dimensionless number. So it is possible to put a pure number of x in ln x. However, many people seem not to know the following expression of lnx as an infinite series expansion. In fact, lnx=2{(x-1)/(x+1)+(1/3)[(x-1)(x-1)(x-1)]/[(x+1)(x+1)(x+1)]+(1/5)[(x-1)(x-1)(x-1)(x-1)(x-1)]/[(x+1)(x+1)(x+1)(x+1)(x+1)]+,,,,,,,} . So now it is clear that we cannot put x in lnx if x has a dimension.

I have published these issues in the three papers:

  1. Mayumi, K. and Giampietro, M. 2010. “Dimensions and Logarithmic Function: A Short Critical Analysis” , Ecological Economics,Vol.69, pp. 1604-1609;
  2. Mayumi, K., and Giampietro, M. 2012. “Response to “Dimensions and logarithmic function in economics: A comment””, Ecological Economics Vol. 75, pp. 12-14;
  3. Mayumi, K., Giampietro, M., And Ramos-Martin, J. 2012. “Reconsideration of Dimensions and Curve Fitting Practice in View of Georgescu-Roegen’s Epistemology in Economics”, Romanian Journal of Economic Forecasting, Vol.15, No.4, pp.17-35.
  • $\begingroup$ Welcome to Physics! Welcome to Physics! Please note that we do have an equation editor built into the site, which can help improve readability of your post. $\endgroup$ – Kyle Kanos Sep 14 '17 at 9:59

protected by Qmechanic Sep 14 '17 at 10:30

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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