Most books on Thermodynamics give intuitive definitions for intensive and extensive thermodynamic variables. The say, for example that the former is independent of the system's size while the latter does depend on the size. Others argue splitting the system into several subsystems and investigating if the quantity of interest changes or remains the same in those. While all approaches make physical sense, I haven't seen someone defining those two concepts in a rigorous manner (even now that I am in uni).

In the question Understanding "natural variables" of the thermodynamic potentials using the example of the ideal gas, one person argued in the answers that an extensive quantity, say $A$ which is dependent from other thermodynamic variables $X_1, X_2,..., X_n$ is called extensive iff $$ A(λX_1,λX_2,...,λX_n)=λA(X_1,X_2,...,X_n) \ \forall λ\in \mathbb R \ \ (1)$$ In other words, a quantity $A $ is intensive if it is described by a first order homogeneous multivariable function. On the other hand, a quantity $B$ is intensive iff $$ B(λX_1,λX_2,...,λX_n)=B(X_1,X_2,...,X_n) \ \forall λ\in \mathbb R \ \ (2)$$ which makes $B$ a 0 degree homogeneous function.

While those definitions make sense, I don't get why an extensive variable should be a homogeneous function of first degree. For instance, An ideal gas has potential energy $ U(N,T) $ which equals $$ U(N,T)=\frac{3}{2}Nk_BT $$ with N being the number of particles and T the temperature of our system. Obviously, the 'test' (1) fails for most choices of $ λ $ (except 0 and 1) as it is:

$$ U(λN,λT)=λ^2 * U(N,T)$$ which is a second order homogeneous function. Should one extend (1) to be:

$$ A(λX_1,λX_2,...,λX_n)=λ^kA(X_1,X_2,...,X_n) \ \forall λ\in \mathbb R \ \ (1)$$ for some $k \in \mathbb N^* $ ? and thus defining an extensive quantity as a homogeneous function of $k$ degree. Are there other rigorous definitions out there ?


I basically agree with the answer of GiorgioP and only have a little more to add to it.

First, statements in words can be just as rigorous as statements in mathematical symbols, because in the end they are two ways of presenting precisely the same assertion.

The assertion in terms of maths is that a quantity $A$ is extensive if it is homogeneous of degree one with respect to the extensive variables. That looks a bit circular, but it turns out that it is not. But in any case I prefer the following simpler but perfectly rigorous approach.

Here is my preferred way of putting it. Take your system and make an exact copy of it. Then place the two copies side by side and consider the result as a single joint system. Let $A$ be some property of the single system, and $A_2$ be the corresponding property of the joint system.

If $A_2 = A$ then $A$ is intensive. (e.g. density, temperature, etc. when boundary effects are negligible)

If $A_2 = 2 A$ then $A$ is extensive. (e.g. energy, particle number)

If neither of the above then $A$ is not intensive, and strictly speaking it is not extensive either.

Finally, we require that there is nothing special about the number 2 for our system, so we require also that similar statements apply to $\lambda$ copies of the system, where $\lambda$ can be any integer, and then we can also imagine gathering the copies in groups which extends the outcome to all rational multiples $\lambda$ , and from there we can take the result for all real $\lambda$ in the limit.


The correct requirement of homogeneity of degree one is a little more precise than what you have cited: Extensive thermodynamic quantities are homogeneous functions of degree one only with respect to their extensive variables.

This solves the problem with your example of $U(V,T)$, but clearly reintroduces the problem of what an extensive variable is.

Personally, I do not think that an apparently more formal definition is actually much more rigorous than apparently simpler definition. The main definition is already in your original statement of size dependence. Maybe it can be sharpened a little more, but almost everything is there.

However, I think I'll add later a second answer to dig a little more on that issue.


I do not believe that there is a general consistent definition that could include cases such as magnetic or electric polarization energy/work. For example, magnetic work density contribution to the internal energy is $\textbf{H}d\textbf{B}$. While it is natural to say that $\textbf{H}$ is intensive, it is much less clear that in what sense $\textbf{B}$ is extensive. The reason is that we have $\textbf{B}=\mu_0\textbf{H}+\mu_0\textbf{M}$ and the polarization $\textbf{M}$ is clearly an extensive density variable, (include volume integrals to turn the work density into work), but if $\textbf{H}$ is intensive then so must be its scalar multiple $\mu_0\textbf{H}$ and then $\textbf{B}$ is the sum of an intensive and an extensive quantity. The same issue is with electric polarization work. Of course, the source of the problem lies in the difference between the concepts of the macroscopic fields $\textbf{H}$ and $\textbf{B}$, i.e., which is the true magnetic field and what does it represent in the presence of polarizable matter, this field or that field, and the arguments have never really ended either way...

  • $\begingroup$ That formula $H dB$ is energy density, not energy. Energy density is intensive. Polarization is dipole moment per unit volume; it is intensive. Total dipole moment is the integral of magnetization with respect to volume, which is extensive. $\endgroup$ – Andrew Steane Mar 6 '19 at 18:45
  • $\begingroup$ @Andrew_Steane that is the reason why I wrote in parentheses "include volume integrals to turn the work density into work", see above; it does not change the argument. $\endgroup$ – hyportnex Mar 6 '19 at 18:49
  • $\begingroup$ @hyportnex It changes the argument. But that is a different story: the difference between "fields" and "densities", once one has moved to use only intensive quantites. $\endgroup$ – GiorgioP Mar 6 '19 at 18:59
  • $\begingroup$ @GiorgioP please tell me then how you would divide, in a unique way, the vacuum energy term (not energy density) the expression of $\int \mu_0 \textbf{H}^2 dV$ so that one part is extensive and the other is an intensive term and there is no confusion which is which. $\endgroup$ – hyportnex Mar 6 '19 at 19:08
  • $\begingroup$ @hyportnex "B is the sum of an intensive and an extensive quantity" is simply untrue. Regarding which is which in the volume integral, a similar issue arises in any formula for work. e.g. the work by tension in a wire of length $L$ is $f dL$ but we can define $x = L/L_0$ for some constant length $L_0$ and then the work is $L_0 f dx$, i.e. a product of a size indicator ($L_0$) multiplied by two intensive terms. In short, expressions for work are not the way to settle the defn of intensive and extensive. $\endgroup$ – Andrew Steane Mar 6 '19 at 22:28

for rigorous treatment/review of this subject see: S.H. Mannaerts (2014), 'Extensive Quantities in Thermodynamics', Eur J Phys 35(3)035017.


it lays bare the difference between 'additivity' and 'proportional to size/mass'.

  • $\begingroup$ Thanks! I'll give it a look (after exams :D) $\endgroup$ – Andreas Mastronikolis Jun 2 '19 at 14:13

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