To determine interesting relationships in the properties/variables of a phenomena, one can often use dimensional analysis.

For example, say we want to determine the dependence of environmental variables, such as the pressure $p$, density $\rho$ and volume $V$ of a gas, on the speed of sound $v$ in that gas. Using dimensional analysis we can determine a relationship by assuming that the environmental dependencies listed above are the only dependencies on the speed of sound in that gas. Then, typically in dimensional analysis, one writes $$v = C p^a \rho^b V^c$$ for some unknown constant $C$ which can be determined by other means, and constants $a,b,c$ which can be determined by matching units on both sides. Doing this, one can obtain the relationship $$ v = C\sqrt{\frac{p}{\rho}}. $$

Now my question is, say we want to determine the dependence of a variable $X$ on a series of variables $Y_i$:

  • In dimensional analysis, why do we always assume that the underlying relationship between $X$ and $(Y_i)$ is multiplicative? There clearly exist laws where the relationship between variables is partially additive.

That is, why do we always assume that $$ X = C\prod_iY_i^{a_i}, $$ for some constants that we wish to solve for being $a_i$.

  • Assuming that one can match dimensions on both sides, why do we not consider partially additive relations?

For example, why can we not assume that $$ X = C\sum_{j} \prod_{i}X_{i(j)}^{a_{i(j)}}? $$ i.e. for simplicity $$ X = C_1Y_1^{a_1}Y_2^{a_2}Y_3^{a_3} + C_2Y_4^{a_4}Y_5^{a_5} + C_3Y_6^{a_6}, $$ where we would assume that the dimensions of all terms match.

  • What makes the multiplicative identity more valid that the partially additive one?

I may be missing some understanding here, and this may have already been answered before. If so, a redirect on where I could find an answer would be appreciated.

  • $\begingroup$ Why do you say that we always assume a multiplicative relationship? I don't think that's true in general $\endgroup$ Feb 12, 2021 at 10:53
  • $\begingroup$ @KarimChahine This is at least, in my limited exposure to dimensional analysis, is what I've seen. If this is not generally the case, it would be nice to understand better how one would obtain relationships in this partially additive case/be redirected to a resource. $\endgroup$
    – spaceman
    Feb 12, 2021 at 10:55
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    $\begingroup$ I like this question. I look forward to an answer. +1 from me. $\endgroup$
    – Gert
    Feb 12, 2021 at 10:58
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    $\begingroup$ @Spaceman All the examples I've seen are of the $\Pi$ variety. $\endgroup$
    – Gert
    Feb 12, 2021 at 10:59
  • $\begingroup$ look up the Bucingham pi theorem $\endgroup$ Feb 12, 2021 at 12:12

2 Answers 2


I will first rewrite your last as $$ \pi_1= C_1 X^{-1} Y_1^{a_1}Y_2^{a_2}Y_3^{a_3} +C_2 X^{-1} Y_4^{a_4}Y_5^{a_5} +C_3 X^{-1} Y_6^{a_6}\, , $$ where $\pi_1$ is dimensionless. As you point out all the monomials in this sum are dimensionless.

It is entirely possible to be able to construct more one dimensionless quantity given a set of variables. In your specific example, they are $ X^{-1} Y_1^{a_1}Y_2^{a_2}Y_3^{a_3}$, $X^{-1} Y_4^{a_4}Y_5^{a_5} $ and $X^{-1} Y_6^{a_6}$.

What Buckingham's $\pi$ theorem states is that $\pi_1$ is a general function $f$ of these three dimensionless quantities: $f(X^{-1} Y_1^{a_1}Y_2^{a_2}Y_3^{a_3},X^{-1} Y_4^{a_4}Y_5^{a_5} ,X^{-1} Y_6^{a_6})$. There is no statement about the form of the function (although they are usually polynomial) and in fact the form of the function is usually determined by experiment. Moreover, the choice of dimensionless variable is not unique as $\frac{X^{-1} Y_1^{a_1}Y_2^{a_2}Y_3^{a_3}}{X^{-1} Y_4^{a_4}Y_5^{a_5}}$ is also dimensionless. It's a matter of which variables (or combinations thereof) one can access that will often dictate which combinations of dimensionless factors should be selected.

There's a nice example of the situation you mention with additive terms in

Goodridge, C.L., Shi, W.T., Hentschel, H.G.E. and Lathrop, D.P., 1997. Viscous effects in droplet-ejecting capillary waves. Physical Review E, 56(1), p.472.

where a dispersion relation is given as \begin{align} \omega^2 = gk+\frac{\sigma}{\rho}k^3\, . \tag{1} \end{align} It is easily shown that, from the wave number $k$, the surface tension $\sigma$ and the water density $\rho$, plus gravity $g$, one can construct more than one combination with dimensions of $\omega^2$. The authors of the paper actually investigate experimentally the two regimes where one or the other term in (1) is the dominant term, and discuss briefly the "cross-over" point where one regime transitions to the other.

(I found the example above in a set of notes by Robert Gilmore of Drexel University: http://www.physics.drexel.edu/~bob/Chapters/dimensional3.pdf )


The reason the aforementioned dimensional analysis trick works is because the only way to combine a pressure $p$, density $\rho$, and volume $V$ to obtain a quantity with dimensions of velocity is $v\sim \sqrt{p/\rho}$.

Dimensions as a Vector Space

To see this more clearly, consider a quantity $Q$ with dimensions $[Q] = \mathrm M^a \mathrm L^b \mathrm T^c$, where $\mathrm M,\mathrm L$, and $\mathrm T$ are mass, length, and time, respectively. To this quantity we associate the 3-tuple of real numbers $$[Q]\rightarrow \pmatrix{a\\b\\c}\in \mathbb R^3$$

From this point of view, we can regard the dimensionality of $Q$ as a point in $\mathbb R^3$ - but we can actually go further here, and think of this "dimensionality space" as a vector space. If another quantity $R$ has dimensions $[R] = \mathrm M^{a'} \mathrm L^{b'} \mathrm T^{c'}$, then $$[Q\cdot R] = \pmatrix{a+a'\\b+b'\\c+c'}$$ $$[Q/R] = \pmatrix{a-a'\\b-b'\\c-c'}$$ Furthermore, a pure number can be associated to the point $(0,0,0)$. You can check that this construction satisfies all the requirements of a vector space.

Note also that there is nothing special about mass, length, and time - I simply chose these as my "basis" dimensions. You could choose different dimensions as your fundamental ones if you'd like, as long as the resulting basis is linearly independent. For example, you could choose mass, velocity, and time, but you couldn't choose velocity, length, and time.

There's also nothing special about only three basis dimensions. You could add e.g. electric charge to this picture, at which point you'd have to consider $\mathbb R^4$, but the same ideas would apply.

We can now frame the dimensional analysis trick as elementary linear algebra. We want to combine $p,\rho$, and $V$ to obtain a quantity with dimensions of velocity. Note that in our chosen basis, $$[p]\rightarrow\pmatrix{1\\-1\\-2}\quad [\rho]\rightarrow \pmatrix{1\\-3\\0}\quad [V]\rightarrow\pmatrix{0\\3\\0}\quad [v]\rightarrow \pmatrix{0\\1\\-1}$$ and so your expression $v\propto p^a\rho^b V^c$ becomes $$\pmatrix{0\\1\\-1} = a\pmatrix{1\\-1\\-2}+b\pmatrix{1\\-3\\0}+c\pmatrix{0\\3\\0}=\pmatrix{1&1&0\\-1&-3&3\\-2&0&0}\pmatrix{a\\b\\c}$$

Because the dimensions of $p,\rho$, and $V$ are linearly independent, that matrix can be inverted, yielding a unique solution $$\pmatrix{a\\b\\c}=\pmatrix{1/2\\-1/2\\0}$$

With this formalism under our belt, we can address your questions.

In dimensional analysis, why do we always assume that the underlying relationship between X and (Yi) is multiplicative? There clearly exist laws where the relationship between variables is partially additive.

The entire premise of this dimensional analysis trick is that there is one unique way to combine your variables to get the dimensions you want. If you have a sum of terms, then either (i) they differ by multiplication by a dimensionless constant, in which case you could trivially combine them into a single term, or (ii) there's more than one way to combine your variables get the proper dimensions. In the latter case, dimensional analysis is not going to work as desired.

Framed in terms of linear algebra, you would find that there is an infinite number of combinations of your variables which yield the correct dimensions. For example, if you're trying to get a length $x$ and your quantities are a velocity $v$, a time $t$, and an acceleration $a$, then you can try $x \propto v^a t^b a^c$. Applying the linear algebra technique above, you would find that

$$\pmatrix{a\\b\\c}=\pmatrix{1\\1\\0} + r\pmatrix{1\\-1\\-1}$$ for any real number $r$. The solution is not unique - you could have $r=0\implies x\propto vt$, or $r=1\implies x\propto v^2/a$, or $r=-1 \implies x\propto a t^2$, or an infinity of other possibilities. The most general solution would be a combination of all of them, but with no further information to go on, dimensional analysis has not been much help.

The Buckingham Pi Theorem seems to a-priori assume a multiplicative relationship, and I cannot find a justification of this assumption.

The only way to get a relationship which isn't multiplicative is to assume from the start that there's no unique product of quantities with the correct dimensions, which means that dimensional analysis isn't going to work. If you'd like to frame it another way, if dimensional analysis is going to be useful then the solution must be unique, which implies that you can't have a sum of different kinds of terms.

One final way to look at it is that you can have a partially additive relationship if and only if it's possible to combine your variables to make a dimensionless quantity $Q$ (in my example above, $[Q]\leftrightarrow (1,-1,-1) \implies Q = \frac{v}{at}$). But if you can do that, then it should be clear that dimensional analysis isn't going to work because for any solution you obtain, you could simply multiply it by an arbitrary power of $Q$ and get another one which works.

  • $\begingroup$ Excellent, thank you for your very detailed explanation. This has been enlightening! $\endgroup$
    – spaceman
    Feb 17, 2021 at 14:34
  • $\begingroup$ @J.Murray Why there is only one way to combine pressure $p$, density $ρ$ and volume $V$ with dimensions of velocity $v$? I mean we can write: $$ v = \sqrt{p/\rho} + \frac{1}{λ} \cdot \left(\sqrt{p/\rho}\right)^2$$ with $λ$ having the appropriate units so that the units of the RHS matches that of the LHS. $\endgroup$ Oct 5, 2022 at 19:28
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    $\begingroup$ @AntoniosSarikas You've added another dimensionful quantity to the list. If we're allowed to do that, then for that matter we could just let $\lambda$ have dimensions of velocity and say that $v = \sqrt{p/\rho}+\lambda$. The point is that $v,p,\rho$ are the only dimensionful quantities which we are assuming to be relevant. $\endgroup$
    – J. Murray
    Oct 5, 2022 at 22:33
  • $\begingroup$ @J.Murray Thank you for the clarification. $\endgroup$ Oct 5, 2022 at 22:43

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