Sign up ×

A common result of theoretical analysis in physics is some sort of relation derived from physical parameters and typically expressed in the form of a non-dimensional parameter. These scale relations are not equalities but proportional relationships. For instance, in turbulence you end with with

$$ \frac{\eta}{l} \sim Re^{-3/4}$$

Assuming that some derivation results in:

$$f(\Pi_1,\Pi_2,\dots,\Pi_i) \sim C$$

where $C$ is the experimental measure, $f$ is a function of non-dimensional parameters, and $\Pi_i$ are the independent non-dimensional parameters, is there a minimum number of experiments to determine the constant of proportionality $a$ such that:

$$f(\Pi_1,\Pi_2,\dots,\Pi_i) = aC$$

sufficiently? I would expect that the number of experiments required is in some way related to the combination of all possible parameters. For instance, if $i = 2$ then I would expect a minimum of 4 experiments would be required. But is the minimum based on combinations sufficient or are more values needed?

share|cite|improve this question
Part of the answer is going to depend on how well you want to characterize the exponents and to exclude non-power-law dependencies. The short short version is that a power law fit is a line fit on a log--log scale and as such takes at least 5 points to characterize both parameters and their errors (or 6 to get the correlation coefficient as well). – dmckee Dec 12 '12 at 18:39

1 Answer 1

Dimensional analysis allows us to write the solution to any physical system in the form $$ f(\Pi_0, \Pi_1, \Pi_2, \dots) = 0 $$ where the $\Pi$s are independent dimensionless constants formed from our dimensionful physical parameters.

Usually, there is a particular physical parameter we are interested in computing, and so, by rescaling our dimensionless parameters against one another we can arrange that our physical parameter of interest appears in only one of the dimensionless constants. At that point, we can imagine solving this equation for that dimensionless parameter obtaining a relation of the form $$ \Pi_0 = g(\Pi_1, \Pi_2, \dots) $$ where $g$ is an unspecified function that you would have to determine by means of experiment.

Sometimes, the problems we consider are simple enough that we have only a single independent dimensionless parameter can be formed, so that we have $ f(\Pi) = 0 $ or equivalently $$ \Pi = C $$ where $C$ is some constant.

As a somewhat trivial example, let's say we wanted to figure out the area of a circle, and we had forgotten how. There are two physical parameters of interest, the area $A$ with dimensions $[L^2]$ and the radius $r$ with dimensions $[L]$. Given two dimensionful parameters in one dimension there is only a single independent dimensional parameter we can form, so we know the physical law has to take the form $$ f(\Pi) = 0 \qquad \Pi = \frac{A}{r^2} $$ which is equivalent to $$ A = C r^2 $$ And we need only do a single experiment to determine the constant of proportionality. (In this case it's $\pi$)

But, it is rare that we have only a single dimensionless parameter and so can reduce the problem to one of direct proportionality. Take for example the problem of determining the period of a pendulum. First let's collect the physical parameters we have think are important, the period itself $T$, the mass $m$, the length $l$ the initial angle $\theta_0$, and gravity $g$. They have their respective dimensions:

$$ T : [T] \quad m : [M] \quad l : [L] \quad \theta_0 : [1] \quad g : [ L T^{-2} ] $$ These are 5 physical parameters in 3 dimensions ($[M], [L], [T]$), so we have only two independent dimensionless parameters that we can form, let's use the decomposition $$ \Pi_0 = \frac{T^2 g}{l} \qquad \Pi_1 = \theta_0 $$ and so the best we can say with dimensional analysis is $$ f(\Pi_0 , \Pi_1 ) = 0 $$ or equivalently $$ \Pi_0 = g(\Pi_1) $$ $$ T = \sqrt{\frac{l}{g}} g(\theta_0) $$ but that is as far as we can go. In principle $g$ can be an arbitrary function and we cannot know how many measurements it would take to specify. In fact, in this particular example we can solve for $g$ numerically and find:

Period of a pendulum

Notice that this is a fully general equation, one that we could not determine in principle with any finite set of experiments. (Notice also that in the limit of low angle, the function is approximately flat, and nearly $2\pi$, which is the answer for a linear pendulum).

But. There are two things can help keep dimensional analysis sane and useful. The first is that when we can reduce a problem to a single dimensionless parameter, it is usually the case that the value of that dimensionless parameter (i.e. the constant of proportionality) is typically of order 1 (probably because as a pure number it doesn't have any reason not to be). In our area example above for instance, the missing constant of proportionality was $\pi \sim 3.14$ which is of order 1.

And the second thing, which is particularly useful, is that the physical world tends to have sane and nearly constant solutions in the extremes. That is to say, one of our physical solutions of the form $$ f(\Pi_0, \Pi_1, \Pi_2, \dots ) = 0 $$ will tend to constant (usually of order one) in the limit that one of the $\Pi$s is either very small ($\ll1$) or very big ($\gg 1$).

This is precisely the behavior we see for the period of our pendulum. In the limit of small angles, the function is well behaved and approaches a constant of order 1 roughly (in this case $2 \pi \sim 6.28$).

This has broader implications that aren't usually shown to students when they are first introduced to dimensional analysis.

For instance, let's rewind and imagine we had done a poor job when constructing our list of possible physical parameters for the pendulum, for instance let's say we were just in our first quantum class and so thought $\hbar$ would be important. This would have introduced another dimensional parameter, so we would have $$ \Pi_2 = \frac{ \hbar^2 }{ m^2 g l^3} $$ and we would have been left with as our most general solution $$ T = \sqrt{ \frac{l}{g} } h( \theta_0, \Pi_2 ) $$ and it would seem we are worse off than we were to begin with. How cruelly we are awarded for trying to get a more accurate answer. But, let's evaluate the actual value of this $\Pi_2$ for a reasonable pendulum, let's say one that is 1 kg, 1 meter long in earths gravity, in this case $\Pi_2 \sim 10^{-69}$ and we expect its functional dependence is as good as a constant and we can take the approximation $$ h(\theta_0 , \Pi_2 ) \sim g(\theta_0) $$ that is some general function $h$ where its second parameter is vanishing, looks like some other function only of its first variable.

In fact, if we wanted to get all philosophical, a real pendulum ought to be best described as a quantum object to begin with, so surely there ought to be some actual dependence on $\Pi_2$, though forgive me if I don't find it analytically. But, since we are interested in a classical pendulum, where classical in this case means the precise statement $$ m^2 g l^3 \gg \hbar^2 $$ we are completely justified in ignoring the quantum contribution to our pendulum, much like in introductory courses, if you are only interested in small angle excitation of a pendulum, you are apt to ignore the full functional dependence on $\theta_0$ and instead say it contributes only a multiplicative constant.

Appendix: Period of pendulum

We have for a real pendulum $$ \frac 12 l^2 \dot \theta^2 - g l \cos \theta = gl \cos \theta_0 $$ We manipulate it into the form $$ dt = \sqrt{\frac{l}{g}} \sqrt 2 \frac{ d\theta }{ \sqrt{ \cos \theta - \cos \theta_0 } } $$ which we can integrate for a quarter period to get the full period $$ T \sqrt{ \frac{g}{l} } = 2 \sqrt 2 \int_{\theta_0}^0 \frac{ d\theta }{ \sqrt{ \cos \theta - \cos \theta_0 } } $$ which is the equation I graphed above in 1.

share|cite|improve this answer

Your Answer


