MA2 - Predicting formulae | AQA A-Level Further Mathematics

Dimensional analysis can narrow a proposed formula to a particular combination of powers. You will equate exponents, interpret the resulting model and recognise what dimensions cannot determine about constants, functions and missing variables.

Finding powers using dimensions

If a resistance force depends only on fluid density, sphere radius and speed, what powers could appear in its formula? A full theory of fluid motion is difficult; dimensions can still constrain a proposed product of powers.

Suppose $F=C\rho^a r^b v^c$ , where $C$ is dimensionless, $\rho$ is density, $r$ is radius and $v$ is speed. Write the dimensions first, then compare the exponents of each independent base dimension. Multiplying powers adds their exponents; $[\rho^a]=(\mathrm{ML}^{-3})^a=\mathrm{M}^a\mathrm{L}^{-3a}$ .

Worked example 1: predicting a force law

Question. Find $a,b,c$ if $F=C\rho^a r^b v^c$ is dimensionally consistent and $C$ is dimensionless.

Plan. Expand the dimensions and match mass, length and time powers separately.

\begin{aligned} \mathrm{MLT}^{-2} &=(\mathrm{ML}^{-3})^a\mathrm{L}^b(\mathrm{LT}^{-1})^c\\ &=\mathrm{M}^a\mathrm{L}^{-3a+b+c}\mathrm{T}^{-c} \end{aligned}

Thus $a=1$ , $-c=-2$ , and $-3a+b+c=1$ . Hence $c=2$ and $-3+b+2=1$ , giving $b=2$ .

Answer. $a=1,b=2,c=2$ : the candidate is $F=C\rho r^2v^2$ .

Check. $(\mathrm{ML}^{-3})(\mathrm{L}^2)(\mathrm{L}^2\mathrm{T}^{-2})=\mathrm{MLT}^{-2}$ , independently recovering force dimensions.

Change only the quantity being modelled, from resistance force to the power needed to overcome it at steady speed. The extra factor of speed should change the result in a predictable way.

Worked example 2: predicting the associated power

Question. Find $a,b,c$ if power $P=C\rho^a r^b v^c$ , with the same variables and dimensionless $C$ .

Plan. Use $[P]=\mathrm{ML}^2\mathrm{T}^{-3}$ and solve the three exponent equations.

\begin{aligned} \mathrm{ML}^2\mathrm{T}^{-3} &=\mathrm{M}^a\mathrm{L}^{-3a+b+c}\mathrm{T}^{-c}\\ a&=1\\ -c&=-3\\ -3a+b+c&=2 \end{aligned}

So $c=3$ and $-3+b+3=2$ , giving $b=2$ .

Answer. $P=C\rho r^2v^3$ .

Check. Multiplying the force candidate from example 1 by $v$ , as in $P=Fv$ , gives the same powers. Doubling speed multiplies candidate force by $4$ but candidate power by $8$ .

The algebra predicts a form only under the stated list of variables and product-of-powers assumption. It has not shown that the model describes every speed or every fluid.

Watch out

Wrong: $[\rho^a]=\mathrm{ML}^{-3a}$ . The power acts on the whole density dimension: $[\rho^a]=\mathrm{M}^a\mathrm{L}^{-3a}$ . Also retain any stated dimensions of a constant.

Try another target quantity using the same building blocks.

Limits on constants and functions

A dimensional prediction cannot choose between $F=\rho r^2v^2$ and $F=\pi\rho r^2v^2$ : both have the same dimensions. The dimensionless coefficient requires physical theory or measurements. It can vary between different shapes or conditions; it is held constant only within the chosen model.

Nor does dimensional analysis automatically establish a product of powers. Suppose a pendulum period depends on length $l$ , gravitational acceleration $g$ and release angle $\theta$ . Since an angle in radians is dimensionless, every expression of the form

\tau=\sqrt{\frac{l}{g}}\,f(\theta)

has the dimensions of time for any dimensionless function $f$ . Dimensions cannot tell you whether $f$ is constant, or how period changes with angle. For the small-angle simple-pendulum model, theory gives the coefficient $2\pi$ ; dimensions alone cannot supply it or establish the small-angle condition.

Inputs to $\sin$ , $\exp$ and $\ln$ in physical formulae must be dimensionless. For example, $\exp(-t/t_0)$ is meaningful when $t_0$ is a time. Writing $\exp(-t)$ as a physical law hides a timescale unless $t$ has explicitly been made dimensionless. A numerical formula using “ $t$ measured in seconds” is a different convention and must state that convention.

If two lengths $l$ and $r$ are allowed, their ratio $r/l$ is dimensionless. A formula may contain an unknown function of that ratio, so dimensions may leave freedom rather than determine unique powers. That is useful information about what further evidence is needed.

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Mechanics

Statistics

Discrete Mathematics

MA2 - Predicting formulae

Finding powers using dimensions

Limits on constants and functions

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