Age 14 to 18

Article by Parin Shah and Matt Wells

Published 2011

The Power of Dimensional Analysis

What is it?
Though dimensional analysis is something you have probably not yet covered, it is actually a very simple and easy way of checking that an expression is consistent. It checks that an expression has the "correct" dimensions. Of course if the dimensions of an expression are not correct, then clearly the expression cannot be correct either. However, an expression with the right dimensions is not necessarily correct but we can say it could be correct.

Take for example these equations for force:

$$\text{F}= \text{ma}$$
$$\text{F}= \text{pA}$$
$$\text{F}= \frac{P}{v}$$
$$\text{F}= \frac{q^2}{4\pi\epsilon_0r^2}$$

Force is given in units of newtons which is a unit in terms of [mass][length][time]$^{-2}$. 1 newton can be expressed in SI form as 1 kgms$^{-2}$. We can check that the dimensions in the above equations are consistent.

$\text{F}= \text{ma}$
$\displaystyle [mass][\frac{length}{time^2}] \equiv [mass][length][time]^{-2}$
SI units of kgms$^{-2}$

$\ $
$\text{F}= \text{pA}$
$\displaystyle \frac{[mass][length][time]^{-2}}{[length]^2}\times [length]^2 \equiv [mass][length][time]^{-2}$

$\ $
$\text{F}= \frac{P}{v}$
$\displaystyle \frac{\frac{[mass][length][time]^{-2}\times[length]}{[time]}}{\frac{[length]}{[time]}}\equiv {[mass][length][time]^{-2}}$

$\ $
$\text{F}= \frac{q^2}{4\pi\epsilon_0r^2}$
$\displaystyle \frac{({[time]}{[current]})^2}{[mass]^{-1}[length]^{-3}[time]^{4}[current]^{2}\times[length]^{2}} \equiv [mass][length][time]^{-2}$

$\ $
$\ $
Check that the units of the following expressions are consistent:

F= $\frac{W}{d}$ where W= work done, d = distance

F= Bqv where B= magnetic flux density, q = charge, v = velocity

F= B I L where I= current and L= length

F= $\frac{\text{-GMm}}{\text{r}^{2}}$ where M and m are different masses and G is the gravitational constant.

$\ $

In dimensional analysis, any unit can be expressed in terms of the dimensions given below:

length [L]

mass [M]

time [T]

temperature [K]

amount of substance [N]

electric current [J]

luminous intensity [I]

To work out the dimensions of a quantity, all you have to do is simply substitute in the basic dimensions listed. All of these dimensions have SI units associated with them. Any factors are just numbers and so have no input in terms of the dimension. For example $\text{E}= \frac{1}{2}\text{mv}^2$ has dimensions of [M][L]$^2$[T]$^{-2}$.

The units of a quantity can often help in dimensional analysis. It is known that the molar heat capacity has units JK$^{-1}$mol$^{-1}$. By considering the units dimensions of each of the units we can find the dimensions of this quantity.

J $\rightarrow$ [M][L]$^2$[T]$^{-2}$

K $\rightarrow$ [K]$^{-1}$

moles $\rightarrow$ [N]$^{-1}$

$\therefore$ molar heat capacity = [M][L]$^2$[T]$^{-2}$[K]$^{-1}$[N]$^{-1}$

In equations where terms are added together, the dimensions of all terms must clearly be the same.

For example, consider the equation: $v^2=u^2 + 2as$, where $s$ is the distance.

$v^2$ and $u^2$ have dimensions of [L]$^2$[T]$^{-2}$

$2as$ has dimensions of [L][T]$^{-2}$ $\times$ [L] = [L]${^2}$[T]$^{-2}$

Check that the following equations are dimensionally consistent:

$v = u + at$

$\displaystyle s = ut + \frac{1}{2} at^2$

E = E$^{o}$ - $\frac{RT}{nF}$ln $\frac{([C]/c^{o})^{c}([D]/c^{o})^{d}}{([A]/c^{o})^{a}([B]/c^{o})^{b}}$ where E and E$^{o}$ are both voltages, R is the ideal gas constant, F is the Faraday constant, n is the number of moles and x$^{o}$ represent standard concentrations.

The last of the above equations is known as the Nernst equation and is used to work out the cell potential in an electrochemical cell. Working through the example you should have come to the conclusion that the logarithm of a dimensionless quantity is also dimensionless. This is true for exponentials of dimensionless quantities too.

In the course of calculations you may be able to determine the dimensions of many physical constants. Some of the dimensions of key constants are listed below:

speed of light (c): [L][T]$^{-1}$

elementary charge ($e$): [T][J]

gas constant ($R$): [M][L]$^2$[T]$^{-2}$[K]$^{-1}$[N]$^{-1}$

permittivity of vacuum ($\epsilon_0$): [T]$^{4}$[J]$^{2}$[L]$^{-3}$[M]$^{-1}$

Avogadro's constant ($N_A$): [N]$^{-1}$

Boltzmann constant (k$_{\text{B}}$): [M][L]$^2$[T]$^{-2}$[K]$^{-1}$

Faraday constant (F): [T][J][N]$^{-1}$

Planck's constant ($h$): [M][L]$^2$[T]$^{-1}$

Consider the expression k = A exp$(\frac{-E_a}{RT})$ which is the Arrhenius law giving the temperature dependence of the rate constant k.

Using the information above:

Express R in terms of two other constants in the table

What dimension does the bracketed quantity in the equation have?

What is the relationship in terms of dimensions between k and A?

The expression for the energy of the 1s AO in a hydrogen atom is given by the formula:

$$ E = -\frac{m_ee^4}{32\pi^{2}\epsilon_0^2\hbar^2}$$

where $\hbar^2$ = $\frac{h}{2\pi}$

Can you verify that the dimensions of the equation are consistent with the correct dimensions of energy?

Further dimensional analysis

The principles introduced thusfar can be used in a way to compose equations. Look at the following example:

Planck time can be expressed using the quantities $\hbar$, the gravitational constant, G, and the speed of light, c.

$\hbar$ can be expressed in SI units as $\frac{m^2kg}{s}$.

G can be expressed in SI units as $\frac{m^3}{kgs^2}$

c can be expressed in SI units as $\frac{m}{s}$

Planck time has units of s.

We can now express Planck time in terms of factors of these other 3 quantities.

s$^1$ = [$\frac{m^2kg}{s}$]$^{\alpha}$[$\frac{m^3}{kgs^2}$]$^{\beta}$[$\frac{m}{s}$]$^{\gamma}$

s$^1$ = $\text{m}^{(2\alpha+3\beta+\gamma)}\text{kg}^{(\alpha-\beta)}\text{s}^{(-\alpha-2\beta-\gamma)}$

so,

2$\alpha$+3$\beta$+$\gamma$ =$0$ (1)

$\alpha$-$\beta$ = $0$ (2)

-$\alpha$-2$\beta$-$\gamma$ = $0$ (3)

$\alpha$ = $\beta$ (4)

5$\alpha$+$\gamma$ = $0$ (4) $\rightarrow$ (1)

-3$\alpha$-$\gamma$ = 1 (4) $\rightarrow$ (3)

2$\alpha$ = 1

$\alpha$ = $\frac{1}{2}$

$\beta$ = $\frac{1}{2}$

$\therefore$ $\gamma$ = -5($\frac{1}{2}$) = $\frac{5}{2}$

so,

Planck time = $\hbar^{\frac{1}{2}}$G$^{\frac{1}{2}}$c$^{\frac{5}{2}}$

Number and algebra

Geometry and measure

Probability and statistics

Working mathematically

Advanced mathematics

For younger learners

The Power of Dimensional Analysis

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