Dimensional Analysis 📂Physics

Dimensional Analysis

Definition¹

Dimensional analysis is a technique based on the principle that each term in an equation containing different physical quantities must have the same dimensions; it is used to check the correctness of equations or to derive new relationships.

Explanation

Verification of equations

Let the Earth’s radius be $R_{e}$, the radius of the orbit of an artificial satellite around Earth be $r_{c}$, and Earth’s gravitational acceleration be $g$. Then the formula for the satellite’s speed $v_{c}$ is as follows.

$$ v_{c} = \left( \dfrac{g R_{e}^{2}}{r_{c}} \right)^{1/2} $$

The dimension of velocity is $\mathsf{L}^{1}\mathsf{T}^{-1}$, and simplifying the dimensions on the right-hand side gives the following.

$$ \mathsf{L}^{1}\mathsf{T}^{-1} = \left( \dfrac{\mathsf{L}^{1}\mathsf{T}^{-2} \cdot \mathsf{L}^{2}}{\mathsf{L}^{1}} \right)^{1/2} = \left( \mathsf{L}^{2}\mathsf{T}^{-2} \right)^{1/2} = \mathsf{L}^{1}\mathsf{T}^{-1} $$

Therefore, since the dimensions of the left- and right-hand sides match, we can see that the above equation is not wrong. Agreement of dimensions means there is no algebraic inconsistency, but it does not guarantee that the physical relationship is actually true. In other words, if the dimensions do not match, the equation is certainly wrong, but if the dimensions match, it does not necessarily mean the equation is true.

Deriving physical relations

Dimensional analysis can also be used to derive relationships between physical quantities. For example, consider a simple pendulum consisting of a mass attached to a string of length $\ell$. Neglect the mass of the string. If the bob of mass $m$ is raised slightly from the equilibrium position and allowed to swing, which physical quantities determine the period of small oscillations $T$? Consider the factors that might affect the period. First, there will be the string length $\ell$ and the bob mass $m$. Because the pendulum moves under gravity, we can also consider the gravitational acceleration $g$. Then the period $T$ can be expressed in the form below.

$$ T \propto \ell^a m^b g^c $$

Here $k$ is a dimensionless constant. Now compare the dimensions of each term.

Dimension of $T$: $\mathsf{T}^1$
Dimension of $\ell$: $\mathsf{L}^1$
Dimension of $m$: $\mathsf{M}^1$
Dimension of $g$: $\mathsf{L}^1\mathsf{T}^{-2}$

Thus the dimensions of the above expression are as follows.

$$ \mathsf{T}^1 = (\mathsf{L}^1)^a (\mathsf{M}^1)^b (\mathsf{L}^1\mathsf{T}^{-2})^c = \mathsf{L}^{a+c} \mathsf{M}^b \mathsf{T}^{-2c} $$

For this expression to hold, the exponents of each base dimension on both sides must be equal, so we can set up the following equations.

$\mathsf{L}$: $a + c = 0$
$\mathsf{M}$: $b = 0$
$\mathsf{T}$: $-2c = 1$

Solving these equations yields $a = \frac{1}{2}$, $b = 0$, and $c = -\frac{1}{2}$. Therefore the period $T$ can be written as follows.

$$ T \propto \ell^{1/2} m^0 g^{-1/2} = \left( \dfrac{\ell}{g} \right)^{1/2} $$

Thus, by dimensional analysis we find that the period $T$ depends only on the string length $\ell$ and the gravitational acceleration $g$, and is independent of the mass. In particular, it is proportional to the square root of the length and inversely proportional to the square root of the gravitational acceleration. The bob’s mass $m$ does not affect the period. Indeed, calculating explicitly shows that the period of a simple pendulum depends only on its length and the gravitational acceleration.