How to Make a Physics Equation Dimensionless

A dimensional equation contains quantities such as metres, seconds, kilograms, or pascals. A dimensionless equation is written in terms of variables and parameters that have no units. This process is called nondimensionalization or dimensionless scaling.

The essential idea is to divide each variable by a suitable characteristic value. For example, if $x$ is a position and $L$ is a characteristic length, define

$$X = \frac{x}{L}.$$

Both $x$ and $L$ have dimensions of length, so $X$ is dimensionless. You do the same for time, velocity, pressure, temperature, and other quantities.

Dimensional homogeneity comes first

Before nondimensionalizing, check that the original equation is dimensionally consistent. Terms that are added or subtracted must have the same dimensions. For example,

$$F = m\frac{d^2x}{dt^2} + cx + kx$$

would not generally be valid because $m\,d^2x/dt^2$ is a force, while $cx$ and $kx$ have different dimensions unless $c$ and $k$ are defined appropriately.

For a damped, driven spring–mass oscillator, a dimensionally consistent equation is

$$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=F_0\cos(\omega t),$$

where:

• $m$ is mass,
• $c$ is the damping coefficient,
• $k$ is the spring constant,
• $F_0$ is the driving-force amplitude, and
• $\omega$ is the driving angular frequency.

Dimensional consistency does not by itself make the equation dimensionless. It only ensures that the equation is physically meaningful.

Choose characteristic scales

Introduce a characteristic displacement $x_0$ and a characteristic time $t_0$. Define

$$X=\frac{x}{x_0}, \qquad \tau=\frac{t}{t_0}.$$

Here, $X$ and $\tau$ are dimensionless. Equivalently,

$$x=x_0X, \qquad t=t_0\tau.$$

Using the chain rule,

$$\frac{dx}{dt}=\frac{x_0}{t_0}\frac{dX}{d\tau},$$

and

$$\frac{d^2x}{dt^2}=\frac{x_0}{t_0^2}\frac{d^2X}{d\tau^2}.$$

The scales $x_0$ and $t_0$ are not unique. A good choice makes important coefficients equal to one and leaves only the physically meaningful ratios.

Worked example: a driven oscillator

Start with

$$m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=F_0\cos(\omega t).$$

Choose the natural oscillator scales

$$t_0=\sqrt{\frac{m}{k}}, \qquad x_0=\frac{F_0}{k}.$$

The time scale $t_0$ is the inverse natural frequency, and $x_0$ is the static displacement that a force $F_0$ would produce.

Substitute $x=x_0X$ and $t=t_0\tau$:

$$m\frac{x_0}{t_0^2}\frac{d^2X}{d\tau^2} +c\frac{x_0}{t_0}\frac{dX}{d\tau} +kx_0X =F_0\cos(\omega t_0\tau).$$

Since $t_0^2=m/k$ and $kx_0=F_0$, divide the entire equation by $kx_0=F_0$. The result is

$$\frac{d^2X}{d\tau^2} +\frac{c}{\sqrt{mk}}\frac{dX}{d\tau} +X =\cos\left(\omega\sqrt{\frac{m}{k}}\,\tau\right).$$

Define the damping ratio and dimensionless driving frequency by

$$\zeta=\frac{c}{2\sqrt{mk}}, \qquad \Omega=\omega\sqrt{\frac{m}{k}}.$$

Then the dimensionless equation becomes

$$\frac{d^2X}{d\tau^2}+2\zeta\frac{dX}{d\tau}+X=\cos(\Omega\tau).$$

Every quantity in this final equation is dimensionless: $X$, $\tau$, $\zeta$, and $\Omega$. The original five parameters have been reduced to two controlling ratios, $\zeta$ and $\Omega$.

Numerical substitution

Suppose

m=2\ \mathrm{kg},\qquad c=2\ \mathrm{N\,s/m},\qquad k=8\ \mathrm{N/m},$$ with $F_0=4\ \mathrm{N}$ and $\omega=3\ \mathrm{s}^{-1}$. Then

x_0=\frac{F_0}{k}=0.5\ \mathrm{m}, \qquad t_0=\sqrt{\frac{m}{k}}=0.5\ \mathrm{s}.

$$The dimensionless parameters are$$

\zeta=\frac{2}{2\sqrt{2\cdot8}}=0.25, \qquad \Omega=3\sqrt{\frac{2}{8}}=1.5.

$$Therefore the equation is$$

\frac{d^2X}{d\tau^2}+0.5\frac{dX}{d\tau}+X=\cos(1.5\tau).

The units have disappeared, but the physical content has not. The numbers $0.25$ and $1.5$ describe damping and forcing relative to the oscillator's natural scales. ## General recipe for any equation For a more complicated model, use this sequence: 1. List every dimensional variable and parameter. 2. Choose characteristic scales such as length $L$, time $T$, velocity $U$, pressure $P$, or temperature difference $\Delta T$. 3. Define dimensionless variables, for example $X=x/L$, $\tau=t/T$, and $\Theta=(T-T_\mathrm{ref})/\Delta T$. 4. Rewrite derivatives using the new variables. 5. Substitute into the original equation. 6. Divide by one common dimensional coefficient. 7. Collect the remaining dimensionless parameter combinations. For example, in fluid mechanics, choosing a length scale $L$ and speed scale $U$ often produces the Reynolds number

\operatorname{Re}=\frac{\rho UL}{\mu},

which compares inertial effects with viscous effects. Such groups are often more informative than the original dimensional constants. ## Common mistakes ### Dividing variables by numbers instead of scales Writing $X=x/2$ does not make $X$ dimensionless unless the $2$ carries the same units as $x$. Use a dimensional reference quantity, such as $x_0=2\ \mathrm{m}$. ### Forgetting derivative factors If $t=t_0\tau$, then $d/dt=(1/t_0)d/d\tau$. A second derivative introduces $1/t_0^2$, not merely $1/t_0$. ### Mixing dimensional and dimensionless variables After defining $X=x/x_0$, use $X$ consistently. Writing $\cos(\omega\tau)$ is usually wrong if $\tau$ is dimensionless, because $\omega$ still has units. The correct argument is typically $\omega t_0\tau$. ### Assuming there is only one correct scaling Different choices of reference scales produce different-looking dimensionless equations. They are equivalent when the variables are related correctly. Choose scales that reflect the regime or simplify the terms you want to study. ### Confusing “dimensionless” with “equal to one” A dimensionless quantity can be $0.2$, $15$, or $\pi$. Dimensionless means it has no units, not that its numerical value is one.