Nondimensionalizing a physical simulation
Choose reference scales, preserve dimensionless physical ratios and translate numerical tolerances consistently.
On this page
The question for today
What you’ll understand
- Derive a scaled equation using the chain rule.
- Distinguish a change of units from a change of physical regime.
- Translate initial conditions, time steps and tolerances into scaled variables.
Before choosing a numerical method, decide what the numbers represent. A simulation of the same falling object can use meters or millimeters without changing its physics, but changing a dimensionless ratio of damping to inertia does change the motion. Nondimensionalization makes that distinction explicit. It also reveals characteristic time scales and helps assign meaningful tolerances when a state vector contains positions, velocities and other quantities with different units. Scaling simplifies a model; it does not repair an incorrect force law.
Transform derivatives as well as coordinates
Take downward displacement x, constant gravitational acceleration g and linear drag with positive coefficient β. For mass m, the drag relaxation time is τ = m/β. Choose a positive reference length L and a positive reference time T, and write the physical coordinate as L times a dimensionless coordinate X. The first derivative brings one factor of inverse T; the second brings two. Substituting these factors before canceling units avoids the common mistake of rescaling displacement while leaving acceleration unchanged.
Choosing the gravitational time T equal to the square root of L over g makes the dimensionless gravitational acceleration unity. One remaining coefficient measures the ratio of that time to the drag relaxation time. Initial position and velocity must be transformed too. A dimensionless initial velocity of one means a physical speed L/T, rather than one meter per second unless those scales happen to agree. The zero-drag limit is obtained by taking the coefficient to zero, not by dividing through a vanishing drag force.
Identify which comparisons preserve the model
Changing reference units changes numerical parameter values but preserves dimensionless ratios describing one physical experiment. Changing L as a chosen reference also changes Γ and the scaled observation interval, even for the same physical trajectory; interpreting such changes requires transforming everything consistently. Conversely, two genuinely different physical systems can share a dimensionless initial-value problem and therefore share its scaled solution. Recovering their dimensional trajectories requires each system's own reference scales.
There is no universally best reference time. A weakly damped oscillator often suggests the inverse natural frequency, whereas an overdamped relaxation may be better described by a slower decay time. A poor choice can conceal a rapid transient or leave extremely disparate coefficients. Scaling can expose stiffness but cannot eliminate the physical separation of time scales. For orbital work, resolving a short periapsis passage remains necessary even when the mean period has been scaled to a convenient number.
Scale numerical decisions and recorded outputs
An absolute position tolerance must be divided by L when solving for X; a relative tolerance remains dimensionless. Velocity tolerances use the velocity scale instead. Combining raw errors in meters and meters per second into an unweighted norm is dimensionally meaningless. Even after scaling, a solver's local tolerance is only an error-control target for its algorithm. Global error, instability, roundoff and inaccuracies in the physical model require separate examination.
A reproducible calculation records the reference scales beside the dimensionless parameters, initial conditions, boundary conditions and output conversion. Otherwise an attractive plot can have an ambiguous time axis. The function lab provides a small scaling check: it evaluates a quadratic and its slope exactly for selected dimensionless inputs. Its controls do not convert units or integrate the falling-body equation, so agreement in that instrument establishes only the arithmetic scaling relationship being tested.
Work it through
Choose gravitational scales with finite drag
A falling-body model has L = 4 m, g = 4 m/s² and drag relaxation time τ = 2 s. Find the gravitational reference time, speed scale and dimensionless damping coefficient. Convert a physical step of 0.10 s.
The gravitational time is second. It is a reference scale, not the predicted duration of a complete fall.
The speed scale is L/T = 4 m/s, and . The scaled acceleration equation is X″ = 1 − 0.5X′.
The time increment becomes Δs = 0.10/1 = 0.10. Changing the length unit to centimeters would leave this scaled equation and increment unchanged if all dimensional data and reference scales were converted consistently.
T = 1 s, Vref = 4 m/s, Γ = 0.5 and the dimensionless step is 0.10.
The living notebook
Explore the model
Interactive lab · function
Test a dimensionless scaling relation
Reset and set BOTH Coefficient a = 1 and Position x = 0.50; predict y = 0.25 and slope 1. Reset and set a = 1 and x = 1; predict y = 1 and slope 2. This quadratic arithmetic checks different scaling powers; it does not convert physical units or solve an equation of motion.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forDifferent derived quantities can scale with different powers of the reference coordinate.
Pause & reason
Check your understanding
Your turn
Try it yourself
For an undamped spring with mass 2 kg and stiffness 8 N/m choose T = √(m/k) and L = 0.10 m. Convert x0 = 0.30 m, v0 = 0.40 m/s and Δt = 0.025 s. State the scaled equation.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use X0 = x0/L and X′0 = v0T/L; the chosen time removes the coefficient k/m from the scaled acceleration equation.
Compare with the solution
The time scale is 0.50 s and the velocity scale is 0.20 m/s. Thus X0 = 3, X′0 = 2 and Δs = 0.05. Substitution into m ẍ + kx = 0 gives X″ + X = 0. The physical initial velocity was not divided by the length scale alone: the time factor is essential. These scaled initial data describe the same spring motion and can be converted back using x = 0.10X meters and physical time equal to 0.50 seconds times the dimensionless time variable s.
Keep the idea
Take it with you
- Derivative transformations determine the dimensionless coefficients.
- Initial conditions and tolerances need compatible reference scales.
- Scaling exposes time-scale separation without removing it.
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