Why Rescale a Model?
A model can contain quantities measured in different units and processes operating at very different magnitudes. Its equation may be correct, yet the relative importance of its terms can be hard to see directly. Rescaling time and the state variable gives a dimensionless equation in which those comparisons are explicit.
The purpose is not to change the model’s predictions. It is to express them in units suited to the question. A useful scaling can show whether replenishment and extraction are comparable, whether their difference is small because of cancellation, or whether a changing input varies slowly or rapidly over the time interval of interest.
State Scales, Time Scales, and Rate Ratios
Suppose a state \(x(t)\) has units \(U\), and its rate of change is described by a sum of processes:
Each \(r_i\) has units \(U\) per unit time. Choose a positive characteristic state scale \(S_*\), with units \(U\), and a positive characteristic time scale \(T_*\), with units of time. Define the dimensionless state \(X\) and dimensionless time \(\tau\) by
If process \(i\) has a characteristic rate magnitude \(R_i\), its dimensionless rate coefficient is
The numerator \(R_iT_*\) is the amount of state that process could change over the chosen time scale. Dividing by \(S_*\) compares that amount with the characteristic state size. Thus, a coefficient much less than \(1\) indicates a small change relative to \(S_*\) over \(T_*\); a coefficient near \(1\) indicates a change comparable to \(S_*\). A coefficient greater than \(1\) indicates a larger change on that scale.
The scales are choices guided by the model and the question, not new physical constants. They should be positive and have the stated units. A useful choice often makes a coefficient of interest equal to \(1\), while leaving other coefficients as transparent ratios.
Why the Rescaling Works
The equation changes form because both the state and the independent variable have been expressed in new units. The chain rule gives the exact conversion; no approximation is involved.
Write \(x=S_*X\) and \(t=T_*\tau\), where \(S_*\) and \(T_*\) are fixed positive constants.
Since \(d\tau/dt=1/T_*\), the chain rule gives \(dx/dt=S_*\,dX/dt=(S_*/T_*)\,dX/d\tau\).
Substitute the derivative into the original equation and divide both sides by \(S_*/T_*\), which has units \(U\) per unit time.
For example, if each process has the constant rate \(r_i\), the resulting equation is
Each coefficient is dimensionless. For a positive process, its coefficient is positive; a removal process written with a minus sign retains that negative sign. The equation therefore preserves the direction of each contribution as well as comparing its size.
Worked Example: Comparing Replenishment and Extraction
Worked Example: Comparing Replenishment and Extraction
A stock \(S(t)\), measured in units, is replenished at \(18\) units per day and extracted at \(12\) units per day. Use a state scale of \(120\) units and a time scale of \(10\) days. Find the dimensionless equation and interpret its coefficients.
The physical balance is
Define \(X=S/120\) and \(\tau=t/10\). The replenishment coefficient is \(18(10)/120=180/120=1.5\). The extraction coefficient is \(12(10)/120=120/120=1\). Therefore,
The replenishment and extraction contributions are both substantial on this scale, while their net dimensionless rate is \(0.5\). To check the conversion, \(dS/dt=(120/10)dX/d\tau=12(0.5)=6\) units per day, exactly the original net rate. For instance, if \(S(0)=60\) units, then \(X(0)=60/120=0.5\); the initial state is half the chosen state scale.
Choosing a Scale When Rates Nearly Cancel
When inflow and outflow are close in size, a time scale based on their net difference can be much longer than a time scale based on their separate magnitudes. Both comparisons may be useful, but they answer different questions. The gross scale indicates how quickly the processes turn over state; the net scale indicates how quickly the stock changes because of the imbalance between them.
Worked Example: Near-Cancelling Rates
Suppose a stock has a constant input of \(51\) units per day and a constant removal of \(49\) units per day. Take \(S_*=100\) units and \(T_*=1\) day. Calculate the separate dimensionless rates, the net rate, and the associated time scales.
The physical balance is \(dS/dt=51-49=2\) units per day. The dimensionless equation is
The individual coefficients, \(0.51\) and \(0.49\), show that both processes act on a scale comparable to the state over one day. Their difference, \(0.02\), shows that the net change over that day is only \(0.02S_*=2\) units.
Using the sum of the two rate magnitudes, the gross turnover time for a \(100\)-unit state is \(100/(51+49)=100/100=1\) day. Using the net rate instead, the time to change by \(100\) units at a constant net rate of \(2\) units per day is \(100/2=50\) days. These are different time scales because one measures gross activity and the other measures net accumulation.
As a check, the dimensionless net rate converts back to physical units as \((S_*/T_*)(0.02)=(100/1)(0.02)=2\) units per day. This agrees with \(51-49=2\). The long net time scale is not evidence that the individual processes are slow: it results from their near-cancellation.
Time Variation and Regime Comparisons
Some rates vary with time. Their relative size is not the only issue: the speed of their variation matters too. If a rate oscillates with angular frequency \(\omega\), then \(1/\omega\) is a characteristic time for its phase to change by one radian, and \(2\pi/\omega\) is its period. Comparing \(\omega T_*\) with \(1\) indicates how much the forcing phase changes during one chosen time scale. A small value means little phase change; a value near or above \(1\) means appreciable change.
Worked Example: An Oscillating Input and Constant Removal
A stock has an input rate \(10+6\cos(0.2t)\) units per day and a constant removal rate of \(8\) units per day, with \(t\) measured in days. Use \(S_*=200\) units and \(T_*=5\) days. Rescale the model and compare the oscillating input with the mean net drift.
The physical equation is
Set \(X=S/200\) and \(\tau=t/5\), so \(t=5\tau\). Then \(0.2t=0.2(5\tau)=\tau\). Multiplying each rate contribution by \(T_*/S_*=5/200=1/40\) gives
The dimensionless mean drift is \(0.05\), whereas the oscillating contribution has amplitude \(0.15\), three times as large. Thus, the instantaneous rate can differ substantially from its mean even though the mean net drift is positive. The phase coefficient is \(1\), since \(\omega T_*=0.2(5)=1\); the forcing changes by one radian over the chosen five-day scale.
To check the scaling, multiply the dimensionless derivative by \(S_*/T_*=200/5=40\) units per day. This gives \(40(0.05+0.15\cos\tau)=2+6\cos\tau\) units per day. The original equation simplifies to \(10+6\cos(0.2t)-8=2+6\cos(0.2t)\), and \(0.2t=\tau\), so the two expressions agree exactly.
A Common Mistake: Comparing Raw Numbers
A common error is to treat the numerical values of dimensional rates as if they were already dimensionless. The comparison can become especially misleading when the separate rates nearly cancel.
Worked Example: Correcting a Raw-Rate Comparison
For a stock with input \(12\) units per day and removal \(10\) units per day, choose \(S_*=100\) units and \(T_*=5\) days. A mistaken calculation might subtract the raw rate values and report a dimensionless net rate of \(12-10=2\). That is incorrect: the subtraction still has units of units per day, and it has not been divided by the state-rate scale.
The correct input coefficient is \(12(5)/100=60/100=0.6\). The correct removal coefficient is \(10(5)/100=50/100=0.5\). Hence \(dX/d\tau=0.6-0.5=0.1\), not \(2\). The selected state-rate scale is \(S_*/T_*=100/5=20\) units per day. Converting back gives \(20(0.1)=2\) units per day, which matches the original net rate \(12-10=2\).
The raw net rate \(2\) is a valid dimensional quantity; it is simply not the dimensionless derivative. A dimensionless comparison must account for both the chosen time scale and the chosen state scale.
Using Scales to Identify Dominant Processes
Once an equation is rescaled, compare coefficients only in relation to the scales used and the regime being studied. A coefficient much smaller than the others may make a small contribution over the selected time interval, provided the state and time variables remain in the range for which the model applies. Comparable coefficients indicate that several processes must be retained. A small net coefficient alongside large opposing terms signals cancellation, not inactivity.
Scaling supports model interpretation, but it does not establish that a model is accurate. The assumed rates, signs, units, and range of applicability still need to be checked. Nor does a coefficient that is small on one chosen time scale have to be small on another. Changing \(T_*\) changes all the rate coefficients proportionally, so state clearly which scale answers the question of interest.
Check Your Understanding
Use the definitions and comparisons in this tutorial to answer the following questions.
- If \(x\) has units of kilograms, \(S_*\) is measured in kilograms, and \(T_*\) is measured in days, what units must a rate \(R_i\) have for \(R_iT_*/S_*\) to be dimensionless?
- For constant input and removal rates, why can the net dimensionless derivative be small even when both individual dimensionless coefficients are near \(1\)?
- What does the comparison between \(\omega T_*\) and \(1\) say about a periodically varying input during one characteristic time?
- For \(dx/dt=7-5\) units per day, \(S_*=20\) units, and \(T_*=2\) days, calculate both dimensionless rate coefficients and check the net derivative by converting back to physical units.
- Why should a statement that one process is negligible specify the chosen state and time scales?
