core29 min studyLesson 103 in Physics

Test equations with dimensions and rescaling

Use dimensional homogeneity, unit invariance, and limiting cases to reject physical models before fitting experimental numbers.

Before you begin

On this page

The question for today

What you’ll understand

Check the dimensions of sums, functions, and derivatives.
Distinguish changing physical scale from changing reporting units.
Combine dimensional checks with experimental scaling and limiting cases.

An equation can fit several data points and still be structurally impossible. Dimensional analysis tests whether its terms can represent the same kind of quantity. A scaling test asks a different question: how should a prediction respond when a physical parameter changes? Together these checks expose errors before elaborate fitting, while also revealing what dimensions alone cannot determine, including numerical factors, signs, and missing dimensionless dependencies.

Every sum compares compatible quantities

The dimensions of a quantity describe its dependence on chosen base dimensions such as length, mass, and time. Metres and centimetres are units of the same length dimension. Every additive term in a quantity equation must have the same dimensions. Products and ratios combine dimensional powers, and differentiation divides by the dimension of the independent variable. Thus the derivative of position with respect to time has dimension length divided by time, regardless of the units used to report it.

Arguments of logarithms, exponentials, and trigonometric functions require dimensionless combinations in physical formulas. An exponential decay written as exp of negative time needs a time scale in the denominator. A logarithm of a pressure should instead refer to a ratio to a specified reference pressure. This matters in fitting: a numerical parameter may hide units, and changing from seconds to milliseconds must not change the physical decay or the meaning of a reference quantity.

[x]=L,\quad[v]=LT^{-1},\quad[a]=LT^{-2},\qquad x=x_0+v_0t+\tfrac12at^2

All three terms in the position sum have dimension L. The formula additionally assumes constant acceleration; dimensional consistency alone does not supply that dynamical assumption.↔ Scroll wide equations horizontally.

A physical scaling is not a unit conversion

Replacing a two-centimetre sample by a four-centimetre sample changes the physical system. Reporting the original sample in millimetres does not. A valid quantity equation survives a consistent unit conversion, whereas a scaling law describes a response to an actual change of inputs. Keep those tests separate. Unit invariance can expose a spreadsheet coefficient that was calibrated for hidden units; physical scaling can expose a missing mechanism even when every unit in the spreadsheet is consistent.

Consider the small-amplitude period of a simple pendulum. If length and gravitational acceleration are the only dimensional inputs, homogeneity fixes the square-root dependence on their ratio. It does not fix the dimensionless coefficient. If amplitude is not negligible, the initial angle is another dimensionless input and the coefficient becomes a function of that angle. A successful scaling observation supports a proposed regime; it does not prove that every omitted parameter is irrelevant under all conditions.

T=C\sqrt{\frac{\ell}{g}},\qquad \frac{T_2}{T_1}=\sqrt{\frac{\ell_2}{\ell_1}}\quad\text{at fixed }g\text{ and }C

Dimensions fix this form if only ℓ and g are relevant dimensional inputs. The ideal small-angle pendulum has C = 2π. The period ratio requires the same applicable dimensionless conditions, including amplitude regime.↔ Scroll wide equations horizontally.

Try to falsify the expression cheaply

After a dimensional check, examine signs, zero limits, large-parameter limits, and independent conservation laws. Two proposed energies proportional to mass times speed squared both pass the unit test, even if one has the wrong factor of two. A negative damping coefficient has the correct dimension but describes energy injection rather than passive friction. State the physical regime before taking a limit; sending a parameter far beyond that regime tests an extrapolation rather than the original model.

In an experiment, compare dimensionless ratios so that an unknown calibration factor may cancel, but verify that it really is shared. Measuring a period at four lengths can test a square-root scaling more clearly than fitting an arbitrary polynomial. Residual trends with amplitude or temperature suggest a missing dependence. Such discrepancies are useful evidence: revise the physical assumptions or the measurement model before adding enough free coefficients to make a structurally wrong relation follow the data.

Work it through

Infer a pendulum scaling without solving its motion

Assume a pendulum period depends only on length ℓ and acceleration g, through T = Cℓᵃgᵇ with dimensionless C. Determine the exponents and predict the period change when length is quadrupled at unchanged conditions.

Insert dimensions: $[T]=L^{a+b}T^{-2b}$ . Matching the time exponent gives $-2b=1$ , so $b=-1/2$ . This equation concerns dimensions, not measured numerical values.
Matching the length exponent requires $a+b=0$ , so $a=1/2$ . Consequently the only permitted power-law form with these inputs is $T=C\sqrt{\ell/g}$ . Nothing here determines C.
For the same g and dimensionless conditions, $T_2/T_1=\sqrt{4}=2$ . If the initial period were $1.5\,\mathrm s$ , the predicted new period would be $3.0\,\mathrm s$ , provided the same idealization remains adequate.

The result

The exponents are one half and minus one half. A fourfold physical increase of length doubles the period in this model; changing the length unit alone changes neither the period nor the apparatus.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Predict two different scaling responses

For y = ax², start at a = 0.5 and x = 0.5. Predict separately the effects of changing a to 1 at fixed x and changing x to 1 at fixed a = 0.5. Then check with the coefficient and position controls: y doubles in the first test and quadruples in the second. The tangent and area readouts belong to this quadratic only, not to a pendulum simulation or arbitrary fitted law.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

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 forChanging a parameter and changing an input can produce different powers even in a simple homogeneous formula.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Dimensions constrain allowable combinations but cannot fix signs, dimensionless coefficients, or functions of dimensionless inputs. Dynamics and experiments must decide those features and establish the range over which a formula applies.

Your turn

Try it yourself

A decay model is written as $A(t)=A_0\exp(-kt)$ . Determine the required dimension of k, then compare the numerical k used with seconds and milliseconds for the same physical decay.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The exponent must be dimensionless, and one second contains one thousand milliseconds.

Compare with the solution

The exponent requires $[k]=T^{-1}$ . If $k=2\,\mathrm{s^{-1}}$ , the same rate is $0.002\,\mathrm{ms^{-1}}$ . A physical duration of $0.5\,\mathrm s$ is $500\,\mathrm{ms}$ , and both representations produce $kt=1$ . Using the numerical value 2 with a time numerical value in milliseconds would instead describe a decay one thousand times faster. The correct dimensions constrain the parameter, but only an actual rate measurement or dynamical model fixes its magnitude.

Keep the idea

Take it with you

Sums need matching dimensions, and elementary function arguments need dimensionless combinations.
Unit changes preserve a system; physical scaling tests change it.
Correct dimensions are a necessary check rather than proof of a physical law.

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