15.3 Dimensional Analysis and Scaling Arguments

The Pendulum Question

How does the period of a pendulum depend on its length, mass, and the gravitational acceleration $g$?

You could solve the differential equation. You could linearize, find the angular frequency $\omega = \sqrt{g/L}$, and extract $T = 2\pi/\omega$. That works, and you have done it before (Chapter 13). But suppose you had never seen the derivation. Suppose you had no equation at all --- just the physical observation that a pendulum swings back and forth with some period $T$, and you suspected it might depend on the length $L$, the mass $m$, and the local gravitational field $g$.

Could you figure out the form of $T$ from units alone?

The period $T$ has dimensions of time: $[T] = \text{s}$.

The length $L$ has dimensions $[L] = \text{m}$.

The mass $m$ has dimensions $[m] = \text{kg}$.

The gravitational acceleration $g$ has dimensions $[g] = \text{m/s}^2$.

Now try to build a quantity with dimensions of time from $L$, $m$, and $g$. Suppose $T \propto L^a \, m^b \, g^c$. Then the dimensions are:

$$[T] = \text{m}^a \cdot \text{kg}^b \cdot \left(\frac{\text{m}}{\text{s}^2}\right)^c = \text{m}^{a+c} \cdot \text{kg}^b \cdot \text{s}^{-2c}$$

For this to have dimensions of seconds, you need:

• Length: $a + c = 0$
• Mass: $b = 0$
• Time: $-2c = 1$

From the time equation: $c = -1/2$. From the length equation: $a = 1/2$. From the mass equation: $b = 0$.

So $T \propto L^{1/2} \, g^{-1/2} = \sqrt{L/g}$.

Mass does not appear. The dimensions forbid it. There is no way to combine $L$, $m$, and $g$ into a quantity with dimensions of time that includes mass. You figured that out without any physics, without any differential equation, without even knowing what a pendulum looks like. Just from units.

The full answer, $T = 2\pi\sqrt{L/g}$, has a numerical factor of $2\pi$ that dimensional analysis cannot determine. But the scaling --- how $T$ depends on $L$ and $g$ --- is exactly right. And the absence of mass is a genuine physical prediction, not an artifact.

Before you read on: A raindrop falls from a cloud. It eventually reaches a terminal velocity where the drag force balances gravity. Does the terminal velocity depend on the radius of the drop? On the density of the water? On the viscosity of air? Try to reason through this using dimensions before we work it out together.

Why Dimensional Analysis Works

Every equation in physics is a statement about physical quantities, not just numbers. When you write $F = ma$, you are asserting that a force equals a mass times an acceleration. The left side has dimensions of $\text{kg} \cdot \text{m/s}^2$. The right side must have the same dimensions. If it did not, the equation would be comparing apples to oranges --- it would be meaningless.

This is not just a bookkeeping rule. It reflects something deep about the nature of physical law: the laws of physics do not depend on the system of units you choose. Whether you measure length in meters or feet, the period of a pendulum is the same physical quantity. The equation that describes it must respect this invariance. That invariance is what dimensional analysis exploits.

Here is the core principle:

Every physically meaningful equation must be dimensionally consistent. Every term in the equation must have the same dimensions.

This principle has two powerful applications:

1. Checking equations. If you derive a result and the dimensions are wrong, the result is wrong. No exceptions. Dimensional checking catches algebraic errors that might otherwise go unnoticed for pages. It is the single most reliable quick check in physics.

2. Predicting functional forms. If you know which physical quantities a result depends on, you can often determine the functional form --- up to a dimensionless constant --- without doing any derivation at all.

The second application is what this section is about. It is called a scaling argument or, more formally, the Buckingham Pi theorem in its general form. The idea is simple: if the answer depends on a certain set of variables, and the answer must have particular dimensions, then the dimensions constrain how the variables can combine.

Building a Velocity from Physical Quantities

Let's start with a concrete exercise.

[Interactive: Dimensional Constructor. The screen displays a target quantity with given dimensions --- for example, "Build a velocity $[\text{m/s}]$." Below the target, a set of available physical quantities is listed, each with its dimensions displayed:

• $g$ (gravitational acceleration): $\text{m/s}^2$
• $L$ (length): $\text{m}$
• $m$ (mass): $\text{kg}$

The student drags quantities into a "workspace" and assigns integer or fractional exponents to each. A dimensional calculator updates in real time, showing the combined dimensions. The student adjusts exponents until the dimensions match the target.

Guided prompts: - "Start with $g$. Its dimensions are $\text{m/s}^2$. You need $\text{m/s}$. What do you need to multiply or divide by?" - "Try $\sqrt{gL}$. What are its dimensions?" - "Can you include mass in any way that preserves the correct dimensions?"

After the student matches the target, a second round presents a new target: "Build an energy $[\text{kg} \cdot \text{m}^2/\text{s}^2]$ from $m$, $g$, and $L$." Then a third: "Build a time from $g$ and $L$ only."]

Let's walk through the first example together. You want to construct a velocity --- dimensions of $\text{m/s}$ --- from $g$, $L$, and $m$.

Suppose $v \propto g^a L^b m^c$. Then:

$$[v] = \left(\frac{\text{m}}{\text{s}^2}\right)^a \cdot \text{m}^b \cdot \text{kg}^c = \text{m}^{a+b} \cdot \text{s}^{-2a} \cdot \text{kg}^c$$

Match to $\text{m}^1 \cdot \text{s}^{-1} \cdot \text{kg}^0$:

• Meters: $a + b = 1$
• Seconds: $-2a = -1 \implies a = 1/2$
• Kilograms: $c = 0$

So $b = 1/2$ and $c = 0$, giving:

$$v \propto \sqrt{gL}$$

Mass cannot appear. There is no way to include kilograms in a quantity that has dimensions of $\text{m/s}$ when the only source of kilograms is $m$ itself. Since neither $g$ nor $L$ carry kilograms, and the target has no kilogram dimension, $m$ is excluded.

Pause and think: Can you build an acceleration from $g$, $L$, and $m$? What is the only possibility? Does $L$ appear? Does $m$ appear?

The answer: the only combination with dimensions of $\text{m/s}^2$ is $g$ itself. Neither $L$ nor $m$ appear. You can verify this by setting up the dimensional equations --- you will find $a = 1$, $b = 0$, $c = 0$. The acceleration of free fall depends only on $g$, which is itself an acceleration. This is almost trivially obvious, but the method would catch it even if it were not.

The Raindrop Problem

Now let's tackle a richer example. A spherical raindrop of radius $r$ and density $\rho$ falls through air with viscosity $\mu$. At terminal velocity, the gravitational force (pulling it down) is balanced by the viscous drag force (pushing it up). What does the terminal velocity $v_t$ depend on?

First, identify the relevant physical quantities and their dimensions:

Quantity	Symbol	Dimensions
Terminal velocity	$v_t$	$\text{m/s}$
Drop radius	$r$	$\text{m}$
Drop density	$\rho$	$\text{kg/m}^3$
Gravitational acceleration	$g$	$\text{m/s}^2$
Dynamic viscosity of air	$\mu$	$\text{kg/(m \cdot s)}$

We assume $v_t = C \, r^a \, \rho^b \, g^c \, \mu^d$, where $C$ is a dimensionless constant. Writing out dimensions:

$$\frac{\text{m}}{\text{s}} = \text{m}^a \cdot \left(\frac{\text{kg}}{\text{m}^3}\right)^b \cdot \left(\frac{\text{m}}{\text{s}^2}\right)^c \cdot \left(\frac{\text{kg}}{\text{m} \cdot \text{s}}\right)^d$$

Expanding:

$$\text{m}^1 \cdot \text{s}^{-1} \cdot \text{kg}^0 = \text{m}^{a - 3b + c - d} \cdot \text{s}^{-2c - d} \cdot \text{kg}^{b + d}$$

This gives three equations in four unknowns:

• Meters: $a - 3b + c - d = 1$
• Seconds: $-2c - d = -1$
• Kilograms: $b + d = 0$

Three equations, four unknowns. That means we have a one-parameter family of solutions. This is important --- it means dimensional analysis alone does not pin down the answer uniquely. You need additional physics to choose among the possibilities.

From the kilogram equation: $d = -b$. From the seconds equation: $-2c + b = -1$, so $b = 2c - 1$. Substituting into the meters equation: $a - 3(2c-1) + c - (-(2c-1)) = 1$, which simplifies to $a - 6c + 3 + c + 2c - 1 = 1$, giving $a - 3c + 2 = 1$, so $a = 3c - 1$.

Let $c$ be the free parameter. Two physically important cases emerge:

Case 1: Viscous drag dominates (small, slow drops). If viscosity is the dominant drag mechanism, we might expect $c$ and $g$ to appear in a particular combination. Setting $c = 1$ gives $a = 2$, $b = 1$, $d = -1$:

$$v_t \propto \frac{\rho g r^2}{\mu}$$

This is correct for small droplets in the Stokes drag regime. The full result (from solving the Stokes drag equation) is $v_t = \frac{2}{9}\frac{\rho g r^2}{\mu}$, and dimensional analysis got the functional form exactly right. The factor of $2/9$ is the dimensionless constant that dimensional analysis cannot determine.

Case 2: Inertial drag dominates (large, fast drops). For larger drops, the drag force depends on the air density $\rho_{\text{air}}$ rather than the viscosity. If we replace $\mu$ with $\rho_{\text{air}}$ in the analysis, we get a different scaling:

$$v_t \propto \sqrt{\frac{\rho g r}{\rho_{\text{air}}}}$$

Both cases are valid physics --- they apply in different regimes. Dimensional analysis identified the possible forms. Knowing which regime applies requires understanding the physics, not just the units. But the method narrowed the possibilities dramatically.

Check your prediction: Did you guess that terminal velocity depends on the radius? It does --- and it depends on $r^2$ in the viscous regime. Larger drops fall faster, and the dependence is strong. A drop twice as wide falls four times faster (in the Stokes regime). Did you guess that mass does not appear directly? It does not --- density and radius carry the information about how much matter is in the drop.

The Method, Step by Step

Here is a systematic procedure for dimensional analysis:

Step 1: Identify the target quantity and write down its dimensions.

Step 2: List all the relevant physical quantities that the target could depend on. This is the hardest step, because it requires physical judgment. Include too many variables and you get an underdetermined system. Include too few and you miss essential physics. The art lies in choosing correctly.

Step 3: Write a power-law ansatz. Assume the target is proportional to each relevant quantity raised to some unknown power.

Step 4: Match dimensions. Write equations for each fundamental dimension (mass, length, time). This gives a system of linear equations for the unknown exponents.

Step 5: Solve the system. If the system has a unique solution, you have the functional form (up to a dimensionless constant). If the system is underdetermined, there are multiple possible forms, and you need additional physical reasoning to choose among them.

Step 6: Interpret the result. Which variables appear? Which are absent? Does the scaling make physical sense?

Pause and think: Step 2 says "list all the relevant physical quantities." But how do you know what is relevant? If you are studying the period of a pendulum, is the color of the string relevant? Obviously not. Is the amplitude relevant? That is less obvious --- and in fact, for large amplitudes it is relevant, but dimensional analysis of $L$, $m$, and $g$ alone cannot reveal this because the amplitude is an angle, which is dimensionless. Dimensional analysis works within the set of variables you provide. Choosing the right set requires physical understanding.

What Dimensional Analysis Cannot Do

Dimensional analysis is powerful, but it has limits. Understanding these limits is just as important as knowing how to use the method.

1. It cannot determine dimensionless constants. The pendulum analysis gives $T \propto \sqrt{L/g}$, but the factor of $2\pi$ is invisible to dimensional analysis. The raindrop analysis gives the correct proportionality, but the factor of $2/9$ requires solving the actual equations. Dimensionless constants like $\pi$, $2/9$, or $1/2$ carry no dimensional information, so the method has no way to find them.

2. It cannot detect dependence on dimensionless variables. The period of a pendulum at large amplitude depends on the initial angle $\theta_0$, which is dimensionless (radians have no dimensions). Dimensional analysis of $L$, $m$, and $g$ will never reveal this dependence because $\theta_0$ can appear in any functional form --- $\sin\theta_0$, $\theta_0^2$, $\log\theta_0$ --- without disturbing the dimensions.

3. It cannot distinguish between different physical regimes. The raindrop example showed two different scaling laws depending on whether viscous or inertial drag dominates. Dimensional analysis produced both but could not tell you which one applies. Physics determines the regime; dimensional analysis determines the scaling within each regime.

4. It requires correct identification of relevant variables. If you forget a relevant variable, the analysis may produce nonsense or misleading results. If you include irrelevant variables, you get an underdetermined system with extra free parameters. The method is only as good as the input.

These are not flaws. They are the boundaries of the method. Dimensional analysis tells you the shape of the answer; detailed physics fills in the rest. The method is most valuable when it tells you something non-obvious --- like the fact that the pendulum period does not depend on mass.

Scaling Arguments: Beyond Dimensional Analysis

Dimensional analysis tells you how a quantity depends on the relevant variables. Scaling arguments go one step further: they tell you what happens when you change one variable while holding others fixed.

For example, dimensional analysis gives $T \propto \sqrt{L/g}$ for a pendulum. From this, you can immediately answer scaling questions:

• If you double the length, the period increases by a factor of $\sqrt{2} \approx 1.41$.
• If you quadruple $g$ (by moving the pendulum to a planet with four times Earth's gravity), the period halves.
• If you change the mass, the period does not change at all.

These are powerful predictions that require no calculation beyond the scaling law. In experimental physics and engineering, scaling arguments like these are used constantly to estimate how systems behave under changed conditions.

Here is another example. From $v_t \propto \rho g r^2 / \mu$ (Stokes regime), you can predict:

• A drop with twice the radius falls four times faster ($v_t \propto r^2$).
• On a planet with half Earth's gravity, the terminal velocity is halved ($v_t \propto g$).
• In a more viscous fluid (say honey instead of air), the terminal velocity is much smaller ($v_t \propto 1/\mu$).

Each of these predictions captures real physics. Each can be tested experimentally. And each was obtained without solving any differential equation.

Pause and think: The speed of sound in an ideal gas depends on the gas pressure $P$ (dimensions $\text{kg/(m \cdot s}^2\text{)}$) and the gas density $\rho$ (dimensions $\text{kg/m}^3$). Use dimensional analysis to find how $v_{\text{sound}}$ depends on $P$ and $\rho$. Then answer: if you double the pressure at constant density, what happens to the speed of sound?

Work through it

Let $v \propto P^a \rho^b$. Then: $$\frac{\text{m}}{\text{s}} = \left(\frac{\text{kg}}{\text{m} \cdot \text{s}^2}\right)^a \cdot \left(\frac{\text{kg}}{\text{m}^3}\right)^b = \text{kg}^{a+b} \cdot \text{m}^{-a-3b} \cdot \text{s}^{-2a}$$ Matching: - kg: $a + b = 0 \implies b = -a$ - s: $-2a = -1 \implies a = 1/2$ - m: $-a - 3b = 1 \implies -1/2 + 3/2 = 1$ (consistent) So $v \propto \sqrt{P/\rho}$. If you double the pressure at constant density, the speed of sound increases by $\sqrt{2} \approx 1.41$. The actual formula is $v = \sqrt{\gamma P/\rho}$, where $\gamma$ is the (dimensionless) adiabatic index, which dimensional analysis cannot determine.

Historical Context: Taylor and the Trinity Test

The most famous application of dimensional analysis in modern history involves the first nuclear explosion --- the Trinity test of July 16, 1945.

In 1950, the British physicist G.I. Taylor published a remarkable paper. Using declassified photographs of the explosion's fireball at various times after detonation, he estimated the energy released by the bomb. The photographs showed the radius of the expanding fireball at known times. Taylor's insight was that the only relevant physical quantities were:

• The energy of the explosion, $E$ (dimensions $\text{kg} \cdot \text{m}^2/\text{s}^2$)
• The density of the surrounding air, $\rho$ (dimensions $\text{kg/m}^3$)
• The time after detonation, $t$ (dimensions $\text{s}$)

He sought the radius of the fireball $R$ as a function of these quantities. Setting $R \propto E^a \rho^b t^c$:

$$\text{m} = \left(\frac{\text{kg} \cdot \text{m}^2}{\text{s}^2}\right)^a \cdot \left(\frac{\text{kg}}{\text{m}^3}\right)^b \cdot \text{s}^c$$

$$\text{m}^1 \cdot \text{kg}^0 \cdot \text{s}^0 = \text{m}^{2a - 3b} \cdot \text{kg}^{a+b} \cdot \text{s}^{-2a+c}$$

• Meters: $2a - 3b = 1$
• Kilograms: $a + b = 0$
• Seconds: $-2a + c = 0$

From kg: $b = -a$. From m: $2a + 3a = 1$, so $a = 1/5$. Then $b = -1/5$ and $c = 2/5$.

$$R \propto \left(\frac{E}{\rho}\right)^{1/5} t^{2/5}$$

Taylor measured $R$ and $t$ from the photographs, used the known density of air, and solved for $E$. His estimate of the bomb's energy yield --- roughly $10^{14}$ joules, equivalent to about 20 kilotons of TNT --- agreed closely with the classified measurements.

This story is often attributed to Lord Rayleigh, who pioneered dimensional analysis methods decades earlier and laid the theoretical foundations that Taylor applied. Rayleigh's general approach --- identifying the relevant variables, writing down their dimensions, and solving for the exponents --- is exactly the method you have been learning in this section.

The point is not just historical drama (though the drama is considerable). The point is that dimensional analysis, applied with good physical judgment about which variables matter, can extract quantitative information from situations that seem impossibly complex. Taylor did not need to solve the equations of fluid dynamics, nuclear physics, or radiation transport. He needed only to identify the right variables and let the dimensions do the work.

Dimensional Analysis as Error Detection

Throughout this course, you have derived equations, simplified expressions, and combined results from multiple steps. Every one of those steps is an opportunity for error. Dimensional analysis is your most reliable error detector.

Here is the practice: after deriving any result, check its dimensions. Every term in every equation must have the same dimensions. If they do not, something is wrong.

This sounds simple, and it is. But students frequently skip it, and then spend an hour tracking down a sign error when a two-second dimensional check would have flagged the problem immediately.

Some examples of what dimensional checking catches:

• A missing factor of $R$. If you write $\omega = v$ instead of $\omega = v/R$, a dimensional check reveals that the left side is $\text{s}^{-1}$ but the right side is $\text{m/s}$. Something is wrong.
• An incorrect exponent. If you write $T = 2\pi\sqrt{L^2/g}$ instead of $T = 2\pi\sqrt{L/g}$, the dimensions of the right side are $\text{m} \cdot \text{s}/\text{m}^{1/2} = \text{m}^{1/2} \cdot \text{s}$, which is not time. Caught.
• Adding quantities with different dimensions. If an intermediate step produces $mgh + mv$ (energy plus momentum), dimensional analysis immediately tells you this is nonsense. You cannot add $\text{kg} \cdot \text{m}^2/\text{s}^2$ and $\text{kg} \cdot \text{m/s}$.

The few seconds that dimensional checking costs will save you far more time in the long run. Make it habitual. Check every final answer and every intermediate result that looks suspicious. If the dimensions work out, the result might still be wrong (a dimensionless factor of 2 could be missing), but if the dimensions don't work out, the result is definitely wrong.

Limits and Consistency Checks

Dimensional analysis also connects naturally to another powerful checking strategy: examining limits. When you derive a formula, ask what happens in extreme cases. Do the limits make physical sense?

For example, the rolling-on-ramp formula from Section 14.1:

$$v_{\text{cm}} = \sqrt{\frac{2gh}{1 + c}}$$

What happens if $c = 0$? Then $v_{\text{cm}} = \sqrt{2gh}$, which is the speed of an object sliding without friction --- all energy goes to translation, none to rotation. Physically sensible.

What happens as $c \to \infty$? Then $v_{\text{cm}} \to 0$. An object with an enormous moment of inertia diverts almost all its energy into rotation and barely translates. Also sensible.

What about the pendulum period $T = 2\pi\sqrt{L/g}$? As $L \to 0$, $T \to 0$ --- a shorter pendulum oscillates faster. As $g \to \infty$, $T \to 0$ --- stronger gravity makes the pendulum swing faster. As $g \to 0$, $T \to \infty$ --- in zero gravity, the pendulum does not oscillate at all. All of these limits are physically correct.

Limit-checking and dimensional analysis work hand in hand. Dimensions tell you whether an equation could be right. Limits tell you whether it behaves right. Together, they form the backbone of rough reasoning --- the physicist's first line of defense against errors and the first tool for building intuition about unfamiliar results.

Practice

Layer 1: Concrete

Problem 1. The frequency $f$ of a vibrating string depends on the string length $L$ (m), the tension $F$ (N = kg $\cdot$ m/s$^2$), and the linear mass density $\mu_L$ (kg/m). Use dimensional analysis to find how $f$ depends on these quantities.

Check your answer

Let $f \propto L^a F^b \mu_L^c$. Frequency has dimensions $\text{s}^{-1}$. $$\text{s}^{-1} = \text{m}^a \cdot \left(\frac{\text{kg} \cdot \text{m}}{\text{s}^2}\right)^b \cdot \left(\frac{\text{kg}}{\text{m}}\right)^c = \text{m}^{a+b-c} \cdot \text{kg}^{b+c} \cdot \text{s}^{-2b}$$ Matching: - s: $-2b = -1 \implies b = 1/2$ - kg: $b + c = 0 \implies c = -1/2$ - m: $a + b - c = 0 \implies a + 1/2 + 1/2 = 0 \implies a = -1$ So $f \propto \frac{1}{L}\sqrt{\frac{F}{\mu_L}}$. The full result is $f = \frac{1}{2L}\sqrt{F/\mu_L}$ (for the fundamental mode). Dimensional analysis got the functional form correct. The factor of $1/2$ is a dimensionless constant that depends on the mode shape (which boundary conditions are imposed). Physical interpretation: higher tension means higher frequency (the string is "tighter"). Greater linear density means lower frequency (the string is "heavier"). Longer string means lower frequency (think of a bass guitar vs. a violin).

Problem 2. The drag force on a sphere moving through a viscous fluid at low speed depends on the fluid viscosity $\mu$ (kg/(m $\cdot$ s)), the sphere radius $r$ (m), and the speed $v$ (m/s). Use dimensional analysis to find how the drag force $F_d$ depends on these quantities.

Check your answer

Let $F_d \propto \mu^a r^b v^c$. Force has dimensions $\text{kg} \cdot \text{m/s}^2$. $$\frac{\text{kg} \cdot \text{m}}{\text{s}^2} = \left(\frac{\text{kg}}{\text{m} \cdot \text{s}}\right)^a \cdot \text{m}^b \cdot \left(\frac{\text{m}}{\text{s}}\right)^c = \text{kg}^a \cdot \text{m}^{-a+b+c} \cdot \text{s}^{-a-c}$$ Matching: - kg: $a = 1$ - s: $-a - c = -2 \implies c = 1$ - m: $-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1$ So $F_d \propto \mu r v$. This is Stokes' law. The full result is $F_d = 6\pi \mu r v$. The dimensionless constant $6\pi$ comes from solving the Navier-Stokes equations for creeping flow around a sphere --- a significant calculation. But dimensional analysis nailed the functional form with far less effort.

Layer 2: Pattern

Problem 3. For each of the following physical quantities, a set of relevant variables is given. Determine which variables can appear in the dimensional formula and which cannot.

(a) The orbital period $T$ of a planet depends on the orbital radius $r$ (m), the planet mass $m_p$ (kg), the star mass $M$ (kg), and the gravitational constant $G$ (m$^3$/(kg $\cdot$ s$^2$)). Can the planet's mass appear?

(b) The speed of waves on deep water depends on the wavelength $\lambda$ (m), the water density $\rho$ (kg/m$^3$), and $g$ (m/s$^2$). Can the density appear?

(c) The energy stored in a capacitor depends on the charge $Q$ (C = A $\cdot$ s) and the capacitance $C$ (F = A$^2 \cdot$ s$^4$/(kg $\cdot$ m$^2$)). What is the dimensional form?

Check your answer

**(a)** Let $T \propto r^a m_p^b M^c G^d$. $$\text{s} = \text{m}^a \cdot \text{kg}^b \cdot \text{kg}^c \cdot \left(\frac{\text{m}^3}{\text{kg} \cdot \text{s}^2}\right)^d = \text{m}^{a+3d} \cdot \text{kg}^{b+c-d} \cdot \text{s}^{-2d}$$ From s: $d = -1/2$. From m: $a = 3/2$. From kg: $b + c = -1/2$. The system gives $b + c = -1/2$ but does not determine $b$ and $c$ individually. If you assume the planet mass is negligible compared to the star mass (which is the standard approximation in Kepler's problem), then $m_p$ does not appear, and $c = -1/2$: $$T \propto r^{3/2} / \sqrt{GM}$$ This is Kepler's third law: $T^2 \propto r^3/(GM)$. Dimensional analysis alone tells you that $b$ and $c$ are not individually determined --- you need the physics (the planet's mass is negligible) to fix them. **(b)** Let $v \propto \lambda^a \rho^b g^c$. $$\frac{\text{m}}{\text{s}} = \text{m}^a \cdot \left(\frac{\text{kg}}{\text{m}^3}\right)^b \cdot \left(\frac{\text{m}}{\text{s}^2}\right)^c = \text{m}^{a-3b+c} \cdot \text{kg}^b \cdot \text{s}^{-2c}$$ From kg: $b = 0$. Density cannot appear. From s: $c = 1/2$. From m: $a + 1/2 = 1$, so $a = 1/2$. $$v \propto \sqrt{g\lambda}$$ The density of the water does not affect the speed of deep-water gravity waves (in this model). This is a non-trivial prediction from dimensional analysis alone. **(c)** Let $E \propto Q^a C^b$. Using SI base units, $[Q] = \text{A} \cdot \text{s}$ and $[C_{\text{cap}}] = \text{A}^2 \cdot \text{s}^4 / (\text{kg} \cdot \text{m}^2)$, and $[E] = \text{kg} \cdot \text{m}^2 / \text{s}^2$. Working through the algebra: $a = 2$, $b = -1$. So $E \propto Q^2/C$. The full result is $E = Q^2/(2C)$, with the dimensionless factor of $1/2$.

Problem 4. A physical quantity $X$ has dimensions of $\text{kg} \cdot \text{m}^2/\text{s}$. You are told it depends on a mass $m$, a velocity $v$, and a length $L$. Find $X$ in terms of these variables. What well-known physical quantity has these dimensions?

Check your answer

$X \propto m^a v^b L^c$. $$\frac{\text{kg} \cdot \text{m}^2}{\text{s}} = \text{kg}^a \cdot \left(\frac{\text{m}}{\text{s}}\right)^b \cdot \text{m}^c = \text{kg}^a \cdot \text{m}^{b+c} \cdot \text{s}^{-b}$$ From kg: $a = 1$. From s: $b = 1$. From m: $1 + c = 2$, so $c = 1$. $$X \propto mvL$$ The dimensions $\text{kg} \cdot \text{m}^2/\text{s}$ are those of angular momentum. The quantity $mvL$ is the angular momentum of a particle of mass $m$ moving with speed $v$ at a distance $L$ from the axis --- specifically, $L = mvr\sin\theta$ in full generality, where $\theta$ is the angle between the velocity and the position vector.

Layer 3: Structure

Problem 5. Why can't dimensional analysis determine numerical coefficients? Give a careful explanation, using examples to support your argument.

Check your answer

Dimensional analysis works by requiring that both sides of an equation have the same dimensions. But a dimensionless number --- $2$, $\pi$, $1/3$, $\sqrt{2}$ --- has no dimensions at all. Multiplying any expression by a dimensionless constant does not change its dimensions. Therefore, dimensional analysis cannot distinguish between $T = \sqrt{L/g}$, $T = 2\pi\sqrt{L/g}$, or $T = 47\sqrt{L/g}$. All three have the same dimensions. More fundamentally, dimensional analysis constrains the *structure* of how physical quantities combine, but the specific numerical value depends on the *details* of the physical law --- the geometry of the system, the exact form of the forces, the boundary conditions. These details are encoded in the equations of motion, not in the dimensions. For example: - The factor of $2\pi$ in $T = 2\pi\sqrt{L/g}$ comes from the geometry of circular motion (the relationship between angular frequency and period). - The factor of $6\pi$ in Stokes' law comes from integrating the pressure and shear stress over the surface of a sphere. - The factor of $2/9$ in the Stokes terminal velocity comes from combining the drag law with the buoyancy force and the volume of a sphere. Each of these constants requires actual mathematical derivation to determine. Dimensions constrain the exponents; physics determines the prefactors. One way to think about it: dimensional analysis tells you the *scaling* of a relationship (how it changes when you change the inputs), but not the *absolute value*. Scaling and absolute value are different kinds of information, and they come from different kinds of reasoning.

Problem 6. In the raindrop terminal velocity problem, we found that the dimensional system was underdetermined (three equations in four unknowns). What does this underdetermination mean physically? Why did it arise, and what additional information resolved it?

Check your answer

The underdetermination arose because there were more relevant variables (four: $r$, $\rho$, $g$, $\mu$) than independent dimensional equations (three: for mass, length, time). With $n$ variables and $k$ independent dimensions, dimensional analysis produces $n - k$ independent dimensionless groups (this is the Buckingham Pi theorem). Here, $4 - 3 = 1$, so there is one free parameter. Physically, the underdetermination reflects the fact that different physics can produce dimensionally consistent scaling laws. In the raindrop problem, the free parameter corresponds to the *type of drag* acting on the drop. Viscous (Stokes) drag and inertial (pressure) drag are both real forces, both dimensionally valid, but they dominate in different regimes. The additional information that resolves the ambiguity is physical knowledge of the drag regime. For small, slow drops (low Reynolds number), viscous drag dominates and the Stokes scaling applies. For large, fast drops (high Reynolds number), inertial drag dominates and a different scaling applies. The Reynolds number --- itself a dimensionless ratio --- is the quantity that distinguishes the regimes. Dimensional analysis identified the possible functional forms; physics selected the correct one for a given situation. This is a general feature: when dimensional analysis leaves a free parameter, it usually signals the existence of multiple physical regimes or the influence of a dimensionless quantity that the analysis cannot constrain.

Layer 4: Creation

Problem 7. Choose a physical system you have studied in this course (or one you encounter in daily life). Identify a key physical quantity and the variables it might depend on. Use dimensional analysis to predict the functional form. Then verify your prediction by looking up or deriving the actual formula. Write up your analysis, including: - What quantity you targeted and why - What variables you included (and what you deliberately excluded) - The dimensional analysis calculation - The actual formula, and how well your prediction matched - What the dimensionless constant turned out to be

Example to guide you

**System:** A mass $m$ on a spring with spring constant $k$ (N/m = kg/s$^2$). **Target quantity:** The oscillation period $T$. **Relevant variables:** $m$ (kg) and $k$ (kg/s$^2$). Deliberately excluded: the amplitude $A$ (it is a length, and for a simple harmonic oscillator, the period is independent of amplitude --- but this is a physical fact, not a dimensional one; we could have included $A$ and dimensional analysis would have allowed it to appear). **Dimensional analysis:** Let $T \propto m^a k^b$. $$\text{s} = \text{kg}^a \cdot \left(\frac{\text{kg}}{\text{s}^2}\right)^b = \text{kg}^{a+b} \cdot \text{s}^{-2b}$$ From s: $b = -1/2$. From kg: $a = 1/2$. $$T \propto \sqrt{m/k}$$ **Actual formula:** $T = 2\pi\sqrt{m/k}$. The scaling matches perfectly. The dimensionless constant is $2\pi$. **Reflection:** Including the amplitude $A$ would have given an underdetermined system (three variables, two independent dimensions), producing a free parameter. Dimensional analysis would say $T \propto \sqrt{m/k} \cdot f(A\sqrt{k/m}/\ldots)$, allowing the period to depend on amplitude through some unknown function. The fact that it does *not* depend on amplitude is an additional physical fact beyond what dimensions alone can tell you. This illustrates both the power and the limitation of the method.

Problem 8. The speed of sound in a solid depends on the Young's modulus $Y$ (Pa = kg/(m $\cdot$ s$^2$)) and the density $\rho$ (kg/m$^3$). Use dimensional analysis to predict $v_{\text{sound}}$. Then look up the actual formula and compare. Finally, use your result to predict: does sound travel faster in steel or in rubber? Faster in lead or in aluminum?

Check your answer

Let $v \propto Y^a \rho^b$. $$\frac{\text{m}}{\text{s}} = \left(\frac{\text{kg}}{\text{m} \cdot \text{s}^2}\right)^a \cdot \left(\frac{\text{kg}}{\text{m}^3}\right)^b = \text{kg}^{a+b} \cdot \text{m}^{-a-3b} \cdot \text{s}^{-2a}$$ From kg: $a + b = 0 \implies b = -a$. From s: $-2a = -1 \implies a = 1/2$. From m: $-1/2 + 3/2 = 1$. Consistent. $$v \propto \sqrt{Y/\rho}$$ The actual formula is $v = \sqrt{Y/\rho}$ (for longitudinal waves in a thin rod). The dimensionless constant is 1 --- dimensional analysis got it exactly right (which is a lucky coincidence, not a guarantee). **Steel vs. rubber:** Steel has $Y \approx 200$ GPa and $\rho \approx 8000$ kg/m$^3$. Rubber has $Y \approx 0.01$ GPa and $\rho \approx 1100$ kg/m$^3$. The ratio $Y/\rho$ is much larger for steel, so sound travels much faster in steel ($\sim$5000 m/s) than in rubber ($\sim$50 m/s). **Lead vs. aluminum:** Lead has $Y \approx 16$ GPa and $\rho \approx 11300$ kg/m$^3$. Aluminum has $Y \approx 70$ GPa and $\rho \approx 2700$ kg/m$^3$. $Y/\rho$ for aluminum is about $70/2.7 \approx 26$, while for lead it is about $16/11.3 \approx 1.4$. Sound travels much faster in aluminum ($\sim$5100 m/s) than in lead ($\sim$1200 m/s). A stiffer, lighter material carries sound faster --- both factors help.

Reflection

Think about how you approach a new problem or check a finished derivation.

How does dimensional analysis change the way you check your work? If you are not already checking the dimensions of every result you derive, consider starting now. It takes seconds and catches real errors.

When might dimensional analysis be the first tool you reach for? Before solving any equation, you can often sketch out the form of the answer using dimensions alone. This gives you a target: when you finish the derivation, the result should match the scaling that dimensions predicted. If it does not, something went wrong.

What does it mean that rough reasoning can be a legitimate scientific tool? Taylor estimated the energy of a nuclear explosion from a photograph. He did not have the equations of nuclear physics or fluid dynamics at hand. He had dimensional reasoning and good physical judgment. The answer was right. This is not a shortcut or a trick --- it is a deep application of the principle that physical laws are dimensionally consistent.

Dimensional analysis will not replace careful derivation. But it stands alongside derivation as a complementary tool: derivation gives you the exact answer, and dimensional analysis gives you the structure of the answer before you start and a check on the answer after you finish.

Looking Ahead

You have just added a powerful tool to your problem-solving arsenal: the ability to extract physical predictions from dimensional reasoning alone, without solving any equation. This is rough reasoning at its best --- quick, principled, and often surprisingly accurate.

In the next section, you will move from theoretical predictions to real-world measurements. Section 15.4 asks: what happens when textbook mechanics meets actual data? Accelerometers, motion sensors, and video analysis all produce measurements that deviate from ideal models. Some deviations are noise. Some reveal missing physics. Some are artifacts of the instrumentation itself. The skill you need there is not calculation but interpretation --- deciding what your data is telling you and what your model is missing. The dimensional intuition you built here will help: when a measurement does not match your prediction, checking the scaling (does the dependence on each variable look right?) is often the fastest way to diagnose where the model breaks down.