Units, Dimensional Analysis & Estimation

Physics is the habit of asking how big something is. Before you can ask that, you have to agree on what "big" means, and that agreement is the whole content of units. It sounds like bookkeeping. It isn't. Units carry real information about how the world is put together, and if you learn to listen to them they will hand you the shape of an answer before you have written a single line of algebra.

Here is the promise, and it is not an exaggeration. You will be able to work out how the period of a swinging pendulum depends on its length without knowing any mechanics, and estimate the number of piano tuners in Chicago to within a factor of two without looking anything up. Both come from the same two habits. This is the first lesson of the course because everything after it quietly assumes you have these.

Dimensions are not units

A unit is a particular agreed-upon amount: the metre, the foot, the second, the hour. A dimension is the kind of thing being measured: length, time, mass. The metre and the foot are different units of the same dimension.

Almost all of mechanics is built from three dimensions. Write them as $M$ for mass, $L$ for length, and $T$ for time. Every other mechanical quantity is a product of powers of these, and you can read the combination straight off the definition. Speed is a distance over a time, so its dimensions are $LT^{-1}$. Acceleration is a speed over a time, so $LT^{-2}$. Force is mass times acceleration, so $MLT^{-2}$. Energy is a force times a distance, so $ML^2T^{-2}$. Pressure is a force over an area, so $MLT^{-2}/L^2 = ML^{-1}T^{-2}$. Density is mass over volume: $ML^{-3}$.

The notation $[X]$ means "the dimensions of $X$". So $[F] = MLT^{-2}$.

Some quantities have no dimensions at all: a pure ratio like $L_1/L_2$, an angle in radians (it's an arc length divided by a radius), $\pi$, or the number $2$. These are called dimensionless, and they are the one thing this whole method is blind to.

Two rules do all the work.

First, both sides of any correct equation have the same dimensions, and so does every term inside a sum. You cannot add a length to a time, any more than you can add three metres to Tuesday. Second, the argument of $\sin$, $\cos$, $\exp$, or $\log$ must be dimensionless. Think about why: $e^x = 1 + x + x^2/2 + \dots$, and that sum adds $x$ to $x^2$. Those can only be added if $x$ has no dimensions at all.

Using dimensions to check

The cheapest use of all this is as an error detector. You derive an expression, you put dimensions in, and if the two sides disagree you have found a mistake without checking a single algebraic step.

Suppose you produce $T = 2\pi\sqrt{g/L}$ for a pendulum period. Check: $g/L$ has dimensions $LT^{-2}/L = T^{-2}$, so the square root is $T^{-1}$: a per-second, not a second. The fraction is upside down. Ten seconds of work, and you caught it.

This check is honest but limited in one way that trips people up. A statement like "$T$ is proportional to $\sqrt{r}$" has a hidden constant of proportionality, and that constant will absorb whatever dimensions it needs to. Proportionalities can never be checked dimensionally. Only full expressions can.

Using dimensions to build an answer

Now the useful part. If you know which quantities an answer can depend on, dimensions often pin down the answer up to a pure number.

The recipe: write the unknown as a product of powers of the given quantities with unknown exponents, demand that the dimensions match, and solve the resulting small linear system: one equation for $M$, one for $L$, one for $T$.

Take a wave on a guitar string. Physically its speed can only be set by the tension $T_s$ in the string and the mass per unit length $\mu$. (Calling tension $T_s$ here so it doesn't collide with $T$ for the time dimension.) Their dimensions are $[T_s] = MLT^{-2}$ and $[\mu] = ML^{-1}$. Write

$v = C\,T_s^{a}\,\mu^{b}$

with $C$ a pure number. Matching dimensions:

$LT^{-1} = (MLT^{-2})^a (ML^{-1})^b = M^{a+b}\,L^{a-b}\,T^{-2a}$

Now read off one equation per dimension. Mass: $0 = a+b$. Length: $1 = a-b$. Time: $-1 = -2a$, so $a = 1/2$ immediately. Then $b = -1/2$ from the mass equation, and the length equation $1 = 1/2 - (-1/2) = 1$ checks out: a free consistency test that the answer exists at all. So

$v = C\sqrt{T_s/\mu}$

and the true answer has $C=1$. With $T_s = 80.0$ N and $\mu = 5.00\times10^{-3}$ kg/m:

$v = \sqrt{\frac{80.0}{5.00\times10^{-3}}} = \sqrt{1.60\times10^{4}} = 126\ \text{m/s}$

Notice what you got for free even before the numbers: doubling the tension does not double the speed, it multiplies it by $\sqrt{2}$. That's physics, extracted from nothing but bookkeeping.

Worked example: a ball dropped from height

Drop a ball from a height $h$. Its speed when it lands should come out of $g$ and $h$ alone, so start from those two and try to build a speed.

Dimensions first: $[g] = LT^{-2}$ and $[h] = L$. A speed has dimensions $LT^{-1}$, so we need a combination that brings the time power from $-2$ to $-1$. Taking a square root does that:

$[\sqrt{gh}] = \sqrt{LT^{-2}\cdot L} = \sqrt{L^{2}T^{-2}} = LT^{-1}$

So a speed of the form $\sqrt{gh}$ passes the check. A tempting wrong answer, $v = gh$, fails straight away: its dimensions are $L^{2}T^{-2}$, which is not a speed.

The energy argument gives the number in front. Energy is conserved, so the height the ball loses becomes kinetic energy: $mgh = \tfrac12 mv^2$. The mass cancels, and

$v = \sqrt{2gh}$

With $h = 5.0$ m and $g = 10$ m/s$^2$, $v = \sqrt{100} = 10$ m/s. The dimension check fixed the form of the answer. The energy argument fixed the $\sqrt{2}$.

Converting units without guessing

Never try to remember whether to multiply or divide. Instead multiply by fractions that equal one, and cancel.

A car at $65$ miles per hour, with $1\ \text{mile} = 1609.344$ m and $1\ \text{h} = 3600$ s:

The "mi" cancels top and bottom, so does the "h", and only m/s survives. If you had set a fraction the wrong way up you would have been left with mi²/h², and you'd see it.

Powers need care. Water at $1.00\ \text{g/cm}^3$: you need $1\ \text{g} = 10^{-3}$ kg and $1\ \text{cm} = 10^{-2}$ m, so $1\ \text{cm}^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3$. Then

The factor is $10^3$, not $10^2$: cubing the length conversion is exactly where people lose it.

Scaling arguments

Often you don't want a number: you want a ratio. Then you don't need any constants at all, because they cancel.

Two hailstones have the same shape and density, but one is eight times heavier. Drag is $F = \tfrac12 C\rho A v^2$. At terminal speed drag balances weight: $mg = \tfrac12 C\rho A v^2$, so $v^2 \propto m/A$.

Same density means mass goes as volume, so $m \propto r^3$ where $r$ is any length across the stone. Eight times the mass means $r$ is doubled, since $2^3 = 8$. Same shape means area goes as $r^2$, so the area is $4$ times larger. Then

The bigger stone falls only $41\%$ faster despite being eight times heavier. The same move handles orbits: if $T^2 \propto r^3$ and $r$ goes up by $4$, then $T^2$ goes up by $4^3 = 64$, so $T$ goes up by $8$.

Estimation and Fermi chains

Sometimes you have no formula and no data, only a question: how many piano tuners work in Chicago? The trick is to break the answer into a chain of factors you can each guess within a factor of a few. Errors in the factors are independent and partly cancel, so the product usually lands within a factor of ten of the truth.

Population $2.7\times10^6$, at $2.5$ people per household, gives $2.7\times10^6/2.5 = 1.08\times10^6$ households. One in twenty owns a piano: $1.08\times10^6/20 = 5.4\times10^4$ pianos. Each tuned once a year, so $5.4\times10^4$ tunings per year. A tuner does $4$ a day for $250$ days, which is $1000$ tunings a year. So

$\frac{5.4\times10^4}{1.0\times10^3} = 54\ \text{tuners}$

The structure matters more than the arithmetic: each step converts one unit into the next (people → households → pianos → tunings per year → tuners), and the units cancel down the chain exactly as they did in the speed conversion.

The habit worth building is to guess the magnitude before you solve. When the algebra later produces a car accelerating at $10^4\ \text{m/s}^2$, you notice.

Anchor numbers

Estimates need somewhere to start. Carry these: $g \approx 10\ \text{m/s}^2$; air density $\approx 1.2\ \text{kg/m}^3$; water density $10^3\ \text{kg/m}^3$; atmospheric pressure $\approx 10^5$ Pa; speed of sound $\approx 340$ m/s; Earth's radius $6.4\times10^6$ m; a year is about $\pi\times10^7$ s (really $3.15\times10^7$); Avogadro's number $6.0\times10^{23}$; a mole of gas at room conditions fills about $0.022\ \text{m}^3$.

Two more habits. Report answers with a sensible number of significant figures: a product carries as many as its least precise input, so $2.50\ \text{m} \times 3.1\ \text{m} = 7.75 \to 7.8\ \text{m}^2$, two figures, because $3.1$ has two. And on a graded exam, an answer with the wrong units earns nothing, so check them last, every time.

How to spot that this is the technique

Reach for dimensional analysis when a problem says an answer "depends only on" a short list of quantities, or gives you an expression like $F = C\rho^a A^b v^c$ and asks for an exponent, or hands you a dimensionless factor ($\sqrt{\gamma}$, $1/\sqrt{2\pi}$, "take the constant to be 1"). That gifted factor is the tell: the problem is telling you it knows you can only get the answer up to a pure number.

Reach for a scaling argument when the problem gives a ratio instead of a value ("eight times the mass", "four times the radius") and asks by what factor something changes. Never compute both cases; divide.

Reach for estimation when the problem supplies a pile of everyday facts with no governing equation in sight, and asks "how many" or "estimate". Build the chain and let the units guide each link.

Next you need the other universal tool: how to handle quantities that have a direction as well as a size. That's vectors.